All Organic

8/14/2019 All Organic

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Summary of Organic Reactions (I. Alkanes) Page 1

I. Alkanes

A. Cracking very long alkane

heat catayst , → long alkane + short alkene

B. Combustion

CxHy + (x +y4 ) O 2(g) → xCO 2(g) +

y2 H2O(l)

C. Chlorination

Chain initiation (chain initiating step)

+Cl ClCl Cl Cl· radical is generated.hυ

Chain propagation (chain propagating step)

Cl CH

H

HH

HCl C

H

HH+ Cl· radical is consumed.

C

H

H

H

ClClC

H

H

HCl Cl + Cl· radical is regenerated.

Chain termination (chain terminating step)

radical is destroyed.

C C

H

H

H

H

H

H

C

H

H

HCH

H

H

Cl C

H

HH C H

HCl

H

+Cl Cl Cl Cl

+ Energy

A mixture of products would be obtained.



Summary of Organic Reactions (II. Alkenes and alkynes) Page 2

II. Alkenes and alkynes

A. Addition reactions

1. Reaction with bromine in CCl 4

CCR

R R

R Br 2 CCR

R R R

Br Br

+in dark

inCCl 4 1,2-di romoalkane

Note : Test for C=C or C ≡ C bond

2. Reaction with bromine water

CCR

R R

R Br 2(aq) CCR

R R

R

Br OH

+in dark

halohydrin Note : Test for C=C or C ≡ C bond

3. Reaction with hydrogen bromide

Br HC CCH

HH

H

H

H propene

+

C C C

H

H

H H

H

HH Br

2-bromopropane

C C C

H

H

H H

H

HHBr

1- romopropane

major product

minor product

Note : Follows Markownikov's orientation

4. Reaction with conc. sulphuric(VI) acid

conc. H 2SO 4(l)CCR

R R

R

CCR

R R

R

H O

S OO

OH

+

alkyl hydrogensulphate(VI) Note : A reversible reaction

CCR

R R

R

H O

S OO

OH

H2O(l) CCR

R R

R

H OH

H2SO 4(aq)++

alkyl hydrogensulphate(VI)

alcohol

Note : Laboratory preparation of alcohol

5. Reaction with water CH2 CH 2 + H2O CH 3 CH 2 OH

H3PO4300 ºC

ethane ethanol Note : Industrial preparation of alcohol

6. Catalytic hydrogenation of oil Note : hardening of oil




B. Ozonolysis of alkene

CH3CHCH CH 2

CH3

CH3CH

CH3

C

O

H +

3-methylbut-1-ene 2-methylpropanal

(1) O 3, CH 2Cl2, -78 ºC

(2) Zn / H 2OH C H

O

methanal Note : Can be used to determine the structure of the alkene with characterization of the derivatives of carbonyl compounds

formed.

C. Oxidation cleavage of double bond

propanone pentan-3-one

+ C CH 3

O

H3CCH3CH 2 C CH 2CH 3

O

3-ethyl-2-methylpent-2-ene

C CCH 3CH 2

CH 3CH 2 CH 3

CH 3

(1) KMnO 4, OH -, hot(2) H +

2-methylpropanoic acid3-methylbut-1-ene

++ H2OCO 2CH3CH

CH3

C

O

OHCH3CHCH CH 2

CH3(1) KMnO 4, OH -, hot

(2) H +

Note : The outcomes are different from the ones from ozonolysis.

a) Reaction with cold acidic or alkaline permanganate solution

C C H

HH

HC C H

HH

H

O O

H H

cold KMnO 4(aq)

H3O+ or OH -

ethane

ethane-1,2-diol

D. Oxymercuration of alkyne

R C C HH

HO

keto-enoltautomerization

Markovinkovorientation

HgSO 4(aq)H2SO 4(aq)

R C C HCR C H

H

O

Hterminalalkyne methyl ketone




E. Polymerization of alkenes

Chain initiating step

2 R C O

O

O C

O

R R R C

O

O2

Diacylperoxide

+ 2 CO 2

alkyl radical

heat

Chain propagating step

C CR

H

H

H

HR C C

H

H

H

H

long chain polymer etc.

C CR

H

H

H

H

C CH

H

H

HR C C C C

H

H

H

H

H

H

H

H

Chain terminating step

R (CH 2 CH 2)n C CH

H

H

H

R (CH 2 CH 2)n C CH

H

H

H

C C (CH 2 CH 2)n R H

H

H

H

C C (CH 2 CH 2)n R H

H

H

H

combination

C C (CH 2 CH 2)n R

H

H

H

H

R (CH 2 CH 2)n C C

H

H

H

H

R (CH 2 CH 2)n C C

H H

H

H C C (CH 2 CH 2)n R

H

H

H

H

disproportionation

oxidation product reduction product



Summary of Organic Reactions (III. Aromatic hydrocarbons) Page 5

III. Aromatic hydrocarbons

A. Substitution reactions of benzene

1. Nitration NO 2

conc. HNO 3(aq)conc. H 2SO 4(l)55°C

benzenenitro enzene

Note : No reaction if conc. H 2SO 4(l) is not added.

2. HalogenationX

X2FeX 3 or AlX 3

enzene halobenzene

3. Sulphonation

benzene

SO

O

O Hconc. H 2SO 4

heat

enzenesulphonic acid

benzenesulphonic acid

OHSO

O

O H O- Na +

NaOH(s)350°C

H3O+

phenolsodium phenoxide/ sodium phenolate

Note : Industrial preparation of phenol

4. Alkylation

enzene

R

RX and AlX 3

alkylbenzene

B. Oxidation of alkylbenzene

alkylbenzene

R C OH

O

1) KMnO 4, OH -, heat2) H 3O+

enzenecar oxylic acid



Summary of Organic Reactions (IV. Halogeno-compounds) Page 6

IV. Halogeno-compounds

A. Nucleophilic substitution

1. Reaction with sodium hydroxideR OHR X

haloalkane

NaOH(aq)

alcohol

NaOH(aq)heat

X OH

halobenzene phenol Note : Prolonged heating is required for the alkaline hydrolysis of halobenzene.

2. Reaction with potassium cyanideR X

haloalkane

KCN(aq)R CN nitrile

3. Reaction with ammoniaR X

haloalkane

NH H

Hammonia

+ NH

H

R + HX

1° amine

R X

haloalkane

NH R

H1° amine

+ NR

H

R + HX

2° amine

R X

haloalkane

N R

H

2° amine

+ NR

R

R + HX

3° amine

R X

haloalkane

N R

R 3° amine

+ NR

R

R

R

X -

quaternary ammonium halide Note : A mixture of products would be obtained.

4. Reaction with alkoxideR X

haloalkane+ O- Na +R'

sodium alkoxideR O R'

ether

5. Reaction with carboxylate ion

R X

haloalkane

+sodium

car oxylate

R' C

O

O- Na + R' C

O

O R

ester



Summary of Organic Reactions (IV. Halogeno-compounds) Page 7

B. Elimination reaction

1. Reaction with alcoholic sodium hydroxide to form alkene NaOH(alc.)

CCR

R R

R CCR

R R

R

X H

haloalkane alkene

2. Reaction with alcoholic sodium hydroxide to form alkyne

CCR

H H

R

X X boiling NaOH(alc.)

R C C R

1,2-dihaloalkane alkyne



Summary of Organic Reactions (V. Hydroxy compounds) Page 8

V. Hydroxy compounds

A. Reactions of alcohols

1. Formation of halide

a) by halogenating agentR OH R Cl

chloroalkanealcohol

PCl 5 / PCl 3 / SOCl 2

b) by heating with halide salt in acidic mediumR OH R X

haloalkanealcohol Na +X-, conc. H 2SO 4(l)

heat

2. Formation of alkoxide2 C 2H5OH (l) + 2 Na (s) → 2 C 2H5O- Na +

(alc.) + H 2(g C2H5OH (l) + NaOH (s) d C2H5O- Na +

(alc.) + H 2O(l)

3. Oxidation of alcohols

aldehyde carboxylic acid1° alcohol

R C O

O

H[O] [O]

R C

O

HR C O H

H

H

ketone2?alcohol

[O]R C

O

R R C O H

H

R

3?alcohol

[O]R C O H

R

R

Resistantto oxidation

4. Dehydration of alcohols

C C

O

H

HH

H

H

H

C CH

H H

Hconc. H 2SO 4

heat

ethanol

ethene

5. Esterification

R OH + R' C

O

OH R' C

O

O R

ester

H2O+conc. H 2SO 4(l)heat

alcohol carboxylicacid




6. Triiodomethane formation

C C

H

H

HO

H

H

R

alcohol with1-hydroxyethyl group

NaOH(aq) + I 2(aq)R C

O

O-

C

I

I

I

H

carboxylateion

iodoform(yellow ppt.)

+

Note : 1. Test for the presence of 1-hydroxyethyl group.

2. Shorten the carbon chain by 1 C.

Details :

oxidation

methyl ketone

2 H 2O2 I -2 OH -I2 ++++ C C

H

H

HR

O

C C

H

H

HO

H

H

R

alcohol with1-hydroxyethyl group

methyl ketone

C C

O H

H

HR R C

O

O -

C

I

I

I

H

carboxylateion


NaOH(aq) + I 2(aq)+

7. Distinction between primary, secondary and tertiary alcohols

a) With potassium dichromate(VI)


R C O

O

H[O] [O]

R C

O

HR C O H

H

H

ketone2?alcohol

[O]R C

O

R R C O H

H

R

3?alcohol

[O]R C O H

R

R


b) Lucas' test

C OH

H

H

conc. HCl(aq)ZnCl 2(aq) no reaction

1° alcohol

C OH

H

R

conc. HCl(aq)ZnCl 2(aq)

CR Cl

H

R 2° alcohol

slow reaction

C OH

R

R

conc. HCl(aq)

ZnCl 2(aq)CR Cl

R

R 3° alcohol

fast reaction

Note : 3º alcohol will turn cloudy quickly, 2º alcohol will turn cloudy slowly but 1º alcohol will not turn cloudy at all.




B. Reactions of phenol

1. Reaction with sodiumOH

O - Na +

+ H2 Na

phenol sodium phenoxide/ sodium phenolate

2. Reaction with sodium hydroxideOH

O- Na +

+ H2O NaOH(aq)

sodium phenoxide/ sodium phenolate phenol

Note : Phenol is soluble in sodium hydroxide solution.

3. EsterificationOH C

O

O R

phenol

R C

O

Cl

R C

O

O C

O

R

or

ester

4. Reaction with bromineOH

Br

Br Br Br 2(aq)

OH

phenol 2,4,6-tribromophenol(a white ppt.)

Note : A test for phenol.



Summary of Organic Reactions (VI. Carbonyl compounds) Page 11

VI. Carbonyl compounds

A. Nucleophilic addition

1. Reaction with hydrogen cyanide

R C R

OHCN+ R C R

O H

C N

2. Reaction with sodium hydrogensulphate(IV)

CO

R

CH 3/H

+ Na +HSO 3- C

R

CH 3/H

OH SO 3- Na +

white precipitate

Note: 1. Sensitive to steric hinderance and reacts with aldehyde and methyl ketone only.2. Can be used to purify aldehyde or methyl ketone.

B. Addition-elimination (condensation)

1. Reaction with hydroxylamine

R C R

O+

NH

H

OH

hydroxylamine

C

R

R

N OH + H2O

oxime (white ppt.)

Note : 1. A test for aldehyde or ketone.2. Identify the carbonyl compound by characterization.

2. Reaction with 2,4-dinitrophenylhydrazine

+

2,4-dinitrophenylhydrazine

NO 2

O2 N

N N

H

CR

R

2,4-dinitrophenylhydrazone(orange ppt.)

+ H2O NO 2

O2 N

N N

H

H

H

C

R

R

O

Note : 1. Test for aldehyde or ketone but no reaction with carboxylic acid and its derivatives2. Identify the carbonyl compound by characterization.

3. Triiodomethane formation

methyl ketone

C C

O H

H

HR R C

O

O-

C

I

I

I

H

carboxylateion


NaOH(aq) + I 2(aq)+

Note : 1. Test for methyl ketone.2. Shortens the carbon chain by 1 C



Summary of Organic Reactions (VI. Carbonyl compounds) Page 12

C. Oxidation and reduction

1. Oxidation of aldehyde with potassium dichromate(VI)

R C

O

H R C

O

OH

aldehyde carboxylic acid

[O]

Note : Orange acidified potassium dichromate will turn to green chromium(III) ion.

2. Oxidation of aldehyde with Tollens' reagent

aldehyde

ammoniumcar oxylate

R C

O

H R C

O

O- NH 4+

+[Ag(NH 3)2]OH(aq)

Tollen's reagentAg(s)

silver mirror

Note : Test for aldehyde

3. Oxidaton of aldehyde with Fehling's reagent

aldehyde

carboxylate ion

R C

O

H R C

O

O- +Cu2O

brick red ppt.Fehling's reagent

Note : Test for aldehyde

4. Resistance of ketones to oxidation

R C

O

R [O]

no reaction

Note : Ketone is stable to general oxidation.

5. Reduction with sodium tetrahydridoborate / lithium tetrahydrioaluminate

R C R/H

O+ R C R/H

O H

H

NaBH 4(aq) R C R/H

O-

H

H3O+

aldehyde

or ketonealkoxide alcohol

R C R/H

O+ R C R/H

O H

H

R C R/H

O-

H

H3O+

aldehydeor ketone alkoxide alcohol

LiAlH 4(ether)

Note : LiAlH4 cannot be used in aqueous medium.

6. Haloform reaction

methyl ketone

C C

O H

H

HR R C

O

O -

C

I

I

I

H

carboxylateion iodoform(yellow ppt.)

NaOH(aq) + I 2(aq)+

Note : 1. Test for methyl ketone.2. Shorten the carbon chain by 1 C.



Summary of Organic Reactions (VII. Carboxylic acids and their derivatives) Page 13

VII. Carboxylic acids and their derivatives

A. Formation of carboxylic acid

1. Hydrolysis of nitrile

R C N R C

O

OHdil. H

2SO

4(aq)

reflux

nitrile car oxylic acid Note : Complete hydrolysis of nitrile

2. Oxidation of alcohol


R C O

O

H[O] [O]

R C

O

HR C O H

H

H

Note : Aldehyde intermediate can be separated from the reacting mixture by distillation.

3. Oxidation of aldehyde

R C

O

H R C

O

OH

aldehyde carboxylic acid

[O]

4. Oxidation of alkylbenzene

alkylbenzene

R C OH

O

1) KMnO 4, OH -, heat2) H 3O+

enzenecar oxylic acid

B. Reactions of carboxylic acid

1. Formation of salt

R C

O

OH NaOH(aq)

car oxylic acid

R C

O

O- Na +

sodium carboxylate

2. Formation of acyl chloride

C OH

O

C Cl

O

R

chloroalkanealcohol

PCl 5 / PCl 3 / SOCl 2




3. Formation of anhydride

O

H

H

O

O

C

C

H

H

C

C

OH

OH

O

O

140°C + H2O

cis-butenedioic acid butenedioicanhydride Note : Intramolecular dehydration

ethanoic acid ethanoic anhydrideethanoic acid

C

O

CH 3 OH C

O

CH 3HO C O C

O O

CH 3CH 3+P2O5 (P4O10)

Note : Intermolecular dehydration

C

O

CH3 O- Na

+C

O

CH3Cl C O C

O O

CH3CH3+ + NaCl

4. Formation of ester R X

haloalkane

+sodium

car oxylate

R' C

O

O- Na + R' C

O

O R

ester

R C

O

ClR OH+

R C

O

O R

acylchloride alcohol ester

R C

O

O C

O

R R OH+

R C

O

O R

acidanhydride alcohol ester

R C

O

OHR OH+

H3O+

heat R C

O

O R

carboxylicacid alcohol ester

5. Formation of amide

R C

O

NH 2amide

R C Ncold dil. H 2SO 4(aq)or hot conc. H 2SO 4(l)

nitrile

Note : Partial hydrolysis of nitrile.

6. Reduction with lithium tetrahydridoaluminate

R C

O

OH

carboxylicacid

R C OH

H

H1° alcohol

(1) LiAlH 4(ether)

(2) H 3O+




C. Reactions of acyl chlorides and anhydrides

1. Reaction with water

R C

O

Clacyl

chloride

R C

O

OH+ HClH2O(l)

carboxylicacid

R C

O

O C

O

R'acid

anhydride

H2O(l)R C

O

OH+

R' C

O

OHcarboxylic

acid

2. Reaction with alcohol

R C

O

Clacyl

chloride

R C

O

OR + HClROH

ester

R C

O

O C

O

R'acid

anhydride

ROHR C

O

OR +

R' C

O

OH

ester

3. Reaction with ammonia

R C

O

Clacyl

chloride

R C

O

NH 2

conc. NH 3(aq)

amide

R C

O

O C

O

R'acid

anhydride

R C

O

NH 2

conc. NH 3(aq)

amide

4. Reaction with amine

R CO

Clacyl

chloride

R CO

NHR

N-substituted amide

RNH 2

R C

O

O C

O

R'acid

anhydride

R C

O

NH

R

N-substituted amide

RNH 2

D. Reactions of ester

1. Acid and base-catalysed hydrolyses

R OH+H3O+

heatR C

O

O R

alcoholester

R C

O

OH

carboxylicacid

R OH+R C

O

O R alcoholester

R C

O

O-

carboxylateion

NaOH(aq)

heat

Note : Rate of hydrolysis : alkaline medium > acidic medium > neutral medium


R C

O

O R

ester

R C OH

H

H1° alcohol

(1) LiAlH 4(ether)

(2) H 3O+




E. Reactions of amide

1. Hydrolysis of amide

R C

O

OHR C

O

NH 2

H3O+ reflux

amide car oxylic acid

R C

O

NH 2 R C

O

O-OH-

reflux

amide car oxylate ion

2. Dehydration of amide

R C NH 2

O P2O5 or (CH 3CO) 2Oor SOCl 2

R C N

3. Hofmann degradation of non-substituted amide

Br 2 / NaOH (aq) N H

H

R

1?aminewith 1 C lessamide

R C NH 2

O

Note : Only applicable to non-substituted amide.


1° amineamide

1) LiAlH 4 / ether

2) H3O+R C

H

H

N H

HR C NH 2

O

Note : Amide is the only acid derivative which will not be reduced to alcohol.



Summary of Organic Reactions (VIII. Nitrogen compounds) Page 17

VIII. Nitrogen compounds

A. Formation of amine

1. From nitrile

R C N R C N

H

H

H

H

(1) LiAlH 4(ether)

(2) H 3O+

nitrile 1° amine Note : LiAlH4 could not be used in aqueous medium.

2. From amide

1° amineamide

1) LiAlH 4 / ether

2) H3O+R C

H

H

N H

HR C NH 2

O

Note : Amide is the only acid derivative which will not be reduced to alcohol.

3. 1º, 2º, 3º amines and quaternary ammonium compounds by alkylation

R X

haloalkane

NH H

Hammonia

+ NH

H

R + HX

1° amine

R X

haloalkane

NH R

H1° amine

+ NR

H

R + HX

2° amine

R X

haloalkane

N R

H

2° amine

+ NR

R

R + HX

3° amine

R X

haloalkane

N R

R 3° amine

+ NR

R

R

R

X-

quaternary ammonium halide Note : A mixture of products would be obtained.

4. Phenylamine from nitrobenzene NO 2

NH 2

1) Sn, conc. HCl(aq)2) OH -(aq)

B. Base properties of amine

1. Salt formationR N H

H+ HCl R N H

H

H

Cl-



Summary of Organic Reactions (VIII. Nitrogen compounds) Page 18

C. Reactions of amine

1. Reactions with ethanoyl chloride and benzoyl chlorideR N H

H+ CH 3 C

O

ClCH 3 C

O

N R

H

amine ethanoyl

chloride N-substituted ethanamide

C

O

ClR N H

H+ C

O

N

H

R

N-subsituted benzenecarboxamide benzoylchlorideamine

2. Reaction with nitric(III) acid

R N

H

H

R + + N2

carbocation nitrogen bubblesaliphatic

primaryamine

HNO 2 0-5°C

NH 2

HNO 2 0-5°C

N N+

aromatic primaryamine

benzenediazoniumion

Note : A test used to distinguish aliphatic primary amine from aromatic primary amine.

3. Coupling reaction of benzenediazonium ion with naphthalen-2-ol and phenol

benzenediazoniumion

N N

HO

OH N N+

red orange ppt.

naphthalen-2-ol

N N OHOH

N N+

phenol

orange ppt. benzenediazonium

ion Note : Tests for benzenediazonium ion. Indirect tests for aromatic primary amine.



Summary of Organic Reactions (IX. Other reactions) Page 19

IX. Other reactions

Radical addtion of X 2 to benzeneX

X

X

XX

X

H

H

HH

H

HX2 uv

Reactions of benzenediazonium ion

CuCN

CN

N N+

H3PO 2

N N+

Preparation of vinyl chloride

C CH

HC C

H

H

H

H

Cl Cl

C CH H

Cl H

Cl2Electrophilicaddition

alcoholic NaOH

Anti-Markownikov's addition

ClHC CCH

HH

H

H

H propene

+

C C C

H

H

H H

H

HH Cl

2-chloropropane

C C C

H

H

H H

HHHCl

1-chloropropane

+ R O O R

alkyl peroxide

minor product

major product

Test for terminal alkyne

R–C ≡ CH Cu NH ( )3 2+

→ R–C ≡ C-Cu +(s) (red ppt.)

R–C ≡ CH Ag NH ( )3 2+

→ R–C ≡ C-Ag +(s) (white ppt.)

Test for phenol [Fe(H 2O) 6]

3+(aq) + 2C 6H5O-

(aq) → [Fe(H 2O) 4(C6H5O) 2]+

(aq) + 2H 2O(l) from neutral from phenol purple complexiron(III) chloride



Nomenclature of Organic compounds

I. Nomenclature of organic compounds

A. Labelling of carbon

1. Methyl, primary, secondary and tertiary carbon2. α carbon, β carbon and γ carbon

B. Skeletal formula / Stick formula

II. Nomenclature of hydrocarbon

A. Alkyl and aryl group

III. Other functional group

A. Halogeno-compoundsB. Alkanols and phenol

C. Aldehydes and ketones

D. Carboxylic acids

1. Carboxylic acid derivatives – esters, acid chlorides, anhydrides and amides

E. Nitriles

F. Amines

G. Amino-acids

H. Ethers

IV. Priority of functional groups

V. Examples

VI. Other common abbreviations



Acid-Base Theory

I. Definition of acid-base A. Arrhenius definition

B. Bronsted-Lowry definitionC. Lewis definition

II. Strength of acid and base A. Leveling effect

III. Factors determining the strength of a Bronsted-Lowry acid A. Relative stability of the conjugate base comparing with the acid

1. Inductive effect(1) Amine vs Amide

2. Resonance effecta) Resonance effect on a benzene ring

(1) Phenol(2) Aromatic amine

3. Intramolecular hydrogen bond4. Solvation / hinderance to solvation

a) Steric hinderance to solvationb) Effect of solvent

5. Strength of the bond between the proton and the conjugate base6. Electronegativity of the atom

a) Effect of hybridization



Isomerism

I. Structural isomerism

A. Chain isomerismB. Position isomerismC. Functional group isomerism

D. Tautomerism / TautomerizationII. Stereoisomerism

A. Diastereomerism / Geometrical isomerism1. Physical properties of cis-/trans-geometrical isomers of 1,2-dichloroethene2. Physical and chemical properties of cis-/trans-geometric isomers of butenedioic

acida) Physical propertiesb) Chemical properties

B. Optical isomerism / Enantiomerism1. Physical properties

2. Optical activitya) Plane polarized light b) Measurement of optical activity by polarimeter c) Specific rotation

3. Chiral centre4. Racemic mixture / Racemic modification / Racemic form / Racemate



Reaction Mechanism Page 1

Reaction Mechanism

I. Bond breaking A. Bond breaking of a bond to carbon

II. Types of reactive species

A. Free radical (electron deficiency)B. Electrophile (electron deficiency)C. Nucleophile (electron rich)

III. Classification of reaction A. Types of reaction

1. Substitution2. Addition3. Elimination




IV. Nucleophilic substitution A. Nucleophilic substitution

1. Nucleophile2. Leaving group

B. S N 2 reaction (1 step reaction)C. S N 1 reaction (2 steps reaction)D. Competition between S N 1 and S N 2 reactionsE. Alkanol from haloalkane (RX → ROH)

1. Alkaline hydrolysis of haloalkane2. Hydrolysis of haloalkane

F. Rate of hydrolysis of haloalkane, haloalkene and halobenzeneG. Other relevant reactions

1. Reactions of haloalkanea) Nitrile from haloalkane (RX → RCN)b) Alkylation of ammonia and amine (NH 3 → RNH 2 )c) Use of S N 2 reaction in organic synthesis

2. Reactions of alkanol

a) Haloalkane from alkanol (ROH → RX)(1) Use of chlorinating and brominating reagentb) Luca's test to distinguish 1º, 2º and 3º alkanol

3. Reactions of aminea) Action of nitric(III) acid on 1º amine

(1) 1º aliphatic amine(2) 1º aromatic amine

b) Laboratory preparation of phenol from benzenamine

V. Elimination reaction A. Elimination reaction

1. Stability of elimination product2. E2 reaction (not required in A-Level)3. E1 reaction (not required in A-Level)

B. Competition between substitution and elimination reaction1. Effect of temperature2. Effect of bulkiness of the substrate and base3. Effect of bascity of the nucleophile

C. Conditions favouring substitution and elimination reactionD. Other relevant reactions

1. Reaction of haloalkanes with alcoholic sodium hydroxide to alkene, diene andalkyne

2. Preparation of vinyl chloride3. Dehydration of alkanol




VI. Nucleophilic addition (Nucleophilic addition-elimination) A. Ad N reaction

1. Addition of HCN to carbonyl compound2. Rate of nucleophilic addition

a) Electronic effect (1) Inductive effect

(a) Effect of protonation(2) Resonance effect

b) Steric effect 3. Other relevant reactions

a) Reactions of carbonyl group(1) Reduction of carbonyl compound by LiAlH 4 and NaBH 4 (2) Addition of NaHSO 3 to carbonyl compound(3) Condensation reaction with hydroxylamine(4) Condensation reaction with 2,4-dinitrophenylhydrazine(5) Haloform reaction / Iodoform reaction

b) Reactions of carboxylic acid and its derivatives(1) Carboxylic acid and its derivatives

(a) Difference between carbonyl compound and carboxylic acid

and its derivatives(b) Reactivities of carboxylic acid and its derivatives

(2) Formation of different acid derivatives(a) Use of chlorinating agent to prepare acyl chloride(b) Formation of ester (Esterification with alkanol and phenol)(c) Formation of acid anhydride

(i) Through intramolecular dehydration by heating(ii) Through intermolecular dehydration by a very strong

dehydrating agent(iii) From acyl chloride

(d) Formation of amide (Acylation and benzoylation of amine)(3) Reaction of ester

(a) Hydrolysis of ester (4) Reaction of amide(a) Reduction of amide and other acid derivatives(b) Hofmann degradation of amide

c) Reactions of nitrile(1) Hydrolysis of nitrile and amide

(a) Dehydration of amide(2) Reduction of nitrile

VII. Electrophilic addition A. Addition of HBr to alkene

1. Markownikoff's rule.2. Reactivity of alkene towards electrophilic addition

B. Other relevant reaction1. Addition of Br 2 to alkene2. Addition of H 2SO 4 to alkene

a) Preparation of alkanol from alkene3. Hydration of alkene4. Ozonolysis of alkene5. Preparation of ethane-1,2-diol

a) Oxidative Cleavage of double bond b) Comparision of ozonolysis and oxidative cleavage

6. Oxymercuration of alkyne




VIII. Electrophilic substitution A. Representation of arenium ionB. Other relevant reaction

1. Sulphonation of benzenea) Preparation of phenol

2. Nitration of benzenea) Reduction of nitrobenzene

3. Halogenation of benzene4. Alkylation of benzene (Friedel-Crafts alkylation)5. Diazocoupling of diazonium ion

a) Colour of a substance6. Bromination of phenol

IX. Free radical Reaction A. Formation of free radical B. Free radical Substitution

1. Chain reaction e.g. chlorination of methanea) Chain initiation (chain initiating step)b) Chain propagation (chain propagating step)c) Chain termination (chain terminating step)

2. Reaction between H 2(g) and Cl 2(g) C. Free radical Addition

1. Chain reaction e.g. Polymerization of alkene2. Anti-Markownikoff orientation



Amino acids

I. Amino acids

A. ZwitterionB. Polypeptides



Oxidation and Reduction

I. Oxidation

A. Combustion of alkane

B. Oxidation of alkanol and aldehyde

C. Oxidation of aromatic side chain

II. Reduction

A. Reduction of nitrobenzene

B. Catalytic hydrogenation (Hydrogenation of alkene)



Uses of Different Compounds

I. Uses of halogeno-compounds A. Use as solvent B. Manufacture of polymer

1. Preparation of vinyl chloride

2. Physical properties of PVC and TeflonII. Uses of alcohols

A. Use as solvent B. Alcoholic drink C. Blending agent D. Ethan-1,2-diol

III. Uses of carbonyl compounds A. Preparation of urea-methanal B. Use of propanone

IV. Uses of carboxylic acids and their derivatives A. Food preservativesB. Manufacture of nylon and teryleneC. Use of ester

V. Uses of amines and their derivatives A. Azo compounds as dyes in dyeing industriesB. Amine derivatives as drugs



Determination of Structure

I. Determination of empirical formula and molecular formula A. Different kinds of formula

B. Determination of empirical formula

C. Determination of molecular formula

II. Degree of unsaturation A. Determination of degree of unsaturation

B. Meaning of degree of unsaturation

III. Sodium fusion test

IV. Test for different functional groups by wet chemistry

V. Introduction to IR and NMR spectroscopy

A. Use of infra-red (IR) spectrum in the identification of functional groups

1. More examples



Organic synthesis

I. Retrosynthetic analysis

II. Structural analysis

A. Chain length

1. Carbon chain

2. Nitrogen and Oxygen containing chain

B. Degree of unsaturation

C. Oxidation or Reduction

D. Position of functional group

III. Systematic approach to organic synthesisIV. Examples



Organic Laboratory Technique

I. Purification of organic compound

A. Solvent extraction

B. Steam distillation

C. Chromatography

D. Recrystallization

E. Filtration and Suction filtration

II. Use of quickfit apparatus

A. Handling of quickfit apparatus

B. Different setup of quickfit apparatus

1. Reflux setup

2. Distillation setup

III. Testing for purity

A. Determination of melting point

B. Determination of boiling point





Nomenclature of Organic compounds

I. Nomenclature of organic compounds


1. Methyl, primary, secondary and tertiary carbon2. α carbon, β carbon and γ carbon

B. Skeletal formula / Stick formula

II. Nomenclature of hydrocarbon

A. Alkyl and aryl group


A. Halogeno-compoundsB. Alkanols and phenol

C. Aldehydes and ketones

D. Carboxylic acids


E. Nitriles

F. Amines

G. Amino-acids

H. Ethers


V. Examples




Nomenclature of organic compounds

Topic Nomenclature of organic compounds Unit 1

ReferenceReading

13Organic Chemistry, Fillans, 3 rd Edition pg. 50–65Guidelines for systematic chemical nomenclature, HK Exam. Authority pg. 7–17Chemical nomenclature, symbols and terminology for use in school science. pg. 58–65

AssignmentReading

Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 380–381, 400–401, 414, 428, 432, 438, 439,446, 453, 467, 484, 498, 516–517Organic Chemistry, 6th Edition, Solomons, pg. 41-47, 65-75, 128-137, 284-286, 288-289, 415-417, 615-617,705-706, 792-793, 797-800, 899-900

Syllabus Nomenclature of organic compoundsClassification of hydrocarbonsPriority of functional group in IUPAC naming

Notes I. Nomenclature of organic compounds

The basic idea of the IUPAC (International Union of Pure and Applied Chemistry) system of naming is that by

following some rules, a name can be assigned to a molecule. And from the assigned name, the original moleculecan be reconstructed unambiguously.

A given structure may have several different names but a name will only give 1 structure.

For example, the structure is called C C C C

H

H

H

O

HH

H

H H H

H is called butan-1-ol or 1-butanol but no matter

butan-1-ol or 1-butanol refer to the same structure.

IUPAC system of naming is a kind of substitutive nomenclature. In substitutive nomenclature, hydrogen atoms ina named 'parent' hydrocarbons are considered to be substituted by other groups.

Actually, the IUPAC system of naming is not very strict. It is possible to have more than 1 name for a givenmolecule. But no matter which name is assigned, only 1 structure can be rebuilt.


To facilitate discussion, carbon atoms on a carbon chain can be labelled by 2 different methods.

1. Methyl, primary, secondary and tertiary carbon

The first way is based on the no. of carbon atoms attached to the carbon to be concerned.If no other carbon is attached to the carbon, the carbon atom is called a methyl carbon.If 1 carbon atom is attached to the carbon, the carbon atom is called a primary carbon, 1º carbonIf 2 carbon atoms are attached to the carbon, the carbon atom is called a secondary carbon, 2º carbonIf 3 carbon atoms are attached to the carbon, the carbon atom is called a tertiary carbon, 3º carbon

C

H

H

H

X

methyl carbon

C

CH 3

H

H

X

1° carbon

C

CH 3

CH 3

H

X

2° carbon

C

CH 3

CH 3

CH 3

X

3° carbon

e.g. CH 3CH 2OH is called a primary alkanol since the hydroxyl group is attached to a 1º carbon.




2. α carbon, β carbon and γ carbon

Another way is according to the position of the carbon relative to a substituent (functional group).If the carbon is immediately next to the substituent, it is called a α carbon.The carbon attached to the α carbon is called a β carbon.The carbon attached to the β carbon is called a γ carbon.

α Cβ C γ C

C C C C

O

carbonylcarbon

C C C X

γ C β C α C

B. Skeletal formula / Stick formula / Bond line formula

In order to make the presentation simple, sometimes skeletal formula is used instead of the full structural formulaor condensed structural formula. In the skeletal formula, all atomic symbols of carbon and hydrogen are omitted.

Full structuralformula

Condensedstructural formula

Skeletal formula /Stick formula /

Bond line formula

Full structural

formulaCondensed

structural formula

Skeletal formula /Stick formula /

Bond line formula

C C

H

H

H

H

H

H

CH 3 CH 3

CH 3CH 3

H3C CH 3

C

C C

C

H

H

H

H

H

H

CH 2

CH 2

CH

CH

C C

H

H

H

H

H

C

H

H

H

CH 3CH 2CH 3 CC

CCC

C

H

H

H

HH

H

C C

H

H

H

H

C

H

H

H

C

H

H H

CH 3CHCH 3

CH 3

CH 3CH(CH 3)CH 3

CC

CCC

CH H

HH

HH

HH

HH

HH

C

C C

C

H

H

H

H

H

H

H

H

CH 2

CH 2

CH 2

CH 2

H C C C C H

H

H

HHH

H H O

H

CH 3CH 2CH(OH)CH 3 OH

H C C C

H

H H

H

H

CH 3CHCH 2

CH 3CH CH 2

C

O

O C C H

H

H

H

H

C

H

H

H CH 3COOCH 2CH 3

O

O

H C C CC

HH

H H

HHH

CH 3C(CH 2)CH 3




II. Nomenclature of Hydrocarbon

Some examples

C C C C

H

H

H

O

HH

H

H H H

H is called butan-1-ol (or 1-butanol). (a primary alcohol)

C

CH3

H3C

H

CH2 CH3

is called 2-methylbutane (or isopentane). iso- means 1 C is attached to the next-to-end C.

C

CH3

H3C CH 2 CH3

CH3

is called 2,2-dimethylbutane (or tert-butylethane).

C

CH3

H3C

CH3

is called the tert-butyl group.

CH2 CH2 CH

CH

CH2 CH2 CH3CH3

CH3H3C

is called 4-(1-methylethyl)heptane (or 4-isopropylheptane).

CH 3 NH 2 is called methanamine (or methylamine).

C H

O

benzenecarbaldehyde (or benzaldehyde)

C

O

OH benzenecarboxylic acid (or benzoic acid)




The basis of IUPAC substitutive nomenclature is that every name consists of a root, one suffix and as many prefixes as necessary.

prefix - root - suffix

Root represents the longest carbon chain containing the principal functional group as indicated by the suffix andalso it should contain the largest no. of substituent.

Suffix represents the functional group with the highest priority among all the functional groups present.Prefix represents other functional group and arranged in alphabetical order.

e.g. CH3 CH CH 2 CH2 OH

NH 2

Cl

3-amino-2-chlorobutan-1-ol

3-amino is a prefix2-chloro is a prefix

butan- is a root-ol is a suffix

The numbers are called locants used to indicate the position of the functional group.The numbers are arranged to give

(a) the lowest possible number to the group cited by a suffix (principal functional group), then(b) the lowest possible individual numbers (not sum) to those cited as prefixes.

CH3 CH2 CH CH CH 2 CH2 CH2 CH2 CH CH 3

CH3CH3 CH3

2,7,8-trimethyldecane (not 3,4,9-trimethyldecane)

N.B. hyphen (-) is used to separate number from letter.comma (,) is used to separate number from number.

Name of the first 10 straight chain hydrocarbons

1. methane (meth-)2. ethane (eth-)3. propane (prop-)

4. butane (but-)5. pentane (penta-)6. hexane (hex-/hexa-)

7. heptane (hept-/hepta-)8. octane (oct-/octa-)9. nonane (nona-)

10. decane (dec-/deca-)

For the parent chain containing double bond or triple bond, suffixes -ene and -yne are used respectively.

C C ethene C C ethyne C C C propyne C C C C but-1-ene C C C C but-2-yne

C C C C C C C hepta-1,3-diene

If the chain contains both double bond and triple bond, the name will be ended with -yne.

C C C C C pent-3-en-1-yne C C C C C pent-1-en-3-yne

C C C C C C hex-3-ene-1,5-diyne

N.B. -e- is omitted if it is followed by a vowel e.g. y, a, o.

If both double bond and triple bond has the same lowest possible number, double bond will be given a higher priority.

C C C C C pent-1-en-4-yne

C C C C C C hexa-1,3-dien-5-yne




Classification of hydrocarbons

cyclohexane

Alicyclic (Cyclic)Aliphatic (Acyclic) Aromatic

benzene

Alkyne

CH CH

ethyne

Alkene

CH2 CH2

ethene

Alkane

CH3 CH3

ethane

Hydrocarbons

(Arene)

Aliphatic (Acyclic) : Hydrocarbon with no ring structure. It can be further classified into straight chain and branched hydrocarbon.

Alicyclic (Cyclic) : Hydrocarbon with ring structure.

Aromatic : A special kind of ring compound with extra stability due to delocalization of electrons.

Benzene and naphthalene are aromatic. Originally, 'aromatic' means having a sweet smell sincethe firstly discovered aromatic compounds possess special smell.

A. Alkyl, vinyl, allyl and aryl group

When a hydrogen is removed from an alkane, the structure is called an alkyl group R–. It is cited by adding -yl.

C

H

H

H

methyl group

When a hydrogen is removed from an alkene, the structure is called a vinyl group. It is cited by adding -enyl.

C

H

H C

H

C

H

H

prop-2-enyl group

When a hydrogen is removed from an alkyne, the structure is called an allyl group. It is cited by adding -ynyl.

C

H

H C C

H

prop-1-ynyl group

N.B. The carbons are counted from the point joining to the main chain.

When a hydrogen is removed from a benzene, the structure is called a phenyl group Ph– or .

phenyl group

However if a phenyl group is attached to a –CH 2 –, it is called the benzyl group.

CH2 benzyl group

Phenyl group and other substituted phenyl group (e.g. chlorophenyl group) are collectively called aryl group Ar–.

NH 2 phenylamine (or more accurate benzenamine).





A. Halogeno-compounds

The presence of halogen (–X) is always cited by the prefixes (fluoro-), (chloro-), (bromo-) and (iodo-).

C

Cl

H H

H

chloromethane C C

Br Br

H

H H

H 1,2-dibromoethane

B. Alkanols and phenol

The presence of hydroxyl group (–OH) is cited by suffix (-ol) or prefix (hydroxy-)

C C OH

H

H

H

H

H ethanol C C

HH

H H

OH OH

ethane-1,2-diol C C

O

OHHO

H

H

hydroxyethanoic acid

N.B. (-e-) of the root is omitted if it is followed by an vowel e.g. -o- or -y-.

When a hydrogen on the benzene is substituted by a hydroxyl group, the compound is called phenol.

OH phenol OHO2 N 4-nitrophenol

C. Aldehydes (Alkanals) and ketones (alkanones)

Aldehyde (Alkanal) contains a terminal carbonyl group C

O

. It has the general formula CR

O

H .It is cited by suffix (-al) or prefix (oxo-).

CH3 C

O

H ethanal C C

O

H

O

OH oxoethanoic acid

Ketone contains a non-terminal carbonyl group CO

. It has the general formula CR R O

It is cited by suffix (-one) or prefix (oxo-) if the C is counted in the root or prefix (carbonyl-) if the C is notcounted in the root.

CH3 C

O

CH2 CH2 CH3 pentan-2-one CH3 C

O

CH2 C

O

OH 3-oxobutanoic acid

D. Carboxylic acids (Alkanoic acid)

Carboxylic acid contains a carboxyl group C

O

OH . It is cited by suffix (-oic acid) or (-carboxylic acid) or by

prefix (carboxy-).

C

O

OHCH3 ethanoic acid C

O

OHCH2C

O

HO propanedioic acid.

C

O

OH benzenecarboxylic acid (common name : benzoic acid)

HO C

O

C

H

H

N+

H

H

H Cl- (carboxymethyl)ammonium chloride

N.B. 1. locant is not necessary for dioic acid since the carboxyl groups must occupy the ends of the C chain.2. When suffix (-carboxylic acid) or prefix (carboxy-) is used, the C in the carboxyl group is not

counted as a part of the parent chain.





Esters, acid chlorides, anhydrides and amides are all derivatives of carboxylic acid.

The derivative is formed by replacing –OH hydroxyl group with other functional groups.

–OH replaced by General formula Example

Carboxylic acid

C

O

R OH C

O

OHCH3 ethanoic acid

Ester –OR

C

O

R O R'

C

O

OCH3 CH2CH2CH3 propyl ethanoate

Acid chloride(acyl chloride)

–Cl

C

O

R Cl

C

O

ClCH3 ethanoyl chloride

Anhydride(formed bydehydration of alkanoic acid)

–OOCR

C

O

R O C

O

R' C

O

O C

O

CH3 CH3 ethanoic anhydride

C

O

O C

O

CH3 CH2CH2CH3

butanoic ethanoic anhydrideAmide –NH 2

C

O

R NH 2 C

O

NH 2CH3 ethanamide

R C

O

is called the acyl group. e.g. C

O

CH3 ethanoyl group.

E. Nitriles

Nitrile contains cyano group –C ≡ N. It is cited by suffix (-nitrile) or (-carbonitrile) for cyclic nitrile or by prefix(cyano-)

C NCH 3 ethanenitrile C N benzenecarbonitrile

C N cyclohexanecarbonitrile CCC OH

OH

H

N cyanoethanoic acid

N.B. When suffix (-carbonitrile) or prefix (cyano-) is used, the C in the cyano group is not counted as a part of the parent chain.




F. Amines

Amine contains amino group –NH 2. Amine is alkyl or aryl derivative of ammonia, with the hydrogen replaced byalkyl group or aryl group. It is cited by suffix (-amine) or prefix (amino-).

CH 3 NH 2 methanamine (methylamine)

CH3 CH CH 2 NH 2

CH3

2-methylpropan-1-amineCH 3 N CH 2CH 3

HN-methylethanamine (ethylmethylamine)

CH3 N CH 2CH3

CH2CH2CH3N-ethyl-N-methylpropan-1-amine (ethylmethylpropylamine)

NH 2 benzenamine (phenylamine).

N

HN-phenylbenzenamine (diphenylamine).

CH 2 CH 2 OHH2 N 2-aminoethanol

N.B. The locant N- is used to indicate the alkyl or aryl group is attached to the N, not the parent C chain.

G. Amino-acids

Amino-acid contains both amino group –NH 2 and carboxyl group –COOH.

CH2 CH2 C OH

O

H2 N 3-aminopropanoic acid

H. Ethers

Ether contains –O–R group. It is cited by prefix (R-oxy-) e.g. methoxy-.

CH 3 O CH 2 CH 3 methoxyethane (ethyl methyl ether)

CH 2CH 3OCH 3CH 2 ethoxyethane (diethyl ether)

CH 2CH 2O O CH 3H3C 1,2-dimethoxyethane

N.B. In some circumstances, -oxy- is used alone to indicate the presence of oxygen.


If a compound has more than 1 functional group. The functional group with the highest priority would be namedas the suffix and all others would be named as prefixes.

e.g.

C

OO

OH

C CH 3

O

C C

O

HO

O

O CH 2 CH3

O CH 2CH3C N

HO

2-ethanoyloxybenzenecarboxylicacid

ethoxycarbonylmethanoic acid 4-ethoxy-2-hydroxybenzenecarbonitrile




V. Examples

Steps in writingIUPAC name

C CH 2 CH CH 3

CH3

CH3 OH OH CHCH 2CH2CHCH 2CHOC

CH3

CH3

CH3

1. Identify the principal functional group hydroxyl group (–OH) alkanal group (–CHO)

2. Identify the parent chain containing the principal functional group

C C C C C

OH OH

C C C C C C C C

O

H

3. Number the parent chain so that the principal functional group is on thelowest numbered carbon.

pentane-2,4-diol oct-6-enal

4. Label other substituents as prefixes. 2-methyl 3-methyl, 7-methyl

5. Group the prefixes, arrange them inalphabetic order and join with the root.

2-methylpentane-2,4-diol 3,7-dimethyloct-6-enal


C2H5C(CH 3)=CHCH 2CH 2CO 2H

C C C CCC CO

OH

C

C C C

OH

1. Identify the principal functional group carboxyl group (–COOH) hydroxyl group (–OH)

2. Identify the parent chain containing the principal functional group C C C CCC C

O

OH

C C C

OH

3. Number the parent chain so that the

principal functional group is on thelowest numbered carbon.

hept-4-enoic acid propan-1-ol

4. Label other substituents as prefixes. 5-methyl 1-phenyl

5. Group the prefixes, arrange them in

alphabetic order and join with the root.

5-methylhept-4-enoic acid 1-phenylpropan-1-ol


C C C C

C

C

C C C C

C

C

O C C

1. Identify the principal functional group double bond (C=C) Nil

2. Identify the parent chain containing the principal functional group

C C C C

C C C C

3. Number the parent chain so that the principal functional group is on the

lowest numbered carbon.

but-1-ene butane

4. Label other substituents as prefix. 3-methyl, 3-methyl 1-ethoxy, 3-methyl, 3-methyl

5. Group the prefixes, arrange them inalphabetic order and join with the root.

3,3-dimethylbut-1-ene 1-ethoxy-3,3-dimethylbutane





Abbreviation Functional group Structure Example

R– Alkyl group

Me– Methyl group C

H

H

H

C H

H

H

H

Me–H

Et– Ethyl group C C

H

H

H

H

H

C C

H

H

H

H

H

H

Et–H

Ph– Phenyl group H

Ph–H

Ar– Alkyl group

H

H

Ar–H

X– Halogeno group Cl–, Br–, I–

Glossary nomenclature IUPAC parent chain prefix root suffix locantfunctional group aliphatic (acyclic) alicyclic (cyclic) alkane alkene alkyne

alkyl group vinyl group allyl group aryl group phenyl group alkanol phenolhydroxyl group aldehyde (alkanal) ketone (alkanone) carbonyl groupcarboxylic acid (alkanoic acid) carboxyl group ester acid chloride (acyl chloride) anhydrideamide acyl group nitrile amine amino group ether

Past PaperQuestion

94 2C 8 c i ii iii96 2C 9 c i ii97 2B 5 d i ii iii98 2B 7 c i ii

94 2C 8 c i ii iii8c Give a systematic name to each of the following compounds.

iCH3CH CH

CH3 CH 3

CH2CH CH 2 1

4,5-dimethylhex-1-ene 1 mark ii

CH3CH2CH2 O C CH 2CH 2CH3

O

1

propyl butanoate 1 miii

CH3CH2 C

O

CHCH 2CH 3

OH

1

4-hydroxyhexan-3-one 1 mark C Badly-answered. Many candidates did not know the general rules for systematic naming.



Nomenclature of organic compounds 96 2C 9 c i ii9c Give a systematic name to each of the following compounds :

(Deduct ½ marks for each minor mistake. Max. Deduction = 2 marks)i

CH3 CH CH 2 CH CH 3

CH3 OH

1

4-methylpentan-2-ol / 4-methyl-2-pentanol 1 mark ii

O

CH3

CH3

CH3

1

2,5,5-trimethylcyclohex-2-enone 1 mark C Disappointing answers for (i). Some candidates did not know the basic rules of nomenclature. Very few gave the

correct name for (ii).

97 2B 5 d i ii iii5d Give a systematic name to each of the following compounds: 3

i N

CH 3CH 3

ii

C CH(CH 3)CH 2CH 3CH 3CH 2

O

iii

C

C

H

H

CH 3

HO 2C

98 2B 7 c i ii7c Give a systematic name to each of the following compounds : 2

i CH 3

CH 2CH 3 ii CH 3CH=CHCH 2CHO



Acid-Base Theory

I. Definition of acid-base A. Arrhenius definition

B. Bronsted-Lowry definitionC. Lewis definition

II. Strength of acid and base A. Leveling effect

III. Factors determining the strength of a Bronsted-Lowry acid A. Relative stability of the conjugate base comparing with the acid

1. Inductive effect(1) Amine vs Amide

2. Resonance effecta) Resonance effect on a benzene ring

(1) Phenol(2) Aromatic amine

3. Intramolecular hydrogen bond4. Solvation / hinderance to solvation

a) Steric hinderance to solvationb) Effect of solvent

5. Strength of the bond between the proton and the conjugate base6. Electronegativity of the atom




Acid-base Theory Unit 1 Page 1

Topic Acid-base Theory Unit 1

ReferenceReading

14.1Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 41–43, 138–140, 185–186, 252–254Organic Chemistry, Solomons, 5th Edition pg. 86–109, 830–834, 942–943Organic Chemistry, Fillans, 3rd Edition pg. 218–220, 301–313

Organic Chemistry, Morrison Boyd, 6th

Edition pg. 33–35, 729–736, 849–852, 903–905Organic Chemistry, 6th Edition, Solomons, pg. 90-93, 96-101

Syllabus Acid-base TheoryExpression of K a and pK a Relative strength of acids and bases

Notes A. Acid-base Theory

At the early stage of development of acid-base theory, acid was onlydescribed as a sour substance. And base is only a substance which neutralizesthe sour taste of an acid.

Nevertheless, scientists find that acids and bases bear some other properties.According to these properties, they developed a series of definitions todescribe acid and base. They are called Arrhenius definition, Brønsted Lowrydefinition and Lewis definition. The coverage of the latter definition is

boarder than the former one.

Common

Sense

A r r h

e n i u s d e f i n i t

i o n

B r φ

n s t

e d L o w r y d e f i n i t

i o n

L e w i s d e f i n i t i o n

I. Definition of acid-base

A. Arrhenius definition

In 1884, Swedish chemist Svante Arrhenius proposed that

Acid – a hydrogen-containing compound that, when dissolved in water, produces hydroxonium ions, H 3O+(aq) .Base – a substance that, when dissolved in water, produces hydroxide ions, OH -.

Examples of Arrhenius acide.g. H 2SO 4(l) + H 2O(l) → H3O+

(aq) + HSO 4-(aq)

HSO 4-(aq) + H 2O(l) → H3O

+(aq) + SO 4

2-(aq)

Examples of Arrhenius basee.g. NaOH (s) + aq → Na+

(aq) + OH -(aq)

KOH (s) + aq → K +(aq) + OH -(aq)

This is the definition learned in certificate level.




B. Brønsted-Lowry definition

Arrhenius theory is criticized that1. Acid is restricted to hydrogen-containing species and base is restricted to hydroxide-containing species.2. The theory is only applicable to aqueous medium where a lot of acid-base reaction takes place in the absence

of water.

In 1923, Danish chemist Johannes Brønsted and British chemist Thomas Lowry proposed, a boarder definition, thatAcid – a proton donor Base – a proton acceptor

e.g. HCl (aq) + CuO (s) → Cu2+Cl -2(aq) + H 2O(l)

proton protondonor acceptor

e.g. NH 3(aq) + H 2O(l) d NH 4+

(aq) + OH -(l)

proton proton proton protonacceptor donor donor acceptor (base) (acid) (conjugate (conjugate

acid of NH 3(aq) ) base of H 2O(l))

e.g. HCl (aq) + H 2O(l) d H3O+(aq) + Cl -

(aq) proton proton proton protondonor acceptor donor acceptor (acid) (base) (conjugate (conjugate

acid of H 2O(l)) base of HCl (aq))

e.g. H 2O(l) + H 2O(l) d H3O+(aq) + OH -

(aq) proton proton proton protonacceptor donor donor acceptor (base) (acid) (conjugate (conjugate

acid of H 2O(l)) base of H 2O(l))

When a Brønsted-Lowry acid, a proton donor, donates a proton, it becomes a potential proton acceptor and iscalled the conjugate base of the acid. Stronger a proton donor, weaker will be the conjugate base.

Conversely, when a Brønsted-Lowry base, a proton acceptor, accepts a proton, it becomes a potential proton donor and is called the conjugate acid of the base. Stronger a proton acceptor, weaker will be the conjugate acid.

All Arrhenius acids and bases can be classified into Brønsted-Lowry acid and base accordingly.




C. Lewis definition

The American chemist Gilbert N. Lewis (1923) proposed an even broader definition for acid and base. He proposed that

Acid – an electron acceptor Base – an electron donor

This definition offers many advantages, includingi. The acids are not limited to compounds containing hydrogen.ii. It works with solvents other than water.iii. It does not require formation of a salt or of acid-conjugate pairs.

Since all chemical species are potential electron acceptor or electron donor, virtually, all chemical species areeither Lewis acid or Lewis base.

e.g. H 3 N: (g) + BF 3(g) → H3 N → BF 3(s) electron electrondonor acceptor (Lewis base) (Lewis acid)

Most of the chemical reaction is caused by redistribution of electrons which leads to rearrangement of atoms andformation of a new substance. Therefore, all kind of reactions may be considered as Lewis acid-base reaction.Concept of Lewis acid-base is very useful in describing reaction mechanisms.

II. Strength of acid and base

In measuring the strength of acid, Brønsted-Lowry definition is usually used.

Strength of an acid can be measured by an equilibrium constant called acidity constant or acid dissociationconstant, K a .

For an acid, HA in water

HA (aq) + H 2O(l) d H3O+(aq) + A -

(aq) Equilibrium constant, K eq =[H 3O+

(aq)][A-(aq)]

[HA (aq)][H 2O(l)]

Acidity constant, K a is defined as

K a =[H 3O

+(aq)][A

-(aq)]

[HA (aq)]= K eq[H2O(l)] where [H 2O(l)] is a constant since water is the solvent in large excess.

For a stronger acid, more HA (aq) molecules will dissociate into H 3O+(aq) ion and A -

(aq) ion and give a larger value for K a.

Similar to concentration of H +(aq) ion, K a can also be expressed in a negative log scale.

pH = - log [H +] A low pH means a high concentration of H +(aq) ion.

pK a = - log K a A low pK a means a larger value for K a and a stronger acid.




Relative strength of selected acids and their conjugate bases

Stronger an acid, weaker will be its conjugate base.Stronger a base, weaker will be its conjugateacid.

N.B.Redox reaction can be considered as acompetition for electron.

oxidizing agent + e - d reducing agent

In redox reaction, only strong oxidizing agentreacts with reducing agent.

Similarly, acid base reaction can be consideredas a compeition for proton.

cojugate base + H + d conjugate acid

In acid base reaction, only strong acid reactwith strong base.

The strength of an acid or a base is also an indicator of their stability. A strong acid or base is less stable thana weak acid or base. A strong acid tends to react with a strong base to form a weak conjugate base and weak conjugate acid.

A. Leveling effect

An acid stronger than hydroxonium ion H 3O+

(aq) , does not show difference in acidity in water. The water willconvert all those acid molecules into H 3O+

(aq) ions.

HCl (aq) + H 2O(aq) → H3O+

(aq) + Cl -(aq)

strong strong weak weak acid base acid base

HBr (aq) + H 2O(aq) → H3O+

(aq) + Br -(aq) strong strong weak weak acid base acid base

A base stronger than hydroxide ion, e.g. NH 2- amide ion, also does not exist in aqueous medium. Water will

convert all those base molecules into OH -(aq) ions.

NH 2- + H 2O(l) → NH 3(aq) + OH -

(aq)

This is known as leveling effect of solvent.

Glossary Arrhenius acid-base Brønsted-Lowry acid-base Lewis acid-base

Past PaperQuestion




Topic Acid-base Theory Unit 2

ReferenceReading

14.0 14.2–14.4Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 41–43, 138–140, 185–186, 252–254Organic Chemistry, Solomons, 5th Edition pg. 86–109, 830–834, 942–943Organic Chemistry, Fillans, 3rd Edition pg. 218–220, 301–313


Edition pg. 33–35, 729–736, 849–852, 903–905

AssignmentReading

Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 471, 499–501, 519–520Organic Chemistry, 6th Edition, Solomons, pg. 102-118, 320, 433-434, 795-796, 903-905, 970-972

Syllabus Factors determining the strength of an acid

Notes III. Factors determining the strength of a Bronsted-Lowry acid

A. Relative stability of the conjugate base comparing with the acid1. Inductive effect2. Resonance effect3. Intramolecular hydrogen bond4. Solvation / hinderance to solvation5. Size of the conjugate base6. Electronegativity of the atom

A. Relative stability of the conjugate base comparing with the acid

The strength of an acid is mainly determined by the relative stability of the conjugate base comparing with theacid.

The equilibrium constant is related to the standard free energy change ( ∆ Go) equation

∆ Go = -2.303 RT log K eq = -RT ln K eq = -RT ln K a[H 2O(l)]

where ∆ Go = ∆ Ho - T∆ So

∆ Go is an indicator of the relative stability of the product comparing with the reactant. A very negative valuemeans the product is more stable than the reactant. For a reaction with small change in entropy, i.e. reactionwithout change in physical state, the ∆ Go is roughly the same as ∆ Ho.

Since K a = K eq [H2O(l)], a very negative value of ∆ Go means a very strong acid. The acidity of the acid increase asthe relatively stability of the conjugate base increase comparing with the acid molecule.




I. Inductive effect

C

H

H

H

C

O

O H H2O C

H

H

H

C

O

O - H3O++ +

Upon ionization in water, an ethanoic acid molecule forms a carboxylate ion and a hydroxonium ion. An electron-withdrawing (-I) groups stabilize the negative charge of the carboxylate anion by dispersing the negative charge.This increases the acidity of the acid.

Inductive effects of halogen atoms on acid strengthAcid Structure pK a (H2O) at 25ºC

Ethanoic CH 3CO 2H 4.76 Relatively weak acid

Iodoethanoic ICH 2CO 2H 3.12Chloroethanoic ClCH 2CO 2H 2.85Trichloroethanoic Cl 3CCO 2H 0.7

Trifluoroethanoic F 3CCO 2H 0.23 Relatively strong acid

Butanoic CH 3CH 2CH 2CO 2H 4.81

2-Chlorobutanoic CH 3CH 2CHClCO 2H 2.81 Relatively strong acid

3-Chlorobutanoic CH 3CHClCH 2CO 2H 4.054-Chlorobutanoic ClCH 2CH 2CH 2CO 2H 4.52 Relatively weak acid

Note : pK a is a log scale, increase by 1 in pK a means 10 folds decrease in acid strength.

As inductive effect is transmitted through σ - bond, it decreases quickly with increasingdistance from the negative centre. It is only ashort range effect.

Electron-donating : positive inductive effect (+I)

Electron-withdrawing :negative inductive effect (-I)

Inductive effects of various Groups




2. Resonance effect

ethanoic acid phenol ethanolAcid CH 3COOH Ph-OH CH 3CH 2OH

pK a 4.76 9.9 16

Experimentally, ethanoic acid is 1.7 × 1011 times more acidic than ethanol. This cannot be explained satisfactorily

by the presence of extra oxygen atom. Comparing choroethanoic acid and ethanoic acid, the presence of thechlorine atom only contribute a 81 times difference in acidity.

The exceptionally high acidity of ethanoic acid can be explained by the resonance stabilized ethanoate ion formedfrom ethanoic acid.

Comparing the structure of ethanoic acid and ethanoate ion, The ethanoate ion has identical resonance structures.In contrast, the two resonance structures of the carboxylic acid molecules are not identical. One of the resonancestructure even involves charge separation and makes it less stable.

Species with identical resonance structure implies a more evenly distributed electron cloud which is relatively morestable. This makes ethanoate exceptionally stable and ethanoic acid exceptionally acidic.

CO

OHCH 3

CO

OHCH 3

CO

O-CH 3

CO-

OCH 3

H++f

pK a = 4.76 As the product is stabilized more than the reactant, theK a of the acid will increase.

Inductive effect and Resonance effect are collectivelycalled Electronic effect.

∆ G'

Free energy

Reaction coordinate

∆ G' is more negative (less positive) than ∆ G

A carboxylic acid yields a resonance-stabilized anion; it is a stronger acid than an alcohol.(The plots are aligned with each other for easy comparison.)




(1) Amine vs Amide

For a similar reason, amine is more basic than amide.

CH 3 CH 2 N

H

HCH 3 CH 2 N

H

H

H+

>+ H+

Alkyl group is an electron-donating group, it will stabilize the conjugate acid formed and make the amine more

basic.

CH 3 C

O

N

H

HCH 3 C

O

N

H

H

H+<+ H+

Just the opposite, acyl group is an electron-withdrawing group, it will destabilize the conjugate base formed andmake the amide less basic.

a) Resonance effect through a benzene ring

(1) Phenol

O

H+

-

+O

H

-

O

H

- H +

O

-

O

-

O-

O

Since the phenoxide ion is more stabilized by resonance than the

phenol molecule, phenol is much more acidic than other alcohols.e.g. pK a of phenol = 9.9

pK a of CH 3CH 2OH = 16(This is a wrong explanation used in HK A-level, for more up-to-date explanation, please read Solomon pg.942–943)

Though phenol is acidic, phenoxide ion does not have identicalresonance structures. Phenol is less acidic than carboxylic acid.Thus, phenol reacts with strong alkali, e.g. NaOH, but not withweak alkali, e.g. Na 2CO 3 or NaHCO 3.

This can be used as a test to distinguish phenol and ethanoicacid.

Resonance effects of various groups

Vorlander's rule of resonance effect on benzene ring1) Saturated groups (containing only single bonds between all atoms) are electron donating (+R).

e.g. F , Cl , O C

O

R (has a lone pair on the atom)2) Unsaturated groups (containing multiple bonds between any atoms) are electron withdrawing (-R).

e.g. C N , C

O

, N

O

O (has a multiple bond between the first and the second atom)




The acidity of phenol increases as the benzene ring is substituted by nitro group, –NO 2. pK a of 4-nitrophenol = 7.15 pK a of 2,4,6-trinitrophenol = 0.42Since the nitro group is very far from the hydroxyl group, the inductive effect is very negligible. Comparing withinductive effect, reasonance effect has a much longer range.

++ +

O

NO O -

-O

NO O- -

-

-

O

NO O

By resonance, the negative charge on the 4-nitrophenoxide ion is more dispersed. This gives extra stability to the4-nitrophenoxide ion and make nitro substituted phenol more acidic.

(2) Aromatic Amine

The basicity of amine can also be explained in the same way.

NH H N H

H

H

pK b 9.44.83.2

CH 2CH 2CH 2CH 3 N

H

H

basicity decreases

The high basicity of butan-1-amine can be explained by the presence of the electron-donating alkyl group. This

stabilizes the conjugate acid CH 3CH 2CH 2CH 2 NH 3.

Consider the equilibrium

NH H NH H

H

H++

is an electron withdrawing group by inductive effective -I. It destabilizes the conjugate acid

NH H

H

and make

NH H

the least basic one. Furthermore,

NH H

is also stabilized by resonance but

NH H

H

is not.

Therefore, the equilibrium position will be lying on the left hand side.

NH H NH H NH H




3. Intramolecular hydrogen bond

Existence of intramolecular hydrogen-bond may have positive or negative effect on the acidity of an acid.

Example 1 – butenedioic acid

C

C

CO 2H

CO 2HH

H

(Z)-butenedioic acid / cis-butenedioic acid (maleic acid) pK a1 = 1.83 pK a2 = 6.07

C

C

CO 2H

HO2C

H

H(E)-butenedioic acid / trans-butenedioic acid (fumaric acid) pK a1 = 3.03 pK a2 = 4.44

For a dibasic acid, there are 2 pK a values. The first dissociation of maleic acid is more complete but the seconddissociation is just the reverse.

Comparatively, only the acid molecule that undergoes 1st dissociation may undergo 2nd dissociation. Therefore,the value of pK a2 is not as important as pK a1 in determining the acidity of an acid. i.e. maleic acid is more acidicthat fumaric acid.

This can be explained by a H-bond stabilized conjugate base of maleic acid.

C

CC

C

O

O

O

O

H

H

H

HC

CC

C

O

O

O

O

H

H

HC

CC

C

O

O

O-

O-

H

H

H+ 2H++ +

δ -

δ -

H-bond stabilized

The first conjugate base is stabilized by formation of intramolecular H-bond which causes a stronger 1stdissociation and weaker 2nd dissociation.

C

CC

O

O HH

HC

O

OH

H+ 2H++ +C

CC

O

OH

HC

O

OHC

CC

O

OH

HC

O

O

Not H-bond stabilized

H2A

2H + + A 2-

H+ + HA -

Dissociation of maleic acid Dissociation of fumaric acid

H2A

2H + + A 2-

H+ + HA -




Example 2 – hydroxybenzenecarboxylic acid

CO

O

OH

H

2-hydroxybenzenecarboxylic acid

OHCO

HO 4-hydroxybenzenecarboxylic acid

2-hydroxybenzenecarboxylic acid is about 40 times more acidic than that of 4-hydroxybenzenecarboxylic acidsince the conjugate base is once again stabilized by intramolecular H-bond.

CO

O

OH

H

CO

O-

-Oδ - δ -

CO

O

OH

4. Solvation / hinderance to solvation

a) Effect of solvent

Acid shows higher acidity in polar solvent. e.g. Hydrogen chloride is a strong acid in water but not acidic inmethylbenzene. Ions can be stabilized in polar solvent but not in non-polar solvent.

b) Steric hinderance to solvation

Bulky acid ⇒ hindered negative centre ⇒ less solvation stabilization ⇒ unstable conjugate base ⇒ lower acidity

The theory regarding the strength of acid is also applicable to the strength of base. The strength of a base is alsodepending on the relative stability of the conjugate acid comparing with the parent base.

Amines exhibit a similar steric hinderance to solvation.In gas phase where there is no solvation, the order of base strength is (CH 3)3 N > (CH 3)2 NH > CH 3 NH 2 [3º > 2º >1º] which is governed by the positive inductive effect (+I) of methyl group only.But in water, the order of base strength of methylamines is (CH 3)2 NH > CH 3 NH 2 > (CH 3)3 N [2º > 1º > 3º]. Theresult is a combination of inductive effect and steric hinderance to solvation. The bulky conjugate acid formed,aminium ion, is less stabilized by solvation.




5. Size of the conjugate base - (Strength of bond between the proton and the conjugate base)

On moving down a group (from F–H to I–H), the bond length increases and bond strength decreases tremendously(from 568 kJmol -1 to 298 kJmol -1).

Acidity H–F < H–Cl < H–Br < H–I pK a 3.2 -7 -9 -10

bond dissociation energy / kJmol-1

568 432 366 298Ionic size of the conjugate base (X -) → g increa sin

As the size of the conjugate base gets larger, the charge density on the X - gets lower and becomes more stable. As aresult HI is most acidic one.

Indeed, acidity is depending on relative stability instead of bond strength. However, an acid molecule possessing aweak bond is usually unstable, it tends to lose the proton to the base (water) to form a more stable system.

H–I (aq) + H 2O(aq) d I-(aq) + H 3O

+(aq)

less stable combination more stable combination

The overall change can be considered as the breaking of a weak H–I bond and formation of a strong H–O bond, asa result the overall stability of the system is enhanced.

N.B. Solely strength of bond would not be accepted as a valid explanation to the difference in acidity.

6. Electronegativity of the atom

Besides the dispersion of the negative charge on the conjugate base, an increase in attraction between the electronsand the nucleus also enhances the stability of the conjugate base.

High electronegativity makes an atom attract the negative charge more strongly and form a more stable anion.Therefore, F - ion is more stable than OH - ion as F is more electronegativity than O.

On moving across a period (from C–H to F–H), the bond strength only increases moderately from 425 kJmol -1 to568 kJmol -1. The effect of difference in size is outweighed by the big difference in electronegativity.

Acidity H 3C–H < H 2 N–H < HO–H < F–H pK a 50 38 15.74 3.2 bond dissociation energy / kJmol -1 425 431 498 568Electronegativity of the atom 2.5 3.0 3.5 4.0(Pauling scale)





hybridization s character pK a

Ethyne C CH H sp 50% 25

EtheneC C

H

H

H

H

sp2 33.3% 44

EthaneC C

H

H

H

H

H

H

sp3 25% 50

The stronger the s character a carbon atom possesses, the closer will be the bonding electron to the nucleus. This is because s orbital is closer to the nuclei than the p orbital. The carbon atom possessing more s character behaveslike a more electronegative atom and leads to a more stable carbanion.

Stability C CH C CH

H

HC CH

H

H

H

H

> >

Consequently, the acidity of hydrocarbon decreases with following order : alkyne > alkene > alkane.

Glossary Arrhenius acid-base Brønsted-Lowry acid-base Lewis acid-base acidity constantleveling effect free energy change inductive effect resonance effect electronic effectVorlander's rule intramolecular hydrogen bond steric hinderance

Past Paper

Question

91 2C 9 a ii92 1A 3 c 92 2C 9 b93 1A 3 f i ii94 1B 4 a95 1A 3 b96 1A 2 e ii iii 96 1A 3 b99 2A 4 c i

91 2C 9 a ii9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs.

Describe what you would observe in each case.ii COOH

OH

2

Add solution of sodium hydrogencarbonate to the compounds. PhCOOH will react to give CO 2 bubble(effervescence observed). PhOH give no observable change.

COOH+ NaHCO 3

COO - Na +

+ CO2

92 1A 3 c3c In aqueous media, why does ammonia act as a base whereas hydrogen fluoride acts as an acid? 2

Higher electronegativity of F than N on HF makes HF acidic because the proton experiences less electron sharingand is easier to release. 1 mark The prominent lone pair on N in NH 3, comparing with the withheld electrons pair on F in HF, donates the lone

pair more readily and favours basic properties. 1 mark C This question required reasons, rather than statements in the form of equations. Various explanations might have

been provided; such as a discussion of the different shielding and ease of release or acceptance of protons whenthere is a change in the electronegativity of the atom to which they are bonded.



Acid-base Theory Unit 2 Page 1092 2C 9 b9b Arrange the following hydroxy-compounds in order of increasing acidity. Explain your order.

OH OH

NO2

OH

5

OHOH

NO 2

OH

< <

1 mark Acidity depends on the equilibrium HA d H+ + A - 1 mark Phenoxide ion is stabilized by resonance , ½ mark

i.e.

-O

-

O

-

O

-

O

½ mark ∴ phenol is more acidic than cyclohexanol.The nitro group in the nitrophenol is electron withdrawing which attracts electron and is capable of resonance,thus further stabilizes the phenoxide ion. 1½ mark

O

NO O- -

O

NO O -

-

½ mark C Many candidates did not use the equilibrium between the acid and its conjugate base to interpret the acid strength.

Candidates described –NO 2 group incorrectly in this answer as a deactivating group rather than an electron-withdrawing group. A few candidates regarded the nitro-group as electron-donating.

93 1A 3 f i ii3f Consider the following compounds:

OH

CH 3

ZOH

CH

O

Y

and

i Account for the fact that both Y and Z are acidic.

Acidity depends on the equilibrium HB (aq) d H+(aq) + B -

(aq) ½ mark Both compounds have phenol functional group which can be involved in above equilibrium ½ mark In cases of phenols, equilibrium lies to the right due to resonance stabilization of the phenoxide ion ½ mark Explanation of resonance stabilization :

O

R

-O

R

-

O

R -

O

R

-

½ mark ii Which compound, Y or Z, is the stronger acid? Explain. 2

Y is the stronger acid. ½ mark

This is because the electron withdrawing C

Ogroup further stabilizes the phenoxide ion. 1 mark

-

OO

HO

O

H

-

½ mark C Most candidates explained the acidity of phenols in terms of the weakening of the O–H bond by resonance,

instead of a more formal consideration in terms of the stability of the phenoxide ion. Some could not writeresonance structures correctly. Few candidates explained acidity in terms of equilibrium.



Acid-base Theory Unit 2 Page 1194 1B 4 a4a Describe a chemical test to show that ethanoic acid is a stronger acid than phenol. (Tests involving the use of a

pH meter or indicator paper are NOT accepted.)2

Addition of Na 2CO 3(aq) or NaHCO 3(aq) to CH 3COOH (aq) gives CO 2(g) / effervescence. 1 mark Addition of Na 2CO 3(aq) or NaHCO 3(aq) to phenol does not give CO 2(g) /effervescent 1 mark Or Only CH 3COOH (aq) reacts with MgInsoluble base oxide e.g. CuO (s) is soluble only in CH 3COOH (aq)

C Some candidates did not read the question with enough care. Instead of giving a chemical test, they employed physical methods (e.g. conductivity measurement) and therefore were not credited with marks. Others erroneouslystated that the stronger acid would require a larger volume of base for neutralization.

95 1A 3 b3b Arrange the following carboxylic acids in the order of increasing acidity. Explain your arrangement.

ClCH 2CO 2H, ClCH 2CH 2CO 2H and FCH 2CO 2H3

ClCH 2CH 2CO 2H < ClCH 2CO 2H < FCH 2CO 2H 1 mark Inductive effect by electron withdrawing group stabilizes the carboxylate anions more than the carboxylic acids.

½ mark F is more electronegative than Cl and closest to the carboxylate / carboxyl group ∴ FCH 2CO 2H is most acidic

½ mark

the Cl substituted carboxylic acids, inductive stabilization operates most effectively with decreasing distance. ∴ the order of increasing acidity follows the decreasing distance. 1 mark

C It was surprising to see that many candidates answered that the stronger the acid, the weaker the O–H bond.Candidates should know that bond strength does not have a direct bearing on acidity and only the relative stabilityof the acid and the conjugate base dictate the acidity.

96 1A 2 e ii iii2e Which is the stronger acid in each of the following pairs of substances ? Briefly explain your choice.

ii HClO 3 (aq) , HClO 4(aq) 2HClO 4 ½ mark the conjugate base ClO 4

- ½ mark possesses a large no. of oxo croup (O atoms) ½ mark

is stabilized to a greater extent by resonance / mesomeric effect. ½ mark Or,In HClO 4, the Cl has a higher oxidation state and ½ mark a larger number of oxo groups (O atoms). ½ mark which can induce a higher positive charge density on H. ♣ ½ mark

iii HMnO 4 (aq) , H 2CrO 4(aq) 2HMnO 4 ½ mark The conjugate base MnO 4

- ½ mark possesses a large no. of oxo group than HCrO 4

- ½ mark is stabilized to a greater extent by resonance / mesomeric effect. ½ mark Or,Mn has a higher O.S. than Cr , ½ mark MnO 4

- is singly charged while CrO 42- is doubly charged, ½ mark

thus the conjugate base MnO 4- is more stable. ½ mark Or,Mn has a higher oxidation state ½ mark and a larger number of oxo groups, ½ mark which can induce a higher positive charge density on H. ♣ ½ mark (For answers marked with ♣ , maximum 1 mark)

C It was mentioned in the 1995 Subject Report that bond strength does not have a direct bearing on the acidity of acompound, and that the relative stability of an acid and its conjugate base dictates the acidity. However, manycandidates still wrote that with the increase in polarization, the O − H bond in the oxoacid weakens and hence itsacid strength increases.Most candidates correctly chose the stronger acid but gave an incorrect explanation. Many did not know thatresonance stabilization in oxoacids is related to the number of oxogroups attached. Some candidates drew wrong

structures for the oxoacids with H bonded to the central atom. As a result, they wrongly explained the acidstrength in terms of the polarization of the Cl–H or Mn–H bond.



Acid-base Theory Unit 2 Page 1296 1A 3 b3b Arrange the following compounds in the order of increasing basic strength. Explain your arrangement.

NH 2

,

NH 2

and NH 3

3

Basicity : NH 2

NH 2

NH 3< <1 mark

The basicity of an amino compound depends on the position of the following equilibrium

RNH 2 + H 2O d RNH 3+ + OH - ½ mark

The primary amine is a stronger base than NH 3

Q Electron donating property / inductive effect of alkyl group stabilizes the RNH 3+ to a greater extent. ∴ 1º

amine is more basic / the e - pair is more available. ½ mark

For the aromatic amine, overlapping of the orbital containing the lone pair in N with the e-electron cloud of the benzene ring makes the lone pair less available. ½ mark Thus, the equilibrium lies on the LHS. ½ mark

∴

NH 2

is the weakest base

C Most candidates explained the relative strengths of the bases in terms of the availability of the lone pair of electrons on the N atom. Only a few candidates completed their answers with a discussion of the equilibrium

position.Some wrongly chose

NH 2

as the strongest base, saying that the cation

NH 3

+

is stabilized by resonance which leads to a dispersion of the positive charge.

99 2A 4 c i4c i Arrange, with explanation, the compounds below in the order of increasing acidity.

HCO 2H, CH 3CO 2H and CF 3CO 2H



Isomerism


A. Chain isomerismB. Position isomerismC. Functional group isomerism

D. Tautomerism / TautomerizationII. Stereoisomerism

A. Diastereomerism / Geometrical isomerism1. Physical properties of cis-/trans-geometrical isomers of 1,2-dichloroethene2. Physical and chemical properties of cis-/trans-geometric isomers of butenedioic

acida) Physical propertiesb) Chemical properties

B. Optical isomerism / Enantiomerism1. Physical properties

2. Optical activitya) Plane polarized light b) Measurement of optical activity by polarimeter c) Specific rotation

3. Chiral centre4. Racemic mixture / Racemic modification / Racemic form / Racemate



Isomerism Unit 1 Page 1

Topic Isomerism Unit 1

ReferenceReading

15.0–15.1 15.2.1Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 26–33Organic Chemistry, Solomons, 5th Edition pg. 164–174, 179–186, 201–203, 175, 302, 836Organic Chemistry, Fillans, 3rd Edition pg. 35–49


Edition pg. 125–138, 144–146, 158–160Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 390–391Organic Chemistry, 6th Edition, Solomons, pg. 59-61

Syllabus IsomerismStructural isomerismStereoisomerismGeometrical isomer (Diastereomer)

Notes Isomerism

Isomers are different compounds sharingthe same molecular formula.

Structural isomers are also calledconstitutional isomers. The atoms of theisomers are connected in a different order.They are said to have a differentconnectivity.

Stereoisomers differs only in arrangementof their atoms in space but have the sameconnectivity.

Isomerism(same molecular formula)

Structural / ConstitutionalIsomerism

(different structural formula / connectivity)

Stereoisomerism(same structural formula /

connectivity)

ChainIsomerism

PositionIsomerism

FunctionalGroup

Isomerism

Tautomerism

Diastereomerism /Geometrical Isomerism /

cis-trans Isomerism(Each isomer is the

mirror images of its own)

Enantiomerism /Optical Isomerism

(Each isomer is not themirror image of

its own)


A. Chain isomerismThey have different carbon chains or skeletons.(Branched and unbranched chain)e.g. C 4H10

CH 3 CH 2 CH 2 CH 3

butane

CH 3 CH CH 3

CH 3

2-methylpropane

B. Position isomerismThey are molecules which have a substituent in different positions on the same carbon skeleton.(Alkanol)e.g. C 3H8O

CH 3 CH 2 CH 2 OH

propan-1-ol

CH 3 CH CH 3

OH

propan-2-ol (Disubstituted benzene) e.g. C 8H10

CH3

CH3

1,2-dimethylbenzene

CH3

CH 3

1,3-dimethylbenzene

CH 3 CH 3

1,4-dimethylbenzene




C. Functional group isomerism

They are isomers in different homologous series and have different functional groups(Ether and alcohol)e.g. C 2H6O

CH 3 O CH 3

methoxymethane (dimethyl ether)

CH 3 CH 2 OH

ethanol

(Acid and Ester) e.g. C 4H8O2

CH 3 CH 2 CH 2 C

O

O H

butanoic acid

CH 3 CH 2 C

O

O CH 3

methyl propanoate

D. Tautomerism / Tautomerization

Tautomerism is a kind of dynamic isomerism. It is a rapid and reversible interconversion of isomers associatedwith the actual movement of electrons as well as of one or more hydrogen atoms.

Tautomerism must not be confused with resonance. Each tautomer is capable of independent existence and potential isolation. Tautomers represent real compounds, whereas individual resonance structure does not.

(Keto-enol tautomerism)e.g. C 3H6O

CH 3 C CH 3

O

keto form> 99 %

CH 2 C CH 3

OH

enol form

1.5 × 10 -4 %

propanone propen-2-ol

Interconvertible keto and enol forms are known astautomers, and their interconversion is calledtautomerizaton.

Usually the enol form contributes only a very small percentage in the equilibrium.

II. Stereoisomerism

A. Diastereomerism / Geometrical isomerism

By definition, a diastereomer is a stereomer which is superposable with its own mirror image.

However, a pair of diastereomers are not mirror images of each other.

cis-trans isomerism is a kind of diastereomerism, where its existence is due to hindered / restricted rotation of double bond. But-2-ene and hexa-2,4-diene are two of the examples.

They have similar but not identical chemical properties and very different physical properties.

(but-2-ene CH 3CH=CHCH 3)

C CCH 3 CH 3

H H

cis-but-2-ene / (Z)-but-2-ene

C CCH3

CH3H

H

trans-but-2-ene / (E)-but-2-ene

For cis-isomer, the two identical groups (usually hydrogen atoms) are on the same side. (Latin: 'cis' = on this side)For trans-isomer, the two identical groups are on the opposite sides. (Latin: 'trans' = across)




For trisubstituted or tetrasubstituted alkene, the terms cis and trans are either ambiguous or do not apply at all sincethere is no two identical groups.

For this kind of structure, the configuration can be denoted by (E-Z) system which is based on the atomic number (bulkness).

(Z) German: 'zusammen' = together (E) German: 'entgegen' = across

Consider

C CF

Br Cl

H

(Z)-2-bromo-1-chloro-1-fluoroethene

Based on the atomic number,Br has a higher priority than HCl has a higher priority than F

Since Br and Cl are on the same side (together), thestructure is called (Z)-1-chloro-1-fluoro-2-bromoethene.

For further details, please read Solomons pg. 302 and 175

1. Physical properties of cis-/trans-geometrical isomers of 1,2-dichloroethene

C CH

Cl Cl

H cis-1,2-dichloroethene

melting point - 80 ºC boiling point 60 ºCdipole moment 1.90 D

C CH

Cl

Cl

H

trans-1,2-dichloroethene

melting point -50 ºC boiling point 48 ºCdipole moment 0 D

The boiling point of the cis-isomer is higher than that of the trans-isomer because the intermolecular forces amongthe former one are stronger, i.e. it is polar. However, the melting point of the later one is higher because the trans-isomer is more symmetrical in shape and can be packed more regularly in the solid crystal.

Because the melting point of a substance is affected by the packing of the molecule on top of the intermolecular forces, boiling point is a better indicator of the strength of the intermolecular forces.




2. Physical and chemical properties of cis-/trans-geometrical isomers of butenedioic acid

Butenedioic acid is another example exhibiting geometrical isomerism.

C

C

CO 2H

CO 2HH

H

(maleic acid) cis-butenedioic acid / (Z)-butenedioic acid

C

C

CO 2H

HO 2C

H

H (fumaric acid) trans-butenedioic acid / (E)-butenedioic acid

a) Physical properties

Maleic acid Fumaric acidDipole moment non-zero zeroSolubility Higher Lower Melting point 131 ºC 287 ºC

Although the dipole moment of fumaric acid is zero, it is capable to form intermolecular hydrogen bond.

Therefore, the melting point is higher.

Dipole moment Maleic acid has a stronger dipole moment than fumaric acid.

C

C

CO 2H

CO 2HH

H

Maleic acid has 2 dipolemoments pointing to thesame side and gives a non-zero resultant. Therefore,it is more soluble in polar solvent. i.e. water.

C

C

CO 2H

HO 2C

H

H

The 2 dipole moments of fumaric acid opposingeach other and give a zeronet dipole moment

Melting pointMaleic acid has a lower melting point than fumaric acid.Although maleic acid has a stronger dipole moment, the formation of intramolecular hydrogen bond reduces theformation of intermolecular hydrogen bond. This reduces the extent of the hydrogen bond formed betweenadjacent molecules in the crystal and causes a lower melting point than that of fumaric acid.




b) Chemical properties

Acidic properties Both acids show acidic properties in water but have different acidity.

C

C

CO 2H

CO 2HH

H

maleic acid pK a1 = 1.83 pK a2 = 6.07

C

C

CO 2H

HO 2C

H

H fumaric acid pK a1 = 3.03 pK a2 = 4.44

The difference is due to the formation of intramolecular H-bond stabilized conjugate base of maleic acid.

Dehydration When maleic acid is heated to 140ºC, water isevolved and anhydride forms.

By contrast, no reaction takes place whenfumaric acid is heated to 140ºC. At 290ºC, thereis sufficient energy to overcome the barrier of rotation about the C=C double bond and the cisconfiguration required for anhydride formationis attained.

Interconversion between maleic acid and fumaric acid

Maleic acid can also be converted to fumaric acid by boiling with hydrochloric acid catalyst. Since fumaric acid ismore stable than maleic acid, the equilibrium lies much towards the fumaric acid side. This is because in fumaricacid, the two bulky carboxyl groups are further apart and experience less steric repulsion.

Glossary structural / constitutional isomer stereoisomerism tautomerism / tautomerizationgeometrical isomerism diastereomerism




Past PaperQuestion

92 1A 1 a i94 2C 7 a96 2C 8 b i97 1A 4 a i ii 97 1A 4 b i98 1A 4 b i 98 2B 5 a ii

92 1A 1 a i

1a There are several isomers of benzenedicarboxylic acid.i Draw the structures of all possible isomers of benzenedicarboxylic acid. 1½

CO2H

CO2H

CO2H

CO2H

CO2H

CO2H

½ mark each

94 2C 7 a7a What do you understand by the terms “structural isomerism” and “stereoisomerism”? 3

Structural isomerism – occurrence of more than one structure for a given molecular formula 1 mark Stereoisomerism – occurrence of more than one configuration (different arrangements of groups in space) for a

give structural formula. 1 mark Examples to illustrate the two types of isomers. 1 mark C Badly-answered. Not many candidates gave precise definitions for 'structural isomerism' and 'stereoisomerism'.

Instead, they just used examples (correctly or incorrectly drawn) to illustrate the various kinds of isomers. Thefollowing terms were often used incorrectly : molecular formula, structural formula, empirical formula.

96 2C 8 b i8b The following compounds can exist in isomeric forms :

In each case, state the type of isomerism and draw suitable representations for the isomers.i butenedioic acid, and 2

Geometrical isomerism / cis-trans isomerism 1 mark (Deduct ½ mark for spelling mistake)

COOH

HOOC

COOH

COOH½ + ½ mark

C Many candidates did not assign trans- and cis-isomers. Some wrongly gave positional isomers as the answer.

97 1A 4 a i ii4a The formula HO 2CCH=CHCO 2H can represent two compounds. 2

i Draw a structure for each compound, clearly showing the difference between them.ii One of the compounds reacts with P 2O5(s) to give compound A . Give the structure of A.

97 1A 4 b i

4b i Draw all possible isomeric structures of dimethylbenzene. 3

98 1A 4 b i4b Alcohol E has the structure CH 3CH(OH)C 2H5.

i Draw the structures of three structural isomers of E, all of which are alcohols. 1½ 98 2B 5 a ii5a Consider the following compound F.

CH CHCH 3 C CH

a b c d

F

5

ii Draw all possible three-dimensional structures for F, indicating the expected bond angles around the carbon atomsa, b, c, and d in one of the structures.




Topic Isomerism Unit 2

ReferenceReading

15.2.2Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 26–33Organic Chemistry, Solomons, 5th Edition pg. 164–174, 179–186, 201–203, 175, 302, 836Organic Chemistry, Fillans, 3rd Edition pg. 35–49


Edition pg. 125–138, 144–146, 158–160Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 392–393, 420, 512Organic Chemistry, 6th Edition, Solomons, pg. 178-185, 188, 193-198, 200

Syllabus Optical isomerism / EnantiomerismOptical activitiesChiral centreRacemic mixture

Notes B. Optical isomerism / Enantiomerism

Optical isomers are also known as enantiomers. Individual enantiomer is not the mirror images of its own.

From another point of view, enantiomers exist in pairs. They are non-superposable mirror images of each other. Inorder to describe this properties, the molecules are said to be chiral. In Greek, 'chiral' means: 'cheir' = hand, sincethe two hands are non-superposable mirror images of each other..

e.g. butan-2-ol

CH 2CH 3

CH3C OH

H

(+)-2-butanol

CH 2CH 3

CH3C H

OHC

CH 2CH 3

CH 3HOH ≡

(-)-2-butanol

Structurally, all chiral molecules do not have a plane of symmetry.

Unlike enantiomer, diastereomer has at least one planeof symmetry. And because of this difference, onlyenantiomer shows optical activity.

Plane symmetry in diastereomer (cis-but-2-ene)

C CCH 3H3C

H H

1st plane of symmetry

2nd plane of symmetry




The 2 enantiomers may be denoted by several different methods

i) (+) or (-) based on the optical activity of the enantiomers.

ii) (R-S) system based on the absolute configuration which is similar to the (E–Z) system for geometrical isomers.For further details, please read Solomons, pg. 174–179

iii) By relative configuration comparing with a standard compound called Glyceraldehyde.For further details, please read Solomons, pg. 199–201

Methods ii) and iii) will not be discussed in the Hong Kong A-level syllabus.

1. Physical properties

They have almost exactly the same chemical and physical properties except optical activity.The mixing of 2 enantiomers causes melting point depression proving that the 2 isomers are indeed different.

Boiling point (1 atm)Density (gcm -3 at 20 ºC)Refractive index (20 ºC)

(R)-butan-2-ol or (-)-butan-2-olor (R)-(-)-butan-2-ol

99.5 ºC0.8081.397

(S)-butan-2-ol or (+)-butan-2-olor (S)-(+)-butan-2-ol

99.5 ºC0.8081.397

2. Optical activity

Optical activity is the properties possessed by certain substances of rotating the plane of polarization of polarizedlight.

a) Plane polarized light

Light is an electromagnetic phenomenon.A beam of light consists of two mutually perpendicular oscillating fields: an oscillating electric field and anoscillating magnetic field.

An ordinary light beam consists of oscillations of the electric field occurring in all possible planes perpendicular tothe direction of propagation.By contrast, a plane-polarized light has an electric field oscillating only in one plane. It can be produced by

passing an ordinary light beam through a polarizer.

Because of the asymmetrical distribution of electron cloud in an enantiomer, an enantiomer is capable to interactwith the electric field and alter the direction of oscillation of the plane-polarized light.




b) Measurement of optical activity by polarimeter

If a substance rotates the plane-polarized light to the right hand side, it is said to be dextrorotatory. (Latin: 'dexter'= right). And the enantiomer is denoted by (+)- prefix. e.g. (+)-butan-2-olSimilarly, if the plane-polarized light is rotated to the left, the substance is said to be levorotatory, denoted by (-)-

prefix. (Latin: 'laevus' = left). e.g. (-)-butan-2-ol

c) Specific rotation (Details not required)

Since the degree of rotation is depending on a lot of factors includingi) length of the polarimeter tubeii) concentration of the sampleiii) wave length of the light sourceiv) temperaturev) nature of solvent

For the purpose of comparison, a specific rotation [ α ] is defined as [ α ] =α

c × l.

where [ α ] is the specific rotationα is the observed rotationc is the concentration of the solution in gcm -3 of solution or density in gcm -3 for pure liquidl is the length of the tube is dm

e.g. [ α ]25D = +3.12º means the specific rotation is 3.12º in a clockwise direction at 25ºC using D line of a sodium

lamp. ( λ = 599.6 nm). In reporting a specific rotation, the solvent used should be quoted.

N.B. The direction of the optical rotation has no direct relationship with absolute configuration of the molecule.It can only be determined by the experiment. e.g. a molecule with (R) configuration may be dextrorotatoryor levorotatory.




3. Chiral centre

The C-2 of butan-2-ol has four different groups attaching to it tetrahedrally. The C-2is called the chiral centre or chiral carbon.

By interchanging any two groups attached to the chiral centre, one enantiomer can beconverted to another.

A molecule contains one chiral centre only must be a chiral molecule.

CH 3 C CH 2CH 3

H

OH

*

4. Racemic mixture / Racemic modification / Racemic form / Racemate

If a mixture contains equal amount (50%-50%) of (+)-isomer and (-)-isomer, the mixture will show no opticalactivity. Owing to the opposing effect of the two enantiomers, optically activity of one cancels that of another one. Such kind of mixture is called racemic mixture and denoted by ( ± )- prefix. e.g. ( ± )-butan-2-ol

Glossary optical isomerism / enantiomerism chiral plane of symmetry optical activity plane polarized light polarizer polarimeter dextrorotatory levorotatory specific rotation

chiral carbon racemic mixture

Past PaperQuestion

90 2C 8 a iv v93 2C 9 c96 2C 8 b ii98 1A 4 a ii 98 1B 8 a i99 2B 6 c i ii iii

90 2C 8 a iv v8a It is suggested that the structure of a compound having the molecular formula C 12H11ClO 4 is either A or B.

A

Cl

COOH

COOH

CH CCH3

CH3

B

COOH

COOH

CH CCH3

CH2Cl

iv Assuming the compound has the structure A, give the structural formula of the hydrogenation product from

reaction with one mole of hydrogen. Would you expect the product to show optical activity? Explain your answer.

2

Cl

COOH

COOH

CH2 CHCH3

CH3

1 mark

not optically active, because there is no chiral center. 1 mark C Most candidates gave correct answers to this part.v Repeat part (iv) assuming the compound has the structure B. 3

COOH

COOH

CH2 CCH3

HCH2Cl

,

COOH

COOH

CH2 CCH3

CH2ClH

1 mark not optically active, because it is a racemic mixture. 2 marks

C Over 90 % of the candidates were not aware that the hydrogenation product was a racemic mixture and henceoptically inactive.



Isomerism Unit 2 Page 593 2C 9 c9c 2-Bromobutane can exist in isomeric forms. Draw a suitable representation for each of these isomers. 2

C

Et

CH 3H

Br C

Et

CH 3H

Br

or

CH 3

H

EtBr

CH 3

H

Et Br

1 mark eachC Candidates should pay more attention in providing a proper 3-dimensional structural formula.

96 2C 8 b ii8b The following compounds can exist in isomeric forms :

In each case, state the type of isomerism and draw suitable representations for the isomers.ii 2-aminopropanoic acid. 2

enantiomerism / optical isomerism 1 mark (Deduct ½ mark for spelling mistake)

H

CO 2H

CH 3

NH 2

CO 2H

CH 3

HH2 N

½ + ½ mark (Accept any appropriate representation of enantiomerism)

C Some candidates showed weakness in three-dimensional representations. Enantiomerism instead of opticalisomerism should have been given. Some wrongly gave stereoisomerism as the answer. Two incorrectrepresentations given by candidates are shown below.

C CO 2H

NH 2

CH 3

H

C

NH 2

CH 3H CO 2H

98 1A 4 a ii

4a Alcohol E has the structure CH 3CH(OH)C 2H5.ii What type of isomerism can be exhibited by E ?

98 1B 8 a i8a Show how you would

i determine whether a sample of C 2H5CH(OH)CH 3 is in the (+) form or ( ± ) form.

99 2B 6 c i ii iii6c State the relationship between each pair of structures shown below:

i CH 2CH 3

ClCl

CH 2CH 3

ClCl

and

ii

C CH

HCH 3

CH 2CH 3

C CCH 3CH 2

H H

CH 3and

iii

CCHCH2

CH3H

Br

CCHCH2

CH3Br

H

and




Reaction Mechanism

I. Bond breaking A. Bond breaking of a bond to carbon


A. Free radical (electron deficiency)B. Electrophile (electron deficiency)C. Nucleophile (electron rich)

III. Classification of reaction A. Types of reaction

1. Substitution2. Addition3. Elimination




IV. Nucleophilic substitution A. Nucleophilic substitution

1. Nucleophile2. Leaving group

B. S N 2 reaction (1 step reaction)C. S N 1 reaction (2 steps reaction)D. Competition between S N 1 and S N 2 reactionsE. Alkanol from haloalkane (RX → ROH)

1. Alkaline hydrolysis of haloalkane2. Hydrolysis of haloalkane

F. Rate of hydrolysis of haloalkane, haloalkene and halobenzeneG. Other relevant reactions

1. Reactions of haloalkanea) Nitrile from haloalkane (RX → RCN)b) Alkylation of ammonia and amine (NH 3 → RNH 2 )c) Use of S N 2 reaction in organic synthesis

2. Reactions of alkanol

a) Haloalkane from alkanol (ROH → RX)(1) Use of chlorinating and brominating reagentb) Luca's test to distinguish 1º, 2º and 3º alkanol

3. Reactions of aminea) Action of nitric(III) acid on 1º amine

(1) 1º aliphatic amine(2) 1º aromatic amine


V. Elimination reaction A. Elimination reaction

1. Stability of elimination product2. E2 reaction (not required in A-Level)3. E1 reaction (not required in A-Level)

B. Competition between substitution and elimination reaction1. Effect of temperature2. Effect of bulkiness of the substrate and base3. Effect of bascity of the nucleophile

C. Conditions favouring substitution and elimination reactionD. Other relevant reactions

1. Reaction of haloalkanes with alcoholic sodium hydroxide to alkene, diene andalkyne

2. Preparation of vinyl chloride3. Dehydration of alkanol




VI. Nucleophilic addition (Nucleophilic addition-elimination) A. Ad N reaction

1. Addition of HCN to carbonyl compound2. Rate of nucleophilic addition

a) Electronic effect (1) Inductive effect

(a) Effect of protonation(2) Resonance effect

b) Steric effect 3. Other relevant reactions

a) Reactions of carbonyl group(1) Reduction of carbonyl compound by LiAlH 4 and NaBH 4 (2) Addition of NaHSO 3 to carbonyl compound(3) Condensation reaction with hydroxylamine(4) Condensation reaction with 2,4-dinitrophenylhydrazine(5) Haloform reaction / Iodoform reaction

b) Reactions of carboxylic acid and its derivatives(1) Carboxylic acid and its derivatives

(a) Difference between carbonyl compound and carboxylic acid

and its derivatives(b) Reactivities of carboxylic acid and its derivatives

(2) Formation of different acid derivatives(a) Use of chlorinating agent to prepare acyl chloride(b) Formation of ester (Esterification with alkanol and phenol)(c) Formation of acid anhydride

(i) Through intramolecular dehydration by heating(ii) Through intermolecular dehydration by a very strong

dehydrating agent(iii) From acyl chloride

(d) Formation of amide (Acylation and benzoylation of amine)(3) Reaction of ester

(a) Hydrolysis of ester (4) Reaction of amide(a) Reduction of amide and other acid derivatives(b) Hofmann degradation of amide

c) Reactions of nitrile(1) Hydrolysis of nitrile and amide

(a) Dehydration of amide(2) Reduction of nitrile

VII. Electrophilic addition A. Addition of HBr to alkene

1. Markownikoff's rule.2. Reactivity of alkene towards electrophilic addition

B. Other relevant reaction1. Addition of Br 2 to alkene2. Addition of H 2SO 4 to alkene

a) Preparation of alkanol from alkene3. Hydration of alkene4. Ozonolysis of alkene5. Preparation of ethane-1,2-diol

a) Oxidative Cleavage of double bond b) Comparision of ozonolysis and oxidative cleavage

6. Oxymercuration of alkyne




VIII. Electrophilic substitution A. Representation of arenium ionB. Other relevant reaction

1. Sulphonation of benzenea) Preparation of phenol

2. Nitration of benzenea) Reduction of nitrobenzene

3. Halogenation of benzene4. Alkylation of benzene (Friedel-Crafts alkylation)5. Diazocoupling of diazonium ion

a) Colour of a substance6. Bromination of phenol

IX. Free radical Reaction A. Formation of free radical B. Free radical Substitution

1. Chain reaction e.g. chlorination of methanea) Chain initiation (chain initiating step)b) Chain propagation (chain propagating step)c) Chain termination (chain terminating step)

2. Reaction between H 2(g) and Cl 2(g) C. Free radical Addition

1. Chain reaction e.g. Polymerization of alkene2. Anti-Markownikoff orientation



Reaction Mechanism Unit 1 Page 1

Topic Reaction Mechanism Unit 1

ReferenceReading

16Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 40–41, 44–46Organic Chemistry, Solomons, 5th Edition pg. 209–211Organic Chemistry, Fillans, 3rd Edition pg. 82–94

Organic Chemistry, Morrison Boyd, 6th Edition pg. 165–167

ReadingAssignment

Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 403–406, 440–441Organic Chemistry, 6th Edition, Solomons, pg. 87-90, 94-96

Syllabus Introduction to reaction mechanismTypes of bond breakingTypes of reactive speciesTypes of reactions

Notes Reaction Mechanism

A chemical reaction is a sequence of bond-breaking and bond-forming steps involving bonding and nonbondingelectrons. A detailed description of a chemical reaction outlining each separate stage is called the Mechanism .

Chemical reaction can be interpreted as redistribution of e - and rearrangement of atoms which leads to formation of new substance.

I. Bond breaking

All kinds of reactions are initiated by a bond breaking process. There are 2 kinds of bond breaking process :

Homolytic fission (Homolysis) – the electrons forming the bond broken are shared equally between thefragments formed. This leads to formation of 2 radicals.

Heterolytic fission (Heterolysis) – the electrons forming the bond broken are shared unequally between the

fragments formed. This leads to formation of 1 cation and 1 anion.

Mode of bond breaking Electronegativity

BA A B+

Homolytic fission

(Symmetrical fission)A ≈ B

BA A- B++

Heterolytic fission

(Unsymmetrical fission)A > B

BA B-A+ +

Heterolytic fission

(Unsymmetrical fission)

A < B

movement of an electron pair

movement of a single electron




A. Bond breaking of a bond to carbon

Homolysis Heterolysis

Carbon radical and carbocation are electron deficient species. Both of them do not have fulfilled octet.Carbanion is an electron rich species. It has surplus electrons available for bonding.


With reference to different modes of bond breaking, 3 different reactive species may form.

A. Free radical (e.g. A· or B·) – An electron deficient species with unpaired electron. e.g. Cl·B. Electrophile (e.g. A + or B +) – An electron deficient species which tends to react with negative centre. e.g.

H+, Cl +, Br +, NO 2+. All electrophile are also Lewis acids.

C. Nucleophile (e.g. A - or B -) – An electron rich species which tends to react with positive centre. e.g. -OH, Cl -

, CN -, H -. Some nucleophile contains a lone pair instead of a negative charge.e.g. H 2O, ROH, NH 3, RNH 2. All nucleophile are also Lewis bases.

Then, the reactive species reacts with the substrate and a new bond is formed.

III. Classification of reaction

According to the reactive species involved, most of the organic reactions can be classified into1. Nucleophilic reaction

2. Electrophilic reaction3. Radical reaction

And according to the outcome of the product, the reactions can be classified into

1. Substitution

2. Addition

3. Elimination S : Substrate R : Reagent

'

A. Types of reaction

1. Substitution

Nucleophilic substitution – e.g. Hydrolysis of haloalkane to form alkanol (RX + H 2O → ROH + HX)Electrophilic substitution – e.g. Nitration of benzene (C 6H6 + NO 2

+ → C6H5 NO 2 + H +)Radical substitution – e.g. chlorination of methane (CH 4 + Cl 2

→ CH 3Cl + HCl → CH 2Cl2 + HCl → etc.)

2. Addition

Nucleophilic addition – e.g. Addition of hydrogen cyanide to carbonyl compound (Formation of cyanohydrin)Electrophilic addition – e.g. Addition of HBr to alkene (CH 2=CH 2 + HBr → CH 3 –CH 2Br)

Radical addition – e.g Polymerization of ethene (CH 2=CH 2 → –[CH 2 –CH 2]n –)

3. Elimination

e.g. Dehydration of ethanol to ethene (CH 3CH 2OH → CH 2=CH 2 + H 2O)




Mechanisms required in HK 97 A-Level syllabus

Nucleophilic Electrophilic Radical

Substitution 97 A-Level Details not required 97 A-Level

Addition 97 A-Level 97 A-Level 97 A-Level

Elimination Details not required Not present Not present

Although Elimination reaction is not known as Nucleophilic Elimination, like the Substitution and Addition, ittakes place in the presence of a strong base e.g. CH 3CH 2O- Na + which is a nucleophile.

Glossary mechanism homolytic fission heterolytic fission carbocation / carbenium ion / carbonium ioncarbanion free radical electrophile nucleophile substitution addition elimination

Past PaperQuestion





ReferenceReading

17.0.1–17.0.2Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 49–53, 158–159Organic Chemistry, Solomons, 5th Edition pg. 212–236, 240–243Organic Chemistry, Fillans, 3rd Edition pg. 174–183

Organic Chemistry, Morrison Boyd, 6th Edition pg. 172–203, 208–210Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 455–458Organic Chemistry, 6th Edition, Solomons, pg. 224-233, 238-252, 256-259

Syllabus Nucleophilic substitutionS N2 reactionS N1 reactionStability of carbocationCompetition between S N2 and S N1 reaction

Notes IV. Nucleophilic substitution

A. Nucleophilic substitution

C L Nu + C Nu L+δ + δ -

Nu: – Nucleophile L: – Leaving group

When an electronegative atom (L) is attached to an C atom, the C–L bond would be polarized. The C atom possesses a partial positive charge and is vulnerable to the attack of a nucleophile.

1. Nucleophile

Nucleophile is an electron rich species which donates itselectrons to the positive centre. The ability of a speciesto be a nucleophile is called nucleophilicity.

Nucleophilicity of a nucleophile depends on 2 factors:

1. Availability of electrons (Lewis basicity)2. Polarizability of electrons

1. Availability of electrons (Lewis basicity)

Nucleophilicity is parallel to Lewis basicity. i.e. theability to donate electron. e.g. OH - is a stronger nucleophile than H 2O.

N atom is more nucleophilic than O atom because N isless electronegative and has less attraction on the lone

pair electrons and makes them more available. e.g. NH 3 is a stronger nucleophile than H 2O.

2. Polarizability of electrons

However, nucleophilicity is something more than just basicity. Besides the availability of electrons, it is alsodepending on the ability to form bond. This involvesdistortion and redistribution of electron cloud. i.e.

polarizability of the species.

In general, polarizability is mainly depending on thesize and electronegativity of the atom.

Relation of nucleophilicity to basicity for atoms inthe same row (similar size) of the periodic table

Relation of nucleophilicity to basicity commonlyobserved for atoms in same family (different size) atthe periodic table




For the atoms with similar size (atoms in the same period), thus similar polarizability, the nucleophilicity is parallelto the basicity.e.g. NH 3 is a stronger nucleophile than H 2O.

For atoms with very different size, the nucleophilicity is depending on both polarizability and basicity.e.g. Though I - is a weaker base than Cl - ion, it is much bigger than Cl - ion and with much higher polarizability.Experimentally, it is found that I - ion is a stronger nucleophile than Cl - ion.

N.B. The actual nucleophicity of a nucleophile has to be determined through measuring the rate of anucleophilic reaction.

2. Leaving group

It is observed that weaker a base, better will be theleaving ability.

Basicity is also a measure of stability of a base. A weak base is more stable than a strong base as the negativecharge on the molecule is less concentrated.

The formation of a more stable leaving group (a weak base) involves a lower activation energy, so this is morefavorable.

Relation of Leaving group ability to the bascity




B. S N2 reaction (1 step reaction)

(Refer to Rate of reaction Unit 4)

S N2 stands for Bimolecular nucleophilic substitution since 2 molecules are involved in the formation of thetransition state.

For an S N2 reaction, the rate of reaction depends on both the concentration of nucleophile and the substrate.

e.g. Rate ∝ [OH -][CH 3CH 2CH 2CH 2Br]

Rate = k[OH -][CH 3CH 2CH 2CH 2Br]

Inversion of configuration

The nucleophile approaches the substrate molecule fromthe opposite side of the leaving group to minimize stericrepulsion.

The nucleophile forms a trigonal bipyramidal transitionstate (pentavalent) with the substrate. The departure of the leaving group results in an inversion of configuration.

The energy profile for the reaction can be depicted asfollows.

Effect of the structure of the substrate

The ease of approaching of the nucleophile affects the rate dramatically. A less sterically hindered C atom will bemore vulnerable to the attack of the nucleophile and involve a lower activation energy.

Simple haloalkanes show the following general order of reactivity in S N2 reactions:

methyl > primary > secondary > (tertiary) almost no reaction




C. S N1 reaction (2 steps reaction)

(Refer to Rate of reaction Unit 4)

S N1 stands for Unimolecular nucleophilic substitution since only 1 molecule is involved in the formation of thetransition state in the rate determining step.

CCH3

CH3

CH3

Cl CCH3

CH3

CH3

OHOH -(aq) Cl-(aq)+ +

Rate = k[(CH 3)3CCl]

Step 1 (slow)

Cl-C

CH 3

CH 3

CH 3

+C

CH 3

CH 3

CH 3

Cl +

Step 2 (fast)

C

CH 3

CH 3

CH 3

+ -OH C

CH 3

CH 3

CH 3

OH

As formation of the transition statefor step 1 involves a higher activation energy E a’, the step 1 would be slower and known as therate determining step.




Carbocation intermediate

C

CH3

CH 3

CH3

+is called a carbocation ion. It is an intermediate of S N1 reaction. Because it is at the bottom of a potential

well, unlike the highly unstable transition state, it is possible to isolate the intermediate from the reaction mixture by cooling.

Since the rate determining step involves only the substrate, therefore the rate of S N1 reaction is independent of thenature and concentration of the nucleophile.

The rate of reaction is governed by the stability of the carbocation intermediate since a more stable carbocationinvolves a lower activation energy of formation.

A carbocation is an electron deficient species, it would be more stable if it is attached to electron releasing groups,either by inductive or resonance effect.

As a result, simple haloalkane shows the following general order of reactivity in S N1 reactions:

tertiary > secondary > primary > methyl

The trend is just the reverse of that of S N2 reaction.

A carbocation attached to an unsaturated functional group is even more stable than 3º carbocation because it isstabilized by resonance.

C C C

H

H

H H

H+

C C C

H

H

H H

H+

C H

H+

C H

H

+

Racemization

The carbocation intermediate is trigonal planar and is achiral. The nucleophile approaches the carbocation fromeither side of the plan with equal probability. Consequently, a racemic mixture of product will be formed.

Thus, S N1 reaction is not very useful in stereospecific synthesis.




D. Competition between S N2 and S N1 reactions

S N2 and S N1 reactions represent 2 extremes of thenucleophilic substitution. Actually, the two reactionscompete with each other.

It is observed that 2º carbon has the slowest rate of

reaction. This is attributed to the crossing of twoopposing curves shown at the right, the mechanism ischanging from S N2 to S N1.

Although structure is the major factor that governs themechanism of substitution reaction, leaving group,nucleophile and solvent also contribute.

Characteristics of S N2 and S N1 reactionS N2 S N1

Mechanism One-stepR–L + Nu: → R–Nu + L:

Two-step

R–Lr d s. . . → R +

Nu :−

→ R–NuKinetic Second order First order

Reagent nucleophilicity Rate-controlling Unimportant to rate

Structure of saturatedcarbon atom

Steric hindrance unfavorablemethyl > 1º > 2º >> 3º

Resonance stabilization favorable3º >> 2º > 1º

Stereochemistry Inversion Racemization

E. Alkanol from haloalkane (RX → ROH)

1. Alkaline hydrolysis of haloalkane

C

H

H

H XHO - C

H

H

H

OH X-+

2. Hydrolysis of haloalkane

deprotonationS N2

r.d.s.H3O+ +C

H

H

H

OH X-+X-

+OH

H

+C

H

H

H

O

H

HC

H

H

H X

O

H

H

Alkanol can be prepared from haloalkane by alkaline hydrolysis or hydrolysis in water. However, thenucleophilicity of H 2O is lower than that of OH -, the rate of hydrolysis is slower in neutral medium.

Glossary nucleophile leaving group nucleophilicity inversion of configuration steric hinderancecarbocation / carbenium ion / carbonium ion racemization




Past PaperQuestion

92 2C 8 c95 1A 3 a96 1A 3 a98 2B 5 c ii iii

92 2C 8 c8c Outline a reaction mechanism for the hydrolysis of 1-bromopropane with dilute aqueous potassium hydroxide and

sketch a labelled energy profile diagram for the reaction.

4

C

HCH3CH 2

H

Br HOδ − δ−

Transition state

CH 3CH 2CH 2OH Br -+-OH

CH 3CH 2CH 2 Br

2½ marks

label of axis ½ mark energy profile 1 mark

C Some candidates gave S N1, Or E2 reaction mechanisms instead of the required S N2 mechanism.Candidates confused "transition state" and "intermediate".Charge distribution at the transition state was incorrectly marked, e.g.

CH 3CH2 C HH

OH

Br (no -ve charge)

95 1A 3 a

3a Arrange the following carbocations in the order of increasing stability. Explain your arrangement.

+

,

+CH2

and

CH2+

3

+

+CH2 CH2

+

< <Stability

1 mark

A 2º

+

carbocation is more stable than a 1º

+CH2

carbocation. ½ mark

The positive inductive / electron donating effect of CH 2 group stabilize the cation to a greater extent than a Hatom. (no mark for +I) ½ mark

CH2+

is the most stable Q the cation is stabilized by the delocalization of e - from the benzene ring /

resonance / mesomeric effect. 1 mark

H

H H

H H

CH

H+

or C H

H

+C H

H+

C Many candidates did not know that cyclohexylmethyl cation, the least stable one, is a primary carbocation, whilecyclohexyl cation is a secondary carbocation and more stable. Most did well in explaining the stability of the

benzyl cation in terms of resonance, but some wrongly stated that its instability is due to the electron withdrawingeffect of the ring, a fact which has only minor influence on its instability.



Reaction Mechanism Unit 2 Page 896 1A 3 a3a Arrange the following carbocations in the order of increasing stability. Explain your arrangement.

CH 3+

,

CH 3

+and

CH2+

3

Stability: 1 mark

CH2

+

<CH 3

+<

CH 3+

CH 3+

is the most stable.

Q It is stabilized by resonance / mesomeric effect / delocalization of e -, ½ mark positive charge on carbocation is dispersed. ½ mark

CH 3+

etc.CH 3

+(1 mark)

(Must show at least 2 resonance structures; deduct ½ mark for wrong structure)

CH 3

+

is a secondary carbocation whileCH 2+

is a primary carbocation ½ mark

Inductive effect / e - donating property of alkyl group stabilizes the secondary carbocation to a greater extent½ mark

C The resonance structures for the carbocation

CH3+

were poorly-drawn, with the positive charge and/or the double bonds arbitrarily placed. Some candidatesattributed the stability of a carbocation to the distance of the positive charge from the benzene ring. As a result,they wrongly explained the stability of the above carbocation in terms of the fact that the positive charge is closestto the benzene ring.

98 2B 5 c ii iii5c Give the structures of the major organic product, H, in (ii) below.

Outline a mechanism for the formation of the major product in each of the two reactions.ii

CH3CH2CH2I NaOH

H

iii

CH 3CH 2 C

CH3

CH 3

Br CH 3CH 2 C

CH3

CH 3

OHH2O

(major product)





ReferenceReading

17.0.3Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 49–53, 158–159Organic Chemistry, Solomons, 5th Edition pg. 212–236, 240–243Organic Chemistry, Fillans, 3rd Edition pg. 174–183

Organic Chemistry, Morrison Boyd, 6th Edition pg. 172–203, 208–210Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 459Organic Chemistry, 6th Edition, Solomons, pg. 260

Syllabus Nucleophilic substitution of haloalkane

Notes F. Rate of hydrolysis of haloalkane, haloalkene and halobenzene

Reagent : a nucleophile – H 2O

C

H

H

H X C

H

H

H

O

H

H +OH

H

+X- + X-C

H

H

H

OH +H+r.d.s.

S N2 deprotonation

The rate of hydrolysis can be monitored by addition of acidified AgNO 3(aq) into the reaction mixture. The X - ion formed will be precipitated by the Ag +

(aq) ion to form AgX (s) ppt. The solution must be kept acidic or neutral if Ag +

(aq) is present because in alkaline medium Ag +(aq) ion will be precipitated as Ag 2O(s).

It is found that haloalkane hydrolyses faster than haloalkene and haloarene do. And iodoalkane hydrolysesfaster than bromoalkane and chloroalkane do.

The difference between haloalkane, haloalkene and haloarene can be explained by the strength of the C–X bond involved. In an haloalkane, the C–X bond is just a simple single bond while in haloalkene andhaloarene, the C–X has certain double bond character and makes it stronger.

C C ClH

H H

C C ClH

H H

-

+

Cl

-

Cl +

H

H

H

Cl

double bond character

H

H H

H H

Cl

double bond character

The difference between chloroalkane, bromoalkane and iodoalkane, once again, can also be explained by the bond strength. Since I atom has the largest size among I, Br and Cl atom, the bonding electron experiencemost shielding effect from the nuclear charge and the C–I bond would be the weakest among the C–I, C–Br and C–Cl bond.Or this can be explained by the difference in leaving goup ability, I - is a better leaving group than Br - and Cl -.

Glossary alkylation elimination stereospecific




Past PaperQuestion

90 2C 8 a iii91 2C 9 a i93 2C 8 a iii95 2C 7 a i96 2C 7 c ii97 1B 8 c ii

90 2C 8 a iii8a It is suggested that the structure of a compound having the molecular formula C 12H11ClO 4 is either A or B.

A

Cl

COOH

COOH

CH CCH3

CH3

B

COOH

COOH

CH CCH3

CH2Cl

iii How would you distinguish A from B in the laboratory? 3

hydrolysis, followed by aq. AgNO 3, B give white ppt. of AgCl. 3 marks

91 2C 9 a i9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs.

Describe what you would observe in each case.i

Cl

CH3

CH2Cl

2

Warm the compound with aqueous KOH, cool, acidify with nitric acid, add AgNO 3.White precipitate of AgCl formed for ArCH 2Cl; clear solution for CH 3ArCl.

CH2Cl+ OH -

CH2OH+ Cl -

; Ag + + Cl - → AgCl ↓ C Some candidates did not realize if alkaline hydrolysis was used, the system should be acidified before addition of

AgNO 3.

93 2C 8 a iii8a Give a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents required, the observation expected, and the chemical equation(s) for each test.iii CH2Cl Cl

and

3

Reagent: Aqueous silver nitrate solution 1 mark Observation: Formation of white ppt. 1 mark Equation:

CH 2Cl+ H2O

CH 2OH+ HCl

Ag + + Cl - → AgCl (white ppt.) 1 mark



Reaction Mechanism Unit 3 Page 395 2C 7 a i7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents used, the observation expected and the chemical equation(s) for each test.i Cl

and

Cl

3

Treat compound with hot water (with or without ethanol) and then with a solution of silver nitrate(V). 1 mark (*If NaOH is used in the hydrolysis, neutralisation must precede addition of silver nitrate.)

Cl OH

+ H2O + +Cl- H+

½ mark Ag +

(aq) + Cl -(aq) → AgCl (s) ↓ ½ mark

(White) precipitate is immediately formed. ½ mark

No ppt. with

Cl

. ½ mark

C Many candidates failed to give the equations for the reactions.

Common mistakes included : using AgNO 3(s) instead of AgNO 3(aq) ; alkaline hydrolysis without subsequentacidification; HCl for acidification instead of HNO 3; alkaline AgNO 3 solution; the wrong colour for AgCl (s).

96 2C 7 c ii7c Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents used and the observation expected.ii CH2Cl

and

CH2I

2

Treat compound with AgNO 3(aq) CH2Cl

will give a white ppt./ ppt. at a slower rate ½ mark

CH2I

will give a yellow ppt./ppt. at a faster rate ½ mark

C Common mistakes included: AgNO 3(s) instead of AgNO 3(aq) ; alkaline hydrolysis without subsequent acidification;HCl for acidification instead of HNO 3; alkaline AgNO 3 solution; Cl 2 or Br 2 to displace benzyl iodide to giveiodine; wrong colour for AgCl (s) and AgI (s). Most did not know the difference in the rate of hydrolysis of the twocompounds.

97 1B 8 c ii8c For each of the following groups of compounds, suggest a chemical test which would enable each compound to be

distinguished from the other(s). In your answer also give the changes that you would expect to observe for eachcompound.

ii C 6H5CH 2I and C 6H5CH 2Cl 2





ReferenceReading

17.1Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 49–53, 158–159Organic Chemistry, Solomons, 5th Edition pg. 212–236, 240–243Organic Chemistry, Fillans, 3rd Edition pg. 174–183

Organic Chemistry, Morrison Boyd, 6th Edition pg. 172–203, 208–210Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 460–461, 518Organic Chemistry, 6th Edition, Solomons, pg. 913-914

Syllabus Nucleophilic substitution of haloalkane

Notes G. Other relevant reactions

1. Reactions of haloalkane

a) Nitrile from haloalkane (RX → RCN)

By using Na +CN -(aq) , CN -

(aq) functions as an nucleophile and haloalkane can be converted to a nitrile.

+ X-C

H

H

H

CNC

H

H

H X N C -

This reaction is an useful method to increase the length of the carbon chain by one. Thereafter, the cyanogroup formed can be converted to amine, amide or carboxylic acid easily.

Since cyanide ion CN - is highly toxic, this experiment should never be attempted in A-Level.

b) Alkylation of ammonia and amine (NH 3 → RNH 2)

C

H

H

XH N

R

R H C

H

H

H N H

R

R

++ +C

H

H

H N R

R

X-

H X

3° amine2° amine

C

H

H

H X N

R

HH C

H

H

H N H

R

H+

+ +C

H

H

H N H

R

X-

H X

1° amine 2° amine

C

H

H

H X NH

H

H C

H

H

H N

H

H

H+

+X

-

+C

H

H

H N

H

H H X

1° amineammonia

C X

H

H

H

N

R

R R C

H

H

H N

R

R

R

++ X-

3° amine 4° quaternary ammonium halide

haloalkane

In the preparation of 1º amine from ammonia and haloalkane, excess NH 3 is necessary to neutralize the HXformed and to minimize further alkylation of the 1º amine.

If excess RX is used instead of excess NH 3, quaternary ammonium halide will be produced.

This method cannot be used to prepare 2º or 3º amine. If neither NH 3 nor RX is in excess, a mix of 1º amine,2º amine, 3º amine and quaternary ammonium halide will be obtained.




A strong base NH 2- is less desirable than NH 3 in this reaction because

1. In hydrolysis of haloalkane, no matter OH -(aq) or H 2O(l) is used as the reagent, the same product, alcohol,

will be produced.In the alkylation of ammonia, if NH 2

- is used instead of NH 3 in the presence of water, NH 2- will react with

water to produce OH - since NH 2- ia a stronger base than OH -. Eventually, the OH - produced will react

with the haloalkane to produce alcohol instead. (Refer to Leveling effect in Acid base theory)

2. NH 2-

is a very strong base which may lead to elimination reaction instead of substitution reaction. (Refer to the Mechanism of elimination reaction)

c) Use of S N2 reaction in organic synthesis

Since S N2 reactions always occur with complete inversion of configuration, S N2 reaction is very useful instereospecific synthesis.

Glossary alkylation elimination stereospecific

Past PaperQuestion

94 2C 8 b ii95 2C 8 a i 95 2C 9 b v97 2B 6 a ii

94 2C 8 b ii8b A cyanoalkane is usually prepared by reacting the corresponding bromoalkane with aqueous potassium cyanide.

ii Is it possible to use hydrogen cyanide instead of aqueous potassium cyanide to carry out this preparation?Explain.

2

No, HCN is a weak acid, it ionizes only slightly in water. In HCN (aq) , [CN -(aq) ] is low, hence rate of S N2 reaction

with R–Br is slow. 2 marksC Many candidates did not realize that HCN is a weaker nucleophile than CN -.

95 2C 8 a i8a In an experiment, 46.3 g of 1-chlorobutane reacts with 30.0 g of sodium cyanide to give 35.2 g of pentanenitrile.

i Calculate the percentage yield of pentanenitrile. 3Formula mass C 4H9Cl = 92.562Formula mass NaCN = 49.01Formula mass C 4H9CN = 83.122 ½ mark

No. of mole of C 4H9Cl =46.3

92.562 = 0.500 ½ mark

No. of mole of NaCN =30.0

49.01 = 0.612 ½ mark

Since NaCN is in excess, C 4H9Cl is the limiting reagent

No. of mole of C 4H9CN =35.2

83.132 = 0.423 ½ mark

% yield =no. of mole of C 4H9CNno. of mole of C 4H9Cl × 100 % =

0.4230.500 × 100 % = 84.6 % 1 mark

C Some candidates did not calculate the relative molecular masses methodically. Generally, they did not show agood grasp of the concept of limiting reagent and percentage yield.



Reaction Mechanism Unit 4 Page 395 2C 9 b v9b Identify K, L, M, N, P, R and S in the following reactions:

vexcess CH 3I

N

H

P

1

P : N

CH3H3C

+

1 mark

C the word 'excess' was overlooked and hence the following structure was given for P

N

CH3

97 2B 6 a ii6a Each of the following conversions can be completed in not more than three steps. Use equations to show how you

would carry out each conversion in the laboratory and for each step, give the reagent(s), conditions, and structureof the product.

9

ii NH 2 NHCH 2





ReferenceReading


Organic Chemistry, Morrison Boyd, 6th Edition pg. 172–203, 208–210Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 470, 472–473Organic Chemistry, 6th Edition, Solomons, pg. 438-443

Syllabus Nucleophilic substitution of alkanol

Notes 2. Reactions of alkanol

a) Haloalkane form alkanol (ROH → RX)

We have seen how alkanol can be prepared from haloalkane by hydrolysis. By changing the reactioncondition, alkanol can also be converted back to haloalkane.

C

H

H

H

OH C

H

H

H XX- HO -slow

+

However, OH - ion is a poor leaving group since it is a strong base. The rate of direct substitution by a halideion is rather slow.

By protonation (by adding acid), OH - can be converted to H 2O which is a much better leaving group. Thus,the rate of reaction can be accelerated considerably.

r.d.s.CH

H

O

H

H

H+ protonation

H2OH+C

H

H

H

O H C

H

H

H XX- +

In case of chloroalkane, HCl is not reactive enough to react with alkanol. This is because Cl - is a much weaker nucleophile than Br - and I -.

In the preparation of chloroalkane, ZnCl 2 is added to catalyze the reaction. ZnCl 2 is a Lewis acid whichincreases the reactivity of the alkanol by enhancing the leaving ability of the OH - group.

-+C O

H

ZnCl 2

R

R

R CR

R

R

+ Cl-ZnCl 2C O H

R

R

C Cl

R

R

R

r.d.s.

N.B. In alkaline or neutral medium, haloalkane will be hydrolysed to alkanol.In acidic medium, alkanol can be converted back to haloalkane.




(1) Use of chlorinating and brominating reagent

However, Cl - is a very weak base, thus a much better leaving group than OH - ion. Chloroalkane is usually lessstable than alkanol. The yield of the conversion of alkanol to chloroalkane using halide ion is rather low. It ismore convenient to use chlorinating agents to convert the –OH group to –Cl group directly.

Chlorinating agent :

SOCl 2 (thionyl chloride), PCl 5 (phosphorus pentachloride) or PCl 3 (phosphorous trichloride)

CH 3CH 2 –OHSOCl PCl PCl 2 5 3, ,

→ CH 3CH 2 –Cl

Brominating agent

Similarly, PBr 3 is a commonly used brominating agent.

CH 3CH 2 –OH PBr

3 → CH 3CH 2 –Br

This can also be done by heating the alcohol with a mixture of red phosphorus (less flammable than white phosphorus) and bromine where P and Br 2 reacts to produce PBr 3..

b) Luca's test to distinguish 1º, 2º and 3º alkanol

In the presence of ZnCl 2 catalyst, a Lewis acid, alkanol can be converted to chloroalkane by conc. HCl (aq) .Alkanol has a reactivity towards this reaction with the following order : tertiary > secondary > primary.

-+C O

H

ZnCl 2

R

R

R CR

R

R

+ Cl-ZnCl 2C O H

R

R

C Cl

R

R

R

r.d.s.

Since the chloroalkane formed is immiscible with water and alkanol, it will form an emulsion with water andalkanol. Depending on the rate of reaction, 3º alkanol will turn cloudy immediately, 2º alkanol will turncloudy gradually while 1º alkanol will not turn cloudy at all. Because 3º alkanol is the most reactive one, it is

believed that this is a S N1 reaction.

Glossary Lewis acid chlorinating / brominating agent Luca's test




Past PaperQuestion

96 2C 7 b i ii iii 96 2C 8 c i99 1B 8 b i ii 99 2B 7 a i ii

96 2C 7 b i ii iii7b In an experiment 25 g of (CH 3)3COH react with 36 g of HCl to give 28 g of (CH 3)3CCl.

i Find the limiting reactant of the reaction showing clearly your calculation. 1½

Formula mass of (CH 3)3COH = 74.12Formula mass of HCl = 36.458Formula mass of (CH 3)3CCl = 92.562

No. of moles of (CH 3)3COH =25

74.12 = 0.337 ½ mark

No. of moles of HCl =36

36.458 = 0.987 ½ mark

HCl is in excess. ∴ (CH 3)3COH is the limiting reactant. ½ mark C Well answered. Some candidates had no idea of a limiting reactant. Candidates calculated the relative molecular

masses carelessly. Some used 35.0 or 36.0 for the relative atomic mass of chlorine without referring to thePeriodic Table given in the examination paper.

ii Calculate the percentage yield of (CH 3)3CCl. 1½

No. of moles (CH 3)3CCl =28

92.562 = 0.302 ½ mark

% yield =0.3020.337 × 100 = 89.6% (90%) 1 mark

C Many did not know how to calculate the percentage yield and they used the mass ratio instead of the mole ratio inthe calculation. The mechanism of reaction was poorly presented.

iii Name the type of the reaction and outline the mechanism of the reaction. (Movement of electron pairs should beindicated by curly arrows.)

3

Unimolecular nucleophilic substitution / S N1 ½ mark Mechanism 2½ marks

H ClOH OH

HCl

Cl -

+ +

C Many candidates did not distinguish between S N1 and S N2 reactions. Many candidates did not know where thearrow should begin and where it should end.Many started the mechanism with protonation of the alcohol by free H + instead of by HCl. The first step is shown

below:

(CH 3)3COH H Cl

Candidates should appreciate that H + does not exist freely in solutions and it is always attached to a conjugate basewhich can be a solvent molecule.




96 2C 8 c i8c Identify K, L, M, N and P in the following reactions :

(Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks)i

(CH3)2CHCH 2CH2ClK PCl5

1

K : OH or (CH3)2CHCH 2CH2OH 1 mark

C Common mistakes made by candidates are listed below: (CH 3)2CHCH 2CH 3 for K;

99 1B 8 b i ii8b In an experiment to prepare 1-bromobutane, a mixture of butan-1-ol, potassium bromide and concentrated

sulphuric(VI) acid was heated under reflux for 30 minutes.i Draw a labelled diagram of the set-up used.ii Suggest how to isolate 1-bromobutane from the reaction mixture.

99 2B 7 a i ii7a Lucas reagent, a mixture of ZnCl 2 and concentrated HCl, can be used to distinguish the following alcohols from

one another :

(CH 3)2CHCH 2OH, CH 3CH 2CH(OH)CH 3 and (CH 3)3COHi State the expected observation when these alcohols are separately treated with Lucas reagent.ii Suggest why these alcohols behave differently towards Lucas reagent.

(Hint: the zinc ion binds strongly with the oxygen atom of an alcohol, weakening the C–O bond and creating a better leaving group.)





ReferenceReading


Organic Chemistry, Morrison Boyd, 6th Edition pg. 172–203, 208–210Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 521–523Organic Chemistry, 6th Edition, Solomons, pg. 923-924, 926-927, 966

Syllabus Nucleophilic substitution of amine

Notes 3. Reactions of amine

NH 2- is a very strong base, thus it is a very poor leaving group. Normally, 1º amine R–NH 2 does not undergo

nucleophilic substitution reaction. However, upon the action of nitric(III) acid, amino group –NH 2 group will beconverted to diazonium group –N +≡ N. Diazonium group is a very good leaving group and leaves the substrate inform of very stable N 2.

a) Action of nitric(III) acid on 1º amine

(1) 1º aliphatic amine

Reagent : nitric(III) acid / nitrous acid (HNO 2) or (H–O–N=O) or (HONO)

HNO 2 + 2 H 2O+ R N N+

R N

H

H

alkyldiazoniumion

R + + N 2

carbocation nitrogen bubblesaliphatic

primaryamine

Since HNO 2(aq) is unstable, it is usually freshly prepared by dissolving NaNO 2(s) in HCl (aq) . Nevertheless, theHNO 2(aq) prepared must be kept below 5ºC otherwise it will decompose into NO 2(g) readily.

NaNO 2(s) + HCl (aq) → NaCl (aq) + HNO 2(aq)

The diazonium salts (diazonium chloride if HCl (aq) is used in the preparation of HNO 2(aq) ) intermediate formedis also highly unstable. Even at a temperature below 5ºC, the diazonium salt will still decompose graduallywith evolution of nitrogen bubbles .

This observation can be used to identify 1º aliphatic amine.

The fate of the carbocation R + depends on whether it will undergo elimination or substitution reaction. A lot

of products may be obtained. e.g. alkene, haloalkane, alcohol and ether.

e.g. R + + Cl - → R–ClR + + H 2O → ROH + H + R + + ROH → ROR + H +




(2) 1º aromatic amine

If a 1º aromatic amine is treated with NaNO 2(s) in HCl (aq) below 5ºC, no bubble will evolve. Comparingwith alkyldiazonium ion, benzenediazonium ion is more stable. This is because benzenediazonium ion isstabilized by resonance and the C–N bond possess certain double bond character.

+

N

N -

+ +

-

N

N

+

N

N

+

H

H H

H H

N N

Benzenediazonium ion can also be converted to a lot of other products. Since N 2 is an extremely good leavinggroup, benzenediazonium ion is capable to form aryl cation by the departure of N 2.

N 2+

aryl cation(not arenium ion)

diazonium ion

NaNO 2HCl, 0-5 °C

N N+

H

H

H

H

H+ NH 2

benzenamine

Nu

Note : arenium ion will be studied inthe chapter of electrophilicsubstitution.

H EH

HH

H

H

+

arenium ion

E : a positive electrophileattaching to the enzene ring

By using different nucleophile, different products can be obtained through a S N1 reaction.


The formation of aryl cation provide a very useful way to prepare phenol in laboratory. Upon heating, thesolution of benzenediazonium ion will be converted to phenol.

benzenamine

NH 2 H

H

H

H

H+ N N+

NaNO 2HCl, 0-5 °C

diazonium ionaryl cation(not arenium ion)

+ N2

O

H

HO

H

HOH

+

In this reaction, water plays the roles of solvent and nucleophile. The water in the acid mixture behaves as annucleophile and reacts with the aryl cation intermediate to form a phenol.

Glossary nitric(III) acid / nitrous acid alkyldiazonium ion benzenediazonium aryl cation




Past PaperQuestion

92 1A 1 d94 1A 3 a iii iv95 1A 3 d i iii

92 1A 1 d1d Give the reagents and reaction conditions required to convert phenylamine into benzene. 2

NaNO , HCl, 0 C3 2

2 Stand in H PO° → →

NaNO 2 ½ mark HCl ½ mark or HONO 1 mark 0ºC ½ mark H3PO 2 ½ mark or EtOH ½ mark

C Few (10-15 %) knew of H 3PO 2 reagent.

94 1A 3 a iii iv3a Consider the two compounds:

CH 3CH 2CH2 NH 2

D

CH 3CH 2CHCH 3

OH

C

and

iii Write equation(s) to show the reaction(s) (if any) of nitric(III) acid with C and D separately. 2

CHNO 2 → no reaction ½ mark

DHNO 2 → CH 3CH 2CH 2OH + N 2 1½ mark

or CH 3CH 2CH 2 N N

+HNO 2D N 2 + CH 3CH 2 C

H

H

+ etc.CH 3CH CH 2

CH 3CH 2CH 2Cl

CH 3CH 2CH 2OHunstable

If C is ignored, no mark for the first part.If “only D reacts” award ½ mark for C.If for D, “CH 3CH 2CH 2 N

+≡ N Cl -”, award ½ mark onlyiv The reaction(s) in (iii) can be used to distinguish between C and D. What observation(s) would you look for in

this test?1

In the case of D, colourless gas bubbles wil be seen or N 2 gas evolved with D 1 mark

95 1A 3 d i iii3d i Give the reactants and conditions for the preparation of benzenediazonium chloride in the laboratory. 1½

Reactant : NaNO 2 / HCl or HNO 2; aniline 1 mark Condition : < 5 ºC or -10 to 5 ºC or ice bath ½ mark

iii If an aqueous solution of benzenediazonium chloride is heated, a solid, which is soluble in dilute NaOH, can beobtained.Suggest a structure for the solid obtained and account for its solubility in dilute NaOH.

2½

OH

(no mark for name) 1 mark Phenol is weakly acidic ½ mark The following equilibrium lies to the right in the presence of NaOH and the ionic phenoxide is more soluble

½ mark OH

+OH - H2OO -

+

½ mark C Most candidates correctly produced the structure of phenol. But many candidates did not know that the solubility

of phenol in NaOH is due to the formation of a water soluble sodium phenoxide.





ReferenceReading

18.0Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 101–102, 108–109, 116, 159–160, 175–176Organic Chemistry, Solomons, 5th Edition pg. 244–251, 317–319, 322–327, 332–333Organic Chemistry, Fillans, 3rd Edition pg. 185–188, 206–210, 429–430

Organic Chemistry, Morrison Boyd, 6th Edition pg. 290–294, 300–304, 306–315Organic Chemistry, Stanley H. Pine, 5th Edition pg. 466–477, 486–491Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 462Organic Chemistry, 6th Edition, Solomons, pg. 260-262, 265-268, 300, 302-305

Syllabus Elimination ReactionCompetition between Substitution and Elimination reaction

Note V. Elimination reaction

A. Elimination reaction

In nucleophilic substitution, the nucleophile attacks the α carbon atom and leads to a substitution product.

CH 3 C C

Br

H

H

H

H

CH 3 C

CH 3

H

O CH 2CH 3

97%

C2H5OH

55°C

But besides this substitution product, another elimination product is also obtained

CH 3 C C

H

H

H

CH 3 C C

Br

H

H

H

H3%

C2H5OH

55°C

This experimental evidence shows that the reaction takes place through 2 competing pathways : substitution vselimination

Substitution

Br - HBr

CH 3 C C

Br

H

H

H

H

CH 3CH 2 O H

CH 3 C

CH 3

H

O CH 2CH 3+ +

ether

+CH 3 C

CH 3

H

O CH 2CH 3

H

α β

In the substitution pathway, the ethanol molecule acts as a nucleophile. It displaces the Br atom from thehaloalkane molecule.

Elimination

CH 3 C C

H

H

H

++ HBr Br - CH 3CH 2 O H

H

++ CH 3CH 2OHCH 3 C C

H

H

H

+

CH 3CH 2 O H

CH 3 C C

Br

H

H

H

H

alkene

In elimination pathway, the ethanol molecule has another role . It acts as a base and subtracts the β H atom fromthe haloalkane.




The role of the base is also supported by another evidence. By using a stronger base, the proportions of the two products change dramatically.

CH 3 C C

Br

H

H

H

H

CH 3 C C

H

H

H

CH 3 C

CH 3

H

O CH 2CH 3

21%79%

C2H5OH

C2H5O - Na +, 55°C+

Sodium ethoxide, C 2H5O- Na +, is a very strong base which can be prepared by dissolved sodium metal in excess

ethanol. The ethanol also serves as a solvent.

2 C 2H5OH (l) + 2 Na (s) → 2 C 2H5O- Na +(alc.) + H 2(g)

Alternatively, solid sodium hydroxide or potassium hydroxide dissolved in ethanol (alcoholic sodium hydroxide)may also be used. This involves an equilibrium of formation of ethoxide ion, the strong base.

C2H5OH (l) + NaOH (s) d C2H5O- Na +(alc.) + H 2O(l)

1. Stability of elimination product

If there are two different kind of β hydrogen available for subtraction, the one which will lead to a moresubstituted alkene will be preferred . This is because double bond (sp 2 hybridized C) and triple bond (sphybridized C) are electron withdrawing by inductive effect, they will be stabilized by electron releasing alkylgroup.

major product(di-substituted)

minor product(mono-substituted)

+αβ γ β

C C C C

H

H

H

H

H

H

H H

Br

H C C C C

H

H

H

HH

H H

H

CH 3CH 2O -

C C C C

H

H

H

H

H

H H

H




Similar to the S N2 and S N1 mechanism, there are also E2 and E1 reactions, depending on the number of moleculeinvolved in the formation of the transition state.

2. E2 reaction (not required in A-Level)

E2 stands for bimolecular elimination reaction.

In E2 reaction, the rate of reaction depends on both the concentration of the substrate and the base. Comparing withS N2 reaction, E2 involves more bond breaking and bond formation, thus a higher activation energy. i.e. S N2reaction only involves 2 bond while E2 reaction involves 4 bonds.

Rate ∝ [substrate][base]

++

transition state of E2 reaction

C C

H

L

B

δ-

δ-

C C

L

HB-:

L-HBC Cr.d.s.

3. E1 reaction (not required in A-Level)

E1 stands for unimolecular elimination reaction. In E1 reaction, the rate of reaction depends on the concentrationof the substrate only.

Rate ∝ [substrate]

r.d.s.C C HB L -+ ++ L-C C

H

+

B-:

intermediate

C C

H

Ltransition state of

E1 reaction

C C

L

H

δ−

δ+

E1 and S N1 reaction share the same intermediate – carbocation.

In S N1 reaction, the nucleophile is joined to the carbocation to yield a substitution product.In E1 reaction, the nucleophile acts as a base rather. It removes the β-H and leads to an elimination product. E1reaction is favoured by a strong and bulky base. e.g. (CH 3)3CO - Na +. Also if the α C is sterically hindered by bulkygroups, e.g. 3º substrate, the E1 reaction will also be favoured over S N1 reaction.




B. Competition between substitution and elimination reaction

1. Effect of temperature Elimination reaction has a higher activation energy than Substitution reaction. This is because more bond breakingand bond formation are involved in elimination than in substitution. As a result, an increase in temperature willcause a larger increase in rate of elimination than in the rate of substitution.

36%64%

47%53%

+CH 2 CH CH 3

100

45

NaOH / CH 3CH 2OHCH 3 CH

CH 3

O CH 2CH 3CH 3 CH CH 3

Br

Elimination reaction is favoured by a higher temperature.

2. Effect of bulkiness of the substrate and baseIf the α carbon is sterically hindered, e.g. in 3º haloalkane and use of bulky base (CH 3)3CO - Na +, elimination will bemore favourable than the substitution reaction.

3. Effect of basicity of the nucleophileIn general, all nucleopiles are also bases. If the basicity of the nucleophile is very high (e.g. CH 3CH 2O- Na +),elimination will compete with nucleophilic substitution.There are only very few strong nucleophile but with low basicity, e.g. I -, CN - and CH 3COO - Na +. They onlyundergoes substitution reaction.

C. Conditions favouring substitution and elimination reaction

Substitution Reaction Elimination Reaction

Temperature Low(substitution has a lower activationenergy)

High(elimination has a higher activationenergy)

Structure of the substrate Non-hindered α carbon(e.g. 1º haloalkane)

Sterically hindered α carbon(e.g. 3º haloalkane)

Bulkiness of thenucleophile

Low High

Basicity of nucleophile Low High

Glossary elimination steric hinderance alcoholic sodium hydroxide

Past PaperQuestion





ReferenceReading

18.1–18.2Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 101–102, 108–109, 116, 159–160, 175–176Organic Chemistry, Solomons, 5th Edition pg. 244–251, 317–319, 322–327, 332–333Organic Chemistry, Fillans, 3rd Edition pg. 185–188, 206–210, 429–430

Organic Chemistry, Morrison Boyd, 6th Edition pg. 290–294, 300–304, 306–315Organic Chemistry, Stanley H. Pine, 5th Edition pg. 466–477, 486–491Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 473–478Organic Chemistry, 6th Edition, Solomons, pg. 308, 312, 318-319, 443-444

Syllabus Formation of alkene, diene and alkyne from haloalkanePreparation of vinyl chloride from dichloroethaneDehydration of alkanol

Note D. Other relevant reactions

1. Reaction of haloalkanes with alcoholic sodium hydroxide (sodium ethoxide) to alkene, diene and alkyne

β β

more substituted alkene(major product)

C C

H

CH 3 CH 3

H

C2H5O -

C C

Br

H

CH 3 CH 3

H

H

By heating with a strong base (e.g. C 2H5O

- Na +), a haloalkane can be dehydrohalogenated to an alkene.

H2 N-

C C

H

CH 3

Br

C

H

H

H

alkene(very unreactive)

C C C CH 3H

H

H

allene(highly unstable)

(very unreactive)alkynealkene

C C CH 3CH 3C C

H

CH 3 CH 3

Br

H2 N-

C2H5O -

C C

Br

H

CH 3 CH 3

H

Br X

For a 1,2-disubstituted haloalkane, use of C 2H5O- Na + will only give a very unreactive haloalkene.C2H5O- Na + is not strong enough to subtract the second hydrogen atom from the substrate. An even stronger

base Na + NH 2- (or boiling C 2H 5O

-Na +) is required to cause further elimination of the molecule. Or, the 1,2-disubstituted haloalkane can be converted into alkyne by Na + NH 2

- (or boiling C 2H 5O-Na +) in one

step.

Further elimination of 1,2-disubstituted haloalkanewill only give alkyne but not allene (1,2-diene).This is because allene is highly unstable where thetwo π bonds are perpendicular to each other without any overlapping.

Orbital representation of allene




Formation of alkyne versus formation of conjugated diene

Normally, a 1,2-disubstituted alkane will undergo elimination to form an alkyne in the presence of a verystrong base, e.g. Na + NH 2

- (or boiling C 2H 5O-Na +). But if the haloalkene does not possess β hydrogen, a

conjugated diene would be produced instead.

Xβ

β

haloalkene(no further elimination)

C C C C

CH 3 H H

H

H

H

H

H Br Br

NaNH 2 C C C C

CH 3 H

H

H

H

H

H Br

C C C C

CH 3

H

H

H

H

H

NaNH 2

formationof alkyneimpossible formation of unstable allen

is not favourable

C C C C

CH 3 H H

H

HH

H Br

C C C C

CH 3 H H

H

H

H

H

H Br Br

C C C C

CH 3 H

H

HH

H

NaNH 2 NaNH 2

formation of more stableconjugated diene

Conjugated 1,3-diene is relatively stable comparingwith allene (1,2-diene). It has an extensivedelocalization of π electrons.

Orbital representation of conjugated 1,3-diene


The major use of vinyl chloride (chloroethene) is to prepare the plastic poly(vinyl chloride), PVC, by polymerization. It can be prepared by chlorination of ethene followed by elimination in the laboratory.

C CH H

H HC C

H

H

H

H

Cl Cl

C CH H

Cl H

Cl2Electrophilicaddition

alcoholic NaOH

CH 3CH 2O -

The mechanism of electrophilic addition of Cl 2 to ethene will be discussed in the section of electrophilicaddition.




3. Dehydration of alkanol

C CH

H H

H +O

H

H

H+C C

O

H

HH

H

H

H

O

H

H

H+C C

O

H

HH

H

H

H

HH2O+

+

C C H

HH

H

H+

H2O

Unlike other elimination reactions, dehydration of alkanol is done by heating with acid instead of base.

In a basic medium, the leaving group of the alkanol would be OH - which is a very strong base and poor leavinggroup. This makes dehydration unfavorable.

In acidic medium, the hydroxyl group –OH is protonated and transformed into a better leaving group water, H 2O.Since there is no strong base present in the reaction medium, the reaction is mainly an E1 reaction. Therefore, thereactivity of alkanol toward dehydration has the following order : 3º alkanol > 2º alkanol > 1º alkanol, parallel toall other reactions with carbocation intermediate.

Use of dehydrating concentrated sulphuric acid also helps to remove the water and shift the equilibrium to the product side. For the more reactive 3º alkanol and 2º alkanol, non-dehydrating and non-oxidizing concentrated phosphoric acid may be used to prevent charring of the substrate.

Besides ethene (elimination product), ethoxyethane (substitution product) is also obtained. By varying thetemperature, the two products can be selected.

CH 3CH 2 OH

CH 2 CH 2

CH 3CH 2 O CH 2CH 3

180 °C (E1)

140 °C (S N1)

conc. H 2SO 4(l)

In the S N1 pathway, CH 3CH 2OH serves as the nucleophile and attacks the carbocation intermediate to form theether (ethoxyethane).

C C

H

H H

H

H+C C

H

H

H H

H

O H

C C

H

H

H H

H

O H

C C

H

H

H H

H

O

H

C C

H

H

H

H

H+

CH 3CH 2 O

H

H+

CH 3CH 2 O CH 2CH 3

Glossary diene allene vinyl chloride electrophilic addition




Past PaperQuestion

91 2C 7 b i92 2C 9 a i94 2C 9 b iii98 1A 4 c ii99 2B 5 b i ii

91 2C 7 b i

7b When compound P is treated with potassium hydroxide in ethanol, two products Q, C 6H12, and R, C 8H18O, areobtained.

C CH 2CH2Cl

CH3

CH3

CH3

P i Give the structures and the systematic names for Q and R. 4

C CH 2 CH2

CH3

CH3

CH3

Cl

P

C CH 2 CH2

CH3

CH3

CH3 O CH 2 CH3

R

C CH CH 2

CH3

CH3

CH3

Q

+

1 mark eachQ 3,3-dimethylbut-1-ene 1 mark R 1-ethoxy-3,3-dimethylbutane 1 mark

C Most candidates could provide the correct structures for Q and R. However, they were generally weak innomenclature, especially for R.

92 2C 9 a i9a Give the structural formula(e) of the major organic product(s) from each of the following reactions :

iCH3CH2CHCHCH 3

Br

Br KOH (fused), 200°C

2

CH3CH2C CCH 3 CH3 CH CH CH CH 2or 2 marks

94 2C 9 b iii9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC)

respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem.

iii Show how ethene can be converted to chloroethene. 2

CH 2 CH 2 CH 2Cl CH 2Cl CH 2 CHCl- HCl NaOH / EtOH

Cl2

Or CH2Cl CH 2Cl CH 2 CHCl

720 - 920 K

2 marksC Some candidates wrongly gave the structure of chloroethene as CH 3CH 2Cl.

98 1A 4 c iiAlcohol E has the structure CH 3CH(OH)C 2H5.

4c On treatment with dilute H 2SO 4(aq) , E gives mainly two isomeric compounds, F and G, both of which have the

formula C 4H8. On treatment with bromine, both F and G give a product H with formula C 4H8Br 2.ii What is the isomeric relationship between F and G ? 1

99 2B 5 b i ii5b The following reaction produces a mixture of K and L.

N CH 3

Br

NaOH(aq)heat C12H17 NO C12H15 N

K

+

L

i Give the structures of K and L. Name the types of reactions leading to their formation.ii Outline a mechanism for the formation of K.





ReferenceReading

19.0Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 220–221, 228–231, 234–237, 239–244,

251, 254–258, 259–263, 265–267Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846

Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740,755–778, 797–799, 802–807,830–831

Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 482, 484–485Organic Chemistry, 6th Edition, Solomons, pg. 719

Syllabus Nucleophilic addition

Notes VI. Nucleophilic addition (Nucleophilic addition-elimination)

Besides substitution reaction with haloalkane, a nucleophile can also react with carbonyl group C

Oand

cyano group C N through addition pathway. This type of reaction is called nucleophilic additionreaction, Ad N.

In carbonyl group, O is more electronegative than C, therefore the C=O bond is polarized by inductive effectthrough the σ bond. The C=O bond is further polarized by mesomeric / resonance effect through the π bond.

The carbonyl group can be represented as

or

-

+C

O

C

O

δ +

δ -

C

O

Similar to the carbonyl group, in a cyano group, the C ≡ N bond is polarized by inductive effect and mesomericeffect / resonance effect. A nitrile group can be represented as

or δ -δ+

C N+ -C NC N

A. Ad N reaction

As a result, the C atom will carry a little positive charge and this makes the atom more vulnerable to the attack of a nucleophile.

Nucleophilic addition

C

O -

Nu +

C

Oδ -

δ +

Nu

C

Nu +

N -C N

δ -δ +

Nu

Nucleophilic substitution

C L Nu + C Nu L+δ + δ -

Comparing with nucleophilic substitution, O and N are joined to C by a multiple bond in carbonyl and cyanogroup. Shifting of a bond pair will not cause departure of them.




1. Addition of HCN to carbonyl compound (Cyanohydrin formation)

δ +

δ -

C

O

C-

N

H C N

C

C N

O -CN -

consumedCN -

regenerated

cyanohydrin

+C

C N

O H

C - N

Since HCN molecule is not a very good nucleophile, the rate of direct addition of HCN will be rather slow.Alternatively, a mixture of cyanide salt and aqueous acid may be used e.g. NaCN (aq) and H 2SO 4(aq) . The acid isadded into the reaction mixture drop by drop to prevent formation of excess HCN (g) which is highly toxic andvolatile. The experiment should be performed in the fume cupboard because of the high toxicity of theHCN (g).

H C H

O NaCN+ H C H

O H

C N

H2SO 4

As CN - ion is regenerated in the reaction, it acts as a catalyst speeding up the reaction.

This reaction has vital importance in organic synthesis. The product contains 1 more C atom than thesubstrate. It provides a useful way to increase the length of a carbon chain. The cyano group (–C ≡ N) can then

be converted to other functional groups, e.g. carboxyl group, amide group, amino group, easily.

2. Rate of nucleophilic addition

The cyanohydrin formation is found to be first order with respect to the carbonyl compound and first order with respect to the cyanide.

Rate = k [ C

O][CN -]

It can be shown that the proton is not involved in the rate determining step of the reaction.

Besides the concentration of the reactants, the susceptibility (reactivity) of the carbonyl carbon to the attack of nucleophile also has key effect on the reaction rate.

The susceptibility depends on 2 factors1. Electronic effect – inductive effect and resonance effect2. Steric effect




a) Electronic effect

Attachment of different substituent affects the positiveness of the carbonyl carbon and easiness of the bondformation between the carbonyl carbon and the nucleophile.

(1) Inductive effect

When an electron donating group is attached to the carbonyl carbon, the carbon is made less positive and will be less susceptible to the attack of the nucleophile. Therefore, propanone is less reactive than methanal.

C

O

H H

δ -

δ+

C

O

H3C CH 3

δ -

δ +

methanal propanone

(a) Effect of protonation

For most of the nucleophilic addition reactions, the reaction is catalyzed by acid. In acidic medium, theoxygen atom of carbonyl group is protonated. This makes the carbonyl carbon more positive and morereactive towards nucleophilic addition. Moreover, if the nucleophile is negatively charged, the product will bea neutral one instead negatively charged.

H++

C

O

HO

C

H

+C

O

(2) Resonance effect

4-methoxybenzenaldehyde is found to be less reactive than benzenaldehyde. This is because methoxy group is

an electron donating group which donates the electron to the carbonyl group. This makes the carbonyl carbonless positive and less susceptible to nucleophilic addition.

CO

H

benzenecarbaldehyde(more reactive)

CO -

HOCH 3+

CO

HOCH 3

4-methoxybenzenecarbaldehyde (less reactive)

(b) Steric effect

The first step of the reaction is the close approach of nucleophile to the carbonyl carbon. If the carbon issterically hindered by bulky groups surrounded, the reaction rate will be slowed down.

Comparing the rate of nucleophilic addition of propanone with that of methanal :

Methanal⇒ less hindered carbonylcarbon⇒ faster reaction rate

C

O

H H

Nu

Propanone⇒ more hindered carbonylcarbon⇒ slower reaction rate

Nu

C

O

CH 3H3C

Besides the electronic effect, the steric effect also favours the reaction of an alkanal over that of a ketone.

Glossary nucleophilic addition (Ad N) carbonyl group nitrile group vulnerable cyanohydrinsusceptibility electronic effect inductive effect resonance effect steric effect

protonation



Reaction Mechanism Unit 9 Page 4Past PaperQuestion

92 2C 9 c i ii93 2C 8 b ii96 1A 3 f ii98 1A 5 b99 1A 5 b ii

92 2C 9 c i ii9c Hydrogen cyanide reacts with butanone to give an addition product.

i Write a mechanism for the reaction. 2

or H+

-CN

OH

CN

OHO

+O

O-

CN

OH

CN-CN

H+

2 marksii The addition product formed can exist in isomeric forms. State the type of isomerism, and draw suitable

representations of the isomers.2

enantiomerism, (optical isomerism or stereoisomerism) 1 mark OH

NC CH2CH3CH3

OH

NC CH3CH2CH3 1 mark

C Candidates were still weak in providing a proper 3-D structural formula, e.g.

C

C2H5

OHCH 3

CN

(no heavy and dotted lines)

93 2C 8 b ii8b Outline a mechanism for each of the following reactions, using curly arrows to show electron pair displacement.

In each case state what type of reaction takes place, and give the expected product.ii butanone with hydrogen cyanide. 2½

Mechanism: Ad N Nucleophilic addition ½ mark

C O H

Me

Et

NCC O

Me

Et

H+

NC -C O

Me

Et

H+

2 marksC Wrong direction of curly arrows; ambiguity in indicating the starting and finishing location of the curly arrows.

Wrong spelling for the terms, "nucleophile".Mixing up the meaning of terms, e.g. nucleophile and electrophile, addition and condensation.



Reaction Mechanism Unit 9 Page 596 1A 3 f i ii3f i Give the structures of compounds D and E in the following organic synthesis :

C

O

HH3O+

heatHCN

D E

2

CN

OH

D :

COOH

OH

E :

1 + 1 marks(Deduct ½ mark for each minor mistake in the structures.)

ii Outline a reaction mechanism for the formation of D. (Movement of electron pairs should be indicated by curlyarrows.)

2

CN

OHOH

H

+

-

CN

O

H

H CN ½

½ ½

½ 4 × ½ marksOr

CN

OHO

HCN

- H+O

H

-CN

½

½

½

½

4 × ½ marksC The reaction mechanism was poorly presented. Many candidates wrongly used curly arrows to represent the

direction of attack of reactants instead of the movement of electron pairs, e.g.

C

CN

H

O - H+

98 1A 5 b5 Consider the five reactions of butanone (C 4H8O) J shown in the reaction scheme below:

5b State the type of reaction involved in the formation of L and outline the mechanism of the reaction. 3



Reaction Mechanism Unit 9 Page 699 1A 5 b ii5b Consider the following reactions :

CHOHCN (1) LiAlH 4

(2) H 3O+D E

ii Name the type of reaction involved in the formation of D and outline the mechanism of the reaction.





ReferenceReading




Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 485–487, 489 505Organic Chemistry, 6th Edition, Solomons, pg. 470-472, 716-719, 729-734, 759-761

Syllabus Nucleophilic addition of carbonyl compoundReduction of carbonyl compoundAddition of NaHSO 3, NH 2OH and 2,4-DNPH to carbonyl compoundIodoform reaction

Notes 3. Other relevant reactions

a) Reactions of carbonyl group

(1) Reduction of carbonyl compound by LiAlH 4 and NaBH 4

LiAlH 4 is called lithium tetrahydridoaluminate or lithium aluminium hydride.

LiAlH 4 is a very powerful reducing agent. It provides hydride ion, H -, for the reaction to take place. Since H - ion is a stronger base than OH - ion, asolvent other than water must be used, otherwise H - ion will react withwater and give H 2. Ether (ethoxyethane or called diethyl either) is the mostcommon solvent for this reaction.

Al- H

H

H

H

Li+

N.B. Since the reaction is very vigorous and potentially explosive, it is not advisable to conduct such anexperiment in the laboratory of a secondary school.

H- is also a very powerful nucleophile. It reacts with and reduces carbonyl compound readily.

C

O

H-C

H

O -

OH H

H

+

C

H

OH

+ H2O

fromLiAlH 4

carbonylcompound

alkoxideion

alkanolfromaqueousacid

An aqueous acid serves as a work up to remove all the unreacted LiAlH 4 and converts the alkoxide ion to analkanol.

NaBH 4 sodium tetrahydridoborate or sodium borohydride is similar to LiAlH 4. NaBH 4 is a weaker reducingagent which does not reacts with water. Therefore, it has the advantage that it can be used with aqueoussolvent.

fromaqueousacid

alkanolalkoxideion

carbonyl

compound

H2O+C

H

OH

C

H

O -

OH H

H

+

C

O

B-H

H

H

H Na +

N.B. Since hydride ion H - is a nucleophile, it does not react with electron rich C=C or C ≡ C .Furthermore H - is a polar reagent while C=C and C ≡C are non-polar substrates.




(2) Addition of NaHSO 3 to carbonyl compound

C

O

S

O

O

HO

- C

O

SO

O

O H

-

C

O

SO

O

O

H

-

hydrogensulphiteaddition product

CO

CH3

CH3

+ Na +HSO 3- C

CH3

CH3

OH SO 3- Na +

white precipitate

In this reaction, sodium hydrogensulphite acts as a nucleophilic reagent. By the reason of the bulkiness of hydrogensulphite ion, addition of NaHSO 3 is very sensitive to steric hinderance. A non-hindered aldehydehave about 90% yield and a slight more bulky methyl ketone have only 10 – 60 % yield. Other higher ketonesdo not form any addition product at all.

R C C

O H

H

HR C

O

HCC C

CH 3

CH 3

CH 3

CH 3

CH 3

CH 3

O

aldehyde methyl ketone 2,2,4,4- tetramethylpropan-3-one

increase in steric hinderance about the carbonyl carbon

Since the addition product is an ionic crystalline compound, this reaction can be used to distinguish aldehydeand methyl ketone from other higher ketones. The latter do not give any precipitate at all.

Furthermore, this reaction is reversible. The equilibrium position can be shifted to left by addition of either an

acid or a base to remove the hydrogensulphite ion.

Addition of acid HSO 3- + H + → SO 2 + H 2O

Addition of base HSO 3- + OH - → SO 3

2- + H 2O

Therefore, this reaction can be used to separate an aldehyde or a methyl ketone from a mixture.The aldehyde or methyl ketone is first precipitated by sodium hydrogensulphite and filtered. It is regeneratedupon addition of an acid or a base.

N.B. Although methanal and ethanal forms addition product with sodium hydrogensulphate(IV), thecrystalline compounds formed are very soluble in water, therefore no ppt. will be obtained.




(3) Condensation reaction with hydroxylamine

Hydroxylamine is a derivative of ammonia. Other amines are alkyl derivatives of ammonia but hydroxylamineis a hydroxyl derivative of ammonia, with a hydrogen replaced by a hydroxyl group.

N

H

H H

ammonia

NH

H

OH

hydroxylamine

NH

H

R

1° amine

Hydroxylamine is another nucleophile. It also reacts with a carbonyl compound through nucleophilic addition pathway. The difference is that the reaction intermediate formed has an acidic hydrogen atom attaching to thenitrogen atom. This makes the intermediate undergo elimination (dehydration) readily with the neighbouringhydroxyl group. The elimination product is called oxime.

The reaction is known as a nucleophilic condensation (or addition-elimination) reaction because it consists of 2 steps, nucleophilic addition followed by elimination.

Condensation reaction = Addition reaction + Elimination reaction

N

H

H

C O -HO + N

H

C OHO H N CHO

hydroxylaminecarbonylcompound

NH

H

OH C O

oximeintermediate

addition reaction

Condensation reaction

- H 2O

elimination




(4) Condensation reaction with 2,4-dinitrophenylhydrazine

NO 2

O2 N

N N

H

H

H

N N

H

H

H

H

hydrazine 2,4-dinitrophenylhydrazine

2,4-dinitrophenylhydrazine, or in short DNPH / DNP, is a derivative of a nitrogen containing compound,hydrazine. Pure DNPH is insoluble in water because it is not very polar. A little conc. H2SO4(l) is used todissolve the orange DNPH solid and then diluted to prepare the DNPH solution.

Reaction of 2,4-dinitrophenylhydrazine with carbonyl compound is similar to that of hydroxylamine. Thecondensation product formed is called hydrazone.

2,4-dinitrophenylhydrazine- H 2O

NO2

O2 N

N N

H H

H

C-O+

NO2

O2 N

N N

H

H

H

CO

NO2

O2 N

N N

H

H

CHO

NO 2

O2 N

N N

H

C

carbonylcompound

2,4-dinitrophenylhydrazone

addition reaction

eliminationreaction

Use of derivative of hydroxylamine and 2,4-dinitrophenylhydrazine for characterization

Both oxime and hydrazone formed in the condensation reactions are high melting solid. By converting anunknown carbonyl compound into the corresponding oxime and hydrazone, the carbonyl compound can beidentified. This is done by measuring the melting points of the oxime and hydrazone and comparing them withthe literature values.

Aldehyde or Ketone m.p. of oxime m.p. of 2,4-dinitrophenylhydrazoneEthanal 46.5 ºC 168.5 ºCPropanone 61 ºC 128 ºCBenzenecarbaldehyde 35 ºC 237 ºC

2-methylbenzenecarbaldehyde 49 ºC 195 ºC3-methylbenzenecarbaldehyde 60 ºC 211 ºC4-methylbenzenecarbaldehyde 79 ºC 233 ºC




(5) Haloform reaction / Iodoform reaction

Formation of enolate ion

Both aldehyde and ketone show tautomerization. The keto form of the molecule coexists with the enol form of the molecule. Keto-enol tautomerization is an dynamic equilibrium involving the movement of a hydrogen

atom.

C C

OH

HC C

OH

C C

H

Oδ-

δ- + H+

- H +

- H +

+ H+

keto form enol formenolate ion

α α α

An intermediate, enolate ion, is formed during the transition between the keto form and the enol form. Enolateion can be considered as conjugate base of the keto or enol form.

C C

H

O -

C C

H

O

- or C C

H

Oδ-

δ-

Since the enolate ion is stabilized by resonance, the acidity of the α -hydrogen is much higher than other hydrogen atoms attached to the carbon chain.

α β

α -hydrogen pK a ≈ 20

β-hydrogen pK a ≈ 50

O

H

HH

CCC

H

It is observed that α -hydrogen is almost 10 30 times more acidic than β-hydrogen. In the presence of a base e.g. NaOH, a carbonyl compound loses its α -hydrogen readily and forms an enolate ion.

α β

O

H

HH

CCC

H

H-OO

H

H

CCC

Hβα

-+ H2O

Enolate ion possesses an electron rich carbon atom and behaves as a nucleophile.

Haloform reaction

Haloform has a general formula CHX 3. Choroform CHCl 3 and bromoform CHBr 3 are liquid insoluble inwater. Iodoform CHI 3 is a bright yellow precipitate. Formation of yellow precipitate of iodoform serves as a

very useful test of methyl ketone C C

O H

H

H and 1-hydroxyethyl C C

H

H

H

OH

Hgroup.




Haloform reaction consists of two steps :1. Formation of trihalo ketone by nucleophilic substitution2. Formation of haloform by nucleophilic addition-elimination

Formation of trihalo ketone from methyl ketone

The presence of a base, e.g. NaOH (aq) , promotes formation of enolate ion. The enolate ion formed acts as a

nucleophile and displace the I-

ion from the I 2 molecule. Once the first α -hydrogen is substituted by an iodineatom, the acidity of the remaining α -hydrogens will be increased due to the negative inductive effect of theiodine atom. Consequently, the remaining two α -hydrogens will also be substituted by iodine atoms.

αC C

O

H

HR -

I I

C C

O H

H

HR α

-O H

C C

O

H

HR

I

α

OH -, I2 OH -, I2 C C

O

R

I

I

Iα

enolate ionmethyl ketone

-I effect of halogen makethe H even more acidic

nuleophilic substitutionenolate ion formation further substitutions

triiodo ketone

Formation of haloform from trihalo ketone

The triiodo ketone formed is much more reactive than the original methyl ketone because –CI 3 group is a better leaving group than the original –CH 3 group, as -CI3 ion is a much weaker base than -CH 3 ion does.

triiodo ketone

-

C CR

I

I

I

O

O

H

CR

O

O

H

C

I

I

I

-R C

O

O -

C

I

I

I

HC C

O

R

I

I

Iα

H O -

carboxylateion

iodoform

For 1-hydroxyethyl group, it is first oxidized to methyl ketone by the iodine in alkaline medium. Then, themethyl ketone formed is converted to carboxylate through the same mechanism mentioned above.

oxidation

methyl ketone1-hydroxyethyl group

2 H 2O2 I -2 OH -I2 ++++ C C

H

H

H

O

C C

H

H

HO

H

H

Besides testing of methyl ketone and 1-hydroxylethyl group, iodoform reaction is also used in organicsynthesis to shorten the carbon chain by one carbon atom. It also converts methyl ketone or 1-hydroxyethyl group to carboxylate ion or carboxylic acid upon acidification.

N.B. 1. Haloform reaction can be used to test for methyl ketone and 1-hydroxyethyl group2. It can be used to cut the carbon chain short by 1 C .3. It oxidizes methyl ketone and 1-hydroxyethyl group to carboxylate .

Some chemistry text books say that NaOI (aq) is used to bring about iodoform reaction. In fact, according to themechanism, the reaction is actually brought about by OH -

(aq) ion and I 2(aq) molecule. NaOI (aq) takes theadvantages that it is prepared from I 2(aq) and NaOH (aq) .

NaOH (aq) + I 2(aq) d NaOI (aq) + NaI (aq) + H 2O(l)

As a personal preference, I prefer I 2(aq) / NaOH (aq) to NaOI (aq) as the reagent used.




Glossary lithium tetrahydridoaluminate / lithium aluminium hydride ether acid work upsodium tetrahydridoborate / sodium borohydride sodium hydrogensulphite hydroxylamineoxime nucleophilic addition-elimination condensation reaction 2,4-dinitrophenylhydrazinehydrazone characterization keto-enol tautomerization enolate β-hydrogen haloformreaction / iodoform reaction methyl ketone 1-hydroxyethyl group trihalo ketone

Past PaperQuestion

90 2C 9 a ii iii

91 1A 1 b i 91 2C 9 a iii92 2C 8 b 92 2C 9 a iii93 2C 8 a i94 2C 9 a i95 2C 7 a ii 95 2C 9 b i96 2C 8 c v97 1A 5 a v 97 2B 5 a i iii 97 2B 5 a i ii iii 97 2B 7 a ii 97 2B 7 b i98 1A 5 a 5 c d i ii99 2B 5 a i

90 2C 9 a ii iii9a Each of the following carbonyl compounds P, Q and R reacts with 2,4-dinitrophenylhydrazine in the presence of a

strong acid to give a phenylhydrazone derivative.

CH3 C CH 2CH2CH3

O

P

CH2CH2CH3CCF3

O

Q

CH3 C

O

CH2 CH CH 3

CH3

R ii Under the same conditions, P, Q and R form phenylhydrazones at different rates. Predict with an explanation

which will react the fastest, and which the slowest.5

Rate Q > P > R (slowest)Taking P as a reference.Q is faster than P because F substituent is electronegative. 1 mark

⇒ carbonyl carbon more electrophilic 1 mark ⇒ more susceptible to nucleophilic attack by -NH 2 1 mark

R is slower than P because the -CH(CH 3)2 group is more bulky and hindered the approach of the -NH 2 to the

carbonyl carbon. (electron donating feature of -CH(CH 3)2 is not significant due to long distance) 2 marksC Many candidates considered the electronic effect but not the steric effect.iii Explain why 2,4-dinitrophenylhydrazones are useful derivatives of carbonyl compounds. 2

2,4-dinitrophenylhydrazones of carbonyl compounds are high melting solids useful for characterizationOR are bright colour ppt useful for identifying carbonyl group 2 marks

C Many candidates confused 2,4-DNPs with azo dyes, and wrongly stated that the former were important dyes inindustry.

91 1A 1 b i1b i

C CH 2

CH3

H

E

C OCH3

H

F Explain why hydroxylamine reacts with F and not with E, and give the product expected from the reaction.

3

In F, the double bond is strongly polarized,C O

due to the electronegativity of oxygen. In E there is no such polarization.OR

The oxygen has a tendency to acquire electrons leading toC O+ -

in which the carbon is electron deficient.1 mark

Hydroxylamine with N having lone electron pair is electron rich and takes part in nucleophilic addition at C.

OR C OHONH 2

(This properties of NH 2OH must be shown in some way) 1 mark Product is Me-CH=NOH (no mark for Me-CHOH-NHOH) 1 mark There are many ways to answer this question:



Reaction Mechanism Unit 10 Page 8For all the 3 marks, answer MUST explain WHY polarization difference is responsible for the difference inreactivity - answer MUST include the fact that NH 2OH reacts as nucleophile or by nucleophilic addition.

C A noticeable number of candidates did not know the structure of hydroxylamine.

91 2C 9 a iii9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs.

Describe what you would observe in each case.iii O

CH 3

C

O

CH3

2

Add iodine and NaOH,

C

O

CH3

will give yellow ppt of iodoform. No ppt observed for

O

CH 3

.

CCH 3

OCOO - Na +

+ + NaOH I2 + CHI 3

2 marksC Many candidates wrote the formula of iodoform incorrectly as CH 3I.

92 2C 8 b8b Describe how 2,4-dinitrophenylhydrazine may be used to identify an unknown ketone such as butanone. Give an

equation for the reaction involved.4

Take the unknown ketone to reacts with 2,4-DNPH; Get the solid derivative; Purified by recrystallization;Determine its m.p.; Compare with literature data. 2½ marks

CH3CH2 C CH 3

O

NHNH 2

NO 2

NO 2

+

NO 2

NO 2

NH N CCH 2CH3

CH3

1½ mark C Candidates gave the wrong structure for 2,4-dinitrophenylhydrazine, e.g.

NH 2

NO 2

NO 2 Many candidates did not know how to make use of the m.p. of hydrazone as a means to identify the ketone. They

just mentioned the use of 2,4-DNP as a test for carbonyl functional group.

92 2C 9 a iii9a Give the structural formula(e) of the major organic product(s) from each of the following reactions :

iii

CCH3

O

I2/NaOH(aq)

2

COO - Na +

+ CHI 3

2 marks



Reaction Mechanism Unit 10 Page 993 2C 8 a i8a Give a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents required, the observation expected, and the chemical equation(s) for each test.i and CH3CH 2OCH 2CH3CH 3CH2COCH 2CH 3

3

Reagent: 2,4-DNPH 1 mark Observation: red/orange/brown/yellow precipitate 1 mark Equation:

NHNH 2

NO 2

NO 2

C OEt

Et+

CEt

Et NNH NO 2

O2 N

1 mark C Incorrect structure for 2,4-dinitrophenylhydrazone.

94 2C 9 a i9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions.

i

CH 3 C CH 2CH 3

O

P NH 2OH

1

P : N

OH

1 mark

95 2C 7 a ii7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents used, the observation expected and the chemical equation(s) for each test.ii

CH 3CH 2CCH 2CH 3

O

and CH 3CH 2CH 2CCH 3

O

3

Treat compound with I 2/OH-

½ + ½ mark

CH 3CH 2CH 2CCH 3

OI2

KOH / NaOHCH 3CH 2CH 2C

O

O -K + + CHI 3 ½ + ½ mark yellow precipitate observed. ½ mark

No ppt. with CH 3CH 2CCH 2CH 3

O

½ mark

C Many candidates failed to give the equations for the reactions.Some candidates erroneously gave the silver mirror test as the answer. One common mistake was the wrongformula for iodoform (CH 3I instead of CHI 3). Some candidates gave the wrong colour for CHI 3. 'No observablechange' instead of 'no reaction' should be given because pentan-2-one does undergo reaction (iodination) under theconditions.

95 2C 9 b i9b Identify K, L, M, N, P, R and S in the following reactions:

i

NHN NO 2

O2 N

K

H2 NHN

O2 N

NO 2

1

K :O

/ cyclobutanone 1 mark

C candidates appeared to be unfamiliar with the cyclic structure and gave an acyclic ketone;



Reaction Mechanism Unit 10 Page 1096 2C 8 c v8c Identify K, L, M, N and P in the following reactions :

(Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks)v

CH3CH2 C CH 3

OP

NaHSO 3

1

P :SO 3

- Na

+

OH CH3CH2 C SO 3

-

Na+

CH3

OHor 1 mark

C wrong structures for P, such as

C CH 3

SO 3H

CH3CH2

ONa

C CH 3

SO 3H

CH3CH2

OH

97 1A 5 a v5a Give a structure for each of the compounds D, E, F. G and H: 5

v I2 / NaOH (aq)triiodomethane + sodium salt of a carboxylic acidC4H8O

H

97 2B 5 a i ii iii5a In an experiment, 10.0 g of butanone reacts with 5.0 g of hydrogen cyanide to give 11.0 g of 2-hydroxy-2-

methylbutanenitrile.6

i Find the limiting reactant of the reaction, showing clearly your calculation.ii Calculate the percentage yield of 2-hydroxy-2-methylbutanenitrile.iii Name the type of reaction and outline the mechanism involved. (Movement of electron pairs should be indicated

by curly arrows.)

97 2B 7 a ii7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test

should include the reagent(s), the expected observation with each compound and the chemical equation(s).

12

ii

CH2CH3

OH

and

CH CH 3

OH

97 2B 7 b i7b Identify J, K, L, M and N in the following reactions. 5

i

C

O

HCH NOHJ



Reaction Mechanism Unit 10 Page 1198 1A 5 a c d i ii5 Consider the five reactions of butanone (C 4H8O) J shown in the reaction scheme below:

5a Suggest a reagent by which K may be formed and give the structure for K. 15c Give structures for compounds M, N, P and R. 45d i S is a structural isomer of J. S also reacts with 2,4-C 6H3(NO 2)2 NHNH 2 to give a red precipitate. Draw the structure

of S.1

ii How may J and S be identified by making use of their reactions with 2,4-C 6H3(NO 2)2 NHNH 2 ? 1

99 2B 5 a i5a Identify D, E, F, G and J in the following reactions.

iO NNH

D





ReferenceReading

19.2.0Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 220–221, 228–231, 234–237, 239–244,



Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505Organic Chemistry, 6th Edition, Solomons, pg. 708-711

Syllabus Carboxylic acid and its derivativesPreparation of acid derivativesReaction of acid derivatives

Notes b) Reactions of carboxylic acid and its derivatives

(1) Carboxylic acid and its derivatives

(a) Difference between carbonyl compound and carboxylic acid and its derivatives

Carboxylic acid consists of a carbonyl group C

O

and a hydroxyl group –OH. Because of the presence of

carbonyl group, in certain extent, it shares the same reaction pattern of ordinary carbonyl compound.

However, carboxylic acid does not resemble the properties of ketone entirely. There are 3 major differences.

1. Carboxylic acid has reaction with sodium, phosphorus trichloride PCl 3 and phosphorus pentachloride PCl 5 while ketone or aldehyde haven't.

Since carboxylic acid also possesses a hydroxyl group, it shows some typical properties of a hydroxycompound.

e.g. 2 H–O–H (l) + 2 Na (s) → H2(g) + 2 NaOH (aq)

2 C

O

CH 3 OH (l) + 2 Na (s) → H2(g) + 2 C

O

CH3 O - Na +(acid)

The ethanoic acid used in this reaction must be dry, in which there is no free hydrogen ion. Otherwise, Na(s) will react with the hydrogen ion vigorously and may cause explosion.

e.g. 3 H–O–H (l) + PCl 3(l) → 3 HCl (g) + H 3PO 3(l)

3 C

O

CH 3 OH (l) + PCl 3(l) → 3 C

O

CH3 Cl (l)+ H 3PO 3(l)

e.g. 4 H–O–H (l) + PCl 5(s) → 5 HCl (g) + H 3PO 4(l) C

O

CH3 OH (l) + PCl 5(s) → C

O

CH3 Cl (l) + POCl 3(l) + HCl (g)

In these reactions, the hydroxyl group –OH is substituted by –Cl atom. Another useful reagent for thesame purpose is thionyl chloride SOCl 2 which has an even higher yield than PCl 3 or PCl 5.




2. Carboxylic acid and its derivatives undergoes condensation (addition-elimination) reaction instead of addition reaction only.

In carboxylic acid, a hydroxyl group is attached to carbonyl carbon. It can be converted to a group leavinggroup by protonation. Therefore, any nucleophilic addition is usually followed by elimination and leads tocondensation reaction.

+ L-

R C

O

NuR C

O

L

Nu-

R C

O

L

Nu

-nucleophilic

addition

elimination

Condensation reaction = Addition reaction + Elimination

3. Carboxylic acid and its derivatives have no reaction with 2,4-dinitrophenylhydrazine.

–OH is a +R and -I group. Since resonance effect is usually stronger than inductive effect, –OH is anelectron donating group in this case. It makes the carbonyl carbon less positive and less reactive thanketone and aldehyde.

C

O

CH 3 O H

-

+C

O

CH 3 O H or CCH 3

O

O H

δ −

δ +

Similarly, all derivatives of carboxylic acid are with the resonance stabilized carbonyl group. They do notreact with 2,4-dinitrophenylhydrazine .

(b) Reactivities of carboxylic acid and its derivatives

Carboxylic acid and its derivative shows the following order of reactivity. The trend is determined by the

leaving ability of the substitutent attaching to the carbonyl group.

Acid chloride(acyl chloride)

Acidanhydride

Carboxylicacid

Ester Amide

C

O

R Cl

C

O

R O C

O

R' C

O

R OH C

O

R O R'

C

O

R NH 2

Most reactive Least reactive

Leaving group Cl - RCOO - -OH RO - NH 2-

Basicity of theleaving group Low –––––––––––––––––––––––––––––––––––––––––––––––––– → High

Glossary




Past PaperQuestion

95 2C 7 a iv97 2B 7 a iii

95 2C 7 a iv7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents used, the observation expected and the chemical equation(s) for each test.iv

CH 3CH 2CNHCH 2CH 3O

and CH 3CH 2CCH 2 NHCH 3O

3

Treat compound with a solution of 2,4-dinitrophenylhydrazine in methanol (or ethanol) 1 mark

NHCH 3

O

NHCH 3

N

NH

O2 N

NO 2 NO 2

O2 N

H2 NNH

1 mark Orange / red / brown / yellow precipitate is observed. ½ mark

No ppt. with CH 3CH 2CNHCH 2CH 3

O

. ½ mark

OR

CH3CH2CNHCH 2CH3

O OH -(aq)

heat CH3CH2C

O

O- C2H5 NH 2+

1 mark

C2H5 NH 2 C2H5 NH 3ClHCl + -

white fume 1½ mark Comparison ½ mark

iv Many candidates failed to give the equations for the reactions.Very badly answered. Few candidates could identify the keto group probably because they were not familiar withcompounds with more than one functional group. Some did not know that secondary amines do not react withHNO

2to give N

2. Most could not recall the chemistry of amides that are susceptible to hydrolysis.

97 2B 7 a iii7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test

should include the reagent(s), the expected observation with each compound and the chemical equation(s).12

iii

C

O

C2H5 O(CH 2)2CH3 and C

O

C2H5 CH2OC 2H5





ReferenceReading

19.2.1Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 220–221, 228–231, 234–237, 239–244,



Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 462–463, 472, 505–508, 510–511, 520Organic Chemistry, 6th Edition, Solomons, pg. 805-817, 820-824

Syllabus Preparation of acid derivativesReaction of acid derivatives

Notes (2) Formation of different acid derivatives

+ L-

R CO

NuR C

O

L

Nu-

R CO

L

Nu

-nucleophilic

addition

elimination

–L = –Cl, –COOR, –OH, –OR, –NH 2 Nu: = RCOO - Na +, Na +OH -, ROH, NH 3

In general, a more reactive acid derivative can be converted to a less reactive acid derivative by using asuitable nucleophilic reagent. The reaction is a simple nucleophilic addition-elimination reaction.

A less reactive derivative can also be converted to a more reactive derivative but the yield of conversion would be rather low.

Exception : 1. Amide cannot be prepared from carboxylic acid and ammonia. This is becausecarboxylic acid is an acid and ammonia is a base, they reacts to give a salt only. Amideis usually prepared from acid chloride with ammonia.

2. Similarly, ester cannot be prepared from carboxylic acid and sodium alkoxide as theywill neutralize each other.

(a) Use of chlorinating agent to prepare acyl chloride

Acyl chloride is the most reactive acid derivative, it cannot be prepared from the other acid derivative. It is prepared by converting the –OH group of carboxylic acid to –Cl group by chlorinating agent.

Chlorinating agent :SOCl 2 (thionyl chloride), PCl 5 (phosphorus pentachloride) or PCl 3 (phosphorous trichloride)

R C

O

OH

SOCl 2 or PCl 5 or PCl 3

R C

O

Cl




(b) Formation of acid anhydride

(i) Through intramolecular dehydration by heating

Certain dioic acid can be dehydrated by heating through intramolecular dehydration to form cyclic acidanhydride. This is only possible if the two hydroxyl groups of the two carboxyl groups are facing each other spatially.

(ii) Through intermolecular dehydration by a very strong dehydrating agent

However, linear acid anhydride, e.g. ethanoic anhydride CH 3COOCOCH 3, cannot be prepared by heating

alone. Since the formation of the anhydride involves intermolecular dehydration, if the acid is heated strongly,the thermal agitation will not allow the two acid molecules to get to each other for the dehydration. Therefore,a very strong dehydrating agent e.g. phosphorus pentaoxide P 2O5(s) (or P 4O10(s) ) should be used instead of heating.

C

O

CH3 OH C

O

CH3HO C O C

O O

CH3CH3+P2O5

(iii) From acyl chloride

Similarly, it can also be prepared from reactive acyl chloride by mixing with a carboxylate salt.

C

O

CH3 O- Na

+C

O

CH3Cl C O C

O O

CH3CH3+ + NaCl




(c) Formation of ester (Esterification with alkanol and phenol)

R' C

O -

O H

O

H

R +

R O H

R' C

O

O H

alcohol

carboxylic acid

charge separationinvolved

r.d.s.R' C

O -

O H

OR

H

+H2O+R' C

O

O R

ester water

Esterification is also a typical condensation reaction. An alcohol molecule is added to a carboxylic acidmolecule which elimination of water.

The nucleophilic oxygen atom is added to the carbonyl carbon. It can be seen that an intermediate with chargeseparation in the structure is involved. This makes the reaction having a high activation energy, thus, a rather slow rate.

Then, the proton possessed by the alcohol is shifted to the hydroxyl group to convert it to a better leavinggroup, water H 2O.

Contrastingly, esterification in the presence of any aqueous acid is found to be catalyzed.

Acid Catalyzed Esterification

C

O

OHH+

fC

O

R OH

H

O H

R'

+f

C

O

R OH

H

+H

O R'

r.d.s.

no chargese a ation involved

C

O

R OR'H2O H++ +f +

C

O

R O

H

O

R'

H

H

f

In the acid catalyzed pathway, the carbonyl oxygen is first protonated. This activates the carbonyl group byincreasing the polarity of the C=O bond and makes the carbonyl carbon more positive and more vulnerable tothe attack of the nucleophile, alcohol. Furthermore, no reaction intermediate with charge separation isinvolved in the acid catalyzed mechanism and a much lower activation energy would be involved.

Esterification of phenol

However, phenol is a weaker base than aliphatic alcohol becauseof benzene ring is electron withdrawing.

O

H

O

H

A substrate more reactive than carboxylic acid is required to give a reasonable yield. Phenol reacts with acylchloride and acid anhydride to form ester.

R C

O

ClOH O C

O

R Cl -+

OH O C

O

R R C

O

O -R C O

O

C

O

R +




(d) Formatioin of amide (Acylation and benzoylation of amine)

Acylation means substitution by an acyl group C

O

R . Two common reagents used in acylation are ethanoyl

chloride C

O

CH3 Cl and ethanoic anhydride C

O

CH3 O C CH 3

O

. Both of them are reactive derivatives of carboxylicacid, which reacts with nucleophile, e.g. NH 3 or RNH 2, readily. When ethanoyl chloride or ethanoic anhydride

are used, the reaction is also called ethanoylation since the acyl group involved is the ethanoyl group CO

CH3 .

C

O

CH 3 Cl

NH

H

H

-

+

C

O

CH 3 Cl

NH H

H

C

O

CH 3 N

H

H

H+

NH 3

Cl-+ NH 4+Cl-+C

O

CH 3 NH 2excess

ethanamide

In ethanoylation of ammonia, excess ammonia is used to neutralize the HCl formed. In the reaction, one H on

ammonia molecule is substituted by an ethanoyl group CO

CH 3 . Unlike esterification, catalysis by protonation is impossible in here. The presence of aqueous acid will neutralize the nucleophile NH 3 to non-nucleophilic species NH 4

+ and stop the reaction. Furthermore, ethanoyl chloride or anhydride also reacts withwater directly.

Similar to acylation, benzoylation means substitution by an benzoyl group C

O

. Benzoyl chloride is thecommonly used reagent. It is prepared by treating benzenecarboxylic acid (benzoic acid) with phosphorustrichloride PCl 3, phosphorus pentachloride PCl 5 or thionyl chloride SOCl 2.

C

O

OH + PCl3 C

O

Cl + H3PO 33 3 benzoyl chloride

The reactive benzoyl chloride reacts with the ammonia or amine through nucleophilic addition-eliminationsimilar to the ethanoylation discussed above.

C

O

Cl

NH H

H

+

-

C

O

Cl

NH

H

H

excessC

O

NH 2 + NH 4+Cl-+ Cl-C

O

N

H

H

H+

NH 3

benzenecarboxamideor benzamide




(3) Reaction of ester

(a) Hydrolysis of ester

Ester can be hydrolyzed in neutral, acidic or alkaline medium. Since OH - is a nucleophile stronger than H 2O,the rate of hydrolysis will be fastest in alkaline medium.

+R' C

O -

O

O

H

R

H O

H

R' CO -

O

O

H

R

H

+R' C

O

O R

-OH

R' C

O

O H O R H R' C

O

O -

OH - is consumed OH - is regenerated

carboxylicacid

alkanol carboxylateion

-OH

OH - ion can be considered as a catalyst in the hydrolysis or it can be considered as a reagent which neutralizesthe carboxylic acid to carboxylate ion.

In acidic medium, hydrolysis is catalyzed by protonation of carbonyl group.

R' CO

O R

H+

OH

H

R' C

O

O R

H

+

+H

O

H

R O

O

CR'H

R' C

O

O H

H

+R' C

O

O R

H

OH H

+ R' C

O

O HH++H O R H O R + +

Therefore, the rate of hydrolysis of ester follows the following order :

alkaline medium > acidic medium > neutral medium.

Glossary Esterification acylation benzoylation

Past PaperQuestion

90 1B 4 c i ii92 1A 1 a iii 92 1B 5 a ii iv93 2C 7 a iii94 2C 9 a ii95 2C 9 b ii96 1A 3 d ii97 1A 5 a iii98 2B 7 a i99 2B 6 a i

90 1B 4 c i ii4c Amides are generally prepared by reacting equimolar amounts of an amine and an acid halide.

(Relative atomic masses : H, 1.0; C, 12.0; N, 14.0; O, 16.0; Cl, 35.5)Calculate:

i the mass (in grams) of Y required to react with 4.0 g of benzoyl chloride (C 6H5COCl), and 1½ Relative molecular mass of Y = 6 × 12.0 + 7 × 1.0 + 14.0 = 93.0

Relative molecular mass of C 6H5COCl = 7 × 12.0 + 5 × 1.0 + 16.0 + 35.5 = 140.5 ½ mark

mass of Y =4.0 × 93.0

140.5 = 2.65 g 1 mark

ii the percentage yield of the same reaction if 4.0 g of the amide product is obtained. 1½ Relative molecular mass of amide (C 13H11 NO) = 13 × 12.0 + 11 × 1.0 + 14.0 + 16.0 = 197.0

100% yield =4.0 × 197 .0

140.5 = 5.6 g

percentage yield =4.05.6 × 100% = 71% 1½ mark

C The calculation of percentage yield was very, very poorly answered. It is clear that such work is rarely done bycandidates. It is also obvious that this very simple concept is unknown to most candidates.



Reaction Mechanism Unit 12 Page 692 1A 1 a iii1a There are several isomers of benzenedicarboxylic acid.

iii On heating, one of the isomers gives a compound W , with formula C 8H4O3. Give the structure of W . 1

CO

C

O

O 1 mark

92 1B 5 a ii iv5a Ethanolylation (acetylation) of phenylamine (C 6H5 NH 2) may be carried out by refluxing for 1 hour with ethanoic

anhydride (acetic anhydride) giving a product which is insoluble in cold water.(Ethanoic anhydride hydrolyses in water to give ethanoic acid. The b.p. of phenylamine and ethanoic anhydrideare 184°C and 140°C respectively; the m.p. and b.p. of the product are 114°C and 304°C respectively.)

ii Give the structure of the product from the ethanoylation reactions. 1

N C CH 3

OH

1 mark iv If 40g of phenylamine is to be ethanoylated, calculate the mass of ethanoic anhydride required.

(Relative atomic masses: H 1.0; C 12.0; N 14.0; O 16.0)2

molecular mass of PhNH 2 = 93molecular mass of (CH 3CO) 2O = 102

Since PhNH (CH CO) O PhNHCOMe CH CO H2 3 2 3 2+ → +

The mass of ethanoic anhydride required = 40102

93

1

1× × g = 43.9g 2 marks

C Calculation was very badly done, suggesting that few do experiments involving quantities.

93 2C 7 a iii7a Give the structural formula(e) of the major organic product(s) formed in each of the following reactions.

iii NH 2CH3COCl

2

NH C CH 3

O

1 mark C This was generally well answered.

94 2C 9 a ii9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions.

ii NH 2

QCH3CH 2COCl

1

Q :

N

O 1 mark

95 2C 9 b ii9b Identify K, L, M, N, P, R and S in the following reactions:

ii

CH 3CH 2CH 2COH

O

CH 3CH 2CH 2CCl

OL

1

L : phosphorus(V) chloride / PCl 5 or phosphorus(III) chloride / PCl 3 or thionyl chloride / SOCl 2 1 mark C Common mistakes included : HCl for L;



Reaction Mechanism Unit 12 Page 796 1A 3 d ii3d Give the structure of the major organic product(s) in each of the following reactions :

ii

C CH 2 CH 2 C

O O

OHHOheat

1

CH 2 CH 2C C OHHO

O OO

O

O

heat

1 mark

97 1A 5 a iii5a Give a structure for each of the compounds D, E, F. G and H: 5

iiiCH3CH2CO 2H

SOCl 2F

98 2B 7 a i7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case,

using the same starting molecule, outline a feasible synthetic route to obtain the target molecule, with no morethan three steps. In each step, give the reagent(s) used, the conditions required and structure of the product.

12

i

C

O

NHCH 3 C

O

OCH 3

CH3OH

99 2B 6 a i6a i Suggest a reagent for the following conversion :

Cl NH 2 Cl NH CCH 3

O

4-chlorophenylamine N-(4-chlorophenyl)ethanamide





ReferenceReading

19.2.2–19.2.3Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 220–221, 228–231, 234–237, 239–244,



Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 505, 519Organic Chemistry, 6th Edition, Solomons, pg. 919-921

Syllabus Reduction of acid and its derivativesHoffmann degradation of amide

Notes (3) Reaction of amide

(a) Reduction of amide and other acid derivatives

Except amide, all other acid derivative are reduced by LiAlH 4 to alcohol. The nucleophilic hydride ion, H -,serves as a strong reducing agent in this reaction.

H

R C

O

L

-

R C

O

L

H

-aldehyde

H -

L-+R C

O

H H2O+CR H

H

OH

CR H

H

O -

OH H

H

+

acidwork up

where L = –Cl, –OOCR, –OH, –OR

The aldehyde intermediate cannot be isolated from the reaction mixture. It will be reduced directly to a primary alcohol.

Exception : Reduction of amide

Amide will be reduced by LiAlH 4 to amine instead of alkanol.

LiAlH 4 can be considered to be comprised of three parts :Li+ : a spectator ionH- : a nucleophile, a very strong reducing agentAlH 3 : a Lewis acid, a catalyst activates the reactivity of a carbonyl group

The actual mechanism of the reaction is still not very clear, the following may be a possible one.

Simplified Proposed Mechanism

Li+-

R C N

H

O

AlH

H

H-

+

H-

R C N

H

O-

Li+

Al

H

H H

R C N H

H

O H-

acid-base reaction

activation reduction

1° amineacid work upimine

(further reduction)

R C N

HH

H H

R C N

HH

H-

R C N

HH

H-

H O

H

H

+

Li+-

R C N

H

O

AlH

H

H

H

-




Depending on the original structure of the amide, it is reduced by LiAlH 4 to 1º amine, 2º amine or 3º amine.Amide is the only acid derivative which will not be reduced to 1º alcohol by LiAlH 4.

R C NH 2

OR C

H

H

N H

H

1) LiAlH 4 / ether

2) H3O+

amide 1° amine

2° amine N-substituted amide

1) LiAlH 4 / ether

2) H 3O+R C

H

H

N H

R'

R C N

O

H

R'

3° amine N.N-disubstitutedamide

1) LiAlH 4 / ether

2) H 3O+R C

H

H

N

R'

R''R C N

O

R'

R''

This reaction is commonly used to prepare 2º and 3º amine where direct alkylation of 1º and 2º amine willyield a mixture of 2º, 3º amine and quaternary ammonium ion.

Reduction selectivity of LiAlH 4 and NaBH 4

In term of reducing power, LiAlH 4 is muchstronger than NaBH 4. LiAlH 4 can reduce all kindsof carbonyl compound to alcohol. This includesketone, aldehyde and all kinds of acid derivativesexcept amide. In contrast, NaBH 4 is only strong

enough to reduce ketone and aldehyde to alcohol, but not any acid derivative.




(b) Hofmann degradation of amide

Reagent : a base – NaOH (aq) and an electrophile – Br 2(l)

R C NH 2

O

N H

H

R + ++Br 2 4 NaOH 2 NaBr Na 2CO 3 2 H2O+ +

non-substitutedamide 1° amine

Hofmann degradation is also known as Hofmann rearrangement. According to the mechanism, the 2 protonson the amino group are first subtracted by the base. Therefore, only a non-substituted amide RCONH 2 canundergo Hofmann degradation.

Comparing with the reduction of amide by LiAlH 4, Hofmann degradation provides a way to shorten thecarbon chain by 1 carbon atom.

1° amineamide

1) LiAlH 4 / ether

2) H3O+R C

H

H

N H

HR C NH 2

O

Br 2 / NaOH (aq) N H

H

R

1?aminewith 1 C lessamide

R C NH 2

O

Glossary reduction selectivity Hofmann degradation / Hofmann rearrangement




Past PaperQuestion

92 2C 8 a i ii93 2C 7 a iv94 2C 9 a iii95 2C 9 b vi vii96 2C 8 a i ii 96 2C 9 a iii97 1A 5 a i ii 97 2B 7 b iii98 1B 8 b ii iii 98 2B 7 a iii

99 2B 5 a iv

92 2C 8 a i ii8a Amines can be prepared from an amide or an alkyl halide.

i Illustrate this statement by the preparation of RCH 2 NHCH 3. Both synthetic reaction sequences must start withRCO 2H. In each sequence, give the reagent(s) for each step and the structures of intermediate compounds.

7

(I) RCOOH RCOCl RCONHCH RCH NHCH PCl CH NH LiAlH 5 3 2 4

3 2 3 → → → 3½ marks

(II) RCOOH RCH OH RCH Br RCH NHCH LiAlH PBr CH NH 4 3 3 2

2 2 2 3 → → → 3½ marks

ii Explain which of your two methods is more suitable for preparing RCH 2 NHCH 3. 2

(II) is less appropriate than (I) in preparation, since RCH Br CH NH

23 2 → also gives (RCH 2)2 NCH 3. 2 marks

C The least well-answered part in IIC.

Common mistakes in conversions include :1. Preparing an amide by direct reaction between an acid with NH 3 or amine,

RCOOH CH 3 NH 2 RCONHCH 3+

RCOOH NH 3 RCONH 2+

2. Converting a secondary amide to a secondary amine employing Br 2/OH - (Hofmann Degradation applicable to primary (non-substituted) amide only)

RCONHCH 3 RCH 2 NHCH 3

Br 2 / OH -

3. Alkylating a primary amide with alkyl halide to obtain a secondary amide

RCONH 2 RCONHCH 3

CH 3Br

93 2C 7 a iv7a Give the structural formula(e) of the major organic product(s) formed in each of the following reactions.

ivCH 3 C CH 2CH2COOCH 3

OLiAlH 4

2

CCH3 CH 2CH2CH2OH

OH

H reduction of ketone – 1 mark, reduction of ester – 1 mark C Many candidates did not realize that esters could be reduced to alcohol by LiAIH 4.

94 2C 9 a iii9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions.

iiiCH 3CH CHCH 2CH 2COOH R

LiAlH 4

1

R : CH 3CH CHCH 2CH2CH2OH 1 mark

C Candidates were casual / careless in writing structural formula, particularly in putting the number of H atoms toeach C atom. Common mistakes included :CH 3CH 2CH 2CH 2CH 2CH 2OH for R (saturating C=C)



Reaction Mechanism Unit 13 Page 595 2C 9 b vi vii9b Identify K, L, M, N, P, R and S in the following reactions:

viLiAlH 4

C N(CH 3)2

O

R

1

R :C

H

H

NCH3

CH3 1 mark

C wrong structure for R. such as

N

OH

CH 3

CH 3 vii

NaBH 4

O

CO 2CH3

S

1

S :

CO 2CH3

OH 1 mark

C wrong structures for 5, such as

OH

CH2OH

OH

COOCH 3

COOCH 3

O Candidates did poorly in the cases of P. R and S. and were thus not familiar with the chemistry of exhaustivemethylation of amine, the hydride reduction of amide, and the mild reducing reagent sodium borohydride.



Reaction Mechanism Unit 13 Page 696 2C 8 a i ii8a i An acyclic compound H of molecular formula C 4H8O2 has a fruity smell. It does not produce a derivative

with 2,4-dinitrophenylhydrazine nor with propanoyl chloride.Deduce the functional group(s) of H. Draw FOUR possible structures for H.

4½

H is not aldehyde / ketone ½ mark because it does not form hydrazone derivative. ½ mark or

NO 2 NH

O2 N

NR

(½ mark)

H does not possess an –OH group / is not an alcohol ½ mark because it does not form ester with propanoyl chloride. ½ mark or

no R O C C 2H5

O

formation (½ mark)

H has a fruity smell, The functional group(s) in F is most likely an ester. ½ mark

Any FOUR of the following structures (½ marks for each structure) 2 marks

H O

O

O

O

H O

O

O

OO O

O

O

(Deduct ½ mark for each extra structure)C Poorly answered. Many candidates did not appear to know the reaction of 2,4-dinitrophenylhydrazine and

propanoyl chloride and hence the formation of the respective 2,4-dinitrophenylhydrazone and ester was omitted.Some did not know the reaction of ester with LiAIH 4 or the structure of ethanoic anhydride.

ii On reduction with an excess of LiAlH 4, H gives only one product, J. which reacts with ethanoic anhydride.Deduce, giving equations, the structure for H and also the structure for J.

3½

ROHO

R

O

O

O

O

1 mark Among the four carbon esters only ethyl ethanoate gives one reduction product ½ mark or, other esters give a mixture of alkanols on reduction (½ mark)

∴ structure of H is

O

O ½ mark

and structure of J is OH ½ mark Equation for the reduction of H

CH3 C OC 2H5

O2 C 2H5OH

LiAlH 4

½ mark



Reaction Mechanism Unit 13 Page 796 2C 9 a iii9a Each of the following conversions can be completed in not more than three steps. Use equations to show how you

would carry out each of these conversions in the laboratory. For each conversion, give the reagent(s), conditions,and structure of the intermediate(s).(For answers with more than 3 steps, deduct 1 mark for each extra step.)

iii NH 2COCl

3

O

NH 2

(1)

(1)

Br 2 / OH - (or OBr -) (½)heat (½)

NH 3

3 marksC Many candidates wrongly thought that acid amide can be prepared directly from the reaction of carboxylic acid

with ammonia. Heating was omitted in the Hofmann degradation step.

97 1A 5 a i ii5a Give a structure for each of the compounds D, E, F. G and H:

iCH3CH2CONH 2 D

(1) LiAlH 4(2) H 2O

iiCH3CH2CONH 2

Br 2 / NaOH (aq)E

97 2B 7 b iii7b Identify J, K, L, M and N in the following reactions.

iiiL

CHC OCH 3

O

CCH3

CH3CHCH 2OHC

CH3

CH3CH3OH+

98 1B 8 b ii iii8b 20.0 g of 4-nitrobenzoic acid (C 7H5 NO 4) reacted with PCl 5 to give a product which reacted

exothermically with NH 3 to give T. After treatment with Br 2(l) and NaOH (aq) , T gave a solid. Crystallization of thesolid from ethanol gave 9.3 g of U (C 6H6 N2O2).

ii Give the structures of T and U.iii Briefly give three reasons which could explain why the yield of U in the above preparation is not quantitative (i.e.

100%).

98 2B 7 a iii7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case,


iii

CH3CH2 C

O

OH CH 3CH2 C

O

HLiAlH 4, H3O+

99 2B 5 a iv5a Identify D, E, F, G and J in the following reactions.

iv CO 2 NH 2

CH3O

Br 2 / NaOHheat G





ReferenceReading




Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 519Organic Chemistry, 6th Edition, Solomons, pg. 711, 825-827, 918

Syllabus Reaction of nitrilePartial and complete hydrolysis of nitrileDehydration of amideReduction of nitrile

Notes c) Reactions of nitrile

Nitrile C N is quite similar to carbonyl group CO

with the N replacing the O.

(1) Hydrolysis of nitrile

Nitrile can be hydrolysed to carboxylic acid or carboxylate by boiling with aqueous acid or alkali, with theamide intermediate. By employing an suitable reaction condition, amide intermediate can be isolated fromfurther hydrolysis to carboxylic acid.

e.g. Heating with dilute sulphuric acid will yield carboxylic acid while heating with concentrated sulphuricacid will only yield amide.Or, amide can be obtained by hydrolysis using cold dilute acid instead of a hot one.

Partial hydrolysis of Nitrile to Amide

O H

H

R C Nδ −δ +

R C N

O

H

H + - O H

H

O H

H

R C N

O

H

HR C N

O

H

H

amide

tautomerization

isoamide

Hydrolysis of Amide to Carboxylic acidTemperatur

e of hydrolysis

Concentrationof sulphuric

acid

Degree of hydrolysis

Complete Low HighPartial High Low

R CO

L

Nu

-

R C

O

L

Nu-

R CO

Nu L-+

Partial Low High

Hydrolysis of amide to carboxylic acid follows the regular pattern of the nucleophilic addition-eliminationreaction of acid derivatives.

OH

H

R C

O

NH 2R C

O

OH

N

H

H

H

-

+

+

-

R C

O

OH

H

N

H

HR C

O

O H NH 3+

-

R C

O

OH

N

H

H

HC

O

O -R NH 4

+




(a) Dehydration of amide

With the appropriate dehydrating agent, phosphorus pentoxide (P 2O5, with molecular formula P 4O10) or boiling ethanoic anhydride (CH 3CO) 2O or thionyl chloride (SOCl 2), amide can be converted back to nitrile.

R C NH 2

O P2O5 or (CH 3CO) 2Oor SOCl 2

R C N

(2) Reduction of nitrile

Similar to the carbonyl group, nitrile group is susceptible to the reduction by hydride ion through nucleophilicaddition pathway. For carbonyl compound, the carbonyl group is reduced to alcohol. A nitrile is reduced toamine by addition of hydride.

R C Nδ

−

δ+

H-

Al

H

H

H Li+

from LiAlH 4

R C N

H

Al H

H

H

Li+-R C N

H

Al H

H

H

Li+-

nitrile

CR N

H

H

H

H

primary amine

H3O+

Glossary hydrolysis of nitrile dehydration of amide reduction of nitrile

Past PaperQuestion

96 1A 3 f i 96 2C 9 a i97 2B 6 a iii 97 2B 7 b iv98 2B 6 d ii99 1A 5 b i 99 2B 7 b i

96 1A 3 f i3f i Give the structures of compounds D and E in the following organic synthesis :

C

O

HH3O+

heatHCN

D E

2

CN

OH

D :

COOH

OH

E :

1 + 1 marks(Deduct ½ mark for each minor mistake in the structures.)

96 2C 9 a i9a Each of the following conversions can be completed in not more than three steps. Use equations to show how you


i CH 2Br CH 2CH 2 NH 2

3

CN

heat

NaCN LiAlH 4 or H 2 / cat (Pd / Pt / Ni)(1)

(1)

(1)

(?

(?

3 marksC Some candidates erroneously used HCN in the first and LiAlH 4/H

+ in the last step. Acid hydrolysis should befollowed by an alkaline work-up as the target molecule is an amine and not a salt.



Reaction Mechanism Unit 14 Page 397 2B 6 a iii6a Each of the following conversions can be completed in not more than three steps. Use equations to show how you


iii

CH 3CH 2C NC

OC

O

O

CH 3CH 2

CH 3CH 2

97 2B 7 b iv7b Identify J, K, L, M and N in the following reactions.

iv

MP2O5heatCH 3CH 2CH 2 C NH 2

O

98 2B 6 d ii6d Identify K in the following reaction.

ii CONH 2 CNK

99 1A 5 b i5b Consider the following reactions :

CHOHCN (1) LiAlH 4

(2) H 3O+D E

i Give the structures of D and E.

99 2B 7 b i

7b With no more than four steps, outline a synthetic route to accomplish each of the following transformations. Ineach step, give the reagent (s) used, the conditions required and the structure of the product.

i

CH 3CHCH 2 C OCH 2CH 3

O

OH

CH 3CHCH 2CO 2H

CO 2H






In 2-chloropropane formation, the ratedetermining step is the attachment of the H + ionto the C-1 to form a 2º carbocation. While for the 1-chloropropane, the H + ion is attached tothe C-2 to form a 1º carbocation.

As 2º carbocation is more stable than the 1ºcarbocation (refer to S N1 reaction), it involves alower activation energy of formation and formsat a faster rate.

Formation of 2-chloropropane

Cl -

2° carbocation(more sta le)

C C C

H

H

H H

H

HH+

2-chloropropane

C C C

H

H

H H

H

HH Cl

C CCH

HH

H

H

H

H+

r.d.s.

Formation of 1-chloropropane

r.d.s.C CCH

HH

H

H

H

H+

1-chloropropane

C C C

H

H

H H

H

HHCl

C C C

H

H

H H

H

HH+

1° carbocation(lesssta le)

Cl -

This observation is generalized and is called Markownikoff's rule. Markownikoff’s rule states that “in theaddition of hydrogen halide to an alkene, the hydrogen atom adds to the carbon with more hydrogenatoms.” or “if an unsymmetrical alkene combines with a hydrogen halide, the halide ion adds to the carbon atom

with the fewer hydrogen atoms.”To be more general, the electrophile joins to the carbon will more hydrogen atoms to yield a more stablecarbocation.




2. Reactivity of alkene towards electrophilic addition

Because the rate of electrophilic addition is depending on the stability of carbocation intermediate. Alkene has thegeneral order of reactivity towards electrophilic addition as follows.

<<<<

C

C

H H

H H

C

CR R

R R C

C

H H

H R

C

C

H H

H

C

CHH

H CF 3

–F is an electron withdrawing group which makes the carbocation intermediate less stable. –R is an electron donating group which make the carbocation intermediate more stable.

C CH

H H

is particularly reactive because the carbocation involved is stabilized by resonance.

+

C CHH

H

HC CHH

H

H

+

C CHH

H

H+

Glossary electrophilic addition Markownikoff’s rule

Past PaperQuestion

91 1A 1 a92 2C 9 a iv93 2C 9 b94 1A 3 b i ii95 1A 3 c i ii96 2C 9 b

98 2B 5 c i



Reaction Mechanism Unit 15 Page 491 1A 1 a1a

C CH 2

CH3

H

E Using E, give reagents and a mechanism to explain what is meant by each of the following :A reaction obeying Markownikoff’s rule.A polymerization reaction.

4

Markownikoff’s rule

The answer must involve “the addition of acid and explain and show the correct direction of addition”.

Me Me

ZH

MeH Z + Z-

Me Me

Acid; could be HI, HCl, H 2SO 4, H 3O

+, HX but not HNO 3 ½ mark Mechanism:

Correct use of arrows or H + in the first step ½ mark Intermediate Carbonium ion (draw or write) ½ mark

Correct product ½ mark

If wrong product MeX

e.g. but use of HX (max. ½ mark)The mechanism is Electrophilic AdditionPolymerization reaction

Reagents: can be radical R·or ion either cationic polymerization (e.g. above using an acid)or anionic polymerization - using a base e.g. KNH 2 ½ mark

Mechanism: Correct intermediate or arrows or indicate chain initiation , propagation and termination. 1 mark Direction of addition is not important.

Product: must clearly show a polymer ½ mark Radical e.g.

R

R R

R CH2 CH

CH 3n

Cationic e.g.

CH 2 CH

CH 3n

H+ CH 2

Me

HCH 2

Me

Hetc. Me CH CH 2 CH

Me Me

Anionic e.g.

CH2

Me

HCH2

Me

H NH 2

- etc. NH 2 CH 2 CH CH 2 CH

Me Me CH2 CH

CH 3n

C Whilst Markownikoff seemed to present few problems, the polymerisation was obviously more difficult -

problems with reagents, mechanism and product.

92 2C 9 a iv9a Give the structural formula(e) of the major organic product(s) from each of the following reactions :

ivCHC CH

H3C

H3Cexcess HBr/CCl 4

2

CH3

CH3 CH3

Br Br

2 marks



Reaction Mechanism Unit 15 Page 593 2C 9 b9b Outline the mechanism for the reaction between but-1-ene and HBr to give bromobutane. Explain why 2-

bromobutane is the major product, rather than 1-bromobutane.3

CH 3CH 2CH CH 3

Br

CH3CH2CH 2 CH 2

Br

CH 3CH2CH CH 3

Br -

CH3CH2CH 2 CH 2

Br -

or

CH 3CH2CH CH 2

H Br

2-bromobutane 1-bromobutane 2-bromobutane is the major product because CH3CH 2CH CH 3

2° carbocationis more stable than CH 3CH2CH 2 CH 2

1° carbocation.

(2 marks for mechanism, 1 mark for explanation)C Many candidates simply stated the Markownikov's rule without offering any explanation.

94 1A 3 b i ii3b i Give the major product from the reaction:

CH 3CH=CH 2 + HBr → 1

CH 3CHCH 3

Br 2-bromopropane or isopropyl bromide 1 mark ii Outline the mechanism of this reaction. Another product is also formed. Give its structure and account for the

fact that it is only the minor product.3

Mechanism:

C CH 3CH 3

H

Br -

CH 3 CH CH 2 δ -δ +H Br C CH 3CH 3

H

Br

1 mark Minor product is CH 3CH 2CH 2Br 1 mark

It is a minor product because its intermediateCH3 CH2 C H

His less stable than the secondary carbocation

C CH 3CH3

H.

The secondary carbocation C CH 3CH3

His more stable than the primary carbocation CH 3 CH2 C H

H because the two

electron releasing methyl groups disperse the charge more and so stabilise the positive ion more effectively.1 mark

Or The carbocation formed exists as an equilibrium

CH 3 CH 2 C H

H

C CH 3CH 3

H

e

more stable less stable Mechanism – 1 mark Minor product – 1 mark Explanation – 1 mark ½ bonus mark – The electron release may be considered to be due to the inductive effect and thehyperconjugative effect.

C In the mechanism of the reaction, the wrong direction of the curly arrows was a common mistake.



Reaction Mechanism Unit 15 Page 695 1A 3 c i ii3c i Give the structure of the major product formed from the following reaction :

CH3+ HBr

1

CH 3

Br

1 mark ii Outline a mechanism for the above reaction. (Movement of electron pairs should be indicated by curly arrows.) 2

CH 3

Br CH 3

H Br δ+ δ -

CH 3

CH 3

H

or

Br -

2 marksC Generally well answered. Some did not use complete arrows to indicate the breaking of covalent bonds.

96 2C 9 b9b Give the structure of the major product in the following reaction and outline the mechanism of the reaction.

(Movement of electron pairs should be indicated by curly arrows.)CH 2

HBr +

3

Major product :

Br CH 3

1 mark

Mechanism :

CH 2 CH 3Br

CH 3

H Br

Br -

(?

(?

(?

(? 2 marksC Candidates were poor in writing curly arrows. Many used H + instead of HBr as the electrophile.

98 2B 5 c i5c Give the structure of the major organic product, G, in (i) below.

Outline a mechanism for the formation of the major product in the reactions.7

i

C C

CH 3H3C

H3C H

HBr G





ReferenceReading

20.1–20.2Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 102–110Organic Chemistry, Solomons, 5th Edition pg. 342–349, 351–356, 360, 366–370Organic Chemistry, Fillans, 3rd Edition pg. 127–141

Organic Chemistry, Morrison Boyd, 6th Edition pg. 317–318, 327–342, 357–362Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 421–422, 424–425,Organic Chemistry, 6th Edition, Solomons, pg. 336-339, 346-351, 422-425

Syllabus Electrophilic addition

Notes B. Other relevant reaction

1. Addition of Br 2 to alkene

Bromine in CCl 4

Normally, Br–Br molecule breaks symmetrically by homolytic fission. But under the influence of electron rich

double bond, Br–Br bond will be polarized when the molecule is approaching the double bond. The polarizationweakens the Br–Br bond and makes it break unsymmetrically.

Owing to the presence of lone pair, halogen atom is capable to form a single bond with the positive centre of thecarbocation formed.

CCBr +

Br -

CC

Br Br

CC

Br

Br

δ +

δ −

carbocation(does not form indeed)

Br -

C C

Br +

bromoniumion

Bromonium ion intermediate is more stable than ordinary carbocation intermediate because the positive charge isdispersed over the 3 membered ring. In ordinary carbocation, the positive charge is concentrated on one carbonatom only.

Finally, the bromide ion Br- attacks the C atom from another side of the molecule and yield a 1,2-disubstituted bromoalkane.

Overall reaction

CCR

R R

R Br 2 CCR

R R

R

Br Br

+in dark




Bromine water

In the presence of water, halohydrin will be formed instead of 1,2-dihalide. This is because H 2O is a better nucleophile than Br -.

bromoniumion

CC

Br O

H

H

CCBr +

OH

H

CC

Br O

H

halohydrin

H2O attacks the bromonium ion to form halohydrin.Anyway, bromine water can also be used to test for double bond or triple bond.

N.B. Hydrin means a hydroxyl compound.

2. Addition of H 2SO 4 to alkene

Alkene doesn't reacts with dil. H 2SO 4(aq) normally. However, it reacts with conc. H 2SO 4(l) to give an addition product.

C C S

O

O

OOH HC C

H O

SO O

OH

C C

H+ S

O

O

O H-O

sulphuric acidmolecule

alkylhydrogensulphate

This reversible reaction can be used to extract alkene from a mixture with alkane. By bubbling impure alkenethrough conc. H 2SO 4(l) , only alkene will dissolve in it. Conc. H 2SO 4(l) can be separated by separating funnel. Byheating the conc. H 2SO 4(l) , alkene will evaporate and can be recovered.

a) Preparation of alkanol from alkene

Since H 2SO 4 is a strong acid, HSO 4- would be a very weak base, therefore, HSO 4

- is a very good leaving group. Byadding water into alkyl hydrogensulphate, it can be converted to alkanol through nucleophilic substitution..

nucleophilicsubstitution

alkylhydrogensulphate

C CH O

H

HC CH O

H

C CH O

SO O

OH

O

H

H

alkanol




3. Hydration of alkene

An alkene can also be converted to alkanol by direct hydration using the catalyst of aqueous sulphuric acid or phosphoric acid. Hydration of alkene proceeds through an ionic mechanism and obeys the Markownikoff’s rule.

N.B. Water is a nucleophile which does not react with electron rich double bond directly.

e.g. Hydration of propene

C C C

H

H

H H

H

HH O

H

OH H

H

+C CC

H

HH

H

H

H

OH H

H

+

C C C

H

H

H H

H

HH+

2º carbocation(more stable)

H

O HH

O H

+C C C

H

H

H H

H

HH OH H

r.d.s.

regenerated propan-2-ol

In the above mechanism, H 3O+

(aq) is consumed in the first step and regenerated in the last step. It only acts as acatalyst.

Since propan-2-ol can be prepared by direct hydration of propene while propan-1-ol cannot be, propan-2-ol ismuch cheaper than propan-1-ol. The cheap propan-2-ol is commonly used as the rubbing alcohol.

For those alkanols involving formation of less stable 1º carbocation, e.g. ethanol, a more vigorous reactioncondition is required.

CH2 CH2 + H2O CH 3 CH2 OHH3PO 4

300 ºC

Comparing with H 2SO 4(l) , H 3PO 4(l) is not an oxidizing agent with will not oxidize CH 3CH 2OH to CH 3COOH.




4. Ozonolysis of alkene

The term ozonolysis means breaking down (-lysis) by ozone (ozon-).Ozone reacts with an alkene vigorously to form a very unstable intermediate initial ozonide which rearrangesimmediately to ozonide. Since the product is potentially explosive, it is carried out at a low temperature.

Upon treatment with zinc and water, the ozonide will be reduced to 2 carbonyl compounds, aldehyde or ketone,depending on the original structure of the alkene.

Since there is a direct relationship between the structure of the carbonyl compound obtained and the structure of the original alkene, ozonolysis is useful in locating the position of the double bond in an alkene.

CH3CHCH CH 2

CH3

CH3CH

CH3

C

O

H +

3-methylbut-1-ene 2-methylpropanal

(1) O 3, CH 2Cl2, -78 ºC

(2) Zn / H 2OH C H

O

methanal

N.B. CH2Cl2 is only the solvent.

In the presence of the reductive zinc metal, the aldehyde formed will not be oxidized to carboxylic acid.

If the alkene is a cyclic one, there will be only one product molecule instead of two.

(1) O3, CH

2Cl

2, -78 ºC

(2) Zn / H 2O CH 2 CH 2 CH 2 CH 2C CH H

O O

cyclohexene hexane-1,6-dial

These characteristics of ozonolysis reaction makes it particularly useful in determining the structure of an organicmolecule.

N.B. The carbonyl compounds formed from ozonolysis can be identified by melting point determination of theoxime and 2,4-dinitrophenylhydrazone derivatives.




5. Preparation of ethane-1,2-diol

A diol can be prepared by oxidation of a double bond using cold dilute basic / acidic potassium permanganate or osmium tetroxide.

N.B. Basic potassium permanganate is more oxidative than acidic potassium permanganate and the reduction product would be brown MnO 2(s) instead of colourless Mn 2+

(aq) .

By cold dilute basic / acid potassium permanganate By osmium tetroxide

Overall reaction

a) Oxidative Cleavage of double bond

If an alkene is treated with hot basic potassium permanganate instead of a cold one, the product will not be adiol. The hot basic potassium permanganate is strong enough to break the double bond by oxidation.

propanone pentan-3-one

+ C CH 3

O

H3CCH 3CH 2 C CH 2CH 3

O

3-ethyl-2-methypent-2-ene

C C

CH3CH 2

CH3CH 2 CH 3

CH 3

(1) KMnO 4, OH -, hot(2) H +

The reaction is similar to ozonolysis of alkene. The difference is that any aldehyde formed will be oxidized tocarboxylic acid immediately.

CH2 CH2 CH2 CH2C C

O O

HO OH

cyclohexene hexane-1,6-dioic acid

(1) KMnO 4, OH -, hot(2) H +

And any terminal =CH 2 group will be oxidized completely to carbon dioxide and water.

2-methylpropanoic acid3-methylbut-1-ene

++ H2OCO 2CH3CH

CH3

C

O

OHCH3CHCH CH 2

CH3(1) KMnO 4, OH -, hot

(2) H +

The products obtained from the oxidation can also be used to deduce the structure of the alkene.




b) Comparison of ozonolysis and oxidative cleavage

Ozonolysis Oxidative cleavageReagent used 1) O 3, CH 2Cl2, -78ºC

2) Zn / H 2OKMnO 4, OH -, heat

=CR 2 (disubstituted terminal group) Ketone

C

O

R R Ketone

C

O

R R

=CHR (monosubstituted terminal group) Aldehyde

C

O

H R

Carboxylate ion / carboxylic acid

C

O

R O-

C

O

R OH

=CH 2 (non-substituted terminal group) Methanal C

O

H H CO 2

6. Oxymercuration of alkyne (Preparation of ketone from alkyne)

Reagent : water, catalyst – Hg 2+ ion (HgSO 4) and strong acid (H 2SO 4)

R C C H

H

HO

keto-enoltautomerization

Markovinkovorientation

HgSO 4(aq)H2SO4(aq)

R C C HCR C H

H

O

H

keto form

In the presence of acid and mercuric ion (Hg 2+) catalyst, water can be added to a triple bond to form an enol. Enolis then rearranged to form a ketone. The addition of the water to the alkyne follows the Markownikoff’s rule.

Glossary bromonium ion hydrin alkyl hydrogensulphate phosphoric acid ozonolysis ozonidediol osmium tetroxide oxidative cleavage oxymercuration mercuric ion

Past PaperQuestion

90 2C 8 a ii91 2C 7 a ii94 1A 3 a ii96 2C 9 a ii iv97 1A 5 a iv 97 2B 6 a i 97 2B 7 a i 97 2B 7 b ii98 1A 4 c i iii 98 2B 6 d iv 98 2B 7 a iv

90 2C 8 a ii8a It is suggested that the structure of a compound having the molecular formula C 12H11ClO 4 is either A or B.

A

Cl

COOH

COOH

CH CCH3

CH3

B

COOH

COOH

CH CCH3

CH2Cl

ii How would you show the presence of a carbon-carbon double bond in the side chain? 2

Decolourization of bromine water / permanganate solution. 2 marksC Some candidates proposed very clumsy ways to detect the presence of a carbon-carbon double bond e.g.

ozonolysis.



Reaction Mechanism Unit 16 Page 791 2C 7 a ii7a State with explanations, what you would observe in each of the following experiments, and write equations for the

reactions.ii Propene is bubbled into aqueous alkaline potassium manganate(VII). 2½

The purple colour of KMnO 4 turns green due to manganate(VI) ion MnO 42-. As the ions are further reduced , the

green turns to a brown suspension of MnO 2. 1½ mark CH3 CH CH 2

OH OH+ MnO2CH3CH CH 2

MnO4-

1 mark C Very few candidates recognized that MnO 4

- is reduced to MnO 2 in alkaline medium; hence they gave wrongobservations.

94 1A 3 a ii3a Consider the two compounds:

CH 3CH 2CH2 NH 2

D

CH 3CH 2CHCH 3

OH

C

and

ii Dehydration of C gives 3 products, E, F, and G all with the formula C 4H8. On treatment with ozone followed

hydrolysis, E gives methanal among other products, but F and G do not give methanal. Give structures for E, Fand G and an equation for the ozonolysis reaction involving E.

4

F and G :C C

H

CH3 CH3

H cis-but-2-ene andC C

H

CH3

CH3

H

trans-but-2-ene 1 mark each

E : CH 2CH 3 CH CH 2 but-1-ene

CH 2CH 3 CH CH 2O3, -78°C

Zn / H 2OCH H

O

CH 3CH 2C

O

H+

1 mark C Some candidates produced very poor drawings of the cis- and trans- isomers of but-2-ene.

96 2C 9 a ii iv9a Each of the following conversions can be completed in not more than three steps. Use equations to show how you


ii

C(CH 3)3

Br

C(CH 3)3C(CH3)3CH

O O

3

(1)

(CH3)3CO -K +

or alc. KOH, heat

(1) O 3(2) (CH 3)2S or Zn or Et 3 N or Ph 3P

(1)

(?

(?

3 marksC Some candidates omitted heating in the first step and wrongly used KMnO 4 in the last step. Some used O 3 and Zn

together as one step instead of as two steps.

iv OH OH

OH

3

MnO 4- / OH -

cold

conc. H 2SO 4 or H 3PO 4or P 2O5 or Al 2O3

heat (?

(?(?

(?(1) 3 marks

C Heating was omitted in the dehydration step. NaOH or NaOEt was wrongly used as the dehydrating agent. Mostcandidates did not give the correct conditions (cold and weakly alkaline) for the dihydroxylation of the alkene.



Reaction Mechanism Unit 16 Page 897 1A 5 a iv5a Give a structure for each of the compounds D, E, F. G and H: 5

iv (1) O 3(2) Zn / CH 3CO 2H butanal onlyC8H16

G

97 2B 6 a i6a Each of the following conversions can be completed in not more than three steps. Use equations to show how you


9

i (CH3)3CCH CH 2 (CH3)3CC CH

97 2B 7 a i7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test

should include the reagent(s), the expected observation with each compound and the chemical equation(s).12

iand

97 2B 7 b ii7b Identify J, K, L, M and N in the following reactions. 5

iiCH 2

OH

CH 2Br

K

98 1A 4 c i iii4c On treatment with dilute H 2SO 4(aq) , E gives mainly two isomeric compounds, F and G, both of which have the

formula C 4H8. On treatment with bromine, both F and G give a product H with formula C 4H8Br 2.i Draw structures for F, G and H. 3iii Outline the mechanism for the formation of H from either F or G. 1

98 2B 6 d iv6d Identify M in the following reaction. 4

iv CH3 (1) O 3(2) Zn, H 2O M

98 2B 7 a iv7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case,


12

iv

CH3CH2CH CH 3

Br

H3CC CCH 3 NaOH





ReferenceReading

21.0Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 133–135, 183, 205, 208, 274, 276Organic Chemistry, Solomons, 5th Edition pg. 636–644, 652, 842–843, 939–940, 946Organic Chemistry, Fillans, 3rd Edition pg. 157–160, 205

Organic Chemistry, Morrison Boyd, 6th Edition pg. 517–518, 525–527, 529, 832, 898Organic Chemistry, Stanley H. Pine, 5th Edition pg. 639, 641–647Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 440–441, 445Organic Chemistry, 6th Edition, Solomons, pg. 617-618, 655-658

Syllabus Electrophilic substitutionAromaticityArenium ion

Notes VIII. Electrophilic substitution

Benzene is another molecule containing multiple bond besides alkene and alkyne. With the presence of π electrons, it is also an ideal substrate for the electrophilic attack by an electrophile.

H

H H

H H

H

E+

However, benzene reacts with bromine less readily than alkene or alkyne does. And, even if there is reaction,substitution product is the only product obtained.

X

substitution product

HBr

Br

H

H

H

H

H

+

H

H

H

HH

H

Br Br

addition product

H

H

H

H

H

H

Br Br

The difference is caused by the aromaticity of the benzene ring.

N.B. Aromaticity – the extra stability associated with the benzene ring through resonance. Or to be more precise, any ring system with (4n + 2) number of π electron will have this kind of extrastability, where n is a natural number. For benzene, n = 1.




When a Br–Br molecule is added to a benzene ring, the hybridization of 2 carbon atoms are change from sp 2 hybridization to sp 3 hybridization. The aromaticity of the benzene ring is lost. This involves a very high activationenergy and is unfavorable. On another hand, the bromobenzene formed from substitution is still aromatic wherethe activation energy involved is relatively lower.

substitution

addition

H

H H

H H

H

Br Br

+

H

H H

H

Br

Br

HH

breaking downof aromaticity

H H

H H

HBr

the aromaticityis retained

+ Br H

sp2

sp3

Although the substitution product is aromatic, inevitably, the aromaticity of a benzene ring is lost temporarily whenan electrophile is added to the ring. This makes substitution reaction of benzene occur only in the presence of acatalyst or under a very rigorous condition.

Aromaticity is retoredArenium ion(temproray lostin aromaticity)

H

H H

H H

H

E

+H

H H

H H

HE+

H

H

H H

H

E

H++

H

H H

H H

H H

H H

H

HH

E

+ H

H

H H

H

E

H++

Arenium ion

r.d.s.

The positive charge possessed by the electrophile is dispersed over the ring of arenium ion. And by losing ahydrogen ion, the aromaticity is restored.

The reaction conditions and mode of reaction can be summarized as follows :

Reaction condition Mode of reaction Normal circumstance no reactionWith Lewis acid catalyst electrophilic substitutionIn the presence of uv or radical initiator radical addition




A. Representation of arenium ion

The arenium ion intermediate can be represented by the drawings of several resonance structures or a singleresonance hybrid.

resonance structuresresonance

hybrid

H E

+

H E

+

H E

+

H E

+ OR

Glossary electrophilic substitution electrophile aromaticity arenium ion

Past PaperQuestion





ReferenceReading

21.1.1–21.1.4Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 133–135, 183, 205, 208, 274, 276Organic Chemistry, Solomons, 5th Edition pg. 636–644, 652, 842–843, 939–940, 946Organic Chemistry, Fillans, 3rd Edition pg. 157–160, 205

Organic Chemistry, Morrison Boyd, 6th Edition pg. 517–518, 525–527, 529, 832, 898Organic Chemistry, Stanley H. Pine, 5th Edition pg. 639, 641–647Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 441–444, 447–448Organic Chemistry, 6th Edition, Solomons, pg. 658-664, 930-931, 974

Syllabus Electrophilic substitution

Notes B. Other relevant reaction

1. Sulphonation of benzene

Benzene reacts with fuming sulphuric acid (H 2S2O7 = SO 3 + H 2SO 4) readily at room temperature or withconcentrated sulphuric acid (H 2SO 4) slowly. If necessary, moderate heating may be employed ( ≈ 80ºC) toaccelerate the reaction.

Indeed, the electrophile is not SO 42- ion since it is negatively charged. It is the sulphur trioxide molecules, SO 3. It

is either present in the fuming sulphuric acid or from the self dissociation of concentrated sulphuric acid molecules.

H2SO 4 + H 2SO 4 d H3O+ + HSO 4

- + SO 3

Overall reaction

benzenesulphonicacid

H2O+SO 3HH2SO 4+

Since the reaction is reversible, benzenesulphonic acid can be desulphonated by heating to a temperature over 100ºC, or by diluting the sulphuric acid, or by passing steam through the sulphonic acid.

Sodium salt of sulphonic acid can be used to prepare phenol by fusion with solid sodium hydroxide.




a) Preparation of phenol

SO 3- Na +

3 NaOHO - Na +

Na 2SO 3 H2O+ + +350°C

sodium benzenesulphonate

phenol

+ Na +Cl-OH

+ HClO - Na +

2. Nitration of benzene

Concentrated nitric acid does not react with benzene at all. Ordinary concentrated nitric acid contains only H + ionand NO 3

- ion. NO 3- ion is a nucleophile which does not react with benzene.

In the presence of concentrated sulphuric acid, concentrated nitric acid reacts with it to produce nitronium ion /nitryl cation / (NO 2

+) which is an electrophile.

Since sulphuric acid is a stronger acid than nitric acid, it protonates the hydroxyl group of the nitric acid andconverts it into a better leaving group. Without sulphuric acid, the concentration of NO 2

+ ion in conc. nitric acid israther low.

2HNO 3 e NO 2+ + NO 3

- + H 2O

Formation of nitronium ion (nitryl cation)

OH NO

O

+- S

O

O

O

OH H

OH NO

OH

+

-

+

HSO 4-+

a better leaving group

H2O NO O++

-

+OH N

O

OH

+nitronium

ion

Electrophilic substitution by nitronium ion

+

H

H H

H H

H H

H H

H

HH

NO O -

+

Arenium ion

r.d.s.

NO O+

Arenium ion

H

H

H H

H

NO -

O

+H

H H

H

HH

NO O -

+

+ +

H2SO 4

HSO 4-




Since nitro group –NO 2 is an electron withdrawing group, it decreases the electron density of the benzene ring.Therefore, further nitration of nitrobenzene is more difficult and requires prolonged heating.

+

- -+

NOO-

+ N

OO N

OO +--

+

Formation of trinitrobenzene takes days of prolonged heating.




3. Halogenation of benzene

At room condition, halogen does not react with benzene at all.

X2no reaction

The reaction is found to be catalyzed by Lewis acid, e.g. FeCl 3, FeBr 3 and AlCl 3. The Lewis acid accepts anelectron pair from the halogen molecule. This polarizes the halogen molecule and promotes the formation of

positive halogen ion.

X+ FeX4-

Lewis acid(an electronacceptor)

electrophile

+FeX3X X

Halogen X 2 and iron can also be used instead because the halogen reacts with the iron readily to produce the Lewisacid e.g. FeX 3 catalyst.

2 Fe + 3 X 2 → 2 FeX 3

Example of bromination of benzene is illustrated as follows :




4. Alkylation of benzene (Friedel-Crafts alkylation)

Similar to halogenation of benzene, benzene can also be alkylated in the presence of anhydrous AlCl 3 andhaloalkane. This method is named after the inventors, Charles Friedel and James M. Crafts.

AlCl 3 is a Lewis acid which catalyzes the reaction by promoting the formation of carbocation (an electrophile).

R Cl Al

Cl

Cl

Cl Al

Cl

Cl

Cl

Cl-

+carbocation(an electrophile)

R +

Then, the carbocation attacks the benzene ring through an electrophilic substitution pathway. Eventually, the benzene ring is alkylated (replacing a hydrogen by an alkyl group) and AlCl 3 is regenerated.

H

H H

H H

H H

H H

H

HH

R

+

Arenium ion

r.d.s.R +

Alkylbenzene

Al

Cl

Cl

Cl

Cl- HCl + AlCl 3

Arenium ion

H

H

H H

H

R

+H

H H

H

HH

R

+

Summary of different electrophilic substitution reaction of benzeneReaction Reagent Catalyst Condition Electrophile

Sulphonation conc H 2SO 4(l) / H 2S2O7(l) Nil 80 ºC SO 3

Nitration conc. HNO3(aq) conc. H 2SO 4(l) 55 ºC NO 2+

Halogenation X 2 FeCl 3 / AlCl 3 X+

Alkylation R–Cl FeCl 3 / AlCl 3 R +

Glossary sulphonation fuming sulphuric acid benzenesulphonic acid fusion nitrationnitronium ion / nitryl cation benzenamine / aniline halogenation Lewis acidFriedel-Crafts alkylation




Past PaperQuestion

92 2B 6 Ab iii 92 2B 6 Bb iii96 2C 8 c iii97 2B 7 b v

92 2B 6 Ab iii Bb iii6Ab What are the products of the reactions between

iii concentrated nitric(V) acid and concentrated sulphuric(VI) acid? 12HNO 3 + H 2SO 4 d NO 2

+ + H 3O+ + NO 3- + HSO 4

- 1 mark 6Bb What are the products of the reactions between

iii concentrated sulphuric(VI) acid and concentrated nitric(V) acid? 12HNO 3 + H 2SO 4 d NO 2

+ + H 3O+ + NO 3- + HSO 4

- 1 mark C The products of this reaction were not widely known.

96 2C 8 c iii8c Identify K, L, M, N and P in the following reactions :

(Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks)iii

NO 2

CH 3

NO 2

NO 2

CH 3

M

1

M : conc. HNO 3, conc.H 2SO 4 (heat) 1 or 0 mark C HNO 3 / H 2SO 4 or HNO 2 or HNO 3 for M. Both nitric(V) and sulphuric(VI) acids used should be concentrated and

the abbreviation conc. should be used instead of c;

97 2B 7 b v7b Identify J, K, L, M and N in the following reactions. 5

v N

(1) conc. H 2SO 4(2) NaOH (aq)





ReferenceReading

21.1.5–21.1.6Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 133–135, 183, 205, 208, 274, 276Organic Chemistry, Solomons, 5th Edition pg. 636–644, 652, 842–843, 939–940, 946Organic Chemistry, Fillans, 3rd Edition pg. 157–160, 205

Organic Chemistry, Morrison Boyd, 6th Edition pg. 517–518, 525–527, 529, 832, 898Organic Chemistry, Stanley H. Pine, 5th Edition pg. 639, 641–647Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 476–477, 521, 523–524Organic Chemistry, 6th Edition, Solomons, pg. 930-931, 974

Syllabus Electrophilic substitution

Notes 5) Diazocoupling of diazonium ion

diazonium ion

NaNO 2HCl, 0-5

NH 2

benzenamine(an aromaticprimary amine)

N N+

When benzenamine is treated with HNO 2(aq) (or NaNO 2(s) / HCl (aq) ) at 0–5 ºC, diazonium ion is formed.

Beside the substrate of nucleophilic substitution, diazonium ion is also an electrophile which is capable to acceptelectron form a benzene ring.

Nucleophilic substitution of diazonium ion (S N)(The high leaving ability of

N2 makes diazonium ion agood substrate of nucleophilic substitution)

diazonium ion

Nu

N 2+

N N+

Nu -

Electrophilic substitution of diazonium ion (S E)(The positive charge and themultiple bond on the azidegroup makes diazonium ion agood electrophile)

+ N N

OHH

naphthalen-2-ol

azo dye(an orange red ppt.)

H2O+ N

N

OH

OH -

N

N

OH

H

+diazonium ion

Phenol compound (e.g. naphthalen-2-ol) is particularly reactive to electrophilic substitution because the electrondensity of the benzene ring is enhanced by the hydroxyl group.

O

H

+

-

+O

H

-

O

H

If napthalen-2-ol is used, the substitution product formed is an orange red ppt. The colour of the ppt. is so intensethat it can be used as a dye. Since the compound contains nitrogen, it is called azo dye.




a) Colour of a substance

Normally, most simple organic molecules are colourless because they absorbs ultra-violet and infra-red only. Asubstance is coloured only if it absorbs light from the visible region.

The colour of azo dye is due to the presence of the extensive π system. This makes the difference between theenergy levels smaller. Therefore, the molecule will absorb visible light during electron transition.

For the same reason, some transition metal ions are coloured in the presence of water. (This will be discussed in thesection of transition metal : d-d transition).

N N

OH

Lewis structure of azo dye

N N

OH

Orbital representation of azo dye

6. Bromination of phenol

Through resonance, the lone pair on the oxygen is donated to the benzene ring. This makes the ortho (2,6) and para (4) positions of the benzene more electron rich than the meta (3,5) position does. Therefore, ortho and para positions of the benzene ring is more vulnerable to the attack of an electrophile. The hydroxyl group is said to beortho-para directing.

-

+ O

H

O

H

-

O

H

+

-

O

H

Phenol reacts with aqueous bromine solution readily to form white precipitate of 2,4,6-tribromophenol. Theformation of the ppt. can be served as a test for phenol. Since the yield of the reaction is almost 100%, the reactioncan also be used in quantitative analysis (e.g. titration).

O

H

Br Br

Br

O

H

2,4,6-tribromophenol(a white precipitate)

+ 3 HBr 3 Br 2+

Because phenol is found to be more reactive than ordinary benzene, hydroxyl group is also known as an activatinggroup in electrophilic substitution of benzene.

Glossary diazocoupling napthalen-2-ol dye azo dye ortho position para positionmeta position ortho-para directing activating group




Past PaperQuestion

90 1B 4 b ii91 1B 5 b 91 2C 7 a iii94 2C 9 a iv95 1A 3 d ii96 1B 4 b 96 2C 8 c ii97 1B 8 a i ii98 2B 7 a ii

90 1B 4 b ii4 A mixture contains equal amounts of X, Y and Z.

OH

X b.p. 182°C

NH 2

Y b.p. 184°C

CH3

Br

Z b.p. 184°C

4b Outline tests involving observable colour changes to show that:ii Y is a primary aromatic amine, and 1

Add NaNO 2 and dil. HCl to Y at 5ºC followed by 2-naphthol ( β-naphthol). An orange ppt. denotes presence of a primary aromatic amine. 1 mark

91 1B 5 b5a CH3

NH 2

X

CH3

NO 2

Y The amine X, C 7 N9 N, may be prepared from Y, C 7H7 NO 2, by reaction with excess hot granulated tin andconcentrated hydrochloric acid.

5b Describe simple chemical tests which would enable a distinction to be made between X and phenylmethylamine(C6H5CH 2 NH 2). Your answer should include chemical equations, reaction conditions and a description of anyvisible change.

4

For X

CH3

NH 20°C

HONO/HCl OH

CH 3

N N Cl -+

no observable change

OH

N N Ar

orange or red or yellow ppt. Basic reaction with HONO/0ºC for both X and phenylmethylamine 1 mark

No observable change in the first step ½ mark Formation of diazonium salt ½ mark Use of 2-naphthol ½ mark Formation of orange or red or yellow ppt. ½ mark For phenylmethylamine

0°C

HONO/HClPh NH 2 Ph OH + N 2

Formation of alkanol ½ mark

Nitrogen gas bubbles observed. ½ mark No coloured ppt. with β-naphthol (2-naphthol).

C Many cases of use of aryl sulphonic acids suggest some very bad teaching – as the use of sulphonic acid todistinguish amines is not in the syllabus.

91 2C 7 a iii7a State with explanations, what you would observe in each of the following experiments, and write equations for the

reactions.iii Aqueous phenol is treated with aqueous bromine. 2

The reddish brown solution of bromine is decolourized. Tribomophenol appears as a white precipitate. 1 mark

OHOH

Br

Br

Br

+ Br 2

1 mark C Most candidates did not give full observations. They only mentioned decolorisation of bromine or formation of

white precipitate, but not both.





Reaction Mechanism Unit 19 Page 598 2B 7 a ii7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case,


12

ii

NH 2 N N OHOH

heat





ReferenceReading

22–23Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 305–306Organic Chemistry, Solomons, 5th Edition pg. 262–263, 267–274, 361–366Organic Chemistry, Fillans, 3rd Edition pg. 127, 141–142

Organic Chemistry, Morrison Boyd, 6th Edition pg. 46–49, 351–354, 356–357, 388–389Organic Chemistry, Stanley H. Pine, 5th Edition pg. 910, 912–914Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 407–409, 426–431, 445–446Organic Chemistry, 6th Edition, Solomons, pg. 334-335, 366-379, 393-397

Syllabus Free radical Substitution (chlorination of methane)Free radical Addition (polymerization of alkene)

Notes IX. Free radical Reaction

Free radical is a species containing 1 or more unpaired electrons. Halogen atom and alkali metal atom are twoexamples of them. Oxygen molecule may also be considered as a free radical. This is because it contains twounpaired electron and is paramagnetic.

Examples of free radicals

As free radical does not have a fulfilled octet, it is an electron deficient species. If possible, it tends to react withan electron rich species e.g. an double bond. Most of the free radicals are very reactive and cannot be isolated.




A. Formation of free radical

Free radical is formed in 3 ways :

1) Homolytic fission of a bond

BA A B+

+Cl ClCl Cl

chlorinefree radical

hυ

or heating

R O O R +OR O R heat

The energy required to break the bond is furnished by exposing the molecule to light (photolysis, h υ ) or byheating of the molecule. A radical forms more readily if atom A and B have similar electronegativity and theA–B bond is weak. Peroxide molecule containing weak O–O bond is usually used as a radical source.

2) By reaction of a molecule with another free radical

a newfree radical

BA

XA + XB

R O HR O H Br + Br

When a free radical gets in contact with another molecule, it may gain an electron from the molecule in order to attain octet. The molecule which loses an electron to the free radical becomes another free radical.

Since the product is still a free radical, the reaction may continue.

3) By oxidation-reduction process

CH 3 C

H

H

+ Br - Na ++CH 3 CH 2 Br Na

Radical can be formed from a molecule by a single electron transfer.




B. Free radical Substitution

1. Chain reaction e.g. chlorination of methane

Most free radicals reacts through a chain mechanism.

A chain mechanism consists of three steps :

1. chain initiation,2. chain propagation,3. chain termination.

Besides free radical reaction, nuclear fission of 235U in nuclear bomb is also a chain reaction.

Chlorination of methane is a typical example of a chain reaction.

a) Chain initiation (chain initiating step)

+Cl ClCl Cl Cl· radical is generated.hυ

In the presence of light or heat, Cl–Cl bond breaks symmetrically and yields 2 Cl· radical.

b) Chain propagation (chain propagating step)

Cl CH

H

H

H

HCl C

H

H

H+ Cl· radical is consumed.

C

H

HH

ClClC

H

HHCl Cl + Cl· radical is regenerated.

Depending on the amount of the reactants present, mono-substituted chloroalkane may undergo further substitution.

Cl CH H

H

Cl

HCl C

H

H

Cl

+ Cl· radical is consumed.

C H

H

Cl

Cl

ClC

H

H

Cl

Cl Cl + Cl· radical is regenerated.




c) Chain termination (chain terminating step)

radical is destroyed.

C C

H

H

H

H

H

H

C

H

H

HCH

H

H

Cl C

H

H

H C H

H

Cl

H

+Cl Cl Cl Cl

+ Energy

When two free radicals meet each other, they may join together to form a molecule. However, free radical isvery reactive, the molecule formed will also be very energetic. It breaks apart instantaneously and forms twofree radicals again.

The molecule formed will become stable only if the energy of the molecule is dissipated immediately . Thishappens when the molecule collides with the wall of the container once it is formed and transfers the energy tothe wall.

Since the occurrence of the chain termination is rather rare, it is estimated that a single Cl· radical may lead tochlorination of thousands of methane molecule.

Apparently, there is more than 1 way of chain propagation and chain termination, the presence of a mixture of products is always a characteristic of radical reaction .

2. Reaction between H 2(g) and Cl 2(g)

Reaction between H 2(g) and Cl 2(g) is also a free radical substitution reaction where the rate of reaction is verysensitive to the presence of light.

Under controlled condition, H 2(g) burns in Cl 2(g) smoothly to produce HCl (g), from which HCl (aq) is prepared.

However, if a mixture of H 2(g) and Cl 2(g) is exposed to strong light, the result may be explosive.

Chain initiation

Cl–Clh υ

→ Cl· + Cl·

Chain propagation

Cl· + H–H → H–Cl + H·H· + Cl–Cl → H–Cl + Cl·

Chain termination

H· + H· → H–H + energyCl· + Cl· → Cl–Cl + energy




C. Free radical Addition

Besides the reaction with saturated compound, free radical reacts with unsaturated compound more readily becauseradical is an electron deficient species.

1. Chain reaction e.g. polymerization of alkene

In polymerization of alkene, an organic peroxide is usually added into the alkene as a radical initiator.Diacylperoxide is commonly used.

Chain initiating step

2 R C O

O

O C

O

R R R C

O

O2

Diacylperoxide

+ 2 CO 2

alkyl radical

heat

When diacylperoxide is heated, it dissociates into alkyl radical and carbon dioxide consequently.

Chain propagating step

C CR

H

H

H

HR C C

H

H

H

H


C CR

H

H

H

H

C CH

H

H

HR C C C C

H

H

H

H

H

H

H

H

The electron deficient alkyl radical reacts with double bond of the alkene repeatedly and results in a very longchain polymer.

Chain terminating step

R (CH 2 CH 2)n C C

H

H

H

H

R (CH 2 CH 2)n C C

H

H

H

H

C C (CH 2 CH 2)n R

H

H

H

H

C C (CH 2 CH 2)n R

H

H

H

H

combination

C C (CH 2 CH 2)n R

H

H

H

H

R (CH 2 CH 2)n C C

H

H

H

H

R (CH 2 CH 2)n C C

H H

H


H

H

H

H

disproportionation

oxidation product reduction product

There are 2 possible terminating steps : combination and disproportionation

In combination, two radicals combine together to form a molecule. In disproportionation, a hydrogen atom istransferred from a radical to another. Thus, the one lost the hydrogen is oxidized and the other radical is reduced.




2. Anti-Makownikoff orientation

ClHC CCH

H H

H

H

H propene

+

C C C

H

H

H H

H

HH Cl

2-chloropropane

C C C

H

H

H H

H

HHCl

1-chloropropane

1 2

In the electrophilic addition of HCl to propene, 2-chloropropane is found to be the major product because itsformation involves a lower activation energy. The mechanism is also known as ionic addition of alkene where ahydrogen ion is added to the propene molecule first.

Cl-

2?carbocation(more stable)

C C C

H

H

H H

H

HH

+

2-chloropropane

C C C

H

H

H H

H

HH Cl

C CCHH

H

H

H

H

H+

r.d.s.

However, in the presence of an peroxide (a radical initiator), 1-choropropane will become the major product. It isobvious that the two reactions have 2 different reaction pathways.

When an peroxide is added, the reaction proceeds through an radical mechanism. Similar to a carbocation, aradical is also an electron deficient species. The influence of the substituent on the stability of a radical is similar to that of a carbocation.

methylradical

3radical

RelativeStability of Free radical

>>> C

H

H

H

C

R

H

HC

R

R

H

C

R

R

R

C

H

H

>

benzylradical




Chain initiation

R O O R +OR O R heat

Cl+R O HR O H Cl

In the chain initiation step, the free radical formed from homolysis of the organic peroxide subtracts an hydrogenatom from the H–Cl molecule. An Cl· radical is formed first.

Chain propagation

Cl

r.d.s.C CCH

HH

H

H

H2º radical

(more stable)

C C C

H

H

H H

H

HCl H

C C C

H

H

H H

H

HCl

H Cl

+ Cl

1-chloropropane

The Cl· radical is added to the propene molecule and a 2º radical is formed. This accounts for the fact that 1-chloropropane is the major product of the reaction.

The main difference between the ionic mechanism and the radical mechanism is that in ionic mechanism aH + ion is added to propene first while in radical mechanism a Cl· radical is added to propene first.

The effect of the peroxide on the orientation of the addition is also known as “peroxide effect”.

Glossary free radical species paramagnetic homolytic fission photolysis free radical

free radical substitution chlorination of methane chain reaction chain initiationchain propagation chain termination free radical addition polymerization of alkenediacylperoxide radical initiator combination / coupling disproportionationanti-Makownikoff’s rule peroxide effect

Past PaperQuestion

91 2C 7 a i94 2C 9 b ii97 1A 4 c ii98 2B 6 d iii99 1A 5 a i ii iii

91 2C 7 a i7a State with explanations, what you would observe in each of the following experiments, and write equations for the

reactions.i A mixture of pentane and bromine in tetrachloromethane is exposed to sunlight. 3½

The reddish brown bromine turns colourless. The reaction is a free-radical chain reaction. Depending on theamount of bromine , a variety of substitution products is obtained. 2½ mark CH 3CH 2CH 2CH 2CH 3 + Br 2 → CH 3CH 2CH 2CH 2CH 2Br + CH 3CH 2CH 2CH 2CHBr 2 etc. 1 mark



Reaction Mechanism Unit 20 Page 894 2C 9 b ii9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC)

respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem.

ii Use equations to show a mechanism of the polymerization of ethene. 3The polymerisation of CH 2=CH 2 is a free-radical addition reaction. An organic peroxide is added as an initiator.Initiation

O OR R heat

O2 R

OR

2 R C O

O

O C

O

R R R C

O

O2

Diacylperoxide

+ 2 CO 2

alkyl radical

heat

1 mark Propagation

C CR

H

H

H

HR C C

H

H

H

H


C CR

H

H

H

H

C CH

H

H

HR C C C C

H

H

H

H

H

H

H

H1 mark

Termination

R (CH 2 CH 2)n C C

H

H

H

H

R (CH 2 CH 2)n C C

H

H

H

H

C C (CH 2 CH 2)n R

H

H

H

H

C C (CH 2 CH 2)n R

H

H

H

H

combination

OR

C C (CH 2 CH 2)n R

H

H

H

H

R (CH 2 CH 2)n C C

H

H

H

H

R (CH 2 CH 2)n C C

H H

H


H

H

H

H

disproportionation

oxidation product reduction product 1 mark

C Mistake in using ' ' instead of ' ' to describe one-electron shift in free radical reaction.

97 1A 4 c ii4c ii Outline a free radical mechanism for the conversion of ethene to poly(ethene). Your answer should include

appropriate arrows to show how the new bonds are made.

98 2B 6 d iii6d Identify L in the following reaction.

iiiH2C CH

Cl

CH 2 CH

Cl n

L

99 1A 5 a i ii iii5a Under certain conditions, methane reacts with chlorine to give chloromethane as the major product.

i State the conditions for the reaction.ii Outline the mechanism and name the mechanistic steps of the reaction.iii Is the reaction of methane with chlorine an appropriate method for the preparation of dichloromethane ? Explain.



Amino acids

I. Amino acids

A. ZwitterionB. Polypeptides



Amino acids Page 1

Topic Amino acids

ReferenceReading

24Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 278–279Organic Chemistry, Solomons, 5th Edition pg. 1096Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 524–525

Organic Chemistry, 6th Edition, Solomons, pg. 1146-1148

Syllabus Amino acidZwitterionPolypeptide

Notes I. Amino acids

Amino acid is a bifunctional compound containing amino group (–NH 2) and carboxyl group (–COOH).

aminoethanoicacid

CH 2 C

O

O HH2 Nα

2-aminopropanoic

acid

CH 3 CH C

O

O H

NH 2

αβ

Both aminoethanoic acid and 2-aminopropanoic acid are also called α -amino acid because the amino group isattached to the α carbon. All the naturally occurring amino acids are α -amino acids, the only difference betweenthem are the –R attached to the α carbon.

α -amino acid

C

O

O HCH2 N

R

H

α

R = H Glycine

CH 3 Alanine

CH 2 OH Tyrosine

CHCH 3

CH 3Valine

Amino acid

A. Zwitterion

Acid pK a Conjugate basestronger acid CH 3COOH 4.76 CH 3COO - weaker conjugate baseweaker acid CH 3 NH 3

+ 10.6 CH 3 NH 2 stronger conjugate base

Since CH 3COOH is a stronger acid than CH 3 NH 3+, thus CH 3 NH 2 is a stronger base than CH 3COO -, the carboxyl

group is capable to react with the amino group by intramolecular proton migration.

- H +

+ H +- H +

+ H +C

O

O HC

R

H

NH

H

H

+ -C

O

OC

R

H

NH

H

H

+ -C

O

OC

R

H

NH

H

Cationic form(Predominant at low pH)

Anionic form(Predominant at high pH)

Zwitterion(Dipolar ion)

At low pH, the concentration of H +(aq) is high, the cationic form of amino acid will be predominant.

At high pH, the concentration of H +(aq) is low, the anionic form of amino acid will be predominant.

At medium pH, beside cationic form and anionic form, amino acid may also exist in form of dipolar ion (alsoknown as zwitterion.

Because of the of the formation of zwitterion, amino acids has1. high melting point2. high solubility in water but low solubility in organic solvent3. large dipole moment..



Amino acids Page 2

B. Polypeptides

Amine is capable to condense with carboxylic acid through nucleophilic pathway to give an amide.(Refer to Mechanism of nucleophilic addition)

amine carboxylic acidamide group

(or peptide linkage)

+ N

H

HR R' C

O

OH R' C

O

N

H

R H2O+

Amino acid is a bifunctional compound which is capable to undergo successive condensation (polymerization)with elimination of water.

CONH CONH CO 2H

R R R

H2 NC

R

COOHH2 N

H successive condensation

amino acid polypeptide(a tripeptide consists of 3 monomers)

The polymer formed is also known as polypeptide (protein) since the amino acids (monomers) are linked together by peptide linkage (amide group).

Moreover, polypeptide can also be hydrolyzed back to amino acid in the presence of enzyme or acid / alkalicatalyst. e.g. reflux with 6M HCl (aq) for 24 hours.

Glossary amino acid bifunctional a-amino acid zwitterion (dipolar ion) predominant polypeptide polymerization amide group (peptide linkage)

Past PaperQuestion

91 1A 1 c93 1A 3 e i iii94 1A 3 c ii98 1C 10 98 2B 6 a ii iii99 1A 6 a 99 2B 5 a v

91 1A 1 c1c Give the structure of a dipeptide.

Give a reagent that could cleave your dipeptide into smaller units.Draw the zwitter-ionic form of one such smaller unit

4

NH 2CHCNHCHCO 2H

R R'O

or NH 3CHCNHCHCO 2

R R'O

2 marksR, R’ can be H, Me, Et, Ph, etc. (must be specified, if no -½)

Note: must be from α -amino acids.(If an amide from NON- α -amino acid, with –CO 2H and –NH 2 1 mark)(If just an amide, e.g. RNHCOR -1½ mark)dil. H 2SO 4 or enzyme by acid hydrolysis (any acid) 1 mark (Alkali causes racemisation, therefore ½ mark only)

Zwitter ion: NH 3CHCO 2

R

1 mark C This was considered by many to be very short and easy for 4 marks. However it was not very well answered

especially since many misused R in such dipeptide structures as NH 2RCO 2H.



Amino acids Page 393 1A 3 e i iii3e A dipeptide, W, yields only valine and glycine on hydrolysis.

CH CHCH 3

CH 3 NH 2

CO 2H

valine glycine

H2 NCH 2CO 2H

i How many different structural isomers are possible for W? Draw one of these structures. 2

2 Structural isomers are possible for W. ½ mark

C N

O H

CH3CHCH CH 2CO 2H

NH 2

CH3or

C

O

CH 3CHCH

CH 3

OH

NHCO CH 2 NH 2

1½ mark iii Draw the structure of the predominant form in which glycine exists in aqueous solution: with acidic conditions

and under basic conditions.2

under acidic conditions

CH CHCH 3

CH 3 NH 3

CO 2H

+ 1 mark

under basic conditionsCH CHCH 3

CH 3 NH 2

CO 2-

1 mark

94 1A 3 c ii3c Hydrolysis of a protein gives rise to a number of amino acids.

ii Draw the structure of an amino acid in its zwitter-ionic form. 1

CHCH3 CO 2-

N+H3

α 1 mark No mark for non-ionic form

97 2B 5 b i ii5b The following equation represents the acid hydrolysis of a dipeptide D to produce compounds E and F. one of

which is a chiral compound.

NHCH 2CO 2HC

O

CHH2 N

CH 3

E + FH3O+

D i Name all functional groups in D .ii Give one structure for E and one for F. Draw a suitable representation for the chiral product.

98 2B 6 a ii iii6a Consider the structures of the two synthetic polymers shown below.

HN(CH 2)6 NHCO(CH 2)4COn

nylon-6.6

CH 2CH 2n

poly(ethene)

ii Briefly explain why aqueous acids can more readily attack nylon-6,6 than poly(ethene) inducing degradation.iii Apart from acidic conditions, state one other condition under which nylon-6,6 degrades more readily than

poly(ethene).

99 1A 6 a6a Consider the amino acids, F and G.

H3 NCH 2CO 2+ -

C CO 2-H3 N

H+

F G

Draw three-dimensional structures of all dipeptides formed from F and G.

99 2B 5 a v



Amino acids Page 45a Identify D, E, F, G and J in the following reactions.

v

NH

O

H3O+

heat J

(Hint: J is a polymer.)



Oxidation and Reduction

I. Oxidation




II. Reduction





Oxidation and Reduction Page 1

Topic Oxidation and Reduction

ReferenceReading

25Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 95, 144–145, 178, 243–244Organic Chemistry, Solomons, 5th Edition pg. 669, 454–456, 458, 714Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 475–476, 518

Organic Chemistry, 6th Edition, Solomons, pg. 165-166, 473-475, 690-691, 915-916

Objectives Oxidation and reductionHydrogenation of double bond

Notes Oxidation and Reduction

I. Oxidation


CxHy + ( x +y4 ) O 2 → x CO 2 +

y2 H2O

CxHyOz + ( x +y4 -

z2 ) O 2 → x CO 2 +

y2 H2O


Reagent : any strong oxidizing agent – e.g. KMnO 4 / H +(aq) or K 2Cr 2O7 / H +

(aq)

Both 1º and 2º alkanol can be oxidized by strong oxidizing agent.

1º alkanol can be oxidized to aldehyde and then carboxylic acid easily.2º alkanol will only be oxidized to ketone.

Since 1º and 2º alkanol turns K 2Cr 2O7/H+(aq) green, K 2Cr 2O7 / H +

(aq) cannot tell which is 1º alkanol and which is2º alkanol. 1º and 2º alkanol can only be distinguished by Luca's test.

3º alkanol can easily distinguished from 1º and 2º alcohol since it has no reaction with common oxidizingagent.


R C O

O

H[O] [O]

R C

O

HR C O H

H

H

ketone2?alcohol

[O]R C

O

R R C O HH

R

3?alcohol

[O]R C O H

R

R


In the oxidation of the primary alkanol, aldehyde can be prevented from further oxidation by distilling itoff the reaction mixture once it is formed. This is

because aldehyde has a lower boiling point than bothalkanol and carboxylic acid.





Reagent : a very strong oxidizing agent – hot KMnO 4 / OH -(aq) or hot H 2CrO 3 (chromic acid)

N.B. chromic acid is the same as K 2Cr 2O7(aq) in acidic medium.

CHH

H

benzylic hydrogen

benzyliccarbon

+ 3 [O]1) KMnO 4/OH-(aq) , heat

2) H +(aq)

CO

OH

benzenecarboxylic acid(benzoic acid)

+ H2O

Benzenecarboxylic acid can be prepared by oxidation of aromatic side chain using a very strong oxidizingagent. The oxidation involves the abstraction of benzylic hydrogen by the oxidizing agent. i.e. all alkylgroups containing benzylic hydrogen can be oxidized to carboxyl group.

C

O

OH

1) KMnO 4/OH-(aq) , heat

2) H+(aq)

benzylic hydrogen

C

H

R R

no benzylic hydrogen

1) KMnO 4/OH -(aq) , heat

2) H+(aq)

CR R

R

resistant tooxidation

Oxidation of aromatic side chain is not limited toalkyl group. Alkenyl, alkynyl and acyl groups arealso oxidized by hot alkaline potassium

permanganate in the same way.




II. Reduction


In organic synthesis, nitrobenzene can be converted to benzenamine / aniline by reduction. Fe in dil. HCl (aq) or Snin conc. HCl (aq) can be employed for this purpose. The ammonium salt –NH 3

+ obtained, is then neutralized by OH -

(aq) .

NO 2 NH 2

(1) Sn / conc. HCl(2) OH -

nitrobenzene enzenamine/ aniline

In Sn / conc. HCl, Sn is oxidized to Sn 2+ by H +(aq) first. Since Sn 2+ is a very strong reducing agent which is oxidized

to Sn 4+ readily, it can reduce the nitro group to amino group. High concentration of Cl - ions form complex [SnCl 4]2-

(aq) with Sn 2+. This makes Sn 2+ a little bit more stable and suitable to be used as a reagent in the laboratory.


H2C=CH 2(g) + H 2(g) → CH 3 –CH 3(g) (Ni catalyst, 400ºC)

As hydrogen is adsorbed to the catalyst surface, the formation of hydrogen–metal bonds provides sufficient energyto dissociate H–H bonds. Hydrogen atoms are thereby made available on the catalyst surface for the reaction withthe π bond. As hydrogen atoms are removed, the catalyst surface is regenerated and becomes available to adsorbmore hydrogen molecules.

The ethene molecule may also be adsorbed on the metal surface. This weakens the π bond in the molecule as well.

Glossary chromic acid benzylic hydrogen alkenyl group alkynyl group




Past Paper 91 1B 5 a ii91 2C 9 a iv93 1A 3 a iii95 2C 9 b iii97 1A 5 b 97 1A 6 a i 97 2B 7 a iv98 2B 6 d i99 2B 5 a ii iii

91 1B 5 a ii5a CH3

NH 2

X

CH3

NO 2

Y The amine X, C 7 N9 N, may be prepared from Y, C 7H7 NO 2, by reaction with excess hot granulated tin andconcentrated hydrochloric acid.

ii If 6 g of X is obtained from 10 g of Y, what is the percentage yield of the reaction?(Relative atomic masses : H, 1.0; C, 12.0; N, 14.0; O, 16.0)

2

relative molecular mass of X, C 7H9 N = 107relative molecular mass of Y, C 7H7 NO 2 = 137

Theoretical yield =107

137× 10 g for 100% yield

Percentage yield = 6 ÷ (107137

× 10) × 100% = 76.8% (76.7 - 76.9%) (-½ mark for wrong no of sig. fig.) 2 marks

91 2C 9 a iv9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs.

Describe what you would observe in each case.iv

CH3CH2CH2 C CH 3

CH 3

OH

CH3CH2CH2 C

CH3

CH2OH

H

2

Warm the compound with acidic Na 2Cr 2O7. The primary alcohol will turn the solution from orange Cr(VI) togreen Cr(III). No change in colour for the tertiary alcohol.

Primary alcohol easily oxidized, not tertiary alcohol.

CH3CH2CH2 C

CH3

CH2OH

H

CH3CH2CH2 C COOH

CH3

H

Na2Cr 2O7

[O]

C Many candidates incorrectly suggested the use of iodoform test to distinguish between the tertiary and the primary

alcohols.Some candidates applying the Lucas test provided the inaccurate observation that tertiary alcohol gave white

precipitate, instead of turbidity, on reaction with Lucas reagent.

93 1A 3 a iii3a Consider the following compound, X:

iii Upon catalytic hydrogenation, one mole of X reacts with two moles of H 2.Draw the structure of the product and give its systematic name.

2

CH 3CH 2CH 2CH 2CO 2H 1 mark Pentanoic acid or n-pentanoic acid 1 mark

95 2C 9 b iii9b Identify K, L, M, N, P, R and S in the following reactions:

iii MCHCO 2CH3CH2 CH3CH2CO 2CH3

1

M : Pd or Ni or Pt / H 2 (½ mark for H 2 / cat. only) 1 mark C Common mistakes included : LiAlH 4 or H 2 without catalyst for M;



Oxidation and Reduction Page 597 1A 5 b5b Consider the following reactions:

CH3CH2CH2OHCH3CH2COCl

conc. H 2SO4heat

Na 2Cr 2O7

H3O+

J

L

M

HBr

NO 2

NO 2

NHNH 2

N(a red precipitate)

K

Give structures for J, K, L, M and N.

97 1A 6 a i6a i Vegetable oils (e.g. peanut oil) can undergo the following chemical reactions to give useful solid products P and

Q.

NaOH (aq)heat

excess H 2 / Pt

Vegetable oil

P

Q

Give one use each for P and Q . Suggest a possible structure for P .

97 2B 7 a iv7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test

should include the reagent(s), the expected observation with each compound and the chemical equation(s).iv

C

O

C2H5C2H5 and C

O

HCH3(CH2)3

98 2B 6 d i6d Identify J in the following reaction.

i

H3C

NH 2(1) Sn / conc. HCl(2) NaOHJ

99 2B 5 a ii iii5a Identify D, E, F, G and J in the following reactions.

ii CH2CH3

O2 N

CO 2H

O2 N

E

iii CH 2CH 2OH

HOCH 3

Na 2Cr 2O7 / H 3O+

heat F



Uses of Different Compounds

I. Uses of halogeno-compounds A. Use as solvent B. Manufacture of polymer


2. Physical properties of PVC and TeflonII. Uses of alcohols

A. Use as solvent B. Alcoholic drink C. Blending agent D. Ethan-1,2-diol

III. Uses of carbonyl compounds A. Preparation of urea-methanal B. Use of propanone

IV. Uses of carboxylic acids and their derivatives A. Food preservativesB. Manufacture of nylon and teryleneC. Use of ester

V. Uses of amines and their derivatives A. Azo compounds as dyes in dyeing industriesB. Amine derivatives as drugs





Uses of compounds with different functional groups Page 2

2. Physical properties of PVC and TeflonPVC is one of the largest consumers of

plasticizers. Pure PVC is rigid, easily crackedand broken, and cannot be readily processed.The rigidity is due to the intermolecular forcewhich occurs between the slightly negativechlorine atoms on one polymer chain and theslightly positive hydrogen atoms on anadjacent polymer chain.By the addition of suitable plasticizers,interaction between the polymer chains can bereduced and a flexible form of PVC can be

produced. Modified PVC is suitable for raincoats, garden hose, and seat covers for automobiles, etc.

Comparing with PVC, molecule of Telfon isnon-polar. And owing to the highelectronegativity and low polarizability of F,the polarizability of Telfon molecule is alsovery low. This makes the intermolecular forcesacting amount the molecules very weak andmakes Telfon to make the non-sticky surfacecoating.

C C C C C

F

F

F

F

F

F

F

F

F

F

II. Uses of alcohols

A. Use as solvent

Many hydroxy compounds are good solvents, for example, methanol and ethanol are commonly used assolvents in laboratory and in industry. Industrial methylated spirit contains 95% ethanol/water mixture and

5% methanol.

B. Alcoholic drink

Ethanol is an essential component in alcoholic beverages. The ethanol in alcoholic drinks is produced bythe fermentation of sugar or starch.

C. Blending agent

Ethanol can also be used as a motor fuel blending agent. As ethanol contains oxygen atom in its molecular structure, adding it to petrol helps the fuel to burn more efficiently. As a consequence, emission of carbonmonoxide is reduced.

D. Ethan-1,2-diol

Ethane-1,2-diol is miscible with water in all proportions and has a high boiling point (hence not easilyvaporized). These properties make it an ideal choice for use as an anti-freeze. Ethane-1,2-diol is also a rawmaterial in the manufacture of a polyester known as Terylene (or Dacron). The polyester is produced bycondensation polymerization of ethane-1,2-diol and benzene-1,4-dicarboxylic acid.

n HOCH 2CH 2OH n HO 2C CO 2H+ C C

O

O

O

CH 2CH 2On

(2n -1) H 2O+




III. Uses of carbonyl compounds

A. Preparation of urea-methanal

Carbonyl compounds play an important role in the manufacture of plastics. Urea-methanal is produced bycondensation polymerization between urea and methanal under heat and pressure.

C

O

n H2 N NH 2 C

O

n H H C

O

N N C

H

H

HH

n

(n - 1) H 2O+ +

In the presence of excess methanal, further heating causes cross-linkage to form between the polymer chains and hence a rigid structure is produced.

B. Use of propanone

Propanone is the raw material in the production of methyl 2-methylpropenoate which is the monomer of poly(methyl 2-methylpropenoate) [Perspex].

Propanone is also an important solvent used in industry and laboratory as it can dissolve a variety of organic compounds.

IV. Uses of carboxylic acids and their derivatives

A. Food preservatives

Benzoic acid and sodium benzoate are commonly used as food preservatives in non-alcoholic beverages,fruit juices, margarine, ketchup, salads, jams and pickled products.

B. Manufacture of nylon and terylene

Nylon 6.6, –[CO(CH 2)4CONH(CH 2)6 NH] n – , is an example of polyamides. It is the synthetic fibre for making ropes, threads, cords, lady's stockings, underwear and dresses.Industrially, nylon 6.6 is prepared by heating hexane-1,6-diamine and hexanedioic acid under heat and

pressure. In laboratory preparation, hexanedioyl dichloride is used instead of hexanedioic acid to increasethe yield. However, it is also more expensive.

Terylene, C C

O

O

O

CH2CH 2On

is an example of polyesters. It is the synthetic fibre for making wash-

and-wear garments.

C. Use of ester

Liquid esters are good solvents and are used in all-purpose adhesives, nail varnish removers and asthinners for paints. Volatile esters have characteristic sweet fruity smells and are therefore generally usedas artificial flavourings in food and drinks.




V. Uses of amines and their derivatives

A. Azo compounds as dye in dyeing industries

Primary aromatic amine is used as a starting material for the manufacture of azo dyes. It reacts withnitric(III) acid to form diazonium salt which can undergo coupling reaction to form azo-compound.

Ar NH 2 Ar N 2+ Ar N N Ar'

HNO 20 - 5

HAr'

azo compound

As azo-compounds are highly coloured, they are widely used in dyeing industry. Some examples of organic dyes are shown below. A recognition of the azo group, –N=N–, is all that is required.

N.B. : Students are NOT expected to reproduce the structures of these dyes.

Direct brown 138 (brown dye for fabrics)

N N N N

N N

SO 3- Na +

NH 2 NH 2

NH 2

NH 2

H2 N

Methyl orange (orange dye for fabrics)

N N N Na+

-

O3S

CH 3

CH 3

Sunset yellow FCF(orange-yellow dye for food product)

N N Na + -O3S

SO 3- Na +

HO

Ponceau (red dye for food product)

N N Na + -O3S

SO 3- Na +

Na +-O3S

B. Amine derivatives as drugs

Students should be made aware of the use of amine derivatives as drugs. A recognition of the amino groupor derived functional group of amine is all that is required. Some examples of these amine drugs areshown below.

N.B. : Students are NOT expected to reproduce the structures of individual drugs.

Chlorpheniramine – an antihistamine that helps to relief allergicdisorders due to cold (runny nose, watery eyes),hay fever, itchy skin, insect bites and stings, etc. Itis present in some over-the-counter drugs such as

Coltalin, Coricidin, Dristan, and Piriton.

N CH

CH 2CH 2 N(CH 3)2

Cl

chlorpheniramine

Chlorpromazine – a tranquillizer that sedates without inducing sleep.It is used to relieve anxiety, excitement,restlessness or even mental disorders.

N

S

CH 2CH 2CH 2 N(CH 3)2

chlorpromazine

Acetaminophen – (also known as paracetamol, or p-acetaminophenol)an analgesic that relieves pains such as headaches.It is believed to be less corrosive to the stomachand is an alternative to aspirin.Acetaminophen is present as the active ingredientin some over-the-counter pain-killers such asPanadol, Saridon, Tylenol and Fortolin.

HO NH C

O

CH 3acetaminophen




Glossary PVC Telfon plasticizer polarizability blending agent anti-freezeTerylene / Darcron urea-methanal perspex

Past PaperQuestion

90 1A 1 c ii91 1A 1 e92 1A 1 a ii95 2C 9 b iv

99 2B 5 c i ii iii

90 1A 1 c ii1c HOOC CH 2CH2CH2CH2 COOH

E ii Show how E, by reaction with suitable reagents, may be converted into an important commercial product. 2

Polyamide

NH(CH 2)6 NH 2CO(CH 2)4COn

∆(CH 2)4

CO2 NH3R

CO2 NH3R E

∆

NH2H2 N

any diamine

1 mark for process, 1 mark for product 2 marksOR via an acid chloride

Polyamide

NH(CH 2)6 NH2CO(CH 2)4COn

H2 N NH 2

∆E

(CH 2)4

COCl

COCl

SOCl 2

Other nylons are possible e.g. Nylon 6,6. Important is salt salt amide∆ → or acid chloride amide∆ → .

91 1A 1 e1e Give all reagents and show how 1,4-dimethylbenzene may be converted into useful polyester. 3

COCH 2CH2O

C

O

O

n

CO 2H

CO 2H

CH3

CH3

oxidation

Reagents:KMnO 4

HOCH 2CH 2OH

H+ or pressure/heat

COCl

COCl

HOCH 2CH 2OHSOCl 2

3 marks

½ mark for KMnO 4 ½ mark for dicarboxylic acid½ mark for the correct diol½ mark for the reaction conditions1 mark for the correct product

C Many candidates failed to give a polyester , giving amides instead, or the correct ester (many guessed quite oddalcohols : HOCH 2(CH 2)4CH 2OH for example.)

92 1A 1 a ii1a There are several isomers of benzenedicarboxylic acid.

ii One of these isomer is used to make terylene. Outline the reaction involved. 2½

Terylene(II)(I)

various methodsCH 2

CH 2

OH

OH

+

CO 2H

CO 2H n

CO 2 O

CO )(

methods (i) SOCl 2 →

COCl

COCl (II) → (ii) Heat (I) and (II) under pressure(iii) H + or acid or H 2SO 4 or conc H 2SO 4

(I) ½ mark (II) 1 mark

Methods ½ mark Terylene formula ½ mark

C Few gave details of reagents for condensation reaction.



Uses of compounds with different functional groups Page 695 2C 9 b iv9b Identify K, L, M, N, P, R and S in the following reactions:

iv N

HOCH 2CH 2OH OCH 2CH 2OCC

OO

n

1

N :C C

O O

ClClor

C C

O O

OHHO1 mark

C Common mistakes included : benzoyl chloride for N;

97 1A 4 b ii4b ii Upon oxidation, one of the isomers of dimethylbenzene produces a compound with the formula C 8H6O4. This

compound on condensation with ethane-1,2-diol gives a useful textile material B . Give the structure of B.

99 2B 5 c i ii iii5c Dacron is the most common of the group of polymers known as polyesters. A segment of the polymer chain is

shown below.

CC OCH 2CH2O C C OCH 2CH2O

O O OO

i Suggest two types of material which can be made from polyesters.ii Draw the structures of the two monomers used in manufacturing Dacron. Name the type of polymerization

involved.iii How can Dacron be degraded in the environment ?



Determination of Structure

I. Determination of empirical formula and molecular formula A. Different kinds of formula



II. Degree of unsaturation A. Determination of degree of unsaturation

B. Meaning of degree of unsaturation

III. Sodium fusion test


V. Introduction to IR and NMR spectroscopy


1. More examples



Structure Determination Unit 1 Page 1

Topic Structure Determination Unit 1

ReferenceReading

27.1–27.2Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 281–382

Syllabus Determination of empirical formula and molecular formula

Degree of unsaturation

Notes I. Determination of empirical formula and molecular formula

A. Different kinds of formula

Formulas are frequently used in chemistry to represent a particle. Like equation, there are many different kindof formulas depending on the usage.

Empirical formula - showing the simplest whole number ratio of atoms present (applicable to all kind of compound).

e.g. sodium chloride (ionic compound) water (molecular compound) ethane (molecular compound)

NaCl H 2O CH 3

Ionic formula - showing the simplest whole number ratio of ions present (only applicable to ioniccompound)

e.g. sodium chloride (ionic compound) Na +Cl -

Molecular formula - showing the actual number of different atoms in a molecule

e.g. water (molecular compound) ethane (molecular compound)H2O C 2H6

Structural formula - showing the connection of the atoms in a molecule

e.g. water (molecular compound) ethane (molecular compound)

OH H

C CH H

H

H

H

H


The word empirical means "based on observation or experiment, not the theory". By determining the relativelyamount of each kind of element in a compound experimentally, the empirical formula can be determined.

Example :A compound X containing only carbon, hydrogen and oxygen burned completely in air to form carbon dioxideand water as the only products. 2.43 g of X gave 3.96 g of carbon dioxide and 1.35 g of water. Find theempirical formula of X.(Given : Atomic mass of H = 1.0, C = 12.0 and O = 16.0)

Answer :

Since all the C is converted to carbon dioxide, the amount of C present = 3.96 g ×12.0

12.0 + 16.0 × 2 = 1.08 g

Similarly, all the H is converted to water, the amount of H present = 1.35 g ×1.0 × 2

1.0 × 2 + 16.0 = 0.15 g

The remaining mass must from O, therefore the amount of O present = 2.43 g - 1.08 g - 0.15 g = 1.20 g




Because empirical formula shows the relative number of atoms present, the mass has to be converted to thenumber of atoms (no. of mole of atoms) first.

C H O

Mass 1.08 g 0.15 g 1.20 g No. of mole of atoms 1.08 g

12.0 gmol-1 = 0.090 mol

0.15 g

1.0 gmol-1 = 0.15 mol

1.20 g

16 gmol-1 = 0.075 mol

The relative no of mole present

0.090 mol0.075 mol = 1.2

0.15 mol0.075 mol = 2

0.075 mol0.075 mol = 1

The simplest whole no.ratio

1.2 × 5 = 6 2 × 5 = 10 1 × 5 = 5

The empirical formula of compound X is C 6H10O5.


From the empirical formula and the molecular mass, molecular formula can also be determined.

Example:Empirical formula of ethane is CH 3 and the molecular mass is 30.0. Find the molecular formula.

Answer :Empirical formula shows the simples whole no. ratio of atoms present but molecular formula shows the actualno. of atoms present. The different between them is only a whole number (an natural no.) factor.

Formula mass of empirical formula × natural no. = Formula mass of molecular formula

(12.0 + 1.0 × 3.0) × n = 30.015.0 × n = 30.0

n =30.0

15.0= 2

Molecular formula of ethane = (CH 3)2 = C 2H6.

Example :Continue from the example given above. If the molecular mass of compound X (with the empirical formulaC6H10O5) is 485, what will be its molecular formula ?

Answer :Formula mass of empirical formula × natural no. = Formula mass of molecular formula

(12.0 × 6 + 1.0 × 10 + 16.0 × 5 ) × n = 485

162.0 × n = 485

n =485

162.0 = 2.99

n ≈ 3

Molecular formula of X is (C 6H10O5)3 = C 18H30O15




II. Degree of unsaturation

Open chain alkane has the general formula C nH2n+2 . It has the maximum no. of hydrogen (2n+2) that can be joined to the C skeleton, it is said to be saturated (with H).

For a compound with the general formula C nH2n-2 , it has 4 H less than the maximum number . 2 moles of hydrogen (H 2) is required to convert 1 mole of C nH2n-2 to a saturated compound. Therefore, the degree of

unsaturation of C nH2n-2 is said to be 2.

A. Determination of degree of unsaturation (or double bond equivalent)

Halogen – Both hydrogen (H) and halogen (X) atom form only 1 covalent bond, they can be treated as thesame in the determination of degree of unsaturation.

e.g. Both C 4H10 and C 4H9Br are saturated compounds with degree of unsaturation zero.

C C C C

H

H

H

H

H

H

H

H

H

H

C C C C

H

H

H

H

H

H

H

H

H

Br

Oxygen – O atom forms 2 covalent bond, it uses up a bond to the C and a bond to the H in a saturatedcompound. It has no effect on the degree of unsaturation.

e.g. Both C 2H6 and C 2H6O are saturated.

C C

H

H

H

H

H H

C C

H

H

H

H

H O H

e.g. Both C 3H6 and C 3H6O2 have 1 degree of unsaturation.

C C

H

H

H

H C

H

H

C C

H

H

H

H

H C

O

O H

Nitrogen – N atom forms 3 covalent bond, it uses up a bond to the C and 2 bonds to the H in a saturatedcompound. The presence of each N makes the molecule require one more H atom to becomesaturated. (Each N increases the degree of unsaturation by ½.)

e.g. Both CH 4 and CH 5 N are saturated.

C

H

H

H H

C

H

H

H N

H

H

Degree of unsaturation can be interpreted as the difference between the no. of H required and the no. of Havailable.

Degree of unsaturation = (no. of C × 2 +2) - (no. of H + no. of X) + no. of N

2




B. Meaning of degree of unsaturation (or double bond equivalent)

Degree of unsaturation cannot tell the actual structure of the molecule, but it provides some information aboutit.

One degree of unsaturation may represent the presence of : 1. 1 double bond, or 2. a ring structure.

e.g. C 4H8

C C C C

H

H

H

H

H

H

H

H

C C

CC

H

H

H

H

H

H

H

H

e.g. C 4H8O

C C C C

OH

H H

H

H

H

H

H

C

C C

CO

H

H

H

H H

H

H

H

Two degree of unsaturation may represent the present of : 1. 2 double bonds, or 2. 1 double bond and a ring structure, or 3. 2 ring structures4. 1 triple bond.

e.g. C 6H10

C C C C CCH

H H H

H

H H H

H

H

C

CC

C

CC H

H

H H

HHH

H

HH

C

C C

CC

C

HHH

H

H

H

HH

H

H

C C C C CC H

H

H

H

H

H

H

H

H

H

Glossary empirical formula ionic formula molecular formula structural formuladegree of unsaturation




Past PaperQuestion

92 2C 7 a94 2C 7 b i ii95 2C 9 a i ii96 2C 8 a i98 2A 1 b99 2B 7 c i ii

92 2C 7 a7 A carboxylic acid P , with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by

mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H 2SO 4 gave the desired ester Q but with much of the starting material P unchanged.(Relative atomic masses : H 1.0; C 12.0; O 16.0)

7a Determine the molecular formula of P . 2P : C 4H6O2 2 marks

94 2C 7 b i ii7b Give, with explanation, THREE possible structures for each of the compounds (G,H and J) below.

For (i) - (iii), 1 mark for the first correct structure, ½ mark for the second and ½ mark for the third structure.i G is a sweet smelling liquid which contains 54.4% carbon, 9.2% hydrogen, and 36.4% oxygen and has a relative

molecular mass between 80 and 100.

5

Mole ratio C : H : O54.4

12.01 9.2

1.008 36.4

16.00

= 4.530 9.127 2.275= 2 4 1 1 mark

Empirical formula of G is C 2H4OSince relative molecular mass of G is between 80 and 100, ∴ molecular formula of G is C 4H8O2. 1 mark G is a sweet smelling liquid, ∴ it is an ester 1 mark Possible structure of G :

OCH 2CH 2CH 3C

O

H OCHC

O CH 3

CH3

H OCH 2CH3C

O

CH3 OCH 3CCH3CH2

O

(any 3) 2 marks

C Generally well-answered. Some candidates missed out the hint in the question for an 'ester'.ii H, C 5H9Cl, is an acyclic chloroalkene which is optically active. 3

H possesses a chiral carbon and a C=C bond 1 mark Possible structure of H

Cl

H

H *

H H

Cl*Cl

*Cl

*(any 3) 2 marks

C A number of candidates did not realize the implication of optical activity on chemical structures. Many candidatesconsidered a pair of enantiomers as two different structures.



Structure Determination Unit 1 Page 695 2C 9 a i ii9a An acyclic hydrocarbon G, with relative molecular mass of between 60 and 80, contains 85.6% carbon and 14.4%

hydrogen by mass.i Determine the molecular formula of G. 2

Mole % of C =85.6

12.01 = 7.13

Mole % of H =14.4

1.008 = 14.29

C : H = 7.13 : 14.29 = 1 : 2∴ Empirical formula of G is CH 2 1 mark Since m.w. is between 60 and 80The molecular formula of G must be C 5H10 1 mark

ii Give all possible structures for the molecular formula determined in (i). 4

½ mark each for the 1 st to 4 th structure1 mark each for the 5 th and 6 th structure

C Some candidates did not appear to understand the term 'acyclic'. Very few could give all six structures. Many didnot realize that pent-2-ene can exist in cis- and trans-forms.

96 2C 8 a i8a i An acyclic compound H of molecular formula C 4H8O2 has a fruity smell. It does not produce a derivative

with 2,4-dinitrophenylhydrazine nor with propanoyl chloride.Deduce the functional group(s) of H. Draw FOUR possible structures for H.

4½

H is not aldehyde / ketone ½ mark because it does not form hydrazone derivative. ½ mark or

NO 2 NH

O2 N

NR

(½ mark)

H does not possess an –OH group / is not an alcohol ½ mark because it does not form ester with propanoyl chloride. ½ mark or

no R O C C 2H5

O

formation (½ mark)

H has a fruity smell, The functional group(s) in F is most likely an ester. ½ mark

Any FOUR of the following structures (½ marks for each structure) 2 marks

H O

O

O

O

H O

O

O

OO O

O

O

(Deduct ½ mark for each extra structure)C Poorly answered. Many candidates did not appear to know the reaction of 2,4-dinitrophenylhydrazine and

propanoyl chloride and hence the formation of the respective 2,4-dinitrophenylhydrazone and ester was omitted.Some did not know the reaction of ester with LiAIH 4 or the structure of ethanoic anhydride.

98 2A 1 b1b A gaseous compound D contains carbon and hydrogen only, and has density of 1.15 gdm -3 at 95.3 kPa pressure

and 298 K. Assuming that D behaves ideally, calculate its molar mass and deduce its molecular formula.(1 kPa = 1 × 10 3 Nm -2)



Structure Determination Unit 1 Page 799 2B 7 c i ii7c Compound M is a hydrocarbon isolated from oranges and lemons.

i M contains 88.2% by mass of carbon and has a relative molecular mass of 136.2. Deduce the molecular formula of M.

ii Based on the reactions given below, deduce the structure of M.

M CH 3 CH(CH 3)2+ 2H2Pt

C(CH 2)2CHCH 2C

CCH3

CH3

O O

O

H + HCHO(1) O 3

(2) Zn dustM





ReferenceReading

27.3 27.4.1–27.4.4Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 61Organic Reactions at Advanced Level, D.G. Davies and T.V.G. Kelly, pg 5–6Work Out Chemistry A-Level, David Burgess, pg 160–162

AssignmentReading

Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 472, 487–489Organic Chemistry, 6th Edition, Solomons, pg. 295-297

Syllabus Sodium fusion testTest for terminal alkyneTest for phenol

Notes III. Sodium fusion test (Ignition test)

Lassaigne sodium fusion test is a qualitative analysis used to test for any Nitrogen, Sulphur and Halogen in anorganic compound.

The organic compound is first decomposed by heating with sodium metal strongly in a ignition tube. Thedecomposed substance is dissolved in water and filtered. The filtrate is then examined for the presence of Nitrogen,Sulphur and Halogen.

Nitrogen – If nitrogen is present, cyanide ions are formed during the fusion. Iron(II) ions is added to form thecomplex hexacyanoferrate(II) ions.Fe2+

(aq) + 6CN -(aq) → Fe(CN) 6

4-(aq)

A drop of FeCl 3(aq) is added to react with hexacyanoferrate(II) ions to give a precipitate of Prussian blue.Fe(CN) 6

4-(aq) + Fe 3+

(aq) → Fe[Fe(CN) 6]- (blue ppt.)

Sulphur – If sulphur is present, sulphide ions are formed during the fusion. In the presence of S 2-(aq) , a few drops

of freshly prepared solution of sodium nitroprusside will turn the solution purple.

Halogen – If halogen is present, halide ions are formed during the fusion. The filtrate has to be boiled with dil.nitric acid in fume cupboard to remove any CN - present because CN - will also form ppt. withacidified AgNO 3(aq) .The presence of any halide ions will form ppt. with acidified AgNO 3(aq) .

Cl - – white ppt. of AgCl which turns purple grey under light.Br - – yellow ppt. of AgBr which turns yellowish grey under light.I- – yellow ppt. which is stable under light.





Criteria of a good test 1. Simple – If possible, a single step reaction is better than a multiple steps reaction.2. Obvious change – The reaction must be accompanied with an obvious change e.g. colour change, evolution of

bubble, formation of ppt..For example, formation of an immiscible layer is usually not a very obvious observation. A

reaction taking place doesn’t mean it can be served as a test.3. Mild reaction condition – Under vigorous condition, there may be other side reactions.

Test for terminal alkyne

For R–C ≡ C–H, because the H on the terminal alkyne is exceptionally acidic, it can be displaced by Cu + or Ag + toform an ionic ppt.

Two common reagents are used are diamminecopper(I) chloride [Cu(NH 3)2]+Cl - and diamminesilver(I) nitrate

[Ag(NH 3)2]+ NO 3

-.

R–C≡

CHCu NH ( )3 2

+

→

R–C≡

C-

Cu+

(s) (red ppt.)

R–C ≡ CH Ag NH ( )3 2+

→ R–C ≡ C-Ag +(s) (white ppt.)

Test for phenol

Phenol turns 'neutral' aqueous iron(III) chloride from yellow-orange to purple in colour.

The phenoxide ion (or called phenolate ion) C 6H5O- acts as a ligand with a lone electron-pair on the oxygen atom

forming a dative bond with the iron(III) cation.

[Fe(H 2O) 6]3+

(aq) + 2C 6H5O-(aq) → [Fe(H 2O)4(C6H5O)2]+

(aq) + 2H 2O(l)

It is important that the test be carried out in neutral or near-neutral conditions. If the conditions are alkaline, therewill be plenty of C 6H5O- ions but the iron(III) ions will precipitate as iron(III) hydroxide.In acidic medium, the concentration of C 6H5O

- ions is very minimal. The molecular phenol molecule C 6H5OH is avery weak ligand which does not complex with iron(III) ions.




Test for 1º, 2º and 3º alcohol

Both 1º and 2º alkanol can be oxidized by strong oxidizing agent.

1º alkanol can be oxidized to aldehyde and than carboxylic acid easily.2º alkanol will only be oxidized to ketone.

Since 1º and 2º alkanol turns K 2Cr 2O7/H+(aq) green, K 2Cr 2O7 / H +(aq) cannot tell which is 1º alkanol and which is2º alkanol. 1º and 2º alkanol can only be distinguished by Luca's test.

3º alkanol can easily distinguished from 1º and 2º alcohol since it has no reaction with common oxidizingagent.


R C O

O

H[O] [O]

R C

O

HR C O H

H

H

ketone2?alcohol

[O]

R C

O

R R C O H

H

R

3?alcohol

[O]R C O H

R

R





D. Test for aldehyde

Aldehyde is easily oxidized to carboxylic acid by any oxidizing agent. This characteristic can be used to testfor the presence of the aldehyde. e.g. aldehyde turns acidified K 2Cr 2O7(aq) from orange to green.

1) KMnO 4 / H+ or OH - (cold)

2) K 2Cr 2O7 / H+

3) HNO 3

4) Pt catalyst / O 2

5) Cu catalyst / O 2

6) Ag 2O

CR O

OH

carboxylic acidaldehyde

CR O

H

1. Silver Mirror Test / Tollen’s Test

Tollen’s reagent, diamminesilver(I) hydroxide [Ag(NH 3)2]OH is prepared by dissolving silver(I) oxide inexcess NH 3. It oxidizes aldehyde slowly to carboxylate. The silver(I) ion is reduced to silver metal anddeposited on the surface of test tube to form a silvery surface.

R–CHO (aq) + 2 [Ag(NH 3)2]OH (aq) → RCOO - NH 4+

(aq) + 2 Ag (s) + H 2O(l) + 3 NH 3(aq) silver mirror

The test tube used must be very clean otherwise the silver will not deposit on the test tube surface to form asilver mirror. Instead, it will deposit in the middle of the solution to form gray ppt.

Potential hazard of using Tollen’s reagent

The Tollen’s reagent is potentially explosive if allowed to evaporate to dryness.

2. Fehling’s Test

Fehling’s reagent consists of

Solution A : CuSO 4(aq)Solution B : Sodium potassium tartrate in excess NaOH (aq) .

Fehling’s solution is freshly prepared by addition of Solution B to Solution A until the blue ppt. just redissolves to give a deep blue solution due to the formation of a Cu(II)complex ion. The tartrate ion serves as the complexing agent in here.The presence of complexing agent keeps the Cu 2+

(aq) soluble even in alkaline medium.

R–CHO (aq) + Cu 2+(aq) + NaOH (aq) + H 2O(l) → RCOO - Na +

(aq) + Cu 2O(s) + 4H +(aq)

blue brick red ppt.

Upon reduction, the Cu(II) ion is reduced to Cu(I) with the formation of brick red ppt. of Cu 2O.

C

C

C

C

O

O

OH

HO

O -

O -

H

H

tartrate ion

3. Benedict’s Test

Benedict’s solution is an aqueous solution of Na 2CO 3, CuSO 4 and sodium citrate. Itsaction is similar to that of Fehling’s solution. But in Benedict’s solution, the complexingagent is the citrate ion instead of the tartrate ion.

R–CHO + Cu 2+ NaOH + H 2O → RCOO - Na + + Cu 2O + 4H + blue brick red ppt.

N.B. Cu+(aq) ion is unstable in aqueous medium. It disproprotionates spontaneously to

Cu 2+(aq) and Cu (s).

2Cu +(aq) d Cu (s) + Cu 2+

(aq) Therefore, the reacting medium is keep alkaline to prepcipitate any Cu + formedto Cu 2O(s)..

C

C

C

COO -

COO -

COO -

HO

citrate ion




Characteristic behaviour of functional groups

Class of compound Tests Comments

Alkene 1. Br 2 rapidly decolorised.2. Brown ppt. with alkaline KMnO 4.

Direct addition.Brown MnO 2(s) ppt.

Alkyne 1. Br 2 rapidly decolorised.2. White ppt. with [Ag(NH 3)2]+ NO 3- and red ppt.

with [Cu(NH 3)2]+Cl -.Direct addition.Terminal alkyne group only.

Alkanoic acid Liberates CO 2 from aqueous NaHCO 3.

Acid halide Hydrolysis to give X - ; Test with AgNO 3(aq) .

Amide Gives NH 3 on warming with NaOH (aq) .

Aliphatic 1º amine Gives N 2 bubbles with cold 'HNO 2'.

Aromatic 1º amine Does not give N 2 bubbles with cold 'HNO 2' butcan be coupled to a phenol to give orange diazodye.

1º alkanol 1. Steamy fumes of HCl with PCl 5.2. Ester with RCO 2H.3. Acidified orange dichromate(VI) goes green

on warming.

Fruity smell, (H + catalyst)Aldehyde and acid product.

2º alkanol 1. Steamy fumes with PCl 5.2. Acidified orange dichromate(VI) goes green

on warming.3. Turns cloudy slowly with ZnCl 2 in conc.

HCl (aq) .

Ketone product.

3º alkanol 1. Steamy fumes with PCl 5.2. No change with acidified orange

dichromate(VI).3. Turns cloudy quickly with ZnCl 2 in conc.

HCl (aq) .

3º alcohol is resistant to oxidation.

Phenol 1. No CO 2 with NaHCO 3 solution.2. White ppt. with bromine water.3. Violet complex with neutral Fe 3+

(aq) .

Aldehyde 1. Reduces Fehling's and Benedict's solutions toorange/red Cu 2O.

2. Ag mirror or grey ppt with Tollens' reagent.3. Alkaline KMnO 4 gives brown ppt.4. DNPH gives orange ppt.5. White ppt. with saturated aqueous

NaHSO 3(aq) .

Ketone 1. No reaction with alkaline KMnO 4.2. DNPH gives orange ppt.3. White ppt. with saturated aqueous

NaHSO 3(aq) .

Ketone is resistant to mild oxidation. Methyl ketone only.

Methyl ketone Yellow ppt. with I 2 / NaOH (aq) . +ve result with methyl ketone,ethanol, ethanol and 2º alcohol with – OH on 2nd C.




Test Reagents for functional groups

Reagent Used to test for

[Ag(NH 3)2]+ NO 3

- Diamminesilver(I) nitrate solution – Tollens' reagent

1. Mirror or ppt. with aldehyde.2. White ppt. with terminal alkynes.

Br 2(aq) Bromine water 1. Decolorised by alkenes, alkynes quickly

and methane slowly under light.2. White ppt. with phenol.

Cr 2O72- Acidified aqueous potassium

dichromate(VI)Orange solution goes green with 1º and 2ºalcohols and with aldehyde.

[Cu(NH 3)2]+Cl - Diamminecopper(I) chloride solution Red ppt. with terminal alkynes.

DNPH 2,4-dinitrophenylhydrazine aqueoussolution

Yellow / orange ppt with aldehydes and ketones

Fe3+ Neutral solution of iron(III) chloride Violet complex with phenols

Fehling's or Benedict's solution

Both contain complexed Cu 2+(aq) Orange to red ppt. with aldehyde and reducingsugar.

'HNO 2' A mixture of HCl (aq) and NaNO 2. Used below 5ºC.

N2 evolved with aliphatic 1º amines. Aromatic1º amines doesn't give N 2 but give diazo dyeswhen coupled with phenols.

I2 / NaOH Iodoform test. Aqueous I 2 in KIsolution, made alkaline with NaOH.

Pale yellow ppt with methyl ketone, ethanol,ethanal and2º alcohol with –OH on 2nd C.

MnO 4- Manganate(VII), i.e. KMnO 4 solution

made alkaline with Na 2CO 3.

Brown ppt. with alkenes, alkyne and aldehydes.

Na metal Used in the absence of water. Evolves H 2 from –OH in alcohols and alkanoic

acid.

NaHCO 3 Saturated sodium hydrogencarbonatesolution

Evolves CO 2 with alkanoic acids.

NaHSO 3 Saturated sodium hydrogensulphitesolution

White ppt. with aldehyde and methyl ketone.

NaOH Cold

Hot

Acids and phenols dissolves. NH 3 evolved from NH 4

+ salts. Amines liberated from amine salts.

NH 3 from NH 4+ salts and amide.

PCl 5 Phosphorous(V) chloride. Used as solid Steamy fumes of HCl from –OH in alcohols,alkanoic acids and phenols.

ZnCl 2 in conc.HCl (aq)

Luca's Test Turns cloudy with 3º alcohol quickly.Turns cloudy with 2º alcohol slowly.Does not turn cloudy with 1º alcohol at all.




Glossary sodium fusion test ignition tube hexacyanoferrate(II) ion Prussian bluesodium nitroprusside wet chemistry

Past PaperQuestion

90 1B 4 b i iii91 2C 7 a iv93 2C 8 a ii iv94 1B 4 b

95 2C 7 a iii96 2C 7 c i97 1A 4 d 97 1B 8 c i98 1B 8 a iii 98 1A 4 b i ii

90 1B 4 b i iii4 A mixture contains equal amounts of X, Y and Z.

OH

X b.p. 182°C

NH 2

Y b.p. 184°C

CH3

Br

Z b.p. 184°C

4b Outline tests involving observable colour changes to show that:i X is a phenolic compound; 1

Add alcoholic FeCl 3 solution. A colour change to red or blue denotes presence of phenol group. 1 mark iii Z contains bromine. 1

Carry out a sodium fusion to resulting solution add HNO 3 and aqueous AgNO 3. A pale yellow ppt. sparinglysoluble in dil. NH 3(aq) denotes presence of Br. 1 mark

C Few knew the Na fusion test. Some were able to devise a feasible procedure e.g. using the Grignard reagent andtesting the inorganic product—showing initiative for which they were rewarded in full.

91 2C 7 a iv7a State with explanations, what you would observe in each of the following experiments, and write equations for the

reactions.iv Hexanal is treated with ammoniacal silver nitrate solution (Tollens’ reagent). 2

A silver mirror is formed on the surface of the test tube or some silver formed. Ag + is reduced by hexanal to Agmetal and deposits on the glass surface. 1 mark

CH3CH2CH2CH2CH2CHO CH 3CH2CH2CH2CH2CO- NH 4+

OAg(NH 3)2 + Ag

1 mark



Structure Determination Unit 2 Page 893 2C 8 a ii iv8a Give a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents required, the observation expected, and the chemical equation(s) for each test.ii and CH3CH2COCH 2CH 3CH 3CH2CH 2CH 2CHO

3

Reagent: Fehling’s reagent / Tollen’s reagent 1 mark Observation: Red ppt. / Formation of silver mirror 1 mark Equation:

CH 3CH 2CH 2CH 2CHO + 2Cu2+

+ 6OH-

→

CH 3CH 2CH 2CH 2COO-

+ 2Cu 2O(red ppt.) + 3H 2Oor CH 3CH 2CH 2CH 2CHO + Ag(NH 3)2

+ + 2OH - → CH 3CH 2CH 2CH 2COO - + 2Ag(silver mirror) + H 2O 1 mark C A few candidates incorrectly stated that Ag 2O was the product formed in the " Silver Mirror" test.

Some candidates could not give the formula of Tollen's reagent or Fehling's solution.iv

CH 3

OH CH2OHand

3

Reagent: Lucas reagent (ZnCl 2 in conc. HCl) 1 mark

Observation: CH 3

OH

gives turbidity / cloudy / misty faster thanCH 2OH

1 mark

Equation:

CH 3

OH CH3+ H2O CH3

ClCl-H+

1 mark C Many candidates did not understand the chemistry involved in the Lucas test and the reason for observing

turbidity.Many candidates incorrectly applied the Iodoform Test to distinguish the given pair of alcohols.

94 1B 4 b4b You are provided with a sample of an aliphatic aldehyde. Describe, with experimental details, the silver mirror

test (Tollens’ test) that you would perform, in order to establish that the sample is an aldehyde. What hazard willyou face if you allow the reaction mixture to evaporate to dryness?

3

Clean test tube ½ mark Add excess NH 3 to AgNO 3 in test tube until initial ppt. dissolves 1 mark

Add aldehyde ½ mark Place the tube in a beaker of warm water (No direct gentle heating) ½ mark Hazard : an explosion may occur ½ mark

C Candidates were reluctant to itemize the procedure, including experimental details. Frequent omissions includedthe need to use a clean test tube; the steps in making Tollens' reagent; and the use of a water bath.



Structure Determination Unit 2 Page 995 2C 7 a iii7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents used, the observation expected and the chemical equation(s) for each test.iii

OH

CH 3and CH2OH

3

Treat compound with MnO 4-/H+ or Cr 2O7

2-/H+ ½ + ½ mark

CH2OH COOH CHOor Cr 2O72- / H+

½ mark

The solution changes colour from orange/yellow to green. ½ + ½ mark

No reaction with OH

CH3

. ½ mark

OR Treat compound with HCl and ZnCl 2 ½ + ½ mark

OH

CH3

gives turbidity at a faster rate thanCH2OH

1 mark

OHCH3

+ HClCH3

Cl+ H2O 1 mark

C The Lucas test was generally used, but most candidates gave wrong observation : precipitation instead of

turbidity; they probably thought that the tertiary chloride was a solid. Some candidates gave incorrect reagentssuch as Zn instead of ZnCl 2; ZnCl 2(aq) instead of anhydrous ZnCl 2(s) ; dilute HCl instead of concentrated HCl. Somegave the wrong formula ZnCl for zinc chloride. The simpler reagent, Cr 2O7

2-/H+ only appeared occasionally. Somecandidates did not give the correct colour change involving this common oxidizing agent.



Structure Determination Unit 2 Page 1096 2C 7 c i7c Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer

should include the reagents used and the observation expected.i

CHOand

O

2

Treat compound with Tollen's reagent / ammoniacal silver(I) oxide / ammoniacal silver nitrate(V) / Ag(NH 3)2+ 1 mark

CHOgives a silver mirror / silver deposit ½ mark

No reaction with

O

½ mark

Or Treat compound with Fehling's reagent / a mixture of CuSO 4,NaOH and sodium potassium tartarate (1 mark)

CHO produces a (bright) red / orange / reddish brown ppt. (½ mark)

No reaction withO

(½ mark)

Or Treat compound with Cr 2O4

2-/H+ (cold MnO 4-/H+) (1 mark)

CHOchanges the colour of the solution from orange to green (½ mark)

(changes colour of MnO 4-/H+ from purple to colourless )

No reaction with

O

(½ mark)

C Some candidates erroneously gave Brady's test for differentiation. One common mistake of the candidateswas to take CuSO 4 as Fehling's reagent.

97 1A 4 d4d Suggest, with explanation, a chemical reaction which would enable 4-methylbenzoic acid and 4-methylphenol to

be distinguished from each other.2

97 1B 8 c i8c For each of the following groups of compounds, suggest a chemical test which would enable each compound to be

distinguished from the other(s). In your answer also give the changes that you would expect to observe for eachcompound.

i CH 3CH 2CH 2OH, (CH 3)2CHOH and (CH 3)3COH 2

98 1A 4 b i iiAlcohol E has the structure CH 3CH(OH)C 2H5.

4b i Draw the structures of three structural isomers of E, all of which are alcohols. 1½ii Describe how the reagent Zn/concentrated HCl can be used to distinguish E from the three structural isomers. 1½

98 1B 8 a iii8a Show how you would

iii distinguish between C 6H5COCl and C 6H5COBr using a chemical test. 1





Notes 27.4.0Past Paper Questions of structural determinationOrganic Chemistry, 6th Edition, Solomons, pg. 359-360, 742

Past PaperQuestion90 2C 7 a b c91 2C 9 b i ii iii92 1A 1 b92 1A 1 f ii 92 2C 7 a b e93 2C 9 a i94 2C 7 b iii95 2C 7 b i ii 95 2C 9 a i iii iv96 2C 7 a i ii98 2B 6 c ii

90 2C 7 a b c7 Consider the following experimental information.

I. X (C 9H11 NO) gives on hydrolysis, two compounds with formulae CH 5 N and C 8H8O2. When the latter

compound is added to sodium hydrogencarbonate solution, a colourless gas is evolved.II. Y (C 9H10) gives on vigorous oxidation, an acid with formula C 7H6O2. Y decolourizes a solution of bromine

in tetrachloromethane.III. Z is an optically active ester containing 64.6% carbon, 10.8% hydrogen and 24.6% oxygen, and has a relative

molecular mass between 120 and 140. Upon hydrolysis, Z gives propanoic acid.(Relative atomic masses: H, 1.0; C, 12.0; O, 16.0)

7a Draw TWO possible structures for each of the compounds X, Y and Z. 9X: Y:

CH3

C NHCH 3

O

CH3

C NHCH 3

O

C CCH3H

HC C

CH3

H H

Z: molecular formula calculated C 7H14O2 3 marksCH3CH2 C

O

O C CH 2CH3

H

CH3

CH3CH2 C

O

O C H

CH2CH3

CH3

1 mark for each structure7b For each pair of possible structures, name the type of isomerism involved. 3

X: Structural isomerism 1 mark Y: Geometric isomerism / cis-trans isomerism 1 mark Z: Enantiomerism / optical isomerism / stereoisomerism 1 mark

7c Using only one of the possible structures for each of X, Y and Z, write equations for the reaction described above. 5

CH3

C NHCH 3

O

CH3

COOH

hydrolysis + CH3 NH2

2 marksC C

CH3H

HCOOH

[O] Br

Br C C

CH3H

H Br 2

1 mark each

CH3CH2 C

O

O C CH 2CH3

H

CH3H2O CH3CH2OOH + C CH 2CH3

H

CH3

HO

1 mark C The majority of the candidates could provide correct structures for X and Y, but not for Z.

One common incorrect structure for X was :

C CH 2 NH 2

O

and the related reaction equation was incorrectly given as :

C CH 2 NH 2

O

hydrolysis C

O

OHCH3 NH 2+

Common incorrect structures for Y were :




Although most candidates could work out the molecular formula for Z using the given analytical data, they failedto provide the correct structures which would display optical activity i.e. a chiral centre.Candidates had difficulty in drawing a 3 dimensional representation for Z properly.Many candidates could not distinguish between different types of isomerism in (b).It is suggested that terms like "chain isomerism", "functional group isomerism", "position isomerism" or "metamerism" should be dropped and the more general term "structural isomerism" should be used to describedifferent structures.One other common mistake in isomerism for Y was regarding structures with the double bond at differentlocations as geometric isomers, e.g.

CH CHCH 3 CH2CH CH 2

91 2C 9 b i ii iii9b Suggest a possible structure for each of the compounds J, K, L, M, and N below and explain briefly your

deductions.i J, C 4H6O, on oxidation gives K, C 4H6O2. 3

CHO COOH[O]

or CHO COOH[O]

or CHO COOH[O]

or CHO COOH[O]

J K 1 mark for J, 1 mark for K

Oxidation increase the number of O atom on J 1 mark ii L, C 5H12O, can exist as a pair of enantiomers, and reacts with phosphorus pentachloride to give hydrogen

chloride.3

OH

CH3H

Cl

CH3H

+ HClPCl 5

or

OH

HCH3

Cl

CH3H

+ HClPCl 5

1 mark L can exist as pair of enantiomers → chiral centre in L 1 mark L gives out HCl on reaction with PCl 5 → -OH group present 1 mark

C A few candidates did not realize the relation between the existence of enantiomers and chiral centre; hence theygave an incorrect structure for L.

iii M, C 6H10, on ozonolysis gives N, C 6H10O2. 3

CHO

CHO

O3

Zn/H 2Oor CHO

OO3

Zn/H 2Oor

CHO

CHO

O3

Zn/H 2O

M N 1 mark for M, 1 mark for NOzonolysis of M without changing the carbon number → double bond forms part of a ring, i.e. cyclic alkene.

1 mark C Many candidates could not account for the ring structure of M by relating the number of carbon atoms in M and

N.

92 1A 1 b1b Compound X with formula C 10H14 is optically active. On treatment with potassium manganate(VII), X gives

benzoic acid (C 6H5COOH). Give the structure of X .2

CHEt

Me

or

C Few commented on the 3-D aspect of the structure.



Structure Determination Unit 3 Page 392 1A 1 f ii1f

CH3 CH NH 3

CO2H

V

ii Give the structure of the compound with a relative molecular mass of approximately 230, which on refluxing withdilute hydrochloric acid gives V as the only product.(Relative atomic masses: H 1.0; C 12.0; N 14.0; O 16.0)

1½

The structure of the compound is C 9H17 N3O4 with m.w. 231.CONH CONH CO 2H

NH 2 Essential features (i) 2 peptide links: CONH

(ii) tripeptide 1½ mark

92 2C 7 a b e7 A carboxylic acid P , with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by

mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H 2SO 4 gave the desired ester Q but with much of the starting material P unchanged.(Relative atomic masses : H 1.0; C 12.0; O 16.0)

7a Determine the molecular formula of P . 2P : C 4H6O2 2 marks

7b Give the structures of four carboxylic acids having the molecular formula you determined in (a). Give thesystematic names for any one of the carboxylic acids and its methyl ester.

6

COOHbut-3-enoic acid or 3-butenoic acid

H

HCH3

COOH

trans-but-2-enoic acidH

COOHCH3

H

cis-but-2-enoic acidCOOH

2-methylpropenoic acid any 4 structures - 4 marksCOOH

cyclopropanecarboxylic acid name of acid - 1 mark ester : methyl but-3-enoate etc. name of ester - 1 mark

C Many candidates did not show the structure of cis and trans isomer clearly, e.g.

C C COOH

HH

CH 3 for cis isomer, and

C C COOH

H

H

CH 3

for trans isomer.

Many candidates treatedCH 3

CCH2 COOH and CH3 C

CH 2

COOH as two different structures.Candidates were generally weak in naming; examples of common mistakes are :CH 3 CH CH COOH butan-2-enoic acidCH 3 CH CH COOCH 3 methy but-2-enoate or methoxy but-2-enoate.

7e Show, using equations, how you would convert Q to P . 2

+ KOHCOOCH 3

Q

COO -K ++ CH3OHHeat

H+

COOH

P 2 marks



Structure Determination Unit 3 Page 493 2C 9 a i9a A compound W of relative molecular mass 121 has the elemental composition: C, 79.3%; H, 9.1%; N, 11.6%.

W reacts with nitric(III) acid to give a derivative X. X reacts with ethanoic anhydride to give a sweet-smellingliquid Y which contains no nitrogen. Oxidation of X with chromic(VI) acid yields Z, C 8H8O. Z reacts with iodineand sodium hydroxide to produce a yellow precipitate and sodium benzoate.

i Deduce the structures of W, X, Y and Z, using all the given data. Explain your reasoning. 10C H N79.3 9.1 11.6

No. of atoms 79.312.0 9.1

1.008 11.614.01

= 6.61 : 9.1 : 0.828≈ 8 : 11 : 1

Empirical formula of W is C 8H11 N Molecular formula is (C 8H11 N) n Molecular mass = 121 ∴ n = 1 1 mark X reacts with ethanoic anhydride to form a sweet-smelling liquid Y.∴ X is an alcohol and Y is an ester 1 mark W reacts with HNO 2 to give an alcohol∴ W carries an –NH 2 group / 1º amine 1 mark Z is obtained from oxidation of X (an alcohol) and it gives positive iodoform test. It is a methyl ketone. 1 mark

W is C

H

NH 2CH3

X is C

H

CH3OH

Y is C

H

CH 3

O C

O

CH3

Z is C CH 3

O

(4 marks for structure, 6 marks for deduction)C A number of candidates were able to give correct structures for the unknowns; however, they were unable to show

clearly and logically how these structures were deduced from the given information. Generally, candidates lackedthe necessary technique in presenting an answer for this type of problem.Many candidates suggested X was a diazonium salt because X was formed from the reaction between an amineand HNO 2. They had ignored the fact that X gave an ester on reacting with anhydride.

94 2C 7 b iii7b Give, with explanation, THREE possible structures for each of the compounds (G,H and J) below.

For (i) - (iii), 1 mark for the first correct structure, ½ mark for the second and ½ mark for the third structure.iii J, C 10H14O, reacts with PCl 5 to give hydrogen chloride. Vigorous oxidation of J gives benzene-1,2-dicarboxylic

acid, but mild oxidation of J gives a ketone.Suggest one chemical test, giving the expected observations, which would allow you to eliminate one of your

possible structures of J in (iii).

6

J reacts with PCl 5 to give HCl, ∴ J carries a –OH group ½ mark Stronge oxidation of J gives , J is a 1,2-disubstituted aromatic compound 1 mark Mild oxidation of J gives a ketone, J is a 2º alcohol ½ mark Possible structures of J

OH

OH

OH

2 marksIodoform test / I 2 in NaOHObservation :

OH and

OH

will give a pale ppt., while

OH

will not. 2 marks

C Poorly-answered. Many candidates did not make use of all information to deduce that J was a secondary alcoholas well as a 1,2-disubstituted benzene derivative.



Structure Determination Unit 3 Page 595 2C 7 b i ii7b An acidic compound D, C 6H6O4, on heating to 140ºC dehydrates to give compound E, C 4H2O3. D gives compound

F upon hydrogenation.i Give the structures of D, E, F. 3

CC

CC

O

O

O

O

H

H

H

H

C

C CO

C

H

H

O

O HOOCCH 2CH2COOH

(1 or 0) mark eachD (must show the cis structure) E F

C Poorly answered. Few candidates gave the correct structure for D. Most candidates erroneously thought thatdehydration only involved the elimination of water from alkanols to give alkenes. Some gave both trans- and cis-isomers for D probably because they did not realize that only cis-butenedioic acid would dehydrate to form ananhydride at 140ºC. Most failed to appreciate geometrical isomerism and less than 5 % named D and Ecorrectly.

ii Give the systematic names for D and E. 2D : cis-butenedioic acid 1 mark E : butenedioic anhydride 1 mark

C Many candidates missed out the stereochemical prefix cis in the naming of D.

95 2C 9 a i iii iv9a An acyclic hydrocarbon G, with relative molecular mass of between 60 and 80, contains 85.6% carbon and 14.4%

hydrogen by mass.i Determine the molecular formula of G. 2

Mole % of C =85.6

12.01 = 7.13

Mole % of H =14.4

1.008 = 14.29

C : H = 7.13 : 14.29 = 1 : 2∴ Empirical formula of G is CH 2 1 mark Since m.w. is between 60 and 80

The molecular formula of G must be C 5H10 1 mark iii G, on ozonlysis, yield H and J which both produce positive iodoform reactions. Give the structures of G, H and J. 3

Structures of H and J : CCH 3 CH 3

O

and CCH 3 H

O

1 + 1 mark

∴ Structure of G :C C

CH3

CH3

CH3

H 1 mark

C Although structures for H and J were given correctly, many candidates could not deduce the structure for G. Somecandidates could present G in (iii) but did not include it as one of the structures in (ii).

iv Give the systematic name of G. 12-methylbut-2-ene 1 mark

C G was frequently but erroneously named as 3-methylbut-2-ene.

96 2C 7 a i ii7a Two compounds F and G. with the same molecular formula C 4H8O, give the following results in four chemical

tests.Reagent F GH2 NOH no reaction no reactionCr 2O7

2-/H+ colour changes from orange to green no reactionBr 2 /CH 3CCl 3 decolorization no reactionI2/NaOH yellow precipitate no reaction

i Give the structure for F. Explain your answer with reference to the result of each of the chemical tests. 5F is CH 2=CHCH(OH)CH 3 1 mark

F does not form any oxime. ∴ F is not aldehyde or ketone ½ + ½ mark or, Aldehyde or ketone will give the following reaction (½ mark)




R C R'

OH2 NOH+

R C R'

N OH

(½ mark)

F can be oxidized by chromic(VI) acid ½ mark ∴ F is a (primary / secondary) alcohol / bears an –OH group ½ mark or, alcohols can be oxidized by chromic(VI) acid (½ mark)

RR'CHOH RR'C Ochromic acid

(½ mark)

F gives addition reaction with Br 2. ∴ C=C / C ≡ C / unsaturation is present ½ + ½ mark or, compounds with C=C / C ≡ C form addition product with Br 2 (½ + ½ mark)

C

C

C

C

Br

Br Br 2+

(1 mark)

F gives positive reaction in iodoform test ∴ CH CH 3

OHor C CH 3

Ois present. ½ + ½ mark

or,C

CH3

OHH

R I2 / OH -

RCOO- + CHI3 (1 mark)

C Candidates were weak in logical deduction. Many candidates missed the important hints and deduced the structurewithout explanation or without referral to related reactions. They should learn how to summarize information,reduce possibilities step by step, and then draw conclusions about the structure. Many candidates failed to giveoxime as the derivative of carbonyl compound with NH 2OH. Some erroneously attributed the colour change withchromic(VI) acid to the double bond. Some did not mention the words addition with Br 2 and oxidation with

chromic(VI) acid. Some wrongly concluded thatC CH 3

OHwas present instead

OH

CH CH 3 for a positive iodoform

test.ii Suggest a structure for G and explain your suggestion. 2

Any ONE of the following:

O O OO OH

1 mark Explanation:

G does not possess C

O, C C , 1º or 2º alcohol / G is 3º alcohol ½ mark

it can only be ether / tertiary alcohol ½ mark C Some candidates erroneously thought that cyclobutanol is a tertiary alcohol and assigned it as compound G.

98 2B 6 c ii6c 2-Ethanoyloxybenzoic acid (aspirin) is one of the most common substances used to relieve pain.

OCOCH 3

CO 2H

2-ethanoyloxybenzoic acid

4

ii Suggest how you could show the presence of the two functional groups in 2-ethanoyloxybenzoic acid.





ReferenceReading

27.5Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 384–390Organic Chemistry, 6th Edition, Solomons, pg. 541-547

AssignmentReading

Syllabus Introduction to IR and NMR spectroscopyInterpretation of IR spectrum

Notes V. Introduction to IR and NMR spectroscopy

Other than wet chemistry, infra-red spectroscopy and nuclear magnetic resonance (NMR) are modern instrumentaltechniques in helping chemists to understand the structures of molecules.

IR spectrum

Infra-red spectroscopy depends on the vibrations of atoms with respect to each other in the molecule. The infra-redspectrum can tell us whether particular functional groups are present in a molecule

IR spectrum of octane

N.B. The actual structure cannot be deduced from the IR spectrum alone, other tests are required to confirm theresult.




NMR spectrum

Nuclear magnetic resonance depends on the magnetic properties of the atomic nuclei (e.g. hydrogen nuclei) in amolecule. The proton NMR spectrum can tell us the number of hydrogen nuclei present in the molecule and giveinformation about the structural environment of the hydrogen.

NMR spectrum of 1,1,2-chloroethane





On an IR spectrum, different functional groups have characteristic absorption wavenumbers, therefore, thefunctional group(s) present in an organic compound can be identified from a given IR spectrum.

The frequency of infra-red spectrum is usually measured in wavenumber (cm -1). This means the number of wavelength in each cm. The wavenumber is directly proportional to the frequency .

An IR spectrum can be divided into several region :

1. 4000 – 2500 cm -1 Absorption of single bonds formed by H and other elements. e.g. C–H, O–H, N–H2. 2500 – 2000 cm -1 Absorption of triple bonds. e.g. C ≡ C, C ≡ N3. 2000 – 1500 cm -1 Absorption of double bonds e.g. C=C, C=O4. 1500 – 500 cm -1 Fingerprint region, consists of complicated bands and is unique to each compound.

IR spectrum of octane CH 3(CH 2)6CH 3

In A-Level, we only focus on the identification of functional groups . Therefore, only the region from4000 cm -1 to 1500 cm -1 will be studied.

Characteristic group absorption wavenumbers of covalent bonds in organic molecules




The following examples show how to make use of the characteristic absorption wavenumbers to identify thefunctional groups present in organic compounds.

Fig. 1 IR spectrum of ethanol

Fig. 1 shows a portion of the IR spectrum of ethanol.

The peak at 2950 cm -1 corresponds to the absorption of C–H group and that at 3340 cm -1 corresponds to the

absorption of O–H group. Hence, a O–H group is present in the ethanol.

Fig. 2 IR spectrum of methanoic acid

Fig. 2 shows a portion of the IR spectrum of methanoicacid.

The peak at 1710 cm -1 corresponds to the absorption of C=O group and that at 3100 cm -1 corresponds to theabsorption of O–H group. Hence. both the C=O andO–H groups are present in the methanoic acid.

It should be noted that the absorption of the O–H groupin alcohols and carboxylic acids does not usually appear as a sharp peak. Instead. a broad band is observed

because the vibration of O–H group is complicated bythe hydrogen bonding.




1. More examples

a. Hex-5-en-2-one C C C C C C

O

H

H H H

H

H

H

H

H

H Peak / cm -1 Bond3080 C–H1720 C=O1635 C=C

b. Hex-1-yne C C C C C CH

H

H

H

H

H

H

H

H

H Peak / cm -1 Bond3300 C–H*2900 to 2950 C–H2200 C ≡ C

* Absorption at

3300 cm-1

is due toC–H bond in terminalC≡ C triple bond, whichis weaker.




c. Butyl ethanoate C C C C O C C

O H

H

HH

HH

H

H

H

H

H

H Peak / cm -1 Bond2900 to 2950 C–H1740 C=O

d. 3-methylbutan-2-one C C

CH3O

CH3CH3 Peak / cm -1 Bond

2880 to 2980 C–H1720 C=O




e. Butylamine C C CC N

H

H

H

H

H

H

H

H

H

H

H Peak / cm -1 Bond3300 to 3400 N–H2850 to 2950 C–H

f. Propanenitrile C C C N

H

HH

H

H Peak / cm -1 Bond2900 to 3000 C–H2250 C ≡ N

Glossary infra-red (IR) spectroscopy nuclear magnetic resonance (NMR) spectroscopy spectrum / spectrawavenumber fingerprint region absorption peak




Past PaperQuestion

97 2B 6 c98 1B 8 a ii99 2B 6 a iii

Sample question7b An organic compound Z, with relative molecular mass below 100, has the following composition by mass.

C 66.7%, H 11.1% and O 22.2%.11

i Determine the molecular formula of Z.no. of mole of C : no. of mole of H : no. of mole of O

=66.712.0 :

11.11.0 :

22.216.0

= 5.56 : 11 : 1.39= 4.00 : 7.91 : 1.00≈ 4 : 8 : 1Empirical formula of Z is C 4H8O.Formula mass of C 4H8O = 12.0 × 4 + 1.0 × 8 + 16.0 = 72.Since the molecular mass of Z is below 100, the molecular mass of Z should be 72.

ii A portion of the infra-red (IR) spectrum of Z is shown below:

Using the IR spectrum and the result from (i), deduce two possible structures of Z, each belonging to a differenthomologous series.Given :Characteristic Group Absorption Wavenumbers of Covalent Bonds in Infra-red Spectra (stretching modes)

Bond Characteristic RangeWavenumber / cm -1

C=C Alkenes 1610 to 1680C=O Aldehydes, ketones, acids, esters 1680 to 1750C–C Alkynes 2070 to 2250C≡ N Nitriles 2200 to 2280O–H Acids (hydrogen-bonded) 2500 to 3300C–H Alkanes, alkenes, arenes 2840 to 3095O–H Alcohols, phenols (hydrogen-bonded) 3230 to 3670

N–H Primary amines 3350 to 3500From the molecular formula C 4H8O, the degree of unsaturation of the molecule is 1.The peak at 1700 cm -1 and absence of a board peak at 3100 cm -1 indicate the presence of C=O bond.Z may be butanal (an aldehyde) or butan-2-one (a ketone).

C C C C

O

H

H

H

H

H

H

H

H

C C C C H

H

H

H

H

H

O H

H butanal butan-2-one

iii Suggest one chemical test, giving the expected observation(s), that can be used to distinguish the two compoundshaving the structures as deduced in (ii) above.By iodoform reaction, when I 2(aq) and NaOH (aq) are added, only butan-2-one will give yellow precipitate of CHI 3(s)

but butanal has no reaction.

Sample question1 The infrared spectra in figure 1 and 2 represent an ester and an alkyne. Identify the peaks which are starred on

each spectrum and so decide which spectrum represents which type of compounds.




Figure 1 Figure 2Figure 1 : an ester Figure 2 : an alkyne

2 An alkene A (C 5H10), on ozonolysis, gave two different compounds, B and C with molecular formulae C 3H6O and

C2H4O respectively. Compound B has an infrared spectrum as shown in figure 3 below.

Figure 3 IR spectrum of compound B

When B and C were separately treated with acidified dichromate(VI), B did not react but C gave D with molecular formula C 2H4O2, D gave effervescence with sodium hydrogencarbonate.Deduce possible structure for A, B, C and D based on above information. A : 2-methylbut-2-eneB : propanoneC : ethanalD : ethanoic acid

97 2B 6 c6c Compound H. C 3H6O2, does not react with NaBH 4 and displays the following infra-red spectrum. Deduce all

possible structures of H.

98 1B 8 a ii8a Show how you would

ii distinguish between propan-2-ol and propanone using spectroscopy,

99 2B 6 a iii



Structure Determination Unit 4 Page 106a

Cl NH 2 Cl NH CCH 3

O


iii Suggest how to show the presence of the amide group in N-(4-chlorophenyl)ethanamide by(I) a chemical test and(II) a spectroscopic method.



Organic synthesis


II. Structural analysis

A. Chain length

1. Carbon chain


B. Degree of unsaturation


D. Position of functional group

III. Systematic approach to organic synthesisIV. Examples



Organic Synthesis Page 1

Topic Organic Synthesis

ReferenceReading

28Organic Chemistry, Solomons, 5th Edition pg. 157Organic Chemistry, Stanley H. Pine, 5th Edition pg. 722–723

ReadingAssignmentChemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 530–534Organic Chemistry, 6th Edition, Solomons, pg. 169-171

Syllabus Organic Synthesis

Notes Synthesis is the real test of our ability to use and control organic reactions. Before attempting any synthesis, youmust be familiar with all kinds of conversion available first.

Besides the knowledge about the conversion (organic reaction), the following analysis may help you to plan thesynthesis.


Retro- means working backwards.

1. It is better to begin with the target molecule and work backwards from it. The no. of ways of preparing atarget molecule is usually fewer than the no. of way that a starting molecule can react.If we start from the starting molecule, the no. of possible synthetic path may be numerous.

The molecule from which the target molecule is prepared is called precursor.

Target molecule ⇒ 1st Precursor ⇒ 2nd Precursor ⇒ ⇒ Starting molecule

For a beginner, it may be quite difficult to plan the complete path at once. The structural analysis mayhelp you to identify some of the precursors (and reagents) required, therefore a long path can be broken

into several shorter pathways.

2. The no. of steps involved should be as few as possible. Try not to use path with more than 4 steps. If theyield of each step is 80%, a 4 steps conversion means merely (80%) 4 = 41% overall yield.

3. Use a precursor (or a reagent) which will give a higher yield.e.g. bromoalkane instead of alcohol in nucleophilic substitution

acyl halide instead of carboxylic acid in nucleophilic addition-elimination

4. Even a conversion seems to be simple may not be so simple.e.g. Direct conversion of acyl bromide (RCOBr) to bromoalkane (RCH 2Br) is impossible.

This is the feasible way : acyl bromide → carboxylic acid → alcohol → bromoalkane




II. Structural analysis(The following guideline is only applicable to A-Level where the choices of conversion are limited.)

Before planning of any organic synthesis, it is important to know the difference between the starting moleculeand the target molecule.

A. Chain length

1. Carbon chain If the no. of C is increased by 1i. Substitution reaction with CN - (haloalkane precursor)ii. Addition reaction with CN - (carbonyl compound precursor)

If the no. of C is decreased by 1i. Haloform reaction (methyl ketone or methyl alcohol precursor)ii. Hoffman degradation (1º amide precursor)


If the chain length is increased.i. Acylation by acylating agent e.g. RCOCl (nucleophile precursor e.g. RNH 2 or ROH)ii. Substitution reaction with RI (nucleophile precursor e.g. RNH 2 or ROH)

B. Degree of unsaturation (consider C=C and C ≡ C bond only)

If the degree of unsaturation is increased → Elimination with strong base e.g. CH 3CH 2ONa + (haloalkane or dihaloalkane precursor)If the degree of unsaturation is decreased → Catalytic hydrogenation (alkene or alkyne precursor)


If it is oxidized toi. Carboxylic acid (methyl ketone, aldehyde, 1º alkanol or alkyl benzene precursor)ii. Ketone (2º alkanol precursor)

If it is reduced toi. Alkanol (any acid derivatives except amide precursor)ii. Amine (amide, nitrile or aromatic nitro compound precursor)

D. Position of the functional group

If there is no change in the position of the functional group, it may be converted by direct conversion.If there is change in the position, the functional group may be converted by elimination followed by addition.

III. Systematic approach to organic synthesis

1. Name all the functional groups on the original and target molecule, must be specific e.g. 1º or 2º.2. Identify if there is any change in length of carbon chain.3. Identify if the molecule is oxidized or reduced.4. Identify if there is any change in degree of unsaturation.5. Figure out the possible intermediate (precursor).6. Repeat step 1 to 5 until all the intermediates are figured out.7. Assign reagent to each conversion.




IV. Examples

Example 1CH3CH2CH2COOH CH 3CH2CH2CH2COOH

Round 1

1 – Naming Starting molecule : CH 3CH 2CH 2COOH (carboxylic acid)Target molecule : CH 3CH 2CH 2CH 2COOH (carboxylic acid)2 – Chain length The no. of C is increased by one.

3 – Redox N.A.

4 – Unsaturation No double or triple bond involved.5 – Intermediate Since the chain is lengthened by 1, nitrile should be the intermediate which can be hydrolyzed

to carboxylic acid.CH3CH2CH2CH2CNCH3CH2CH2COOH CH 3CH2CH2CH2COOH

Round 2 1 – Naming Starting molecule : CH 3CH 2CH 2COOH (carboxylic acid)

Target molecule : CH 3CH 2CH 2CH 2CN (nitrile)

2 – Chain length No change on the main C chain.3 – Redox Carboxylic acid is reduced.

4 – Unsaturation No double or triple bond involved.5 – Intermediate Carboxylic acid has to be reduced to alkanol first.

CH3CH2CH2CH2CNCH3CH2CH2CH2OHCH3CH2CH2COOH

Round 3 1 – Naming Starting molecule : CH 3CH 2CH 2CH 2OH (1º alkanol)

Target molecule : CH 3CH 2CH 2CH 2CN (nitrile)2 – Chain length No change on the main C chain.

3 – Redox N.A.

4 – Unsaturation No double or triple bond involved.5 – Intermediate –OH group is not a good leaving group. This makes alkanol not a good substrate for

nucleophilic substitution by CN - ion. –OH has to be converted to a good leaving group first,e.g. –Br.

CH3CH2CH2CH2CNCH3CH2CH2CH2OH CH 3CH2CH2CH2CH2Br

7 Overall conversion

LiAlH 4 H3O+KCN / H +PBr 3 CH 3CH 2CH 2CH2CNCH 3CH 2CH2CH 2OHCH3CH 2CH2COOH CH 3CH2CH 2CH2CH2Br CH 3CH 2CH2CH 2COOH




Example 2

(CH3)2CHCH 2COOH (CH3)2CHCH 2CH2 N+(CH3)3I

-

Round 1

1 – Naming Starting molecule : (CH 3)2CHCH 2COOH (carboxylic acid)Target molecule : (CH 3)2CHCH 2CH 2 N

+(CH 3)3I- (quartenary ammonium iodide)2 – Chain length No change on the main C chain.

3 – Redox Carboxylic acid is reduced.4 – Unsaturation No double or triple bond involved.

5 – Intermediate Quartenary ammonium salt can only be prepared from alkylation of amine, therefore, theintermediate may be a 1º amine, (CH 3)2CHCH 2CH 2 NH 2.

(CH3)2CHCH 2CH2 NH 2(CH3)2CHCH 2COOH (CH3)2CHCH 2CH2 N+(CH3)3I

-

Round 2 1 – Naming Starting molecule : (CH 3)2CHCH 2COOH (carboxylic acid)

Target molecule : (CH 3)2CHCH 2CH 2 NH 2 (1º amine)2 – Chain length No change on the main C chain.

3 – Redox Carboxylic acid is reduced.

4 – Unsaturation No double or triple bond involved.5 – Intermediate Amine has 3 possible precursors :

Nitrile (by reduction), amide (by reduction or Hoffman degradation) or haloalkane (byalkylation of ammonia)Since there is no change in the length of C chain, nitrile intermediate and Hoffman degradationof amide can be ruled out.1. (CH3)2CHCH 2COOH (CH 3)2CHCH 2CH2 NH 2(CH3)2CHCH 2CH2 Br

(Remark : Only 1º amine can be prepared by alkylation of ammonia)

2. (CH 3)2CHCH 2C

O

NH 2(CH 3)2CHCH 2COOH (CH 3)2CHCH 2CH 2 NH 2

Round 3 1 – Naming Starting molecule : (CH 3)2CHCH 2COOH (carboxylic acid)

Target molecule : 1. (CH 3)2CHCH 2CH 2 –Br (haloalkane)2. (CH 3)2CHCH 2CONH 2 (1º amide)

2 – Chain length No change on the main C chain.

3 – Redox 1. haloalkane – Carboxylic acid is reduced.2. 1º amide – Carboxylic acid is neither oxidized nor reduced.

4 – Unsaturation No double or triple bond involved.

5 – Intermediate 1. Carboxylic acid has to be reduced first before it can be converted to haloalkane.Therefore, the intermediate may be alkanol.(CH 3)2CHCH 2COOH (CH 3)2CHCH 2CH 2 Br (CH 3)2CHCH 2CH 2 OH

2. Carboxylic acid can be converted to amide through acid chloride intermediate. Because

ammonia is a base which forms a salt RCOO - NH 4+ with carboxylic acid, it does not form

amide with acid directly.

(CH3)2CHCH 2C

O

NH 2(CH3)2CHCH 2COOH (CH3)2CHCH 2C

O

Cl

7 Overall conversion

Path 1(CH 3)2CHCH 2COOH

LiAlH 4 (CH 3)2CHCH 2CH 2 Br (CH 3)2CHCH 2CH 2 OH PBr 3(CH 3)2CHCH 2CH2 N

+(CH3)3I-(CH 3)2CHCH 2CH2 NH 2excess CH 3Iexcess NH 3

Path 2

(CH 3)2CHCH 2COOH (CH 3)2CHCH 2COCl(CH3)2CHCH 2C

O

NH 2

SOCl 2 NH 3(CH 3)2CHCH 2CH2 N +(CH 3)3I-(CH 3)2CHCH 2CH 2 NH 2

LiAlH 4 excess CH 3I

Glossary organic synthesis retrosynthetic analysis target molecule precursor structural analysis




Past PaperQuestion

90 2C 9 b i ii91 2C 8 a b d93 2C 7 b i ii94 2C 8 a i ii95 1A 3 e ii 95 2C 8 b i ii iii98 2B 6 c i99 2B 7 b ii

90 2C 9 b i ii9b Using equations, show the reactions you would employ for the following conversions in the laboratory. Give the

reagent(s) for each step and structures of all intermediate compounds.i

CH2OHCCH 3

O

3

CH2OHC

CH3

O

2. H+(aq)

1. I2, NaOH COOHLiAlH 4

3 marksii

C CHCH2CH 2Br

3

C CHCH 2CH2Br NaOEt Br 2 NaOEt

Br

Br

3 marks

91 2C 8 a b d8 Using equations, show the reactions you would employ for the following conversions in the laboratory. Give the

reagent(s) for each step and the structures of all intermediate compounds.8a OH

3

HNO2

NH2[H+]

Sn, HCl

NO 2

conc. H 2SO4

conc. HNO 3OH

3 marks8b (CH 3)2CHCH 2COOH → (CH 3)2CHCH 2CH 2COOH 4

(CH 3)2CHCH 2COOH (CH 3)2CHCH 2CH2OH (CH 3)2CHCH 2CH2Br KCNPBr 3LiAlH 4 (CH 3)2CHCH 2CH2CN (CH 3)2CHCH 2CH2COOHH2O, H+

hydrolysis 4 marksC Some common mistakes were : ROH

KCN → RCN (without going through RX)

RXHCN

→ RCN (wrong reagent; should use KCN)8d

CH3CH2CH2 C NH 2

O

CH3CH2CH2CH2 N

CH3

CH3 3

CH3CH2CH2CH2 N

CH3

CH3CH3CH2CH2 C NH 2

OCH3CH2CH2CH2 NH 2LiAlH 4 CH3I

3 marksC

Some common mistakes were : CR NH 2

OZn / Hg

HCl RCH 2 NH 2 (wrong reagent)



Organic Synthesis Page 693 2C 7 b i ii7b Show how you would carry out the following conversions in the laboratory, giving the structures of intermediate

compounds and the reagents for each step.i

CH3C

O

OC

O

3

C

O

OC

O

CH 3 COOH COCl

COOH

PCl5MnO4- / OH -, heat

3 marksC Common mistakes revealed in the candidates' synthetic conversions were :

RCOOHheat

→ OC C

O O

R R (without dehydrating agent);

ii CH 3CH2CH 2COOH CH 3CH2CH 2CH2COOH

3LiAlH 4 H3O+

KCN / H +PBr 3 CH 3CH 2CH 2CH2CNCH 3CH 2CH2CH 2OHCH3CH2CH2COOH CH 3CH2CH2CH2CH2Br CH 3CH 2CH2CH2COOH 3 marksC Common mistakes revealed in the candidates' synthetic conversions were :

ROHKCN

→ RCN (without going through RX)

RCOOHLiAlH 4 → RCHO

94 2C 8 a i ii8a Show how you would carry out the following multi-step conversions. For each step, give the structure of the

intermediate compound and the necessary reagent(s).i CH2CH 2OH C CH

3

C CHCH 2CH2OHconc. H 2SO 4 Br 2 KOH / EtOHC C

Br Br

H H

H

3 marksC Common mistakes of the candidates included :

RCH 2CH 2OHalc NaOH .

→ RCH=CH 2

RCH=CH 2 → RC ≡ CH directlyii

CH2CH 2 O C CH 3

OCH 2OH

3

CNBr CH 2OH PBr 3

(or PCl 5)H3O+, heatKCN

COOHCH 2CH2 O C CH 3

O

Cl

O

LiAlH 4

OH

3 marksC Common mistakes of the candidates included :

ROHKCN or HCN

→ RCN



Organic Synthesis Page 795 1A 3 e ii3e Use equations to show how you would carry out the following conversions in the laboratory. For each

conversion, give the reagent(s), conditions and structure of the intermediate compound(s) formed.ii

CH 2=CH 2 to COCH 2CH 2OC

O O

3

COCH 2CH 2OCO O

CH 2 CH 2CH 2 CH 2

OH OH

MnO 4-

OH -

C Cl

O

3 marksC Some candidates did not state that concentrated sulphuric(VI) acid is needed for the acid-catalyzed esterification if

benzenecarboxylic acid is used.

95 2C 8 b i ii iii8b Each of the following conversions can be completed in not more than three steps. Use equations to show how you

would carry and each of these conversions in the laboratory. For each conversion, give the reagent(s), conditions,and structure of the intermediate compound(s)1 mark for reagents1 mark for intermediate

C All the conversions could be accomplished in two steps. Many candidates did not state the experimentalconditions for the transformations, e.g., reaction temperature, concentration of the acid or base etc. Commonmistakes were as follows :

i NO 2

CH 3

OH

CH 3

3

OH

CH 3

NO 2

CH 3

Fe or Zn in dil. HClSn in conc. HCl

LiAlH 4 Na or K in EtOHH2 with Pt or Ni CH 3

NH 2

NaNO 2, HCl

or HNO 20 - 5 °C

3 marksC HNO 3 instead of HNO 2 was given. Direct conversion of the nitrobenzene into the phenol without going through

the amine stage was erroneously given.ii

CH 3CH 2CH 2OHCH 3CHCH 3

OH

3

CH 3CHCH 3

OHCH 3CH 2CH 2OH

H3O+ or H + or H 2O, heat

or conc. H 2SO 4, H2O

conc. H 2SO 4 or Al2O3

heatCH 2CHCH 3

3 marksC NaOH was wrongly used as the dehydrating agent. Aqueous KOH or NaOH was wrongly used for the

dehydrohalogenation, but alcoholic metal hydroxide for the substitution reaction. Nucleophilic displacement of secondary alkyl halides is not of high preparative value because the reaction is generally low-yielding due thecompeting yet facile elimination. Heating is necessary in the dehydration of alkanol with concentrated H 2SO 4 andin the hydrolysis of alkene. Some candidates wrongly used H + as the reagent for acid hydrolysis instead of H 2O/H + or H 3O+. They did not realize that the reactant for the hydrolysis is water and H + is just a catalyst. The followingexamples should illustrate the point.

CH3CH CH 2 CH3CHCH 3

OHH2O / H +

CH3CH CH 2 CH 3CHCH 3

OCH 3CH3OH / H +

iii

CHO

NO 2

C CO 2H

OH

H

NO 2

3




C CO 2H

OH

H

NO 2

CHO

NO 2

NaCN / H +

or HCN

H3O+ or H+ / H2O

heat

C

OH

H

NO 2

CN

3 marksC Again H + was used as the reagent for acid hydrolysis of the nitrile group and the important condition, heating, was

omitted.

98 2B 6 c i6c 2-Ethanoyloxybenzoic acid (aspirin) is one of the most common substances used to relieve pain.

OCOCH 3

CO 2H

2-ethanoyloxybenzoic acid

i Outline a synthetic route for preparing 2-ethanoyloxybenzoic acid from 2-methylphenol.

99 2B 7 b ii7b With no more than four steps, outline a synthetic route to accomplish each of the following transformations. In

each step, give the reagent (s) used, the conditions required and the structure of the product.ii

CH 3O

C

OH

CH 3

HCH 3O

CH 2 NH 2



Organic Laboratory Technique

I. Purification of organic compound


B. Steam distillation

C. Chromatography



II. Use of quickfit apparatus


B. Different setup of quickfit apparatus

1. Reflux setup

2. Distillation setup

III. Testing for purity





Organic Laboratory Techniques Unit 1 Page 1

Topic Organic Laboratory Techniques Unit 1

ReferenceReading

29.1.1Modern Physical Chemistry ELBS pg.Chemistry in Context, 3rd Edition ELBS pg.Physical Chemistry, Fillans pg.

AssignmentReading

Syllabus solvent extraction

Notes I. Purification of organic compound


Technique of solvent extraction is commonly used to purify organic compound. It has anadvantage that no heating is required. This is particularly important to some heat sensitivecompounds which decompose easily upon heating.

Water and ether are the solvents commonly used. They are immiscible with each other.Moreover, ether is a very volatile solvent which can be evaporized by a warm water bath / atreduced pressure to obtain the pure solute dissolved in it.

Precaution : Since ether is very volatile, a pressure may build up inside the separating funnelwhile being shaken. The pressure should be released occasionally by openingthe tap, especially, at the beginning of the extraction process when there is a lotof air inside the flask.

Theoretically, in solvent extraction, a solute will partition between the two solvents.Practically, the solute is usually more soluble in one solvent than another.

For example, if a brown iodine solution in KI (aq) is shaken with ether. The brown aqueouslayer will turn to pale yellow while the ethereal layer will turn to deep purple. This is becauseiodine is much more soluble in ether than in KI (aq) .

The following are the solubilities of some organic compounds :

More soluble in aqueous (polar) solvent More soluble in organic (non-polar) solvent

Salt of an organic acide.g. sodium benzoate, sodium phenoxide

Organic acide.g. benzoic acid, phenol

Salt of an organic basee.g. benzenammonium chloride

Organic basee.g. benzenamine

Small alkanole.g. ethanol Large alkanole.g. phenol, decan-1-ol




Solvent extraction separates different components in asolution by their difference in solubility.

For an aqueous solution containing ethanol(more water soluble) and benzoic acid(more ether soluble), benzoicacid can be extracted by shaking the solution with ether.Eventually, the ether layer is evaporated to obtain pure

solid benzoic acid.

However, if the two solutes have similar solubility in both aqueous and organic solvent, the solutes must beconverted to other derivatives with different solubilityfirst.

For example, both benzoic acid and phenol are very soluble in ether but not very soluble in water. However, benzoic acid is more acidic than phenol, benzoic acid reacts with sodium carbonate but phenol does not. Benzoicacid and phenol in an ethereal solution can be separated by converting benzoic acid to sodium benzoate first. Thiscan be done by shaking the ethereal solution with aqueous sodium carbonate solution.

An insoluble organic acid caneasily be converted to asoluble salt by reacting withan alkali. For a relativelystrong acid, e.g. benzoic acid,a weak base, like sodiumhydrogencarbonate, is usuallyused to neutralized the acid.

For a very weak acid, e.g. phenol, a very strong base,like sodium hydroxide, must

be used.

After the separation have been done, the benzoic acidcan be regenerated by addinghydrochloric acid to sodium

benzoate. And the benzoicacid is then extracted withether again.

Glossary solvent extraction partition ethereal layer aqueous layer

Past PaperQuestion

90 1B 4 a91 1B 5 a i92 1B 5 b94 1A 3 a i96 1A 3 e97 1B 8 b98 1B 8 c99 2B 6 a ii



Organic Laboratory Techniques Unit 1 Page 390 1B 4 a4 A mixture contains equal amounts of X, Y and Z.

OH

X b.p. 182°C

NH 2

Y b.p. 184°C

CH3

Br

Z b.p. 184°C

4a Outline a chemical method to separate X, Y and Z from each other. 4

Two methods are possible:Method IDissolve the mixture in diethyl ether, place in a separating funnel and shake with dil. HCl and separate. The ether layer will contain X and Z, the aqueous layer will contain Y. 1 mark To aqueous layer add dil. alkali (NaOH), ether and separate - the ether layer on evaporation will yield the amineY . 1 mark To ether layer containing X and Z, add dil. alkali, and shake, separate. Ether layer will contain Z . 1 mark

Neutralize alkaline layer with acid (dil. HCl) and extract with ether which on evaporation will yield X . 1 mark

Method IIDissolve the mixture in diethyl ether using a separating funnel. Shake with NaOH and separate. 1 mark Acidify (HCl) the aqueous layer and extract with ether to obtain X . 1 mark Acidify (dil. HCl) the other layer (from NaOH extraction), shake and separate. The ether layer will contain Z .

1 mark Add NaOH or alkali to this aqueous layer and extract with ether. The ether layer contains Y . 1 mark Answers which separate by fractional distillation of X, Y, Z. (0 mark)If no solvent (-1 mark)If in doubt, ask ‘will separation succeed?”. If yes award mark, if maybe award ½ mark, if no then 0 mark.

C Some extraordinary answers were given (e.g. using Na or HNO 2, or not using any solvent in the separation process) showing absolutely no training in this basic technique.

91 1B 5 a i5a CH3

NH 2

X

CH3

NO 2

Y

The amine X, C 7 N9 N, may be prepared from Y, C 7H7 NO 2, by reaction with excess hot granulated tin andconcentrated hydrochloric acid.

i If, in this reaction, some Y remains unreacted, suggest a procedure to isolate pure X from the reaction mixture. 4The reaction mixture would contain X, Y, HCl and Sn which need to be separated. 1 mark Firstly remove Sn by filtration and Y by extraction with ether. 1 mark Followed by basification with NaOH of the aqueous layer and extraction with ether 1½ mark (Instead of ether extraction, distillation / steam distillation may be used.)Evaporation of ether gives X. ½ mark

C This kind of basic organic process is still not well answered.

92 1B 5 b5b An incomplete reaction of an amine with ethanoic anhydride gave a liquid mixture on cooling. Briefly outline a

procedure, using a separating funnel and appropriate reagent(s), to isolate the ethanoylation product.3

The mixture contains phenylamine, ethanoic acid, ethanoic anhydride and N-phenylethanamide.Use acid/base separation technique and hydrolyse all anhydride to acid.Procedure:Shake mixture with dil. HCl and ether. Remove ether layer, it contains N-phenylethanamide and ethanoic acid.All ethanoic anhydride have been converted to ethanoic acid with phenylamine forming a salt which is water soluble.Shake resultant ether layer with dil. alkali (or Na 2CO 3 or NaHCO 3). The alkali forms a salt with ethanoic acidwhich is water soluble. Remove ether layer which contains only N-phenylethanamide. In this way,

N-phenylethanamide is isolated from other parts present in the mixture. ½ mark for each underlinedC Use of separating funnel - very weak. This basic organic technique had obviously never been experienced by

students. Every student should do at least one separation. Most answers had no solvent (ether) for example.

Candidates can learn more about the separation of mixtures either through laboratory work or by the use of flowcharts.





Organic Laboratory Techniques Unit 1 Page 56a

Cl NH 2 Cl NH CCH 3

O


ii Based on the fact that the amino group of 4-chlorophenylamine is more basic than the amide group of N-(4-chlorophenyl)ethanamide, outline a procedure to separate a mixture of the two compounds.





ReferenceReading

29.1.2–29.1.5Modern Physical Chemistry ELBS pg.Chemistry in Context, 3rd Edition ELBS pg.Physical Chemistry, Fillans pg.

AssignmentReading

Syllabus Steam distillationChromatographyRecrystallizationFiltration and suction filtration

Notes B. Steam distillation (Refer to Phase Equilibria – Unit 4)

Setup of steam distillation

Steam distillation is used in the purification of organic compound which is insoluble in water. The presence of water lowers the boiling point of the mixture to avoid decomposition of the compound.

C. Chromatography (Refer to Phase Equlibria – Unit 5)

Paper chromatography is usually used to separate a mixture of compounds for identification purpose. Nevertheless,chromatography can also be used for separation purpose in larger scale. However, gas chromatography or columnchromatography would be used instead.





A solute starts to crystallize from a solution if the solution is saturated. The amount of solute that can be dissolvedin a solvent is depending on the amount of solvent and the temperature of the solvent. Therefore, it is possible tomake a solution saturated by reducing the amount of solvent (by evaporation) or by lowering the temperature(lowering the solubility by cooling).

The size of the crystal is depending on the speed of crystallization, not the method of crystallization. Although alarge crystal are usually prepared by evaporation, indeed, if a saturated solution is allowed to cool down veryslowly, a large crystal can also be obtained.

Recrystallization enhances the purity of a crystal. For example, a crystal containing 95% sodium chloride and 5%sugar can be purified by using a minimium amount of hot water to dissolve the crystal and recrystallizing thecrystal. The solution prepared would have a very high concentration of sodium chloride (95%) and a very lowconcentration of sugar (5%). Upon recrystallization, the solution will become saturated with sodium chloride butnot with sugar. The sugar will remain in solution form. And the purity of the sodium chloride crystal obtainedwould be higher, say (99.5%).


Filtration is used to separate an insolublesubstance from a liquid. However, thespeed of filtration may be very slow if thesolid to be filtered is very fine. The processcan be speeded up by using suction i.e.suction filtration.

The suction flask is connected to a filter pump to generate a partial vacuum. A filter paper is placed onto the Buchner funnelwith a few drops of solvent sticking thefilter paper in place. The pump is switchedon and the mixture to be filtered is placedin the Buchner funnel.

Suction filtration

When the filtration is done, it is important to disconnect the filter pump from the suction flask before switching off the water tap. Otherwise, the partial vacuum in the suction flask may draw the water from the tap into the flask andcontaminate the filtrate.

Glossary steam distillation chromatography gas chromatography column chromatographyrecrystallzation suction filtration filter pump Buchner flask




Past PaperQuestion

92 1B 5 a iii95 1B 4 b i

92 1B 5 a iii5a Ethanolylation (acetylation) of phenylamine (C 6H5 NH 2) may be carried out by refluxing for 1 hour with ethanoic


iii Suggest one method for the isolation of the product from the reaction mixture. 2I Throw the mixture into ice-water. The product with m.p. 114ºC which is insoluble in water will precipitate.

Filter and wash with cold water. 2 marksor II. Boil off all the reagent and by-product, leaving the high boiling (304ºC) product. 2 marks

95 1B 4 b i4b i Describe the experimental procedure by which you would recrystallize a sample of impure benzoic acid 3

Dissolve in minimium quantity of hot (boiling) water using boiling tube / conical flask. ½ + ½ mark Filter the solution while it is hot through fluted filter paper into beaker / conical flask. ½ mark Leave for crystals to form (scratch side, seed..) ½ mark Filter crystals off and wash with minimium amount of cold water. ½ mark Dry between filter papers until not damp ½ mark

C The recrystallization of benzoic acid requires the minimum quantity of hot water. Many candidates . confusedsolvent extraction and fractional distillation with recrystallization.





ReferenceReading

29.2Modern Physical Chemistry ELBS pg.Chemistry in Context, 3rd Edition ELBS pg.Physical Chemistry, Fillans pg.

AssignmentReading

Notes II. Use of quickfit apparatus

Quickfit apparatuses of the same set have identical junction size. All joints are finished with ground glass. Whendifferent pieces are connected, the junctions are air tight.

Names of different pieces quickfit apparatus

1. Pear shaped flask 2. Stillhead3. Liebig condenser 4. Screwcap adapter 5. Receiver adapter 6. Thermometer 7. Dropping funnel with Rotaflo tap8. Stopper 9. Air leak / steam inlet tube10. Round bottom flask 11. Air condenser / drying tube12. Sintered glass funnel13. Drying tube14. Pear shaped flask with angled side neck

15. Air leak / steam inlet tube16. Adapter with 'T' connection17. Screwcap adapter





CleaningSee that the ground surfaces are absolutely free from dirt before use.

Lubrication For many preparations ground glass joints can be used unlubricated, but for reactions at high temperature apply a

small amount of lubricant to the top portion of the cone only to ensure easily disassembly of the setup.

Methods of HeatingUse direct heating with a small flame whenever possible. Adjust the bunsen flame to a height of about 2 cm. Keepthe tip of the flame about 3 cm from the base of the flask so that heaing is, in effect, by an air bath.Smooth boiling can be induced, if necessary, by adding antibumping granules.

Clamping Supporting rather than clamping should always be the principle with assemblies of glassware. The clamping shouldnot be too tight since the glassware will expand upon heating. Where more than one clamp has to be used, a hint isalways to adjust and tighted the bosshead on the clamp, and finally on the stand.It is advisable is fix the flask to be heated first because its position is depending on the position of the bunsen

burner.

Seizure – sticking of joints Strong alkali should never be allowed to remain in contact with a joint face. Its attack on the matt surface is quiterapid and it left between two ground surfaces sticking them together very effectively.

Disassembly of the joints while apparatus is still warm and application of a little lubricant are recommended for prevention of seizure.

If the joints are stuck together, they may be separated by i) rocking the cone in the socket,ii) tapping the socket flange on a wooden surface,iii) heating the socket and not the cone in a localized

flame.

C. Different setup of quickfit apparatus

1. Reflux setup 2. Distillation setup




Glossary quickfit apparatus ground surface seizure

Past PaperQuestion

92 1B 5 a i

92 1B 5 a i5a Ethanolylation (acetylation) of phenylamine (C 6H5 NH 2) may be carried out by refluxing for 1 hour with ethanoic


i With the aid of a labelled diagram, show the laboratory apparatus required to carry out the above reaction. 2

½ mark for each piece of apparatusC Many did not know the difference between reflux and fractional distillation. Generally very poorly answered.

Experimental knowledge and labelling poor. It is obvious that few have used the apparatus in a laboratory.





ReferenceReading

29.3Modern Physical Chemistry ELBS pg.Chemistry in Context, 3rd Edition ELBS pg.Physical Chemistry, Fillans pg.

AssignmentReading

Syllabus Melting point determinationBoiling point determination

Notes III. Testing for purity


A pure substance has a sharp melting point but a mixturedoes not melts sharply. According to this behaviour, the

purity of a substance can be tested. Furthermore, the presence of impurity will also lower the melting point of themixture.

Example 1Pure ice melts at 0ºC but a mixture of salt and ice melts atabout -20ºC to 25ºC . This phenomenon is called depressionof melting point by an impurity.

Example 2Pure octadecan-1-ol melts at 58ºC and pure napth-1-ol meltsat 120ºC.

Experimentally , pure octadecan-1-ol will start to melts at57.5ºC and when the temperature reaches 58.5ºC, all solidwill be melted.If it is mixed with a little napth-1-ol, the mixture will start tomelts at 50ºC and will melts completely at 55ºC.

By knowing whether a substance melts sharply or not, the purity of the substance can be tested. A pure substanceonly has a very narrow melting range.





Unlike melting point determination, a sharp boiling point may not represent a pure sample. A constant boiling liquid may be an azeotropic mixture of twosubstances.

However, boiling point can be used as a kind of positiveconfirmation of the identity of an unknown sample.

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