Download - Chemistry Benzene

Hassan Subhie GCE A-level Chemistry Unit 5 1

Benzene the basis of aromatic compounds:

o The study of organic chemistry is divided into 2 main sections:

1. Aliphatic compounds: compounds based on carbon chains.

2. Aromatic compounds: compounds based on benzene ring structure.

Structure of benzene:

o The molecular formula of benzene is C6H6.

o The molecule of benzene has a regular hexagonal shape and is planar.

o The benzene ring doesn’t consist of alternate double and single bonds.

o All the bond lengths are the same are between single C-C and and double C=C.

o Each C-C bond in benzene is therefore neither a single nor a double bond but and

intermediate type lying between the 2 forms.

o Each C-atom in benzene forms 2σ bonds with 2 carbon atoms and 1 σ bond with

hydrogen atom and will still have one p orbital with single electron.

o These p orbitals overlap below and above the plane of the molecule forming a

continuous or deloecallised π system.

o This gives stability to the benzene structure which is why it undergoes substitution

reaction rather than addition reactions.

o Benzene can be represented either by:

Thermo chemical evidence for the structure of benzene:

o The enthalpy change of hydrogenation of cyclohexene is KJmol120

H2+

o The enthalpy of hydrogenation of benzene if it has 3 C=C double bonds is

KJmol360)120(3

Organic Chemistry III


3 H

2+ KJmol120

o The actual of hydrogenation of benzene is KJmol208 .

Benzene can’t have cyclohexatriene structure. It has a more stable structure

which is the delocalized structure.

o The extra stability of benzene gained by delocalization is equal to

360-208 =152 KJmol- which is called delocalization enthalpy or stabilization

enthalpy.

152 KJ

-360 KJ

-208 KJ

Cyclohexane

Naming compounds of benzene:

NO2

Cl

CH3

Benzene Nitro Benzene Chloro benzene methyl benzene

(Toluene)


C2H

5

NH2

OH

CH2Cl

Ethyl benzene phenyl amine Phenol Chloro methylbenzene

SO3H

CO2H

CHO

NO2

NO2

Benzene sulphonic acid Benzoic acid Benzyldehyde 1,4 dinitrobenzene

Benzene carboxylic acid

NO2

NO2

NO

2

NO2

CH2CO

2H

1,2 dinitrobenzene 1,3 dinitrobenzene Phenyl ethanoic acid

CO2H

C2H

5

NH2

CHO

Br

BrBr

2-ethyl benzoic acid 2-amino benzyldehyde 1,3,5 tri bromo benzene


COCH3

OCOCH3

Br

BrBr

OH phenyl ethanone Phenyl ethanoate 2,4,6 tri bromo phenol

Physical properties of benzene:

o Colorless liquid with boiling point 80

0C and melting point 5.5

0C

o Immiscible with water.

o It burns with a luminous smoky flame.

o It is toxic and its inhalation over a period of time leads to anaemia and leukemia.

Types of reactions of benzene:

o Benzene makes electrophilic substitution reaction rather than addition reaction

although the benzene ring is unsaturated.

Reason:

Benzene ring is thermodynamically very stable due to the delocalization of

the pi molecular orbitals.

Addition reactions to benzene would result in disruption of this

delocalization and so reduces the stability of the benzene ring.

Substitution reactions can occur with only temporary disruption and so the

stability of the benzene ring is maintained.

Reactions of Benzene:

Benzene reacts mainly by electrophilic substitution reactions.

1. Nitration: Benzene reacts with a mixture of nitric acid and sulphuric acid to form

nitro benzene.

Reaction: OHNOHCHNOHCSOH

225636642

HNO3

H2SO

4

NO2

H 2O

Nitro benzene

+ +


Reagent: concentrated HNO3

Condition: mix with concentrated H2SO4 at 50 0C

2. Bromination: Benzene reacts with liquid Bromine in the presence of a catalyst of

anhydrous iron(III) bromide.

Reaction: HBrBrHCBrHC l 56)(266

Br2

Br

H Br

Bromo benzene

+ +

Reagent: Br2 (l)

Condition: Liquid Br2 with iron catalyst.

3. Friedel-Craft alkylation: Benzene will react with halogeno alkanes in the

presence of a catalyst of anhydrous aluminium

chloride.

Reaction: HClRHCRClHC 5666

Reagent: Halogeno alkane RCl

Condition: Anhydrous AlCl3 catalyst.

E.g. HClHCHCClHCHC 52565266

C2H

5Cl

C2H

5

H Cl

Ethyl benzene

+ +

4. Friedel-Craft acylation: Benzene will react with acid chlorides in the presence

of a catalyst of anhydrous aluminium chloride.

Reaction: HClCORHCRCOClHC 5666

Reagent: Acyl chloride RCOCl

Condition: Anhydrous AlCl3 catalyst.


C

H ClRC//O

__Cl

//O

__R

+ +

E.g. HClCOCHHCCOClCHHC 356366

Phenyl Ethan one

C

H Cl//O

__Cl

//O

__CH

3

CH3C+ +

Reactions of compounds with a carbon containing side chain:

o Compounds such as methylbenzene C6H5CH3 and ethyl benzene C6H5C2H5 can be

oxidized, and the product will contain a COO- group on the benzene ring,

regardless of the number of carbon atoms in the chain. On acidification benzoic

acid is produced.

Reaction: 225656 3][6 COOHCOOHCOHORHC

R

6 [O] OH

_

_

COO

3 H

2O CO

2+ + + +

Reagent: Potassium manganate (VII) + NaOH

Condition: heat under reflux

E.g. 22565256 3][6 COOHCOOHCOHOHCHC


Phenols:

o These contain the –OH group connected directly to the benzene ring.

OH

Phenol

Naming derivatives of Phenols:

OH|

|Cl

OH|

CH3

OH|

|NO

2 4-chloro phenol 2-methyl phenol 4-nitro phenol

Br

BrBr

OH 2,4,6 tri bromo phenol

Physical properties of phenols:

o Colorless solid

o Its vapor is toxic

o Burns the skin

o Although it forms some hydrogen bonds with water, its only partially soluble and

forms two layers when added to water at room temperature.

o It is a weaker acid than carbonic acid, and so it will not liberate CO2 from sodium

hydrogen carbonate, unlike carboxylic acid which will.


o Acidity of Phenol

Phenol is weakly acidic dissociating in water

C6H5OH C6H5O- + H

+ Ka=1.3 x 10

-10 mol dm

-3

Phenol is too weak an acid to affect litmus paper or liberate carbon dioxide from

carbonates and hydrogencarbonates, but it dissolves readily in alkalis.

o It is soluble in alkali as NaOH, as it is a weak acid and reacts with NaOH to form

colorless solution.

)(2)(56)()(56 laqaql OHOHCOHOHHC

H 2O

OH

OH

_

_

O

+ +

Phenate ion

Reactions of Phenol:

o The presence of the OH group makes substitution into the benzene ring easier.

No catalyst is required.

1. With Bromine water:

Reaction: HBrBrHOHCBrOHHCaq

33 326)(256

OH| |

+3 Br

2+

|

OH

Br

Br

Br

3 HBr

2,4,6- tribromo phenol

Reagent: Bromine.

Condition: aqueous

Observation: When red-brown bromine in H2O is added, a white ppt. rapidly forms and

bromine decolorizes .


2. With acid chlorides: Phenol reacts as an alcohol, and an ester is formed.

Reaction: HClHRCOOCRCOClOHHC 5656

OH

RCOCl

|OOCR

H Cl+ +

Reagent: Acyl chloride.

E.g. HClHCOOCCHCOClCHOHHC 563356

Phenyl ethanoate

OH|OOCCH

3

H ClCH3COCl+ +

Phenyl amine :( Aniline)

o Phenyl amine has a NH2 group attached to the benzene ring.

|NH

2

Preparation:

o Nitrobenzene is reduced by tin and concentrated hydrochloric acid when

heated under reflux.

o Then sodium hydroxide is added to liberate the phenyl amine, which is

removed by steam distillation.

Reaction: Nitrobenzene→ Phenyl amine

OHNHHCHNOHC 2256256 2][6

Reagent: Tin and concentrated hydrochloric acid.


Condition: heat under reflux, and then add sodium hydroxide

Reactions of Phenyl amine:

1. Its weakly basic and reacts with acids to form salts.

ClNHHCHClNHHC 356256

2. Reacts with acyl chlorides and and benzoyl chlorides

HClNHCOCHHCCOClCHNHHC 3563256

Phenylethanamide

HClHNHCOCHCCOClHCNHHC 565656256

o These substituted amides are sometimes used as pharmaceuticals,

e.g. paracetamol.

NHCOCH

3

|OH

|

3. Reacts with nitrous acid to form a diazonium salt.

Reaction: Phenyl amine → diazonium salt.

Reagents: nitrous acid at 5 0C

Conditions: add dilute hydrochloric to phenylamine and cool the solution to

5 0C.Sodium nitrite solution is then added, keeping the temperature close

to 5 0C and above 0 0C

OHNHCNOHNHHC 22562256 22

|NH

2

+ +

|

______N N

2 H

2O+NO

2-H

2 +

+

Diazonium salt


o The solutions of diazonium ions react with phenol. A yellow precipitate of diazo

compound is obtained. This reaction is called coupling.

N+

OH__+N __N=N__ __OH

Azo dye

O Na+

N2Cl H

+

__N=N

__

HO

aromatic diazonium chloride

red precipitate

+

+

o The main use of this reaction is to produce synthetic dyes some acid indicator.

Questions:

1. State the structural formulae of the organic product obtained by the reaction of:

a) Propylbenzene with alkaline potassium manganate (VII).

b) Phenol with sodium hydroxide solution.

c) Phenol with propanoyl chloride

2. What would you observe if bromine water were added to aqueous phenol?

3. State the conditions for the conversion of nitrobenzene to phenylamine.

4. (a) Describe the mechanism for the reaction of bromine with benzene .

(b) Why does benzene undergo a substitution reaction with bromine rather than an

addition reaction?

5. In the reaction between propanone and hydrogen cyanide no reaction occurs unless a

small amount of a base such as sodium hydroxide is added. Explain these observations.

6. When benzenediazonium chloride is prepared, the reaction is carried out at about 5 0C.

(a) Why are these conditions chosen?

(b) What is the formula of the product obtained by the reaction of the benzene

diazonium chloride solution with phenol in alkaline solution?


7. Robert Lister was the first to use phenol as an antiseptic, saving the life of a young

boy who had been in a traffic accident. At that time phenol was known as “carbolic

acid”.

(a) (i) Draw the structural formula of phenol.

[1]

(ii) When phenol is added to water, a weakly acidic solution is formed. Write an

equation explaining this observation.

[2]

(b) Phenol reacts with bromine to make tribromophenol, which is analogous to the

commercially available antiseptic trichlorophenol (TCP). Write the equation for the

reaction between bromine and phenol.

[2]

(c) Vinegar (ethanoic acid solution) also has medicinal properties. Explain how a

solution of sodium hydrogencarbonate may be used to distinguish between a solution

of vinegar and one of carbolic acid.

[2]

8. (i) What is observed when phenol reacts with bromine water

[2]

(ii) Write a balanced equation for the reaction and name the organic product.

[3]


Mechanisms of Reactions:

Reaction mechanisms tell us:

1. Which bonds are broken and how they are broken.

2. Which bonds are made and in what order they are made.

A curly arrow ( ) represents the movement of a pair of electrons, either from a

bond or a lone pair.

A half-headed arrow ( ) represents the movement of a single electron.

Types of bond fission:

o There are 2 types of bond fission:

1. Homolytic bond fission: When a bond breaks evenly and one electron

moves to each atom forming a free radical.

A__

B A B +

2. Heterolytic bond fission: When a bond breaks unevenly and the 2

electrons go to one atom forming a nucleophile

and an electrophile.

A B

__A

+B:+

Note: Ensure that the curly arrow starts from a bond or an atom with alone pair of

electrons.

Ensure that the curly arrow points towards an atom either forming a negative ion or

a new covalent bond.

Types of mechanisms:

I. Homolytic free radical substitution

1. Reaction between an alkane and chlorine or bromine.

e.g. reaction of CH4 and Cl2.

There are three stages in this type of mechanism.


1st

Initiation: Light energy causes homolytic fission of Cl2.

Cl

2 Cl

2nd

Propagation: each propagation step involves a radical reacting with a molecule to

produce a new molecule.

.CH4

Cl CH3 H Cl+ +

. Cl CH

3then : Cl2

CH3Cl+ +

3rd

Termination: Involves two radicals joining with no radicals being produced.

. Cl CH

3CH

3Cl+

Cl Cl

2Cl +

.CH

3.CH

3CH

3CH

3+

2. Homolytic free radical addition: ( Homolytic polymerization)

o The conditions are very high pressure (100 atm) and a trace of O2.

o The conditions produce radicals R , which attack the ethene molecule.

Initiation: .CH

2=CH

2R

R CH

2CH

2+

Propagation:

.CH2=CH

2 R CH2

CH2

.R CH2

CH2 CH

2CH

2__+

etc….

Note: Another way of polymerization of ethene is by using titanium tetrachloride and

aluminium tri ethyl, but this is not a hemolytic mechanism.

II. Heterolytic electrophylic addition.

Reaction of alkenes with halogens or hydrogen halides.

The reaction takes place in 2 steps:


1. The electrophile accepts the electrons to form a new bond with one of

the carbon atoms and the same time π the Br-Br bond breaks.

2. The intermediate carbon cation then bonds with Br- ion.

C C

BrBrBr

Br

CC

Br:Br-

CC

BrBr

+

With HBr:

C C

Br

CC

:Br-

CC

Br

H

H H

+

Note: With the addition of a hydrogen halide to unsymmetrical alkene, the hydrogen atom adds

to the carbon which already has more hydrogen atoms directly bonded to it.

This is because:

30 carbocation is more stable then 20 carbocation then 10 carbocation.

C C C

C____|

C C C C C C

most stable least stable

> >+ +

+


III. Heterolytic, electrophilic substitution:

In general

E+

+

HE

E

+ H+

+

E+: is the electrophile which could be NO2

+, Br

+ C2H5

+ or RCO

1. Nitration of benzene:

i. H2SO4 is a stronger acid than nitric acid and so protonates it

432342 HSONOHHNOSOH

2232 NOOHNOH

ii. The NO2+ is the electrophile and attacks the benzene ring.

iii. The HSO4- ion pulls off a H

+ and reforms H2SO4

+

H

+

NO2+NO2+ NO2

NO2

HSO4_

H2SO4

The loss of H+ results in the reforming of the benzene ring and the gain in

stability associated with the ring.

2. Bromination of benzene:

i. The catalyst FeBr3 reacts with Br2 to form the electrophile

Br+

ii. Br+ attacks the benzene ring

iii. FeBr4- pulls off a H

+ and reforms FeBr3

BrFeBrFeBrBrBr 43


+

H

+

Br+ Br Br

FeBr4-

HBr FeBr3+

3. Friedel’s craft alkylation:

452352 AlClHCAlClClHC

+

H

+ +

C2H5+ C2H5

C2H5

AlCl4-

HCl AlCl3

IV. Heterolytic, nucleophilic substitution:

Reaction between the halogeno alkanes and hydroxide or cyanide or cyanide ions

There are two mechanisms depending on the type of halogenoalkanes.

With primary (10) halogenoalkanes, SN2 mechanism is dominant.

Reaction proceeds via a transition state when the lone pair of electrons on the

nucleophile attacks the halogeno alkanes.

CHO:

_

Br

CH3

HH

H

CHO Br

H

CH3

_

C

HH

CH3

HO

With tertiary (30) halogenoalkanes the SN1 mechanism is dominant.

This happens in 2 steps:

1. The halogeno alkane ionizes in the relatively slow rate-determining step to

form an intermediate carbocation.


C___ ___

|

|CH

3

CH3

CH3

Br C___

|

|CH

3

CH3

CH3

:Br_

+ +

2. This rapidly forms a bond with the nucleophile OH-

C___

|

|CH

3

CH3

CH3

OH.. _

C___

|

|CH

3

CH3

CH3

___OH Br:+ +

V. Heterolytic, nucleophilic addition:

Reaction between carbonyl compound and hydrogen cyanide.

Reaction between carbonyl compound and hydrogen cyanide.

Although the reaction is the addition of HCN, the 1st step is the nucleophilic

attack by the CN- ion the carbon atom in the carbonyl compound.

The negatively charged oxygen in the anion then removes an H+ from the HCN

molecule. This produces another CN- ion, which reacts with another carbonyl

group and so

C//

R ___

H

O

CN:_

CR

O

H______

|

|CN

_:

H___

CN

CR

O

H______

|

|CN

___H

CN_

+

The conditions of this reaction are either HCN and trace of base or KCN and

small amount of dilute H2SO4.

Note: -Any optical activity is maintained in a SN2 reaction

-If the mechanism is SN1, a reactant which is an optical isomer will produce a

racemic mixture as carbocation can be attacked from either side.


Summary of the reactions of aromatic substances:

benzene

C6H

6

concHNO3

conc H2SO

4

C6H

5NO

2

Br2

Fe catalyst

C6H

5Br

C2H

5Cl

AlCl3 catalyst

C6H

5C

2H

5

CH3COCl

AlCl3 catalyst

C6H

5COCH

3

Phenol

C6H

5OH

OH_

(aq)

Br2 (aq)

CH3COCl

C6H

5O

_

C6H

2Br

3OH

CH3COOC

6H

5

H+

(aq)

HNO2

PhyenylamineC

6H

5NH

2

C6H

5NH

3+

C6H

5N

2+

At5 CO


Tests for alkenes or unsaturation: (C=C) Test: Shake the alkene with red-brown bromine solution

Observation: it gets decolorized.

Test: Shake with cold alkaline purple potassium manganate (VII),

Observation: it rapidly goes green

Test for Halogeno alkanes:

Test: Warm the halogeno alkane under reflux with aqueous sodium

hydroxide then acidify with dilute nitric acid to remove excess of OH- ions

then add aqueous silver nitrate.

Observation: If white ppt. forms which is soluble in dilute ammonia Chloro

alkane

If creamy ppt. forms which is partially soluble in dilute ammonia

Bromo alkane

If yellow ppt. forms which is insoluble in concentrated ammonia

Iodo alkane

Test for OH group:( in alcohols and acids)

Test: Add dry PCl5

Observation: Steamy acidic fumes of hydrogen chloride are produced (HCl)

Test for the type of alcohol:

Test: Warm with dilute sulphuric acid and potassium dichromate (VII)

solution.

Organic Tests


Observation: -Primary (10) and secondary (20) alcohols will reduce the orange

dichromate ions to green.

-Tertiary (30) alcohols don’t affect the color

To distinguish between 10and 20 repeat the experiment and distil the product

into ammonical silver nitrate solution:

10 alcohols are oxidized to aldehydes, which give silver mirror

20 alcohols are oxidized to ketones, which don’t react.

Test for Carbonyl group C=O: (Aldehyde and Ketone)

Test: Add solution of 2,4-dinitro phenyl hydrazine (Brady’s Reagent).

Observation: An orange or red precipitate is seen.

Test for CHO group aldehydes:

There are two different tests to differentiate between aldehydes and ketones:

Test 1: Add ammonical silver nitrate solution and warm.

Observation: silver metal is precipitated in the form of a mirror on the side of a

test tube with aldehydes , with ketones no reaction.

Test 2: Add Fehling’s or Benedict’s solution and warm.

Observation: A red precipitate of copper (I) oxide is formed with aldehydes, no

reaction with ketones.

Iodoform reaction:

Test: an organic compound is added to a mixture of iodine and dilute NaOH

(or a mixture of NaI and NaOCl).

Observation: a pale-yellow precipitate of iodoform, CHI3

This test works with compounds containing the CH3C=O and

alcohols containing the CH3CH(OH) group.


Test for the carboxylic acids, COOH group:

Test: Add Na2CO3 or NaHCO3 solution.

Observation: Colorless gas is evolved which turns limewater milky.

In synthesizing organic compounds, chemists normally chose the shortest possible

route.

Organic reactions never give 100% yield.

The more the number of stages in the synthesis of a compound, the lower is the %

yield.

E.g. To prepare butanol from propanol 2 routes are possible.

Route 1:

propanol

OHKHSOBrHCSOHKBrOHHC 24734273

OHHCNHHCCNHCBrHCHClNaNOLiAlHKCN

94273737324

toxic reagent Expensive reagent Very low yield

Route 2:

propanol

OHKHSOBrHCSOHKBrOHHC 24734273

C3H

7Br step

1Mg/dry ether

C3H

7MgBr

step 2

methanol followed by hydrolysisC

4H

9OH

Step 1 gives excellent yield

Step 2 gives good yield.

Route 2 is better as it gives better yield.

Ex. 1. How can you convert propan-2-ol into 2-methyl propan-2-ol ?

2. How would you make hexan-3-ol from propan-1-ol using no other

reagents except solvents?

Organic Synthesis


Answer:

1.

propan-2-ol

[O]propanone

CH3MgBr

CH3C CH

3

CH3

OMgBr

H+

CH3C CH

3

CH3

OH

2.

CH3CH

2CH

2OH

Propan-2-ol

K2Cr

2O

7 H+

/

CH3CH

2CHO

KBr H2SO

4/CH

3CH

2CH

2Br

Mg/dryether

CH3CH

2CH

2MgBr

H+

(aq)

CH3CH

2CH(OH)CH

2CH

2CH

3

Hexan-3-ol

Main groups of synthetic reactions:

1. No change in carbon skeleton.

2. If number of carbon atoms in the chain increases by one carbon

The step could be:

-Halogeno alkanes +KCN

-Carbonyl+HCN

-Gringard reagent CH3MgBr with a carbonyl compound

-Gringard reagent with CO2 or with methanal.

3. If number of carbon atoms in the chain increases by more than one carbon

-Gringard reagent with more than one carbon with carbonyl compounds.

-Friedel- craft reaction for aromatic substances.

4. If the number of carbon atoms is decreases by one.

-Hoffman’s degradation of amides with NaOH and Br2

222 CONaBrRNHNaOBrRCONH


-The iodoform reaction of methyl ketone and methyl secondary alcohols

RCOCH

3

X2 / NaOH

RCOOH

-The oxidation of the aromatic side chain in alkaline KMnO4

C

6H

5C

2H

5

KMnO4 / OH

-

C6H

5COOH

No Change in the carbon Skeleton:

Reactions of this type include

1. Simple substitution reactions involving non-carbon containing groups.

e.g. OH , Br , I , NH2 ………..

2. Oxidation of alcohols, aldehydes, reduction of aldehydes, ketones, nitriles and

carboxylic acids.

3. Elimination yielding C=C double bonds, amides and nitrilels, esters formation

and hydrolysis.

Example:

RCH2CH

2OH

PCl5

RCH2CH

2X

NH3 Ethanol/

RCH2CH

2NH

2

K2Cr

2O

7/ H

+

RCH2CHO

K2Cr

2O

7/ H

+

RCH2COOH

NaNO2 / H Cl


RCH2CH

2OH

K2Cr

2O

7 / H+

conc. H2SO

4 RCH=CH2

H2SO

4/H

2O

RCH(OH)CH3

RCOCH3

LiAlH4

/dry ether

RCHXCH3

HX

KOH/ethanol

RCH(NH2)CH

3

NH3/ethanol

KOH(aq)

NaNO2/HCl

conc. H2SO

4

Exercise: How would you convert?

1- bromo propane into 2- bromo propane

Number of carbon atoms in the skeleton is increased by one or more than one:

RCH2X RCH

2MgX

RCH2COH|

|R

R,,

,

RCOR

,,,

HCHO

RCH2CH

2OH

CO2

RCH2COOHRCH

2CN

KCN

H+

(aq)

LiAlH4 /ether

RCH2CH

2NH

2

HNO2


Problems: 1. How would you convert proapn-2-ol into?

a. 2,3 dimethyl butan-2-ol

b. Propene.

2. How would you convert phenyl ethanone into

a. Benzoic acid

b. Phenyl amine

c. 2,4,6 tribromo phenol

Polymerization:

There are two types of polymers:

1. Addition polymers: The monomers contain one or more C=C groups. In the

formation of these polymers the pi bond in the

monomer is broken.

E.g. poly (ethene), Poly (propene), poly (vinychloride) and poly (tetra fluorine)

o The monomer and the polymer have the same empirical formula.

2. Condensation Polymerization: Both the monomers have two functional

groups, one at each end. Polymerization

involves the loss of a simple molecule (usually

H2O or HCl) as each link forms.

o Examples are nylon (a polyamide) and terylene (a polyester)

Nylon: It consists of two types of monomers

(i) di carboxylic acid HOOC-----COOH and diamine H2N -----NH2

(ii) di acid di chloride ClOC-----COCl and diamine H2N -----NH2

C

O

___OH -----------C\ \

O

___ OH//

C__ C------- ____N H

|H

N

|__

H

H

C

O

___-----------C\ \

O

___//

C__ C------- ____N

|H

N

|H

+

C

O

___-----------C\ \

O

___Cl

//C__ C------- ____N H

|H

N

|__

H

H

C

O

___-----------C\ \

O

___//

C__ C------- ____N

|H

N

|H

Cl +


(CH2)4

______CC

ClCl

OO

(CH2)6________ NN

H

HH

H

di acid chloride di amine

H Cl

(CH)4

___ ___C||

C||

______N|

(CH2)6

___N|

n

OO H H

__________

+

Tyrelene: It consists of two types of monomers

(i) di carboxylic acid HOOC-----COOH and diol HO -----OH

(ii) di carboxylic acid dichloride ClOC-----COCl and diol HO -----OH

C

O

___ O

H

-----------C\ \

O

___Cl

//C__ C------- ____ H|H

|__

H

H

C

O

___O

H

-----------C\ \

O

___//

C__ C------- ____

H

O

H|

H

OCl|

|| ||H

+

C

O

___ O

H

-----------C\ \

O

___//

C__ C------- ____ H|H

|__

H

H

C

O

___O

H

-----------C\ \

O

___OH

//C__ C------- ____

H

O

H|

H

O|

|| ||H

HO +

____COOHHOOC HOCH

2CH

2OHN N

______

||C||

C___

O___

CH2CH

2

___O

n

O O

________

+


Synthetic polymers are not easily biodegraded and can cause an environmental

problem of disposal. One answer is for them to be recycled.

Polymers usually soften over a range of temperature rather than have a sharp

melting temperature .The reason for this is that the polymer is a mixture of

molecules of different chain lengths.

Practical techniques:

1. Recrystallisation: It is a technique used to purify solids.

Steps: - Dissolve the solid in a minimum of hot solvent.

- Filter the hot solution through a preheated funnel using a fluted

filter paper.

- Allow to cool.

- Filter under reduced pressure using Buchner funnel.

- Wash with a little cold solvent and allow to dry.

- The solid should have a sharp melting point.

2. Mixing:

Reactions of organic chemistry often use immiscible liquid reactants and its thus

necessary to shake or stir the mixture.

Shaking is done in a stoppered flask for reactions that don’t build up pressure or

in flask fitted with condenser for reflux.

When there is difference in densities of immiscible liquids as in nitration of

benzene or reduction of nitrobenzene, vigorous shaking is the only method of

bringing the reactants together.

3. Temperature control:

Most organic reactions are relatively slow and therefore heating is needed for the

reaction to occur. This is done on a steam bath or by using an electric heating

mantle.

Since many organic reactants are volatile and heating causes them to escape , thus

heating is done under reflux.

4. Heating under reflux:

This process boiling under reflux is used in organic reactions where the reactions

are often slow and reagents are volatile.


The process consists of a condenser mounted vertically on top of the reaction

vessel so that any vapors escaping during the heating process will condense to

liquid and run back into the flask.

By this way the heating process can be carried out for a longer period.

1

2

3

45 6

7

8

9

1 10

2

3

45 6

7

8

9

11

5. Safety precaution:

No naked flame when chemicals are flammable i.e. use water bath or electric

heating mantle.

Do the reactions in a fume cupboard when a reactant or a product is toxic ,irritant

or carcinogenic.

6. Simple distillation:

Used to separate chemicals that have big difference in their boiling points.

E.g. bromoethane has boiling point 38 oC and ethanol boiling point is 78

oC.

The liquid with the lower boiling point distils first.


1

2

3

45 6

7

8

9

1 10

2

3

45 6

7

8

9

11

7. Fractional distillation:

This is used to separate a mixture of 2 liquids.

This mixture can usually be separated into pure samples.

But if the boiling points are too close, a good separation will be difficult.


Temperature /composition diagram:

If a liquid of composition X is heated, it will boil at T1

oC to give a vapor of

composition Y. If this is condensed and reboiled, it will boil at a temperature T2 oC

giving a vapor of composition Z. Eventually pure B will distill off the top and pure A

will be left in the flask.

8. Measuring melting point:

The solid is crushed on a clean porous tile in order to put it in a capillary tube.

Capillary tube is attached to a thermometer and placed in water or oil.

The liquid is heated gently and stirred continuously.

The temperature when the solid starts to melt is recorded.

Heating is continued and temperature at which the solid is completely melted is

recorded.

The difference between the temperatures should be ± 3 oC or less.


Errors in the method:

- Solid may be impure/damp.

- Heating is too rapid.

- Water is a poor conductor and therefore temperature of solid is different to

that on the thermometer.

- Insufficient stirring.

- Difficult to see solid and thermometer together.

- Difficult to decide when melting starts.

- Melting point not sharp.

Improvement for the method:

- Purify the solid.

- Use a heating block (Thiele tube)

- Stir more or heat more slowly.

- Use a magnifying glass.

- Repeat and take the average temperature.

9. Measuring boiling point:

- The liquid is placed in an ignition tube and the tube is fixed to the thermometer.

- Place a capillary tube with a closed end inside the ignition tube with the opened end

inside the liquid.

- Clamp the thermometer in a beaker of water or oil.

- Heat the water gently with stirring until rapid stream of continuous are produced.

Record the temperature.

- Cool the water bath and record the temperature at which bubbles stop escaping.

- The difference between the 2 temperatures should be less than or equal to 5 oC.


Errors in the method:

- Can’t see the temperature together.

- Heating is too rapid.

- No enough stirring.

- Water is a poor conductor and therefore temperature of the liquid is different to that

on the thermometer

Improvement for the method:

- Use fractional distillation to observe the steady boiling point when the boiling occurs.

Applied Organic chemistry:

Targeting pharmaceutical compounds:

Pharmaceutical compounds which are ionic or have several groups that can form

hydrogen bonds with water, will tend to be retained in aqueous (non fatty) tissues.

These water-soluble compounds will be excreted from the body.

Compounds with long hydrocarbon chains and with few hydrogen-bonding

groups will be retained in fatty tissues. These compounds will be stored in the

body.

Chemists chemically modify the structure of pharmaceuticals, in order to change

the solubility and retention time of the material in the body.

Nitrogenous fertilizers:

There are three types of fertilizers:

1. Quick release, such as those containing NO3

- and NH4

+

2. Slow release, such as urea, NH2CONH2

3. Natural, such as slurry, compost and manure.

The first two are soluble and can be leached out. The last is bulky and contains

little nitrogen.

Esters, oils and fats:

Simple esters are used in food flavorings, perfumes and as solvents for glues,

varnishes and spray paints.

Fats are esters of propan-1, 2,3-triol and saturated acids such as stearic acid

C17H35COOH.


When fats are hydrolyzed by boiling them with aqueous sodium hydroxide, they

produce soap which is the sodium salt of the organic acid.

Vegetable oils are liquids, which are esters of propan-1, 2,3-triol and

unsaturated acids.

Margarine is made from polyunsaturated vegetable oils by partial

hydrogenation (addition of hydrogen) so that only a few double bonds remain

Questions:

1. How would you distinguish between:

a. Pentan-3-one and pent anal

b. 2-bromo propane and 2-chloro propane

c. Methyl propan-2-ol, methyl propan-1-ol and butan-2- ol?

2. How would you prepare from propene:

i. 2-bromopropne

ii. propanone

iii. 2,3-dimethyl butn-2-ol?

3. How would you make

i. Phenol

ii. Benzoic acid

iii. Phenyl benzoate,

Using benzene and methylbenzene as only organic materials?

4. How would you convert oleic acid CH3 (CH2) 7CH=CH (CH2) 7CO2H into

a. Stearic acid, CH3 (CH2) 16CO2H

b. Plamitic acid, CH3 (CH2) 14CO2H

c. Octadeca-8, 10-dienoic acid CH3 (CH2) 6 CH=CH (CH2) 6 CO2H?

5. Starting from butan-1-ol and methanol as only organic materials, how would you

make

i. Butanone

ii. Propanoic acid

iii. Methyl butanoate

iv. Propylamine

v. Butene

vi. 2-methylbutne?


Interpret simple fragmentation patterns from a mass spectrometer

An organic compound produces ions in a mass spectrometer. The ions generate

pulses of electric current which are sent as signals to a computer (or chart plotter)

to be displayed as a mass spectrum.

On the spectrum the large peak on the right is the parent molecular ion and this

indicates the relative molecular mass of the compound.

In the spectrometer the molecules are fragmented into positive ions which form a

pattern which depends on the structure of the molecule. e.g. ether CH3OCH3

Example 1: Another compound of relative molecular mass 46 also contains carbon,

hydrogen and oxygen has the mass spectrum below. Identify each fragment

and the structural formula.


Problem 2: Sketch the mass spectrum of propanal showing the masses of each fragment

Interpret simple infrared spectra

The bonds in organic molecule absorb infra-red radiation. This happens when the

frequency of the radiation matches the natural frequency of vibrations in the bonds.

These might be stretching, or bending vibrations.

A spectrometer shines infra-red light at a sample of an organic material and measures

how much of the light is absorbed. A measure of the frequency (wave number) is

displayed in the spectrum. Each bond has its own frequency (wave number) and this

can be used to identify the bonds present in a compound.

bond wavenumber/cm-1 seen on spectrum

C-H 2840 - 3095

C-C 1610 - 1680

C=O 1680 - 1750

C-O 1000 - 1300

C-Cl 700 - 800

O-H 3233 - 3550

2500 - 3300

N-H 3100 - 3500

Infra-red spectrum for propanone

o Is substance A in the infra-red spectrum below most likely to be ethyl

ethanoate or butane?


5.5a(iii)(c) Low resolution nuclear magnetic resonance spectra (NMR) Hydrogen atoms can be detected using this sort of spectrometry. The nucleus

of a hydrogen atom, the proton, spins and so has a magnetic moment. This can

be aligned or not aligned with a magnetic field . When electromagnetic

radiation of the right frequency is applied resonance occurs and the protons flip

from one state to the other and absorb energy. This absorption of energy is

used to detect protons in organic compounds. The exact resonance frequency

for a proton (hydrogen atom) depends on its environment. For example the

frequency is different for hydrogen atoms in CH3, CH2 , C6H5- and in O-H.

Trimethylsilane, TMS, is used as a standard. The distance in the spectrum

from the TMS peak is called the chemical shift.

Type of proton Chemical Shift (ppm)

R-CH3 0.9

R-CH2 1.3

R-CH2-O- 4.0

C6H5- 7.5

-O-H 5.0

-CHO 9.5


To which functional groups do the protons in the following NMR spectrum

belong? Identify the compound.

Sketch an NMR spectrum for propanal.

An excellent guide to low resolution nmr.

5.5a (iii) (d) The interpretation of simple ultra-violet/visible spectra.

Some chemical structures absorb electromagnetic radiation in the ultra violet

part of the spectrum. These include conjugated (contain alternate double and

single bonds) dienes. E.g. 1,3-butadiene. The ultraviolet absorption spectrum

http://www.chemguide.co.uk/analysis/nmr/lowres.html


for 2,5-dimethyl-2,4-hexadiene is shown below.

Ultra-violet wavelengths are from about 200nm to about 400nm. Visible light

has wavelength between 400nm and 800nm. -carotene, which gives carrots

their orange colour absorbs at 497nm. Lycopene, which gives tomatoes their

red colour, absorbs at 505nm. Both of these compounds have 11 conjugated

double bonds.