AL-CHEM Chemistry of Carbon Compounds(97-02)

AS/AL-CHEM (1997-2002) Part XI Chemistry of carbon compounds

1

NOTE: Questions with an * are NOT required by the HKDSE syllabus. They are included so as to enrich your knowledge and broaden your understanding in chemistry.

1997-AL-CHEM 1

5. (b) Consider the following reactions:

CH3CH2CH2OHCH3CH2COCl L

conc.H2SO4heat J

HBr K

Na2Cr2O7H3O

+ M N(a red precipitate)

NHNH2

NO2

NO2

Give structures for J, K, L, M and N. (5 marks)

Mark Scheme .............................................................................................................................................. 5. (b) J : CH3CH=CH2 1 K : CH3CHCH3

Br

1

L : CH3CH2COOCH2CH2CH3 1

M : CH3CH2CHO 1

N :

NC NH NO2

O2NH

CH3CH2 (a condensation reaction by elimination of H2O: equation DSE)

1

Please identify the types of reactions involved in forming J, K, L and M.

[5]

1997-AL-CHEM 2

5. (b) The following equation represents the acid hydrolysis of a dipeptide D to produce compounds E and F, one of which is a chiral compound.

H2NCHC NHCH2CO2H

H3C OH3O+ E + F

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.

(iii) Suggest a method to separate E and F from the reaction mixture. (5 marks)

5. (d) Give a systematic name to each of the following compounds:

(ii) CH3CH2 C CH(CH3)CH2CH3O

(iii) CC

HHO2C

CH3H

(2 marks)

Served as a chemical test for carbonyl group

EEEExam Practice Functional groups Do Brilliantly

2

Marking Scheme Mark

5. (b) (i) amino group / amine amide group / peptide linkage 1

carboxyl group / carboxylic acid (deduct marks for each extra functional group) (ii) E:

H2N CH CO

OHCH3

and F: H2NCH2CO2H (Also accept protonated forms of the amino acids) ,

Stereo-structure of E: CCH3

NH2 CO2HH (L or D isomer)

1

(iii) paper chromatography / thin layer chromatography / column chromatography / electrophoresis 1 [5] 5. (d) (ii) 4-methylhexan-3-one 1 (iii) trans-but-2-enoic acid / E-but-2-enoic acid 1 [2]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (a) The following conversion 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 structure of the product.

NHCH2NH2(ii)

(3 marks)

6. (c) Compound H, C3H6O2, does not react with NaBH4 and displays the following infra-red spectrum. Deduce all possible structures of H.

(4 marks) Mark Scheme .............................................................................................................................................. 6. (a) (ii)

NOTE: Step 1 amide formation (in basic medium); Step 2 Reduction of amide to give amine (in dry ether)

3


3

6. (c) IR peak at ~ 1750 cm-1 indicates the presence of a carbonyl group (C=O) , No broad OH absorption band implies that H is not a carboxylic acid

No reaction with NaBH4 implies H is not ketone or aldehyde / may be an ester

Possible structures of H:

CH3 CO

OCH3 H CO

OCH2CH3;

1+1

(Deduct 1 mark for each extra structure.) [4] NOTE: NaBH4 is a milder reducing agent compared with LiAlH4. It is more selective in use and can only

reducing carbonyl compounds to alcohols but not carboxylic acids (and their derivatives, e.g. esters).

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) 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).

(i) and

(iii) C2H5 C O(CH2)2CH3O

and C2H5 C CH2OC2H5

O

(iv) C2H5 C C2H5O

and CH3(CH2)3 C HO

(9 marks) Mark Scheme .............................................................................................................................................. 7. (a) (i) Treat compound with Br2 (in CCl4 or in CH3CCl3) 1

rapidly decolorizes the red-orange Br2 solution while does not ,

+ Br2CCl4

Br

Br 1

(Also accept Baeyers test or using acidified potassium permanganate solution)

(iii) Treat compound with a solution of 2,4-dinitrophenylhydrazine in methanol/Bradys reagent 1

Red/orange/yellow ppt. is observed with CH3CH2CCH2OCH2CH3O

No ppt. with CH3CH2CO

OCH2CH2CH3

OO H2NNH NO2

O2N

ON

NH NO2

O2N

1

(iv) Treat compound with Tollens reagent/ammoniacal silver nitrate 1

A silver mirror / silver deposit is observed with CH3(CH2)3CHO;

No silver mirror / silver deposit with CH3CH2COCH2CH3

CH3(CH2)3CHO[Ag(NH3)2]+ / OH-

CH3(CH2)3COO- + Ag(s)silver mirror

1 9 marks

( + H2O )


4

1998-AL-CHEM 1

4. Alcohol E has the structure CH3CH(OH)C2H5.

(a) (i) Draw a three-dimensional representation of E.

(ii) What type of isomerism can be exhibited by E? (2 marks)

(b) (i) Draw the structures of three structural isomers of E, all of which are alcohols.

*(ii) Describe how the reagent Zn/concentrated HCl can be used to distinguish E from the three structural isomers.

(3 marks) (c) When heating with concentrated H2SO4(l), E gives mainly two isomeric compounds, F and G, both of which

have the formula C4H8. On treatment with bromine, both F and G give a product H with formula C4H8Br2.

(i) Draw structures for F, G and H.

(ii) What is the isomeric relationship between F and G?

*(iii) Outline the mechanism for the formation of H from either F or G. (5 marks)

Marking Scheme Mark

4. (a) (i) (Accept any correct representation of butan-2-ol showing the arrangement of groups at the chiral carbon)

C

H

CH3 OHC2H5

1

(ii) optical isomerism / enantiomerism 1 2 marks

(b) (i) + +

CH3CH2CH2CH2OH CH3CHCH2OHCH3

CCH3

CH3

CH3

OH

(ii) Upon reaction with Zn/conc.HCl, E gives turbidity (or cloudiness) at a slower rate than (CH3)3COH, but at a faster rate than CH3(CH2)3OH and (CH3)2CHCH2OH. ,

NOTE: This test is a modified version of Lucas test (conc.HCl/ZnCl2) used to distinguish 1, 2 and 3 ROH. Please refer to 1999-AL-CHEM 2 Q.7(a) for further information.

ROH RCl (RCl formed is insoluble in water)

3 marks

(c) (i) F and G: 1+1

and C CHH

CH3 CH3

C C

CH3H

CH3 H

and

H:

Br

Br

CH3CCCH3

HBr

BrH 1

(ii) Geometrical isomerism / cis-trans isomerism 1

(iii) ( ) ( ) 1

Br BrBr

+ Br

Br

Bror, Br

5 marks

Electrophilic addition mechanism DSE

conc.HCl/ZnCl2


5

1998-AL-CHEM 2

6. (a) Consider the structures of the two synthetic polymers shown below:

CH2CH2n

poly(ethene)

HN(CH2)6NHCO(CH2)4COn

nylon-6,6

(i) Suggest an explanation for the fact that nylon-6,6 has a higher tensile strength than 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).

(5 marks)

(c) 2-Ethanoyloxybenzoic acid (aspirin) is one of the most common substances used to relieve pain.

2-ethanoyloxybenzoic acid

COOH

OCOCH3

*(i) Outline a synthetic route for preparing 2-ethanoyloxybenzoic acid from 2-methylphenol.

(ii) Suggest how you could show the presence of the two functional groups in 2-ethanoyloxybenzoic acid.

(4 marks) (d) Identify J, K, L and M in the following reactions.

CNK

CONH2(ii)

n

CH2 CH

Cl

H2C=CH

Cl

L(iii)

(4 marks) Mark Scheme .............................................................................................................................................. 6. (a) (i) The interaction between polymer chain of PE is van der Waals forces 1

while that between Nylon 6.6 is hydrogen bond.

The higher tensile strength of nylon is due to its stronger intermolecular attraction.

(ii) CC linkage in PE is non-polar. not readily attacked by acids/has no reaction with acids. 1

The amide linkage in nylon hydrolyse in acids to give NH3+ and COOH groups causing shortening of the chain. ,

(iii) alkaline conditions / enzyme 1 5 marks


6

6. (c) (i) 2

OHCH3 KMnO4/H+

heat

(1/2)

(1/2)

OHCO2H CH3COCl

or, (CH3CO)2O

OCOCH3CO2H(

1/2)

(1/2)

Note: In carrying out the ethanoylation of phenol, we usually convert the phenol to phenoxide ion (by adding NaOH) before adding ethanoyl chloride (or ethanoic anhydride) for the ethanoylation.

(ii) Use IR spectroscopy: Absorption at 2500 to 3300 cm-1 indicates the presence of OH of the carboxylic acid ,

[Note that extensive H-bonding in the acid gives rise to the very broad characteristic absorption band.]

Absorption at 1680 to 1750 cm-1 indicates the presence of C=O of the acid and the ester. ,

or, Chemical test: treat compound with Na2CO3(aq) / NaHCO3(aq) ()

evolution of (colourless) gas bubbles, CO2(g), indicates the presence of CO2H ()

or, warm compound with dilute acid. Smell of vinegar indicates the presence of an ethanoyl ester. (,) 4 marks

(d) (ii) P2O5 (a dehydrating agent), heat NOTE: P2O5 is the empirical formula of P4O10. 1 Although this dehydration reaction is NOT mentioned in DSE, the reagent is worth learning.

(iii) H2O2, or other initiators (e.g. ROOR, i.e. organic peroxide) 1

4 marks

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) 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 more than three steps. In each step, give the reagent(s) used, the conditions required and structure of the product.

Starting molecule Target molecule

C OCH3

OCH3OHC NHCH3

O(i)

(iii) CH3CH2 C OHO

LiAlH4, H3O+ CH3CH2 C HO

(iv) CH3CH2CH CH3Br

NaOH H3CC CCH3

(9 marks)

(c) Give a systematic name to each of the following compounds.

(ii) CH3CH=CHCH2CHO

(1 mark)


7

Mark Scheme .............................................................................................................................................. 7. (a) (i) No reaction between the reactants

A feasible synthetic route is shown below: 5

NOTE: The functional group transformation required is from amide to ester.

or,

CO

NHCH3 H3O+CO

OH SOCl2CO

Cl CH3OHCO

OCH3

NOTE: Step 1 acid hydrolysis of amide; Step 2 conversion to acyl chloride from acid; Step 3 ester formation

(iii) LiAlH4 will react with H3O+; or, LiAlH4 reduces carboxylic acids to alcohols, not aldehydes

NOTE: LiAlH4 + 4H2O LiOH + Al(OH)3 + 4H2 The bases formed will be neutralised by the acid H3O+

A feasible synthetic route is shown below:

2

or, using Cr2O72-/H+ to oxidize the alcohol and distilling the aldehyde out from the reaction mixture as the aldehyde is being produced.

or,

CH3CH2COOHSOCl2 CH3CH2COCl

H2, Pd/BaSO4 CH3CH2CHO (for reference only)

(iv) Only alkene / but-2-ene will be formed. NOTE: Elimination reaction or, butan-2-ol will be formed. NOTE: Substitution reaction ()

A feasible synthetic route is shown below: 4

Bralcoholic NaOH heat

Br2 / CCl4Br

Br

alcoholic NaOH heat C C CH3CH3or, using a stronger base, e.g. NaOC2H5 or NaNH2 instead of NaOH

(1/2) (1/2) (1/2) (1/2)

9 marks 7. (c) (ii) pent-3-enal / 3-pentenal 1

1 mark

1999-AL-CHEM 1

5. (a) 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.

(5 marks)

Step 1: Reduction of COOH to CH2OH Step 2: Mild oxidation of 1 ROH to aldehyde

in dry ether


8

Marking Scheme 1999-AL-CHEM 1 Mark

5. (a) (i) diffused light; excess methane ,

(ii) mechanism: 3

(iii) No. The reaction of CH4 with Cl2 gives a mixture of CH3Cl, CH2Cl2, CHCl3 and CCl4. 1 5 marks

1999-AL-CHEM 2

5. (a) Identify D, E, F, G and J in the following reactions.

(i) O D NNH

(ii)CH2CH3

O2N

ECO2H

O2N

(iii)

CH2CH2OH

HOCH3

Na2Cr2O7/H3O+

heatF

(v)NH

O

H3O+

heat

(Hint: J is a polymer.)

J

(4 marks)

(c) Dacron is the most common of the group of polymers known as polymers. A segment of the polymer chain is shown below.

CO

CO

OCH2CH2O CO

CO

OCH2CH2O

(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? (5 marks)


9


5. (a) (i) D: NHNH2

1

(ii) E: MnO4-/H3O+; heat 1 (iii) F: CO2H

HOCH3

CHO

HOCH3

OR

1

(v) J: HN (CH2)5 C

O

n

1

[4] (c) (i) Any TWO of the following: fibres, plastics, coatings, clothing, etc. (Accept other correct uses of Dacron) , (ii)

HO CO

CO

OH or CH3O CO

CO

OCH3

1

and HOCH2CH2OH condensation polymerization

1 1

(iii) Hydrolysis of the ester group by enzymes /bacteria 1 [5]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (a) (i) Suggest a reagent for the following conversion:

Cl NH2 Cl NH CCH3

O

4-chlorophenylamine N-(4-chlorophenyl)ethanamide

(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.

(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. (6 marks)

6. (c) State the relationship between each pair of structures shown below:

(i)

CH2CH3

ClCl

and

CH2CH3

ClCl

(ii) C CH

CH2CH3

CH3

Hand C C

CH3

H

CH3CH2

H

(iii) CBr

CHCH2 CH3Hand C

H

CHCH2 CH3Br (3 marks)


10


6. (a) (i) (CH3CO)2O / CH3COCl

(ii) Shake the mixture with dilute HCl and ether in a separating funnel. 1, The amine will form the aminium salt and dissolve in the aqueous solution. The amide is not basic enough to be protonated by HCl and dissolves in the ether.

Separate the aqueous layer from the ethereal layer. Treated the aqueous layer with NaOH(aq) to liberate the free amine. , Use ether (or an appropriate solvent, e.g. THF, hexane, etc.) to extract amine from aqueous layer. (Distil off the organic solvent to obtain amine.) Distil off ether from the ethereal layer to obtain the amide.

(iii) (I) Chemical test: Heat the amide with acid, smell of vinegar can be detected. ,

(II) Spectroscopic method: IR: strong absorption at 1680 1750 cm-1 indicates the presence of C=O functional group /

at 3350 3500 cm-1 indicates the presence of NH group. , 6 marks

(c) (i) identical compound 1

(ii) geometrical isomers /cis-trans isomers / Z-E isomers 1

(iii) enantiomers /optical isomers 1 3 marks

-------------------------------------------------- Take a break ----------------------------------------------------- 7. *(a) Lucas reagent, a mixture of ZnCl2 and concentrated HCl, can be used to distinguish the following alcohols from

one another: (CH3)2CHCH2OH, CH3CH2CH(OH)CH3 and (CH3)3COH

(i) 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 CO bond and creating a better leaving group.) (4 marks)

7. (c) 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 + 2H2Pt CH3 CH(CH3)2

M (1) O3(2) Zinc dust CH3 C(CH2)2CHCH2C HOO

CO CH3

+ HCHO

(5 marks) Marking Scheme 1999-AL-CHEM 2 Mark

7. (a) (i) (CH3)2CHCH2OH : 1 ROH clear solution CH3CH2CH(OH)CH3 : 2 ROH cloudiness/turbidity appears after about five minutes (CH3)3COH : 3 ROH cloudiness/turbidity appears immediately ( marks for the appearance of cloudiness; 1 mark for the difference in rate.)

(ii) The appearance of cloudiness is due to the formation of haloalkanes which are immiscible with water. Due to the formation of the relatively stable 3 carbocations, 3 alcohols reacts rapidly with Lucas reagent. The reaction mechanism is SN1. 1 carbocation is relatively unstable. The reaction of 1 alcohol with Lucas reagent proceeds through a mechanism which is not favoured in acidic conditions. (The reaction rate of 2 alcohols with Lucas reagent is moderate.)

4 marks


11

7. (c) (i) 1 C H

Mass (g) 88.2 11.8 no. of moles

01.122.88

= 7.34 008.1

8.11= 11.71

Mole ratio 34.734.7

= 1 34.771.11

= 1.60

Simple whole no. ratio 5 8

Empirical formula of M is C5H8.

Relative molecular mass of M = 136.2.

Molecular formula of M is C10H16. 1

(For the empirical/molecular formula, award 1 mark for the answer + deduction)

(ii) M reacts with H2 in mole ratio of 1 : 2. M possesses two C=C bond /one CC bond. Upon ozonolysis, M gives HCHO and a compound with 3 carbonyl groups.

M possesses a terminal C=C bond and a cycloalkene/cyclohexene structure. ,

Structure of M: 1 NOTE:

limonene

Note that limonene is optically active; the (+) form being present in citrus oils whilst the (-) form is found in pine needle oil. The racemic mixture, called dipentene, is a constituent of turpentine.

7 marks

2000-AL-CHEM 1 5. (a) Consider the reaction:

HBr

Br

(i) Name the type of the reaction.

*(ii) Outline a mechanism of the reaction.

(iii) Draw the structures of all possible stereoisomers of the product.

(iv) Would the product rotate the plane of a beam of plane polarized light? Explain your answer. (5 marks)

isoprene

limonene


12

5. (b) (i) Give the reagents and conditions for the two steps of the conversion:

CO

CH3CH2 N(CH3)2Step 1

CHCH3CH2 N(CH3)2OH

CH N(CH3)2CH3CH

Step 2

Step 1:

Step 2:

(ii) Suggest a chemical test for the carbonyl group in compound A.

(iii) State the expected observation when HCl(aq) is added to compound A. (4 marks)


5. (a) (i) electrophilic addition 1 (ii)

2

(iii)

,

(iv) No. The product is a mixture of the enantiomers in mole ratio of 1 : 1 / a racemic mixture

1

[5] (b) (i) Step 1: LiAlH4/ether, followed by acid hydrolysis 1 or, NaBH4 / H2, Pd; high pressure (,)

Step 2: Heat with concentrated H2SO4 / Al2O3 / H3PO4 / P4O10 ,

(ii) Reagent: 2,4-dinitrophenylhydrazine / Bradys reagent Observation: yellow / orange / red precipitate

or Reagent: hydroxylamine / phenylhydrazine / semicarbazide () Observation: white (colourless) crystals ()

(iii) A colourless solution is formed / only one layer is observed. 1 4 marks

2000-AL-CHEM 2

5. (d) A mixture of two organic products, G and H, was obtained in the reaction :

The mixture was separated by paper chromatography using a mixture of hexane and ether as solvent. The Rf values of G and H were found to be 0.84 and 0.55 respectively.

(i) Deduce the structures of G and H.

(ii) Suggest a chemical test to distinguish between G and H.

(iii) State how the use of infra-red spectroscopy can provide further proof for the structures of G and H. (8 marks)


13


5. (d) (i) G: C

OO(CH2)5O C

O

H: C

OO(CH2)5OH

1,1

G is less polar than H (or H has a polar OH group) and is therefore less tightly bound to the stationary phase and hence moves faster.

1,1

(ii) Reagent: PCl5 / PCl3 / SOCl2 Only H will give misty fumes of HCl NOTE: ROH + SOCl2 RCl + SO2 + HCl

1 1

or, Reagent: Na Only H will react with Na to give gas bubbles (H2) NOTE: 2ROH + 2Na 2RO-Na+ + H2

(1) (1)

or, Reagent: Cr2O72-/H+ Only H will induce a colour change from orange to green

or, Reagent: MnO4-/H+ Only H will induce a colour change from purple to colourless

(iii) Compound H shows an IR absorption at 3200-3700 cm-1 but G does not, because H has an OH functional group whereas G does not.

1,

[8]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. *(d) The conversion below can be completed in not more than four steps.

Use equations to show how you would carry out the conversion in the laboratory. For each step, give the reagent(s), conditions and structure of the product.

(4 marks) (e) Identify L, M, N, and P in the following reactions:

(4 marks)


14


6. (d)

4

4 marks

6. (e) (i) L: Cl2 and light or hv / N-chlorosuccinimide (NBS) and light 1

(ii) M: H2 and transition metal catalyst (e.g. Pt or Ni) + pressure 1 *(iii) N:

(elimination reaction)

1

*(iv) P: HOOC(CH2)4COOH (hydrolysis of ester group and nitrile group (CN) 1 5 marks

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) An acyclic compound R (C7H12O) has a linear structure. R can be converted to S and then to T:

Given that R exists as a mixture of geometrical isomers, S has a chiral carbon centre, and T does not have any chiral carbon centre, deduce all possible structures of R, S and T.

(8 marks)

7. (b) Aspartame, a sweetener, has the structure below:

(i) Name all functional groups in aspartame.

(ii) Upon hydrolysis, aspartame gives two amino acids. Draw their structures.

(iii) Two electrodes are dipped into an aqueous solution containing the two amino acids in a pH 12 buffer, and are connected to the two poles of a battery. Which species derived from these amino acids will move faster towards the anode? Explain your answer.

(7 marks)

heat

heat


15

7. (c) The compounds below are the monomers of some polymers.

COClO2N CO2HH2N

These compounds react in the presence of an inorganic reagent U to give compound V.

(i) What is reagent U?

(ii) The flow diagram below illustrates the reaction sequence for the transformation of V to compound W with four aromatic rings. What are reagents X and Y?

(iii) In the reaction sequence as in (ii), V is replaced by W, i.e., W is allowed to react with reagents X and Y, and the products react in the presence of U to form a polymer Z. Draw the structure Z.


7. (a) T (C7H16O) is an alkanol / alcohol. 1 R has 2 double bonds and S has 1 double bond. 1

R can be reduced by LiAlH4 and S can undergo catalytic hydrogenation.

R possesses the >C=O and >C=C< functional groups and S is an unsaturated alcohol. ,

S has a chiral carbon centre while T has not.

Possible structures of R:

O

or

O

2 Possible structures of S:

OH

* or

OH

* 2+1

(2 marks for structures; 1 mark for a clear indication of the chiral centre.) T is heptan-4-ol:

OH

1

(For each of R, S and T, deduct marks for each additional structure.) 8 marks


16

7. (b) (i) (max. 4 guessing, phenyl group not counted as a guess) amine/amino carboxylic acid/carboxyl/alkanoic acid amide/peptide (link) ester/alkyl alkanoate

(ii)

COOH

COOHH2N

COOHH2N 1+1

(accept structures in form of Zwitterion.) (iii) At pH 12, the amino acids exist as anions:

COO-

COO-H2N

COO-H2N 1+1

(I) (II) (I) is doubly (more highly) charged, and will moves faster towards the anode. ,

7 marks

7. (c) (i) U: NaOH/KOH/Na2CO3/K2CO3 1

(ii) X: SOCl2 / PCl3 / PCl5 1

Y: Sn/HCl or Fe/HCl, or Zn/HCl or H2/Pd-C or H2/PtO2 or Na + alcohol 1

(iii) Z: or

O2N CO

NH

COOH

7

7

O2N CO

NH

COOH

2 5 marks

2001-AL-CHEM 1

5. (b) Give the structures of the organic compound E.

(ii)

(1 mark)

6. (b) After some lessons in organic chemistry, a student remarked, Alkanes are more stable than alkenes, therefore alkanes do not react with chlorine but alkenes do.

Do you agree with the student ? Explain. (3 marks)

Marking Scheme 2001-AL-CHEM-1 Mark

5. (b) (ii) E:

1

NOTE: IT is a condensation polymerisation reaction. [1] 6. (b) Alkanes are thermodynamically/energetically more stable than alkenes because hydrogenation of alkenes

is exothermic. 1

or,

H < OC C + H2 C CHH

Pt

(1)

Alkanes do react with chlorine under sunlight to give chloroalkanes. 1


17

e.g. CH3CH3 + Cl2

hv CH3CH2Cl + HCl

Comment on the relationship of the first and second statement.

(1) 1

[3]

2001-AL-CHEM 2

5. (a) Compound A has the following structure :

*(i) The carbon atom indicated by arrow a is more reactive towards nucleophilic substitution (SN2) reaction with C2H5ONa than that indicated by arrow b. Explain.

(ii) 5.56 g of A is treated with 1.36 g of C2H5ONa. Deduce the structure of the major substitution product B.

(iii) 5.56 g of A is treated with 3.00 g of C2H5ONa. The reaction gives two products, D and E. E has the molecular formula C11H14O.

(I) Give the structure of D.

(II) If the yield of D is 82%, calculate the mass of D that can be obtained.

(III) Give the structure of E and name the types of reactions involved in the formation of E.


5. (a) (i) The reaction involves the attack of C2H5O- on the C atom of the substrate. Carbon atom b is more crowded. it is less reactive. (Accept equivalent answers, such as steric hindrance, etc.)

1

(ii) No. of moles of A = )29.7910008.1901.12(

56.5++

= 0.02

No. of moles of C2H5ONa = )99.22165008.1201.12(36.1

+++= 0.02

A : C2H5ONa = 1 : 1 1

Structure of B: O

Br

1

(iii) (I) Structure of D: O

O

1

(II) C2H5ONa is in excess. Molar mass of D (C13H20O2) = 12.01 13 + 1.008 20 + 16 2 = 208.3 g mol-1 Mass of D obtained = 0.02 208.3 0.82 = 3.42 g (Deduct marks for wrong/no unit.)

1

1

(III) Structure of E: O

1

Types of reactions: elimination and substitution ,

[8]


18

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (b) *(i) Give the reagent(s) and conditions for Step 1 and Step 2 in the reaction sequence below :

(ii) 4-nitrobenzoic acid can be converted to benzocaine, an anaesthetic, in not more than three steps.

Use equations to show how you would carry out the conversion. In each step, give the reagent(s), conditions and structure of the product.


5. (b) (i) Step 1: conc. HNO3 + conc. H2SO4 (); 0-60C / heat/warm () 1 Step 2: MnO4-/H3O+ (); heat () 1 ( marks for reagents; marks for conditions.) (ii) 3

c: H2/Pt, or Na/EtOH, or NaBH4 (1) Alternative answers: (3)

d: H2/Pt, or Sn/HCl + heat e: EtOH/H3O+ + heat f: OH-(aq)

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (c) Consider the information below concerning the production of low density polyethene from ethene.

*(i) Outline a mechanism for the polymerization and name each mechanistic step.

(ii) Explain why benzoyl peroxide is used.

(iii) Why is high pressure needed for the polymerization ?

(iv) Is the product a single compound ? Explain. (7 marks)


19


5. (c) (i) Initiation ( ) 2

Propagation ( ) 1

Termination ( ) 1

(Fish-hooks should be shown in the mechanism; otherwise max. deduction 1 mark)

(ii) Benzoyl peroxide readily undergoes homolytic cleavage / decompose to give free radicals 1

or, Benzoyl peroxide is a free radical initiator (1)

(iii) Concentration of reactants is high.

chance of collision between ethene and free radical increases /higher reaction rate. 1

(iv) No, because the product is a mixture of polymer molecules with different chain lengths. 1

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) For each of the following reactions, suggest a possible reactant :


20


7. (a) (i)

O COCl O COOCOR

1

(ii) (iii) (iv) (v) O

OHCl

OH

O

CO2CH3

Each: 1

mark

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (b) Study the following information:

Aromatic compounds T and U have the same molecular formula C10H12O.

T gives positive result in iodoform test.

U reacts with hydrogen in the presence of a palladium catalyst to give compound V (C10H14O). V can also be obtained from the reaction of T with LiAlH4.

Treatment of U with ozone followed by acidified KMnO4 gives two products, one of which is benzoic acid.

(i) Deduce the structures of T, U and V. For each compound, give all possible stereoisomers.

(ii) Suggest how infra-red spectroscopy can be used to distinguish between T and U. (9 marks)


7. (b) (i) T gives positive result in the iodoform test and has the structural characteristic

CO

CH3 or CHCH3

OH

( marks for each characteristic.)

1

U reacts with H2-Pd U is an alkene / has a C=C group

T can be reduced by LiAlH4 T is a ketone / has a C=O group

Reaction of U with O3 following by KMnO4 gives benzoic acid.

The double bond in U is adjacent to the benzene ring. 1 Structures: (Four possible structures; 1 mark each for the 1st and 2nd structures, and marks for

each of the 3rd and 4th structures)

T: O

U: OH

*

1,1

V: OH

*

(two stereoisomers; marks for each structure)

1

NOTE: Also accept the following structures for T, U and V:

(ii) T has a sharp absorption peak in the wavenumber range from 1680 to 1750 cm-1 but U has not. , U has a broad absorption peak in the wavenumber range from 3230 to 3670 cm-1 but T has not. ,

[9]


21

2002-AL-CHEM 1

5. (a) Two acyclic structural isomeric compounds have the molecular formula C3H6O3. Both are optically active and have infrared absorptions at 3400 cm-1 and 1700 cm-1. Neither possesses an alkoxy group.

(i) Draw the structure of each of the compounds. Label one of these structures A and the other B.

(ii) Suggest a chemical test, giving the expected observation, to distinguish between the compounds represented by A and B.

(iii) Give the structure of the major organic product formed when the two compounds are treated separately with excess LiAlH4.

Product from the compound represented by A Product from the compound represented by B

(6 marks)


5. (a) (i) CO2H

OH

CHOHO

OH

1,1

(ii) Treat compounds with Na2CO3(aq). 1 Only the carboxylic acid will give colourless gas bubbles, CO2(g). 1 Treat compounds with 2,4-dinitrophenylhydrazine / Fehlings solution (1) Only the aldehyde will give a yellow precipitate / red precipitate (1) (Accept other tests for the carbonyl group and the corresponding expected observations) (iii) product from the carboxylic acid product from the aldehyde

OH

OH

OH

HOOH

1,1

[6]

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (b) Suggest reagent(s) to accomplish each of the following single-step transformations:

(i) CH3C(CH2)4COCH3OO

CH3CH(CH2)4COCH3OOH

(iii)


5. (b) (i) NaBH4 or H2/Pt 1 (iii) Cr2O72-/H3O+ 1 (Accept other correct answers.)

3 marks

CHO

COCH3OH

CO2H

COCH3OH


22

2002-AL-CHEM 2

5. (b) Suggest a synthetic route, in not more than four steps and using LiAlD4 or NaBD4 as a source of deuterium (isotope of hydrogen), for the following transformation.

DOH


-------------------------------------------------- Take a break ----------------------------------------------------- 5. (c) Consider the substances listed below:

butane benzoic acid dichlorodifluoromethane

ethanoic acid hexane polystyrene

propanone tetrachloromethane triethylamine

For each of the descriptions of physical properties from (i) to (viii) below, choose from the above list, one substance which best fits the description.

(i) a colourless, flammable gas

(ii) a colourless liquid with a sour odour

(iii) a colourless, water miscible, flammable liquid

(iv) a colourless, non-flammable liquid

(v) a colourless liquid with a fishy smell

(vi) a colourless, water immiscible, flammable liquid

(vii) a white solid which is insoluble in both cold and hot water

(viii) a white solid which is insoluble in cold water, but soluble in hot water (8 marks)


23


5. (c) (i) butane - a colourless, flammable gas 1 (ii) ethanoic acid - a colourless liquid with a sour odour 1 (iii) propanone / ethanoic acid - a colourless, water miscible, flammable liquid 1 (iv) tetrachloromethane - a colourless, non-flammable liquid 1 (v) triethylamine - a colourless liquid with a fishy smell 1 (vi) hexane - a colourless, water immiscible, flammable liquid 1 (vii) polystyrene - a white solid which is insoluble in both cold and hot water 1 (viii) benzoic acid - a white solid which is insoluble in cold water, but soluble in hot water 1 [8]

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (b) For each of the following pairs of molecules, identify their relationship as identical, enantiomeric, geometrical

isomeric or structural isomeric.


7.


24

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (c) A mixture of 2.8 g of butane-1,4-diol and 6.3 g of benzene-1,3-dicarbonyl chloride was heated at 215C for 30

minutes to give 6.4 g of a polymer M.

(i) Draw the repeating unit of M. (ii) What type of polymerization is involved in the formation of M ?

(iii) Calculate the percentage yield of M. (5 marks)


7.

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (d) Suggest a synthetic route, in not more than three steps, for the transformation of 3-methylbenzoic acid to

N,N-diethyl-3-methylbenzamide, a substance commonly used in mosquito repellent.


7.

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AL-CHEM Chemistry of Carbon Compounds(97-02)

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