Answer Paper 1 962/1

PAGE 7

SMK TENGKU INTAN ZAHARAH

23000 DUNGUN TERENGGANUMARKING SCHEME

STPM CHEMISTRY 962/2MARKING SCHEME SECTION AQuestion No.AnswerExplanation

1CThis compound has one chiral carbon with four different groups. Thus it can show optical isomers.

COOH COOH

| | C C

H CH3 H3C H NH2 NH2

2APhenol reacts with bromine water to produce a white precipitate.

OH

OH Br Br

Room temp

+ 3Br2 (aq) Br

2, 4, 6-tribromophenol

(white precipitate)

whereas phenylmethanol CH2OH does not react with bromine

water

3DThe reaction between methane and chlorine in the presence of light involves a free radical mechanism.

The propagation step is

Cl + CH4 HCl + CH3 Cl2 + CH3 CH3Cl + Cl

4

B (-

O

(+ (-

CH2Br (answer A) and C CH3 (answer C) are susceptible to

(+

nucleophilic attack because the carbon atom in these carries a partial positive charge, CH3 is susceptible to free radical substitution reactions. Benzene ring (answer B) and carbon-carbon double bonds contain ( electrons. Hence, they are susceptible to electrophilic attack.

Remember : The electron density of an electrophile is very low. Therefore, it will attack an electron rich centre, such as, the benzene ring since the electron density is very high in the benzene ring.

5D OH OH

Cl Cl

+ 3Cl2 + 3HCl

Cl

2, 4, 6 trichlorophenol

Phenol undergoes electrophilic substitution.

6B O Cl

|| |CH3 C CH + PCl5 CH3 C CH3 + POCl3 | Cl

7CEster produces alcohol and carboxylic acid when acidic hydrolysis is performed.

CH3(CH2)24COO(CH2)29CH3 + H2O

CH3(CH2)24 COOH + CH3(CH2)29OH

Carboxylic acid Alcohol

8DRemember: The amino group activates the benzene ring towards substitution, at the ortho and para positions. Therefore, bromine will replace hydrogen at the ortho and para positions.

9DCarbonyl group is found in aldehydes and ketones.

10AThe polymer is produced from the addition polymerisation reaction. Molecules with double bonds are joined together to form a long

chain of the polymer.

11D(1, 2, 3)Optical isomers are only different in the aspect of polarized light rotation.

12A

(1 only)Butan-2-ol is a secondary alcohol. Oxidation of the alcohol produces a ketone.

H H H

| | |CH3 C C CH3 + [O] CH3 C C CH3 + H2O

| | | ||

H OH H O

Butanone

13D(1, 2, 3)The C = C undergoes electrophilic addition.

The carbonyl group undergoes nucleophilic addition.

The benzene ring undergoes electrophilic substitution.

14B(1,2)The particular polymer can be drawn as follows...

Thus this polymer can be synthesised from

This polymer is polyamide because it contains

The group in this molecule makes it a polyester. It is not resistant against alkaline hydrolysis because esters and amides can easily be hydrolysed.

15A(1)Poly(2-methylbuta-l ,3-diene)

CH3

-(-CH2-C=CH-CH2)nis unsaturated, with C=C double bond and can be oxidised by MnO4- /H+. Poly(ethene) and poly(phenylethene) are saturated, cannot be oxidised.[STPM 2010]

STPM CHEMISTRY 962/2

MARKING SCHEME

SECTION B ( Structured Question )

QUESTION NOSUGGESTED ANSWERSMARK

16(a)(i)

(i) -Br + NH3

+ 4Br-

1

16(a)(ii) (CH3)2CBr2 + NaOH CH3- C - CH3

O1

!6(a)(iii) -CHO CH-CH2CH3

OH

1

16(b)(i)

1

16(b)(ii)

1

16(b)(iii) O

and

1

16(b)(iv)

1

TOTAL

8

MARKS

17(a)(i)An electrophile is an electron deficient species which accepts a lone pair electrons from another species to form a coordinate bond, Electrophile are Lewis acids1

17(a)(ii)CH2=CH2 + Br2 BrCH2-CH2Br

2

17(b)(i)X : HOCH2CH(CH3)OH or

HOOCCH2CH2COOHY : CH2=C(CH3)COOCH3

1

1

1

17(b)(ii) Polymer X: Condensation polymerization

Polymer Y: Addition polymerisation1

1

Total8 marks

SECTION C (Essay Question)QUESTION NOSUGGESTED ANSWERSSUGGESTED MARKS

18(a)2-butanol Optical isomerism.

It has a chiral carbon atom bonded to four different groups.

It has a pair of enantiomers that are non superimposable mirror images.

1

1

1

Butenedioic acid geometrical isomerism

This isomerism arises due to restricted rotation about the C=C bond.

HOOC

COOH HOOC H

C = C

C = C

H

H

H COOH

Cis- butenedioic acid

Trans- butenedioic acid

1

1

1

18(b)Molecular formula : C5H12O

P : CH3CH2CH2CH(OH)CH3 O

Q : CH3CH2CH2CCH3

CH3CH2CH2CH(OH)CH3 + [O] CH3CH2CH2COCH3P is an alcohol with the CH3CH(OH) group which reacts with alkaline iodine to produce CHI3.

CH3CH2CH2CH(OH)CH3 + 4I2 + OH- CH3CH2CH2COO- + CHI3 + 5HI

Q is a ketone with the (CH3CO-) which reacts with alkaline iodine to produce CHI3.

CH3CH2CH2COCH3 + 3I2 + OH- CH3CH2CH2COO- + CHI3 + 3HI

Q ia a carbonyl compound which forms a precipitate with 2,4 dinitrophenylhydrazine.

O2NCH3CH2CH2COCH3 + H2NNH NO2

O2N

CH3CH2CH2C=NNH NO2 + H2O CH3

Total

15 marks

19(a)(i)Bimolecular nucleophilic substitution (SN2)

1+1

1

19(a)(ii)The rate of reaction increases because the breaking of C-Br bond needs less energy than that of C-Cl bond. This is due to the length of C-Br bond is less than the C-Cl bond length making it weaker and easier to break.

1

1

1

19(b)(i)

1+1

1+1

1

19(b)(ii)Catalytic cracking is used to provide extra petrol and as a source of alkenes for the manufacture of many petrochemicals.1

1

19(b)(iii)Catalytic cracking is carried out at 450C using the catalysts silicon (IV) oxide and aluminium oxide.

1

1

TOTAL15 MARKS

20(a)Add solid phosphorous (V) chloride to ethanol at room temperature.

C2H5OH + PCl5 C2H5Cl3 + HCl

Bubbled chloroethane into hot potassium cyanide dissolved in ethanol.

C2H5Cl + KCN C2 H5CN + KCl

Boiling propanenitrile with dilute sulphuric acid.

C2H5CN + 2H2O + H+ C2H5COOH + NH4 +

Correct steps

1 m

(1 m for correct reagent and condition of each step.)

(1+1+1) m

20(b)Heat 1-propanol with excess of concentrated sulphuric acid at 180 oC.

CH3CH2CH2OH CH3 CH = CH2 + H2O

Pass propene together with steam over heated phosphoric acid (adsorbed on the surface of silicon dioxide).

Correct steps

1 m

(1 m for correct reagent and condition of each step.)

(1+1) m

20(c)Boiling 2-bromopropane with aqueous sodium hydroxide.

Oxidise 2-propanol with hot acidified potassium dichromate.

Correct steps

1 m

(1 m for correct reagent and condition of each step.)

(1+1) m

20(d)Heat 1-propanol with excess of concentrated sulphuric acid.

Reaction with hydrogen chloride at room temperature.

Heat 2-choloropropane with ethanolic potassium cyanide.

Boiling 2 cynopropane with dilute sulphuric acid.

Correct steps

1 m

(1 m for correct reagent and condition of each step.)

(1+1+1+1) m

END OF MARKING SCHEME

CHEMISTRY

One hour and a half hours

SUMMATIVE ASSESSMENT

THIRD TERM

STPM

2013

COOH

Ethanol

Na+O-

CH2CCH3

OH

Na+O-

CH2CHCH3

O-Na+

Na+O-

Mirror plane

CH2CH3

CHI3

CH2CO-Na+

(i) CH3CH2MgBr

(ii) H3O+

N+

H

C

CH3

OH

H

CH3

C

OH

H3CH2C

HO