VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY ...score-in-chemistry.weebly.com/.../vjc_2014_prelim.pdfVJC 2014 9647/01/PRELIM/14 [Turn over VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATIONS

VJC 2014 9647/01/PRELIM/14 [Turn over

VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATIONS Higher 2

CHEMISTRY

9647/01

Paper 1 Multiple Choice

24 September 2014

1 hour

Additional Materials: Multiple Choice Answer Sheet Data Booklet

READ THESE INSTRUCTIONS FIRST Write in soft pencil.

Do not use staples, paper clips, highlighters, glue or correction fluid.

Write your NRIC/FIN number, name and CT group on the Answer Sheet.

There are forty questions. Answer all questions. For each question there are four possible answers A, B, C and D.

Choose the one you consider correct and record your choices in soft pencil on the separate Answer Sheet.

Read the instructions on the Answer Sheet very carefully.

Each correct answer will score one mark. A mark will not be deducted for a wrong answer.

Any rough working should be done in this booklet.

This document consists of 17 printed pages and 1 blank page.

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VJC 2014 9647/01/PRELIM/14

Section A For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 Compound X reacts with ethanoic acid in the presence of an H+ catalyst to produce the

compound below.

What is the molecular formula of compound X?

A C2H6O2 B C2H6O3 C C4H8O D C4H8O2 2 When Tl+(aq) ions are reacted with VO3

–(aq) ions, Tl3+(aq) ions and V2+(aq) ions are formed.

Assuming the reaction goes to completion, how many moles of Tl+(aq) and VO3

–(aq) would result in a mixture containing equal number of moles of VO3

–(aq) and V2+(aq) once the reaction had taken place?

moles of Tl+(aq) moles of VO3–(aq)

A 1 2

B 1 3

C 2 3

D 3 4

3 Which of the following contains only one unpaired electron?

A Ga– B Se– C Te+ D As+ 4 Which of the following substances contains a central atom having eight electrons only?

A NO2 B SO2 C F2O D ClO2 5 For which of the following pairs does the first species have a smaller bond angle?

A SnCl2, OCl2

B H2O, H2S

C I3–, N3

–

D NF3, NCl3

3


6 The value of PV is plotted against P at the same temperature for four gases, where P is the pressure and V is the volume of the gas.

Which of the following represents the possible identities of gases W, X, Y and Z?

W X Y Z

A CH3CH2OH HCO2H CH3CHO CH3CH2CH3

B CH3CH2OH CH3CHO HCO2H CH3CH2CH3

C HCO2H CH3CH2OH CH3CHO CH3CH2CH3

D HCO2H CH3CHO CH3CH2OH CH3CH2CH3

7 Use of the Data Booklet is relevant to this question. Two cells are connected in series as shown in the diagram where E, F, G and H are

electrodes.

After the current is passed for some time, at which electrode will there be the greatest

change in mass?

A E B F C G D H

PV

P

W

X

Y

Z

Al2(SO4)3(aq) CuSO4(aq)

Cu Al

E F G H

Cu Ag

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VJC 2014 9647/01/PRELIM/14

8 Use of the Data Booklet is relevant to this question.

A student carried out an experiment to determine the enthalpy change for the combustion of methanol.

The following results were obtained by the student.

initial temperature of the water: 20 oC final temperature of the water: 53 oC mass of alcohol burner before burning: 259.65 g mass of alcohol burner after burning: 259.15 g mass of beaker plus water: 150.00 g mass of beaker: 50.00 g

How much of the heat energy produced by the burning of methanol went into the water?

A 209 J B 13800 J C 20700 J D 22200 J

9 If N2O4 gas is placed in a sealed vessel, the following equilibrium is established.

N2O4(g) ⇌ 2NO2(g)

The forward reaction is endothermic.

Which of the following is incorrect when the temperature is increased?

A The equilibrium constant increases.

B The partial pressure of NO2 increases.

C The enthalpy change increases.

D The activation energy is unchanged. 10 Which of the following equilibria is not affected by the addition of aqueous ammonia?

A BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq)

B AgCl(s) ⇌ Ag+(aq) + Cl–(aq)

C Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH–(aq)

D [Cu(H2O)6]2+(aq) + 4Cl–(aq) ⇌ [CuCl4]

2–(aq) + 6H2O(l)

5


11 Human plasma is buffered mainly by dissolved CO2 which has reacted to form carbonic acid.

H2CO3(aq) ⇌ H+(aq) + HCO3

–(aq) Usually the pH of human plasma is 7.4. The acid dissociation constant, Ka, of carbonic

acid is 7.90 x 10–7 mol dm–3. Which of the following statements is incorrect?

A This buffer system can be prepared by mixing suitable amounts of sodium hydrogencarbonate and hydrochloric acid.

B The ratio of [HCO3–] to [H2CO3] in human plasma is 20 : 1.

C Its pH value remains constant when diluted with water.

D It has a pH value equal to the pKa value of carbonic acid. 12 The mechanism below has been proposed for the reaction of CHCl3 with Cl2.

Step 1: Cl2(g) ⇌ 2Cl(g) fast

Step 2: Cl(g) + CHCl3(g) CCl3(g) + HCl(g) slow

Step 3: CCl3(g) + Cl(g) CCl4(g) fast

Which of the following rate equations is consistent with this mechanism?

A rate = k[CHCl3][Cl]

B rate = k[CHCl3][Cl2]

C rate = k[CHCl3][Cl2]1/2

D rate =

13 Consecutive elements X, Y and Z are in the third period of the Periodic Table. Element

Y has the highest first ionisation energy and the lowest melting point of these three elements.

What could be the identities of X, Y and Z?

A sodium, magnesium, aluminium

B magnesium, aluminium, silicon

C aluminium, silicon, phosphorus

D silicon, phosphorus, sulfur

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VJC 2014 9647/01/PRELIM/14

14 The nitrates of beryllium and aluminium can undergo thermal decomposition similarly as magnesium nitrate. Using relevant data from the Data Booklet, which of the following shows the correct trend of the decomposition temperatures for these three compounds?

A Mg(NO3)2 > Al(NO3)3 > Be(NO3)2

B Mg(NO3)2 > Be(NO3)2 > Al(NO3)3

C Be(NO3)2 > Al(NO3)3 > Mg(NO3)2

D Al(NO3)3 > Be(NO3)2 > Mg(NO3)2 15 The table below shows the results of experiments in which the halogens, X2, Y2 and Z2

were added to separate aqueous solutions containing X–, Y– and Z– ions.

X–(aq) Y–(aq) Z–(aq)

X2 no reaction no reaction Z2 formed

Y2 X2 formed no reaction Z2 formed

Z2 no reaction no reaction no reaction

In which sequence is the solubility of the silver halides in aqueous ammonia arranged in increasing order?

A AgX, AgY, AgZ

B AgY, AgX, AgZ

C AgY, AgZ, AgX

D AgZ, AgX, AgY 16 Which of the following reactions of first row transition metal ions is correct?

A Addition of NaOH(aq) to K2CrO4(aq) produces an orange solution of K2Cr2O7.

B Addition of KI(aq) to Fe2(SO4)3(aq) produces a brown precipitate of FeI3.

C Addition of concentrated HCl to CuSO4(aq) produces a yellow solution of H2[CuCl4].

D Addition of Na2CO3(aq) to CrCl3(aq) produces a green precipitate of Cr2(CO3)3. 17 Use of the Data Booklet is relevant to this question.

Which set of reagents, when added in the order shown below, would convert Fe3+(aq) to [Fe(CN)6]

−(aq)?

step 1 step 2 step 3

A Zn(s) CN−(aq) SO2(g)

B I−(aq) CN−(aq) Cl2(g)

C H2O2(aq) I−(aq) CN−(aq)

D Ag(s) I−(aq) CN−(aq)

7


18 Which of the following reagents, upon reaction with compound E, gives only one sp2 hybridised carbon atom in the product molecule?

compound E

A NaBH4 in methanol

B LiAlH4 in dry ether

C HCl(aq), heat

D SOCl2 19 The dehydration of propan–2–ol to form propene is thought to involve the following

steps.

Which of the following statements is incorrect? A Propan–2–ol acts as a base in step 1.

B H2SO4 is a catalyst in this reaction.

C A possible side product of the reaction is CH3CH(OSO3H)CH3.

D It is more likely for primary alcohols to proceed via this mechanism than tertiary alcohols.

20 Compound S upon reaction with hot acidified potassium manganate(VII) yields

CH3COCH3, CH3COCH2CH2CO2H and CH2(CO2H)2.

Which compound could be S?

A CH3CH=C(CH3)CH2CH2C(CH3)=CHCH2CH2OH

B (CH3)2C=CHCH2CH2C(CH3)=CHCH2CH2OH

C (CH3)2C=CHCH2CH2C(CH2OH)=CHCH2CH3

D (CH3)2C=C(CH3)CH2C(CH3)=CHCH2CH2OH

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VJC 2014 9647/01/PRELIM/14

21 P, Q and R are three isomeric aromatic compounds with the molecular formula, C8H8O2. P and Q are monobasic acids whereas R is a neutral compound. The pKa of P and Q are 3.90 and 4.31 respectively. When R is heated strongly with NaOH(aq), one of the organic products yielded is C6H5O

−Na+.

Which of the following shows the correct identities of P, Q and R?

P Q R

A

B

C

D

22 A key molecule in the chemistry of vision is the highly conjugated rhodopsin, which is

synthesised in the rod cells of the eye from 1 1–cis–retinal and a primary amine in the protein opsin.

Which of the following statements is correct?

A 1 mol of 1 1–cis–retinal reacts with 5 mol of hydrogen gas.

B The reaction between 1 1–cis–retinal and NH2–opsin is a condensation reaction.

C The reaction of 1 1–cis–retinal with LiAlH4 produces a product that is optically active.

D The reaction between 1 1–cis–retinal and excess hot acidified potassium manganate(VII) produces a total of five different organic products.

9


23 Which of the following compounds are arranged in decreasing order of their solubility in water?

A CH3CH2CO2Na, CH3CH2CH2NH2, CH3CH2CH2Cl

B CH3CH2CO2Na, CH3CH2CH2Cl, CH3CH2CH2NH2

C CH3CH2CH2NH2, CH3CH2CO2Na, CH3CH2CH2Cl

D CH3CH2CH2Cl, CH3CH2CH2NH2, CH3CH2CO2Na 24 Which of the following compounds cannot be prepared using fumaric acid as a starting

material?

CH(CO2H)=CHCO2H fumaric acid

A (CO2H)2

B CH2(CO2H)CH2CO2H

C CH(OH)(CO2H)CH(OH)CO2H

D 25 Aldehydes and ketones undergo addition in the presence of a strong base. A reaction

involving propanal, CH3CH2CHO, and barium hydroxide is shown below.

Which product could be formed when barium hydroxide is added to a mixture of ethanal, CH3CHO, and propanone, CH3COCH3?

A

B

C

D

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VJC 2014 9647/01/PRELIM/14

26 Which of the following compounds will react with SOCl2 but not react with HCl(g)?

A CH2(OH)CO2CH3

B CH2(OH)CO2H

C

D

27 A derivative of morphine has the structure shown below.

What is the product formed when the above compound is heated with excess HBr(aq)?

A B

C D

11


28 The diagram shows an experimental setup used in a laboratory.

Which of the following preparations could this setup be used for?

A 2,4,6–tribromophenol from phenol and bromine in tetrachloromethane

B ethanol from ethene and concentrated sulfuric acid

C ethylamine from ethanamide and lithium aluminium hydride in dry ether

D methyl methanoate from methanol, methanoic acid and concentrated sulfuric acid 29 Aspartic acid was first discovered in 1827 by Plisson. It is found in animal sources such

as luncheon meat and sausages as well as vegetable sources such as sprouting seeds, oat flakes, avocado and asparagus.

aspartic acid

There are three pKa values associated with aspartic acid: 2.10, 3.86, 9.82. Using the pKa values, what is the major species present in solutions of aspartic acid at

pH 7?

A B

C D

ice water magnetic stirrer bar

magnetic stirrer

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VJC 2014 9647/01/PRELIM/14

30 Electrophoresis is an analytical method frequently used in molecular biology to analyse and separate proteins and amino acids. Amino acid samples are placed in a gel slab containing a buffer. The ends of the gel slab are connected to charged electrodes and the charged amino acid particles move towards the oppositely charged electrodes.

Which amino acid will move most readily towards the anode at pH 10?

A

B

C

D

13


Section B For each of the questions in this section, one or more of the three numbered statements 1 to 3 may be correct. Decide whether each of the statements is or is not correct (you may find it helpful to put a tick against the statements that you consider to be correct). The responses A to D should be selected on the basis of

A B C D

1, 2 and 3 are

correct

1 and 2 only are correct


1 only is

correct

No other combination of statements is used as a correct response. 31 The table shows the electronic configurations of two d–block elements in the Periodic

Table.

element electronic configuration

X [Ar]3d74s2

Y [Ar]3d104s1

Which statements about X and Y are incorrect?

1 X is likely to exist as K2X2O7.

2 The Eo value of X3+/X2+ is more positive than that of Y3+/Y2+.

3 Upon reduction from YCl2(aq) to [YCl2]–(aq), the solution turns colourless.

32 Which of the following graphs have the same general shape according to the ideal gas

law for a fixed mass of gas?

1 PV against P (at constant T) and V/T against T (at constant P)

2 V against T (at constant P) and V against 1/P (at constant T)

3 T against 1/P (at constant V) and T against V (at constant P)

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VJC 2014 9647/01/PRELIM/14

The responses A to D should be selected on the basis of

A B C D

1, 2 and 3 are

correct



1 only is

correct

No other combination of statements is used as a correct response. 33 The table below shows the results of three experiments conducted for the reaction

P + Q products.

[P] / mol dm–3 [Q] / mol dm–3 rate / mol dm–3 s–1

0.012 0.005 0.9 x 10–4

0.024 0.010 2.0 x 10–4

0.048 0.010 4.0 x 10–4

If the rate constant doubles for each 10 oC rise in temperature, which pairs of

experimental conditions will result in the same rate?

[P] / mol dm–3 [Q] / mol dm–3 temperature / oC

condition 1 0.10 0.20 40

condition 2 0.20 0.20 30

condition 3 0.30 0.30 20

1 conditions 1 and 2



15


34 Use of the Data Booklet is relevant to this question.

An experiment is set up as shown below. A steady current of 2 A was passed through the electrolyte for 10 min at s.t.p.

Which statements regarding the electrolysis are correct?

1 The anode becomes pinkish in colour.

2 140 cm3 of gas will be collected at the cathode.

3 A straight line graph showing the mass of anode decreasing with time can be obtained.

35 A Group II metal, M, undergoes two reaction routes.

Which sets contain at least two identical compounds?

1 P Q Y

2 Q W Y

3 R W X

Bronze (alloy of copper and tin)

electrode

Platinum electrode

100 cm3 of 0.01 mol dm–3 dilute sulfuric acid

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VJC 2014 9647/01/PRELIM/14

The responses A to D should be selected on the basis of

A B C D

1, 2 and 3 are

correct



1 only is

correct

No other combination of statements is used as a correct response. 36 The properties of astatine (At), a Group VII element, can be deduced from the trends of

other Group VII elements. Which of the following statements about astatine are likely to be correct?

1 HAt is a stronger acid compared to HCl.

2 At2 disproportionates in hot dilute sodium hydroxide to yield AtO3– and At– as

major products.

3 Potassium astatide forms a precipitate with silver nitrate quickly and can dissolve in dilute aqueous ammonia solution.

37 Diphenol P reacts with Q to give a polymer. A segment of the polymer with two repeat

units is shown below.

Which of the following statements are correct?

1 Condensation reaction takes place.

2 One HCl molecule is removed for each repeat unit of the polymer formed.

3 A polymer made up of n number of repeat units will contain n number of chiral centres.

38 Which of the following compounds will give a precipitate of triiodomethane (iodoform)

when reacted with iodine and aqueous sodium hydroxide?

1 CH3CH2CH(OH)CH3

2 C6H5COCH3

3 CH3OCOCHI2

17


39 Below are the structures of compounds X and Y.

Which sets of reagents and conditions can be used to distinguish between compounds

X and Y?

1 acidified K2Cr2O7, heat

2 aqueous alkaline iodine, heat

3 NaOH(aq), followed by excess HNO3(aq) and AgNO3(aq) 40 Which of the following are involved in the mechanism of the reaction between aqueous

sodium hydroxide and 2–bromo–2–methylbutane?

1 a curly arrow from a lone pair on the OH– ion to the C+ atom of 2–bromo–2–methylbutane

2 the heterolytic fission of the C–Br bond

3 an attack by a nucleophile on a carbocation

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VJC 2014 9647/01/PRELIM/14

Level of difficulty Percentage No of questions

Easy 20 8

Moderate 60 24

Hard/Tedious 20 8

MCQ Questions for H2 Chemistry

Qn No Topic Remarks Answer

1 Stoichiometry (molecular formula) E D

2 Stoichiometry (redox calculation) T D

3 Atomic Structure (no. of unpaired e–) M B

4 Bonding (no. of e–) M C

5 Bonding (bond angle) M D

6 Gases (intermolecular forces) E C

7 Eo (change in mass of electrode) M C

8 H (calculation) E B

9 Chemical Eqm E C

10 Eqm position (complex ion, LCP) M A

11 Ionic Eqm (buffer) M D

12 Kinetics (mechanism) M C

13 Periodic Table (physical properties) M D

14 Group II (thermal decomposition) M A

15 Group VII (oxidising power of X–, solubility of AgX) M D

16 TM (reactions) T C

17 TM (feasibility of reactions) T B

18 Organic (RCN/RCO/RCO2H reactions)+ hybrid orbitals M A

19 Organic (mechanism) M D

20 Organic (oxidation of alkenes) M B

21 Organic (acidity + hydrolysis of ester) M C

22 Organic (C=C/RCHO reactions) T B

23 Organic (solubility of organic compounds) E A

24 Organic (synthesis) M A

25 Organic (pattern) T A

26 Organic (reaction with SOCl2 and HCl) E D

27 Organic (amine/amide reactions) M D

28 Organic (preparations) M C

29 Amino acid (pKa) M B

30 Amino acid (electrophoresis) M C

31 Atomic Structure, Eo, TM T B

32 Gases (Graphs) M B

33 Kinetics (initial rate method, effect of temperature) M D

34 Electrolysis T A

35 Group II (reactions) M C

36 Group VII (reactions) E B

37 Organic (nucleophilic substitution reaction, polymer) T A

38 Organic (iodoform reaction) E B

39 Organic (distinguishing tests) M A

40 Organic (SN1 mechanism) M C

9 A, 10 B, 11 C, 10 D

2014 JC2 H2 Chemistry Preliminary Examination Paper 1 Detailed Answers

1 D 2 D 3 B 4 C 5 D 6 C 7 C 8 B 9 C 10 A 11 D 12 C 13 D 14 A 15 D 16 C 17 B 18 A 19 D 20 B 21 C 22 B 23 A 24 A 25 A 26 D 27 D 28 C 29 B 30 C 31 B 32 B 33 D 34 A 35 C 36 B 37 A 38 B 39 A 40 C 1 D

Molecular formula of compound X is C4H8O2.

2 D

[O]: Tl+(aq) Tl

3+(aq) + 2e

–

[R]: VO3–(aq) + 6H

+(aq) + 3e

– V

2+(aq) + 3H2O(l)

Redox: 3Tl+(aq) + 2VO3

–(aq) + 12H

+(aq)

2V2+

(aq) + 3Tl3+

(aq) + 6H2O(l)

Add 2 mol of VO3

– to the redox equation will lead to

equal number of moles of VO3–(aq) and V

2+(aq):

3Tl

+(aq) + 4VO3

–(aq) + 12H

+(aq)

2VO3–(aq) + 2V

2+(aq) + 3Tl

3+(aq) + 6H2O(l)

3 B

Option Valence shell electronic

configuration of ion No. of

unpaired e–

A ns2np

2 2

B ns2np

5 1

C ns2np

3 3

D ns2np

2 2

4 C

Option Molecule No. of electrons on central atom

A

7

B

10

C

8

D

11

5 D

For A, SnCl2: ~118o (2 bp, 1 lp), OCl2: 105

o (2 bp, 2 lp)

For B, both are bent in shape. The electronegativity of O atom is greater than that of S atom. Hence the bond pair of electrons is closer to the nucleus of O atom resulting in stronger bp–bp repulsion in H2O and a larger bond angle for H2O. For C, both are linear in shape. For D, both are trigonal pyramidal in shape. The electronegativity of F atom is greater than that of Cl atom. Hence the bond pair of electrons is further away from the N atom resulting in weaker bp–bp repulsion in NF3 and a smaller bond angle for NF3.

6 C Weak dispersion forces between CH3CH2CH3 molecules, dipole–dipole attractions between CH3CHO molecules, hydrogen bonding between CH3CH2OH molecules and hydrogen bonding between HCO2H molecules. The stronger the attractive forces between the molecules, the greater the deviation from ideal gas behaviour. HCO2H is able to form more extensive hydrogen bonding between its molecules. Hence it has a greater deviation from ideal gas as compared to CH3CH2OH.

7 C The same current flows through all the electrodes. Assuming 6 mol of e

- is passed, the reactions at the

various electrodes are as shown.

E (oxidation): 2Al 2Al3+

+ 6e–

F (reduction): 6H2O + 6e– 3H2 + 6OH

–

G (oxidation): 6Ag 6Ag+ + 6e

–

H (reduction): 3Cu2+

+ 6e– 3Cu

Electrode F has no change in mass of electrode. Comparing the number of moles of electrode involved in the electrolysis, G has the greatest change in mass of electrode compared to E and H.

8 B

Mass of water = 150.00 – 50.00 = 100.00 g Change in temperature of water = 53 – 20 = 33

oC

Heat gained by water = mc∆T = (100)(4.18)(33) = 13800 J 9 C

For A and B, forward endothermic reaction is favoured at higher temperature, leading to higher Kc and increase in partial pressure of NO2. For C, enthalpy change (∆H) does not change with temperature. For D, activation energy is unchanged unless in the presence of a catalyst.

10 A

NH3 + H2O ⇌ NH4+ + OH

–

For A, BaSO4 or its ions do not react with NH3(aq). For B, AgCl reacts with NH3(aq) to give [Ag(NH3)2]

+Cl

–.

Thus equilibrium is affected. For C, increase in [OH

–] will cause equilibrium position

to shift left. For D, decrease in [Cu(H2O)6

2+] due to Cu

2+(aq) +

2OH–(aq) ⇌ Cu(OH)2(s) will cause equilibrium position

to shift left. 11 D

For A, excess sodium hydrogencarbonate with hydrochloric acid will produce the buffer consisting of a mixture of H2CO3 and HCO3

–.

For B, [H

+] = 10

–pH = 3.98 x 10

–8 mol dm

–3

Ka = [H+][HCO3

–] / [H2CO3]

[HCO3–] / [H2CO3] = (7.90 x 10

–7) ÷ (3.98 x 10

–8)

≈ 20

For C, dilution with water affects the [HCO3–]/[H2CO3]

equally. Hence there is no change in its pH value. For D, pH = pKa when [HCO3

–] = [H2CO3], i.e.

maximum buffer capacity. Since [HCO3-] / [H2CO3] ≈ 20,

pH ≠ pKa.

12 C Step 2 is the rate–determining step and rate = k’[CHCl3][Cl]. Since Cl is an intermediate, it cannot be in the rate equation and [Cl] is dependent on [Cl2] in step 1.

Kc =

2

2 [Cl] = (Kc)

1/2 [Cl2]

1/2

Thus, rate = k’[CHCl3] (Kc)1/2

[Cl2]1/2

= k[CHCl3] [Cl2]1/2

where k = k’(Kc)

1/2.

13 D

For A, Mg has the highest 1st IE among the three

elements but its melting point is not the lowest. For B, Al has the lowest 1

st IE among the three

elements and its melting point is higher than Mg. For C, Si has a lower 1

st IE than P and its melting point

is the highest. For D, P has the highest 1

st IE among the three

elements and its melting point is the lowest. Si has the highest melting point due to strong covalent bonds in a giant molecular structure. S8 has a higher melting point than P4 as it has a greater number of electrons leading to stronger dispersion forces.

14 A From Data Booklet, ionic radii of Be

2+ is 0.031 nm,

Mg2+

is 0.065 nm and Al3+

is 0.050 nm. Hence Be2+

has the highest charge density and Mg

2+ has the

lowest charge density. The higher the charge density, the greater the distortion of the electron cloud of nitrate ion leading to lower thermal decomposition temperature. Thus, decomposition temperature of Mg(NO3)2 > Al(NO3)3 > Be(NO3)2.

15 D

Y2 can undergo displacement reactions with X– and Z

–.

X2 can undergo displacement reaction with Z– only.

Hence on going down Group VII it will be Y, X and Z as oxidising power of halogen decreases down the group. AgY is expected to be the most soluble in aqueous ammonia and AgZ the least soluble in aqueous ammonia.

16 C

For A (incorrect), addition of acid (such as H2SO4(aq)) instead of NaOH(aq) to K2CrO4(aq) will produce an orange solution of K2Cr2O7. For B (incorrect), the redox reaction between Fe

3+(aq)

and I–(aq) is feasible from the Data Booklet:

2Fe3+

+ 2I– 2Fe

2+ + I2

For C (correct), [Cu(H2O)6]

2+ + 4Cl

– ⇌ CuCl4

2– + 6H2O,

yellow solution of H2[CuCl4] is produced. For D (incorrect), Na2CO3(aq) hydrolyses to form HCO3

–(aq) and OH

–(aq), and CrCl3(aq) hydrolyses to

form Cr(H2O)5(OH)2+

(aq) and H3O+(aq). Effervescence

of CO2 and green ppt of Cr(OH)3 instead of Cr2(CO3)3 are produced when Na2CO3(aq) is added to CrCl3(aq).

17 B Step 1: redox

2Fe3+

+ 2I– 2Fe

2+ + I2 E

ocell = +0.77 – (+0.54) > 0 V

Step 2: ligand exchange

Fe(H2O)62+

+ 6CN– Fe(CN)6

4– + 6H2O

Step 3: redox

Fe(CN)64–

+ ½Cl2 Fe(CN)63–

+ Cl–

Eo

cell = +1.36 – (+0.36) > 0 V For A, step 3 is incorrect as SO2 cannot be reduced by Fe(CN)6

4–.

For C, step 2 is incorrect as Fe2+

is oxidised to Fe3+

while I– cannot be reduced.

For D, step 1 is incorrect as reaction is not feasible since E

ocell = +0.77 – (+0.80) < 0 V. Step 2 is incorrect

as Fe2+

is oxidised to Fe3+

while I– cannot be reduced.

18 A

Compound E contains nitrile, ketone and carboxylic acid. For A, NaBH4 is a weaker reducing agent than LiAlH4 and will reduce nitrile and ketone to sp

3 hybridised

carbon atoms, leaving behind the C=O of the carboxylic acid as the only sp

2 hybridised carbon atom

in the product. For B, LiAlH4 will reduce all the functional groups to sp

3 hybridised carbon atoms, leaving no sp

2 hybridised

carbon atom in the product. For C, nitrile undergoes hydrolysis in the presence of HCl(aq) to give carboxylic acid. There are three sp

2

hybridised carbon atoms in the product. For D, carboxylic acid reacts with SOCl2 to give acid chloride. There are two sp

2 hybridised carbon atoms in

the product. 19 D

For A (correct), propan–2–ol is a proton acceptor in step 1 and hence it acts as a base. For B (correct), H2SO4 is a catalyst as it is regenerated at the end of step 3. For C (correct), the reactants in step 3 can undergo addition reaction to give CH3CH(OSO3H)CH3. For D (incorrect), the formation of carbocation in step 2 is not likely for primary alcohols as compared to tertiary alcohols since primary carbocation is less stable than tertiary carbocation.

20 B

Option Products

A CH3CO2H CH3COCH2CH2COCH3 CH2(CO2H)2

B CH3COCH3

CH3COCH2CH2CO2H CH2(CO2H)2

C CH3COCH3

CH2(CO2H)CH2COCO2H CH3CH2CO2H

D CH3COCH3

CH3COCH2COCH3

CH2(CO2H)2

21 C

P is which is a stronger acid with a

smaller pKa value compared to Q, .

R is which undergoes base hydrolysis to form C6H5O

−Na

+ and CH3CO2

–Na

+.

is a less stable anion than as the electron donating inductive effect of

is greater than that of , intensifying the negative charge on –CO2

–.

22 B

For A (incorrect), five C=C and one RCHO reduced by 6 mol of H2(g). For B (correct), 1 1–cis–retinal and NH2–opsin undergo condensation reaction to form 1 mol of H2O. For C (incorrect), C=C is not reduced by LiAlH4. RCHO

is reduced to a primary alcohol which does not have a chiral centre. For D (incorrect), only three different organic products are formed. Hot H

+/KMnO4 will cleave the C=C bonds

to form five products. However, the 2 mol of ethanedioic acid (see circled) will be further oxidised to CO2 and H2O.

23 A

CH3CH2CO2Na exists as ions in solution and form strong ion–dipole interactions with water, making it the most soluble. CH3CH2CH2NH2 molecules form hydrogen bonds with water which are weaker than ion–dipole interactions but stronger than dipole–dipole interactions formed between CH3CH2CH2Cl molecules and water.

24 A

For A, the C=C bond is cleaved by hot H+/KMnO4.

However, the strong oxidation will oxidise (CO2H)2 to CO2 and H2O. For B, compound can be formed by reducing fumaric acid using H2/Pt catalyst. For C, compound can be formed by oxidising fumaric acid using cold acidified KMnO4. For D, compound can be formed by hydrating fumaric acid using cold concentrated H2SO4 followed by warming with water, and then oxidising the 2

o alcohol

to ketone using hot H+/KMnO4.

25 A From given pattern,

Thus,

26 D

Aliphatic alcohols undergo nucleophilic substitution with both SOCl2 and HCl. Carboxylic acids undergo nucleophilic substitution with only SOCl2. Phenol does not undergo nucleophilic substitution with either SOCl2 or HCl.

27 D

HBr(aq) acts as an acid. It hydrolyses the amide linkage and neutralises the amine groups to form salts.

28 C

For A, phenol with Br2 in CCl4 will result in monosubstitution instead of trisubstitution. For B, ethene is a gas and hence setup is unsuitable. For C, ethanamide is a liquid at rtp and can be placed in the conical flask while lithium aluminium hydride in dry ether (liquid) can be added dropwise from the funnel to the conical flask at cold temperature. For D, formation of ester from alcohol and carboxylic acid in the presence of concentrated sulfuric acid catalyst requires reflux, not cold temperature.

29 B

The fully protonated form of aspartic acid, , is a tribasic acid. Deprotonation occurs when the given pH is above the pKa value. At pH 7, deprotonation

occurs at ① (α–CO2H) where pKa = 2.10 and ②

(–CO2H at R group) where pKa = 3.86. No

deprotonation at ③ (α-amino group) where pKa = 9.82. 30 C

At pH 10, the acidic carboxylic and phenolic groups will be deprotonated. B and C both have a charge of –2 but C will move more readily to the positive anode as the R group has a smaller molecular mass.

At pH 10:

A:

B:

C:

D:

α

α

①

② ③

α

α

31 B Based on the given electronic configurations, X is Co and Y is Cu. For 1, the maximum oxidation state of X is +5 (no. of unpaired 3d + 4s electrons) and does not exist as K2X2O7 with an oxidation state of +6. For 2, Y

3+ lies further down the period compared to X

3+

and has a higher effective nuclear charge. Hence Y3+

is less stable than X

3+ and has a higher tendency to be

reduced compared to X3+

. Eo value of Y

3+/Y

2+ should

be more positive. For 3, Y

2+ in YCl2(aq) has an incomplete 3d subshell in

which d–d transition can occur while Y+ in [YCl2]

–(aq)

has a complete 3d subshell in which d–d transition cannot occur. Hence [YCl2]

–(aq) is colourless.

32 B

For 1, both graphs are horizontal lines, i.e. Y = constant. For 2, both graphs are straight lines passing through the origin, i.e. Y = kX. For 3, T against 1/P (at constant V) has the shape of Y = k/X while T against V (at constant P) is a straight line passing through the origin, i.e. Y = kX.

33 D

Reaction is first order with respect to P and zero order with respect to Q. Hence rate is dependent on temperature (doubles for every 10

oC rise) as well as [P].

Let the rate of [P] = 0.10 mol dm

–3 at 20

oC be r.

Condition 1: T increases by 20 oC, rate = (2)

2r = 4r

Condition 2: [P] doubles and T increases by 10 oC,

rate = (2)2r = 4r

Condition 3: [P] increases by 3 times, rate = (2)3r = 8r

34 A

For 1, at the anode Sn is oxidised to Sn2+

, leaving behind the pink copper metal as E

o(Sn

2+/Sn) is less

positive than Eo(Cu

2+/Cu).

For 2, at the cathode: 2H+ + 2e

– H2

Q = I t = 2 10 60 = 1200 C

No. of moles of electrons = 1200 96500 = 0.0124 mol

Volume of H2 gas = 0.0124 2 22400 140 cm3

For 3, since current is steady at 2 A, the mass of anode will decrease at a constant rate.

35 C

Solid P: MO, solution of Q: M(OH)2, solid R: MO Solution of W: M(OH)2, solid X: MO, solution of Y: MSO4

36 B

For 1, H–At has a weaker covalent bond than H–Cl due to larger atomic size of At. Hence HAt is a stronger acid than HCl as it undergoes dissociation more

readily to produce a larger [H+].

For 2, since I2 disproportionates in hot dilute sodium hydroxide to yield IO3

– and I

– as major products, At2

which is below I2 in Group VII would also behave likewise. For 3, solubility of AgX in dilute aqueous ammonia solution decreases down Group VII. Since AgI is

insoluble in dilute aqueous ammonia solution, AgAt would also be insoluble in dilute aqueous ammonia solution.

37 A Condensation reaction takes place between P and Q to form the polymer and HCl molecules. One HCl molecule is removed for each repeat unit of the polymer formed. The repeat unit is

and each repeat unit contains one chiral centre.

38 B To have a positive iodoform test, the molecule must

contain either alcohol with group or

carbonyl with group. For 3, the ester is hydrolysed by NaOH(aq) to form CH3OH and CHI2CO2

–Na

+. CH3OH does not undergo

positive iodoform test. CHI2CO2–Na

+ can undergo

nucleophilic substitution with NaOH(aq) to form CH(OH)2CO2

–Na

+ which then eliminate water to form

. 39 A

For 1, ester group in X undergoes acid hydrolysis to form 3

o alcohol which cannot be oxidised by K2Cr2O7.

There is no change in the colour of solution. However, the acid hydrolysis of Y formed 2

o alcohol which can

be oxidised by K2Cr2O7. The colour of solution changes from orange to green. For 2, both the ester groups in X and Y undergo base hydrolysis. However, only Y shows a positive iodoform test due to presence of CH3CH(OH)– group after hydrolysis. For 3, base hydrolysis of R–Br in X followed by excess HNO3(aq) and AgNO3(aq) results in a cream ppt of AgBr. No cream ppt is observed for Y as the Br atom is bonded to C=C bond where the p orbital of Br interacts with the p orbital of C in C=C, resulting in partial double bond character in the C–Br bond.

40 C

Since 2–bromo–2–methylbutane is a tertiary bromoalkane, it undergoes SN1 mechanism.

slowCR

R'

R''

X CR

R'

R''

+ + X

fastCR

R'

R''

OHCR

R'

R''

+ + OH-

*


CANDIDATE NAME CT GROUP


……………………………………………….…………..

……………………………..

CHEMISTRY

9647/02

Paper 2 Structured

Candidates answer on the Question Paper.

16 September 2014

2 hours

Additional Materials: Data Booklet

READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams, graphs or rough working.


Answer all questions.

A Data Booklet is provided.

The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

1

2

3

4

5

Total


VJC 2014 9647/02/PRELIM/14

2

Answer all the questions in the space provided.

1 Planning (P)

Aqueous chlorine, Cl2, displaces iodine, I2, from aqueous potassium iodide. If a solution of chlorine is mixed with an excess of aqueous potassium iodide, iodine is displaced in a 1:1 molar ratio with chlorine.

The concentration of chlorine in the original solution can therefore be calculated from the concentration of the displaced iodine.

I2(aq) + 2Na2S2O3(aq) 2NaI(aq) + Na2S4O6(aq)

(a) Write an equation for the reaction between aqueous chlorine, and an excess of aqueous potassium iodide.

…………………………………………………………………………………………………..

[1]

(b) You are to plan an experiment to determine as accurately as possible the

concentration of a saturated aqueous solution of chlorine by titration. The approximate solubility of chlorine is 5 g dm–3 at 25 oC.

You are provided with the following materials:

saturated aqueous chlorine;

solid sodium thiosulfate Na2S2O3.5H2O;

concentrated aqueous potassium iodide. This will be used in excess.

the apparatus and chemicals normally found in a school or college laboratory. Your plan should include details of:

a calculation of the approximate concentration of saturated aqueous chlorine in mol dm–3 at 25 oC;

[Ar: Cl, 35.5]

a detailed description of the method for preparing a solution of aqueous sodium thiosulfate that can be used in the titration. In a titration, it is usual for the titre volume to be approximately equal to that of the volume of solution pipetted. Calculate the mass of sodium thiosulfate, Na2S2O3.5H2O, which will produce a solution suitable for use in this titration. The relevant calculations and reasoning must be shown in full;

[Ar: H, 1.0; O, 16.0; Na, 23.0; S, 32.1]

a step–by–step description of how you would carry out sufficient titrations using a suitable indicator to allow an accurate end–point to be obtained;

an outline calculation to show how the results are to be used to determine the accurate concentration of the aqueous chlorine. Assume that the titre volume used is w cm3.

…………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

3


………………………………………………………………………….............................

………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

VJC 2014 9647/02/PRELIM/14

4

…………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ ………………………………………………………………………............................... ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ ………………………………………………………………………….............................

[10]

(c) State one hazard that must be considered when planning the experiment and describe a precaution that should be taken to keep risks from this hazard to a minimum.

…………………………………………………………………………………………………... ……………………………………………………………………………………….................

[1]

[Total: 12]

5


2 (a) On Planet Uranus, it is postulated that the number of subshells associated with each

principal quantum number and the respective energy levels of the subshells are similar to that on Earth. Each orbital contains a maximum of two electrons. However, the number of orbitals that make up a subshell may or may not be identical to that on Earth.

Figure 1 represents the sketch of the successive ionisation energies of all the electrons of an element T on Planet Uranus.

Figure 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19No. of ionisation

log

IE

(i) By interpreting figure 1, suggest with reasoning, which period of the Periodic Table element T belongs to.

…………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

(ii) Using figure 1, deduce the number of 2p orbitals present in element T. …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

I I I I I I I I I I I I I I I I I I I

Figure 1 No. of electrons removed

lg IE

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

VJC 2014 9647/02/PRELIM/14

6

(iii) State the full electronic configuration of element T.

…………………………………………………………………………………................

[3]

(b) The hydrolysis of ester to form acid and alcohol is an equilibrium reaction. Using ethyl ethanoate as an example, the reaction for the hydrolysis is shown below.

CH3CO2CH2CH3(l) + H2O(l) ⇌ CH3CO2H(l) + CH3CH2OH(l)

Draw two labelled Maxwell–Boltzmann distribution curves to illustrate clearly how the addition of concentrated sulfuric acid and heating of reaction mixture can increase the rate of hydrolysis. (Written explanation is not required.)

Addition of concentrated sulfuric acid: Heating of reaction mixture:

[3]

7


(c) The rate of hydrolysis (in mol dm–3 s–1) of ethyl ethanoate at 25 oC is given by the

following equation.

rate = (4.0 x 10–6 [ester][H+]) + (2.5 x 10–1 [ester][OH–])

(i) Calculate the rate of hydrolysis at pH = 1.0 and 13.0 respectively, given [ester] = 0.0100 mol dm–3.

pH 1.0 pH 13.0: (ii) Comment on your answers in (c)(i) by relating to the rate equation.

…………………………………………………………………………............................. ………………………………………………………………………................................

[3] (d) Tetracycline is a broad–spectrum polyketide antibiotic indicated for use against many

bacterial infections. It is commonly used to treat acne today. Compound M, a derivative of tetracycline, has the structure shown below.

compound M

Draw the structures of the organic products formed when compound M is treated with

(i) K2Cr2O7(aq), H2SO4(aq), heat

VJC 2014 9647/02/PRELIM/14

8

compound M

(ii) aqueous bromine, room temperature

(iii) LiAlH4 in dry ether, followed by addition of water

(iv) Suggest a simple chemical test to distinguish between compound M and compound N. You should state clearly the reagents and conditions used, and the observations you would expect to make with each compound.

compound N …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

[6]

[Total: 15]

9


3 Pseudohalogens are a class of compounds that show chemical reactivity similar to

halogens. An example of a pseudohalogen and its anion is (SCN)2 and SCN– respectively. thiocyanogen thiocyanate

(a) (i) Draw the shape of the hybrid orbitals around one of the carbon atoms in thiocyanogen.

(ii) Draw a diagram to illustrate the shape of thiocyanogen, indicating clearly the

covalent bonds and lone pairs of electrons. (iii) Thiocyanogen can be synthesised by reacting sodium thiocyanate, NaSCN, with

chlorine gas. Write an equation for the formation of thiocyanogen. …………………………………………………………………………………..................

[4]

(b) Thiocyanogen can react with hydrocarbons such as methylbenzene via free radical substitution.

(i) Write equations that describe the following steps. Initiation Propagation

N C S S C N S C N

HSCN

CH3

+ (SCN)2uv

CH2SCN

+

VJC 2014 9647/02/PRELIM/14

10

Termination (Write down all the possible termination steps.)

(ii) An isomer of the product shown below was also isolated. Suggest a reason for the formation of this isomer.

………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

[4]

(c) The standard reduction potential of (SCN)2 is given as

½(SCN)2 + e– ⇌ SCN– E = +0.43 V

With the aid of the Data Booklet, predict the reactions, if any, that might occur when the following reagents are mixed. Write a balanced equation for the reaction that occurs.

(i) Na2S2O3 and (SCN)2

(ii) acidified K2Cr2O7 and (SCN)2

[3]

CH2 N C S

11


(d) SCN– acts as a ligand in the formation of complex ions. The formation of a complex

ion from the hydrated metal ion is shown in the following equilibrium:

[M(H2O)6]m+(aq) + nL–(aq) ⇌ [M(H2O)6–n Ln]

(m–n)+(aq) + nH2O(l) where m and n are whole numbers.

The equilibrium constant is called the stability constant, Kstab. The table below provides information for the formation of two complexes.

Complex ion Mm+ L– n Kstab Colour

Iron(III) thiocyanate

Fe3+ SCN– 1 90 Red

Cobalt(II) thiocyanate

Co2+ SCN– 4 100 Blue

(i) Write an equation for the formation of cobalt(II) thiocyanate and hence write an

expression for its equilibrium constant, Kstab and state its units.

(ii) Calculate the concentration of the complex ion formed when 0.0050 mol of solid cobalt(II) chloride was added to a 2 dm3 solution of 0.95 mol dm–3 KSCN solution.

(iii) In an experiment to study the stoichiometry of cobalt(II) thiocyanate, a solution of 1.0 x 10–3 mol dm–3 cobalt(II) chloride is mixed with different proportions of a solution of 1.0 x 10–3 mol dm–3 KSCN(aq) such that the total volume of each mixture is kept constant at 10 cm3. The absorbance due to the complex ion in each mixture is measured with a colorimeter. The absorbance is directly proportional to the concentration of the complex.

VJC 2014 9647/02/PRELIM/14

12

In the grid below, sketch the graph you would expect to obtain.

absorbance

0 2 4 6 8 10 Volume of cobalt(II) chloride / cm3 10 8 6 4 2 0 Volume of KSCN / cm3

[6]

(e) KSCN is used to standardise a solution of silver nitrate. A white precipitate of AgSCN (Ksp = 1.1 x 10–12 mol2 dm–6) is formed during the titration. The indicator used in the titration is yellow aqueous iron(III) solution.

Ag+(aq) + SCN–(aq) ⇌ AgSCN(s)

In an experiment, 5 cm3 of 5 mol dm–3 nitric acid, 25 cm3 of water and 1 cm3 of iron(III) solution are added to 25.0 cm3 of silver nitrate solution in a conical flask. End-point of titration is reached when 15.00 cm3 of 0.100 mol dm–3 KSCN solution is added.

(i) Calculate the concentration of the original silver nitrate solution.

(ii) Calculate the concentration of Ag+(aq) in the solution at the equivalence point.

(iii) With reference to the table given in (d), explain why iron(III) solution can be used as an indicator and state the observation at the end-point of the titration.

………………………………………………………………………………................... ……………………………………………………………………………….................... …………………………………………………………………………………..................

[4]

[Total: 21]

13


4 (a) Haemoglobin is an important protein present in the red blood cell, which is responsible for transport of oxygen in the body. It contains a haem group and a metal centre as shown below. The four N atoms and the Fe atom are planar.

(i) State the types of hybridisation at N1 and N2, and the nature of the bonding around Fe atom.

Hybridisation at N1: ………………… Hybridisation at N2: …………………

Type of bond between Fe atom and N1: ……………………………………………… Type of bond between Fe atom and N2: ………………………………………………

(ii) One molecule of haemoglobin (Hb) binds up to four oxygen molecules to form oxyhaemoglobin (Hb(O2)4), as shown in the following equation.

Hb(aq) + 4O2(g) ⇌ Hb(O2)4(aq) (I)

By applying Le Chatelier’s Principle, explain how oxygen molecules are transported from the lungs to the body tissues.

…………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ ……………………………………………………………………………………………..

(iii) The percentage saturation of haemoglobin is defined as the percentage of haemoglobin present in the blood that has been converted to oxyhaemoglobin.

With reference to equilibrium (I), state and explain how the percentage saturation of haemoglobin changes when temperature increases.

…………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

VJC 2014 9647/02/PRELIM/14

14

(iv) The relationship between the percentage saturation of haemoglobin and concentration of oxygen in blood at different temperatures is shown below.

Suggest the optimum temperature at which haemoglobin operates. Explain your

answer. …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

[8] (b) (i) Draw the dot-and-cross diagram of ozone, O3. (ii) Oxygen–oxygen bond lengths of 3 species are given in the table below.

Species Oxygen–oxygen bond length / nm

O3 0.128

O22– 0.149

O2 0.120

With the aid of suitable structures, explain why the two oxygen-oxygen bonds in the ozone molecule have the same bond length which are very different from the peroxide ion and oxygen molecule.

[3]

[Total: 11]

Percentage Saturation / %

[O2] / mol dm–3

37 °C

50 °C

10 °C

[O2]lungs [O2]tissues

15


5 (a) Eugenoxyacetic acid is a colourless and odourless compound. It has shown anti-viral and anti-bacterial properties and is therefore used as antioxidant food preservative in food industry.

Eugenoxyacetic acid can be synthesised from eugenol found in clove essential oil as shown in the procedure below via SN2 mechanism.

(i) How can you tell that the solution has just turned alkaline in step 1? ………………………………………………………………………................................ ………………………………………………………………………….............................

(ii) Explain the role of NaOH(s) in step 2. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

(iii) Write a balanced equation for the reaction in step 3.

(iv) Comment if an excess amount of NaOH(s) can be used in step 2. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

Preparation of solution A

1. Dissolve 1.0 g of 2-chloroethanoic acid in 5.0 cm3 of distilled water in a beaker. Add Na2CO3(s) slowly until the solution just becomes alkaline.

Preparation of eugenoxyacetic acid

2. Dissolve 0.6 g of NaOH(s) in 3.0 cm3 of distilled water in a conical flask. Add 2.0 cm3 eugenol to the flask.

3. Add solution A to the flask in step 2 and keep stirring the mixture for 60 min.

4. Cool the reaction mixture to room temperature and add HCl(aq). Collect the solid formed by suction filtration. Wash the solid with water to obtain a pale yellow solid which is the crude product.

VJC 2014 9647/02/PRELIM/14

16

(v) Give a simple chemical test to confirm that eugenol is absent in the final product. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

[6]

(b) The structure of a small protein is given below, along with the abbreviated names of its constituent amino acids: val, gln etc.

(i) Use the abbreviated names of the amino acids to state the primary structure of the above protein.

………………………………………………………………………………….................. (ii) This protein is ionised at pH 7. Identify one functional group that could be ionised

at pH 7 by circling it on the structure above. (iii) The biological function of the protein depends on its three dimensional shape.

Describe how the particular amino acids in the above protein are likely to be involved in maintaining its three-dimensional shape.

…………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................ …………………………………………………………………………............................. ………………………………………………………………………................................

17


(iv) The activity of the above protein is destroyed by heating it under reflux with

6 mol dm–3 sulfuric acid for several hours.

Draw the structures of the organic products formed under these conditions from the part of the above protein containing the amino acids: val-gln.

[7]

[Total: 13]


2014 Preliminary Examination H2 Chemistry Paper 2 1 Planning (P)

Aqueous chlorine, Cl2, displaces iodine, I2, from aqueous potassium iodide. If a solution of chlorine is mixed with an excess of aqueous potassium iodide, iodine is displaced in a 1:1 molar ratio with chlorine. The concentration of chlorine in the original solution can therefore be calculated from the concentration of the displaced iodine.

I2(aq) + 2Na2S2O3(aq) 2NaI(aq) + Na2S4O6(aq)

(a) Write an equation for the reaction between aqueous chlorine, and an excess of aqueous potassium iodide.

Cl2(aq) + 2KI(aq) I2(aq) + 2KCl(aq) [1] [Accept ionic equation, state symbols are optional.] (b) You are to plan an experiment to determine as accurately as possible the

concentration of a saturated aqueous solution of chlorine by titration. The approximate solubility of chlorine is 5 g dm–3 at 25 oC.

You are provided with the following materials:

saturated aqueous chlorine;

solid sodium thiosulfate Na2S2O3.5H2O;

concentrated aqueous potassium iodide. This will be used in excess.

the apparatus and chemicals normally found in a school or college laboratory. Your plan should include details of:

a calculation of the approximate concentration of saturated aqueous chlorine in mol dm–3 at 25 oC;

[Ar: Cl, 35.5]

a detailed description of the method for preparing a solution of aqueous sodium thiosulfate that can be used in the titration. In a titration, it is usual for the titre volume to be approximately equal to that of the volume of solution pipetted. Calculate the mass of sodium thiosulfate, Na2S2O3.5H2O, which will produce a solution suitable for use in this titration. The relevant calculations and reasoning must be shown in full;

[Ar: H, 1.0; O, 16.0; Na, 23.0; S, 32.1]

a step–by–step description of how you would carry out sufficient titrations using a suitable indicator to allow an accurate end–point to be obtained;

an outline calculation to show how the results are to be used to determine the accurate concentration of the aqueous chlorine. Assume that the titre volume used is w cm3.

Approximate concentration of saturated Cl2(aq)

=

= 0.0704 mol dm–3 Assume 10.0 cm3 of saturated Cl2(aq) is pipetted in each titration. No. of moles of saturated Cl2(aq) used

= 0.0704 x

= 7.04 x 10–4 mol = no. of moles of I2 formed

VJC 2014 9647/02/PRELIM/14

2

No. of moles of Na2S2O3 needed for titration = 2 x 7.04 x 10–4 = 1.41 x 10–3 mol

Assume 10.0 cm3 of Na2S2O3 is needed for each titration and 100 cm3 volumetric flask is used to prepare Na2S2O3 solution.

Concentration of Na2S2O3 needed

= 1.41 x 10–3

= 0.141 mol dm–3 No. of moles of Na2S2O3.5H2O needed

= 0.141 x

= 0.0141 mol Mass of Na2S2O3.5H2O needed = 0.0141 x [2(23.0) + 2(32.1) + 3(16.0) + 5(18.0)] = 3.50 g

[Accept other combinations of pipette and volumetric flask volumes with correct justification.]

Procedure [mass and volumes ecf from above justification]: Step 1: Accurately weigh 3.50 g of Na2S2O3.5H2O in a weighing bottle and completely

dissolve it in distilled water using a beaker. Pour the solution and the washings into a 100 cm3 volumetric flask and make up to the mark with distilled water. Shake well to obtain a homogenous solution. Label this solution FA1 and pour it into a burette.

Step 2: Pipette 10.0 cm3 of saturated aqueous chlorine into a conical flask. Step 3: Using a measuring cylinder, add 10 cm3 of concentrated KI(aq) into the

conical flask. Step 4: Titrate the iodine liberated in this solution with FA1. When the colour of the

solution turns pale yellow, add about 1 cm3 of starch indicator. The solution will turn blue–black. Continue the titration and the end–point is reached when the colour just disappears. Record the results.

Step 5: Repeat steps 2 to 4 as many times as necessary to obtain consistent results

to within ±0.10 cm3. Calculations:

Assume 10.0 cm3 of Cl2(aq) reacted with exactly w cm3 of FA1 of concentration 0.141 mol dm–3.

No. of moles of FA1 used

= 0.141 x

= 1.41w x 10–4 mol No. of moles of I2 formed

= 1.41w x 10–4 2 = 7.05w x 10–5 mol = no. of moles of Cl2(aq) used


3

Concentration of Cl2(aq) used

= 7.05w x 10–5

= 7.05w x 10–3 mol dm–3 [10]

(c) State one hazard that must be considered when planning the experiment and

describe a precaution that should be taken to keep risks from this hazard to a minimum.

Chlorine gas escapes from saturated aqueous chlorine which is poisonous. Perform the titration in a fume cupboard. OR Chlorine gas escapes from saturated aqueous chlorine which causes eye irritation. Perform the titration in a fume cupboard. OR Saturated aqueous chlorine is acidic/oxidising. Wear safety gloves when performing the titration.

[1]

[Total: 12] 2 (a) On Planet Uranus, it is postulated that the number of subshells associated with each

principal quantum number and the respective energy levels of the subshells are similar to that on Earth. Each orbital contains a maximum of two electrons. However, the number of orbitals that make up a subshell may or may not be identical to that on Earth.

Figure 1 represents the sketch of the successive ionisation energies of all the electrons of an element T on Planet Uranus.

Figure 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19No. of ionisation

log

IE

I I I I I I I I I I I I I I I I I I I

Figure 1 No. of electrons removed

lg IE

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

VJC 2014 9647/02/PRELIM/14

4

(i) By interpreting figure 1, suggest with reasoning, which period of the Periodic Table element T belongs to.

Since there are a total of three large increments between successive ionisation

energies, this implies that the 3rd, 12th and 18th electrons are removed from different principal quantum shells.

Hence, T belongs to the fourth period of the Periodic Table. (ii) Using figure 1, deduce the number of 2p orbitals present in element T. The 18th and 19th IE belong to 1s subshell. OR The 16th and 17th IE belong to 2s subshell. Thus, 12th to 15th IE, i.e. four electrons are removed from a 2p subshell. Since each orbital can accommodate a maximum of 2 electrons, there are two 2p

orbitals. (iii) State the full electronic configuration of element T. 1s2 2s2 2p4 3s2 3p4 3d3 4s2

[3]

(b) The hydrolysis of ester to form acid and alcohol is an equilibrium reaction. Using ethyl ethanoate as an example, the reaction for the hydrolysis is shown below.

CH3CO2CH2CH3(l) + H2O(l) ⇌ CH3CO2H(l) + CH3CH2OH(l)

Draw two labelled Maxwell–Boltzmann distribution curves to illustrate clearly how the addition of concentrated sulfuric acid and heating of reaction mixture can increase the rate of hydrolysis. (Written explanation is not required.)

Addition of concentrated sulfuric acid: Heating of reaction mixture: Correct axes for both Correct graph for addition of concentrated sulfuric acid with illustration/shading Correct graphs for heating of reaction mixture

[3]

number of molecules with a given energy

energy Ea

T1

T2

T2 > T1

number of molecules

with energy Ea at T1

number of molecules

with energy Ea at T2

number of molecules with a given energy

Ea’ Ea (cat.) (no cat.)

T1

energy


5

(c) The rate of hydrolysis (in mol dm–3 s–1) of ethyl ethanoate at 25 oC is given by the following equation.

rate = (4.0 x 10–6 [ester][H+]) + (2.5 x 10–1 [ester][OH–])

(i) Calculate the rate of hydrolysis at pH = 1.0 and 13.0 respectively, given [ester] =

0.0100 mol dm–3. pH 1.0: rate = (4.0 x 10–6)(0.0100)(10–1) + (2.5 x 10–1)(0.0100)(10–13) = 4.00 x 10–9 mol dm–3 s–1 pH 13.0: rate = (4.0 x 10–6)(0.0100)(10–13) + (2.5 x 10–1)(0.0100)(10–1) = 2.50 x 10–4 mol dm–3 s–1

(ii) Comment on your answers in (c)(i) by relating to the rate equation. The rate of hydrolysis of ethyl ethanoate is slower when the pH is low, i.e. at high

[H+], due to the much smaller rate constant associated with [H+]. OR The rate of hydrolysis of ethyl ethanoate is faster when the pH is high, i.e. at low

[H+], due to the much larger rate constant associated with [OH–]. [3]

(d) Tetracycline is a broad–spectrum polyketide antibiotic indicated for use against many

bacterial infections. It is commonly used to treat acne today. Compound M, a derivative of tetracycline, has the structure shown below.

compound M

Draw the structures of the organic products formed when compound M is treated with (i) K2Cr2O7(aq), H2SO4(aq), heat

VJC 2014 9647/02/PRELIM/14

6

(ii) aqueous bromine, room temperature

(iii) LiAlH4 in dry ether, followed by addition of water

(iv) Suggest a simple chemical test to distinguish between compound M and compound N. You should state clearly the reagents and conditions used, and the observations you would expect to make with each compound.

compound N

Reagents: KMnO4, NaOH(aq) / H2SO4(aq) Condition: cold or room temperature M will decolourise purple KMnO4. N will not decolourise purple KMnO4.

[6]

[Total: 15]


7

3 Pseudohalogens are a class of compounds that show chemical reactivity similar to halogens. An example of a pseudohalogen and its anion is (SCN)2 and SCN– respectively.

thiocyanogen thiocyanate

(a) (i) Draw the shape of the hybrid orbitals around one of the carbon atoms in thiocyanogen.

(ii) Draw a diagram to illustrate the shape of thiocyanogen, indicating clearly the

covalent bonds and lone pairs of electrons.

bent shape at sulphur (bond angle of 105o) linear at CN Lone pairs on N Lone pairs on S (iii) Thiocyanogen can be synthesised by reacting sodium thiocyanate, NaSCN, with

chlorine gas. Write an equation for the formation of thiocyanogen.

2NaSCN + Cl2 (SCN)2 + 2NaCl [4]

(b) Thiocyanogen can react with hydrocarbons such as methylbenzene via free radical

substitution.

(i) Write equations that describe the following steps. Initiation Propagation

N C S S C N S C N

CH2SCNCH2

++ (SCN)2 SCN

+

CH3

SCN

CH2

+ HSCN

NCS SCN SCN2

sp

CH3

+ (SCN)2uv

CH2SCN

+ HSCN

VJC 2014 9647/02/PRELIM/14

8

Termination (Write down all the possible termination steps.)

(ii) An isomer of the product shown below was also isolated. Suggest a reason for the formation of this isomer.

Delocalisation of electrons in produced at the initiation step allows to be formed, resulting in such a side product.

[4]

(c) The standard reduction potential of (SCN)2 is given as

½(SCN)2 + e– ⇌ SCN– E = +0.43 V

With the aid of the Data Booklet, predict the reactions, if any, that might occur when the following reagents are mixed. Write a balanced equation for the reaction that occurs.

(i) Na2S2O3 and (SCN)2

½(SCN)2 + e– ⇌ SCN– E = +0.43 V S4O6

2– + 2e– ⇌ 2S2O32– E = +0.09 V

E

cell = 0.43 – 0.09 = +0.34 V > 0. Hence reaction is feasible.

2S2O32– + (SCN)2 S4O6

2– + 2SCN–

(ii) acidified K2Cr2O7 and (SCN)2

Cr2O72– + 14H+ + 6e– ⇌ 2Cr3+ + 7H2O E = +1.33 V

Both Cr2O7

2– and (SCN)2 are oxidising agents. Hence reaction cannot proceed.

[3]

2 SCN (SCN)2

CH2

+ SCN

CH2SCN

+

CH2CH2

CH2 N C S

NCS

NCS


9

(d) SCN– acts as a ligand in the formation of complex ions. The formation of a complex ion from the hydrated metal ion is shown in the following equilibrium:

[M(H2O)6]

m+(aq) + nL–(aq) ⇌ [M(H2O)6–n Ln](m–n)+(aq) + nH2O(l)

where m and n are whole numbers.

The equilibrium constant is called the stability constant, Kstab. The table below provides information for the formation of two complexes.

Complex ion Mm+ L– n Kstab Colour

Iron(III) thiocyanate

Fe3+ SCN– 1 90 Red

Cobalt(II) thiocyanate

Co2+ SCN– 4 100 Blue

(i) Write an equation for the formation of cobalt(II) thiocyanate and hence write an

expression for its equilibrium constant, Kstab and state its units. [Co(H2O)6]

2+(aq) + 4SCN–(aq) ⇌ [Co(H2O)2(SCN)4]2–(aq) + 4H2O(l)

Kstab = [[Co(H2O)2(SCN)4]

2–] [Co(H2O)6

2+][SCN–]4 Units : mol–4 dm12

(ii) Calculate the concentration of the complex ion formed when 0.0050 mol of solid cobalt(II) chloride was added to a 2 dm3 solution of 0.95 mol dm–3 KSCN solution.

[Co(H2O)6]

2+(aq) + 4SCN–(aq) ⇌ [Co(H2O)2(SCN)4]2–(aq) + 4H2O(l)

Initial 0.0050/2 0.95 Change – z – 4z + z Eqm 0.0025 – z 0.95 – 4z z [Co(H2O)2(SCN)4]

2– = 100 x (0.95 – 4z)4 x (0.0025 – z) z = 100 x 0.8145 x (0.0025 – z) since 0.95 >> 4(0.0025) z = 0.2036 – 81.45 z 82.45 z = 0.2036 z = 2.47 x 10–3 mol dm–3

(iii) In an experiment to study the stoichiometry of cobalt(II) thiocyanate, a solution of 1.0 x 10–3 mol dm–3 cobalt(II) chloride is mixed with different proportions of a solution of 1.0 x 10–3 mol dm–3 KSCN(aq) such that the total volume of each mixture is kept constant at 10 cm3. The absorbance due to the complex ion in each mixture is measured with a colorimeter. The absorbance is directly proportional to the concentration of the complex.

In the grid below, sketch the graph you would expect to obtain.

VJC 2014 9647/02/PRELIM/14

10

absorbance

0 2 4 6 8 10 Volume of cobalt(II) chloride / cm3 10 8 6 4 2 0 Volume of KSCN / cm3

Inverted V-shape Point : 0 cm3 of cobalt(II) chloride and 10 cm3 of KSCN Point : 10 cm3 of cobalt(II) chloride and 0 cm3 of KSCN Max absorbance at 2.0 cm3 of CoCl2 and 8.0 cm3 of KSCN

[6]

(e) KSCN is used to standardise a solution of silver nitrate. A white precipitate of AgSCN (Ksp = 1.1 x 10–12 mol2 dm–6) is formed during the titration. The indicator used in the titration is yellow aqueous iron(III) solution.

Ag+(aq) + SCN–(aq) ⇌ AgSCN(s)

In an experiment, 5 cm3 of 5 mol dm–3 nitric acid, 25 cm3 of water and 1 cm3 of iron(III) solution are added to 25.0 cm3 of silver nitrate solution in a conical flask. End-point of titration is reached when 15.00 cm3 of 0.100 mol dm–3 KSCN solution is added.

(i) Calculate the concentration of the original silver nitrate solution.

No of moles of KSCN = 0.100 x 15.00/1000 = 1.50 x 10–3 mol No of moles of Ag+ = 1.50 x 10–3 mol [Ag+] = 1.50 x 10–3 ÷ 25.0/1000 = 6.00 x 10–2 mol dm–3

(ii) Calculate the concentration of Ag+(aq) in the solution at the equivalence point.

Ksp = [Ag+][SCN–] Since [Ag+] = [SCN–] [Ag+] = √ 1.1 x 10–12 = 1.05 x 10–6 mol dm–3

(iii) With reference to the table given in (d), explain why iron(III) solution can be used as an indicator and state the observation at the end-point of the titration.

Additional SCN– added just after the end-point will react with Fe3+ to form

Fe(H2O)5(SCN)2+ which will produce a red coloration. [4]

[Total: 21]


11

4 (a) Haemoglobin is an important protein present in the red blood cell, which is responsible for transport of oxygen in the body. It contains a haem group and a metal centre as shown below. The four N atoms and the Fe atom are planar.

(i) State the types of hybridisation at N1 and N2, and the nature of the bonding around Fe atom.

Hybridisation at N1: sp2 Hybridisation at N2: sp2 Type of bond between Fe atom and N1: covalent bond Type of bond between Fe atom and N2: dative bond / co-ordinate bond

(ii) One molecule of haemoglobin (Hb) binds up to four oxygen molecules to form oxyhaemoglobin (Hb(O2)4), as shown in the following equation.

Hb(aq) + 4O2(g) ⇌ Hb(O2)4(aq) (I)

By applying Le Chatelier’s Principle, explain how oxygen molecules are transported from the lungs to the body tissues.

In the lungs, the concentration of O2 is high.

Thus equilibrium position shifts to the right to form oxyhaemoglobin. OR Oxygen is transported in the body as

oxyhaemoglobin. In the tissues, the concentration of O2 is low. Thus equilibrium position shifts to the left and oxyhaemoglobin starts to dissociate to release oxygen.

(iii) The percentage saturation of haemoglobin is defined as the percentage of haemoglobin present in the blood that has been converted to oxyhaemoglobin.

With reference to equilibrium (I), state and explain how the percentage saturation of haemoglobin changes when temperature increases.

The forward reaction of the above equilibrium is exothermic

due to formation of dative/coordinate bond between the oxygen and Fe atom in the haemoglobin.

When temperature increases, endothermic reaction is favoured OR the backward reaction is favoured so as to remove excess heat.

Oxyhaemoglobin dissociates and the percentage saturation decreases.

VJC 2014 9647/02/PRELIM/14

12

(iv) The relationship between the percentage saturation of haemoglobin and concentration of oxygen in blood at different temperatures is shown below.

Suggest the optimum temperature at which haemoglobin operates. Explain your answer.

37 °C Optimum temperature occurs at high percentage saturation at lungs and low percentage saturation at tissues OR The difference in the percentage saturation at lungs and tissues is the largest.

[8] (b) (i) Draw the dot-and-cross diagram of ozone, O3.

(ii) Oxygen–oxygen bond lengths of 3 species are given in the table below.

Species Oxygen–oxygen bond length / nm

O3 0.128

O22– 0.149

O2 0.120

With the aid of suitable structures, explain why the two oxygen-oxygen bonds in the ozone molecule have the same bond length which are very different from the peroxide ion and oxygen molecule.

O3 forms resonance structures as shown below.

OR (dative bonds acceptable) Thus, the two oxygen-oxygen bonds are equivalent, shorter than oxygen-oxygen

single bond in O22– and longer than oxygen-oxygen double bond in O2.

[3]

[Total: 11]

Percentage Saturation / %

[O2] / mol dm–3

37 °C

50 °C

10 °C

[O2]lungs [O2]tissues


13

5 (a) Eugenoxyacetic acid is a colourless and odourless compound. It has shown anti-viral and anti-bacterial properties and is therefore used as antioxidant food preservative in food industry.

Eugenoxyacetic acid can be synthesised from eugenol found in clove essential oil as

shown in the procedure below via SN2 mechanism.

(i) How can you tell that the solution has just turned alkaline in step 1? No more effervescence is observed on addition of Na2CO3(s).

(ii) Explain the role of NaOH(s) in step 2. NaOH(s) acts as a base to react with phenol. The phenoxide formed is a stronger nucleophile to react with ClCH2CO2Na.

(iii) Write a balanced equation for the reaction in step 3.

(iv) Comment if an excess amount of NaOH(s) can be used in step 2.

An excess amount of NaOH(s) should not be used as the hydroxide ions can compete with the phenoxide to react with ClCH2CO2Na.

(v) Give a simple chemical test to confirm that eugenol is absent in the final product. Dissolve the solid in ethanol and add neutral FeCl3 at room temperature. Absence of violet colouration would indicate that eugenol is absent.

[6]

Preparation of solution A

1. Dissolve 1.0 g of 2-chloroethanoic acid in 5.0 cm3 of distilled water in a beaker. Add Na2CO3(s) slowly until the solution just becomes alkaline.

Preparation of eugenoxyacetic acid

2. Dissolve 0.6 g of NaOH(s) in 3.0 cm3 of distilled water in a conical flask. Add 2.0 cm3 eugenol to the flask.

3. Add solution A to the flask in step 2 and keep stirring the mixture for 60 min.

4. Cool the reaction mixture to room temperature and add HCl(aq). Collect the solid formed by suction filtration. Wash the solid with water to obtain a pale yellow solid which is the crude product.

VJC 2014 9647/02/PRELIM/14

14

(b) The structure of a small protein is given below, along with the abbreviated names of its constituent amino acids: val, gln etc.

(i) Use the abbreviated names of the amino acids to state the primary structure of the above protein.

cys-pro-val-gln-tyr-pro-val-gln-cys-tyr (ii) This protein is ionised at pH 7. Identify one functional group that could be ionised

at pH 7 by circling it on the structure above. circle the phenol group (iii) The biological function of the protein depends on its three dimensional shape.

Describe how the particular amino acids in the above protein are likely to be involved in maintaining its three-dimensional shape.

Disulfide linkages between the cys amino acid residues. Van der waals’ forces of attraction between the CH(CH3)2 group of val residues

(other plausible answers acceptable including between val and pro). Hydrogen bonding between the CH2CH2CONH2 group of gln residues (or gln

with CH2C6H4OH group of tyr). (iv) The activity of the above protein is destroyed by heating it under reflux with

6 mol dm–3 sulfuric acid for several hours.

Draw the structures of the organic products formed under these conditions from the part of the above protein containing the amino acids: val-gln

[7]

[Total: 13]



CHEMISTRY

9647/03

Paper 3 Free Response

Candidates answer on separate paper.

22 September 2014

2 hours

Additional Materials: Answer Paper Data Booklet Graph Paper

READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams, graphs or rough working.


Answer any four questions.

A Data Booklet is provided.

You are reminded of the need for good English and clear explanation in your answers.

The number of marks is given in brackets [ ] at the end of each question or part question.


2

VJC 2014 9647/03/PRELIM/14

Answer any four questions.

1 (a) Use of the Data Booklet is relevant in this question.

The following cell has been used to light up a light bulb.

(i) Explain how the cell potential changes when the current has passed through for a while.

The electrode potential of the cell is related to the equilibrium constant by the following equation,

Eo= ln Kc

where T is the temperature measured in Kelvin, z is the number of moles of electrons transferred during the redox reaction and F is the Faraday constant.

(ii) Determine the equilibrium constant Kc for the reaction between Cu2+ and Mg. (iii) Hence, determine the ratio [Mg2+]:[Cu2+], when there is no current flowing in this

cell.

(iv) Suggest the significance of the magnitude of your answer in (a)(iii).

[5] (b) Copper minerals often contain suifides of magnesium and silver as impurity. After

minerals are reduced with carbon, the solid impure copper is purified by electrolysis.

(i) Describe the electrode reactions that take place during the electrolysis and explain in detail how each of the two impurity metals is removed from copper.

An impure copper rod is purified by electrolysis using a constant current. After 1 hour, mass of one electrode decreased by 19.0 g while mass of the other electrode increased by 17.8 g. The electrolyte was further analysed and found that the amount of Cu2+ ions is reduced by 0.020 mol. (ii) Calculate the current used for this process.

(iii) Determine the percentage composition by mass of the three elements. (iv) Suggest how magnesium can be recovered as magnesium oxide near the end

of the electrolytic process. Write balanced equations for any reactions involving magnesium that might occur.

[11]

salt bridge

Cu electrode

Cu2+(aq), 1 mol dm–3

light bulb X

Mg electrode

Mg2+(aq), 1 mol dm–3

RT zF

3


(c) The methods of synthesis shown below are faulty. In each case

Explain why this is so.

Suggest how the synthesis from the initial reactant to the final product could be

satisfactorily achieved.

(i)

(ii)

[4]

[Total: 20]

4

VJC 2014 9647/03/PRELIM/14

2 (a) (i) When CHI2CHO was warmed with alkaline aqueous iodine, two organic

products are formed. Construct an equation for this reaction.

(ii) There are suggestions on using CF3CH2F to replace CFCs. One problem

however is that CF3CH2F can be converted to CF3CO2H, which is toxic when

ingested. When CF3CH2F eventually reaches the atmosphere, it is thought that

CF3CH2F is initially attacked by •OH radicals.

A student suggested the following equation to represent the above reaction:

CF3CH2F + •OH CF3CH2 + HOF

By considering the

bonds broken and

bonds formed

in the above reaction, explain why the above reaction will not occur.

(iii) In light of your answers to (a)(ii), suggest a more appropriate equation for the

reaction between CF3CH2F and •OH.

(iv) Ethanedioic acid, HO2C-CO2H is a dibasic acid, with pKa1 = 1.3 and pKa2 = 4.3.

Methanoic acid is a monobasic acid with pKa = 3.7.

Account for the lower pKa1 value of ethanedioic acid as compared to the pKa of methanoic acid.

[6]

(b) But-2-yne, H3CC≡CCH3 may undergo the addition of hydrogen at the surface of a metal catalyst to produce an alkene X. Draw relevant diagrams to illustrate how the metal catalyses this reaction. Give the structure of X.

[3]

(c) (i) Describe briefly how gaseous HCl can be made in the laboratory from NaCl and concentrated H2SO4.

(ii) Explain with relevant equations, if HBr and HI can be prepared in a similar

method as in (c)(i).

(iii) Silver chloride is soluble in aqueous ammonia while silver bromide is soluble in

concentrated aqueous ammonia. Explain these observations.

[6]

•

5


(d) A is one of the elements from Na to P in the third period of the Periodic Table. When

A is boiled with aqueous sodium hydroxide, a colourless gas, B is formed together

with a sodium salt, C. Gas B is weakly basic and reacts with hydrogen iodide to form

a solid compound, D, which contains 19.1% of A, 2.5% of hydrogen and 78.4% of

iodine by mass. The sodium salt C contains 26.1% of sodium, 2.27% of hydrogen,

35.2% of A and 36.4% of oxygen by mass. A also forms two chlorides which dissolve

readily in water to produce strongly acidic solutions of pH 2.

Identify the element A and the compounds B, C and D. Write equations for all the reactions involved.

[5]

[Total: 20]

6

VJC 2014 9647/03/PRELIM/14

3 Alcohols are very useful in organic synthesis and can be used to produce a wide variety

of compounds. The structural formula of one such alcohol is shown below.

compound A (C13H20O2)

(a) When compound A is heated with an excess of concentrated sulfuric acid, a mixture of two isomeric hydrocarbons, P and Q are produced. Draw the structures of P and Q.

[2]

(b) Compound D can be formed from compound A via a five-step synthesis.

(i) State the reagents and conditions required for steps 2 to 4 and the structural formulae of the two intermediates B and C.

(ii) Step 1 involves the ‘protection’ of one of the alcohol groups to convert it into an ether, which is inert. Suggest a reason why this step is necessary.

(iii) Suggest the type of reaction in step 5.

[6]

(c) Nitrosyl bromide, NOBr, is a red gas and can be formed by the reversible reaction of nitric oxide with bromine.

(i) Explain why nitrosyl bromide has a boiling point of 14 oC whereas nitrosyl

chloride, NOCl boils at −5 oC. NOBr decomposes to NO and Br2 as shown below.

2NOBr(g) 2NO(g) + Br2(g)

(ii) Given that the bond energy of N−Br is 120 kJ mol−1, use the data from the Data Booklet to calculate the enthalpy change of decomposition of nitrosyl bromide, NOBr.

http://en.wikipedia.org/wiki/Nitric_oxide

http://en.wikipedia.org/wiki/Bromine

7


(iii) Standard enthalpy changes of formation of nitrosyl bromide and nitrogen monoxide are given below.

compound Hf / kJ mol−1

NOBr(g) +82

NO(g) +90

With the aid of an energy cycle, use your answer in (c)(ii) and data above to calculate the enthalpy change of vaporisation of bromine.

(iv) Predict and explain the ‘sign’ of entropy change for the decomposition of

nitrosyl bromide. [6]

(d) A calorimeter containing 300 cm3 of water at 25 °C was calibrated as follows.

(i) 10 kJ of heat energy from a heating coil was used to increase the temperature

of water by 7.5 °C. Calculate the heat capacity of the empty calorimeter.

(ii) A solution containing 0.040 mol Ag+(aq) was mixed with a second solution containing 0.050 mol Br–(aq) in this calorimeter, causing AgBr(s) to precipitate. The temperature increased by 2.4 oC. Calculate the equilibrium concentrations of Ag+(aq) and Br–(aq) present in the final voiume of 320 cm3. [Ksp of AgBr = 5 x 10-13 mol2 dm–6]

(iii) Hence, calculate the enthalpy change of solution of AgBr(s).

[You may assume that the density of all solutions is 1 g cm–3.] [6]

[Total: 20]

8

VJC 2014 9647/03/PRELIM/14

4 (a) Sulfuryl dichloride, SO2Cl2, is a compound of industrial, environmental and scientific interest, widely used as a chlorinating or sulfonating agent. At room temperature, SO2Cl2 is a colourless liquid with a pungent odour.

An empty container is filled with SO2Cl2. Its decomposition to SO2 and Cl2 is followed by monitoring the change in total pressure at 375 K. The following data are obtained.

time/ hour 0 1 3 6 10 14 18

Ptotal / atm 1.000 1.076 1.211 1.378 1.547 1.670 1.760

22CSOP l / atm 1.000 p q 0.622 r 0.330 0.240

(i) Determine the values of p, q and r. Show your working clearly. (ii) Hence, by plotting a suitable graph, show that the decomposition of SO2Cl2 is a

first order reaction. (iii) Calculate the rate constant, stating its unit.

(iv) Calculate the concentration of SO2Cl2, in mol dm-3 after 18 hours.

You may assume ideal gas conditions.

(v) In another experiment, the decomposition reaction is repeated at 375 K with initial total pressure of 2 atm of SO2Cl2.

Use the graph drawn in (a)(ii) to estimate the time taken for partial pressure of SO2Cl2 to fall by 20% in this experiment. Explain your answer.

(vi) There will be a negligible amount of SO2Cl2 (g) in the reaction vessel after a

long period of time. Therefore, the content of the vessel might be considered to be a mixture of SO2 and Cl2 gases.

A possible way to separate SO2 and Cl2 from each other is to pass the mixture over solid CaO which will convert all the SO2 to CaSO3, a strong electrolyte. Calculate the pH of a 0.020 mol dm−3 CaSO3 solution. [For H2SO3 Ka1 = 1.7×10−2 mol dm-3 and Ka2 = 6.4×10−8 mol dm−3.]

[12]

9


(b) Compound G has a sweet smell. When heated with an excess of acidified

concentrated manganate(VII) ions, it forms initially three products H, J and K.

G H + J + K C13H20O2 C6H8O3 C5H10O2 C2H2O4

When compound G was mixed with liquid bromine at room temperature,

C13H20O2Br4 was obtained.

Compound J evolves CO2 with NaHCO3. J also reacts with LiAlH4 to give a

compound which does not react with concentrated sulfuric acid.

Compound H gives an orange precipitate with 2,4-dinitrophenylhydrazine

reagent. H has no reaction with Tollens’ reagent. When 1 mol of H reacts with

an excess of alkaline aqueous iodine, 2 mol of yellow precipitate are produced.

Use the information above to deduce the structures for compounds G, H, J and K. [8]

[Total: 20]

MnO4- / H

+

10

VJC 2014 9647/03/PRELIM/14

5 Amino acids are critical to life, and they serve as building blocks of proteins.

(a) Lysine is an alpha amino acid which shows both acidic and basic properties. The R

group of lysine is –(CH2)4NH2. In acidic solution, lysine is completely protonated.

10.0 cm3 of completely protonated lysine is titrated with 1.00 mol dm−3 sodium

hydroxide solution. Its titration curve is shown below.

(i) Calculate the first dissociation constant Ka1 of fully protonated lysine.

(ii) Identify the two species at point B. Write an equation to show how the species

present at point B can resist change in pH when a small amount of base is added.

(iii) Calculate the pH when 10.0 cm3

of 0.0200 mol dm−3 sodium hydroxide is added

to the solution at point B. (You may represent the fully protonated lysine as HA.)

(iv) State the predominant species at point C. Explain why it has a high melting

point.

(v) State the type of isomerism in the species at point C and draw its isomers.

[8]

(b) The first known synthesis of an amino acid occurred in 1850 in the laboratory of Adolf Strecker.

(i) Name the type of reaction in step 1.

(ii) Suggest suitable reagents and conditions for step 3.

(iii) In step 2, the reaction proceeds via two stages: (I) acid-base reaction between the N atom in imine and HCN (II) nucleophilic attack on C by CN−

Write the mechanism for step 2, showing clearly the movement of electrons

and partial charges.

11


(iv) An imine intermediate is also formed in preparing secondary amines from

ketones.

With reference to the Adolf Strecker synthesis, suggest the synthetic route by giving the reagents, conditions and intermediates for the preparation of (CH3)2CHNH(CH2CH3).

(v) Another amine X, CH3CH(NH2)CH2NH2 reacts with ethanedioyl dichloride (COCl)2 to produce compound Y with molecular formula C5H8N2O2. Suggest the structure for compound Y and write the equation between X and ethanedioyl dichloride.

[8]

(c) A polypeptide H was analysed and found to contain the following amino acids.

amino acid aspartic acid glycine serine tyrosine valine

abbreviation asp gly ser tyr val

number of residues

1 1 2 1 1

Analysis gave the following results: ● The N-terminus was shown to be ser. ● On reaction with the enzyme chymotrypsin, which hydrolyses at the carboxylic

acid end of tyr, H gave two tripeptides. ● On reaction with a special reagent which digests at the carboxylic end of val, H

gave two peptides. One of these two was a dipeptide of sequence gly-ser. Determine the amino acid sequence of polypeptide H. Briefly justify your answer. [You should use the same 3–letter abbreviations as shown above to write out the amino acid sequence].

[2]

(d) Papain, an enzyme in fresh papaya juice is used as a meat tenderiser. The three main amino acids involved in the catalytic activity of papain are his, asp and ser. Part of the mechanism of the action of papain is illustrated in the figures below.

With reference to figures (I) and (II) above, explain why the action of this enzyme would be inhibited if the pH was too low.

[2]

[Total: 20]

(I) (II)

1 VJC 2014 9647/03/PRELIM/14

2014 Preliminary Examination H2 Chemistry Paper 3

Answer any 4 questions.

1 (a) Use of the Data Booklet is relevant in this question.

The following cell has been used to light up a light bulb.

(i) Suggest, with a reason, how the cell potential changes when the current has passed through for a while.

Anode: Mg(s) Mg2+(aq) + 2e– Cathode: Cu2+(aq) + 2e– Cu(s)

OR Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq)

As [Mg2+] increases, E(Mg2+/Mg) will be less negative.

As [Cu2+] decreases, E(Cu2+/Cu) will be less positive.

The cell potential will become less positive.

The electrode potential of the cell is related to the equilibrium constant by the following equation,

Eo= c

RTln

zFK

where T is the temperature measured in Kelvin, z is the number of moles of electrons transferred during the redox reaction and F is the Faraday constant.

(ii) By using relevant data from the Data Booklet, determine the equilibrium constant Kc for the reaction between Cu2+ and Mg.

Ecell = Ered – Eox

= (+0.34) – (–2.38)

= +2.72 V

ln Kc = 298 x 8.31

2.72 x 96500 x 2 (ecf)

= 212

Kc = 1.16 x 1092 (ecf)

salt bridge

Cu electrode

Cu2+(aq), 1 mol dm–3

light bulb X

Mg electrode

Mg2+(aq), 1 mol dm–3

2 VJC 2014 9647/03/PRELIM/14

(iii) Hence, determine the ratio [Mg2+]:[Cu2+], when there is no current flowing in this cell.

When there is no current flowing in the electric cell, it has reached equilibrium.

[Mg2+]:[Cu2+] = Kc = 1.18 x 1092

(iv) Suggest the significance of the magnitude of your answer in (a)(iii).

Almost all the Cu2+ has been consumed. OR Reaction goes to completion when the equilibrium has been reached. [5]

(b) Copper minerals often contain suifides of magnesium and silver as impurity. After

minerals are reduced with carbon, the solid impure copper is purified by electrolysis.

(i) Describe the electrode reactions that take place during the electrolysis and explain in detail how each of the two impurity metals is removed from copper.

Two electrodes + Electrolyte

OR diagram above

At the anode, oxidation takes place and Cu dissolves.

Cu(s) Cu2+(aq) + 2e– Mg with Eo which is less positive than that of Cu dissolves as ions. Ag with Eo which is more positive than that of Cu remains undissolved and drop to

the bottom of the vessel as ‘anode sludge’. At the cathode, reduction of copper ions occurs due to more positive Eo (Cu2+/Cu)

compared to Eo (H2O/H2) and Eo (Mg2+/Mg)

Cu2+(aq) + 2e– Cu(s) Hence, only pure Cu is formed at the cathode.

An impure copper rod is purified by electrolysis using a constant current. After 1 hour, mass of one electrode decreased by 19.0 g while mass of the other electrode increased by 17.8 g. The electrolyte was further analysed and found that the amount of Cu2+ ions is reduced by 0.020 mol.

Power source

electron flow

Anode (+)

Impure Cu

2Cl-(l) Cl2(g) + 2e–

Cathode (–)

Pure Cu

Na+(l) + e– Na(s)

+ –

CuSO4(aq)

3 VJC 2014 9647/03/PRELIM/14

(ii) Calculate the current used for this process.

Amount of Cu formed = 5.63

8.17

= 0.280 mol

Amount of electron transferred = 0.280 x 2

= 0.560 mol (ecf)

Charge transferred = 0.560 x 96500

= 54100 C (ecf)

Current = 3600

54100

= 15.0 A (ecf)

(iii) Determine the percentage composition by mass of the three elements.

Amount of Mg oxidised = amount of Cu2+ changed

= 0.020 mol

Mass of Mg = 0.020 x 24.3

= 0.486 g

Percentage composition by mass of Mg = %100 x 0.19

486.0

= 2.56%

Amount of Cu oxidised = 0.280 – 0.020

= 0.260 mol

Mass of Cu = 16.51 g

Percentage composition by mass of Cu = %100 x 0.19

51.16

= 86.9 %

Percentage composition by mass of Ag = 10.54 %

(iv) Suggest how magnesium can be recovered as magnesium oxide near the end of the electrolytic process. Write balanced equations for any reactions involving magnesium that might occur.

Filter the electrolyte to remove the metallic silver.

Add aqueous ammonia until excess.

Filter the mixture to obtain the residue which is magnesium hydroxide.

Heat the solid until no more change in mass.

Mg2+ + 2OH– Mg(OH)2

Mg(OH)2 MgO + H2O

[Accept reaction of Mg(OH)2 with HNO3 and followed by thermal decomposition of

Mg(NO3)2].

[11]

4 VJC 2014 9647/03/PRELIM/14

(c) The methods of synthesis shown below are faulty. In each case

Explain why this is so.

Suggest how the synthesis from the initial reactant to the final product could be

satisfactorily achieved.

(i)

(ii)

[4]

(c) (i) When HCl is used:

Dehydration will not occur with dilute HCl. Ester will be hydrolysed to give alcohol and carboxylic acid. Correct reaction reagent and condition: Al2O3 catalyst with heating. (Conc. H2SO4 is not a good choice as it may lead to hydrolysis of ester.) (ii) NaOH causes hydrolysis of CN instead of reduction. Amine is a stronger nucleophile as nitrogen is less electronegative, thus it will

react with CH3COCl first to give amide instead. OR Both alcohol and amine will react with CH3COCl, when excess amount of

CH3COCl is used. React with CH3COCl at room temperature and followed by reduction with H2 with Pt catalyst at room temperature.

[4]

[Total: 20]

2 (a) (i) When CHI2CHO was warmed with alkaline aqueous iodine, two organic products

are formed. Construct an equation for this reaction.

CHI2CHO + I2 + 2OH– CHI3 + HCO2– + I– + H2O

(ii) There are suggestions on using CF3CH2F to replace CFCs. One problem however

is that CF3CH2F can be converted to CF3CO2H, which is toxic when ingested.

When CF3CH2F eventually reaches the atmosphere, it is thought that CF3CH2F is

initially attacked by •OH radicals.

A student suggested the following equation to represent the above reaction:

CF3CH2F + •OH CF3CH2 + HOF

By considering the

•

5 VJC 2014 9647/03/PRELIM/14

bonds broken and

bonds formed

in the above reaction, explain why the above reaction will not occur.

Bond broken: C-F bond is very strong / high bond energy

due to the extensive overlap between the C and F atomic orbitals.

Bond formed: O-F bond is very weak / low bond energy

O and F atoms are small / strongly electronegative, and the electron pairs will be

close to each other and repel strongly.

(iii) In light of your answers to (a)(ii), suggest a more appropriate equation for the

reaction between CF3CH2F and •OH.

CF3CH2F + •OH CF3CHF + H2O

(iv) Ethanedioic acid, HO2C-CO2H is a dibasic acid, with pKa1 = 1.3 and pKa2 = 4.3.

Methanoic acid is a monobasic acid with pKa = 3.7.

Account for the lower pKa1 value of ethanedioic acid as compared to the pKa of methanoic acid.

Lower pKa1 value implies stronger acid.

HO2C-CO2– is more stable than HCO2

–.

CO2H group exerts an electron withdrawing inductive effect

leading to better charge dispersal in HO2C-CO2– compared to HCO2

–.

Hence, greater tendency for HO2C-CO2H to dissociate

OR equilibrium position of HO2C-CO2H HO2C-CO2– +H+ lies more to the

right.

HO2C-CO2– can also be stabilised by intramolecular hydrogen bonding.

OR equivalent diagram

[6]

(b) But-2-yne, H3CC≡CCH3 may undergo the addition of hydrogen at the surface of a metal catalyst to produce an alkene X. Draw relevant diagrams to illustrate how the metal catalyses this reaction. Give the structure of X.

X is (cis isomer shown)

•

6 VJC 2014 9647/03/PRELIM/14

[3]

(c) (i) Describe briefly how gaseous HCl can be made in the laboratory from NaCl and concentrated H2SO4.

Add conc H2SO4 to solid NaCl with heating.

Collect the gas produced in a gas syringe / gas jar.

OR equivalent diagram

(ii) Explain with relevant equations, if HBr and HI can be prepared in a similar method

as in (c)(i).

No for both HBr and HI

HBr and HI are strong reducing agents and will further react with conc H2SO4.

2HBr + H2SO4 Br2 + SO2 + 2H2O

8HI + H2SO4 4I2 + H2S + 4H2O

Adsorption of reactants on

surface of catalyst

Bonds weakened in reactants

and less energy is required to

form the activated complex.

After reaction, products desorb from the surface and diffuse away.

(award this mark even if the correct

isomer is not shown)

7 VJC 2014 9647/03/PRELIM/14

(iii) Silver chloride is soluble in aqueous ammonia while silver bromide is soluble in

concentrated aqueous ammonia. Explain these observations.

Ag+ + 2NH3 ⇌ [Ag(NH3)2]+

The Ksp value of AgCl is larger than Ksp value of AgBr.

AgCl is soluble in aqueous NH3 due to complex formation from Ag+(aq), enabling

the ionic product [Ag+][Cl–] to be less than the Ksp of AgCl.

For AgBr, due to the lower Ksp value, concentrated NH3 has to be used to lower the

[Ag+(aq)] sufficiently so that the ionic product [Ag+][Br–] is less than its Ksp value.

[6]

(d) A is one of the elements from Na to P in the third period of the Periodic Table. When A is

boiled with aqueous sodium hydroxide, a colourless gas, B is formed together with a

sodium salt, C. Gas B is weakly basic and reacts with hydrogen iodide to form a solid

compound, D, which contains 19.1% of A, 2.5% of hydrogen and 78.4% of iodine by

mass. The sodium salt C contains 26.1% of sodium, 2.27% of hydrogen, 35.2% of A and

36.4% of oxygen by mass. A also forms two chlorides which dissolve readily in water to

produce strongly acidic solutions of pH 2.

Identify the element A and the compounds B, C and D. Write equations for all the reactions involved.

A is phosphorus.

Phosphorus has two chlorides: PCl3 and PCl5.

PCl3 + 3H2O → H3PO3 + 3HCl

OR PCl5 + 4H2O → H3PO4 + 5HCl

Element P H I

Mass (g) 19.1 2.5 78.4

No. of moles

0.31

1.19= 0.616

0.1

5.2 = 2.5

127

4.78= 0.617

Ratio

616.0

616.0= 1

616.0

5.2 = 4

616.0

617.0= 1

D is PH4I.

B is PH3.

PH3 + HI → PH4I

Element Na H P O

Mass (g) 26.1 2.27 35.2 36.4

No. of moles

0.23

1.26= 1.13

0.1

27.2= 2.27

0.31

2.35= 1.14

0.16

4.36= 2.275

Ratio

13.1

13.1= 1

13.1

27.2= 2

13.1

14.1 = 1

13.1

275.2 = 2

8 VJC 2014 9647/03/PRELIM/14

C is NaH2PO2.

P4 + 3H2O + 3NaOH → PH3 + 3NaH2PO2 (accept 4P instead of P4)

[5]

[Total: 20]

3 Alcohols are very useful in organic synthesis and can be used to produce a wide variety of

compounds. The structural formula of one such alcohol is shown below.

compound A (C13H20O2)

(a) When compound A is heated with an excess of concentrated sulfuric acid, a mixture of two isomeric hydrocarbons, P and Q are produced. Draw the structures of P and Q.

P: Q:

Structures for P and Q can be interchanged. [2]

(b) Compound D can be formed from compound A via a five-step synthesis.

(i) State the reagents and conditions required for steps 2 to 4 and the structural

formulae of the two intermediates B and C. Step 2:

Reagents: K2Cr2O7 / dil H2SO4 [Do not accept KMnO4 /dil H2SO4] Conditions: Reflux

B:

9 VJC 2014 9647/03/PRELIM/14

Step 3: Reagents: HCN / NaCN cat. [Note: If use KMnO4 to form B, ecf for Conditions: Room temperature steps 3 and 4]

C:

Step 4:

Reagents: Dil. H2SO4

Conditions: Reflux (ii) Step 1 involves the ‘protection’ of one of the alcohol groups to convert it into an

ether, which is inert. Suggest a reason why this step is necessary. Protection is necessary to prevent the primary alcohol from being oxidised in step

2. (iii) Suggest the type of reaction in step 5. Reduction

[6]

(c) Nitrosyl bromide, NOBr, is a red gas and can be formed by the reversible reaction of nitric oxide with bromine.

(i) Explain why nitrosyl bromide has a boiling point of 14 oC whereas nitrosyl chloride,

NOCl boils at −5 oC. NOBr has more electrons compared to NOCl. Hence, there are stronger dispersion forces between NOBr molecules.

Thus NOBr has a higher boiling point. NOBr decomposes to NO and Br2 as shown below.

2NOBr(g) 2NO(g) + Br2(g)

(ii) Given that the bond energy of N−Br is 120 kJ mol−1, use the data from the Data Booklet to calculate the enthalpy change of decomposition of nitrosyl bromide, NOBr.

Hdecomposition = ½ [2 BE(N−Br) – BE(Br−Br)] = ½ [2 (120) – 193]

= +23.5 kJ mol-1

http://en.wikipedia.org/wiki/Nitric_oxide

http://en.wikipedia.org/wiki/Bromine

10 VJC 2014 9647/03/PRELIM/14

(iii) Standard enthalpy changes of formation of nitrosyl bromide and nitrogen monoxide are given below.

compound Hf / kJ mol−1

NOBr(g) +82

NO(g) +90

With the aid of an energy cycle, use your answer in (c)(ii) and data above to calculate the enthalpy change of vaporisation of bromine.

N2(g) + O2(g) + Br2(l) 2NOBr(g)

2NO(g) + Br2(l)

2NO(g) + Br2(g)

Complete state symbols

Hvap = –2(90) + 2(82) + 47 (ecf) = +31 kJ mol–1 (ecf)

(iv) Predict and explain the sign of entropy change for the decomposition of nitrosyl bromide.

Entropy change is positive as number of moles of gaseous particles increases from

2 mol to 3 mol, which increase the disorderliness of the system. [6]

(d) A calorimeter containing 300 cm3 of water at 25 °C was calibrated as follows.

(i) 10 kJ of heat energy from a heating coil was used to increase the temperature of

water by 7.5 °C. Calculate the heat capacity of the empty calorimeter.

Heat used to heat water = 300 x 4.18 x 7.5 = 9405 J Heat used to heat calorimeter = 10000 – 9405

= 595 J

Heat capacity of calorimeter = 7.5

595

= 79 J K–1

Hvap

2(+82)

2(+90)

+47

11 VJC 2014 9647/03/PRELIM/14

(ii) A solution containing 0.040 mol Ag+(aq) was mixed with a second solution containing 0.050 mol Br–(aq) in this calorimeter, causing AgBr(s) to precipitate. The temperature increased by 2.4 oC. Calculate the equilibrium concentrations of Ag+(aq) and Br–(aq) present in the final voiume of 320 cm3. [Ksp of AgBr = 5 x 10-13 mol2 dm−6]

Let x be the no. of moles of Ag+ that do not precipitate.

Ag+(aq) Br-(aq) ⇌ AgBr(s)

Initial / mol 0.040 0.050 0

Change / mol –0.040 –0.040 0.040

Final / mol x 0.010 + x 0.040 – x

Ksp = [Ag+][Br–]

= )32.0

x+010.0)(

32.0

x(

Since Ksp is small, x is small and 0.010 + x 0.010

[Br–] = 32.0

010.0

= 0.031 mol dm-3

[Ag+] = 031.0

10 x 5 -13

= 1.6 x 10-11 mol dm-3

(iii) Hence, calculate the enthalpy change of solution of AgBr(s). [You may assume that the density of all solutions is 1 g cm–3.]

Heat evolved = (320 x 4.18 x 2.4) + (79 x 2.4) = 3400 J

Since precipitation is exothermic, enthalpy change of solution is endothermic.

Hsol of AgBr = +040.0

3400

= +85000 J mol–1 = +85 kJ mol–1 (ecf)

[6]

[Total: 20]

12 VJC 2014 9647/03/PRELIM/14

4 (a) Sulfuryl dichloride, SO2Cl2, is a compound of industrial, environmental and scientific interest, widely used as a chlorinating or sulfonating agent. At room temperature, SO2Cl2 is a colourless liquid with a pungent odour.

An empty container is filled with SO2Cl2. Its decomposition to SO2 and Cl2 is followed by monitoring the change in total pressure at 375 K. The following data are obtained.

time/ hour 0 1 3 6 10 14 18

Ptotal / atm 1.000 1.076 1.211 1.378 1.547 1.670 1.760

22CSOP l / atm 1.000 p q 0.622 r 0.330 0.240

(i) Determine the values of p, q and r. Show your working clearly.

SO2Cl2(g) SO2(g) + Cl2(g)

Initial partial pressure / atm 1 0 0 Change in partial pressure / atm -x +x +x Final pressure / atm 1– x x x Ptotal = 1 – x + x + x = (1 + x) atm

22ClSOP = (1 – x) atm

p = 0.924 atm, q = 0.789 atm, r = 0.453 atm

Note: 1m for 3 correct values [0.5m for 2 correct values]

(ii) Hence, by plotting a suitable graph, show that the decomposition of SO2Cl2 is a

first order reaction.

t ½ = 8.50 hr

t ½ = 8.50 hr

t ½ = 8.50 hr

13 VJC 2014 9647/03/PRELIM/14

x-axis with units y-axis with units correct shape

shows three t1/2 shows two constant t1/2 two correct t1/2 values

(iii) Calculate the rate constant, stating its unit.

k = 1/2t

2 ln

= 0.0815 h–1

(iv) Calculate the concentration of SO2Cl2, in mol dm-3 after 18 hours.

You may assume ideal gas conditions.

PV= nRT

22ClSOP = (n/V)RT

[SO2Cl2] = 22ClSOP / RT

= 375 x 31.8

10 x 1.01 x 24.0 5

(ecf)

= 7.78 mol m–3

= 7.78 x 10-3 mol dm–3

(v) In another experiment, the decomposition reaction is repeated at 375 K with initial total pressure of 2 atm of SO2Cl2.

Use the graph drawn in (a)(ii) to estimate the time taken for partial pressure of SO2Cl2 to fall by 20% in this experiment. Explain your answer.

For a first order reaction, time taken for 20% fall in partial pressure

(concentration) is independent of initial partial pressure (concentration). Partial pressure in (a)(ii) with 20% decrease = 0.80 atm. Time taken for partial pressure to fall by 20% in (a)(ii) = 3 hr

(vi) There will be a negligible amount of SO2Cl2 (g) in the reaction vessel after a long

period of time. Therefore, the content of the vessel might be considered to be a mixture of SO2 and Cl2 gases.

A possible way to separate SO2 and Cl2 from each other is to pass the mixture over solid CaO which will convert all the SO2 to CaSO3, a strong electrolyte. Calculate the pH of a 0.020 mol dm−3 CaSO3 solution. [For H2SO3 Ka1 = 1.7×10−2 mol dm-3 and Ka2 = 6.4×10−8 mol dm–3.]

14 VJC 2014 9647/03/PRELIM/14

SO32− + H2O ⇌ HSO3

− + OH− Kb = 10−14 / 6.4×10−8

= 1.56 x 10−7 mol dm-3

Kb = ]SO[

]OH][HSO[-2

3

--3

= 02.0

x2

(since [HSO3−] = [OH−] and x << 0.02)

[OH−] = 5.59 × 10−5 mol dm−3

pH = 14 – pOH = 14 + lg(5.59 x 10−5)

= 9.75

[12] (b) Compound G has a sweet smell. When heated with an excess of acidified concentrated

manganate(VII) ions, it forms initially three products H, J and K.

G H + J + K C13H20O2 C6H8O3 C5H10O2 C2H2O4

When compound G was mixed with liquid bromine at room temperature, C13H20O2Br4 was obtained.

Compound J evolves CO2 with NaHCO3. J also reacts with LiAlH4 to give a compound which does not react with concentrated sulfuric acid.

Compound H gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent. H has no reaction with Tollens’ reagent. When 1 mol of H reacts with an excess of alkaline aqueous iodine, 2 mol of yellow precipitate are produced.

Use the information above to deduce the structures for compounds G, H, J and K.

G undergoes acid hydrolysis to give a carboxylic acid and an alcohol. G is an ester. G undergoes oxidative cleavage with MnO4

-/H+ to give H, J and CO2. G undergoes electrophilic addition with Br2. G is an alkene and contains two C=C bonds due to the addition of four Br atoms. J undergoes neutralisation with NaHCO3. Thus J is a carboxylic acid. J undergoes reduction with LiAlH4 to give a primary alcohol. Primary alcohol cannot undergo elimination with conc H2SO4 as it has no H on the beta-

C (C atom next to C with –OH group has no available H atom). H undergoes condensation with 2,4-DNPH to give an orange ppt. Thus H is a ketone

compound. Aldehyde group is absent since there is no reaction with Tollen’s reagent. H undergoes oxidative cleavage with I2/OH- to give CHI3, the yellow ppt. H contains two CH3CO structural units.

[8]

[Total: 20]

MnO4- / H

+

15 VJC 2014 9647/03/PRELIM/14

5 Amino acids are critical to life, and they serve as building blocks of proteins.

(a) Lysine is an alpha amino acid which shows both acidic and basic properties. The R

group of lysine is –(CH2)4NH2. In acidic solution, lysine is completely protonated.

10.0 cm3 of completely protonated lysine is titrated with 1.00 mol dm−3 sodium hydroxide

solution. Its titration curve is shown below.

(i) Calculate the first dissociation constant Ka1 of fully protonated lysine.

pKa1 = 2.2 Ka1 = 10–2.20

= 6.31 x 10–3 mol dm–3

(ii) Identify the two species at point B. Write an equation to show how the species present at point B can resist change in pH when a small amount of base is added.

When base is added,

(iii) Calculate the pH when 10.0 cm3

of 0.0200 mol dm−3 sodium hydroxide is added to

the solution at point B. (You may represent the fully protonated lysine as HA.)

Number of moles of HA at point A= 1.00 x 1000

10= 1.00 x 10–2 mol

At point B, HAn = -An

= 2

10 x 00.1 -2

= 5.00 x 10–3 mol

No of moles of NaOH added = 0.0200 x 1000

10.0

= 2.00 x 10–4 mol

16 VJC 2014 9647/03/PRELIM/14

HA + OH– → A– + H2O

HAn present after NaOH is added= (5.00 x 10–3) – (2.00 x 10–4)

= 4.80 x 10–3 mol

-An present after NaOH is added = (5.00 x 10–3) + (2.00 x 10–4)

= 5.20 x 10–3 mol

Ka = [HA]

]A][[H -+

6.31 x 10–3 = 3-

-3+

10 x 4.80

10 x 5.20 x ][H

[H+] = 5.82 x 10-3 mol dm-3

pH = 2.24

(iv) State the predominant species at point C. Explain why it has a high melting point.

The predominant species at point C is a zwitterion, as given by the structure below.

A large amount of energy is required to overcome the strong ionic bonds.

(v) State the type of isomerism in the species at point C and draw its isomers.

Optical isomerism

[8]

(b) The first known synthesis of an amino acid occurred in 1850 in the laboratory of Adolf Strecker.

(i) Name the type of reaction in step 1.

Condensation

(ii) Suggest suitable reagents and conditions for step 3.

17 VJC 2014 9647/03/PRELIM/14

H2SO4(aq), reflux (iii) In step 2, the reaction proceeds via two stages:

(I) acid-base reaction between the N atom in imine and HCN (II) nucleophilic attack on C by CN−

Write the mechanism for step 2, showing clearly the movement of electrons and

partial charges.

(iv) An imine intermediate is also formed in preparing secondary amines from ketones.

With reference to the Adolf Strecker synthesis, suggest the synthetic route by giving the reagents, conditions and intermediates for the preparation of (CH3)2CHNH(CH2CH3).

Note: Can accept LiAlH4 in dry ether, r.t. as reducing agent in second step

(v) Another amine X, CH3CH(NH2)CH2NH2 reacts with ethanedioyl dichloride (COCl)2 to

produce compound Y with molecular formula C5H8N2O2. Suggest the structure for compound Y and write the equation between X and ethanedioyl dichloride.

Suggest the structure for compound Y and write the equation between X and ethanedioyl dichloride.

18 VJC 2014 9647/03/PRELIM/14

[8]

[8]

(c) A polypeptide H was analysed and found to contain the following amino acids.

amino acid aspartic acid glycine serine tyrosine valine

abbreviation asp gly ser tyr val

number of residues

1 1 2 1 1

Analysis gave the following results: ● The N-terminus was shown to be ser. ● On reaction with the enzyme chymotrypsin, which hydrolyses at the carboxylic acid

end of tyr, H gave two tripeptides. ● On reaction with a special reagent which digests at the carboxylic end of val, H gave

two peptides. One of these two was a dipeptide of sequence gly-ser. Determine the amino acid sequence of polypeptide H. Briefly justify your answer. [You should use the same 3–letter abbreviations as shown above to write out the amino acid sequence].

ser–asp–tyr–val–gly–ser

[Any 2 valid points from below]

H has 6 amino acids OR first amino acid on left is ser Enzyme hydrolyses at carboxylic acid end of tyr gives 2 tripeptides, hence third amino acid from left is tyr.

Reagent digests at carboxylic end of val gave a tetrapeptide ending with val. OR Tripeptide val–gly–ser is present.

[2]

(d) Papain, an enzyme in fresh papaya juice is used as a meat tenderiser. The three main amino acids involved in the catalytic activity of papain are his, asp and ser. Part of the mechanism of the action of papain is illustrated in the figures below.

19 VJC 2014 9647/03/PRELIM/14

With reference to figures (I) and (II) above, explain why the action of this enzyme would be inhibited if the pH was too low.

When pH is low, [H+] is high. CO2

– of asp is protonated to form –CO2H. Proton on his cannot be abstracted.

O– of ser in figure II cannot be formed. –OH is a weaker nucleophile than O–. Hence, absence of O– on ser prevent attack on

peptide to occur. [2]

[Total: 20]

(I) (II)

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