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CATHOLIC JUNIOR COLLEGE

JC2 PRELIMINARY EXAMINATIONS

Higher 2

CHEMISTRY 9647/01

Paper 1 Multiple Choice Wednesday 2 September 2015

1 hour

Additional Materials: Multiple Choice Answer Sheet

Data Booklet

READ THESE INSTRUCTIONS FIRST

Write your name and HT group on the Answer Sheet in the spaces provided.

Write in soft pencil.

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

There are forty questions on this paper. 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 choice 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.

The use of an approved scientific calculator is expected, where appropriate.

This document consists of 16 printed pages and 0 blank page.

2

9647/01/CJC JC2 Preliminary Exam 2015

Section A

For each question there are four possible answers, A, B, C and D. Choose the one you

consider to be correct and record your choice in soft pencil on the separate Answer Sheet

provided.

1 When 20 cm3 of a gaseous hydrocarbon was completely burnt in 130 cm3 of oxygen,

the volume of gas remaining after the reaction was 100 cm3. This volume was

decreased to 40 cm3 when the resulting mixture was passed through aqueous sodium

hydroxide. All measurements were made at room temperature and pressure.

What is the formula of this hydrocarbon?

A C2H2 B C3H6 C C3H8 D C4H10

2 Consider the following half-equations:

C2O42– → 2CO2 + 2e–

Fe2+ → Fe3+ + e–

MnO4– + 8H+ + 5e– → Mn2+ + 4H2O

What volume of 0.01 mol dm–3 potassium manganate(VII) is required to oxidise

completely 25.0 cm3 of an acidified solution of 0.01 mol dm–3 FeC2O4?

A 10 cm3 B 15 cm3 C 25 cm3 D 42 cm3

3 X and Y are elements with atomic numbers between 6 and 15.

Their first seven ionisation energies in kJ mol–1 are shown below.

X 580 1800 2700 11600 14800 18400 23300

Y 1310 3400 5300 7500 11300 13300 71300

Which of the following best describes the compound formed between X and Y?

A basic

B acidic

C neutral

D amphoteric

3

9647/01/CJC JC2 Preliminary Exam 2015 [Turn over

4 Which pair of compounds fits the following descriptions?

(i) The first compound has a larger bond angle about the central atom than the

second compound and,

(ii) The second compound is more polar than the first compound.

first compound second compound

A

B

C

D

BCl3

CF4

HCN

PH3

SO2

XeF4

BeCl2

NH3

5 Which of the following statements regarding covalent compounds is true?

A Tetrachloromethane is a volatile liquid because the C–Cl bond can be broken

easily.

B CHCl3 has a higher boiling point than CHF3 because it has more electrons than

CHF3.

C Hydrogen chloride is soluble in water because it can form permanent dipole-

permanent dipole forces of attraction with water molecules.

D Each hydrogen bond formed between H2O molecules is stronger than that formed

between HF molecules.

6 Which of the following diagrams does not describe the behaviour of a fixed mass of

an ideal gas?

( = density of the gas, T = temperature measured in K)

A B

C D

constant T

0

p

1/p

constant T

0

V

constant T pV

0

constant V

T 0

p

4


7 A nitrogen–hydrogen mixture, initially in the mole ratio of 1:3, reached equilibrium with

ammonia when 50 % of the nitrogen had reacted. The total final pressure was p.

N2(g) + 3H2(g) ⇌ 2NH3 (g)

What was the partial pressure of ammonia in the equilibrium mixture?

A

B

C

D

8 The heat liberated in the neutralisation given below is -57 kJ mol–1.

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

By using this information, what is the most likely value for the heat liberated in the

following neutralisation?

H2SO4 (aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

A -57 kJ mol–1 B -86 kJ mol–1 C -114 kJ mol–1 D -228 kJ mol–1

9 The value of ionic product Kw, at 35oC is 2.04 x 10-14.

What is correct for pure water at 35 oC? A pH < 7

B pH = 7

C [H+] < [OH-]

D [OH-] = 1.02 x 10-14 mol dm-3

10 A solution of 20.0 cm3 of 0.10 mol dm–3 sulfuric acid was titrated with

0.10 mol dm–3 ammonia.

Which statement regarding the titration is correct?

A The equivalence point is at pH 7.

B The initial pH of the solution is 1.

C 20.0 cm3 of ammonia is required for titration for equivalence point to be reached.

D 80.0 cm3 of ammonia is required for titration for maximum buffer capacity to be

reached.

5


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

Soda is slightly acidic due to dissolved carbon dioxide.

By considering the relevant Eo values, which metal will not be dissolved by the soda?

A V

B Ag

C Mg

D Sn

12 15.7 g of the metal gadolinium (Gd) was deposited in electrolysis by a current of 5 A

for 96.5 minutes. What is the formula of gadolinium ions?

[Ar of Gd = 157]

A Gd+

B Gd2+

C Gd3+

D Gd4+

13 Barium is the final product formed by a series of changes in which the rate-determining

step is the radioactive decay of caesium-137. This radioactive decay is a first-order

reaction with a half-life of 30 years.

How long would it take for a rock sample, originally barium-free, to contain a molar

proportion of caesium-137 to barium of 1:7?

A 15 years

B 30 years

C 60 years

D 90 years

6


14 The reaction between chlorite ions, ClO2–, and iodide ions, I–, in acid solution may be

represented by the following equation.

ClO2– + 4I– + 4 H+→ 2I2 + Cl– + H2O

The rate-time graph for this reaction is as shown below.

What does the shape of the graph suggest about this reaction?

A The reaction is endothermic.

B The reaction is overall first order.

C The reaction produces its own catalyst.

D The reaction rate is independent of iodide ions.

15 The graph below shows the variation in melting points for eight consecutive elements,

A - H, in the Periodic Table, all with atomic number between 10 and 20.

What can be deduced from the graph?

A The chloride of element C fumes in moist air.

B Element F exists as a diatomic molecule.

C The oxide of element B dissolves in water to give an acidic solution.

D Element A has a higher first ionisation energy than the element preceding it.

melting point

atomic number

A

B

C

D

E

F

G

H

rate

time

G

H

7


16 Which of the following statements about the Group II elements (from Mg to Ba) or its

compounds is incorrect?

A The ease of thermal decomposition of Group II nitrates decreases down the

group.

B The reactivity of the elements with chlorine gas increases down the group.

C The melting point of the oxides increases down the group.

D The volume of gases formed per gram of carbonate decomposed decreases

down the group.

17 Some properties of two salts, M and N, are given below.

When salt M was mixed with concentrated sulfuric acid, a colourless gas was

produced. An aqueous solution of this colourless gas gave a white precipitate with

silver nitrate. The precipitate readily dissolved in aqueous ammonia.

When salt N was heated with concentrated sulfuric acid, two gases were evolved. One

gas has a red-brown colour while the other turned acidified potassium dichromate(VI)

paper from orange to green.

Which of the following are possible identities of salt M and N?

salt M salt N

A NaCl KBr

B NaBr KI

C KI NaBr

D KCl NaI

8



Peroxodisulfate(VI), S2O82–, is capable of oxidising the tartrate ion, [CH(OH)CO2]2

2–, to

carbon dioxide and methanoate as shown in the following equation.

[CH(OH)CO2]22– + 3S2O8

2– + 2H2O → 2CO2 + 2HCO2– + 6H+ + 6SO4

2–

The reaction can be catalysed by a homogenous catalyst. Given that the standard

electrode potential for the following half-equation is

2CO2 + 2HCO2– + 6H+ + 6e– ⇌ [CH(OH)CO2]2

2– + 2H2O Eo = + 0.56 V

Which metal ion can be a suitable catalyst for this reaction?

A Cu2+

B Cr2+

C Mn2+

D V2+

19 The diagram shows the structure of Spironolactone, which is an anti-androgen used in

hormone therapy.

How many stereoisomers does Spironolactone have?

A 23 B 24 C 26 D 27

20 What is the major product of the reaction between but-1-ene and gaseous DBr?

(D = 2H)

A CH3CH2CH(OD)CH2Br B CH2DCHBrCH2CH3

C CH2BrCHDCH2CH3 D CH3CHDCH2CH2Br

9


21 Terfenadine (sold under the trade name Seldane) alleviates seasickness and asthma

in the same way as the older drugs, but it does not cause drowsiness as a side effect.

terfenadine

What deductions about terfenadine can be made from this structure?

A It is soluble in aqueous sodium hydroxide.

B It is chiral and contains a primary amine group.

C It can be oxidised to form a product which can react with ammoniacal silver

nitrate solution.

D One mole of terfenadine will give out one mole of hydrogen on reaction with two

moles of sodium.

22 An iodoalkane and sodium react when heated under reflux in an inert organic solvent

to form an alkane according to the equation:

2RI + 2Na → R–R + 2NaI

When a mixture of equal number of moles of the two different iodoalkanes, CH3CH2I

and CH3CHICH3, is used, which of the following alkanes will not be formed?

A (CH3)2CHCH(CH3)2 B CH3CH2CH(CH3)2

C CH3CH2CH2CH3 D CH3CH2CH2CH2CH2CH3

10


23 Ferulic acid, is an antioxidant that is reactive towards free radicals implicated in DNA

damage, cancer and accelerated cell aging.

Assuming the CH3O− grou is inert, which statement about ferulic acid is correct?

A One mole of ferulic acid reacts with only one mole of bromine.

B An ester is formed when ferulic acid is heated with propanoic acid in the

presence of concentrated sulfuric acid.

C One mole of ferulic acid reacts with only one mole of thionyl chloride.

D Ferulic acid reacts with lithium aluminium hydride in dry ether to form a product

that does not exhibit geometric isomerism.

24 Bromoalkanes are commonly prepared by reacting either a primary or secondary

alcohol with phosphorus tribromide, PBr3. The reaction usually involves two steps, and

the mechanism is shown below:

Which of the following statements regarding the reaction shown is correct?

A The rate of reaction for step 1 is independent of the concentration of the alcohol.

B Both step 1 and step 2 involve the formation of a 5-coordinated transition state.

C The reaction can be carried out in aqueous medium.

D The alcohol acts as a nucleophile in step 1.

25 Which compound will not produce yellow precipitate on warming with alkaline aqueous

iodine?

A CH3CH2CH(OH)CH3

B ICH2COCH2CH=CH2

C CH3CO2CHI2

D I3CCOCI3

11


26 The reaction between compound R and phenylhydrazine gives rise to a precursor

compound S, used in the formation of Phenazone, an anti-inflammatory drug.

Which of the following is the correct structure of S?

A B

C D

27 1 g of each of the following compounds was heated with NaOH(aq). After cooling, HNO3(aq) was added followed by AgNO3(aq).

Which compound will produce the largest mass of white precipitate?

A B

C D

phenylhydrazine

12


28 Deuterium, D, is a heavier isotope of hydrogen. Which of the following shows the

correct product when Compound K reacts as shown below?

A B

C D

29 Liquid Q is sparingly soluble in water but dissolves readily in cold sulfuric acid. 1 mol of

Q reacts with 3 mol of Br2(aq).

Which of the following is a possible identity of Q?

A C6H5CO2H B C6H5NH2 C CH3(CH2)5NH2 D C6H5CH=CH2

30 Consider the following compounds below:

I C6H5NH2 II CH3CH2CONH2 III CH3CH(Cl)NH2 IV CH3CH2NH2

Which of the following shows the correct order of decreasing pKb values for the above

compounds?

A II, III, IV, I

B II, I, III, IV

C III, IV, I, II

D IV, III, I, II

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

2 and 3 only are correct

1 only is

correct

No other combination of statements is used as a correct response. 31 Use of the Data Booklet is relevant to this question. Compound L (Mr = 179) has the structure as shown:

Which of the following statements about compound L are correct?

1 The empirical formula of L is C3H2ClO.

2 179 g of L requires 319.6 g of bromine dissolved in an inert organic solvent for

complete reaction.

3 Upon complete reaction with aqueous AgNO3, 358 g of L forms 287 g of white

precipitate.

32 The enthalpy change of reaction between calcium and water is measured in the

laboratory and found to be x kJ mol–1.

Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g) ΔH = x kJ mol–1

What other information is needed to calculate the ΔHf of Ca(OH)2(s)?

1 enthalpy change of combustion of hydrogen

2 lattice energy of calcium hydroxide

3 enthalpy change of atomisation of calcium

14


33 The following graph is obtained from an experiment.

Which of the following are not correct statements about the reaction?

1 At time t, dynamic equilibrium is achieved.

2 Addition of a catalyst increases the magnitude of t.

3 Upon addition of a catalyst, at time t, the [product] to [reactant] ratio increases.

34 Which is not a trend from left to right across the elements of the third period of the

Periodic Table?

1 melting points of the elements decrease steadily

2 compounds of the elements change from ionic to covalent

3 oxides of the elements change from basic to acidic.

35 During anodising of aluminium, 1 mol dm–3 aqueous H2SO4 is electrolysed between Al

anode and graphite cathode, using a constant current for 60 s.

What affects the mass of oxide layer deposited on the Al anode?

1 increasing the time taken

2 changing the electrolyte to 1 mol dm–3 aqueous HCl

3 increasing the concentration of the electrolyte solution

t Time

Concentration

reactant

product

15


36 The Diels-Alder reaction is a cycloaddition that occurs between an “electron-rich”

conjugated diene and “electron-poor” alkene, forming a cyclohexene ring.

Which of the following statements about this reaction are correct?

1 During the cycloaddition, carbons 1 and 4 of the diene and the two alkene

carbons will change from being sp2 hybridised to being sp3 hybridised.

2 The presence of a –CHO substituent on the alkene molecule will increase the

rate of the Diels-Alder reaction.

3 The presence of a –CH3 substituent on the diene molecule will increase the rate

of the Diels-Alder reaction.

37 Methylparaben is an anti-fungal agent that is commonly used in a variety of cosmetic

and personal care products.

Which deductions about the reactions of methylparaben can be made from its

structure?

1 It can decolourise hot acidified KMnO4 solution.

2 It can react with PCl5 to form steamy fumes.

3 It can react with hydrogen cyanide in the presence of trace amount of NaCN.

16


38 Keratin is a fibrous and insoluble protein found in hair, nails and horns of mammals. It comprises mainly the three amino acids below:

NH2CH(CH2SH)CO2H NH2CH(CH2OH)CO2H NH2CH(CH2CH2CO2H)CO2H

cysteine serine glutamic acid

Based on the above information, what type of bonding would stabilise the tertiary structure of keratin?

1 disulfide bonds 2 hydrogen bonding 3 van der Waals’ forces

39 Prostaglandins are biomolecules with a wide range of biological effects which include lowering blood pressure and gastric secretion.

Which of the following statements correctly describe the two prostaglandin molecules shown above?

1 Both prostaglandin E2 and F α are optically active molecules. 2 Prostagladin E2 can be converted into prostaglandin F α by reacting with

hydrogen in presence of nickel catalyst. 3 Both prostaglandin molecules will form a silver mirror with Tollens’ reagent.

40 Which of the following transformations involve a nucleophile?

1

2

3

1

[Turn over

CATHOLIC JUNIOR COLLEGE

JC2 PRELIMINARY EXAMINATIONS

Higher 2

CHEMISTRY 9647/01

Paper 1 Multiple Choice Wednesday 2 September 2015

1 hour

Additional Materials: Multiple Choice Answer Sheet

Data Booklet


Write your name and HT group on the Answer Sheet in the spaces provided.

Write in soft pencil.


There are forty questions on this paper. 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 choice 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.

The use of an approved scientific calculator is expected, where appropriate.


WORKED SOLUTIONS

2


Section A

For each question there are four possible answers, A, B, C and D. Choose the one you

consider to be correct and record your choice in soft pencil on the separate Answer Sheet

provided.

1 When 20 cm3 of a gaseous hydrocarbon was completely burnt in 130 cm3 of oxygen,

the volume of gas remaining after the reaction was 100 cm3. This volume was

decreased to 40 cm3 when the resulting mixture was passed through aqueous sodium

hydroxide. All measurements were made at room temperature and pressure.

What is the formula of this hydrocarbon?

A C2H2 B C3H6 C C3H8 D C4H10

Answer: B

Since temperature and pressure are constant, mol ratio of gas ∝ vol ratio

CxHy (g) + )(4

yx O2 (g) → xCO2 (g) + 2

y H2O (l)

Initial vol/ cm3: 20 130 0 0

Final vol/ cm3: 0 40 (100-40)

Vol reacted or formed/ cm3: 20 (130-40) 60

= 90

1 vol 4.5 vol 3 vol

By inspection, x = 3.

2

9)

4

y(3 y = 6

molecular formula of the hydrocarbon is C3H6

2 Consider the following half-equations:

C2O42– 2CO2 + 2e–

Fe2+ Fe3+ + e–

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

What volume of 0.01 mol dm–3 potassium manganate(VII) is required to oxidise

completely 25.0 cm3 of an acidified solution of 0.01 mol dm–3 FeC2O4?

A 10 cm3 B 15 cm3 C 25 cm3 D 42 cm3

Answer: B

Both Fe2+ and C2O42– ions can be oxidised by MnO4

– ions which is reduced.

Hence, there are two oxidation half-equations to consider:

Fe2+ Fe3+ + e– --- 1 and

C2O42– 2CO2 + 2e– --- 2

So combine 1+2 first by summing up both half-equations to get:

Fe2+ + C2O42– Fe3+ + 2CO2 + 3e– --- 3

3


Combine this equation with the reduction half-equation of MnO4–

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

To get: 3MnO4– ≡ 5 FeC2O4

Amount of FeC2O4 = 25/1000 x 0.01 = 2.50 x 10–4 mol

Amount of KMnO4 reacted = (2.50 x 10-4 / 5) x 3 = 1.50 x 10–3 mol

Vol. of KMnO4 required = (1.50 x 10–3) / 0.01 = 1.50 x 10–2 dm3 = 15 cm3

3 X and Y are elements with atomic numbers between 6 and 15.

Their first seven ionisation energies in kJ mol–1 are shown below.

X 580 1800 2700 11600 14800 18400 23300

Y 1310 3400 5300 7500 11300 13300 71300

Which of the following best describes the compound formed between X and Y?

A basic

B acidic

C neutral

D amphoteric

Answer: D

Calculate the difference between successive I.E.

For X, there is a sharp increase from the 3rd to 4th I.E. This indicates that the 4th

electron is removed from an inner electron shell, closer and more tightly bound by the

positively-charged nucleus. Thus, there are 3 valence electrons; X belongs to Group III,

and is Al.

Using the same concept, there is a sharp increase from the 6th to 7th I.E. for Y. Y

belongs to Group VI, and is O.

The compound formed from X and Y has the formula Al2O3 which is amphoteric in

nature.

Alternatively, you can match the I.E. given to those given in the data booklet to

determine the identity of X and Y.

4


4 Which pair of compounds fits the following descriptions?

(i) The first compound has a larger bond angle about the central atom than the

second compound and,

(ii) The second compound is more polar than the first compound.

first compound second compound

A

B

C

D

BCl3

CF4

HCN

PH3

SO2

XeF4

BeCl2

NH3

Answer: A

If necessary, draw the dot-and-cross diagrams of the compounds first.

Option A:

SO2 has a bent shape and a smaller bond angle than 120˚ due to the presence of the

lone pair of electrons. (i) is satisfied.

(ii) is also satisfied as BCl3 is non-polar while SO2 has a net dipole moment and is

polar.

Option B:

(i) is satisfied but (ii) is not as both are symmetrical and have no net dipole moment

and are hence non-polar.

Option C:

(i) is not satisfied. (ii) is not satisfied as HCN is polar while BeCl2 is not.

Option D: The more electronegative the central atom, the greater the repulsion

between the bond pairs. N is more electronegative than P and pulls the bonded

electron pairs closer to itself. Hence, the bonded electron pairs get closer to each other

and repel much more, resulting in a larger bond angle for NH3 (even though NH3 and

PH3 are both trigonal pyramidal in shape).

5


5 Which of the following statements regarding covalent compounds is true?

A Tetrachloromethane is a volatile liquid because the C–Cl bond can be broken

easily.

B CHCl3 has a higher boiling point than CHF3 because it has more electrons than

CHF3.

C Hydrogen chloride is soluble in water because it can form permanent dipole-

permanent dipole forces of attraction with water molecules.

D Each hydrogen bond formed between H2O molecules is stronger than that

formed between HF molecules.

Answer: B

Option A: For a simple covalent molecule like CCl4, covalent bonds are not broken

when a liquid boils/vaporizes. CCl4 is non-polar and the weak intermolecular van der

Waals’ forces of attraction are overcome instead.

Option B: boiling point ∝ strength of van der Waals’ forces ∝ no. of electrons

Option C: HCl when dissolved in water undergoes complete dissociation to form H+

and Cl- ions which form ion-dipole attractions with water molecules.

Option D: strength of hydrogen bond ∝ difference in electronegativity of the 2 atoms

involved in the hydrogen bonding. O–H bond has a smaller electronegativity difference

than H–F bond as O is less electronegative than F. Note: H2O has a higher boiling

point than HF as there are, on average, more hydrogen bonds between H2O molecules

than there are between HF molecules (H2O has more extensive intermolecular

hydrogen bonding than HF).

6 Which of the following diagrams does not describe the behaviour of a fixed mass of

an ideal gas?

( = density of the gas, T = temperature is measured in K)

A B

C D

1/p

T

0

p

pV

0 0

V

constant T constant T

constant T constant V

0

p

6


Answer: C

For a given mass of gas, the no. of moles of gas, n, will be constant since Mr

unchanged.

pV = nRT

Option A: V = nRT (

)

V = constant x (

) , at constant T for a given mass of gas.

Hence, V (

)

Option B: pV = nRT

pV = constant, at constant T for given mass of gas, regardless of changes in P, V or

Option C: pV = nRT =

RT

ρ =

=

(P) = constant x (P), at constant T for a given mass of gas.

Hence, ρ P and it should be a Y = mX + C linear graph, passing through origin.

Option D:

P =

(T)

P= constant x (T), at constant V for a given mass of gas.

Hence, P T

7 A nitrogen–hydrogen mixture, initially in the mole ratio of 1:3, reached equilibrium with ammonia when 50 % of the nitrogen had reacted. The total final pressure was p.

N2(g) + 3H2(g) ⇌ 2NH3 (g)

What was the partial pressure of ammonia in the equilibrium mixture?

A p

B

p

C

p

D

p

2

Answer: C

Let the initial amount of N2 and H2 be 1 mol and 3 mol respectively.

N2(g) + 3H2(g) 2NH3 (g)

Initial amount/mol 1 3 -

Change/mol –(50/100) x 1 –(0.5x3) +(0.5x2)

Equilibrium

amount/mol 0.5 1.5 1.0

At equilibrium, total amount of gas = 0.5 + 1.5 + 1.0 = 3.0 mol

PNH3 = (1.0/3.0) x p = p/3 (pA = xA . pT) where x is the mole fraction of gas A

7


8. The heat liberated in the neutralisation given below is -57 kJ mol-1.

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

By using this information, what is the most likely value for the heat liberated in the

following neutralisation?

H2SO4 (aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

A -57 kJ mol-1 B -86 kJ mol-1 C -114 kJ mol-1 D -228 kJ mol-1

Both neutralisations involve strong acid and strong base. The heat liberated in

the neutralization to produce 1 mol of water = - 57 kJ mol-1

The neutralisation between H2SO4 and NaOH produces 2 mol of water.

Thus, heat liberated = 2 (-57) = -114 kJ mol-1

9 The value of ionic product Kw, at 35oC is 2.04 x 10-14.

What is correct for pure water at 35 oC? A pH < 7

B pH = 7

C [H+] < [OH-]

D [OH-] = 1.02 x 10-14 mol dm-3

Option A Since water is neutral, [H+] = [OH-] = (2.04 x 10-14)1/2 = 1.43 x 10-7 mol dm-3

pH = -log (1.43 x 10-7) = 6.85 <7

8


10. A solution of 20.0 cm3 of 0.10 mol dm-3 sulfuric acid was titrated with

0.10 mol dm-3 ammonia.

Which statement regarding the titration is correct?

A The equivalence point is at pH 7.

B The initial pH of the solution is 1.

C 20.0 cm3 of ammonia is required for titration for equivalence point to be reached.

D 80.0 cm3 of ammonia is required for titration for maximum buffer capacity to be

reached.

Statement A is incorrect: Since this is a weak base-strong acid titration, the salt

obtained is acidic. Thus the equivalence point is less than pH 7.

Statement B is incorrect: Since sulfuric acid is a dibasic acid, the pH of the

solution = -lg (0.10 x 2) = 0.700

Statement C is incorrect: since 2 mol of NH3 is required to neutralize 1 mol

H2SO4, thus 40 cm3 of NH3 is required to reach equivalence point.

Statement D is correct:

Volume required to reach maximum buffer capacity

= 2 x (Volume of NH3 required for equivalence point)

= 2 x 40

= 80 cm3

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

Soda is slightly acidic due to the dissolved carbon dioxide.

By considering the relevant Eo values, which metal will not be dissolved by the soda?

A V

B Ag

C Mg

D Sn

Since soda is acidic, 2H+ + 2e– H2 Eredo = 0.00V

To determine whether metal will not dissolve by soda, Ecello <0. This will indicate

the reaction is non-spontaneous.

Equation Eoxio Overall equation Ecell

o<0

V2+ + 2e– V -1.20 V + 2H+

H2 + V2+ +1.20

Ag+ + e– Ag +0.80 Ag + 2H+

H2 + Ag+ -0.80

Mg2+ + 2 e– Mg -2.38 Mg + 2H+

H2 + Mg2+ +2.38

Sn2+ + 2 e– Sn -0.14 Sn + 2H+

H2 + Sn2+ +0.14

Hence, Ag cannot dissolve in an acidic solution.

9


12. 15.7 g of the metal gadolinium (Gd) was deposited in electrolysis by a current of 5 A

for 96.5 minutes. What is the formula of gadolinium ions?

[Ar of Gd = 157]

A Gd+

B Gd2+

C Gd3+

D Gd4+

Q = It = (5)(60)(96.5) = 28950C

= (28950/96500) F = 0.3F

No. of moles of Gd = 15.7/157 = 0.1

Gdn+ + n e- Gd

Quantity of electricity required to deposit 0.1 mol of Gd = 0.3F

no. of mol of electrons: no. of mol of Gd = 0.3: 0.1

n = 3

Thus, formula of Gd ion is Gd3+.

13. Barium is the final product formed by a series of changes in which the rate-determining

step is the radioactive decay of caesium-137. This radioactive decay is a first-order

reaction with a half-life of 30 years.

How long would it take for a rock sample, originally barium-free, to contain a molar

proportion of caesium-137 to barium of 1:7?

A 15 years

B 30 years

C 60 years

D 90 years

Cs: x ½ x ¼ x 1/8 x

Ba: 0 ½ x ¾ x 7/8 x

n = 3 half-lives

Time taken = 30 x 3 = 90 years

OR

10


Using o

t

c

c= (

2

1)n

n = 3 half-lives

Time taken = 30 x 3 = 90 years

14. The reaction between chlorite ions, ClO2-, and iodide ions, I-, in acid solution may be

represented by the following equation.

ClO2- + 4I- + 4 H+→ 2I2 + Cl- + H2O

The rate-time graph for this reaction is as shown below.

What does the shape of the graph suggest about this reaction?

A The reaction is endothermic.

B The reaction is overall first order.

C The reaction produces its own catalyst.

D The reaction rate is independent of iodide ions.

This is an example of autocatalytic reaction. The rate of reaction is slow at first

but increases as the catalyst (product) is produced.

rate

time

11


15 The graph below shows the variation in the melting points for eight consecutive

elements, A - H, in the Periodic Table, all with atomic number between 10 and 20.

What can be deduced from the graph?

A The chloride of element C fumes in moist air.

PCl3 and PCl5 hydrolyse in water to give HCl, which appears as white fumes.

PCl3 + 3H2O → H3PO3 + 3HCl

PCl5 + 4H2O → H3PO4 + 5HCl

B Element F exists as a diatomic molecule.

F is a noble gas and exists as monoatomic gas.

C The oxide of element B dissolves in water to give an acidic solution.

SiO2 has a giant covalent structure and is insoluble in water.

D Element A has a higher first ionisation energy than the element preceding it.

Al has a lower first ionisation energy than Mg since the electron removed

from Al is from a higher energy 3p orbital, as compared to that of Mg which

involves 3s orbital. (Note that the exceptions to the trend occur in Groups

III and VI.)

melting point

atomic number

A

B

C D

E F

G

H

Al

Si

P4

S8

Cl2 Ar

K

Ca

12


16 Which of the following statements about the Group II elements (from Mg to Ba) or its

compounds is incorrect?

A The ease of thermal decomposition of Group II nitrates decreases down the

group.

Ease of thermal decomposition decreases down the group. From Mg2+ to

Ba2+, the ionic charge remains the same but ionic radius increases. Thus,

charge density of the cation decreases down the group. The nitrate ion is

less polarised and thermal decomposition occurs less readily.

B The reactivity of the elements with chlorine gas increases down the group.

Group II metals undergo redox reactions with Cl2. The reducing strength of

Group II metals increases down the group. Down the group, the atomic

size of atom and shielding effect increases and thus it is easier for the atom

to lose two electrons. The reactivity of the metals with Cl2 would increase

down the group. (Recall from Group VII that Cl2 is an oxidising agent.)

C The melting point of the oxides increases down the group.

Group II oxides are ionic compounds and lattice energy is an indication of

the ionic bond strength.

Lattice energy ∝ -

-

Since ionic radius of the metal cation increases down the group, the lattice

energy of the metal oxide becomes less exothermic. The ionic bond

strength becomes weaker down the group and thus melting point of the

oxides should decrease.

D The volume of gases formed per gram of carbonate decomposed decreases

down the group.

MCO3 → MO + CO2

The Mr of the metal carbonate increases down the group. Thus the number

of mol per gram of metal carbonate decreases down the group. Amount of

CO2 formed and thus volume of gas formed would decrease.

13


17 Some properties of two salts, M and N, are given below.

When salt M was mixed with concentrated sulfuric acid, a colourless gas was

produced. An aqueous solution of this colourless gas gave a white precipitate with

silver nitrate. The precipitate readily dissolved in aqueous ammonia.

Since the precipitate is readily dissolved in aqueous ammonia, precipitate

formed is AgCl. Thus salt A must contain a Cl¯

ion. The colourless gas evolved

is HCl.

When salt N was heated with concentrated sulfuric acid, two gases were evolved. One

gas has a red-brown colour while the other turned acidified potassium dichromate(VI)

paper from orange to green.

The red-brown gas is Br2 and thus salt B must contain Br¯. The other gas is SO2.

KBr + H2SO4 → HBr + KHSO4

HBr is then oxidised by H2SO4:

2HBr + H2SO4 → B 2 + 2H2O + SO2

Which of the following are possible identities of salt H and J?

salt M salt N

A NaCl KBr

B NaBr KI

C KI NaBr

D KCl NaI

14



Peroxodisulfate(VI), S2O82-, is capable of oxidising the tartrate ion, [CH(OH)CO2]2

2-, to

carbon dioxide and methanoate as shown in the following equation.

[CH(OH)CO2]22- + 3S2O8

2- + 2H2O → 2CO2 + 2HCO2- + 6H+ + 6SO4

2-

The reaction can be catalysed by a homogenous catalyst. Given that the standard electrode potential for the following half-equation is

2CO2 + 2HCO2- + 6H+ + 6e- ⇌ [CH(OH)CO2]2

2- + 2H2O E = + 0.56 V

Which metal ion can be a suitable catalyst for this reaction?

A Cu2+

B Cr2+

C Mn2+

D V2+

Ans: C

S2O82- + 2e ⇌ 2SO4

2- E = +2.01 V

2CO2 + 2HCO2- + 6H+ + 6e- ⇌ [CH(OH)CO2]2

2- + 2H2O E = + 0.56 V

Cu2+ + 2e ⇌ Cu+ E = +0.15 V

Cr3+ + e ⇌ Cr2+ E = -0.41 V

Mn3+ + e ⇌ Mn2+ E = +1.49 V

V3+ + e ⇌ V2+ E = -0.26 V Catalyst chosen must have a value between +0.56 V and +2.01 V in order to reduce S2O8

2- and to oxidize [CH(OH)CO2]22-.

Catalysed reaction using Mn2+ 2Mn2+ + S2O8

2- ⇌ 2Mn3+ + 2SO42-

Ecell = 2.01 – (1.49) = + 0.52V

6Mn3+ + [CH(OH)CO2]2

2- + 2H2O ⇌ 6Mn2+ + 2CO2 + 2HCO2- + 6H+

Ecell = 1.49 - 0.56 = + 0.93 V

15


19 The diagram shows the structure of Spironolactone, which is an anti-androgen used in

hormone therapy.

How many stereoisomers does Spironolactone have?

No. of stereoisomers = 2n = 27

A 23 B 24 C 26 D 27

20 What is the major product of the reaction between but-1-ene and gaseous DBr?

(D = 2H)

A CH3CH2CH(OD)CH2Br B CH2DCHBrCH2CH3

C CH2BrCHDCH2CH3 D CH3CHDCH2CH2Br

This reaction is similar to the electrophilic addition reaction with HBr.

Apply Markovnikov’s ule.

General guide to get the Markovnikov’s p oduct: The elect ophile (in this case,

+ D) will go to the carbon in the C=C with more H directly attached to it.

16


21 Terfenadine (sold under the trade name Seldane) alleviates seasickness and asthma

in the same way as the older drugs, but it does not cause drowsiness as a side effect.

terfenadine

What deductions about terfenadine can be made from this structure?

A It is soluble in aqueous sodium hydroxide.

Alcohols do NOT react with NaOH.

B It is chiral and contains a primary amine group.

There is one chiral carbon but it contains a tertiary amine instead.

C It can be oxidised to form a product which can react with ammoniacal silver

nitrate solution.

Secondary alcohols are oxidised to ketones while tertiary alcohols cannot

be further oxidised.

Recall that only aldehydes eact with Tollens’ eagent..

D One mole of terfenadine will give out one mole of hydrogen on reaction with

two moles of sodium.

1 – OH ≡ ½ H2 ≡ 1 Na

1 mole of terfenadine contains 2 –OH groups which will react with 2

moles of Na and form 1 mole of H2 gas.

22 An iodoalkane and sodium react when heated under reflux in an inert organic solvent

to form an alkane according to the equation:

2RI + 2Na → R–R + 2NaI

When a mixture of equal number of moles of the two different iodoalkanes, CH3CH2I

and CH3CHICH3, is used, which of the following alkanes will not be formed?

A (CH3)2CHCH(CH3)2 B CH3CH2CH(CH3)2

C CH3CH2CH2CH3 D CH3CH2CH2CH2CH2CH3

Pattern recognition question.

*

tertiary amine

secondary alcohol tertiary alcohol

17


23 Ferulic acid, is an antioxidant that is reactive towards free radicals implicated in DNA

damage, cancer, and accelerated cell aging.

Assuming the CH3O− group is inert. Which statement about ferulic acid is correct? A One mole of ferulic acid reacts with only one mole of bromine. B An ester is formed when ferulic acid is heated with propanoic acid in the

presence of concentrated sulfuric acid. C One mole of ferulic acid reacts with only one mole of thionyl chloride. D Ferulic acid reacts with lithium aluminium hydride in dry ether to form a product

that does not exhibit geometric isomerism.

Answer: C A Aside from electrophilic addition of bromine, ferulic acid will also undergo

electrophilic substitution with bromine. B Phenols will not react with carboxylic acids to form esters. Phenyl esters involve

the reaction of the phenolic functional group with acyl chlorides. C Only carboxylic acid reacts with SOCl2. Phenol does not react with SOCl2. D Lithium aluminium hydride does not react with alkene to form alkane.

18


24 Bromoalkanes are commonly prepared by reacting either a primary or secondary

alcohol with phosphorus tribromide, PBr3. The reaction usually involves two steps, and

the mechanism is shown below:

Which of the following statements regarding the reaction shown is correct?

A The rate of reaction for step 1 is independent of the concentration of the alcohol. B Both step 1 and step 2 involve the formation of a 5-coordinated transition state. C The reaction can be carried out in aqueous medium. D The alcohol acts as a nucleophile in step 1.

Answer: D A The alcohol is involved in the formation of the intermediate species in step 1. B Only step 2 involves the formation of a 5-coordinated transition state via SN2. C PBr3 will react with water to form H3PO3 and HBr. D The oxygen atom in the alcohol donates a lone pair of electrons to the electron

deficient phosphorous atom in PBr3 to form a dative covalent bond,

25 Which compound will not produce yellow precipitate on warming with alkaline aqueous

iodine?

A CH3CH2CH(OH)CH3

B ICH2COCH2CH=CH2

C CH3CO2CHI2

D I3CCOCI3

Aside from , aldehydes and ketones with and and

will also undergo positive iodoform test as the iodine substitution of the -CH3 group

attached to the carbonyl group occurs stepwise.

Answer: C

A Alcohols that have structure give a positive iodoform test.

B structure is present.

C CH3CO2CHI2 is an ester that will not give a positive iodoform test.

D structure is present.

19


26 The reaction between compound R and phenylhydrazine gives rise to a precursor

compound S, used in the formation of Phenazone, an anti-inflammatory drug.

Which of the following is the correct structure of S?

A B

C D

R, which contains the ketone functional group, undergoes condensation reaction with

phenylhydrazine, similar to 2,4-DNPH.

Answer: C

A The ester functional group in R does not undergo condensation with

phenyhydrazine.

B The formation of a 2o amine with loss of NH3 from the reaction between

phenyhydrazine and R does not take place.

C Condensation takes place between the ketone functional group in R with

phenylhydrazine.

D The amide shown is not formed from the reaction between R and

phenylhydrazine.

20


27 1 g of each of the following compounds was heated with NaOH(aq). After cooling,

HNO3(aq) was added followed by AgNO3(aq).

Which compound will produce the largest mass of white precipitate?

A B

C D

Answer: C

Halogenoalkane and acyl chlorides can undergo hydrolysis but not halogenoarenes.

Compound No. of moles of compound

in 1 g No. of moles of AgCl formed

A 0.0127 0.0127 x 1 = 0.0127

B 0.00550 0 (no hydrolysis due to partial double

bond character)

C 0.00619 0.00619 x 3 =0.0185

D 0.00645 0.0645 x 2 =0.0129

21


28 Deuterium, D, is a heavier isotope of hydrogen. Which of the following shows the

correct product when Compound K reacts as shown below?

A B

C D

Answer: A

There are 3 functional groups (phenol, ester and carboxylic acid) present in K

that can react with NaOD in D2O under heating.

After reacting with NaOD, -OH in phenol and carboxylic acid loses a proton to

give –O- and –CO2- respectively. The ester group (–OCOCH3) attached to the

benzene ring is hydrolysed to give –O- and CH3CO2-.

22


29 Liquid Q is sparingly soluble in water but dissolves readily in cold sulfuric acid. 1 mol of

Q reacts with 3 mol of Br2(aq).

Which of the following is a possible identity of Q?

A C6H5CO2H B C6H5NH2 C CH3(CH2)5NH2 D C6H5CH=CH2

Answer: B

Compound Solubility in water Solubility in H2SO4 Reaction with Br2(aq)

C6H5CO2H Soluble as benzoate

ion is formed during

dissociation

Does not react with

H2SO4

No reaction

C6H5NH2 Sparingly soluble

due to formation of

H-bonding

Soluble as C6H5NH3+

ions are formed

Reacts with 3 moles of

Br2(aq)

CH3(CH2)5NH2 Sparingly soluble

due to formation of

H-bonding

Soluble as

CH3(CH2)5NH3+ ions

are formed

No reaction

C6H5CH=CH2 Insoluble Does not react with

H2SO4

Reacts with 1 mole of

Br2(aq)

Since only B fits all 3 criteria, B is a possible identity of Q.

30 Consider the following compounds below:

I C6H5NH2 II CH3CH2CONH2 III CH3CH(Cl)NH2 IV CH3CH2NH2

Which of the following shows the correct order of decreasing pKb values for the above

compounds?

A II, III, IV, I

B II, I, III, IV

C III, IV, I, II

D IV, III, I, II

Answer: B

The stronger the base, the lower the pKb value. Basicity depends on the availability of

lone pair of electrons on N atom to attract a proton.

IV is the most basic as it has an electron-donating group (CH3CH2-) which increases

the availability of lone pair of electrons on N atom to attract a proton.

III is a weaker base than IV due to the presence of an electron-withdrawing Cl atom

that decreases the availability of lone pair of electrons on N atom to attract a proton.

I is the weakest base as the lone pair of electrons on N can be delocalised into the

benzene ring thus making the lone pair of electrons on N atom unavailable to attract a

proton.

II is an amide which is neutral so it would have the highest pKb value.

23


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

2 and 3 only are

correct

1 only is correct

No other combination of statements is used as a correct response.


Compound L (Mr = 179) has the structure as shown:

Which of the following statements about compound L are correct?

1 The empirical formula of L is C3H2ClO.

2 179 g of L requires 319.6 g of bromine dissolved in an inert organic solvent for

complete reaction.

3 Upon complete reaction with aqueous AgNO3, 358 g of L forms 287 g of white

precipitate.

Answer: A (1, 2 and 3)

Compound L has the molecular formula of C6H4Cl2O2

1: Empirical formula of L is C3H2ClO. Statement 1 is correct.

2: 179 g of L gives 1 mol of L. L contains 2 C=C therefore 1 mol of L will react with 2

mol of Br2 in CCl4.

2 mol of Br2 = 2 (79.9x2) = 319.6 g. Statement 2 is correct.

3: Compound L contains 1 acyl chloride and 1 Cl that is directly attached to a C

across the C=C, therefore, only the acyl chloride can be hydrolysed to produce Cl-.

358 g of L gives 2 mol of L. 2 L ≡ 2Cl- (from 2 mol of acyl chloride)

Mass of AgCl produced = 2 (108+35.5) = 287 g Statement 3 is correct.

24


32 The enthalpy change of reaction between calcium and water is measured in the

laboratory and found to be x kJ mol-1.

Ca(s) + 2H2O(l) Ca(OH)2(s) + H2(g) ΔH = x kJ mol-1

What other information is needed to calculate the ΔHf of Ca(OH)2(s)?

1 enthalpy change of combustion of hydrogen

2 lattice energy of calcium hydroxide

3 enthalpy change of atomisation of calcium

Answer: D (1 only)

1: Equation for combustion of hydrogen : H2(g) +1/2 O2 (g) H2O(l)

Enthalpy change of formation of H2O = 2x(enthalpy change of combustion of

hydrogen.) This information is required to find the value for y in the energy cycle

above.

From the energy cycle above, information 2 and 3 are not required.

Hf =0

Ca(s) + 2H2O(l) Ca(OH)2(s) + H2(g)

Ca (s) + 2H2 (g) + O2 (g)

Hf =0

H = x kJ mol-1

Hf =y

Hf = 2 Hc of H2

25


33 The following graph is obtained from an experiment.

Which of the following are not correct statements about the reaction above?

1 At time t, dynamic equilibrium is achieved.

2 Addition of a catalyst increases the magnitude of t.

3 Upon addition of a catalyst, at time t, the [product] to [reactant] ratio increases.

Answer: C (2 and 3)

At time t, there are no changes to the reactant and product concentrations respectively

and the system has reached dynamic equilibrium. Statement 1 is correct.

A catalyst speeds up the reaction rate by providing an alternative reaction pathway

with lower activation energy and thus equilibrium is reached faster. Magnitude of t

should decrease when catalyst is added. Statement 2 is not correct.

A catalyst has no effect on the position of equilibrium, and so, does not alter the

composition of the equilibrium mixture. At time t, the [product] to [reactant] ratio stays

the same. Statement 3 is not correct.

34 Which is not a trend from left to right across the elements of the third period of the

Periodic Table?

1 melting points of the elements decrease steadily

2 compounds of the elements change from ionic to covalent

3 oxides of the elements change from basic to acidic.

Answer: D

Option 1 is correct only; recall melting point trend:

t Time

Concentration

reactant

product

26


Melting points for Na to Ar

35 During anodising of aluminium, 1 mol dm-3 aqueous H2SO4 is electrolysed between Al

anode and graphite cathode, using a constant current for 60 s.

What affects the mass of oxide layer deposited on the Al anode?

1 increasing the time taken

2 changing the electrolyte to 1 mol dm-3 aqueous HCl

3 increasing the concentration of the electrolyte solution

Answer: D

Option 1 is correct.

Recall that mass of elements liberated during electrolysis only depends on (i)

quantity of electricity (time of passing steady current & magnitude of steady

current passed) & (ii) charge on the ion of element.

Option 1: Increasing time will increase the mass of oxide layer deposited on the

anode.

Option 2: Changing the anion in the electrolyte from SO42- to Cl

- does not change

the anode reaction (water is still being oxidised to O2.)

Option 3: Increasing concentration of the H2SO4 (aq) solution will not cause

further change in mass of oxide layer deposited.

Na

Mg Al

Si

P S

Cl Ar

-400

-200

0

200

400

600

800

1000

1200

1400

1600

10 11 12 13 14 15 16 17 18

melt

ing

po

int

/ oC

proton number

27


36 The Diels-Alder reaction is a cycloaddition that occurs between an “electron-rich”

conjugated diene and “elect on-poor” alkene, forming a cyclohexene ring.

Which of the following statements about this reaction are correct?

1 During the cycloaddition, carbons 1 and 4 of the diene and the two alkene

carbons will change from being sp2 hybridised to being sp3 hybridised.

2 The presence of a –CHO substituent on the alkene molecule will increase the

rate of the Diels-Alder reaction.

3 The presence of a –CH3 substituent on the diene molecule will increase the rate

of the Diels-Alder reaction.

Answer: A

Option 1: Correct;

Option 2: Correct; –CHO is an electron-withdrawing substituent which will reduce the

electron density of the alkene, causing it to be more “electron-poor”, thus increasing

the rate of reaction.

Option 3: Correct; the –CH3 substituent is an electron-donating substituent which will

increase the electron density of the diene, causing it to be more electron-rich, thus

increasing the rate of reaction.

28


37 Methylparaben is an anti-fungal agent that is commonly used in a variety of cosmetic

and personal care products.

Which deductions about the reactions of methylparaben can be made from its

structure?

1 It can decolourise hot acidified KMnO4 solution.

2 It can react with PCl5 to form steamy fumes.

3 It can react with hydrogen cyanide in the presence of trace amount of NaCN.

Answer: D

Option 1: Correct; ester functional group will first undergo acid hydrolysis to

form CH3OH which in turn undergoes oxidation in presence of KMnO4.

Option 2: Incorrect; methylparaben can only undergo electrophilic substitution

but not nucleophilic substitution (due to partial double bond character between

C and O atoms of phenolic group).

Option 3: Incorrect; reaction with HCN in presence of NaCN not possible due to

absence of carbonyl functional group.

38 Keratin is a fibrous and insoluble protein found in hair, nails and horns of mammals. It

comprises mainly of the three amino acids below:

NH2CH(CH2SH)CO2H NH2CH(CH2OH)CO2H NH2CH(CH2CH2CO2H)CO2H

cysteine serine glutamic acid

Based on the above information, what type of bonding would stabilise the tertiary

structure of keratin?

1 disulfide bonds

2 hydrogen bonding

3 van der Waals’ forces

Answer: B Options 1 & 2 are correct. Option 1: Correct; due to presence of -CH2SH Option 2: Correct; due to presence of -CH2OH Option 3: Incorrect

29


39 The prostaglandins are biomolecules with a wide range of biological effects which

include lowering blood pressure and gastric secretion.

Which of the following statements correctly describe the two prostaglandin molecules

shown above?

1 Both prostaglandin E2 and F2α are optically active molecules.

2 Prostagladin E2 can be converted into prostaglandin F2α by reacting with

hydrogen in presence of nickel catalyst.

3 Both prostaglandin molecules will form a silver mirror with Tollens’ reagent.

Answer: D

Option 1 is correct only.

Option 1: True; both prostaglandin molecules consist of chiral carbons as

highlighted by * below:

Option 2: False; the alkene functional groups in prostaglandin E2 will also

undergo reduction in presence of H2/Ni thus prostaglandin F2α will not form.

Option 3: False; both do not contain the aldehyde functional group and thus will

not give silve mi o with Tollens’ eagent.

30


40 Which of the following transformations involve a nucleophile?

1

2

Answer: A

Options 1, 2 & 3 are all correct.

Option 1: Nucleophilic substitution (SN2) of CH3Br; nucleophile: H-C≡C¯

Option 2: Nucleophilic addition of CH3OH to ketone; nucleophile: CH3OH

Option 3: Nucleophilic acyl substitution; nucleophile: CH3NH2

1 B 11 B 21 D 31 A

2 B 12 C 22 D 32 D

3 D 13 D 23 C 33 C

4 A 14 C 24 D 34 D

5 B 15 A 25 C 35 D

6 C 16 C 26 C 36 A

7 C 17 A 27 C 37 D

8 C 18 C 28 A 38 B

9 A 19 D 29 B 39 D

10 D 20 B 30 B 40 A

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2

CANDIDATE NAME

CLASS 2T

CHEMISTRY 9647/02 Paper 2 Structured Questions Monday 24th August 2015 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class in the spaces provided above. Write in dark blue or black pen in the spaces provided, on the Question Paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. You are reminded of the need for good English and clear presentation in your answers.

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

This document consists of 18 printed pages.

For Examiner’s Use

Paper 1

40

Paper 2

Q 1 12

Q 2 15

Q 3 15

Q 4 15

Q 5 15 72

Paper 3

Q 1 20

Q 2 20

Q 3 20

Q 4 20

Q 5 20 80

Total 192

[Turn over

2

[Turn over

1 Planning (P)

Cream of tartar (potassium hydrogen tartrate, KHC4H4O6) is one of the ingredients in

baking powder which is used to make cakes rise. Potassium hydrogen tartrate is a weak

acid that is not very soluble in water.

KHC4H4O6(s) ⇌ K+(aq) + HC4H4O6–(aq)

The HC4H4O6–(aq) ion contains one acidic hydrogen, so the quantity of potassium

hydrogen tartrate in solution can be determined by titration with a base.

HC4H4O6–(aq) + OH–(aq) H2O(l) + C4H4O6

2–(aq)

You are to design an experiment to determine the solubility of potassium hydrogen

tartrate in two solvent systems: pure water and 0.100 mol dm–3 potassium chloride.

In addition to the standard apparatus available in a school laboratory, you are provided

with the following materials:

potassium hydrogen tartrate

distilled water

methyl orange indicator

phenolphthalein indicator

standard sodium hydroxide solution

(a) Write an expression for the solubility product, Ksp, of potassium hydrogen tartrate,

stating its units.

…………………………………………………………………………………………………. [1]

(b) When a solution is saturated, the undissolved solid is in equilibrium with its aqueous

solution.

X(s) + aq ⇌ X(aq)

Describe how you would prepare a saturated solution of potassium hydrogen tartrate

at room temperature.

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

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

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

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

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

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

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

…………………………………………………………………………………………………. [3]

For Examiner’s

Use

3


(c) The concentration of a saturated solution of potassium hydrogen tartrate may be

determined by titration with standard sodium hydroxide.

(i) Given that, at room temperature, Ksp of potassium hydrogen tartrate has a

numerical value of 6.27 × 10–4, suggest an appropriate concentration of the

standard sodium hydroxide solution to be prepared. Show your working.

(ii) Name the indicator used in this titration.

……………………………………………………….………………………………….. [3]

(d) Describe what further experiments you would carry out to determine the solubility of

potassium hydrogen tartrate in 0.100 mol dm–3 potassium chloride. State clearly the

volume of solution used and the expected observation.

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

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

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

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

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

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

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

State and explain how the titration result of this further experiment would compare

with that of the solution prepared in (b).

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

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

………………………………………………………………………...……………………….. [5]

[Total: 12]

For Examiner’s

Use

[Turn over

4


2 Carbon is the fourth most abundant element in the universe by mass after hydrogen,

helium and oxygen. It is present in all forms of carbon-based life, and in the human body.

This abundance, together with the unique diversity of organic compounds, makes this

element the chemical basis of all known life.

(a) (i) Organic compound P contains only the elements carbon, hydrogen and

oxygen in the following composition by mass: C, 40 %; H, 6.7 %.

Calculate the empirical formula of P.

(ii) When a 0.102 g sample of compound P was vapourised in a suitable

apparatus, the vapour occupied 59 cm3 at 150 °C and 101 kPa.

Calculate the relative molecular mass of compound P, and hence determine

its molecular formula.

(iii) Compound P gives a silver mirror with Tollens’ reagent. Draw the displayed

formula of compound P.

[5]

For Examiner’s

Use

5


(b) Starch is another organic compound consisting of carbon, hydrogen and oxygen.

The hydrolysis of starch into simpler sugars can be catalysed by the enzyme

amylase. Amylase is present in the saliva of humans and acts as a biological

catalyst in the digestion of starch.

(i) A few experiments are carried out with different concentrations of starch

solution and the initial rates of hydrolysis in the presence of the enzyme,

amylase, are measured.

The graph below shows how the initial rates of the hydrolysis reaction in the

presence of amylase vary with the concentrations of starch solution.

Explain the relationship between the concentration of substrate and the initial

rate of reaction, making reference to the order of reaction with respect to the

substrate concentration.

………………………………………………………….……………………………….

………………………………………………………….……………………………….

.………………………………………………………………………………………….

………………………………………………………….……………………………….

.………………………………………………………………………………………….

.………………………………………………………………………………………….

.………………………………………………………………………………………….

(ii) On the same graph in (b)(i), sketch another graph to show how the initial rates

will vary with the concentrations of starch solution if the concentration of

amylase is halved. Label this graph as (ii).

[3]

[starch]

initial rate

For Examiner’s

Use

[Turn over

6


(c) Allotropes are pure forms of the same element that differ in structure. There are

several allotropes of carbon such as graphite and diamond.

A mixture of carbon dioxide reacts with hot graphite according to the equation below:

CO2(g) + C(s) ⇌ 2CO(g) H = +170 kJ mol–1

The mixture was allowed to reach dynamic equilibrium at 817 °C in a 2 dm3 closed

vessel. The amount of CO2, C and CO at equilibrium is given below.

CO2 C CO

Equilibrium amount / mol 0.1 0.7 1.7

(i) What do you understand by the term dynamic equilibrium?

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

…….……………………………………………………………………………………..

…….……………………………………………………………………………………..

….………………………………………………………………………………………..

(ii) Write an expression for the equilibrium constant, Kc, of this reaction, stating

the units.

…………………………………………………………………………………………...

(iii) Use the data given and your answer to (c)(ii) to calculate a value for Kc of this

reaction.

For Examiner’s

Use

7


(iv) State, with reasoning, the effect on the position of equilibrium if

(I) the total volume of the system is increased at 817 °C.

…………………………………………………………………………………….

……………………………………………………………………………….……

.……………………………………………………………………………………

.……………………………………………………………………………………

.……………………………………………………………………………………

(II) some of the graphite is removed from the equilibrium mixture.

…………………………………………………………………………………….

…………………………………………………………………………………….

…………………………………………………………………………………….

…………………………………………………………………………………….

…………………………………………………………………………………….

(v) The activation energy for the forward reaction is 370 kJ mol–1. Calculate the

activation energy, in kJ mol–1, for the reverse reaction.

[7]

[Total: 15]

[Turn over

For Examiner’s

Use

8


3 (a) Manganese sulfate, MnSO4, which is soluble in water, is commonly used to make

fungicide. When the sulfate has dissolved, the anions and cations are each surrounded by a number of water molecules.

(i) Draw a simple diagram to show how a water molecule can be attached to a

sulfate anion. Label the type of interaction involved.

The aqueous solution of MnSO4 is pale pink in colour due to the presence of complex ion, [Mn(H2O)6]

2+.

(ii) Draw the structure of the complex ion, [Mn(H2O)6]2+.

(iii) State the coordination number of the complex ion, [Mn(H2O)6]2+.

……………

(iv) Explain why [Mn(H2O)6]2+(aq) is pale pink in colour.

…..……………………………………………………………………………………….

……..…………………………………………………………………………………….

………..………………………………………………………………………………….

…………..……………………………………………………………………………….

……………..…………………………………………………………………………….

………………..………………………………………………………………………….

…………………..………………………………………………………………………. [6]

For Examiner’s

Use

9


(b) When aqueous ammonia is added gradually to a pale pink solution of [Mn(H2O)6]2+,

an off-white precipitate is observed which rapidly turns brown on contact with air.

(i) Suggest the identity of the off-white precipitate.

…………………………….

(ii) Write relevant equations to explain the formation of the off-white precipitate.

………………………………………………………….………………………………..

………………………………………………………….………………………………..

………………………………………………………….………………………………..

………………………………………………………….……………………………….. [3]

(c) MnO2 acts as a heterogeneous catalyst in the decomposition of H2O2.

2H2O2(l) → 2H2O(l) + O2(g) ∆Ho = -196 kJ mol–1

∆Go = -224 kJ mol–1

(i) Use the data to calculate ∆So for this reaction and comment on its sign with

respect to the equation for the reaction.

(ii) Draw the displayed formula to show the bonding in a hydrogen peroxide

molecule. Predict and indicate clearly the bond angle in your diagram.

[3]

For Examiner’s

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[Turn over

10


(d) Alkaline batteries are often used in portable radios and electric torches. In an alkaline

battery, the reaction at the cathode involves the reduction of MnO2 to Mn2O3 while the

reaction at the anode is as follows:

Zn + 2OH– ZnO + H2O + 2e–

(i) State the oxidation number of manganese in Mn2O3.

……………………

(ii) Write a half-equation for the reaction that occurs at the cathode.

…………………………………………………………………………………………...

(iii) Hence, construct the equation for the overall reaction.

…………………………………………………………………………………………... [3]

[Total: 15]

For Examiner’s

Use

11


4 (a) Silk emitted by the silkworm consists mainly of the insoluble protein fibroin

surrounded by a layer of glue-like substance. The protein secondary structure of

fibroin consists mostly of layers of anti-parallel beta-pleated sheets and its primary

structure is a recurrent 6-amino acid residue sequence consisting of only serine,

glycine and alanine.

Partial hydrolysis of fibroin results in the following fragments.

ala-gly-ala , ser-gly-ala, gly-ser

(i) Using the same 3-letter abbreviations as above, write out the recurrent

6-amino acid residue sequence of fibroin based on the information provided.

…………………………………………………………..………………………………

(ii) From your answer in (a)(i) and the proportion of individual amino acids in the

protein structure, suggest why close packing is possible in the polypeptide

chain.

…………………………………………………………..………………………………

……………………………………………………………………………………..……

…………………………………………………………..………………………………

(iii) With reference to the secondary structure of fibroin, describe what is meant by

denaturation, clearly making reference to any bonding or interaction involved.

…………………………………………………………………………………..………

……………………………………………………………………………………..……

……………………………………………………………………………………..……

……………………………………………………………………………………..……

……………………………………………………………………………………...……

[3]

For Examiner’s

Use

[Turn over

12


(b) Alanine is a non-essential amino acid and can be manufactured by the human body.

Alanine can also be chemically synthesised from 2-bromopropanoic acid.

(i) State the reagents required to synthesise alanine from 2-bromopropanoic acid.

Reagents: ……………………………………………………………………………...

(ii) Identify the inorganic by-product from the reaction shown above.

By-product: …………………………………………………………………………….

The pKa associated with 2-bromopropanoic acid is 3.00. The two pKa associated

with alanine are 2.35 and 9.69 and its isoelectric point is 6.01.

(iii) Arrange the following compounds in order of decreasing acidity. Explain your

answer in terms of their structural properties.

Order of decreasing acidity: .……….………………………………………………..

Explanation:

…………………………………………………………...………………………………

..…………………………………………………………………………………………

..…………………………………………………………………………………………

…..………………………………………………………………………………………

……..……………………………………………………………………………………

………..…………………………………………………………………………………

…………..………………………………………………………………………………

For Examiner’s

Use

13


(iv) Write an equation to show how alanine can act as a buffer when a small

amount of OH–(aq) is added at pH 2.35.

(v) Calculate the pH of the resulting solution when 7.5 cm3 of 0.10 mol dm–3

NaOH is added to 10.0 cm3 of 0.100 mol dm–3 of the protonated form of alanine.

(vi) Sketch the pH-volume curve you would expect to obtain when 30 cm3 of 0.10

mol dm–3 NaOH is added to 10 cm3 of 0.10 mol dm–3 of the protonated form of

alanine. Show clearly on your curve where the two pKa values occur and the

isoelectric point.

[12]

[Total: 15]

For Examiner’s

Use

[Turn over

pH

Volume of NaOH added / cm3

14


5 (a) Compound C may be synthesised from ethanal as follows:

State the reagents and conditions required for Steps 3 – 5 and give the structures of

the intermediate organic compounds A and B in the boxes provided.

Reagents and conditions

Step 3 …………………………………………………………………………………

Step 4 …………………………………………………………………………………

Step 5 …………………………………………………………………………………

[5]

For Examiner’s

Use

15


(b) Aldol condensation reaction is a useful method for making new carbon–carbon

bonds in organic chemistry. In the presence of a strong base, two carbonyl

compounds can undergo the aldol condensation reaction.

The same product, compound C, from part (a) can also be obtained from ethanal

and methanal via the aldol condensation. The reaction proceeds as follows.

(i) Identify the type of reaction occurring in Reaction 2.

…………………………………………………………………………………

Reaction 1 of the aldol condensation consists of three steps. The first step involves

the removal of hydrogen from the alpha carbon of one carbonyl compound to form

an enolate ion. The alpha carbon is the carbon adjacent to the carbonyl group.

(ii) State the type of reaction occurring in Step I.

…………………………………………………………………………………….……

For Examiner’s

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[Turn over

16


(iii) Steps II and III are as follows:

Step II: nucleophilic attack on methanal to form an alkoxide ion

Step III: protonation of the alkoxide ion by water

Using the above information, suggest the mechanism for Steps II and III,

showing all charges and using curly arrows to show the movement of electron

pairs.

(iv) Given that the negative charge on the enolate ion can be delocalised, draw

another possible structure in which the enolate ion can exist as.

For Examiner’s

Use

17


(v) Complete the reaction sequence for the following aldol condensation, by giving

the structural formulae for the compounds involved in the boxes provided.

[8]

For Examiner’s

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[Turn over

18


(c) Cinnamaldehyde is an organic compound that gives cinnamon its flavour and odour.

Cinnamaldehyde can also be synthesised in the laboratory via the aldol

condensation reaction.

(i) State the type of stereoisomerism that cinnamaldehyde can exhibit.

………………………………………………………………………………..…

(ii) Jasmone is the active ingredient in jasmine which is used in the perfume

industry. The structure of jasmone is shown below.

State the reagents and conditions for a simple chemical test that could be

used to distinguish between cinnamaldehyde and jasmone.

……………………………………………..……………………………………………

…………………………………………...………………………………………..……. [2]

[Total: 15]

For Examiner’s

Use

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2

CANDIDATE NAME

CLASS 2T

CHEMISTRY 9647/02 Paper 2 Structured Questions Monday 24th August 2015 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class in the spaces provided above. Write in dark blue or black pen in the spaces provided, on the Question Paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. You are reminded of the need for good English and clear presentation in your answers.

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

For Examiner’s Use

Paper 1

40

Paper 2

Q 1 12

Q 2 15

Q 3 15

Q 4 15

Q 5 15 72

Paper 3

Q 1 20

Q 2 20

Q 3 20

Q 4 20

Q 5 20 80

Total 192

ANSWER SCHEME

2

1 Planning (P)

Cream of tartar (potassium hydrogen tartrate, KHC4H4O6) is one of the ingredients in

baking powder which is used to make cakes rise. Potassium hydrogen tartrate is a weak

acid that is not very soluble in water.

KHC4H4O6(s) ⇌ K+(aq) + HC4H4O6–(aq)

The HC4H4O6–(aq) ion contains one acidic hydrogen, so the quantity of potassium

hydrogen tartrate in solution can be determined by titration with a base.

HC4H4O6–(aq) + OH–(aq) H2O(l) + C4H4O6

2–(aq)

You are to design an experiment to determine the solubility of potassium hydrogen

tartrate in two solvent systems: pure water and 0.100 mol dm–3 potassium chloride.

In addition to the standard apparatus available in a school laboratory, you are provided

with the following materials:

potassium hydrogen tartrate

distilled water

methyl orange indicator


standard sodium hydroxide solution

(a) Write an expression for the solubility product, Ksp, of potassium hydrogen tartrate,

stating its units.

…………………………………………………………………………………………………. [1]

(b) When a solution is saturated, the undissolved solid is in equilibrium with its aqueous

solution.

X(s) + aq ⇌ X(aq)

Describe how you would prepare a saturated solution of potassium hydrogen tartrate

at room temperature.

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

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

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

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

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

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

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

…………………………………………………………………………………………………. [3]

Dissolve solid KHC4H4O6 in water until some solid remains undissolved.

Allow solution to stand for few hours (to establish equilibrium).

Filter to remove undissolved solid. The saturated solution of KHC4H4O6 is

collected as the filtrate.

OR

Dissolve solid KHC4H4O6 in hot water until some solid remains undissolved.

Cool the solution to room temperature to saturate. Excess solid would crystallise

out.

Filter to remove excess solid/crystals. The saturated solution of KHC4H4O6 is

collected as the filtrate.

Ksp = [K+ ] [HC4H4O6– ] mol2 dm–6

3


(c) The concentration of a saturated solution of potassium hydrogen tartrate may be

determined by titration with standard sodium hydroxide.

(i) Given that, at room temperature, Ksp of potassium hydrogen tartrate has a

numerical value of 6.27 × 10–4, suggest an appropriate concentration of the

standard sodium hydroxide solution to be prepared. Show your working.

(ii) Name the indicator used in this titration.

……………………………………………………….…………………………………..

[3]

(d) Describe what further experiments you would carry out to determine the solubility of

potassium hydrogen tartrate in 0.100 mol dm–3 potassium chloride. State clearly the

volume of solution used and the expected observation.

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

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

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

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

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

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

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

State and explain how the titration result of this further experiment would compare

with that of the solution prepared in (b).

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

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

………………………………………………………………………...……………………….. [5]

[Total: 12]

for a saturated solution of KHC4H4O6, [K+ ] [HC4H4O6

– ] = Ksp

[HC4H4O6– ] = spK (since [K+ ] = [HC4H4O6

– ])

= 41027.6 = 0.0250 mol dm–3

To avoid too small a titre value which has high percentage error, or too large a titre value which requires more than one burette full of titrant, [titrant] is chosen such that volume of titrant at end-point is about equal to volume of analyte.

Since, HC4H4O6– + OH– C4H4O6

2– + H2O

[NaOH ] = [HC4H4O6– ]

= 0.0250 mol dm–3

a suitable [NaOH ] = 0.0250 mol dm–3

[Accept answers between 0.0200 and 0.0300 mol dm–3 ]


Prepare a saturated solution of KHC4H4O6 in 0.100 mol dm–3 KCl;

i.e. repeat procedure stated in (b) but use 0.100 mol dm–3 KCl instead of

water.

Pipette 25.0 cm3 of the saturated solution of KHC4H4O6 prepared above

(into a 250 cm3 conical flask). Add (2-3 drops of) phenolphthalein indicator.

Titrate with standard / 0.0250 mol dm–3 NaOH (placed in a 50 cm3 burette)

until the solution in the conical flask changed from colourless to pink.

[1] correct indicator + end-point colour change

vol of NaOH less than that for saturated solution of KHC4H4O6 in water.

Presence of common ion, K+, lowers/depresses the solubility of KHC4H4O6.

4


2 Carbon is the fourth most abundant element in the universe by mass after hydrogen,

helium and oxygen. It is present in all forms of carbon-based life, and in the human body.

This abundance, together with the unique diversity of organic compounds, makes this

element the chemical basis of all known life.

(a) (i) Organic compound P contains only the elements carbon, hydrogen and

oxygen in the following composition by mass: C, 40 %; H, 6.7 %.

Calculate the empirical formula of P.

(ii) When a 0.102 g sample of compound P was vapourised in a suitable

apparatus, the vapour occupied 59 cm3 at 150 °C and 101 kPa.

Calculate the relative molecular mass of compound P, and hence determine

its molecular formula.

(iii) Compound P gives a silver mirror with Tollens’ reagent. Draw the displayed

formula of compound P.

[5]

% by mass of O = 100 – 40 – 6.7 = 53.3%

Let mass of compound P be 100 g.

C H O

Mass / g 40 6.7 53.3

Amount / mol 40 ÷ 12.0 = 3.33 6.7 ÷ 1.0 = 6.7 53.3 ÷ 16 = 3.33

Simplest ratio 1 2 1

Hence, the empirical formula of P is CH2O. with working

Using ideal gas equation, pV = nRT

= rM

m RT

Mr = pV

mRT = )1059)(10101(

)273150(31.8)102.0(63

correct conversions

= 60.2 correct value to 1 dp; NO units Let the molecular formula of P be CnH2nOn.

Since Mr of CnH2nOn = 60.2

n(12.0) + 2n(1.0) + n(16.0) = 60.2

n = 2

Hence, molecular formula of P is C2H4O2.

structure & all bonds shown

5


(b) Starch is another organic compound consisting of carbon, hydrogen and oxygen.

The hydrolysis of starch into simpler sugars can be catalysed by the enzyme

amylase. Amylase is present in the saliva of humans and acts as a biological

catalyst in the digestion of starch.

(i) A few experiments are carried out with different concentrations of starch

solution and the initial rates of hydrolysis in the presence of the enzyme,

amylase, are measured.

The graph below shows how the initial rates of the hydrolysis reaction in the

presence of amylase vary with the concentrations of starch solution.

Explain the relationship between the concentration of substrate and the initial

rate of reaction, making reference to the order of reaction with respect to the

substrate concentration.

………………………………………………………….……………………………….

………………………………………………………….……………………………….

.………………………………………………………………………………………….

………………………………………………………….……………………………….

.………………………………………………………………………………………….

.………………………………………………………………………………………….

.………………………………………………………………………………………….

(ii) On the same graph in (b)(i), sketch another graph to show how the initial rates

will vary with the concentrations of starch solution if the concentration of

amylase is halved. Label this graph as (ii).

[3]

[starch]

initial rate

(ii) max. initial rate is about half that of original graph & graph plateau earlier

At low substrate concentration, initial rate increases linearly as

substrate/starch concentration increases. The reaction is first order with

respect to substrate concentration.

At higher substrate concentrations, initial rate is constant. All the

active sites of amylase are in use/saturated. Adding more substrate will

not increase rate of reaction. Hence, the reaction is zero order with

respect to substrate concentration.

6


(c) Allotropes are pure forms of the same element that differ in structure. There are

several allotropes of carbon such as graphite and diamond.

A mixture of carbon dioxide reacts with hot graphite according to the equation below:

CO2(g) + C(s) ⇌ 2CO(g) H = +170 kJ mol–1

The mixture was allowed to reach dynamic equilibrium at 817 °C in a 2 dm3 closed

vessel. The amount of CO2, C and CO at equilibrium is given below.

CO2 C CO

Equilibrium amount / mol 0.1 0.7 1.7

(i) What do you understand by the term dynamic equilibrium?

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

…….……………………………………………………………………………………..

…….……………………………………………………………………………………..

….………………………………………………………………………………………..

(ii) Write an expression for the equilibrium constant, Kc, of this reaction, stating

the units.

…………………………………………………………………………………………...

(iii) Use the data given and your answer to (c)(ii) to calculate a value for Kc of this

reaction.

Dynamic equilibrium refers to a reversible reaction in which the rate

of forward reaction is equal to the rate of reverse reaction, and there

is no change in the concentrations of reactants and products

although the reaction is still proceeding.

Not acceptable: “No change in amount” or “Concentration of

reactants and products are the same”.

Kc = 2

2

CO

CO mol dm–3 both correct expression and units

The reaction was carried out in a 2 dm3 closed vessel.

CO2(g) + C(s) ⇌ 2CO(g)

Equilibrium conc/ moldm–3 21.0

27.1

= 0.050 = 0.850

Kc = 2

2

CO

CO =

0.05

0.8502

= 14.5 mol dm–3 correct value and for dividing eqm conc

by 2

7


(iv) State, with reasoning, the effect on the position of equilibrium if

(I) the total volume of the system is increased at 817 °C.

…………………………………………………………………………………….

……………………………………………………………………………….……

.……………………………………………………………………………………

.……………………………………………………………………………………

.……………………………………………………………………………………

(II) some of the graphite is removed from the equilibrium mixture.

…………………………………………………………………………………….

…………………………………………………………………………………….

………………………………………………………………………………….

…………………………………………………………………………………….

…………………………………………………………………………………….

(v) The activation energy for the forward reaction is 370 kJ mol–1. Calculate the

activation energy, in kJ mol–1, for the reverse reaction.

[7]

[Total: 15]

No effect on (or no change in) the position of equilibrium. This is

because graphite exists as a solid and its concentration remains

unchanged no effect and for reason

Pressure 1/volume.

If total volume of system is increased, total pressure of the system

decreases. By Le Chatelier’s Principle, the position of equilibrium

shifts to the right [link to pressure] so as to increase the pressure by

increasing the number of molecules of gas. mention of gaseous

molecules

Given H = +170 kJ mol–1, the forward reaction is endothermic.

Hence, activation energy for the reverse reaction = 370 – 170

= 200 kJ mol–1

Note:

energy/ kJ mol

–1

reaction pathway CO2 +C

CO

ΔH = +170

transition state

Ea (forward) = 370 Ea (backward) = ?

8


3 (a) Manganese sulfate, MnSO4, which is soluble in water, is commonly used to make

fungicide. When the sulfate has dissolved, the anions and cations are each surrounded by a number of water molecules.

(i) Draw a simple diagram to show how a water molecule can be attached to a

sulfate anion. Label the type of interaction involved.

The aqueous solution of MnSO4 is pale pink in colour due to the presence of complex ion, [Mn(H2O)6]

2+.

(ii) Draw the structure of the complex ion, [Mn(H2O)6]2+.

(iii) State the coordination number of the complex ion, [Mn(H2O)6]2+.

……………

(iv) Explain why [Mn(H2O)6]2+(aq) is pale pink in colour.

…..……………………………………………………………………………………….

……..…………………………………………………………………………………….

………..………………………………………………………………………………….

…………..……………………………………………………………………………….

……………..…………………………………………………………………………….

………………..………………………………………………………………………….

…………………..………………………………………………………………………. [6]

partial charge & charges on SO42- & interaction labeled

Note: When an ionic compound dissolves in water, ion-dipole interactions

are formed between the cation/anion(ion) and water(dipole).

lone pairs on H2O ligands, arrows showing dative bonds, dotted lines showing plane, Roman numeral on Mn2+ and square brackets with overall charge on complex

6

Mn2+ [Ar] 3d5 ; Mn2+ has incompletely filled (or partially filled) 3d-orbitals.

In the presence of ligands, the 3d-orbitals become non-degenerate and

split into two groups of slightly different energy levels (d-d* splitting).

When electrons from the lower energy level are promoted to the higher

energy level (d-d* electronic transition), radiation in the green/blue region

of the visible spectrum is absorbed. Light energy not absorbed is thus

seen as the colour of the complex, which is pale pink.

Ion-dipole

attraction

9


(b) When aqueous ammonia is added gradually to a pale pink solution of [Mn(H2O)6]2+,

an off-white precipitate is observed which rapidly turns brown on contact with air.

(i) Suggest the identity of the off-white precipitate.

…………………………….

(ii) Write relevant equations to explain the formation of the off-white precipitate.

………………………………………………………….………………………………..

………………………………………………………….………………………………..

………………………………………………………….………………………………..

………………………………………………………….……………………………….. [3]

(c) MnO2 acts as a heterogeneous catalyst in the decomposition of H2O2.

2H2O2(l) → 2H2O(l) + O2(g) ∆Ho = -196 kJ mol–1

∆Go = -224 kJ mol–1

(i) Use the data to calculate ∆So for this reaction and comment on its sign with

respect to the equation for the reaction.

(ii) Draw the displayed formula to show the bonding in a hydrogen peroxide

molecule. Predict and indicate clearly the bond angle in your diagram.

[3]

Mn(OH)2 or [Mn(H2O)4(OH)2]

NH3 + H2O ⇌ NH4+ + OH-

2OH– + [Mn(H2O)6]2+ [Mn(H2O)4(OH)2] + 2H2O OR 2OH– + Mn2+

Mn(OH)2

NH3 acts as a base and gives OH– ions in the presence of water. These OH–

ions will then react with [Mn(H2O)6]2+ to give an off-white precipitate of

[Mn(H2O)4(OH)2].

∆Go = ∆Ho – T∆So

-224 = -196 – (25 + 273) ∆So

∆So = +0.0940 kJ K-1 mol-1 [with workings]

∆So is positive because there is an increase in disorder due to the an

increase in the number of moles in gaseous particles from 0 to 1.

correct structure and correct bond

angle (accept 94.8o-105o)

10


(d) Alkaline batteries are often used in portable radios and electric torches. In an alkaline

battery, the reaction at the cathode involves the reduction of MnO2 to Mn2O3 while the

reaction at the anode is as follows:

Zn + 2OH– ZnO + H2O + 2e–

(i) State the oxidation number of manganese in Mn2O3.

……………………

(ii) Write a half-equation for the reaction that occurs at the cathode.

…………………………………………………………………………………………...

(iii) Hence, construct the equation for the overall reaction.

…………………………………………………………………………………………... [3]

[Total: 15]

2MnO2 + H2O + 2e– Mn2O3 + 2OH–

2MnO2 + Zn Mn2O3 + ZnO

+3

11


4 (a) Silk emitted by the silkworm consists mainly of the insoluble protein fibroin

surrounded by a layer of glue-like substance. The protein secondary structure of

fibroin consists mostly of layers of anti-parallel beta-pleated sheets and its primary

structure is a recurrent 6-amino acid residue sequence consisting of only serine,

glycine and alanine.

Partial hydrolysis of fibroin results in the following fragments.

ala-gly-ala , ser-gly-ala, gly-ser

(i) Using the same 3-letter abbreviations as above, write out the recurrent

6-amino acid residue sequence of fibroin based on the information provided.

…………………………………………………………..………………………………

(ii) From your answer in (a)(i) and the proportion of individual amino acids in the

protein structure, suggest why close packing is possible in the polypeptide

chain.

…………………………………………………………..………………………………

……………………………………………………………………………………..……

…………………………………………………………..………………………………

(iii) With reference to the secondary structure of fibroin, describe what is meant by

denaturation, clearly making reference to any bonding or interaction involved.

…………………………………………………………………………………..………

……………………………………………………………………………………..……

……………………………………………………………………………………..……

……………………………………………………………………………………..……

[3]

(b) Alanine is a non-essential amino acid and can be manufactured by the human body.

Alanine can also be chemically synthesised from 2-bromopropanoic acid.

Close packing is possible due to the high proportion (50%) of glycine

where glycine's R group is only a hydrogen atom and so is not as

sterically constrained.

Denaturation is the process that alters the hydrogen bonding between

the C=O group of one polypeptide strand and the N-H group of the

adjacent polypeptide strand in the beta-pleated sheets of the secondary

structure of fibroin.

gly-ser-gly-ala-gly-ala

https://en.wikipedia.org/wiki/Glycine

https://en.wikipedia.org/wiki/Serine

https://en.wikipedia.org/wiki/Alanine

12


(i) State the reagents required to synthesise alanine from 2-bromopropanoic acid.

Reagents: ……………………………………………………………………………...

(ii) Identify the inorganic by-product from the reaction shown above.

By-product: …………………………………………………………………………….

The pKa associated with 2-bromopropanoic acid is 3.00. The two pKa associated

with alanine are 2.35 and 9.69 and its isoelectric point is 6.01.

(iii) Arrange the following compounds in order of decreasing acidity. Explain your

answer in terms of their structural properties.

Order of decreasing acidity: .……….………………………………………………..

Explanation:

…………………………………………………………...………………………………

..…………………………………………………………………………………………

..…………………………………………………………………………………………

…..………………………………………………………………………………………

……..……………………………………………………………………………………

………..…………………………………………………………………………………

…………..………………………………………………………………………………

ammonia

HBr or NH4Br

Alanine is the strongest acid as it contains the most electronegative /

electron-withdrawing amino group increasing the polarisation of O–H

bond, and so weakening the O–H bond, thus making it easier for the

acid to lose the H+ . 2-bromopropanoic acid is the second strongest acid

as it contains slightly less electronegative Br atom compared to the

amino group and propanoic acid is least acidic due to the absence of

any electronegative / electron-withdrawing substituent group present.

Credit was given to candidates who were able to clearly explain why 2-

bromopropanoic acid is a stronger acid than propanoic acid, even if the

order between alanine and 2-bromopropanoic acid was wrongly given.

Also accept answers mentioning stability of conjugate ion and

formation of intramolecular H-bonding within the conjugate base ion of

alanine.

13


(iv) Write an equation to show how alanine can act as a buffer when a small amount of OH–(aq) is added at pH 2.35.

(v) Calculate the pH of the resulting solution when 7.5 cm3 of 0.10 mol dm–3

NaOH is added to 10.0 cm3 of 0.100 mol dm–3 of the protonated form of alanine.

(vi) Sketch the pH-volume curve you would expect to obtain when 30 cm3 of 0.10

mol dm–3 NaOH is added to 10 cm3 of 0.10 mol dm–3 of the protonated form of

alanine. Show clearly on your curve where the two pKa values occur and the

isoelectric point.

[12]

[Total: 15]

Start: 10

= 0.001 mol 7.5

= 0.00075 mol 0

(excess reagent) (limiting reagent)

end: 0.00025 mol 0 mol 0.00075 mol

Hence, resulting solution contains a weak acid and its conjugate base (i.e. forms an acidic buffer)

pH = pKa + lg

= 2.35 + lg

= 2.83

pH

Volume of NaOH added / cm3

correct shape

correctly labeled pKa values and volumes

isoelectric pt correctly labelled

0

2.35

9.69

5 10 15 20 25 30

6.01 X

14


5 (a) Compound C may be synthesised from ethanal as follows:

State the reagents and conditions required for Steps 3 – 5 and give the structures of

the intermediate organic compounds A and B in the boxes provided.

Reagents and conditions

Step 3 …………………………………………………………………………………

Step 4 …………………………………………………………………………………

Step 5 …………………………………………………………………………………

[5]

(excess) concentrated H2SO4, 170˚C

LiAlH4 in dry ether

K2Cr2O7, dilute H2SO4, heat with distillation

15


(b) Aldol condensation reaction is a useful method for making new carbon–carbon

bonds in organic chemistry. In the presence of a strong base, two carbonyl

compounds can undergo the aldol condensation reaction.

The same product, compound C, from part (a) can also be obtained from ethanal

and methanal via the aldol condensation. The reaction proceeds as follows.

(i) Identify the type of reaction occurring in Reaction 2.

……………………………………………………………………………

Reaction 1 of the aldol condensation consists of three steps. The first step involves

the removal of hydrogen from the alpha carbon of one carbonyl compound to form

an enolate ion. The alpha carbon is the carbon adjacent to the carbonyl group.

(ii) State the type of reaction occurring in Step I.

…………………………………………………………………………………….……

[Turn over

Acid-base reaction or neutralisation

Elimination (also accept dehydration)

16


(iii) Steps II and III are as follows:

Step II: nucleophilic attack on methanal to form an alkoxide ion

Step III: protonation of the alkoxide ion by water

Using the above information, suggest the mechanism for Steps II and III,

showing all charges and using curly arrows to show the movement of electron

pairs.

all 2 correct arrows and partial charges indicated (Step II)

all 2 correct arrows and partial charges indicated (Step III)

correct alkoxide given

17


(iv) Given that the negative charge on the enolate ion can be delocalised, draw

another possible structure in which the enolate ion can exist as.

correct structure

18


(v) Complete the reaction sequence for the following aldol condensation, by giving

the structural formulae for the compounds involved in the boxes provided.

[8] (c) Cinnamaldehyde is an organic compound that gives cinnamon its flavour and odour.

Cinnamaldehyde can also be synthesised in the laboratory via the aldol

condensation reaction.

(i) State the type of stereoisomerism that cinnamaldehyde can exhibit.

………………………………………………………………………………..…

(ii) Jasmone is the active ingredient in jasmine which is used in the perfume

industry. The structure of jasmone is shown below.

State the reagents and conditions for a simple chemical test that could be

used to distinguish between cinnamaldehyde and jasmone.

……………………………………………..……………………………………………

…………………………………………...………………………………………..……. [2]

[Total: 15]

geometrical or cis-trans isomerism

Add Tollens’ reagent and heat

or Fehling’s and heat

or K2Cr2O7, dilute H2SO4 and heat

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS

Higher 2

CHEMISTRY 9647/03

Paper 3 Free Response Thursday 27 August 2015

2 hours

Additional Materials: Data Booklet

Answer Paper


Write your name and class on all the work you hand in.

Write in dark blue or black pen on both sides of the paper. [PILOT FRIXION ERASABLE PENS ARE

NOT ALLOWED]

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


Answer any four questions. Write your answers on the answer paper provided.

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

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

At the end of the examination, fasten all your work securely together.


[Turn over


2


Answer any four questions.

1 Butanone is an industrial organic solvent used in the manufacture of plastics, lacquers and

varnishes.

(a) Butanone reacts with iodine in the presence of an acid according to the equation:

CH3CH2COCH3 + I2 CH3CH2COCH2I + HI

In a series of experiments, the reaction was carried out with different concentrations of

reagents, and the following initial rates were obtained:

Expt. [CH3CH2COCH3]

/ mol dm–3 [I2]

/ mol dm–3

[H+] / mol dm–3

Initial rate / mol dm–3 min–1

1 0.010 0.010 0.010 8.31 x 10–8

2 0.010 0.010 0.013 10.8 x 10–8

3 0.013 0.010 0.013 14.0 x 10–8

4 0.018 0.013 0.018 26.9 x 10–8

(i) Using the data given, deduce the order of reaction with respect to each of the three

reagents, clearly showing how you arrive at your answers. Hence, write a rate

equation for the reaction.

(ii) Explain how the rate of reaction would change if the solution was diluted with an

equal volume of inert solvent, using r to represent the original rate before dilution.

(iii) The reaction can also be carried out in the presence of a strong base instead of an

acid. State what would be observed when a strong base is used and write an

equation for the reaction that occurs.

[7]

(b) Elemental iodine, I2, has limited solubility in water. Its solubility is greatly increased in a

solution containing iodide ion, I–, due to the formation of triiodide ion, I3–, as shown:

I2(aq) + I–(aq) ⇌ I3–(aq)

(i) Write an expression for the equilibrium constant, Kc, of this reaction, stating its

units.

(ii) Experiments have shown that when [I– ] = 1.4 x 10–3 mol dm–3, the concentrations

of I2 and I3– are equal. Use this information to calculate a value for Kc.

(iii) Using your answer from (b)(ii), calculate the concentration of iodide ions

necessary for 99 % of the iodine to be dissolved.

[4]

3

[Turn over 9647/03/CJC JC2 Preliminary Exam 2015

(c) Explain, using Maxwell-Boltzmann distribution curves, how the rate of a reaction would

change if the temperature was increased.

[3]

(d) Grignard reagents are organo-magnesium halides, commonly used in synthesis to

prepare a variety of organic compounds.

The carbon-magnesium bonds in Grignard reagents are highly polar and this makes it

extremely useful in organic synthesis as it is able to react with other polar organic

molecules to form carbon-carbon bonds. An example of the use of a Grignard reagent is

the two-step reaction of CH3CH2MgBr with butanone, CH3CH2COCH3, to form

3-methylpentan-3-ol.

(i) Suggest the type of reaction that has taken place in step 1.

(ii) Suggest the structural formula of the organic compound to be used with

CH3CH2MgBr to form propan-1-ol, if the reaction undergoes a similar two-step

process.

(iii) Suggest a suitable Grignard reagent and another organic compound to be used if

propan-2-ol is to be prepared using a similar two-step process.

[4]

(e) There are certain practical issues with the use of Grignard reagent, notably slow

formation of the Grignard reagent and its susceptibility to reaction with water in air.

The slow formation of the Grignard reagent may be offset by cutting the precursor

magnesium into smaller pieces or scraping the sides of the metal before use. Using a

balanced equation, explain why magnesium, which has been stored in dry ambient

conditions, may not react as quickly as expected.

[2]

[Total: 20]

4


2 The direct oxidation of alcohols in a fuel cell represents potentially the most efficient method of

obtaining useful energy from renewable fuel.

The diagram below illustrates the functional parts of a direct-ethanol fuel cell (DEFC). The

anode and cathode compartments are separated by the proton exchange membrane. The

protons are transported across the proton exchange membrane to the cathode where they

react with oxygen to produce water.

Diagram of a DEFC

(a) (i) Explain, in terms of the change in average oxidation number of the carbon atoms,

whether electrode 1 or 2 acts as the anode.

(ii) Write the half-equations for the reactions which take place at the electrodes of the

DEFC, and hence an overall equation for the cell reaction.

(iii) The cell is capable of producing an e.m.f. of 1.47 V. By using suitable data from

the Data Booklet, calculate a value for the E

of the CO2/CH3CH2OH electrode

reaction.

(iv) By quoting relevant E

values from the Data Booklet, explain why an acidic

electrolyte is preferred to an alkaline electrolyte in the DEFC.

(v) Describe and explain the effect on the cell potential when the concentration of

ethanol is increased.

[9]

H2SO4(aq)

External circuit

CO2

CH3CH2OH H2O

O2

Electrode 2 Electrode 1

5


(b) Chromium plating is an important industrial application of electrolysis, where a steel

object is plated with a thin layer of chromium to make it resistant to rusting and scratches.

In an electroplating experiment, a researcher connected the DEFC to an electrolytic cell

as shown below. The aqueous solution containing chromium(III) chloride was

electrolysed using a piece of pure chromium and the steel object to be electroplated as

the electrodes.

(i) The total time taken for 10 g of chromium to be plated was 4 hours. Calculate

the current produced by the DEFC.

(ii) Calculate the volume of ethanol used when 10 g of chromium metal was plated

onto the steel object under room conditions, given that the density of ethanol is

0.789 g cm–3.

[3]

(c) Chromium is a transition element.

(i) Explain what is meant by the term transition element.

(ii) Chromium(III) chloride, CrCl3, reacts with water to form a violet hydrate

[Cr(H2O)6]Cl3. This violet hydrate is isomeric to the dark green

[CrCl2(H2O)4]Cl.2H2O and the pale green [CrCl(H2O)5]Cl2.H2O

Based on the above information, give one property of transition elements or their

compounds that is not shown by s-block elements.

(iii) When 0.03 mol of aqueous silver nitrate was added to 0.01 mol of one of the

chromium-containing compounds in (c)(ii), a white precipitate weighing 1.44 g was

formed. Deduce the formula of the compound to which the silver nitrate was added.

[3]

CrCl3 (aq) pure chromium

steel object

DEFC

6


(d) The following reaction scheme shows the chemistry of some chromium-containing

species in aqueous solution.

(i) Identify the chromium-containing species present in A. Describe how this species

accounts for the formation of carbon dioxide in Step III. Include any relevant

equations in your answer.

(ii) Give a balanced equation for step VI.

(iii) Similar to some organic molecules, the complex [Cr(C2O4)3]3– formed in step II can

exist as a pair of optical isomers. One of the isomers is as shown below:

where is used to represent the C2O42– ion.

Draw the structure of the other optical isomer.

[5]

[Total: 20]

3–

Cr

L

L

L

L

7


3 (a) Aspartame is made from the amino acids aspartic acid and phenylalanine. It is one of

the most commonly used artificial sweeteners as it is about 200 times sweeter than

sugar so lesser amount is required to achieve the same level of sweetness. The

structure of aspartame is shown below:

(i) Aspartame can exist as a zwitterion. Using the information given above, draw the

structure of the zwitterion and suggest one possible pH value at which aspartame

exists predominantly as a zwitterion.

(ii) Draw the structural formulae of the organic products formed when aspartame

reacts with

I an excess of hot NaOH(aq)

II cold HCl(aq)

(iii) Draw the structure of compound A in the following reaction.

8


(iv) Describe a simple chemical test to distinguish between aspartame and compound

B, clearly stating the reagents, conditions and observations for each compound.

[9]

(b) Aspartic acid and phenylalanine can be found in insulin, which consists of two

polypeptide chains. Insulin is responsible for the regulation of blood glucose level in the

human body. The following table shows some amino acid residues that can be found in

insulin:

Amino acid Abbreviation R group

aspartic acid asp –CH2CO2H

phenylalanine phe

lysine lys –(CH2)4NH2

cysteine cys

glycine gly

(i) When mercury ions, Hg2+, are added to a sample of insulin, the protein is found to

be denatured. State which aspects of the protein structure are affected during this

denaturation.

(ii) By considering all the interactions formed by the above amino acid residues,

explain clearly how Hg2+ can interact with the insulin protein to bring about

denaturation.

(iii) Upon partial hydrolysis of insulin, the polypeptide chain is broken down into several

fragments. One of the fragments formed is cys-gly-asp. Draw the displayed

structure of this tripeptide.

[4]

9


(c)

Aspartic acid, C4H7NO4, is heated gently to produce a cyclic compound C, C4H5NO3, and

H2O as by-product. C does not react with PCl5 and does not give any orange precipitate

with 2,4-dinitrophenylhydrazine. The addition of water to C re-forms aspartic acid. In a

similar reaction, C reacts with methanol in the presence of an organic solvent to give

compound D, C5H9NO4. 2 mol of D reacts completely with 1 mol of Na2CO3 to produce

effervescence of CO2.

(i) Deduce structures of C and D. Give your reasoning.

(ii) Write a balanced equation for the reaction of D with sodium carbonate.

[7]

[Total: 20]

10


4 Hydrazine is a hydride of nitrogen with the formula N2H4. It is a colourless, flammable liquid

with an ammonia-like odour. It behaves as a weak monoacidic base, forming salts with

common mineral acids.

(a) Draw the structure of hydrazine molecule to show the shape of the molecule and

indicate the bond angle. [2]

(b) Write a balanced equation for the reaction between hydrazine and sulfuric acid. [1]

(c) One of the industrial uses of hydrazine is as a low-power monopropellant in spacecraft.

In all hydrazine monopropellant engines, the hydrazine is passed over a catalyst which

causes it to decompose into ammonia, nitrogen and hydrogen according to the following

reactions:

reaction 1: 3N2H4(l) 4NH3(g) + N2(g) Ho = -335.4 kJ mol–1

reaction 2: 4NH3(g) + N2H4(l) 3N2(g) + 8H2(g)

Given that Hof of NH3(g) = -45.9 kJ mol–1, calculate

(i) Hof of N2H4(l)

(ii) Ho for reaction 2

[3]

(d) By considering the entropy and enthalpy changes in reaction 1 and reaction 2, suggest

how the standard Gibbs free energy change of the two reactions will compare in sign

and in magnitude.

Hence predict which reaction will be more spontaneous. Give your reasoning. [3]

(e) Diphosphine, P2H4, is the phosphorus analogue of hydrazine. Unlike hydrazine,

diphosphine is not basic. It disproportionates into phosphine, PH3, and phosphorus, P4,

in UV light. Both phosphine and diphosphine have lower boiling points than their nitrogen

analogues.

Compound b.p./ °C compound b.p./ °C

NH3 -33 PH3 -87

N2H4 113 P2H4 (decompose at 30)

(i) Suggest, with a reason, the bond angle in diphosphine.

(ii) Estimate what the boiling point of diphosphine would be if it did not decompose.

Explain your answer.

[3]

11


(f) In a similar reaction to that of ammonia, phosphine can react with bromoethane as

follows.

Use this information to help explain the following reaction.

When heated with a catalytic quantity of bromoethane, triethyl phosphite isomerises to

diethyl ethylphosphonate:

The isomerisation of triethylphosphite in the presence of bromoethane is thought to

proceed in two steps.

The first step involves the formation of an intermediate upon reaction with

bromoethane.

The second step involves the cleavage of a C–O bond in the intermediate to form

the final product, diethylethylphosphonate. Bromoethane is also regenerated.

Use the information given above to draw out the full mechanism for the isomerisation of

triethylphosphite. You are advised to use the structural formulae for all species so that it

is clear which bonds are broken and which are formed. Include all charges and use curly

arrows to show the movement of electron pairs.

[3]

(g) Bromoethane is formed when ethane and bromine are irradiated with light at room

temperature.

Outline the mechanism of this reaction. [3]

12


(h) The reaction described in (g) is seldom used for synthesis of halogenoalkanes as it

results in the formation of several isomeric products.

It was observed that methylpropane reacts with bromine to form a mixture of two

monobrominated alkanes, A and B, as shown below.

(i) State the expected ratio of products A and B in the mixture, assuming equal rate of

substitution of H atom.

The ratio of the isomeric products is more accurately determined if the relative rates of

substitution of H atoms are taken into account. The types of hydrogen atoms in alkanes,

together with their relative rates of substitution, are shown in the following table.

Types of H atoms

Structure Relative rate of

substitution

primary

1

secondary

4

tertiary

6

(ii) By taking into account the relative rates of substitution of H atoms given above,

show that products A and B are formed in the ratio of 3:2.

[2]

[Total: 20]

13


5 (a) Explain the following observation and give balanced equations, where appropriate, to

support your answers.

When chlorine is bubbled through cold sodium hydroxide and acidified silver nitrate

solution is added, only half of the chlorine which has dissolved is precipitated as silver

chloride. When the sodium hydroxide is hot, up to five-sixth of the chlorine can be

precipitated. [3]

(b) In film photography, light strikes an emulsion containing silver bromide grains. The

grains which are exposed turn partially into silver.

AgBr hv

Ag + 21

Br2

A developer is then added to turn all the silver bromide in the affected grains to silver.

(i) Name the type of reaction taking place in the action of light on AgBr.

The remaining silver bromide in the unaffected grains is then removed by a fixer,

containing thiosulfate ions, which is a monodentate ligand.

AgBr(s) + 2S2O32–(aq) [Ag(S2O3)2]

x–(aq) + Br–(aq)

(ii) State the value of x in [Ag(S2O3)2]x–(aq) and suggest the shape of this ion.

(iii) A student suggests adding concentrated sulfuric acid to the photographic emulsion

to test for the presence of bromide ions. Describe and explain, with the aid of

balanced equations, what the student would observe as a result of a successful

test.

(iv) If a photographic film is over-exposed, a reducer is used to diminish the intensity

of the image. One such reducer is 'Farmer's reducer' which contains aqueous

hexacyanoferrate(III), [Fe(CN)6]3–. This dissolves some of the silver metal and

hexacyanoferrate(II) ions, [Fe(CN)6]4–, are formed.

Write an ionic equation to represent the action of 'Farmer's reducer'. Hence

explain why, to a chemist, 'reducer' is a wrong word in view of what happens to the

silver.

[7]

(c) An organic compound A, C9H11Br, on treatment with hot aqueous potassium hydroxide

gave compound B, C9H12O.

B responded to oxidation in three different ways. With acidified potassium

dichromate(VI), it yielded C, C9H10O. With sodium hydroxide and iodine, it yielded D,

C8H7O2Na, and a yellow precipitate. With hot, acidic potassium manganate(VII), it

yielded E, C7H6O2.

(i) Identify compounds A to E.

(ii) Suggest a possible mechanism for the formation of compound B from compound

A, include all charges and curly arrows to show movement of electrons.

[8]

14


(d) Bromoethane and bromomethane, present in 1:1 molar ratio, can both react with

ethylamine to form three different tertiary amines.

Suggest the structures of these three tertiary amines and hence state the ratio in which

they are formed. [2]

[Total: 20]

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS

Higher 2

CHEMISTRY 9647/03

Paper 3 Free Response Thursday 27 August 2015

2 hours

Additional Materials: Data Booklet

Answer Paper


Write your name and class on all the work you hand in.

Write in dark blue or black pen on both sides of the paper. [PILOT FRIXION ERASABLE PENS ARE

NOT ALLOWED]

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


Answer any four questions. Write your answers on the answer paper provided.

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

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

At the end of the examination, fasten all your work securely together.

[Turn over


ANSWER SCHEME &

EXAMINERS’ COMMENTS

2


Answer any four questions.

1 Butanone is an industrial organic solvent used in the manufacture of plastics, lacquers and

varnishes.

(a) Butanone reacts with iodine in the presence of an acid according to the equation:

CH3CH2COCH3 + I2 CH3CH2COCH2I + HI

In a series of experiments, the reaction was carried out with different concentrations of

reagents, and the following initial rates were obtained:

Expt. [CH3CH2COCH3]

/ mol dm–3 [I2]

/ mol dm–3

[H+] / mol dm–3

Initial rate / mol dm–3 min–1

1 0.010 0.010 0.010 8.31 x 10–8

2 0.010 0.010 0.013 10.8 x 10–8

3 0.013 0.010 0.013 14.0 x 10–8

4 0.018 0.013 0.018 26.9 x 10–8

(i) Using the data given, deduce the order of reaction with respect to each of the

three reagents, clearly showing how you arrive at your answers. Hence, write a

rate equation for the reaction.

Let rate = k[CH3CH2COCH3]a[I 2]

b[H+]c

Using data from experiments 1 and 2,

cba

cba

k

k

)010.0(]010.0()010.0(

)013.0()010.0()010.0(

1031.8

108.108

8

c = 1

rate is first order with respect to H+


)013.0()010.0()010.0(

)013.0()010.0()013.0(

108.10

100.148

8

ba

ba

k

k

a = 1

rate is first order with respect to CH3CH2COCH3


]013.0[]010.0][013.0[

]018.0[]013.0][018.0[

100.14

109.268

8

b

b

k

k

b = 0

rate is zero order with respect to I2

rate = k[CH3CH2COCH3][H+]

3


(ii) Explain how the rate of reaction would change if the solution was diluted with an

equal volume of inert solvent, using r to represent the original rate before dilution.

Concentration of all solutions would be halved.

Since rate = k[CH3CH2COCH3][H+] = r (original rate)

New rate = k {[CH3CH2COCH3]/2} {[H+]/2} = r /4.

Hence, the rate would be ¼ the original rate.

(iii) The reaction can also be carried out in the presence of a strong base instead of

an acid. State what would be observed when a strong base is used and write an

equation for the reaction that occurs.

Yellow precipitate of CHI3 observed.

CH3CH2COCH3 + 4OH- + 3I2 →CH3CH2CO2- + CHI3 + 3I- + 3H2O

(OR CH3CH2COCH3 + OH- + 3I2 →CH3CH2CO2- + CHI3 + 3HI) (Less accurate

as HI further reacts with OH-)

[7]

(b) Elemental iodine, I2, has limited solubility in water. Its solubility is greatly increased in a

solution containing iodide ion, I-, due to the formation of triiodide ion, I3-, as shown:

I2(aq) + I-(aq) ⇌ I3-(aq)

(i) Write an expression for Kc for this reaction, stating its units.

Kc = ]][I[I

][I-

2

-

3 mol-1 dm3

(ii) Experiments have shown that when [I-] = 1.4 x 10-3 mol dm-3, the concentrations

of I2 and I3- are equal. Use this information to calculate a value of Kc.

Since [I2] = [I3-],

Kc = ]][I[I

][I-

2

-

3 = )10 x ](1.4[I

][I3-

2

-

3 = )10 x (1.4

13-

= 714 mol-1 dm3

(iii) Using your answer from (ii), calculate the concentration of iodide ions necessary

for 99 % of the iodine to be dissolved.

][I

][I

2

-

3 = 1

99

Kc = ]][I[I

][I-

2

-

3 = ][I

99-

= 714

[I-] = 714

99= 1.39 x 10-1 mol dm-3

[4]

4


(c) Explain, using a Maxwell-Boltzmann distribution curve, how the rate of a reaction would

change if the temperature was increased.

From the Maxwell-Boltzmann distribution of molecular energies, an increase in temperature results in more molecules having greater kinetic energy. This means

that the fraction of particles with energy Ea would increase. Thus, more molecules would have sufficient energy to overcome the energy barrier. Hence, the frequency of effective collisions increases, leading to increase in rate.

[3]

(d) Grignard reagents are organo-magnesium halides, commonly used in synthesis to

prepare a variety of organic compounds.

The carbon-magnesium bonds in Grignard reagents are highly polar and this makes it

extremely useful in organic synthesis as it is able to react with other polar organic

molecules to form carbon-carbon bonds. An example of the use of a Grignard reagent is

the two-step reaction of CH3CH2MgBr with butanone, CH3CH2COCH3, to form

3-methylpentan-3-ol.

T2 > T1

Number of particles

with energy ≥ Ea at 273k

Number of particles

with energy ≥ Ea at 298k

Number of molecules with energy E

5


(i) Suggest the type of reaction that has taken place in step 1.

Nucleophilic Addition

(ii) Suggest the structural formula of the organic compound to be used with

CH3CH2MgBr to form propan-1-ol, if the reaction undergoes a similar two-step

process.

(iii) Suggest a suitable Grignard reagent and another organic compound to be used if

propan-2-ol is to be prepared using a similar two-step process.

[4]

(e) There are certain practical issues with the use of Grignard reagents, notably slow

formation of the Grignard reagent and its susceptibility to water in air.

The slow formation of the Grignard reagent may be offset by cutting the precursor

magnesium into smaller pieces or scraping the sides of the metal before use. Using a

balanced equation, explain why magnesium, which has been stored in dry ambient

conditions, may not react as quickly as expected.

Mg(s) + ½ O2(g) MgO(s)

Mg reacts with the oxygen in air to form magnesium oxide which acts as a

protective layer and so, prevents the magnesium from reacting with the organic

compounds.

[2]

6


2 The direct oxidation of alcohols in a fuel cell represents potentially the most efficient method of

obtaining useful energy from renewable fuel.

The diagram below illustrates the functional parts of a direct-ethanol fuel cell (DEFC). The anode and cathode compartments are separated by the proton exchange membrane. The protons are transported across the proton exchange membrane to the cathode where they react with oxygen to produce water.

Diagram of a DEFC

(a) (i) Explain, in terms of the change in average oxidation number of the carbon atoms,

whether electrode 1 or 2 acts as the anode.

Electrode 1 acts as the anode as oxidation occurs since there is an

increase in average oxidation state for carbon from -2 to +4.

(ii) Write the half-equations for the reactions which take place at the electrodes of

the DEFC, and hence an overall equation for the cell reaction.

Anode: CH3CH2OH + 3H2O → 2CO2 + 12H+ + 12e-

Cathode: O2 + 4H+ + 4e- → 2H2O

Overall: CH3CH2OH + 3O2 → 2CO2 + 3H2O

(iii) The cell is capable of producing an e.m.f. of 1.47 V. By using suitable data from

the Data Booklet, calculate a value for the E

of the CO2/CH3CH2OH electrode

reaction.

E

O2/H2O = +1.23 V

E

cell = E

O2/H2O – E

CO2/ CH3CH2OH

1.47 = 1.23 – E

CO2/ CH3CH2OH

E

CO2/C2H5OH = – 0.24 V

H2SO4(aq)

External circuit

CO2

CH3CH2OH H2O

O2

H2O

Electrode 2 Electrode 1

7


(iv) By quoting relevant E

values from the Data Booklet, explain why an acidic

electrolyte is preferred to an alkaline electrolyte in the DEFC.

In an acidic medium, E

O2/H2O = +1.23 V while in an alkaline medium, E o

O2/OH

= +0.40 V.

Since E

cell = E

reduction -– E

oxidation and E

reduction becomes more positive,

E

cell would be more positive and hence a higher emf is obtained.

(v) Describe and explain the effect on the cell potential when the concentration of

ethanol is increased.

When [ethanol] is increased, position of equilibrium for the following reaction

shifts to the left. There is greater tendency for ethanol to be oxidised and thus

ECO2/ CH3CH2OH becomes more negative.

2CO2 + 12H+ + 12e- CH3CH2OH + 3H2O ECO2/ CH3CH2OH

Since Ecell = EO2/H2O - ECO2/ CH3CH2OH

A more negative ECO2/ CH3CH2OH will cause Ecell to be more positive.

[9]

(b) Chromium plating is an important industrial application of electrolysis, where a steel

object is plated with a thin layer of chromium to make it resistant to rusting and scratches.

In an electroplating experiment, a researcher connected the DEFC to an electrolytic cell

as shown below. The aqueous solution containing chromium(III) chloride was

electrolysed using a piece of pure chromium and the steel object to be electroplated as

the electrodes.

(i) The total time taken for 10 g of chromium to be plated was 4 hours. Calculate

the current produced by the DEFC.

Cr3+ + 3e- → Cr

Amount of Cr =

2 = 0.192 mol

Amount of e¯ = 0.192 x 3 = 0.577 mol

Q = 0.577 x 96500 = 55700 C

CrCl3 (aq)

pure chromium

steel object

DEFC

8


Q = It

55700 = I x (4 x 60 x 60)

I = 3.87 A

(ii) Calculate the volume of ethanol used when 10 g of chromium metal was plated

onto the steel object at room conditions, given that the density of ethanol is

0.789 g cm-3.

Cr3+ + 3e- → Cr

CH3CH2OH + 3H2O → 2CO2 + 12H+ + 12e- (from aii)

Cr ≡ 3e- ≡

CH3CH2OH

No of mol of CH3CH2OH = 0.192 x ¼ = 0.0480 mol

Mass of CH3CH2OH = 0.0480 x 46 = 2.208 g

Volume = 2 2

= 2.80 cm3

[3]

(c) Chromium is a transition element.

(i) Explain what is meant by the term transition element.

A transition element is defined as a d-block element which forms at least one stable ion with partially filled d-subshell of electrons (or incompletely filled d-orbitals).

(ii) Chromium(III) chloride, CrCl3, reacts with water to form a violet hydrate

[Cr(H2O)6]Cl3. This violet hydrate is isomeric to the dark green

[CrCl2(H2O)4]Cl.2H2O and the pale green [CrCl(H2O)5]Cl2.H2O

Based on the above information, give one property of transition elements or its

compounds that is not shown by s-block elements.

Transition elements form complexes that are coloured.

(iii) When 0.03 mol of aqueous silver nitrate is added to 0.01 mol of one of the

chromium-containing compounds in (ii), a white precipitate weighing 1.44 g is

formed. Deduce the formula of the compound that the silver nitrate was added to.

Amt of AgCl = Amt of Cl- =

= 0.0100 mol

There is 1 mol of free Cl- per mol of the compound.

Formula of the compound: [CrCl2(H2O)4]Cl.2H2O

[3]

9


(d) The following reaction scheme shows the chemistry of some chromium-containing

species in aqueous solution.

(i) Identify the chromium-containing species present in A. Describe how this species

accounts for the formation of carbon dioxide in Step III. Include any relevant

equations in your answer.

Cr3+ ion has a high charge density and high polarising power. Hence [Cr(H2O)6]

3+ can undergo hydrolysis in water, polarising and weakening the O-H bonds of surrounding H2O molecules, to release H+ ions. H+ ions react with carbonate ions via acid-base reaction to form CO2.

[Cr(H2O)6]3+(aq) ⇌ [Cr(OH)(H2O)5]

2+(aq) + H+(aq)

2H+(aq) + CO3

2-(aq) CO2(g) + H2O(l)

(ii) Give a balanced equation for step VI.

2CrO42- + 2H+ Cr2O7

2- + H2O

(iii) Similar to some organic molecules, the complex [Cr(C2O4)3]3- formed in step II

can exist as a pair of optical isomers. One of the isomers is as shown below:

3–

Cr

L

L

L

10


is used to represent the C2O42- ion.

Draw the structure of the other optical isomer.

[5]

[Total: 20]

L

11


3(a) Aspartame is made from the amino acids aspartic acid and phenylalanine. It is one of the

most commonly used artificial sweeteners as it is about 200 times sweeter than sugar so

lesser amount is required to achieve the same level of sweetness. A molecule of

aspartame is shown below:

(i) Aspartame can exist as a zwitterion. Using the information given above, draw the

structure of the zwitterion and suggest one possible pH value at which aspartame

exists predominantly as a zwitterion.

For aspartame to exist as a zwitterion, the pH of the solution should be

between 3.5 and 8.5. Acceptable range: 3.5 ≤ pH ≤ 8.5.

(ii) Draw the structural formulae of the organic products formed when aspartame reacts

with

I an excess of hot NaOH(aq).

12


II cold HCl(aq).

(iii) Draw the structure of compound A in the following reaction.

(iv) Describe a simple chemical test to distinguish between aspartame and compound B,

clearly stating the reagents, conditions and observations for each compound.

13


To separate test tubes containing aspartame and compound B, add

aqueous alkaline Iodine and heat.

For the test tube containing aspartame, no yellow crystals of CHI3 will be

formed while for the test tube containing compound B, yellow crystals of

CHI3 (CH3CH2OH is produced upon hydrolysis of ester) will be seen.

[9]

(b) Aspartic acid and phenylalanine can also be found in insulin, which consists of two polypeptide chains. It is responsible for the regulation of blood glucose level in the human body. The following table shows some amino acid residues that can be found in insulin:

Amino acid Abbreviation R group

aspartic acid asp –CH2CO2H

phenylalanine phe

lysine lys –(CH2)4NH2

cysteine cys glycine gly

(i) When mercury ions, Hg2+, are added to a sample of insulin, the protein is found to

be denatured. State which aspects of the protein structure are affected during this denaturation. The tertiary and quaternary structures of insulin protein will be affected as the R-group interactions will be disrupted.

(ii) By considering all the interactions formed by the above amino acid residues, explain

clearly how Hg2+ can interact with the insulin protein to bring about denaturation. Addition of Hg2+ which disrupts the ionic interaction formed between –CO2

- in asp and –NH3

+ in lys as Hg2+ will form ionic interactions with –CO2- in asp instead.

Addition of Hg2+ will also disrupts the disulfide linkages as Hg2+ binds permanently with S, thus breaking the covalent disulfide linkages formed between 2 cysteine residues.

14


(iii) Upon partial hydrolysis of insulin, the polypeptide chain is broken down into several fragments. One of the fragments formed is cys-gly-asp. Draw the displayed structure of this tripeptide.

[4] c)

Aspartic acid, C4H7NO4, is heated gently to produce a cyclic compound C, C4H5NO3, and

H2O as by-product. C does not react with PCl5 and does not give any orange precipitate

with 2,4-dinitrophenylhydrazine. The addition of water to C re-forms aspartic acid. In a similar reaction, C reacts with methanol in the presence of an organic solvent to give compound D, C5H9NO4. 2 moles of D reacts completely with 1 mole of Na2CO3 to produce effervescence of CO2.

(i) Deduce structures C and D and give your reasoning. (ii) Write a balanced equation for the reaction of D with sodium carbonate.

Observation Type of reaction Deduction

Aspartic acid, C4H7NO4, is heated gently to produce compound C, C4H5NO3, and H2O as by-product.

- A cyclic compound is formed after condensation

C does not react with PCl5 No nucleophilic substitution

No alcohol group and –CO2H present in C.

C does not give any orange precipitate with 2,4-dinitrophenylhydrazine.

No condensation No carbonyl group present in C. C could be an acid anhydride

C reacts with methanol in the presence of an organic solvent to give compound D, C5H9NO4.

- An ester bond and a –CO2H group is formed.

2 moles of D reacts completely with 1 mole of Na2CO3 to produce effervescence of CO2

Acid-base reaction of carbonate and acid

1 –CO2H group is present in D.

15


(i) Write a balanced equation for the reaction of D with sodium carbonate.

16


4 Hydrazine is a hydride of nitrogen with the formula N2H4. It is a colourless, flammable

liquid with an ammonia-like odour. It behaves as a weak monoacidic base, forming salts

with the common mineral acids.

(a) Draw the structure of hydrazine molecule to show the shape of the molecule and

indicate the bond angle.

[2]

(b) Write a balanced equation for the reaction between hydrazine and sulfuric acid.

[1]

(c) One of the industrial uses of hydrazine is as a low-power monopropellant in spacecraft. In all hydrazine monopropellant engines, the hydrazine is passed by a catalyst which causes it to decompose into ammonia, nitrogen and hydrogen according to the following reactions:

reaction 1: 3N2H4(l) → 4NH3(g) + N2(g) Ho = -335.4 kJ mol-1 reaction 2: 4NH3(g) + N2H4(l) → 3N2(g) + 8H2(g)

Given that Hof of NH3(g) = -45.9 kJ mol-1, calculate

(i) Hof of N2H4(l)

2N2H4 + H2SO4 [N2H5]+

2SO42- [OR]

N2H4 + H2SO4 [N2H5]+ HSO4

-

3N2H4(l) 4NH3(g) + N2(g)

-335 = - 3(Hof of N2H4(l)) + 4(-45.9) + 0

Hof of N2H4(l) = (4(-45.9) + 335) / 3 = + 50.5 kJ mol-1

-335.4

3N2(g) + 6H2(g)

0

3 (Hof of N2H4(l)) 4(-45.9)

17


(ii) Ho for reaction 2

[3]

(d) By considering the entropy and enthalpy changes during reaction 1 and reaction 2, suggest how the standard Gibbs free energy change of the two reactions will compare in sign and in magnitude.

Hence predict which reaction will be more spontaneous. Explain your reasoning. [3]

4NH3(g) + N2H4(l) 3N2(g) + 8H2(g)

Ho = 4(+45.9) - 50.5 = + 133 kJ mol-1

3N2(g) + 8H2(g)

0 0

+50.5 4(-45.9)

Ho = ?

Entropy changes during reaction 1 and reaction 2 are positive due to increase in disorder of the systems as there is an increase in the number of moles of gaseous molecules formed in both reactions.

reaction 1: G = H -TS (Thus, G < 0 at all T values) (-) (-)(+)

reaction 2: G = H –TS (Thus, G < 0 only at higher T values) (+) (-)(+)

As H for reaction 2 is positive, its G would only become negative at a higher

temperatures whereas since the H for reaction 1 is negative, the G for this reaction is negative at all temperatures. Hence reaction 1 will be more spontaneous.

18


(e) Diphosphine, P2H4 is the phosphorus analogue of hydrazine. Unlike hydrazine,

diphosphine is not basic. It disproportionates into phosphine, PH3 and phosphorus, P4 in UV light. Both phosphine and diphosphine have lower boiling points than their

nitrogen analogues.

Compound b.p./ °C compound b.p./ °C

NH3 -33 PH3 -87

N2H4 113 P2H4 (decompose at 30)

(i) Suggest, with a reason, the bond angle in diphosphine.

(ii) Estimate what the boiling point of diphosphine would be if it did not decompose. Explain your answer.

[3]

(f) In a similar reaction to that of ammonia, phosphine can react with bromoethane as

follows.

Use this information to help explain the following reaction.

When heated with a catalytic quantity of bromoethane, triethylphosphite isomerises

to diethylethylphosphonate:

The isomerisation of triethyl phosphite in the presence of bromoethane is thought to

proceed in two steps.

The bond angle in P2H4 is ______ (accept any angle < 107.5° but > 90°). P atoms in P2H4 are less electronegative than N atoms in hydrazine. Hence bond pair-bond pair repulsion in P2H4 will be less than that in N2H4, resulting in a smaller bond angle in P2H4 as compared to that in N2H4.

Boiling point of diphosphine would be (any value quoted between 30°C & 113°C acceptable). Diphosphine would have a lower b.p. than hydrazine due to less energy required to overcome its weaker intermolecular permanent dipole-permanent dipole attractions than the stronger intermolecular hydrogen bonding in hydrazine.

19


The first step involves the formation of an intermediate upon reaction with

bromoethane.

The second step involves the cleavage of a C–O bond in the intermediate to

form the final product, diethyl ethylphosphonate. Bromoethane is also

regenerated.

Use the information given above to draw out the full mechanism for the isomerisation of triethylphosphite. You are advised to use the structural formulae for all species so that it is clear which bonds are broken and which are formed. Include all charges and use curly arrows to show the movement of electron pairs.

[3]

(g) Bromoethane is formed when ethane and bromine are irradiated with light at room

temperature. Outline the mechanism of this reaction.

[3]

20


(h) The reaction described in (g) is seldom used for synthesis of halogenoalkanes as it

results in the formation of several isomeric products.

It was observed that methylpropane reacts with bromine to form a mixture of two

monobrominated alkanes, A and B, as shown below.

(i) State the expected ratio of products A and B in the mixture, assuming equal

rate of substitution of H atom.

The ratio of the isomeric products is more accurately determined if the relative rates

of substitution of H atoms are taken into account. The types of hydrogen atoms in

alkanes, together with their relative rates of substitution, are shown in the following

table.

Types of H

atoms

Structure Relative rate of

substitution

Primary

1

Secondary

4

Tertiary

6

(ii) By taking into account the relative rates of substitution of H atoms given above,

show that the products A and B are formed in the ratio of 3:2.

[2]

Expected ratio of A: B = 9 : 1

21


[Total: 20]

5 (a) Explain the following observation and give balanced equations, where appropriate, to

support your answers.

When chlorine is bubbled through cold sodium hydroxide and acidified silver nitrate

solution is added, only half of the chlorine which has dissolved is precipitated as silver

chloride. When the sodium hydroxide is hot, up to five-sixth of the chlorine can be

precipitated. [3]

When Cl2 is bubbled through cold NaOH,

Cl2 + 2NaOH NaCl + NaClO + H2O

The products, NaCl and NaClO, are formed in the ratio of 1:1. Thus, only one-

half of the chlorine can be precipitated as AgCl from the NaCl formed.

When Cl2 is bubbled through hot NaOH,

3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O

The products, NaCl and NaClO3, are formed in the ratio of 5:1. Thus, only

five-sixth of the chlorine can be precipitated as AgCl from the NaCl formed.

(b) In film photography, light strikes an emulsion containing silver bromide grains. The

grains which are exposed turn partially into silver.

AgBr hv

Ag + 21

Br2

A developer is then added to turn all the silver bromide in the affected grains to silver.

(i) Name the type of reaction taking place in the action of light on AgBr.

redox/decomposition

The remaining silver bromide in the unaffected grains is then removed by a fixer,

containing thiosulfate ions, which is a monodentate ligand.

AgBr(s) + 2S2O32–(aq) [Ag(S2O3)2]

x–(aq) + Br–(aq)

(ii) State the value of x in [Ag(S2O3)2]x–(aq) and suggest the shape of this ion. [1]

9 possible (primary) hydrogen (Ha) can be substituted to form A 1 possible (tertiary) hydrogens (Hb) can be substituted to form B

Assuming equal probability of substitution, Ratio of A : B = 9:1 However since a tertiary H is substituted 6 times faster than a primary H, Ratio of A : B = 9×1 :1×6 = 9 : 6 = 3: 2

22


3, linear

(iii) A student suggests adding concentrated sulfuric acid to the photographic

emulsion to test for the presence of bromide ions. Describe and explain, with the

aid of balanced equations, what the student would observe as a result of a

successful test. [3]

red/reddish brown fumes/red liquid of Br2 and some steamy fumes of HBr

Br– reacts with conc. H2SO4 to give HBr, which is further oxidised by conc. H2SO4 to Br2.

Br– + H2SO4 HBr + HSO4–

2HBr + H2SO4 Br2 + SO2 + 2H2O

(iv) If a photographic film is over-exposed, a reducer is used to diminish the

intensity of the image. One such reducer is 'Farmer's reducer' which contains

aqueous hexacyanoferrate(III), [Fe(CN)6]3-. This dissolves some of the silver

metal and hexacyanoferrate(II) ions, [Fe(CN)6]4-, are formed.

Write an ionic equation to represent the action of 'Farmer's reducer'. Hence

explain why, to a chemist, 'reducer' is a wrong word in view of what happens to

the silver. [2]

Ag + [Fe(CN)6]3– Ag+ + [Fe(CN)6]

4–

Ag is oxidised to Ag+

(c) An organic compound A, C9H11Br, on treatment with hot aqueous potassium hydroxide

gave compound B, C9H12O.

B responded to oxidation in three different ways. With acidified potassium

dichromate(VI), it yielded C, C9H10O. With sodium hydroxide and iodine, it yielded D,

C8H7O2Na, and a yellow precipitate. With hot, acidic potassium manganate(VII) it

yielded E, C7H6O2.

(i) Identify compounds A to E. [5]

23


24


(ii) Suggest a possible mechanism for the formation of compound B from compound A, include all charges and curly arrows to show movement of electrons. [3] Accept SN1 or SN2 mechanism

SN1 Nucleophilic Substitution

SN2 Nucleophilic Substitution

(d) Bromoethane and bromomethane, present in 1:1 molar ratio, can both react with

ethylamine to form three different tertiary amines.

Suggest the structures of these three tertiary amines and hence state the ratio in which they are formed. [2]

25


The three tertiary amines are:

N CH3

CH3

CH3CH2 N CH3CH3CH2

CH2CH3

NCH3CH2

CH2CH3

CH2CH3

ratio 1 2 1

[Total: 20]

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