2012 Gce a Level h2 p1 p2 p3 Qns Ans

GCE Advanced Level Worked Solution 9647 Higher 2 Chemistry 2012 Paper 1 (Multiple Choice Questions) 1 2 3 4 5

C C D B C

11 12 13 14 15

D C B A C

21 22 23 24 25

D B B D C

31 32 33 34 35

B D A B B

6 7 8 9 10

A D C A D

16 17 18 19 20

D D C C C

26 27 28 29 30

A C B B A

36 37 38 39 40

B D A B B

MCQ worked solutions Please use the table format for our internal exams for keying in the 2012 questions. Formatting wise, not so strict, just try your best =) 1

In the body, cellular respiration produces energy from the oxidation of glucose. The diagram shows the structure of glucose. CH2OH O H

OH

OH

H

OH

H

H

OH

H

A new artificial sweetener has been produced by replacing all of the hydroxyl groups attached directly to the ring carbon atoms in glucose with chlorine atoms. What is the empirical formula of this chlorinate glucose? A

CHClO

B

CH2Cl

C

C3H4Cl2O

D

C6H7Cl5O

Ans :C Keyword: only replaced all hydroxyl groups attached directly to the ring carbon atoms

H2 Chemistry 9647 2012

CH2OH O H

Cl

Cl

H

Cl

H

H

Cl

H

molecular formula = C6H8Cl4O2 empirical formula = C3H4Cl2O

2

When Fe is reacted with Fe3+(aq) ions, Fe2+(aq) ions are formed. Assuming the reaction goes to completion, how many moles of Fe and of Fe3+(aq) would result in a mixture containing equal numbers of moles of Fe3+(aq) and Fe2+(aq) once the reaction had taken place? moles of Fe

moles of Fe3+(aq)

A

1

2

B

1

3

C

1

5

D

2

3

Ans: C [this qn also came out in N2007] Thinking process: write balance equation for the reaction of Fe and Fe3+ to form Fe2+. Fe + 2Fe3+ 3 Fe2+ At the end of reaction, equal amount of Fe3+ and Fe2+ are in solution mixture. This implies that Fe is the limiting reagent and Fe3+ is in excess. (thus, option D is out) When 1 mol of Fe areacts with 5 mol of Fe3+, Number of mol of Fe3+ required = 2 Number of mole of Fe3+ left = 5 – 2 = 3 No. of mole of Fe2+ produced = 3

3

Which one of the following does not contain either an unpaired s electron or an unpaired p electron?


A

Cr

B

Ge

C

S−

D

Sc

Ans: D Cr: [Ar]3d54s1 Ge: [Ar]3d10 4s2 4p2 S− : [Ne]3s23p5 Sc: [Ar]3d1 4s2 Thus, Sc does not contain either an unpaired s electron or an unpaired p electron.

4

The graph shows the first thirteen ionisation energies for element X.

What can be deduced about element X from the graph? A

It is a d-block element.

B

It is in Group II of the Periodic Table.

C

It is in Group III of the Periodic Table.

D

It is in the second period (Li to Ne) of the Periodic Table.

Ans: B Element has 2 valence electrons (required the least amount of energy to remove), because 2 electrons removed from outermost shell. Hence it is a Group II element, not Group III. After removing 2 valence electrons, the inner shell contains 8 electrons, and not 18 electrons. Thus element is not d-block element. [thus option A is out] Element has more than 2 occupied quantum shells, hence, not in the second period. [thus, option D is out]

5

The boiling point of water (100 °C) is greater than that of HF (20 °C). Which statement is a correct explanation of this?


A

Each hydrogen bond formed between water molecules is stronger than that formed between HF molecules.

B

There are more atoms in a water molecule than there are in an HF molecule, so van der Waals’ forces are stronger in water.

C

There are, on average, more hydrogen bonds between water molecules than there are between HF molecules.

D

The water molecule has a greater permanent dipole than the HF molecule.

Ans: C Both water and HF have intermolecular hydrogen bonding. Each Water molecule has 2 Hydrogen atoms & 2 lone pair of electrons on the oxygen atom, so on average water molecule can form 2 hydrogen bonds. Each HF molecule has 1 hydrogen atom & 3 lone pair of electrons on the fluorine atom, so on average HF molecule can only form 1 hydrogen bond. Thus, more energy needed to overcome greater number of hydrogen bonds in water that in HF.

6

One suggestion for the reduction of greenhouse gas emissions from coal-fired power stations is to separate the CO2 from the flue gases and pump it into the sea bed, where it will dissolve in water under pressure. CO2(g) + nH2O(l)

CO2(aq)

During this process, hydrogen bonds are formed between CO2 and H2O molecules. What will be the signs of ∆H and ∆S for this equilibrium for the forward reaction? ∆H

∆S

A

−

−

B

−

+

C

+

−

D

+

+

Ans: A Since hydrogen bonds are formed, there is energy released & ∆H is negative. From the equation, it can be seen that there is a decrease in number of gas particles from reactants to products. Thus, ∆S is negative.

7

Two separate electrolyses were performed as follows, under the same conditions of temperature and pressure. 1


When molten copper(II) chloride was electrolysed for five minutes, 100 cm3 of

chlorine were collected from the anode. When aqueous sulfuric acid was electrolysed for five minutes, 200 cm3 of oxygen were collected from the anode.

2

If current used in electrolysis 1 was I, what was the current used in electrolysis 2? A

0.5I

B

2I

C

I

D

4I

Ans: D 2Cl- Cl2 + 2e 2H2O O2 + 4H+ + 4e Q = It = ne F I=

ne F t

Thus, I α

ne

No. of mol of electrons (ne) in electrolysis 1: (ne) in electrolysis 2 100(2) : 200 (4) 1: 4 Thus, current used in reaction 2 is 4I

8

The gas-phase reaction of carbon monoxide with hydrogen forming methanol is an example of an equilibrium. The reaction was investigated by mixing 2.0 mol of H2(g) with 1.0 mol of CO(g) and allowing equilibrium to be established. 2H2(g) Initial moles

2.0

+ CO(g) 1.0

CH3OH(g) 0

At equilibrium, x mol of H2 had reacted with CO. What are the amounts, in moles, of each of the components of the equilibrium mixture?

A B C

H2(g)

CO(g)

CH3OH(g)

1 2.0 − x 2 1 2.0 − x 2

1.0 – x

1 x 2

D


2.0 – x 2.0 – x

1.0 – 2x

1 1.0 − x 2 1.0 – x

x

1 x 2 1 x 2

Ans : C 2H2(g)

9

+ CO(g)

CH3OH(g)

Initial moles

2.0

1.0

0

Change in mole

-x

-1/2 x

-1/2 x

Eqm moles

2.0 – x

1 1.0 − x 2

1 x 2

Pyruvic acid is a weak acid that is an important intermediate in many biochemical processes. CH3COCO2H Pyruvic acid Which graph best represents the change in pH that occurs when a sample of pyruvic acid is titrated with NaOH(aq)? A

B

C

D

Weak acid – strong base titration At equivalence point, an alkaline salt is formed. Hence, pH > 7 Maximum buffer capacity will be when both weak acid and the salt is present hence it is before the equivalence point. 10

Stomach juice have a pH of 1.0. Aspirin is a monobasic (monoprotic) acid represented by HA (Ka = 1 x 10-4) which dissociates into ions H+ and A-. What are the relative concentrations of H+, A- and HA when aspirin from a tablet enters the stomach? A B C D

[HA] > [H+] = [A-] [H+] = [A-] > [HA] [H+] > [A-] > [HA] [H+] > [HA] > [A-]

Answer: D HA is a weak acid which dissociates partially into H+ and A-. HA

Ini conc Eqm conc H2 Chemistry 9647 2012

c c–x

H+ + A0 0 x x

[H+] = [A-] and [HA] > [H+] However, in the stomach of pH 1.0, [H+] is high [H+] > [HA] > [A-] 11

A 1 dm3 solution was made by mixing 0.0040 mol of HCl(aq) and 0.0025 mol of NaOH(aq). What was the pH of the resulting solution? A B C D

2.19 2.40 2.60 2.82

Answer: D HCl + NaOH NaCl + H2O 0.0040 0.0025 0.0015 mol of HCl is in excess. Concentration of H+ after neutralization = 0.0015 mol dm-3 pH = - log 0.0015 = 2.82 12

Lead is the final product formed by a series of changes in which the rate-determining step is the radioactive decay of uranium – 238. This radioactive decay is a first order reaction with a half-life of 4.5 x 109 years. How long would it take for a rock sample, originally lead-free, to contain a molar proportion of uranium to lead of 1 : 3 A B C D

1.5 x 109 years 2.25 x 109 years 9.0 x 109 years 13.5 x 109 years

Answer: C Number of half - life

Uranium 1 1/2 1/4

4.5 x 109 4.5 x 109

Lead 1/2 3/4

Total number of years = 4.5 x 109 x 2 = 9.0 x 109 years

13

The decomposition of hydrogen peroxide in aqueous solution is a first order process. If 10% of the hydrogen peroxide in a solution of concentration 0.1 mol dm -3 decomposes in 5 minutes at a certain temperature, what percentage of the hydrogen peroxide in a solution of concentration 0.2 mol dm-3 would decompose in 5 minutes at the same temperature? A

5%


B

10%

C

19%

D

20%

Since decomposition is a first order process, the half-life of H2O2 at the same temperature remains constant. n x t1/2 = 5 mins

 0.09   1   =   0.1   2 

n

 x  1  =   0.2   2 

Since time taken is the same,

n

0.09 x = 0.1 0.2

[H2O2] left = 0.18 mol dm-3 [H2O2] decomposed = 0.02 mol dm-3 Percentage decomposed = 0.02/ 0.2 x 100% = 10% Examiner’s Comment: More than half the candidates did not appreciate that the same percentage of hydrogen peroxide would decompose under these conditions. 14

The uncatalysed reaction between SO2 and O2 is slow. 2SO2 + O2 2SO3 The reaction is speeded up in the presence of the homogeneous catalyst NO, which participates as follows. 2NO + O2 2NO2 2NO2 + 2SO2 2NO + 2SO3 Which reaction pathway diagram is most appropriate for describing the enthalpy changes occurring during the catalyzed reaction? In each case the reaction pathway for the uncatalysed reaction is shown as a dashed line.


A

B

C

D

Answer: A Catalyst will lower the activation energy by providing an alternative pathway. However, it does not alter the initial and final states of the system and hence do not affect ∆H of the reaction. In the catalyzed reaction, the reaction pathway undergoes two steps hence there should be two humps for the reaction. 15

Adding concentrated HCl(aq) to CuSO4(aq) causes the colour of the solution to change from blue to green. Which row best explains this observation?

A B C D

number of d-electrons around copper changes changes remains the same remains the same

energy gap between the d-orbitals changes remains the same changes remains the same

Answer: C Adding concentrated HCl to CuSO4 causes ligand exchange reaction to occur.The identity of the green solution is CuCl42-. There is no change in the oxidation number of Cu hence there should be no change in the number of d – electrons around copper. However, the colour of the solution is due to the magnitude of the energy gap between the d – orbitals. Since there is a colour change, this shows that the magnitude of the energy gap differs, resulting in the absorbance of a different wavelength from the visible light spectrum. Examiner’s Comment: The most common incorrect response to Question 15 was A, showing that candidates thought that the number of copper d orbital electrons changed in the reaction. 16

Which reaction will not take place readily under the conditions specified by the state symbols? [Note: H2O(g) = steam] A B C D

Ba(s) + 2 H2O(l) Ba(OH)2(aq) + H2(g) CaO(s) + H2O(l) Ca(OH)2(s) Mg(s) + H2O(g) MgO(s) + H2(g) Mg(s) + 2H2O(l) Mg(OH)2(aq) + H2(g)

Answer: D Mg(s) has slow or no reaction with cold water, does not produce Mg(OH)2(aq). H2 Chemistry 9647 2012

CaO(s) dissolves in water to give a white precipitate (of calcium hydroxide) is formed, together with an alkaline solution (also of calcium hydroxide – calcium hydroxide is slightly soluble).

17

The data below refers to a particular element. density 10.5g cm-3 first ionisation energy 730 kJmol-1 second ionisation energy 2070 kJmol-1 third ionisation energy 3360 kJmol-1 melting point 1235K Where in the Periodic Table is this element most likely to be found? A

group I

B

group II

C

group VII

D

the transition elements

Ans: D Thinking process: There is no drastic increase from the 1st to the 3rd ionisation energies. A high melting point indicates the element has a giant structure. The element has a high density. Hence, the element is most likely to be the transition element.

18

When chlorine reacts with hot aqueous sodium hydroxide, the reaction below occurs. chlorine + sodium hydroxide

water + sodium chloride + X

What is compound X? A

sodium chlorate(I)

B

sodium chlorate(III)

C

sodium chlorate(V)

D

sodium chlorate(VII)

Ans: C Thinking process: Under hot aqueous NaOH, chlorine undergoes disproportionation to form NaCl and NaClO3. 3Cl2(g) + 6NaOH(aq)

5NaCl(aq) + NaClO3(aq) + 3H2O(l)

NaClO3(aq) is known as sodium chlorate(V), where Cl atom is in the +5 oxidation state.


19

Covalent bonds are formed by orbital overlap. The shape of unsaturated hydrocarbon molecules can be explained in terms of hybridisation of orbitals. Which bond is not present in HC≡CCH2CH=CH2? A

a π bond formed by 2p–2p overlap

B

a σ bond formed by 1s–2sp overlap

C

a σ bond formed by 2sp–2sp2 overlap

D

a σ bond formed by 2sp2–2sp3 overlap

Ans: C Thinking process:

Between C1 and C2, and between C4 and C5, there are π bonds formed by a 2p–2p overlap. A σ bond is formed between 1s of H and 2sp of C1. A σ bond is formed between 2sp2 of C4 and 2sp3 of C3.

20

When heated with chlorine, the hydrocarbon 2,2-dimethylbutane undergoes free radical substitution. In a propagation step, the free radical X• is formed by the loss of one hydrogen atom.

How many different forms of X• are theoretically possible?

A

1

C

3

B

2

D

4

Ans: C Thinking process:

There are 3 different types of H atoms as indicated by the different shapes.


21

PGE2 and PGF2α are two prostaglandins, both with pharmacological activity.

PGF2α

PGE2 Which reagent will convert PGE2 into PGF2α efficiently?

A

H2 / Ni

C

H2O / H+

B

H2 / Pt

D

NaBH4

Ans: D Thinking process: H2/ Ni and H2/Pt will reduce both C=C bond in alkenes and C=O bond in carbonyl compounds. H2O / H+ is the reagent and condition for hydration. NaBH4 is a specific reducing agent for reduction of carbonyl group.


22

The diagram shows the structure of the naturally-occuring molecule cholesterol.

cholesterol Cholesterol reacts with cold, dilute acidified KMnO4. Four students made claims about this reaction. W X Y Z

The cyclohexene ring becomes saturated. The cyclohexene ring is broken open. The number of chiral centres increases by 1. The number of chiral centres increases by 2.

Which students’ claims are correct?

A

W and Y

C

X and Y

B

W and Z

D

X and Z

Ans: B Thinking process: H3C CH3

CH3

H

H

H

H

* *

HO OH

OH

Mild oxidation of alkene by cold, dilute acidified KMnO4 to form diol. The cyclohexene ring becomes saturated, and does not break open. The number of chiral carbons increases by 2 (indicated by *).


23

The diagram represents the transition state of an SN2 reaction in which N is an anionic nucleophile and L the leaving group, but without any indication of the location of positive or negative charges.

Which is a possible combination of charges? charge on N

charge on L

A

δ–

δ+

B

δ–

δ–

C

δ+

δ+

D

δ+

δ–

Ans: B Thinking process: Both N and L are δ–. It is the carbon atom that is δ+.

24

Chlorofluorocarbons (CFCs) have been widely used in aerosol sprays, refrigerators and in making foamed plastics, but are now known to destroy ozone in the upper atmosphere. What will not destroy ozone, and therefore can be used as a replacement for CFCs? A

CHBr3

B

CCl3CBr3

C

CHClFCCl2F2

D

CH3CH2CH2CH3

Ans: D Thinking process: A safe replacement for CFCs is to use compounds that do not contain C–X bond where X is a halogen. Note: repeated question from N2007/P1/23 25

Rosmaric acid occurs in culinary herbs such as rosemary, sage and thyme. OH O O O HO OH

rosmaric acid H2 Chemistry 9647 2012

OH

OH

How many moles of NaOH(aq) will react with one mole of rosmaric acid when the acid is heated under reflux with an excess of NaOH(aq)? A

4

B

5

C

6

D

7

Thinking process:

• •

5 moles of NaOH is required to react with acidic functional groups like RCOOH and phenol, via acid-base reaction. 1 mole of NaOH is required to react with the carboxylic acid functional group formed after the alkaline hydrolysis of the ester functional group.

Answer: C 26

Cattle suffering from stress produce 6-methylhept-5-en-2-one, which repels mosquitoes. (CH3)2C=CHCH2CH2COCH3 6-methylhept-5-en-2-one Which property will 6-methylhept-5-en-2-one possess? A

Addition of Br2(aq) produces a chiral compound.

B

Prolonged heating with acidified concentrated KMnO4 produces HO2CCH2CO2H.

C

Reduction by a methanolic solution of NaBH4 produces a compound C8H18O.

D

Warming with alkaline aqueous I2 produces CH3CO2H.

Thinking process: •

Option A: Addition of Br2(aq) forms a compound with a chiral carbon.

•

Option B: Prolonged heating with acidified concentrated KMnO4 will form (CH3)2C=O and HO2CCH2CH2COCH3 • Option C: Methanolic solution of NaBH4 only reduces carbonyl compounds. Hence the product will be (CH3)2C=CHCH2CH2CH(OH)CH3. Molecular formula is C8H16O. • Option D: Warming with alkaline aqueous iodine produces (CH3)2C=CHCH2CH2COO- and CHI3. Answer: A 27

What stabilises the α-helix of proteins?


A

disulfide bridges and hydrogen bonding

B

disulfide bridges only

C

hydrogen bonding only

D

van der waals’ forces only

Thinking process: • • •

The α-helix is a regular coiled configuration of the polypeptide chain. There are 3.6 amino acids per helical turn. The C=O is hydrogen bonded to the N-H associated with the fourth peptide linkage further down the chain.

Answer: C

28

After the reduction of nitrobenzene to phenylamine using tin and concentrated hydrochloric acid, an excess of sodium hydroxide is used. What is the purpose of the sodium hydroxide? A

to dry the product

B

to liberate the phenylamine

C

to lower the boiling point for subsequent distillation

D

to precipitate tin(II) hydroxide

Thinking process: • Recall: Tin, conc HCl, heat with reflux, followed by aqueous NaOH, room temperature

•

Addition of NaOH in step 2 is to react with the acidic ammonium salt, and liberate phenylamine.

Answer: B 29

Why are amides, RCONH2, less basic than amines, RNH2? A

Amides form a zwitterion in which the nitrogen atom carries a positive charge.

B

Amides have resonance structure involving the movement of a pair of electrons from the nitrogen atom to the oxygen atom. Electrons on the nitrogen atom move on to the C-N bond giving it some double bond character so that it is more difficult to break.

C


The amide carbonyl group withdraws electrons from the NH2 group to make the hydrogen atoms acidic. Thinking process: • Option A: Amino acids, not amides, form zwitterions • Option B: In amides, the lone pair of electrons on N atom is delocalised over the C=O bond. The presence of the highly electronegative O atoms decreases the electron density on N, further making the lone pair of electrons less available to accept H+. • Option C: This is a correct conclusion but it does not explain the concept of basicity D

Answer: B

30

When 2-methylphenylamine reacts with an excess of Br2(aq), one of the intermediates is cation Z.

2-methylphenylamine

Z

What is the final product of this reaction? A

B

NH2

C Br

D Br

Br

Thinking process: This is the typical reaction in which phenylamine react with aqueous Br2 under r.t.p conditions. The type of reaction is electrophilic substitution. Answer: A Section B 31

Which statements correctly describe the graphite lattice? 1

The lattice contains delocalised electrons.

2

Each carbon atom in the lattice has three closest neighbours.

3

The valency of each carbon atom in the lattice is 3.

Thinking process: H2 Chemistry 9647 2012

•

In graphite, each carbon atom is sp2 hybridised (each C has 1 unhybridised 2p orbital). Hence, out of the 4 valence electrons that C has, the 4th electron is not used in bonding but is delocalised parallel to the layer of graphite. Hence option A is correct. Note: The unhybridised 2p orbitals of the C atom overlaps to form a delocalised π-electron cloud. The π electrons act as mobile charge carriers to conduct electricity along the plane.

•

Valency is the number of bonds an atom can form.

•

Answer: B

32

A 1 mol dm-3 aqueous copper(II) salt is electrolysed between copper electrodes, using a constant current for 60 s. What affects the mass of copper deposited on the cathode? 1

decreasing the time taken

2

Increasing the concentration of the solution

3

the nature of the anion present

Thinking process: • • • • • •

33

The mass of copper formed is dependent on the quantity of charge passed. Q = It = neF Hence ne = It/F Mass of copper deposited is dependent on ne, which is in turn dependant on I (current) and t (time). Only option 1 is correct Answer: D

After the closure of a chemical plant in Switzerland, the reaction between 1-bromobutane and hydrogen sulfide ions (HS-) was found to be taking place in the ground. The following results were obtained. Initial concentration of 1bromobutane /mol dm-3 0.1 0.2 0.3 0.1 0.1

Initial concentration of hydrogen sulphide ions /mol dm -3 0.1 0.1 0.1 0.2 0.3

Initial rate of reaction /mol dm-3 s-1 1.5 x 10-5 3.0 x 10-5 4.5 x 10-5 3.0 x 10-5 4.5 x 10-5

Which conclusions can be drawn about the reaction? 1

The reaction is first-order with respect to 1-bromobutane.

2

Both 1-bromobutane and hydrogen sulfide ions are involved in the rate-determining step. The rate constant is 1.5 x 10-3 mol-1 dm3 s-1.

3

Thinking process: H2 Chemistry 9647 2012

Comparing expt 1 & 2: When [1-bromobutane] increases from 0.1 to 0.2 mol dm-3, keeping the [HS-] constant, the initial rate increases from 1.5 x 10-5 to 3.0 x 10-5 mol dm-3 s-1. Hence, order with respect to 1bromobutane is first order. [Option 1 is correct] Comparing expt 4 & 5: When [HS-] increases from 0.1 to 0.2 mol dm-3, keeping [1-bromobutane] constant, the initial rate increases from 1.5 x 10-5 to 3.0 x 10-5 mol dm-3 s-1. Hence, order with respect to HS- is first order. Thus, rate equation is rate=k[1-bromobutane][HS-]. Both 1-bromobutane and HS- are involved in the rate determining step. [Option 2 is correct] To find the rate constant, k : 1.5 x 10-5 = k(0.1)(0.1) K = 1.5 x 10-3 mol-1 dm3 s-1 [Option 3 is correct] Answer : A

34

The number of moles of chlorine that react with 1 mol of X is twice the number of moles of chlorine that react with 1 mol of Y. Which of these pairs could be X and Y? X

Y

1

Mg(s)

Na(s)

2

H2(g)

KBr(aq)

3

Cold NaOH(aq)

Hot NaOH(aq)

Thinking process: Mg(s) + Cl2(g) MgCl2(s) Na(s) + ½ Cl2(g) NaCl(s) Ratio of Cl2 needed = 1 : ½ = 2 : 1 (Option 1 is correct) H2(g) + Cl2(g) 2 HCl(g) KBr(aq) + ½ Cl2(g) KCl(aq) + ½ Br2(aq) Ratio of Cl2 needed = 1 : ½ = 2 : 1 (Option 2 is correct) 2 NaOH(aq) (cold) + Cl2(g) NaCl(aq) + NaClO(aq) + H2O(l) ⇒ NaOH(aq) (cold) + ½ Cl2(g) ½ NaCl(aq) + ½ NaClO(aq) + ½ H2O(l) 6 NaOH(aq) (hot) + 3Cl2(g) 5NaCl(aq) + NaClO3(aq) + 3H2O(l) ⇒ NaOH(aq) (hot) + ½ Cl2(g) 5/6 NaCl(aq) + 1/6 NaClO3(aq) + ½ H2O(l) Ratio of Cl2 needed = ½ : ½ = 1 : 1 (Option 3 is wrong) Answer : B 35

In which reactions does the oxidation state of chlorine change by one? 1 Electrolysis of brine (anode reaction) 2

Hydrogen + chlorine

3

Sodium chloride + concentrated sulfuric acid

Thinking process:


Electrolysis of brine : Cl- Cl2; oxidation no. increases by 1 (Option 1 is correct) Hydrogen + chlorine : H2 + Cl2 HCl; oxidation no. decreases by 1 (Option 2 is correct) Sodium chloride + concentrated sulfuric acid : NaCl + H2SO4 NaHSO4 + HCl; oxidation no. remains unchanged (Option 3 is wrong) Answer : B 36

Which properties of transition elements are not shown by s-block elements such as calcium? 1 They have variable oxidation states. 2

Their ions can act as oxidising agents in aqueous solution.

3

Their ionic radii are less than their atomic radii.

Thinking process: s-block elements do not have variable oxidation states. s-block elements have only one oxidation number. Eg. +1 for Group I elements; +2 for Group II elements (Option 1 is correct) The ions of s-block elements cannot act as oxidising agents in aqueous solution. The ions of s-block elements do not undergo reduction in aqueous solutions due to very negative Eored. Eg. Na+ Na (Eored = -2.71V; not feasible) (Option 2 is correct) The ionic radii of s-block elements are less than their atomic radii. The ions has one less occupied shell as compared to their atoms. (Option 3 is wrong) Answer : B Examiner’s comments: The majority of candidates gave D, rather than B, thinking that calcium ions can act as oxidising agents in aqueous solution. 37

During the preparation of many organic compounds by-products are formed. This usually occurs because the reagents can react in more than one way depending on the conditions used or because the products formed may react with the reactants. Propan-1-ol is produced by the reaction between 1-bromopropane and aqueous sodium hydroxide. CH3CH2CH2Br + NaOH CH3CH2CH2OH + NaBr What could be a by-product of this reaction? 1 CH3CH=CH2 2

CH3CH2CH2ONa

3

CH3CH(OH)CH3

Thinking process: ** It is important that students read the question carefully especially on the part “the products formed may react with the reactants” Although aqueous sodium hydroxide is used ie. the predominant reaction is nucleophilic substitution, a possible competing reaction ie. elimination could still occur. Thus, alkene could be a by-product. (Option 1 is correct) The product, CH3CH2CH2OH formed, does not react with NaOH. (Option 2 is wrong) H2 Chemistry 9647 2012

In order for CH3CH(OH)CH3, the initial reactant has to be CH3CH2BrCH3.(Option 3 is wrong) Answer : D

38 The mould Phytophthora damages many plants, destroying agricultural crops such as potatoes. A hormone-like compound called alpha 1 regulates the reproduction of all species of Phytophthora. The structure of alpha 1 is now known, giving scientists a key to the possible future eradication of the mould.

Which will react with alpha 1? Cl2(g) / light 1 2

SOCl2

3

H2 / Pt

Thinking process: Identify the functional groups present in alpha 1 :alcohol, ketone Cl2(g) / light reacts with alkane chain via FRS SOCl2 reacts with the alcohol via nucleophilic substitution H2 / Pt reacts with ketone to form secondary alcohol via reduction Answer : A

39

Bromoethene, CH2=CHBr, is unreactive to nucleophiles CH2=CHCH2Br, is very reactive by comparison.

whereas

3-bromopropene,

What could be a reason for the lack of reactivity of CH2=CHBr? 1

The electrons of the bromine atom delocalise into the π bond.

2

The electrons of the π bond repel the attacking nucleophile.

The presence of the π bond prevents free rotation of the C-Br bond thus decreasing the reactivity. Thinking process: 3

In CH2=CHBr, the p orbital of the Br can overlap with the π bond, causing the electrons of the Br atom to delocalise into the π bond. This decreases the partial positive charge on the carbon. Hence, the carbon is less susceptible for nucleophilic attack. Also, the overlapping of the p orbital of the Br atom causes the C-Br bond to have double bond character, making the bond stronger. (Option 1 is correct)


The π electrons of the C=C bond are electron rich and it repels the attacking nucleophiles are nucleophiles are also electron rich. (Option 2 is correct) The presence of π bond which prevents free rotation does not affect the reactivity. (Option 3 is wrong) Answer : B Examiner’s comments: Unfamiliar with the reaction mechanism details 40

In an industrial process, heptane vapour is passed over a heated catalyst to make methylbenzene.

CH3CH2CH2CH2CH2CH2CH3

CH3

+ 4H2

Under similar conditions, which of the C8H18 isomers could give 1,4-dimethylbenzene?

1

2

3

Thinking process: To form methylbenzene, six carbons (CH3CH2CH2CH2CH2CH2-) are linked up to form the benzene ring.


For Option 1:

(correct)

For Option 2:

(correct) For Option 3,

(wrong) Answer : B Examiner’s comments: Not confident in deducing the organic reactant


Paper 2 Please use the table format for our internal exams for keying in the 2012 questions. Formatting wise, not so strict, just try your best =) You can use this table for a start. • Please type in your relevant question and answer, and assign marks or ticks whenever you can. We can refine it when the examiner’s report is out. • Please also insert a border around the solution, for ease of reading/finding the answer. • Please also leave a row of one line spacing between the answer and the next question. • Thank you! ☺ 1

When an acidified solution of hydrogen peroxide is mixed with a solution containing potassium iodide, starch and sodium thiosulfate, the colourless mixture suddenly turns dark blue after a few seconds. This is one version of what are known as iodine clock experiments. A teacher demonstrates this iodine clock experiment as follows. She prepares two separate solutions, A and B. Solution A

contains starch (0.20 g), potassium iodine (50.00 g), sodium ethanoate (4.10 g) and sodium thiosulfate (9.40 g) in 1.0 dm3 water.

Solution B

is a mixture of 500 cm3 of 1.5 mol dm-3 hydrogen peroxide and 500 cm3 of 1.0 mol dm-3 ethanoic acid.

She mixes 50 cm3 of solution A with 50 cm3 of solution B and, after about 20 seconds, the mixture suddenly turns dark blue.

(a)

In acidic solutions, hydrogen peroxide and potassium iodide react to produce iodine. Write an ionic equation for this reaction. [1] H2O2 + 2H+ + 2I- I2 + H2O [1m]

(b)

Suggest why the reaction mixture contains both ethanoic acid and sodium ethanoate. [1] To form a buffer solution that keeps the [H+] relatively constant. [1m]

(c)

Consider the description of the iodine clock experiment given above. The order of the reaction with respect to the concentration of hydrogen peroxide may be determined by performing a number of experiments and then graphically analysing the results. Write a plan for such a series of experiments. In your plan, you should use the same proportions of solution A and aqueous ethanoic acid as are described above. Your plan should ensure that at least one of your experiments would be expected to


take significantly less than 20 seconds. You may assume that you are provided with the following. • • • • •

solution A, having the concentrations of reagents given above 1 mol dm-3 ethanoic acid 7.5 mol dm -3 hydrogen peroxide solution deionised water the apparatus normally found in a school or college laboratory

Your plan should contain the following. • • • • •

the intended concentration of the stock solution of hydrogen peroxide to be used details for the preparation of this stock solution from the hydrogen peroxide solution provided appropriate quantities and concentrations of solutions all essential experimental details an outline of how the results would be used to determine the order with respect to the concentration of hydrogen peroxide

You may find it useful to use

1 to represent the reaction rate. time [9]

Dilution required to dilute 7.5 mol dm-3 hydrogen peroxide solution to 1.5 mol dm-3 From 7.5 mol dm-3 hydrogen peroxide solution to 1.5 mol dm -3 , it’s dilution by 5 times So using a 250 cm3 volumetric flask, burette out 50 cm3 of 7.5 mol dm -3 hydrogen peroxide solution to do dilution by 5 times. Preparation of 1.5 mol dm-3 stock solution 1

Using a 50 cm3 burette, measure 50 cm3 of 7.5 mol dm -3 H2O2 solution.

2

Transfer this H2O2 solution & its washings into a 250 cm3 volumetric flask.

3

Make up to the mark using distilled water.

4

Stopper, invert, shake to ensure homogeneous solution.

Experiment Keeping total volume of mixture at 100 cm3 (because Question say so above: “She mixes 50 cm3 of solution A with 50 cm3 of solution B and, after about 20 seconds, the mixture suddenly turns dark blue”) vol of solution A at 50 cm3 & volume of CH3COOH at 25 cm3, max vol of H2O2 in mixture = 25 cm3. 1

Use a 25 cm3 measuring cylinder, measure 25 cm3 of H2O2 into a 250 cm3 conical flask.


2

Use a 50 cm3 measuring cylinder, measure 50 cm3 of solution A into the same conical flask & start the stop watch immediately.

3

Stop the stop watch when the colourless mixture suddenly turns dark blue.

4

Record the time in Table 1.

5

Calculate the reaction rate.

6

Repeat step 1 to 5 for reaction mixture 2 to 5 using another 25 cm3 measuring cylinder to measure the volume of water required & pour it into the conical flask before step 2. Ensure that the total volume of mixture is 100 cm3.

Reaction mixture

Vol of Vol of Vol of Vol of Time take solution H2O2/ CH3COOH water solution to A/ cm3 cm3 / cm3 / cm3 dark blue /s

1 2 3 4 5

50 50 50 50 50

25 20 15 10 5

25 25 25 25 25

for Reaction turn rate, 1/t (s-1)

0 5 10 15 20

Determine the order wrt [H2O2] 1

Plot a graph of reaction rate against volume of H2O2 used.

2

Check the graph against table 2 to obtain the order wrt H2O2.

Zero order

s-1 Rate /mol dm-3 s-1

1st order

2nd order -1

–3 –1 s dm s Rate/ mol

Rate / mol s-1dm–3 s–1

k Rate = constant

slope = k

3 Vol [A] H2 O 2/ cm mol dm-3

(d)

[A] /3mol Vol H2O2/Cm

dm–3

3 –3 mol dm Vol[A] H/2O 2/ cm

Identify one potential safety hazard in this experiment and state how you would


minimize this risk. [1] Iodine is eye-irritant, wear goggles. OR Ethanoic acid or concentrated hydrogen peroxide solution used initially is very high concentration & corrosive, wear gloves.

2

Fossil fuels which re found in deposits beneath the ground have been formed from plants and animals that lived up to 300 million years ago. When fossil fuels are burned in internal combustion engines, CO2 and H2O are the main products, with NO and SO2 also being formed. (a)

(i)

Which class of compound, present in fossil fuels, produces CO2 and H2O on combustion? Hydrocarbons [1] Examiner’s Comment: This was generally well answered with the majority of candidates referring to alkanes or hydrocarbons.

(ii)

Briefly explain how NO is formed in an internal combustion engine. N2 and O2 in the air react under high temperature in the combustion engine.[1] High temperature provides sufficient energy to overcome the strong N≡N. Limited Must emphasise to students to read self –read What is in the fuel (alkanes) must go through Examiner’s Comment: While many candidates understood that in the car engine nitrogen from the air will react with oxygen to form NO, some believed that the nitrogen came from organic nitrogen compounds present in the fuel.

(iii)

Suggest which class of compound, originally present in the living matter that produced the fossil fuel, contained the sulfur that produces SO2 on combustion. Protein / Amino acid [1] Disulfide linkages R group Examiner’s Comment: This part was less well answered with relatively few candidates correctly stating that the sulfur comes from proteins or amino acids that are present in the hydrocarbon fuel. [3]

Most of the sulfur present in fossil fuels is removed at oil refineries. It is used to produce SO2 which is converted, in the Contact process, into SO3 which is used for the manufacture of sulfuric acid. The Contact process involves the oxidation of SO2 using a vanadium (V) oxide catalyst. H2 Chemistry 9647 2012

(b)

(i)

SO2(g) + ½ O2(g) SO3(g) reaction 1 At 298K, ∆Hfº(SO2) = -297 kJ mol-1 and ∆Hfº(SO3) = -396 kJ mol-1. Use these data to calculate ∆Hº for reaction 1. State the units. Weak in hess law and practice Burn in O2 combustion Memorise formula without understanding BE of O2 More eyes to look through tutorial ∆Hrθ =

∑ n∆ H

θ f

(products) –

∑ m∆ H

θ f

(reactants)

∆Hº = -396 – (-297) = -99.0 [1] Units : kJ mol-1 [1] Examiner’s Report: This was generally well answered. The most common incorrect answer included the bond energy of the O2 molecule in the calculation. (ii)

For reaction 1, ∆Sº = -93.5 JK-1mol-1 at 298K. Use this value and your answer to (i) to calculate ∆Gº for reaction 1 at 298K. State the units. ∆G θ = ∆H θ − T∆S θ ∆Gº = -99 –(298) (-93.5 x 10-3) = -71.1 [1] Units: kJ mol-1 [1] Examiner’s Comment: While there were many correct answers to this part, a significant number of candidates did not appreciate that their ∆Ho was in kJ mol –1 while ∆So was in J K –1 mol –1. [4]

For any reaction, ∆Gº and the equilibrium constant Kc, are related according to the equation ∆Gº = -2.303 RT log Kc. (c)

(i)

Use this equation to calculate a value of Kc for the oxidation of SO2 at 298K. ∆Gº = -2.303 RT log Kc (-71.1 x103 ) = -2.303 (8.31) (298) log Kc log Kc = 12.4669 Kc = 2.93 x 1012 [1] Examiner’s Comment: This was less well answered than the previous calculation.

(ii)


Suggest what is the significance of the magnitude of your answer in (i).

The magnitude of Kc indicates that the reaction reaches a state of equilibrium where both reactants and products are present in significant amount[1]. Examiner’s Comment: Many candidates found it difficult to explain the significance of the magnitude of their value of Kc. A significant number of those who correctly calculated Kc simply said that ‘the equilibrium position lies towards the right’. Few were able to explain clearly that the very large numerical value indicated that the reaction would be effectively complete. Answers often contained irrelevant material with reference to the relative rates of the forward and backward reactions or to the energetic stability of SO3. Many of those who calculated the value of Kc to be 1.03 also stated that ‘this value shows that the equilibrium position lies to the right’. Very few candidates made any attempt to explain that this value indicated that significant quantities of both reactants and products would be present. (iii)

Suggest why SO2 and not SO3 is formed when sulfur is heated in air. The activation energy for the conversion of SO2 to SO3 is high.[1] Examiner’s Comment: Answers to this part question were often confused. Incorrect answers included reference to the difference in oxidation number of sulfur in SO2 and SO3, the effect of temperature on the value of Kc, and the relative rates of the forward and backward reactions. Candidates needed to concentrate on the focus of the question; why SO2 and not SO3 is formed when sulfur is burned in air. Although the answers to part (iv) demonstrated that candidates knew that the Contact Process requires a catalyst, they did not apply this knowledge to suggest that the activation energy for the formation of SO3 is very high.

(iv)

Use your answer to (iii) to suggest why a catalyst is used industrially in the Contact process. Explain your answer. A catalyst increases the rate of reaction[1] by providing an alternative pathway of lower activation energy[1], allowing the equilibrium to be established faster[1]. Examiner’s Comment: This part was more successfully answered by many candidates, who knew that catalysts speed up reactions. Most of them were able to explain that catalysts provide a reaction with a lower activation energy.

3

Fifty years ago, students of A-level chemistry carried out ‘qualitative inorganic analysis’ in their practical sessions. This method included precipitating cations as chlorides, sulfides, etc., and used the concept of solubility product, Ksp. (a)

(i)

Write an expression for Ksp of PbCl2, stating the units. Ksp = [Pb2+(aq)][Cl–(aq)]2

(ii)


[1m]

The solubility of PbCl2 is 4.7 gdm-3. Calculate the solubility product of PbCl2. Give your answer to two significant figures.

units: mol3dm-9

[1m]

Solubility of PbCl2 =

4.7 = 0.169 mol dm-3 [1m] 207.0 + 2(35.5)

Ksp = (0.169)(2 x 0.169)2 [1m] = 1.93 x 10-5 mol3 dm-9 [1m] [5] Examiners’ Comments: Most candidates answered this correctly, with calculations set out clearly. The most common mistake seen was the expression of the answer to the incorrect number of significant figures. When 2.0 moldm-3 hydrochloric acid is added to separate samples of aqueous silver nitrate and aqueous lead(II) nitrate, AgCl and PbCl2 are precipitated. (b)

(i)

When the HCl(aq) is added to AgNO3(aq), what will be seen in the container immediately: white precipitate [1m] Examiners’ Comments: This part was well-answered.

after one hour in daylight: grey solid/ precipitate [1m] Examiners’ Comments: Very few candidates could suggest what happens when this precipitate is left exposed to daylight. (ii)

How would your confirm that the original precipitate was AgCl by adding a single reagent to it? Describe what you would see and give an equation with state symbols for the reaction that occurs with the reagent you have chosen. Add aqueous ammonia. [1m] White precipitate dissolves in aqueous ammonia. [1m] AgCl(s) + 2NH3(aq)

[Ag(NH3)2]+(aq) + Cl–(aq) [1m, with state symbols]

Examiners’ Comments: The use of NH3 (aq) to confirm the identity of AgCl was known by about half of the candidates, although many of their responses did not give a correct equation for the corresponding reaction. (iii)

When concentrated hydrochloric acid is added to the precipitate of PbCl2, the precipitate dissolves. Suggest, with an ionic equation, an explanation for this observation. PbCl2(s) + 2Cl–(aq)

[PbCl4]2–(aq) [1m]

When concentrated HCl(aq) is added, a soluble complex ion [PbCl4]2–(aq) is formed. [1m] Examiners’ Comments: Only about one third of the candidates was able to suggest what happens in the reaction between HCl(aq) and a precipitate of PbCl2. [7] H2 Chemistry 9647 2012

When concentrated hydrochloric acid is added to aqueous copper(II) chloride, the solution eventually becomes yellow. When water is added to this yellow solution, it becomes blue. Suggest by means of ionic equations, an explanation for these observations. [3]

(c)

[Cu(H2O)6]2+(aq) + 4Cl–(aq) blue

[CuCl4]2–(aq) + 6H2O(l) [1m] yellow

When concentrated HCl(aq) is added, [Cl–] increases. Position of equilibrium shifts to the right [√], resulting in ligand exchange [√] of H2O to Cl–, forming yellow CuCl42– solution. When water is added, position of equilibrium shifts to the left [√], resulting in ligand exchange [√] of Cl– to H2O, forming blue [Cu(H2O)6]2+(aq). Any 2 [√] – 1m Examiners’ Comments: There was a wide range of responses to this part. Some explained very clearly what happens when concentrated HCl is added to aqueous copper(II) chloride. A significant number did poorly. Selective precipitation of sulfides may be used to separate the cations present in a solution containing Cu2+, Ni2+ and Zn2+. The reagent used is hydrogen sulfide which behaves as a dibasic (diprotic) weak acid when in aqueous solution. H+(aq) + HS–(aq)

H2S(g) + aq HS–(aq)

and

Ka = 9.1 x 10-8 moldm -3

H+(aq) + S2–(aq)

Ka = 1.0 x 10-15 moldm-3

Metal sulfides are precipitated by the following reaction. M2+(aq) + S2–(aq)

MS(s)

Relevant Ksp values are given in the table. salt CuS NiS ZnS

Ksp/ mol2dm-6 6.3 x 10-36 4.0 x 10-21 1.6 x 10-24

The results of passing hydrogen sulfide through separate solutions of Cu2+, Ni2+ and Zn2+ are shown in the table below. M

2+

being tested

2+

Cu 2+ Ni 2+ Zn

(d)

H2S passed through an acidic solution of M2+ black precipitate of CuS No precipitate No precipitate

H2S passed through a neutral solution of M2+ black precipitate of CuS No precipitate white precipitate of ZnS

H2S passed through an alkaline solution of M2+ black precipitate of CuS black precipitate of NiS white precipitate of ZnS

Use the data on the opposite page to give an explanation for each of the following questions. You should give equations where appropriate. You are not expected to do any calculations in this part.


(i)

Why is CuS the only sulfide of these three precipitated in acid solution? In acid solution, [H+] is high. [S2–] is low [√]. Since Ksp of CuS is the lowest, the ionic product of CuS > Ksp of CuS [1m]. However, the ionic product of NiS and ZnS < Ksp of NiS and ZnS respectively [√]. Hence, only CuS precipitates. 2 [√] – 1m

(ii)

Why is NiS precipitated in alkaline solution and not in neutral or acidic solution? In alkaline solution, [H+] is low. [S2–] is high [√]. Ksp of NiS is highest. Only in alkaline solution where the [S2–] is high, the ionic product of NiS > Ksp of NiS [1m]. In both acidic and neutral solution, [S2–] is not high enough for the ionic product of NiS to exceed Ksp of NiS [√]. 2 [√] – 1m

(iii)

Why is ZnS precipitated in both neutral and alkaline solution? [S2–] in both neutral and alkaline solution is sufficient for the ionic product of ZnS to exceed Ksp of ZnS [1m], hence precipitation occurs. However, the Ksp value is high enough such that in acidic solution, [S2–] is insufficient for the ionic product of ZnS to exceed Ksp of ZnS [1m], hence ZnS does not precipitate in acidic solution. Examiners’ Comments: In this part too there was a wide range of responses. In some cases candidates used the data very well and gave clear explanations of what happens in each of the three examples. In contrast, others found it difficult to give an explanation. This was usually because they did not use the data to determine the relative values of [S2–] in acidic and alkaline solution and how this would affect precipitation of each of the metal sulfides. [6] [Total:21]

4

Chlorogenic acid occurs naturally in coffee and an edible species of bamboo.


chlorgenic acid The chlorogenic acid molecule contains a saturated ring, an aromatic ring and an alkene group. (a)

(i)

Identify as fully as you can the other functional groups present in chlorogenic acid. Ester [1], secondary alcohol [1], tertiary alcohol [1], phenol [1], carboxylic acid [1].

(ii)

How many chiral carbon atoms are there in a molecule of chlorogenic acid? 4 chiral carbons [1] [6]

(b)

When chlorogenic acid is reacted under suitable conditions with NaOH(aq), followed by acidification, two crystalline solids may be obtained, one non-aromatic and one aromatic. Draw the displayed formula of each of these compounds in the boxes below.

[1] non-aromatic compound

Aromatic compound R [1] (Note: need to draw out the “C” atom on the alkene)

Examiners’ comments: Most common mistake was to draw the sodium salts of the carboxylic acids! Question want the structure of the cpds itself, NOT when they have reacted with NaOH! [2] (c)

Draw the structural formula of the compound formed with aromatic compound R you have drawn in part (b) is added to an excess of Br2 dissolved in an organic solvent.

OR Note the difference in drawing the alkene portion [1] alkene group reacting [1] phenol group reacting


Examiners’ comments: Many students did not notice that the side-chain alkene can react too! [2] (d)

When the aromatic compound R is reacted with an excess of NaHCO3(aq), compound S is formed which has the molecular formula C9H7O4Na. (i)

Suggest the structural formula of S. Thinking process: NaHCO3 is a weak base can only react with stronger acids such as carboxylic acids. Phenol is a weaker acid than carboxylic acid hence cannot react with NaHCO3.

[1] (ii)

Suggest why this product contains only one Na atom/ion although the NaHCO3 is in excess. NaHCO3 is a weak base can only react with stronger acids such as carboxylic acids. [1] Phenol is a weaker acid than carboxylic acid [1] hence cannot react with NaHCO3. 1 mole of carboxylic acid group can only react with 1 mole of NaHCO3. Note: Must explain in terms of difference in acid strength of phenol and carboxylic acid. [3] [Total: 13]

5

Lactic acid (2-hydroxypropanoic acid), CH3CH(OH)CO2H, is present in milk. Lactic acid can also be produced in the laboratory from ethanal, CH3CHO, by the following reactions.

(a)

(i)

Name the reagent and conditions used and the type of reaction that occurs in step 1. Reagent …………………………. Conditions ………………………. Type of reaction ……………………… Reagent : HCN [1] Conditions : trace amount of NaOH, 10-20oC [1] Type of reaction : nucleophilic addition [1] Examiners’ comments: Most candidates knew the relevant chemistry for the production of lactic acid.


(ii)

Describe the mechanism of this reaction. In your answer you should show all charges and lone pairs and show the movement of electrons by curly arrows. Name of mechanism : Nucleophilic addition HCN

H+ +CN-

H+ + OH-  → H2O

H H3C

Partial charges & arrow

δ+ C

δ−

H

slow

O

H3C

C CN

Lone pair & arrow

OProduct

CN Partial charges & arrow Slow & fast

Lone pair & arrow

Both products

6 to 8 [2] 3 to 5 [1] 0 to 2 [0] Examiners’ comments: Some struggled to describe the mechanism. Some answers did not gain credit because candidates omitted to check that they had included relevant charges and lone pairs. One other common incorrect answer was not regenerating the CN– ion at the end of the reaction sequence. [5] Lactic acid present in milk rotates the plane of polarised light. Lactic acid produced in a laboratory by the reactions above does not rotate the plane of polarised light. (b)

(i)

What feature of the lactic acid present in milk causes the rotation of the plane of polarised light? The presence of chiral carbon in lactic acid causes the rotation of the plane of polarised light with only one optical isomer present [1] Examiners’ comments: Candidates performed poorly in part (i) Very few candidates were able to explain clearly and unambiguously that lactic acid present in milk rotates the plane of polarised light because all of the molecules present are of the same chirality, that is, only one enantiomer is present.

(ii)

Suggest, in terms of the mechanism of step 1 in the above reaction scheme, and the shape of the CH3CHO molecule, why synthetic lactic acid does not rotate the plane of polarised light. As the nucleophile CN- can attack the planar carbonyl carbon from either side of the


plane with equal chance [1] , it will result in the formation of a racemic mixture [1] of the products. The optical activity of the 2 optical isomers cancel out, hence synthetic lactic acid shows no optical activity [1] (ie. does not rotate the plane of polarised light).s

[4] The effect of plane polarised light on tartaric acid (2,3-dihydroxybutanedioic acid), HO2CCH(OH)CH(OH)CO2H was investigated by Louis Pasteur.

Pasteur identified three different types of tartaric acid molecule. molecule A rotated plane polarised light to the right molecule B rotated plane polarised light to the left molecule C had no effect on plane polarise light (c)

Suggest an explanation for these observations. molecule explanation A B C Thinking process: • This question is talking about meso compounds (optically INACTIVE cpds) • Recall the criteria to be a meso cpd is: o Have at least 2 chiral centres o Have an internal plane of symmetry molecule A

explanation

Molecule A has 2 chiral carbons, and the compound is asymmetrical (ie. no internal line of symmetry) [1] Hence it is optically active. B


Molecule B is the mirror image of compound A. Hence, it rotates plane polarised light in the opposite direction of molecule A. [1] C

Or

Molecule C has 2 chiral carbons, but the compound is symmetrical (ie. has internal line of symmetry). The rotating capacity of one isomer cancels off the other. (Molecule C is a meso compound) [1] Examiners’ comments: The performance of candidates for this part was very poor. Many answers attempted to explain the chirality of tartaric acid in terms of just one of the chiral carbon atoms present in the molecule. Such answers usually stated that this carbon atom alone was responsible for the different rotations observed with molecule A and molecule B. Having got this far the majority of answers then simply said that there was no chiral carbon atom present in molecule C, despite the structural formula of tartaric acid being given in the question. [3] [Total: 12]


Paper 3 1

(a)

The acid-base behaviour of aluminium oxide, Al2O3, shows similarities to that of magnesium oxide, MgO, on the one hand, and phosphorus(V) oxide, P4O10, on the other. Describe what these similarities are, and explain why aluminium oxide occupies the in-between position. Write equations for all the reactions you choose to illustrate your answer. [5] Aluminium oxide behave as an acid like that of covalent oxide P4O10, to give salt & water Al2O3 (s) + 2 NaOH (aq) + 3H2O (l) 2NaAl(OH)4 (aq) P4O10 + 12NaOH 4Na3PO4 + 6H2O Aluminium oxide behave as a base like that of ionic oxide MgO, to give salt & water Al2O3 (s) + 3H2SO4 (aq) Al2(SO4)3 (aq) + 3H2O (l) MgO + H2SO4 MgSO4 + H2O

Aluminium oxide has a giant ionic structure with strong ionic bonds between the Al3+ & O2- ions.

high charge density of Al3+ can polarise the electron cloud of O2- to some extent to give rise to some covalent character.

Hence aluminium oxide is amphoteric,.showing both acidic & basic, properties [9 ‘ √‘ – 5 mk, 7-8 ‘ √‘ – 4 mk, 5-6 ‘ √‘ – 3 mk, 3-4 ‘ √‘ – 2 mk, 2 ‘ √‘ – 1 mk] (b)

The following equilibrium exists in a sample of aluminium chloride vapour. Al2Cl6(g) (i)


2AlCl3(g)

∆H = +63 kJ mol-1

Draw a dot-and-cross diagram of the Al2Cl6 molecule including its co-ordinate (dative covalent) bonds.

2 dative bonds [1m] 6 Al-Cl dot-and-cross bond [1m] (ii)

Suggest the shape of the AlCl3 molecule. Shape of AlCl3 molecule: trigonal planar (3 bond pair, 0 lone pair) [1m]

(iii)

Predict the effect of increasing the temperature on the above equilibrium. Explain your answer. When temperature is increased, the above equilibrium position will shift to right side to favour endothermic reaction, to absorb the extra heat. Dimer of aluminium chloride will dissociate to its monomer, AlCl3 molecule. [2 ‘√ ‘ – 1 m]

When 1.50 g of aluminium chloride was introduced into an evacuated flask of 250 cm3 capacity and heated to 500 K, the pressure inside the flask rose to 1.16 x 105 Pa. (iv)

Assuming the gaseous mixture behaves ideally, use the general gas equation to calculate the average Mr of the mixture. pV = nRT

(1.6 × 10 )( 250 × 10 ) = 1.50 M ( 8.31)( 500 ) 5

−6

r

Average Mr = 214.9 (v)

Using the following relationships, calculate the mole fraction of Al2Cl6, x, and the mole fraction of AlCl3, y, in the mixture. x+y=1 Mr(average) = 267x + 133.5y Rearrange: x = 1 – y 214.9 = 267(1-y) + 133.5y y = 0.390


x = 1 – 0.390 = 0.610 Hence calculate the partial pressures of Al2Cl6 and AlCl3 in the mixture.

(vi)

0.610 (1.16 × 105 ) = 7.08 x 104 Pa 1

Partial pressure of Al2Cl6 =

Partial pressure of AlCl3 = 0.390/1 (1.16 x 105) = 4.52 x 104 Pa (vii)

Write an expression for Kp for the reaction, and calculate its value. Include units in your answer. [11]

(PAlCl ) = ( 4.52 × 10 ) = 2

Kp

PAl Cl 2

(c)

4

3

6

7.08 × 104

2

= 2.89 x 104 Pa

Aluminium chloride is used as a catalyst in electrophilic substitution reactions. The chlorination of benzene is represented by the following overall equation.

The reaction occurs in several steps. •

The first step is the reaction between Cl2 and AlCl3. Cl2 + AlCl3 Cl+ + AlCl4−

•

The benzene ring is then attached by the Cl+ cation in the second step.

AlCl3 reacts in a similar way with acyl chlorides, producing a carbocation that can then attack a benzene ring. (i)


Predict the structure of the product of each of the following reactions. Both compounds A and B react with 2,4-dinitrophenylhydrazine.

Compound A:

Compound B:

(ii)

Suggest the reagents and conditions for a reaction that could be used to distinguish between A and B. you should state how each of A and B react. Draw the structures of the products of any reactions that occur. [4] [Total: 20] Add I2(aq) & NaOH(aq) to A and B separately & warm.[1m] Observation: Compound A: yellow ppt [1m] Compound B: no yellow ppt

Products of compound A: CHI3 [1m] and

2

(a)

[1m]

Chlorine and iodine differ in their relative reactivity as oxidising agents. Chlorine is a gas under standard conditions. Although iodine is not very soluble in water, it is freely soluble in KI(aq), according to the following equilibrium.

I-(aq) + I2(s)

I3-(aq)

A solution containing I3-(aq) reacts as if it were a solution of iodine in water. (i)


Draw a fully labeled diagram of the electrochemical cell you would set up in order to measure the relative oxidising powers of chlorine and iodine under standard conditions.

Pt

Cl2(g)

-

I3- ions and I- ions

Cl (aq)

Concentration = 1 mol dm-3

Pt

Salt bridge [1] voltmeter [1] Half cell of Cl2/Cl- with correct electrode and ions of correct concentration [1] Half cell of I3-/I- with correct electrode and ions of correct concentration [1]

Examiner’s Comment: • Despite the question asking for only one electrochemical cell, many candidates chose to draw two cells, each with a standard hydrogen electrode on one side. • Few candidates are able to describe correctly the iodine electrode (a Pt electrode dipping into a mixture of 1 moldm-3 I–(aq) and I2(s) or I3 –(aq)). Many suggested a solid electrode made of (non conducting) iodine, or suggested passing I2(g) over a Pt electrode. • Other incorrect answers were the omission of the salt bridge, or the inclusion of a battery in the circuit. (ii)

Describe the measurements you would make, and explain their significance. First measurement: The electromotive force of the cell [√] at zero current is measured. E ∅cell = E ∅(reduction half cell) – E ∅(oxidation half cell) A positive cell potential indicates that the reaction is spontaneous[√]. Second measurement: Titrate the I3- solution against S2O32- to determine the concentration of I3- [√] after some time. If the concentration of I3- falls below 1 mol dm-3, it will indicate that I2 has been reduced and I3-/ I- half cell is the cathode. [√] 2 [√] – 1 mark Alternative: - Using colorimeter to measure the colour intensity of I2. The increase in the intensity of the red-brown colour indicates that I- has been oxidised to I2.


Examiner’s Comment: • Candidates were expected to suggest measuring the Eo of the cell, and to explain that the more positive of the two electrodes would contain the stronger oxidising agent. • Candidates should be aware of the difference between ‘more positive’ and ‘larger’ or ‘bigger’. (iii)

Write a balanced equation for the reaction that would take place if the electrode of the cell were connected together by an external circuit. Cl2(g) + 2I-(aq) 2Cl-(aq) + I2(aq) [1] Examiner’s Comment: Most candidates who answered (i) in terms of a single cell gained credit here. [7]

(b)

When solutions of I-(aq) and Fe3+(aq) are mixed, an equilibrium is set up. (i)

Write a balanced equation for this equilibrium. 2Fe3+ + 2I-

2Fe2+ + I2 [1]

Examiner’s Comment: • Some candidate suggested incorrectly the ‘Ksp’-like equilibrium: 3I–(aq) + Fe3+(aq) ⇌ FeI3(s) • Others suggested that I- could reduce Fe3+ to Fe metal.

(ii)

Use the Data Booklet to suggest how the position of equilibrium, involving I- and Fe3+, would change in the presence of CN-(aq) ions. Explain your answer. Fe3+ + e [Fe(CN)6]3- + e

Fe2+ [Fe(CN)6]4-

E∅ = +0.77 V E∅ = +0.36 V[√]

In the presence of CN-, E∅ value is more negative[√], hence equilibrium position lies more to the left[√], this means iron (III) (as [Fe(CN)6]3-) is predominant and more stable[√]. There is less tendency for it to be reduced to the +2 oxidation state. 2[√] – 1 mark Examiner’s Comment: • The majority of candidates gave answers written in terms of the CN– ions causing the [Fe3+] to decrease, and thus moving the equilibrium to the left. Thes candidates did not appreciate that the CN– ion forms complexes with both Fe3+and Fe2+and therefore adding CN– to a solution containing Fe3+ and Fe2+ will cause the concentrations of both cations to decrease. • Only a minority of candidates made use of the Eo value for the [Fe(CN)6]3-/[Fe(CN)6]4- electrode (Eo = +0.36V) in the Data Booklet to correctly explain that this indicates that the Fe(III) is less oxidising than I2 (Eo = +0.54V) under standard conditions. The equilibrium will still move to the left hand side, but this is caused by the decrease in oxidising power of the complexed Fe(III). [3] I

Using monohalogenoethanes, C2H5X, as examples, describe and explain the relative


reactivities of chloro- and iodo-compounds with respect to hydrolysis. Rate of hydrolysis is dependent on the bond strength of the C – halogen bond. The C – halogen bond becomes progressively stronger from C – I to C – Cl[√] due to decreasing bond length[√]. Hence, ease of breaking of the C – halogen bond decreases in the order: C – I > C – Cl. The rate of hydrolysis decreases in the order: C2H5-I > C2H5-Cl. 2[√] – 1 mark Examiner’s Comment: • Most candidates gained credit for this application of the decrease in C–X bond strength down the group. • Some candidates incorrectly compared the H–X rather than the C–X bond energies. [1] (d)

The different reactivities of organic halogeno compounds, and of amines and alcohols, can be exploited in successive reactions of dihalogeno compounds Predict the outcomes of the following transformations, drawing the structures of the intermediates C and E and the products D and F. Both D and F are cyclic compounds. (i)

C:

D:

[1]

[1]

Examiner’s Comment: A large number of candidates correctly predicted the outcomes of these two reactions: the C–I bond is hydrolysed first, giving the chloroalcohol C, which is then aminolysed to the aminoalcohol D. A frequent incorrect answer was the inclusion of an –NH3 group rather than an –NH2 group in the aminoalcohol.


(ii)

O H2 C

C

E:

F:

CH2Cl

N H

OH C H2

[1]

[1]

Examiner’s Comment: • Fewer candidates achieved full credit for this part; often suggesting that E was the chloroester rather than the chloroamide: they did not relate the higher basicity of amines compared to alcohols to their higher nucleophilicity. • Often the structure of F was given as the alkoxide anion, but better responses gave its structure as the cyclic amide-ether. [4] (e)

(i)

Describe the conditions needed to hydrolyse amides in the laboratory. Acidic hydrolysis: HCl(aq) or H2SO4(aq), heat with reflux[1] OR Alkaline hydrolysis: NaOH(aq), heat with reflux[1] Examiner’s Comment: Most candidates answered this question well.

(ii)


Predict the products of the hydrolysis of the amide groups in the following two drugs.

Acidic hydrolysis: Phenylbutazone OH

O

[1]

OH

O

[1]

Phenobarbital

[1] NH3 and CO2[1] Examiner’s Comment: This was also a well answered part question. Whilst most candidates suggested the correct 2-ethyl,2-phenylpropanedioic acid as one of the products from the hydrolysis of phenobarbital, fewer realised that any urea formed would also be hydrolysed to NH3 (or NH4+) and CO2 (or CO3 2– - but not H2CO3, which would decompose to CO2 + H2O). Candidates need to take care to write the correct H2 Chemistry 9647 2012

ionic structures to fit with the hydrolysing agent (acid or alkali). [5] [Total: 20]

3

(a)

When heated, the carbonates of the Group II elements can lose carbon dioxide. MCO3(s)

MO(s) + CO2(g)

Under the influence of the metal cation, the carbonate ion is thought to break up as follows.

M2+

M2+

O2–

+

The nitrates of the Group II elements also decompose on heating. (i)

Write an equation for the decomposition of Mg(NO3)2. Mg(NO3)2(s)

MgO(s) + 2NO2(g) + ½O2(g) [1m]

Examiners’ Comments: This part is well-answered by most candidates. (ii)

State and explain the trend observed in the thermal stability of the Group II nitrates. Down the group, • radius of cation increases • charge density of cation decreases [√] • polarising power of cation decreases [√] • ease of distortion of electron cloud of NO3– decreases [√] • N–O bond is less weakened • more energy required to break N–O bond [√] • stability of nitrates to heat increases [1m] 4[√] – 1m Examiners’ Comments: The explanations of the trend were sometimes lacking steps in their logic; mentioning the reduced charge density, and then not explaining that this reduces the tendency for the anion to distort. [3]

(b)

Dinitrogen pentoxide, N2O5, exists as covalent molecules O2N–O–NO2 in the gas phase, but as the ionic [NO2+ NO3–] in the solid or in solution. N2O5 has been suggested as a possible intermediate in the thermal decomposition of Group II nitrates M(NO3)2. (i)


Using this information, and that provided in (a), suggest a mechanism for the formation of N2O5 from a Group II nitrate, M(NO3)2.

Examiners’ Comments: This part question required candidates to apply knowledge in an unfamiliar context. Few candidates made full use of the information about the decomposition of the carbonate ion in their answers. These candidates did not appreciate that the first step in (i) would be NO3– NO2+ + O2–, which + – was followed by NO2 + NO3 N2O5. N2O5 decomposes on heating in a first order reaction to give NO2 and O2. The mechanism of this thermal decomposition is thought to involve three steps. • There is an initial homolytic breaking of an N–O bond to give NO2• and NO3• radicals. • This is followed by dimerisation of the NO3• radical to give N2O6, which contains a –O–O– peroxide bond. • The third step involves the dissociation of this dimer via further N–O bond cleavage to form the products. (ii)

Write a balanced equation for the thermal decomposition of N2O5. N2O5

2NO2 + ½O2 [1m]

Examiners’ Comments: The equation in (ii) was usually correct. (iii)

Use the information given above to draw out the full mechanism for the thermal decomposition of N2O5. You are advised to use structural formulae for all

species, such as for N2O5, so that it is clear which bonds are broken and which are formed. Indicate any unpaired electrons by a dot (•).


2 half arrows – [√], correct structural formulae of NO2• and NO3• – [√]

2 half arrows – [√],correct structural formulae of N2O6– [√]

2 half arrows –[√] , 4 half arrows – [√] any 2 [√] – 1m (iv) Suggest which of the steps you have drawn is the slowest step. Explain your reasoning. Since the reaction is first order with respect to [N2O5], the slow step is step 1 as it is the only step that involves just one molecule of N2O5. Examiners’ Comments: Some candidates pointed out that the first step would be the slowest (ratedetermining) one, then did not explain that this is because it is the only one involving just one molecule of N2O5. [7]

(c)

When benzene is added to a solution of N2O5 in CCl4, an excellent yield of nitrobenzene is obtained in the absence of any added catalyst. (i)

Suggest the other product formed during this reaction. HNO3 [1m] Examiners’ Comments: There were only a few completely correct responses here. Many candidates performed poorly. Some suggested that the other product was chlorobenzene, others that it was dinitrobenzene. Only a handful correctly suggested HNO3.

(ii)

Suggest a mechanism for this reaction, showing all charges and using curly arrows to show the movements of electron pairs. Electrophilic substitution


NO2+ + NO3– [√]

N2O5

NO2

H

NO2

fast

+

NO3– + H+

H+

+

HNO3 [√]

both curly arrows – [√] arenium ion intermediate – [√] 4[√] – 2m Negate 1[√] if slow and fast is not indicated Examiners’ Comments: The mechanisms often did not show the production of the electrophile NO2+ from N2O5, or showed an incorrect equation (e.g. N2O5 2NO2+ + ½ O2). The – last step, the abstraction of the proton by NO3 to produce HNO3, was usually missing. [3] The following table lists some ∆HƟf values. ∆HƟf / kJmol-1 +33.2 +5.0

compound NO2(g) N2O5(g)

The enthalpy change for the dissociation of N2O5(g) is given below. reaction 1: (d)

N2O5(g)

NO2(g) + NO3(g)

∆HƟ = +92.5 kJmol-1

Use the data above to calculate (i)

∆HƟf of NO3(g) ∆HƟ = ∆HƟf [NO2(g)] + ∆HƟf [NO3(g)] – ∆HƟf [N2O5(g)] +92.5 = +33.2 + ∆HƟf [NO3(g)] –(+5.0) [1m] ∆HƟf [NO3(g)] = +64.3 kJmol-1 [1m]

(ii) H2 Chemistry 9647 2012

∆HƟ for reaction 2.

∆HƟ = 2∆HƟf [NO2(g)] – 2∆HƟf [NO3(g)] ∆HƟ = 2(+33.2) – 2(+64.3) = –93.1 kJmol-1 [1m] Examiners’ Comments: Part (d) was generally well-answered. The most common incorrect answers resulted from not multiplying by 2 in (ii), or using positive/negative signs incorrectly. reaction 2:

2NO3(g)

2NO2(g) + O2(g)

(e) 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 the more spontaneous. Explain your reasoning. Reaction 2 would be more spontaneous. [1m] ∆G = ∆H + (–T∆S) For both reactions, there is an increase in number of gaseous particles from reactants to products by 1, hence ∆S is positive [1m]. For reaction 2, ∆G will be negative at all temperatures since ∆H and the term (–T∆S) is negative [1m]. For reaction 1, ∆G will be only be negative at higher temperatures where magnitude of (–T∆S) > ∆H [1m]. Examiners’ Comments: Many candidates did well in this part. Most recognised that the ∆S of both reactions was positive, though fewer pointed out that the ∆G for reaction 2 would be negative at all temperatures because of the negative ∆H, whereas that for reaction 1 will only be negative at higher temperatures, so reaction 2 would be the more spontaneous. Some candidates did not give clear arguments for ∆G values based on those for ∆H and ∆S, and hence explaining the degree of spontaneity. [4] [Total: 20]


4

Beams of charged particles are deflected by an electric field. If the particles are all travelling at the same speed, though an electric field of constant strength, the angle of deflection is proportional to their charge/mass ratio. In a particular experimental set-up, protons are deflected through an angle of +15o.

(a)

Assuming an identical set of experimental conditions, by what angles will the following particles be deflected? (D is deuterium, 2H, and T is tritium, 3H) • • •

D– T+ He2+

Protons are 1H+. deflection of 1H+ α

charge charge = k mass mass

1 1

15 = k  

k = 15 All the other ions have the same k value in the same electric field. •

D–: charge: -1, mass: 2

 -1  o  = –7.5 2  

deflection of D– = 15  •

[1]

T+: charge: +1, mass: 3

 +1  o  = +5.0 [1] 3  

deflection of T+ = 15  •

He2+: charge: +2, mass: 4

 +2  o  = +7.5  4 

deflection of He2+ = 15 

[1]

Examiners’ comments: The question gave the deflection angle with a “+” sign. This means that the direction of deflection is important. Hence, students need to present the deflection angle in terms of “+” or “–”. Need to show working. No working means no marks awarded. (b)

Under identical conditions, a beam of particles, R, each having 12 times the mass of a proton, was deflected by an angle of +5o. (i)

Suggest the overall charge on a particle of R. Since the condition is identical, it means that the value of k is still 15. Hence, particle R: charge: x+, mass: 12

 x+    12 

deflection of R = 15 


 x+    12 

+5 = 15 

x = 4+

[1]

Note: Sign on the charge is important, need to put “+” behind the number. (For oxidation state, the “+” sign is in front) (ii)

Given that a particle of R contains 6 protons, deduce the number of neutrons and electrons in a particle of R. Particle R: 12R4+ Mass number = 12 = no. of protons and neutrons No. of neutrons = 12 – 6 = 6 [1] No. of electrons = 6 – 4 = 2 [1]

(c)

(i)

Some ionic radii are listed in the Data Booklet. State and explain the differences between the ionic radii of Na, Si, P and Cl. Na exist as Na+, Si can exist as Si4+, P exist as P3-, Cl exist as Cl-. 1st trend: • Anions (P3- and Cl-) are larger than cations (Na+ and Si4+) [√] • Anions have one more occupied quantum shell [√], hence valence electrons are further away from the nucleus. 2nd trend: • Ionic radius of Si4+ and Cl- is smaller than that of Na+ and P3- respectively. [√] • The cations, Na+ and Si4+ are isoelectronic and anions, P3- and Cl-, are isoelectronic. [√] • For each isoelectronic series, Si4+ and Cl- have larger nuclear charge than Na+ and P3- respectively [√]. Each isoelectronic series have the same shielding effect [√]. Hence the effective nuclear charge of Si4+ and Cl- is larger than Na+ and P3- respectively. Valence electrons are more strongly attracted to the nucleus, pulled closer to the nucleus. [√] 6 – 7 [√]: 3 marks 3 – 5 [√]: 2 marks 2 [√]: 1 mark Examiners’ comments: Many candidates did not mention the concept of isoelectronic ions in part (i), and those that did implied that all four ions were isoelectronic, rather than being split into two isoelectronic pairs. Few mentioned that P3- and Cl- each had one extra shell of electrons than Na+ and Si4+.

(ii)

These four elements also differ greatly in their melting points. Describe how their melting points differ, and explain this variation. • Na has giant metallic structure [√] and possess strong electrostatic forces of attraction between cations and sea of delocalised electrons [√]. o Large amount of energy is needed to overcome the strong metallic bonds. [√] Hence Na has a high melting point. [√] •


Si has giant covalent structure [√] and has strong and extensive covalent bonds between Si atoms [√]. o More amount of energy is needed to overcome the strong and extensive covalent bonds than the metallic bond [√], hence melting point of Si is higher than Na. [√]

•

P and Cl have simple molecular structures [√] with weak van der waals’ forces of attraction between molecules. [√] o Hence they have the lowest melting points as least amount of energy is needed to overcome the intermolecular forces of attraction. [√]

•

However, P4 has a larger electron cloud size than Cl2 [√], more easily polarised, hence stronger van der waals’ forces of attraction between P4 molecules [√], and more energy is needed to overcome these forces of attraction [√]. Hence P4 has higher melting point than Cl2 [√].

10 – 15 [√]: 3 marks 5 – 9 [√]: 2 marks 2 – 4 [√]: 1 mark Note: “Variation” indicates that the anomalies/details need to be included. Student need to know that P has molecular formula of P4. Examiners’ comments: In part (ii) several described the melting point trend correctly, then did not explain the trend well in terms of bonding. For example, that the giant covalent bonding in silicon was stronger than the metallic bonding in sodium. Some suggested that metallic bonding was stronger than covalent bonding, or that sodium was giant ionic, whereas the other three were simple covalent. (d)

Ionic charges are important in determining the 3-dimensional structure of the haemoglobin molecule. The four protein chains of deoxyhaemoglobin are held in position by 8 ionic ‘salt’ links. Some of these interactions are between the C-terminal end of one chain and the N-terminal of another chain, whereas other interactions are between aspartic acid residues on one chain and arginine residues on another chain. (i)


Using the tetrapeptide whose structure is given below, and taking into account of the ionic forms that the amino acid residues exist at pH7, draw the structures of two of these tetrapeptide chains to illustrate both of the ionic interactions mentioned above.

) 2 3 (CH

[1] for ionic interaction between COO- and NH3+ of the C- and N-terminus [1] for ionic interaction between COO- of aspartic acid and NH3+ of arginine. [1] for converting COOH to COO- and NH2 to NH3+ [discuss whether the 1 mark should go to this point or to (dii) “describe” in greater detail] Note: Student should show and label the “ionic interaction” but the red words are for illustration purposes during teaching, student need not draw them. Examiners’ comments: Students MUST show the ionisation of the groups at both ends of the chain and their subsequent ionic bonding with complementary groups in another chain. Common mistake was to label “hydrogen bonding” instead of “ionic bonding”

(ii)

Describe two further ways, other than ionic linkages, by which the tertiary structures of proteins can be stabilised. Any of these 2 answers can be accepted: • • •

Hydrogen bonding between C=O and N-H groups [1] Disulfide linkages between cysteine R groups [1] Van der waals’ forces between non-polar R groups [1]

Examiners’ comments: Most performed well on part (ii), but candidates must be specific about what type of R-group interactions there are, instead of just saying 'R-group interactions' in general. [5]


(e)

The antibiotic valinomycin acts by allowing the uncontrolled movement of ions across cell membranes. Valinomycin is a cyclic compound containing six amino acid and six hydroxyacid residues.

(i)

Apart from the peptide group, what other functional group is present in the valinomycin molecule? Ester [1]

When treated with a peptidase enzyme, the peptide bonds in the ring of valinomycin are hydrolysed. Only two products are formed. (ii)

Draw the structural formulae of the two compounds produced.

[1]

and

[1]

Note: Ester linkage must be intact as the question used “peptidase” and peptide bonds are hydrolysed. Peptide = amide linkage! Examiners’ comments: Several candidates omitted part (ii), or thought the ester bonds would also be hydrolysed. Common wrong products (CH3)2CHCH(OH)CO2H, (CH3)2CHCH(NH2)CO2H and CH3CH(OH)CO2H. [3] [Total: 20]


5

Malic acid occurs in green apples and grapes. It is often added to beverages and confectionery to confer a sour taste.

(a)

Suggest the reagents and conditions you would use in a three-step synthesis of malic acid from bromoethanal, identifying the intermediates G and H.

[5] Step 1: [1] Reagent : HCN Condition : trace amount of NaOH, 10-20oC G:

[1] Step 2: [1] Reagent : ethanolic aq KCN Condition : heat with reflux H:

[1] Step 3: [1] Reagent : dilute HCl Condition : heat with reflux Malic acid :

Examiners’ comments: Some did not state clearly the conditions for the hydrolysis of nitriles (dilute HCl or H2SO4 and heat), and some did not write the structures of the intermediates G and H.


For the cyanohydrin, HCN + NaCN, HCN + NaOH or KCN + H+ (less than one equivalent) were acceptable. For the nucleophilic substitution reaction, the KCN in ethanol had to be heated under reflux. Subsequent hydrolysis with hot aqueous H2SO4 would produce malic acid. A number of candidates did not represent the structure of G (or H) accurately enough: CN–CH2CHO is an isocyanide (isonitrile), rather than the correct nitrile NC–CH2CHO. Some unusual reagents were suggested for use in syntheses: CH2COCl; NaOCN; LiAlH4; Na2Cr2O7. (b)

Malic acid can be dehydrated to give a mixture of two isomeric alkenedioic acids with the molecular formula C4H4O4. (i) Draw the structures of the isomers, and state the type of isomerism they show.

[1] mark for both correct isomers [1] indicating geometric/cis-trans isomerism

(ii)

Suggest reagents and conditions for the dehydration reaction. Reagent : excess concentrated sulfuric acid Condition : 170oC [1] for both Examiners’ comments: Almost all knew the conditions for dehydration (conc. H2SO4 and heat). Most candidates chose to use concentrated H2SO4 at 180 °C to dehydrate malic acid. A number wrongly suggested that the concentrated H2SO4 was aqueous. Those who suggested Al 2O3 needed to use a higher temperature, between 300 °C and 500 °C.

(iii)

The pKa values of the two acidic groups in one isomer, J, are 3.0 and 4.4, whereas in the other isomer, K, they are 1.9 and 6.2. Use the pKa values to suggest which isomer, J or K, produces the more stable mono-anion on treatment with 1 mol of NaOH? Explain your answer. Recall : Higher pKa values imply lower Ka values, hence less acidic. The second pKa value of isomer K is larger than 4.4 indicates that the second – COOH group in K is less acidic.[] Hence, K produces the more stable mono-


anion.[ ] [1] for both Examiners’ comments: There was some confusion as to what was meant by the term ‘mono-anion’, with a number of candidates thinking that the question referred to the di-anion, – O2C-CH=CH-CO2–. Some took the average of the two pKs values, and concluded that isomer J (average 3.7), was more acidic than isomer K (average 4.05), and so its anion would be more stable. A significant number, however, appreciated that a low first pKa, coupled with a high second pKa, indicated a stable mono-anion, and used this reasoning to correctly identify K as the acid producing the more stable mono-anion.

(iv) Draw the displayed formula of the mono-anion produced in (iii), and use your formula to suggest an explanation for why it is more stable than the mono-anion of the other isomer. [6]

[1] The cis-mono-anion would be the more stable, due to hydrogen bonding between the -COO- group and the unionised -CO2H group. [1] Thus, it is more difficult for the H in the –COOH group to be hydrolysed to release H+. Examiners’ comments: Only a minority appreciated that its stability arises from the ability to form a hydrogen bond between the –COO- group with the adjacent non ionised –CO2H.

Gentle heating the anhydrous crystals of one of the isomers J or K produces a neutral compound L, C4H2O3, which does not react with sodium metal or give a precipitate with 2,4-dinitrophenylhydrazine. Compound L reacts with ammonia to give a compound M, C4H5NO3, which gives a salt on reaction with NaOH but not with HCl. Compound M decolourises Br2(aq). (c)

Suggest structures for compounds L and M, and identify which isomer, J or K, produces L. Write equations for all reactions, and explain the observations. [6] Heating isomer K [1] forms L:


[1] Eqn []

Due to the close proximity of the 2 –COOH groups in K, K can undergo intramolecular dehydration (losing a H2O molecule) to form a cycloester, L.

Since L does not have an –OH group, it does not react with Na metal.

Since the C=O group in L is not a carbonyl functional group, L does not give a precipitate with 2,4-dinitrophenylhydrazine.

L reacts with NH3 to form M. The carbon of the C=O group in L is highly electron deficient, and is thus susceptible for nucleophilic attack by NH3.

[1] Eqn []

M has both an amide group and an acidic group. The lone pair of electrons on the nitrogen atom of the amide group is delocalized into the C=O group and hence is not able to accept H+ from the HCl.

The –COOH group in M is acidic and hence reacts with NaOH to form a salt.

Eqn []

M decolourises aqueous Br2 as it undergoes electrophilic addition with the alkene group.

Eqn [] 8 to 11 [3] 5 to 7 [2] 2 to 4 [1] H2 Chemistry 9647 2012

0 to 1 [0] Examiners’ comments: Many responses correctly related the observations to the functional groups present, although only a few suggested structures for the products of the reactions. Some candidates did not relate their inference to the test results: they merely stated that “L does not contain an OH group or a carbonyl group”. There were very few candidates who gave equations for the reactions of M with NaOH(aq) or with Br2(aq). Most worked out that L was maleic anhydride. Fewer drew the correct structure for M, the half-amide of maleic acid. When malic acid is heated with acidified K2Cr2O7, compound N is formed. Heating N in an inert solvent for several hours produces P, C3H4O3 and a gas Q. Both N and P react with 2,4-dinitrophenylhydrazine. Compound P also gives a yellow precipitate with aqueous alkaline iodine. (d)

Suggest structures for N and P, and use the information given to suggest the identity of gas Q. [3] [Total: 20] Information from question When malic acid is heated with acidified K2Cr2O7, compound N is formed. Both N and P react with 2,4dinitrophenylhydrazine. Compound P also gives a yellow precipitate with aqueous alkaline iodine.

Deduction Malic acid undergoes oxidation. 2o alcohol oxidised to ketone. N is ketone Both undergo condensation. N and P contain carbonyl group P undergoes mild oxidation.

P contains

N:

[1] P:

[1] Molecular formula of Q = Molecular formula of N – Molecular formula of P = C4H4O5 - C3H4O3 = CO2 H2 Chemistry 9647 2012

Q : CO2 [1] Examiners’ comments: Where it was attempted, this part was answered better than (c). Most responses gave the structure of N, and many also recognised P as CH3COCO2H and Q as CO2. Most candidates deduced that gas Q was CO2, although there were some H2, H2O, O2, CO and even ‘HCO’ among the answers.


2012 Gce a Level h2 p1 p2 p3 Qns Ans

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