Table 1 Amine Boiling point / K pKb (approx.) P 329.3 3.00 Q 310.5 9.83 R 350.8 3.22

A. Amine P is more volatile than amine R, but less basic.

B. Amine R will give a white precipitate with Hinsberg reagent, which is soluble in NaOH.

C. Amine Q is more soluble in water as compared to R.

D. Amine P and Q can be prepared by the Gabriel phthalimide process.

ANSWER

B. Amine R will give a white precipitate with Hinsberg reagent, which is soluble in NaOH.

Solution

On the basis of the boiling points, P is a secondary, Q a tertiary and R a primary amine. A primary amine gives a white precipitate with Hinsberg reagent (benzenesulphonyl chloride); the N-H of that sulphonamide is acidic, so the product dissolves in NaOH.

Marking Scheme

1 mark for the correct option.

A. [Ni(CN)4]2−

B. [Cu(NH3)4]2+

C. [NiCl4]2−

D. [Co(en)3]3+

ANSWER

C. [NiCl4]2−

Solution

[NiCl4]2− is tetrahedral because Cl− is a weak field ligand and does not cause pairing of electrons, so the metal uses sp3 hybridisation. The other complexes are square planar or octahedral.

Marking Scheme

1 mark for the correct option.

A. 1F

B. 6F

C. 3F

D. 2F

ANSWER

C. 3F M3+ + 3e− → M quantity of charge =

∴ 3F

Marking Scheme

1 mark for the correct option.

A. Chloromethane is treated with AgCN.

B. Nitrobenzene is reduced.

C. Benzene diazonium chloride is treated with Cu/KCN.

D. Aniline reacts with acetyl chloride.

ANSWER

B. Nitrobenzene is reduced.

Solution

Reduction of nitrobenzene gives aniline. The foul smell is that of the isocyanide produced when an aliphatic or aromatic primary amine undergoes the carbylamine reaction. Only option B supplies a primary amine.

Marking Scheme

1 mark for the correct option.

Consider a system in a state of dynamic equilibrium, as shown in Fig. 1(a). The lower part is solution and the upper part is a gaseous system at pressure p and temperature T. The pressure is increased over the solution phase by compressing the gas to a smaller volume, as shown in Fig. 1(b). Which of the following statements is true about figure 1(b)?

Figure 1

A. Solubility of the gas will increase until dynamic equilibrium is reached.

B. Solubility of the gas will decrease until dynamic equilibrium is reached.

C. Solubility of the gas will increase until static equilibrium is reached.

D. Solubility of the gas will decrease until static equilibrium is reached.

For Visually Challenged Learners - According to Henry’s law, the partial pressure P of a gas above the liquid at constant temperature is directly proportional to

A. volume of the gas

B. mole fraction of the gas in the solution

C. atmospheric pressure

D. vapour pressure of the solution

ANSWER

A. Solubility of the gas will increase until dynamic equilibrium is reached.

For Visually Challenged Learners: B. mole fraction of the gas in the solution

Solution

By Henry’s law, p = KHx: raising the partial pressure of the gas raises its mole fraction in solution, so more gas dissolves until the rates of dissolution and escape become equal once more - a new dynamic equilibrium.

Marking Scheme

1 mark for the correct option.

CH3COCl + H2O CH ⟶ 3COOH + HCl She carries out the reaction with (i) equimolar amounts of acetyl chloride and water and (ii) excess of water. The rate law obeyed in the two cases will be:

A. (i) First order (ii) Second order

B. (i) First order (ii) First order

C. (i) Second order (ii) Second order

D. (i) Second order (ii) First order

ANSWER

D. (i) Second order (ii) First order

Solution

In case (i) the rate depends on both reactants. In case (ii), water is in large excess so its concentration is effectively constant and the rate becomes independent of it; such reactions are called pseudo first order (pseudo unimolecular) reactions.

Marking Scheme

1 mark for the correct option.

A. an acidic and non-essential amino acid

B. a neutral and non-essential amino acid

C. an acidic and essential amino acid

D. a neutral and essential amino acid

ANSWER

A. an acidic and non-essential amino acid

Solution

The amino acid can be synthesised in our body, so it is non-essential. It has two -COOH groups and only one - NH2 group, so it is acidic.

Marking Scheme

1 mark for the correct option.

(i) C2H5COC2H5 + CH3MgCl

(ii) C2H5COCH3 + C2H5MgCl

(iii) CH3CHO + C2H5-CH(MgCl)-C2H5

(iv) C2H5CHO + CH3-CH(MgCl)-C2H5

A. (i) and (iii)

B. (ii) and (iv)

C. (iii) and (iv)

D. (i) and (ii)

ANSWER

D. (i) and (ii)

Solution

3-Methylpentan-3-ol, CH3CH2-C(OH)(CH3)-CH2CH3, is a tertiary alcohol, so it must come from a ketone plus a Grignard reagent. Routes (i) and (ii) both supply the required CH3, C2H5 and C2H5 groups on the carbinol carbon. Routes (iii) and (iv) use aldehydes, which can give only secondary alcohols.

Marking Scheme

1 mark for the correct option.

A. Cr

B. Mn

C. Ti

D. Fe

ANSWER

B. Mn

Solution

Mn3+ (3d4 ) is reduced to Mn2+ (3d5 ), gaining the extra-stable half-filled d5 configuration; E° = +1.51 V. It therefore acts as a strong oxidising agent.

Marking Scheme

1 mark for the correct option.

A. The reaction follows the SN2 mechanism and is accompanied by inversion of configuration.

B. The reaction follows the SN1 mechanism and is accompanied by inversion of configuration.

C. The reaction follows the SN2 mechanism and is accompanied by retention of configuration.

D. The reaction follows the SN1 mechanism and is accompanied by retention of configuration.

ANSWER

A. The reaction follows the SN2 mechanism and is accompanied by inversion of configuration.

Solution

A secondary halide with a strong nucleophile reacts by SN2. Attack occurs from the side opposite the leaving group through a single transition state, giving Walden inversion. An SN1 route would proceed through a planar carbocation and give racemisation.

Marking Scheme

1 mark for the correct option.

A. 16 min

B. 32 min

C. 48 min

D. 64 min

ANSWER

B. 32 min

Solution

50% of the reaction is complete in 16 min. For the remaining 50% to fall to half (i.e. 25% left) a further 16 min is needed. Hence 50% + 25% = 75% is complete in 32 min, that is two half-lives.

Marking Scheme

1 mark for the correct option.

A. 0.1 M HCl

B. 0.1 M CH3COOH

C. 0.1 M glucose

D. 0.1 M NH4OH

ANSWER

A. 0.1 M HCl

Solution

HCl is a strong acid and dissociates almost completely into H+ and Cl− ; more ions mean higher conductivity, and H+ has the highest ionic mobility. CH3COOH and NH4OH are weak electrolytes and only partially ionised; glucose is a non-electrolyte and furnishes no ions.

Marking Scheme

1 mark for the correct option.

A statement of Assertion (A) is followed by a statement of Reason (R). Select the most appropriate answer from the options given below. Assertion (A): Nitration of aniline gives a significant amount of the meta derivative. Reason (R): -NH2 is an ortho and para directing and highly activating group.

A. Both A and R are true, and R is the correct explanation of A.

B. Both A and R are true, but R is not the correct explanation of A.

C. A is true but R is false.

D. A is false but R is true.

ANSWER

B. Both A and R are true, but R is not the correct explanation of A.

Solution

In the strongly acidic nitrating mixture aniline is largely protonated to the anilinium ion, in which the positively charged -NH3 + group is deactivating and meta directing. The meta product therefore arises in spite of -NH2 being o/p directing, not because of it.

Marking Scheme

1 mark for the correct option.

A statement of Assertion (A) is followed by a statement of Reason (R). Select the most appropriate answer from the options given below. Assertion (A): Permanganate and manganate ions have a tetrahedral structure. Reason (R): In permanganate and manganate ions, pi bonding takes place which is due to overlap of p orbitals of oxygen with p orbitals of manganese.

A. Both A and R are true, and R is the correct explanation of A.

B. Both A and R are true, but R is not the correct explanation of A.

C. A is true but R is false.

D. A is false but R is true.

ANSWER

C. A is true but R is false.

Solution

Both MnO4 − and MnO4 2− are tetrahedral, so A is true. The Reason is false: the bonding is π p -d π π, formed by overlap of the p orbitals of oxygen with the d orbitals of manganese.

Marking Scheme

1 mark for the correct option.

A statement of Assertion (A) is followed by a statement of Reason (R). Select the most appropriate answer from the options given below. Assertion (A): Miscibility of ethers with water resembles those of alcohols of the same molecular mass. Reason (R): The oxygen of ether can also form hydrogen bonds with water.

A. Both A and R are true, and R is the correct explanation of A.

B. Both A and R are true, but R is not the correct explanation of A.

C. A is true but R is false.

D. A is false but R is true.

ANSWER

A. Both A and R are true, and R is the correct explanation of A.

Solution

The ether oxygen carries two lone pairs and can act as a hydrogen-bond acceptor towards the O-H of water. It is exactly this hydrogen bonding that makes an ether as miscible as the corresponding alcohol.

Marking Scheme

1 mark for the correct option.

A. Both A and R are true, and R is the correct explanation of A.

B. Both A and R are true, but R is not the correct explanation of A.

C. A is true but R is false.

D. A is false but R is true.

ANSWER

A. Both A and R are true, and R is the correct explanation of A.

Solution

Co3+ is 3d6 . NH3 is a strong field ligand, so Δo > P and the electrons pair up as t2g 6 eg 0 . With no unpaired electrons the complex is diamagnetic, and the pairing caused by the strong field ligand is precisely the cause.

Marking Scheme

1 mark for the correct option. Questions 17 to 21 are very short answer questions carrying 2 marks each.

I Complete and balance the following reaction: [1] SO3 2− + MnO4 − + H+ →

Answer

2MnO4 − + 5SO3 2− + 6H+ → 2Mn2+ + 5SO4 2− + 3H2O

II Why does titanium melt at a lower temperature than chromium? [1]

Answer

Chromium has a larger number of unpaired electrons than titanium, which leads to stronger interatomic attractions holding its atoms together. More energy is therefore needed to break them, so chromium melts at a higher temperature than titanium.

Marking Scheme

I - 1 mark for the correctly balanced equation. II - 1 mark.

OR

Alternative (B): 1 + 1 marks

I Why do titanium and scandium form complex compounds, whereas potassium and calcium do not, although they belong to the same period? [1]

Answer

The transition metals form a large number of complex compounds because of the comparatively smaller size of their metal ions, their high ionic charge and the availability of d orbitals for bond formation. Titanium and scandium possess all three; potassium and calcium are large, of low charge, and have no d orbitals of suitable energy, so they do not form complexes.

II Complete and balance the following reaction: [1] Fe2+ + Cr2O7 2− + H+ →

Answer

Cr2O7 2− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O

Marking Scheme

I - 1 mark. II - 1 mark for the correctly balanced equation.

“The rate of reaction remains constant during the course of reaction.” Is this statement always true, or is it true under certain conditions? Reflect on the statement.  The statement is true for a zero order reaction, as the rate is independent of the concentration of the reactants.  For reactions of any other order - first, second and so on - the rate of reaction decreases as the concentration of the reactants decreases.

Marking Scheme

1 mark for each point.

II Milk contains a protein called casein. How does the structure of casein change on addition of a few drops of lemon juice to milk? [1]

For Visually Challenged Learners - (I) Name the nitrogenous bases in DNA. (II) Milk contains a protein called casein. How does the structure of casein change on addition of a few drops of lemon juice to milk?

Answer

 I. The complementary bases follow Chargaff’s rule - G pairs with C (three hydrogen bonds) and A pairs with T (two hydrogen bonds):

 II. Addition of lemon juice lowers the pH, which breaks the hydrogen bonds; the globules of casein unfold and the protein is denatured.  For Visually Challenged Learners - I. Adenine, Guanine, Cytosine and Thymine. II. As above. Completed strand

Marking Scheme

I - 1 mark. II - 1 mark. (Same allocation for the Visually Challenged Learners’ version.)

Chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at 623 K and 300 atm; however, the presence of a substituent group affects its activity towards this nucleophilic substitution reaction. Which substituent, (i) nitro or (ii) methoxy, will increase the reactivity of chlorobenzene, and why?

Answer

The nitro group will increase the reactivity.  The nitro group is a strong electron-withdrawing group, by both the inductive (-I) and the resonance (- M) effect, whereas the methoxy group is electron donating in nature.  When the nitro group is present at the ortho or para position it withdraws electron density from the ring and stabilises the negative charge of the intermediate carbanion through resonance, which increases reactivity towards nucleophilic substitution.

Marking Scheme

1 mark for identifying the nitro group; ½ + ½ marks for the reasoning.

I Underground iron pipelines are often protected by connecting them to a more reactive metal like magnesium. [1]

Answer

Magnesium, being more reactive than iron, acts as a sacrificial anode and undergoes oxidation, Mg → Mg2+ + 2e− , supplying electrons to the iron pipe, which then behaves as the cathode. The iron is thus prevented from corroding.

II Write the reactions occurring at the cathode and the anode during the electrolysis of molten magnesium chloride. [1] Cathode: Mg2+ + 2e− → Mg(l) Anode: 2Cl− → Cl2(g) + 2e−

Marking Scheme

I - 1 mark. II - ½ mark for each electrode reaction. Questions 22 to 28 are short answer questions carrying 3 marks each.

Answer

Fe2+ has the highest magnetic moment, because it contains the largest number of unpaired electrons of the three. Ion Configuration Unpaired electrons (n) μ = √[n(n+2)] / BM Ti2+ [Ar] 3d2 2 2.83 Co2+ [Ar] 3d7 3 3.87 Fe2+ [Ar] 3d6 4 4.90

II Which characteristic of iron makes it suitable to (a) act as a catalyst in Haber’s process, and (b) form an interstitial compound with oxygen? [2]  (a) It provides a large surface area for adsorption and shows variable oxidation states, so it can form unstable intermediates and offer a path of lower activation energy.  (b) Iron has a large crystal lattice containing interstitial sites big enough to accommodate small non-metal atoms such as oxygen, carbon and nitrogen.

Marking Scheme

I - ½ mark for the ion and ½ mark for the justification. II - 1 mark for (a) and 1 mark for (b).

The values of log k and 1/T for a reaction are given below. The plot of log k versus 1/T obtained for this data is shown in Figure 3. Here k is the rate constant and T is the absolute temperature. Calculate Ea for the reaction. (Given R = 8.314 J K−1 mol−1 ) S. No. log k 1/T (/K)

1. −4.46 0.00333

2. −3.91 0.00323

3. −3.37 0.00313

4. −2.82 0.00303

5. −2.31 0.00294 Figure 3

For Visually Challenged Learners - The rate constant k of a reaction varies with temperature T according to log k = log A − Ea/(2.303RT), where Ea is the activation energy. When a graph is plotted for log k versus 1/T, a straight line with a slope of −4250 K is obtained. Calculate Ea for the reaction. (Given R = 8.314 J K−1 mol−1 )

Answer And Solution

log k = log A − Ea / (2.303 RT) slope = − E

∴ a / (2.303 R) Taking two points from the plot:

slope = (y2 − y1) / (x2 − x1) = [−4.46 − (−3.91)] / (0.00333 − 0.00323) = −5500 K Ea = 5500 × 2.303 × 8.314 = 1.053 × 105 J mol−1 Ea = 105.31 kJ mol−1 *The value of the slope may vary slightly if different pairs of points are chosen; any slope between −5400 K and −5670 K, giving Ea ≈ 103-108 kJ mol−1 , is acceptable.*

For Visually Challenged Learners − Ea / (2.303 R) = −4250 E ⇒ a = 4250 × 2.303 × 8.314 Ea = 81 375 J mol−1 = 81.37 kJ mol−1

Marking Scheme

½ mark for the formula, ½ mark for the slope, 1 mark for the substitution and 1 mark for the correct answer with the correct unit.

Identify each name reaction and complete the missing reactant or product in the given reaction sequence. (I) 2CH3CHO - - ? - → CH3CH(OH)CH2CHO (II) Toluene + __?__ / CS2 → C6H5CH(OCrOHCl2)2 - H3O+ → benzaldehyde (III) >C=O - NH2NH2, −H2O→ __?__ - KOH / ethylene glycol, heat→ >CH2 + N2

Answer

 (I) Aldol condensation - the missing reagent is dilute NaOH.  (II) Étard reaction - the missing reagent is chromyl chloride, CrO2Cl2.  (III) Wolff-Kishner reduction - the missing intermediate is the hydrazone, >C=N-NH2.

Marking Scheme

½ + ½ marks for each part - name of the reaction and the missing species.

I Aman prepared a solution by dissolving 10 mL of acetone and 10 mL of carbon disulphide. The net volume of the resulting solution was not found to be 20 mL. (a) What would be the observation in this case? Comment. (b) If carbon disulphide is replaced by chloroform, what would be the expected change in the volume of the resulting solution? Support your answer with an appropriate reason. [2]  (a) The net volume of the resulting solution is found to be slightly more than 20 mL. Carbon disulphide and acetone have weak intermolecular interactions between them, which leads to an increase in the volume of the solution (positive deviation from Raoult’s law).  (b) When carbon disulphide is replaced by chloroform, the net volume shows a slight decrease from 20 mL, as chloroform and acetone have strong intermolecular interactions owing to hydrogen bonding (negative deviation).

II Calculate the concentration of oxygen (in ppm) present in 1 litre of sea water, given that a litre of sea water weighs 1030 g and contains about 6 × 10−3 g of dissolved oxygen. [1] ppm = (mass of the component / total mass of all components) × 106 = (6 × 10−3 × 106 ) / 1030 Concentration of oxygen = 5.8 ppm

Marking Scheme

I - ½ + ½ marks for (a) and ½ + ½ marks for (b). II - ½ mark for the formula and ½ mark for the answer.

(a) Nitrobenzene to fluorobenzene C6H5NO2 - Sn / HCl→ C6H5NH2 - NaNO2 / HCl, 273-278 K→ C6H5N2 + Cl− C6H5N2 + Cl− - NaBF4→ C6H5N2 + BF4 − - heat→ C6H5F + N2 + BF3 The last step is the Balz-Schiemann reaction. The Sandmeyer reaction cannot be used for fluorine.

(b) Benzoic acid to benzylamine C6H5COOH + NH3 - heat→ C6H5CONH2 - LiAlH4→ C6H5CH2NH2

II Aniline undergoes electrophilic substitution reactions readily. However, acetylation of aniline is required to obtain monosubstituted products. Give a reason for your answer. [1]

Answer

The -NH2 group in aniline is strongly activating in electrophilic substitution, so the reaction does not stop at one substitution and gives 2,4,6-trisubstituted products. Acetylation converts -NH2 to -NHCOCH3, which is less activating because the nitrogen lone pair is delocalised towards the carbonyl group. The reaction can then be controlled and stops at the monosubstituted (mainly para) stage.

Marking Scheme

I - 1 mark for each conversion. II - ½ + ½ marks.

I Write the IUPAC name of the following compound. [1]

Answer

1-Bromo-2-(1-methylpropyl)benzene - equivalently, 1-bromo-2-(butan-2-yl)benzene in current IUPAC nomenclature.

II When CH3CH2CH2Cl reacts with AgCN and with KCN, compounds “X” and “Y” respectively are formed. Identify “X” and “Y”. Support your answer with an appropriate reason. [2]  With AgCN: X = CH3CH2CH2NC (propyl isocyanide).  With KCN: Y = CH3CH2CH2CN (propyl cyanide / butanenitrile).  KCN is predominantly ionic; it dissociates in solution to give free CN− , an ambident nucleophile, which attacks through the carbon atom, so the cyanide is formed.  AgCN is predominantly covalent; only the lone pair on nitrogen is available, so bond formation occurs through nitrogen and the isocyanide is formed.

Marking Scheme

I - 1 mark. II - ½ mark for each product and ½ + ½ marks for the reasons.

OR

Alternative (B): 1 + 2 marks

I What is the IUPAC nomenclature of the compound (CH3)3C-C(CH3)Br-CH2CH3? [1]

Answer

3-Bromo-2,2,3-trimethylpentane

II Which of the following will be hydrolysed more readily? CH2=CH-CH2Cl or (CH3)2CHCl. Support your answer with a reason. [2]  CH2=CH-CH2Cl (allyl chloride) will be hydrolysed more readily than (CH3)2CHCl.  The allyl carbocation CH2=CH-CH2 + is more stabilised than the secondary carbocation (CH3)2CH+ .  This is due to resonance in the allyl carbocation, in which the positive charge is delocalised over two carbon atoms, making it more stable.

Marking Scheme

I - 1 mark. II - 1 mark for the choice and 1 mark for the reason.

I On adding 100 g of non-volatile solute A to 1000 g of water, the vapour pressure is reduced by 50 per cent. Calculate the molar mass of solute A. [2] (p° − ps) / p° = χB = nB / (nA + nB) = 0.5 nA = 1000 / 18 = 55.56 mol 0.5 = nB / (nB + 55.56) 0.5 n ⇒ B + 0.5 × 55.56 = nB n ⇒ B = 55.56 mol MB = wB / nB = 100 / 55.56 MB = 1.8 g mol−1

II It has been observed that a nitric acid solution containing 32% of water by mass cannot be separated into its components through fractional distillation. Why? [1]

Answer

Such a solution forms an azeotrope (a constant boiling mixture). It has the same composition in both the liquid and the vapour phase, so the components cannot be separated by fractional distillation.

Marking Scheme

I - ½ mark for the relation, ½ mark for nA, ½ mark for nB and ½ mark for the molar mass. II - 1 mark. Questions 29 and 30 are case-based / data-based questions carrying 4 marks each.

Acidity of substituted phenols. Phenol and substituted phenols are relatively more acidic than aliphatic alcohols. The pKa value is a quantitative measure of the acidic strength. The effect of substituents on the acidity of phenols (pKa values) is given in Figure 4. Study the graph and answer the following questions. Figure 4

I The pKa value of 2,4,6-trinitrophenol will be: [1]

A. more than 8.36

B. between 7.2 and 8.3

C. between 4.09 and 2.3

D. less than 4.09

ANSWER

D. less than 4.09

Solution

Each additional -NO2 group withdraws electron density and delocalises the negative charge of the phenoxide ion further. With three nitro groups the phenoxide is more stable than that of 2,4-dinitrophenol (pKa = 4.09), so the pKa must be lower still.

II The order of acidity of the p-halophenols is: [1]

A. p-bromophenol > p-chlorophenol > p-fluorophenol > phenol

B. phenol > p-fluorophenol > p-chlorophenol > p-bromophenol

C. p-chlorophenol > p-bromophenol > p-fluorophenol > phenol

D. phenol > p-fluorophenol > p-bromophenol > p-chlorophenol

ANSWER

A. p-bromophenol > p-chlorophenol > p-fluorophenol > phenol

Solution

Reading the pKa values from the graph - p-bromophenol 9.34, p-chlorophenol 9.38, p-fluorophenol 9.90, phenol 10.0 - and remembering that a lower pKa means a stronger acid. Attempt either (IIIA) or (IIIB). IIIA Predict which one of the following two compounds will have the lower value of pKa. Give a reason in support of your answer. [2]

Answer

Phenol is a stronger acid than cyclohexanol, so it has the lower pKa value.  Phenol loses a hydrogen ion to form its conjugate base, the phenoxide ion, which is stabilised by resonance.  In cyclohexanol the conjugate base formed is less stable, because the negative charge remains localised on the oxygen atom.

OR IIIB Predict whether 4-chloro-2,6-dinitrophenol is more or less acidic than 2,4,6-trinitrophenol. Use the values in the graph to support your answer. [2]

Answer

4-Chloro-2,6-dinitrophenol is less acidic than 2,4,6-trinitrophenol.  From the graph, the pKa of p-nitrophenol is 7.2 while that of p-chlorophenol is 9.38, so a para -NO2 group lowers pKa far more than a para -Cl group does.  Following this trend, the more the nitro groups, the lower the pKa. Replacing the para nitro group by chlorine therefore raises the pKa of 4-chloro-2,6-dinitrophenol above that of 2,4,6-trinitrophenol, and a higher pKa means lower acidic strength.

For Visually Challenged Learners - Read the passage and answer the questions that follow. The presence of electron withdrawing groups such as the nitro group enhances the acidic strength of phenol. It is due to the effective delocalisation of negative charge in the phenoxide ion. On the other hand, electron releasing groups, such as alkyl groups, in general do not favour the formation of the phenoxide ion, resulting in a decrease in acid strength. Cresols, for example, are less acidic than phenol. The pKa value is a quantitative measure of the acidic strength; the higher the pKa value, the lower is the acidic strength.

I Which of the following will form the least stable phenoxide ion? [1]

A. phenol

B. o-cresol

C. o-nitrophenol

D. o-chlorophenol

ANSWER

B. o-cresol

II Which group causes the maximum increase in the pKa value of phenol? [1]

A. -CH3

B. -OCH3

C. -NO2

D. -Cl

ANSWER

B. -OCH3

Solution

An increase in pKa means a decrease in acidity, which requires an electron-releasing group; this rules out -NO2 and -Cl. Of the two remaining groups, -OCH3 releases electron density into the ring most strongly through its +R (resonance) effect, destabilising the phenoxide ion to the greatest extent. Attempt either (IIIA) or (IIIB). IIIA Which of the following will be more acidic - nitrophenol or 2,4-dinitrophenol? Support your answer with an appropriate reason. [2]

Answer

2,4-Dinitrophenol is more acidic in nature.  The nitro group is a strong electron-withdrawing group, exerting both the -I and the -M effect, and so withdraws electron density from the ring; the phenoxide ion formed is stabilised.  The presence of two nitro groups makes the phenoxide ion more stable than the presence of only one.

OR IIIB Predict whether the pKa value of p-cresol or of p-nitrophenol will be higher or lower than that of phenol. Give a reason to support your answer. [2]  The pKa of p-cresol will be higher than that of phenol. The pKa value increases due to the presence of an electron-donating group.  The pKa of p-nitrophenol will be lower than that of phenol. The pKa value decreases due to the presence of an electron-withdrawing group.

Marking Scheme

I - 1 mark. II - 1 mark. III - 1 + 1 marks. Total 4 marks. (Same allocation for the Visually Challenged Learners’ version.)

Identification of carbohydrate. Students of class 12 are given a sample of carbohydrate obtained from a plant extract. They find that the carbohydrate is soluble in water. They take about 1-2 mL of an aqueous solution of the carbohydrate in a test tube and add 2 mL of Benedict’s reagent. The test tube is heated in a water bath for 3-5 minutes and a brick red precipitate is obtained. The sample, on hydrolysis with dilute acid, gives only one type of monosaccharide.

I The sample gives a positive Benedict’s test. We can conclude that the given carbohydrate is a [1]

A. monosaccharide

B. reducing sugar

C. non-reducing sugar

D. polysaccharide

ANSWER

B. reducing sugar - reducing sugars give a brick red precipitate with Benedict’s solution.

II Which of the following is a plant-based carbohydrate? (i) sucrose (ii) lactose (iii) starch (iv) glycogen [1]

A. (i) and (ii)

B. (i) and (iii)

C. (iii) and (iv)

D. (i) and (iv)

ANSWER

B. (i) and (iii) - sucrose and starch are plant-based carbohydrates; lactose is milk sugar and glycogen is animal starch. Attempt either (IIIA) or (IIIB). IIIA Considering all the observations reported in the passage, the sample is most likely (i) glucose,

(ii) fructose, (iii) sucrose, (iv) lactose or (v) maltose. Justify your answer. [2]

Answer

The sample is maltose.  The sugar is reducing in nature, as it gives a positive Benedict’s test - so it is not sucrose.  The sugar undergoes hydrolysis, so it is not a monosaccharide - it is neither glucose nor fructose.  The sugar gives only one type of monosaccharide on hydrolysis: lactose would give glucose and galactose, whereas maltose gives two molecules of glucose. Therefore it is maltose.

OR IIIB Ravi reports that the given sample is starch. What could be his justification, based on the observations in the passage? Do you agree with Ravi? Why or why not? [2]  Ravi’s justification: the carbohydrate gives only one monosaccharide on hydrolysis, and starch on hydrolysis gives only glucose.  No, the carbohydrate is not starch. The carbohydrate is soluble in water, while starch is insoluble; moreover, starch is not a reducing sugar and would not give a positive Benedict’s test.

Marking Scheme

I - 1 mark. II - 1 mark. III - 1 + 1 marks. Total 4 marks. Questions 31 to 33 are long answer type questions carrying 5 marks each.

I Account for the following: (a) The pH should be around 3.5 during the addition of ammonia derivative compounds to an aldehyde or ketone. (b) Alcohols are added to carbonyl compounds in the presence of dry HCl gas. (c) The reaction of aldehydes with sodium hydrogen sulphite is commonly used for the separation and purification of aldehydes. [3]  (a) A highly acidic medium protonates the ammonia derivative, reducing its nucleophilicity. In a basic medium the carbonyl compound is not protonated, reducing its electrophilicity for nucleophilic attack.

Hence a mildly acidic pH of about 3.5 gives the best rate for derivative formation.  (b) Dry hydrogen chloride protonates the oxygen of the carbonyl compound and therefore increases the electrophilicity of the carbonyl carbon, facilitating nucleophilic attack by the alcohol.  (c) The hydrogen sulphite addition compound is water soluble and can be converted back to the original carbonyl compound by treatment with dilute mineral acid or alkali. These adducts are therefore useful for the separation and purification of aldehydes.

II A student has three colourless samples: methanol, acetaldehyde and acetic acid. They have only NaHCO3 and 2,4-DNP reagent. Devise a flowchart to identify each compound using only these two reagents. [2]

Marking Scheme

I - 1 mark for each of (a), (b) and (c). II - ½ + ½ marks for the first test and ½ + ½ marks for the second.

OR

Alternative (B): 3 + 2 marks

I Arrange the following according to the property indicated: (a) CH3CHO, HCHO, C6H5CHO - decreasing order of the ease of oxidation; (b) CH3COCl, CH3COOH, CH3CONH2, CH3COOCH3 - increasing order of nucleophilic acyl substitution; (c) CH3CHO, CH3COCH3, CF3CHO - increasing order of electrophilicity of the carbonyl carbon. [3]  (a) HCHO > CH3CHO > C6H5CHO  (b) CH3CONH2 < CH3COOH < CH3COOCH3 < CH3COCl  (c) CH3COCH3 < CH3CHO < CF3CHO

II Write any two possible isomers of the compound having molecular formula C3H6O. Identify which of these isomers will give a positive Tollens’ test and explain the reason. [2]  Propanal: CH3-CH2-CHO (an aldehyde)  Propanone: CH3-CO-CH3 (a ketone)  Propanal gives a positive Tollens’ test (a silver mirror). Tollens’ reagent oxidises aldehydes to carboxylates, reducing Ag+ to metallic silver. Ketones have no hydrogen on the carbonyl carbon and are not oxidised under these mild conditions.

Marking Scheme

I - 1 mark for each correct order. II - ½ + ½ marks for the isomers and ½ + ½ marks for the identification with the reason.

I Two test tubes, X and Y, contain solutions of the complexes [Cr(H2O)3Cl3] and [Cr(en)3]3+ respectively. If plane-polarised light is passed through these solutions, which solution will rotate the plane of polarisation, and why? [1]

Answer

[Cr(en)3]3+ (test tube Y) will rotate the plane of polarisation. It contains three bidentate ethylenediamine ligands arranged in a chiral octahedral geometry, so it is optically active. [Cr(H2O)3Cl3] exists as fac and mer isomers, but neither is optically active because each possesses a plane of symmetry. II [Ni(CO)4] and [Ni(H2O)6]2+ are two complexes of Ni. Which complex will have the higher crystal field splitting energy, and which factor determines this difference? [1]

Answer

[Ni(CO)4] has the higher crystal field splitting, because CO is a strong field ligand whereas H2O is a weak field ligand. The nature of the ligand - its position in the spectrochemical series - determines the difference.

III A solution of the coordination compound PdCl2·4NH3 is treated with AgNO3, and 2 mol of AgCl precipitate. Determine the primary valency and the secondary valency of Pd in the compound. [1]

Answer

Two chloride ions are precipitated, so they lie outside the coordination sphere and the compound is [Pd(NH3)4]Cl2. The primary valency is the oxidation number of the metal, which is 2; the secondary valency is the coordination number, which is 4, satisfied by the four NH3 ligands.

IV The complex [Fe(CN)6]4− exhibits d2 sp3 hybridisation. Based on this, comment on the strength of the ligand CN− and its effect on the geometry of the complex. [1]

Answer

CN− is a strong field ligand. It forces the 3d electrons of Fe2+ to pair, vacating two inner d orbitals for d2 sp3 hybridisation and giving a low-spin octahedral complex that is diamagnetic.

V Why is anhydrous CuSO4 white, but CuSO4·5H2O blue in colour? [1]

Answer

In anhydrous CuSO4 there is no ligand around Cu2+ , so the d orbitals remain degenerate, crystal field splitting does not occur and no d-d transition is possible; the solid therefore appears white. In CuSO4·5H2O the water molecules act as ligands and split the d orbitals, so the d9 ion absorbs light in the visible region and the complementary blue colour is transmitted.

Marking Scheme

1 mark for each part - ½ mark for the statement and ½ mark for the reason where applicable.

OR

Alternative (B): 5 × 1 marks

Answer

the following questions.

I Three coordination compounds of cobalt(III) are given - A: [Co(NH3)6]Cl3; B: [Co(NH3)5Cl]Cl2; C:

[Co(NH3)4Cl2]Cl. Which of the three compounds will show maximum electrolytic conductivity, and why? [1]

Answer

Compound A shows maximum conductivity. A → [Co(NH3)6]3+ + 3Cl− (4 ions) B → [Co(NH3)5Cl]2+ + 2Cl− (3 ions) C → [Co(NH3)4Cl2]+ + Cl− (2 ions) The greater the number of ions furnished in solution, the higher is the conductivity.

II On the basis of crystal field theory, write the electronic configuration for a d5 ion if Δo > P. [1]

Answer

t2g 5 eg 0 - a strong field causes pairing within the t2g set, so only one unpaired electron remains.

III Using IUPAC norms, write the formula of diaquadiammine(oxalato)chromium(III) chloride.[1]

Answer

[Cr(NH3)2(H2O)2(C2O4)]Cl

IV Compare the optical activity of [Co(NH3)2Cl2(en)]Cl and [Co(NH3)4Cl2]Cl. Which one is optically active and why? [1]

Answer

[Co(NH3)2Cl2(en)]Cl is optically active; [Co(NH3)4Cl2]Cl is not. The bidentate ligand en can produce a chiral, non-superimposable octahedral arrangement with no internal mirror plane, giving optical isomerism. Complexes with only monodentate ligands, of the type MA4B2, are symmetric and achiral.

V Explain why [Cu(NH3)4]2+ shows a blue colour while [Cu(H2O)4]2+ shows a greenish colour. Which factor influences the colour difference? [1]

Answer

The different ligands (NH3 and H2O) produce different crystal field splitting , so the two ions absorb light of Δ different wavelengths and transmit different colours. The factor that controls the difference is the nature, that is the field strength, of the ligand, which fixes and hence the energy of the d-d transition. Δ

Marking Scheme

1 mark for each part.

I A student investigates the effect of concentration on the cell potential of a Zn-Cu galvanic cell at 298 K. The cell is Zn(s) | Zn2+ (aq) Cu ‖ 2+ (aq) | Cu(s), with E°(Zn2+ /Zn) = −0.76 V, E°(Cu2+ /Cu) = +0.34 V and E°cell = 1.10 V. The following experimental data were collected. [3] Experiment [Zn2+ ] / M [Cu2+ ] / M Measured EMF / V 1 1.00 1.00 1.10 2 0.1 1.00 1.16 3 1.00 0.10 1.04 4 0.01 1.00 1.22 5 1.00 0.01 0.98

(a) Why does experiment 4 show the maximum EMF?

(b) For experiment 3, calculate the theoretical EMF using the Nernst equation. Suggest one possible reason why the measured EMF differs slightly from the calculated value.

Answer And Solution

ANSWER

(a) The cell potential given by the Nernst equation increases as the reaction quotient Q = [Zn2+ ]/[Cu2+ ] decreases. In experiment 4, [Zn2+ ] = 0.01 M (very low) and [Cu2+ ] = 1.00 M (high), so Q is very small, log Q is large and negative, the −(0.059/2) log Q term is positive and large, and E is therefore the highest.

(b) E = E° − (0.059 / 2) log ([Zn2+ ] / [Cu2+ ]) = 1.10 − (0.059 / 2) log (1.00 / 0.10) = 1.10 − 0.02955 × 1 E = 1.07 V The measured EMF differs because, when current flows, the cell has an internal resistance - of the electrolyte, the electrodes and the salt bridge - which lowers the observed voltage.

II If is expressed in S m κ −1 and the concentration c in mol m−3 , the units of molar conductivity are S m2 mol−1 . However, if S cm−1 is used for and mol cm κ −3 for c, the units of molar conductivity are S cm2 mol−1 . Show how 1 S m2 mol−1 is related to 1 S cm2 mol−1 . [1] 1 m = 100 cm 1 m ⇒ 2 = 104 cm2 1 S m2 mol−1 = 104 S cm2 mol−1 or 1 S cm2 mol−1 = 10−4 S m2 mol−1

III What happens to the specific gravity of the electrolyte in a lead storage battery during charging and discharging of the battery? [1]

Answer

During charging the specific gravity of the electrolyte increases, because H2SO4 is regenerated; while discharging it decreases, because H2SO4 is consumed and water is produced.

Marking Scheme

I - 1 mark for (a); ½ + ½ marks for the calculation in (b) and ½ mark for the reason. II - 1 mark. III - ½ + ½ marks.

OR

Alternative (B): 3 + 2 marks

I For aqueous NaCl at 25 °C the molar conductivities at two concentrations are measured as Λm(0.010 M) = 123.9 S cm2 mol−1 and Λm(0.040 M) = 120.7 S cm2 mol−1 . Molar conductivity for strong electrolytes can be represented as Λm = ° Λ m − A√c. Determine ° Λ m (molar conductivity at infinite dilution) and the constant A for NaCl solution. [3] For c1 = 0.010 M: 123.9 = ° Λ m − A(0.100) …(1) For c2 = 0.040 M: 120.7 = ° Λ m − A(0.200) …(2) Subtracting (2) from (1): 3.2 = 0.100 A A = 32.0 ⇒ Substituting in (1): ° Λ m = 123.9 + 32 × 0.100 = 123.9 + 3.2 Λ°m = 127.1 S cm2 mol−1 and A = 32.0 S cm2 mol−1 M−1/2

II Write the electrode reactions that occur in a hydrogen-oxygen fuel cell. Also state two advantages of fuel cells compared to conventional batteries. [2] Anode: 2H2 → 4H+ + 4e− Cathode: O2 + 4H+ + 4e− → 2H2O Overall: 2H2 + O2 → 2H2O *In the alkaline cell described in NCERT, where the electrolyte is aqueous KOH or NaOH, the equivalent equations are anode: 2H2 + 4OH− → 4H2O + 4e− ; cathode: O2 + 2H2O + 4e− → 4OH− . Either form is acceptable.* Two advantages of fuel cells  Higher efficiency: fuel cells convert chemical energy into electrical energy more efficiently than conventional batteries.  Continuous operation: as long as fuel (hydrogen) and oxidant (oxygen) are supplied, a fuel cell can operate indefinitely, unlike a battery, which needs recharging.

Marking Scheme

I - ½ mark for each equation, ½ mark for A and ½ mark for Λ°m. II - ½ mark for each electrode reaction and ½ mark for each advantage.