9701_s24_qp

Question 1

Topic: 29.1

(a) Describe the trend in the solubility of the sulfates of magnesium, calcium and strontium. Explain your answer.

(b) Define lattice energy, \(\Delta H_{latt}\).

(d) Table 1.1 shows some energy changes.

(i) Born–Haber cycles can be used to determine the lattice energies of ionic compounds. Complete the Born–Haber cycle in Fig. 1.1 for potassium sulfide, \(K_2S\)(s). Include state symbols for all of the species.

(ii) Calculate the lattice energy, \(\Delta H^{\Theta}_{latt}\), of \(K_2S\)(s) using relevant data from Table 1.1. Show your working.

▶️ Answer/Explanation

Solution

(a)

Explanation: The solubility trend is magnesium > calcium > strontium. This is because both lattice energy (\(\Delta H_{latt}\)) and hydration energy (\(\Delta H_{hyd}\)) become less exothermic down the group. The hydration energy dominates, making the overall \(\Delta H_{sol}\) less exothermic (or more endothermic) for larger ions.

(b)

Explanation: Lattice energy (\(\Delta H_{latt}\)) is the energy change when 1 mole of an ionic solid is formed from its gaseous ions under standard conditions.

(c)

Explanation: The magnitude of lattice energy depends on (1) ionic radii (larger ions → less exothermic \(\Delta H_{latt}\)) and (2) ionic charge (higher charge → more exothermic \(\Delta H_{latt}\)).

(d)(i)

Explanation: The Born-Haber cycle includes steps for atomization, ionization, electron affinity, and formation of \(K_2S\)(s) from its elements.

(d)(ii)

Explanation: Using Hess’s Law, the lattice energy is calculated as follows:

\(\Delta H_f^\Theta = 2 \times \Delta H_{atom}(K) + 2 \times IE(K) + \Delta H_{atom}(S) + EA(S) + \Delta H_{latt}\)

Substituting values: \(-381 = (2 \times 89) + (2 \times 419) + 279 + (-200) + 640 + \Delta H_{latt}\)

Solving gives \(\Delta H_{latt}^\Theta = -2116 \text{ kJ mol}^{-1}\).

Question 2

Topic: 23.1

(a) (i) Lithium nitrate, LiNO₃, decomposes on heating in a similar way to Group 2 nitrates to give the metal oxide, a brown gas and oxygen. Write an equation for the decomposition of LiNO₃.

(ii) The other Group 1 nitrates, \(MNO_3\), decompose on heating to form the metal nitrite, \(MNO_2\), and oxygen. The thermal stability of these nitrates increases down the group. Suggest why the thermal stability of \(MNO_3\) increases down the group.

(b) Acidified manganate(VII) ions, MnO₄⁻, can be used to analyse solutions containing nitrite ions, NO₂⁻, by titration. X is a solution of NaNO₂. 250.0 cm³ of X is added to 50.0 cm³ of 0.125 moldm⁻³ acidified MnO₄⁻ (aq). The MnO₄⁻ (aq) ions are in excess; all the NO₂⁻ ions are oxidised in the reaction. The unreacted MnO₄⁻ (aq) required 22.50 cm³ of 0.0400 moldm⁻³ Fe²⁺ (aq) to reach the end‑point. The relevant half-equations are shown.

Calculate the concentration, in mol/dm³, of NaNO₂ in X

(i) Aqueous manganate(VI) ions, MnO₄²⁻, are unstable in acidic conditions and undergo a disproportionation reaction. The \(E^o_{cell}\) for this reaction is +1.14V. Construct an overall ionic equation for this disproportionation reaction

(ii) Suggest and explain how the \(E_{cell}\) value of the disproportionation reaction changes with an increase in pH.

▶️ Answer/Explanation

Solution

(a) (i) \( 4LiNO_3 \rightarrow 2Li_2O + 4NO_2 + O_2 \)

Explanation: Lithium nitrate decomposes similarly to Group 2 nitrates, forming lithium oxide (\(Li_2O\)), nitrogen dioxide (\(NO_2\), the brown gas), and oxygen (\(O_2\)).

(a) (ii) The thermal stability increases down the group because the larger cations (e.g., \(K^+\), \(Rb^+\)) have lower charge density and polarize the nitrate ion less, making it harder to decompose.

Explanation: Larger alkali metal ions stabilize the nitrate ion due to weaker polarizing power, reducing the tendency to break the \(NO_3^-\) structure.

(b) The concentration of NaNO₂ in X is \(0.0125 \, \text{mol/dm}^3\).

Explanation: 1. Moles of MnO₄⁻ initially = \(50.0 \times 0.125 / 1000 = 0.00625 \, \text{mol}\).
2. Moles of Fe²⁺ used = \(22.50 \times 0.0400 / 1000 = 0.0009 \, \text{mol}\).
3. Moles of MnO₄⁻ remaining = \(0.0009 / 5 = 0.00018 \, \text{mol}\) (since 1 MnO₄⁻ reacts with 5 Fe²⁺).
4. Moles of MnO₄⁻ reacted with NO₂⁻ = \(0.00625 – 0.00018 = 0.00607 \, \text{mol}\).
5. Moles of NO₂⁻ = \(0.00607 \times 2 = 0.01214 \, \text{mol}\) (from the half-equations).
6. Concentration of NaNO₂ = \(0.01214 / (250/1000) = 0.0125 \, \text{mol/dm}^3\).

(c) (i) \( 3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O \)

Explanation: Manganate(VI) disproportionates into manganate(VII) (\(MnO_4^-\)) and manganese(IV) oxide (\(MnO_2\)) in acidic conditions.

(c) (ii) The \(E_{cell}\) decreases as pH increases because \(H^+\) ions are consumed in the reaction. Higher pH reduces the driving force for disproportionation.

Explanation: The Nernst equation shows \(E_{cell}\) depends on \([H^+]\). Lower \([H^+]\) (higher pH) reduces the cell potential.

Question 3

Topic: 25.2

(a) Carbon disulfide, CS₂, is flammable and reacts readily with oxygen, as shown in reaction 1.

reaction 1 \(CS_2(g) + 3O_2(g) \to CO_2(g) + 2SO_2(g)\)
Table 3.1 shows the standard enthalpy of formation, \(\Delta H^o_f\), and the standard entropy, \(S^o\), for some substances.

Calculate the standard Gibbs free energy change, \(\Delta G^o\), in \(kJmol^{–1}\), for reaction 1 at 25°C.

(b) Carbon disulfide reacts with chlorine to form tetrachloromethane, as shown in reaction 2.

Calculate the maximum temperature, in K, for reaction 2 to be feasible.

▶️ Answer/Explanation

Solution

(a) \(\Delta G^o = -1074.2 \text{ kJmol}^{-1}\).

Explanation:

Calculate \(\Delta H^o\) using \(\Delta H^o = \sum \Delta H^o_f(\text{products}) – \sum \Delta H^o_f(\text{reactants})\): \[ \Delta H^o = [(-393.5) + 2(-296.8)] – [117.4 + 3(0)] = -1104.5 \text{ kJmol}^{-1} \]
Calculate \(\Delta S^o\) using \(\Delta S^o = \sum S^o(\text{products}) – \sum S^o(\text{reactants})\): \[ \Delta S^o = [213.6 + 2(248.1)] – [237.8 + 3(205.0)] = -101.8 \text{ Jmol}^{-1}\text{K}^{-1} \]
Convert \(\Delta S^o\) to kJ: \(-101.8 \times 10^{-3} \text{ kJmol}^{-1}\text{K}^{-1}\).
Calculate \(\Delta G^o\) using \(\Delta G^o = \Delta H^o – T\Delta S^o\) at 298K: \[ \Delta G^o = -1104.5 – (298 \times -0.1018) = -1074.2 \text{ kJmol}^{-1} \]

(b) Maximum feasible temperature = 629 K.

Explanation:

Calculate \(\Delta H^o\) for reaction 2: \[ \Delta H^o = [(-106.7) + 2(-296.8)] – [117.4 + 2(0)] = -817.7 \text{ kJmol}^{-1} \]
Calculate \(\Delta S^o\) for reaction 2: \[ \Delta S^o = [309.4 + 2(248.1)] – [237.8 + 2(223.0)] = 122.8 \text{ Jmol}^{-1}\text{K}^{-1} \]
Convert \(\Delta S^o\) to kJ: \(122.8 \times 10^{-3} \text{ kJmol}^{-1}\text{K}^{-1}\).
For feasibility, \(\Delta G^o \leq 0\). Solve for \(T\): \[ T \leq \frac{\Delta H^o}{\Delta S^o} = \frac{-817.7}{-0.1228} = 629 \text{ K} \]

Question 4

Topic: 24.1

(a) (i) Explain why transition elements have variable oxidation states.

(ii) Sketch the shape of a \(3d_z^2\) orbital in Fig. 4.1.

(b) Samples of [Cu(H₂O)₆]²⁺(aq) are reacted separately with an excess of solution A and with an excess of solution B. The reaction of [Cu(H₂O)₆]²⁺(aq) with solution A is a precipitation reaction. The reaction of [Cu(H₂O)₆]²⁺(aq) with solution B is a ligand substitution reaction. Suggest a possible identity for solution A and for solution B. Give relevant observations and the formula of the copper‑containing product for each reaction.

(d) Two bidentate ligands are shown in Fig. 4.2.

Explain what is meant by a bidentate ligand.

(e) Ruthenium(III) ions, Ru³⁺, form an octahedral complex, [Ru(dpys)₂Cl₂]⁺, with the ligands dpys and chloride ions. This complex shows the same kind of stereoisomerism as [Ru(NH₃)₄Cl₂]⁺ but also shows a different type of stereoisomerism.
(i) Complete the three‑dimensional diagrams in Fig. 4.3 to show the three different stereoisomers of \([Ru(dpys)_2Cl_2]^+\). The dpys ligand can be represented using

(ii) State the different types of stereoisomerism shown by [Ru(dpys)₂Cl₂]⁺.
(iii) Deduce which stereoisomers in (e)(i) are non-polar. Explain your answer.

▶️ Answer/Explanation

Solution

(a)(i) Transition elements have variable oxidation states because their 3d and 4s orbitals are close in energy, allowing electrons to be lost from both subshells.

Explanation: The small energy difference between 3d and 4s orbitals enables multiple electron loss/gain, e.g., Fe²⁺ (3d⁶) and Fe³⁺ (3d⁵).

(a)(ii)

Explanation: The \(3d_z^2\) orbital has a dumbbell shape along the z-axis with a doughnut-shaped ring in the xy-plane.

(b)
Solution A: NaOH (aq)
Observations: Blue solution forms a pale blue precipitate.
Product: Cu(OH)₂
Solution B: NH₃ (aq)
Observations: Blue solution turns deep blue.
Product: [Cu(NH₃)₄(H₂O)₂]²⁺

Explanation: NaOH precipitates Cu(OH)₂, while NH₃ substitutes H₂O ligands to form a deep blue tetraammine complex.

(c) [Ag(NH₃)₂]⁺ is colourless because the d-d transition requires energy that corresponds to UV light (not visible light).

Explanation: The energy gap between d orbitals in Ag⁺ (d¹⁰) is too large for visible light absorption; only UV light is absorbed.

(d) A bidentate ligand donates two lone pairs to a metal ion, forming two coordinate bonds (e.g., ethanediamine or oxalate).

Explanation: Bidentate ligands like dpys bind via two atoms (e.g., N in dpys), creating chelate rings for stability.

(e)(i)

(e)(ii) Cis-trans (geometric) and optical isomerism.

Explanation: The cis isomer has two types (optical isomers), while the trans isomer is non-polar.

(e)(iii) The trans isomer is non-polar because the dipoles of the dpys ligands cancel out.

Explanation: In the trans isomer, symmetric arrangement nullifies net dipole moment, unlike the polar cis isomers.

Question 5

Topic: 28.1

(a) Nitrosyl chloride, NOCl, can be formed by the reaction between nitrogen monoxide and chlorine, as shown.

\(2NO + Cl_2 \to 2NOCl\)

The initial rate of this reaction is investigated, starting with different concentrations of NO and Cl₂. The results obtained are shown in Table 5.1.

(i) Use the data in Table 5.1 to deduce the rate equation for this reaction. Explain your reasoning.

(ii) Use your rate equation from (a)(i) and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k.

(b) NO₂Cl is another compound containing nitrogen, oxygen and chlorine. In sunlight, NO₂Cl can undergo homolytic fission to release chlorine radicals which can catalyse the conversion of ozone, \(O_3\), into oxygen. Complete the mechanism for this process.

(c) Ozone reacts with nitrogen dioxide, as shown.
\(O_3+ 2NO_2 \to N_2O_5 + O_2\)
The rate of reaction is first order with respect to \(O_3\) and first order with respect to \(NO_2\). Suggest equations for a two‑step mechanism for this reaction.

▶️ Answer/Explanation

Solution

(a)(i) Rate equation: \(\text{rate} = k[\text{NO}]^2[\text{Cl}_2]\)

Explanation:
1. Comparing experiments 1 and 2: Doubling [NO] (while keeping [Cl₂] constant) quadruples the rate → second order with respect to NO.
2. Comparing experiments 1 and 3: Doubling [Cl₂] (while keeping [NO] constant) doubles the rate → first order with respect to Cl₂.

(a)(ii) Rate constant calculation:
Using experiment 1 data: \(1.14 \times 10^{-5} = k(0.250)^2(0.100)\)
Solving for \(k\): \(k = \frac{1.14 \times 10^{-5}}{0.00625} = 1.82 \times 10^{-3}\)
Units: \(\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\)

(b) Mechanism completion:
Step 1: \(\text{NO}_2\text{Cl} \rightarrow \text{NO}_2 + \text{Cl}\cdot\) (homolytic fission)
Step 2: \(\text{Cl}\cdot + \text{O}_3 \rightarrow \text{ClO}\cdot + \text{O}_2\)
Step 3: \(\text{ClO}\cdot + \text{O}_3 \rightarrow \text{Cl}\cdot + 2\text{O}_2\)
Explanation: Chlorine radicals (\(\text{Cl}\cdot\)) catalyze the breakdown of ozone in a cyclic process.

(c) Two-step mechanism:
Step 1 (slow): \(\text{O}_3 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{O}_2\)
Step 2 (fast): \(\text{NO}_3 + \text{NO}_2 \rightarrow \text{N}_2\text{O}_5\)
Explanation: The rate-determining step involves one molecule of each reactant, consistent with the observed first-order dependence on both \(O_3\) and \(NO_2\).

Question 6

Topic: 31.2

(a) Aqueous solutions of methanoic acid, HCOOH, and propanoic acid, CH₃CH₂COOH, are mixed together. An equilibrium is set up between two conjugate acid–base pairs.

(i) Define conjugate acid–base pair.

(ii) The pKa of HCOOH is 3.75 and of CH₃CH₂COOH is 4.87. Complete the equation for the Brønsted–Lowry equilibrium between the stronger of these two acids and water.

(b) (i) Write an expression for the acid dissociation constant, \(K_a\), for butanoic acid, CH₃CH₂CH₂COOH.

(ii) The \(pK_a\) of CH₃CH₂CH₂COOH is 4.82. A solution of CH₃CH₂CH₂COOH(aq) has a pH of 3.25. Calculate the concentration, in mol/dm³, of CH₃CH₂CH₂COOH in this solution.

(ii) A buffer solution containing a mixture of CH₃COOH and CH₃COONa is prepared as follows. A solution of 600 cm³ of CH₃COOH is mixed with 400 cm³ of 0.125 mol/dm³ CH₃COONa. The buffer solution has pH 5.70. The Ka of CH₃COOH is 1.78 × 10⁻⁵ moldm⁻³. Calculate the initial concentration, in mol/dm³, of CH₃COOH used.

(d) A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel. Methanoic acid, HCOOH, is being investigated as a fuel in fuel cells. When the cell operates, HCOOH is oxidised to carbon dioxide. The half‑equation for the reaction at the cathode is: \(O_2 + 4H^+ + 4e^– \to 2H_2O\).
In this fuel cell, the overall cell reaction is the same as that for the complete combustion of HCOOH.
(i) Deduce the half‑equation for the reaction at the anode.

(ii) Calculate the volume, in cm³, of oxygen used when a current of 3.75A is delivered by the cell for 40.0 minutes. Assume the cell operates at room conditions.

▶️ Answer/Explanation

Solution

(a)(i) A conjugate acid–base pair consists of two species that differ by the presence or absence of a proton (\(H^+\)). For example, \(HCOOH\) (acid) and \(HCOO^-\) (base) form a conjugate pair.

(a)(ii) The stronger acid is HCOOH (lower pKa = 3.75). The equilibrium equation is: \[ HCOOH + H_2O \rightleftharpoons H_3O^+ + HCOO^- \] Explanation: The stronger acid donates a proton to water, forming hydronium ions (\(H_3O^+\)) and the conjugate base (\(HCOO^-\)).

(b)(i) The \(K_a\) expression for butanoic acid is: \[ K_a = \frac{[H^+][CH_3CH_2CH_2COO^-]}{[CH_3CH_2CH_2COOH]} \]

(b)(ii) Given pH = 3.25, \([H^+] = 10^{-3.25} = 5.62 \times 10^{-4} \ \text{mol/dm}^3\). Using \(K_a = 10^{-4.82} = 1.51 \times 10^{-5}\): \[ 1.51 \times 10^{-5} = \frac{(5.62 \times 10^{-4})^2}{[CH_3CH_2CH_2COOH]} \] Solving gives \([CH_3CH_2CH_2COOH] = 0.0209 \ \text{mol/dm}^3\). Explanation: The concentration is derived from the \(K_a\) expression, assuming \([H^+] \approx [CH_3CH_2CH_2COO^-]\).

(c)(i) A buffer solution resists changes in pH when small amounts of acid or base are added. It contains a weak acid and its conjugate base (or vice versa).

(c)(ii) Using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \] Substituting pH = 5.70, \(K_a = 1.78 \times 10^{-5}\) (pKa = 4.75), and \([CH_3COO^-] = 0.125 \times \frac{400}{1000} = 0.05 \ \text{mol}\): \[ 5.70 = 4.75 + \log \left( \frac{0.05}{[\text{CH}_3\text{COOH}]} \right) \] Solving gives \([CH_3COOH] = 0.0562 \ \text{mol/dm}^3\). Explanation: The initial concentration is calculated using the buffer equation and mole ratios.

(d)(i) The anode half-reaction is: \[ HCOOH \to CO_2 + 2H^+ + 2e^- \] Explanation: Methanoic acid is oxidised to CO₂, releasing protons and electrons.

(d)(ii) Total charge \(Q = 3.75 \ \text{A} \times 2400 \ \text{s} = 9000 \ \text{C}\). Moles of electrons = \(9000 / 96485 = 0.0933 \ \text{mol}\). Since \(O_2 + 4e^- \to 2H_2O\), moles of \(O_2 = 0.0933 / 4 = 0.0233 \ \text{mol}\). Volume at RTP = \(0.0233 \times 24000 = 560 \ \text{cm}^3\). Explanation: The volume of oxygen is calculated using Faraday’s laws and stoichiometry.

Question 7

Topic: 33.3

Methyl red can be synthesised as shown in Fig. 7.1.

(a) (i) Give the systematic name of P.

(ii) P can be synthesised as shown in Fig. 7.2.

Suggest reagents and conditions for this reaction.
(iii) A student attempts to synthesise P by an alternative route, as shown in Fig. 7.3. Compound T is the major product in this reaction rather than P.

Explain why T is the major product in this reaction.

(b) S reacts in a similar way to phenol in step 3.
(i) Draw the structures of Q, R and S in the boxes in Fig. 7.1.
(ii) Suggest reagents and conditions for steps 1 and 2 in Fig. 7.1.

▶️ Answer/Explanation

Solution

(a)(i)

Answer: 2-nitrobenzoic acid OR 2-nitrobenzenecarboxylic acid.

Explanation: The systematic name of P is derived from the parent structure benzoic acid with a nitro group at the 2-position (ortho to the carboxyl group).

(a)(ii)

Answer: Hot/reflux conditions with acidified/alkaline \( KMnO_4 \) (potassium permanganate).

Explanation: The oxidation of a methyl group (\( -CH_3 \)) to a carboxyl group (\( -COOH \)) requires a strong oxidising agent like \( KMnO_4 \) under heating or reflux.

(a)(iii)

Answer: The carboxyl group is electron-withdrawing and meta-directing, favouring nitration at the 3- and 5-positions (T) over the 2-position (P).

Explanation: Electron-withdrawing groups like \( -COOH \) deactivate the benzene ring and direct electrophilic substitution (nitration) to the meta positions, making T the major product.

(b)(i)

Answer: Structures of Q, R, and S

Explanation: Q is the diazonium salt formed from P, R is the intermediate after coupling, and S is the final methyl red structure (as shown in the image).

(b)(ii)

Answer: Step 1: \( NaNO_2 \) and dilute HCl at 0–5 °C (diazotisation).
Step 2: \( N,N \)-dimethylaniline in acidic/alkaline medium (coupling reaction).

Explanation: Step 1 converts the amine to a diazonium salt, while Step 2 involves an azo coupling reaction with \( N,N \)-dimethylaniline to form methyl red.

Question 8

Topic: 30.1

(a) State the relative basicities of phenylamine, C₆H₅NH₂, benzylamine, C₆H₅CH₂NH₂, and ammonia, NH₃, in aqueous solution. Explain your answer.

(b) An excess of \(Br_2(aq)\) is added to separate samples of \(C_6H_5NH_2\) and benzene, \(C_6H_6\).
(i) \(C_6H_5NH_2\) reacts readily with \(Br_2 (aq)\) to form organic product M. State the expected observations for this reaction. Draw the structure of M.

(ii) \(C_6H_6\) does not react with \(Br_2 (aq)\). Suggest why \(Br_2 (aq)\) reacts with \(C_6H_5NH_2\) but not with \(C_6H_6\).

(d) C₆H₅CONH₂ is formed by reacting benzoyl chloride, \(C_6H_5COCl\), with \(NH_3\). Complete the mechanism in Fig. 8.1 for the reaction of \(C_6H_5COCl\) with \(NH_3\). Include all relevant lone pairs of electrons, curly arrows, charges and dipoles. Draw the structure of the organic intermediate.

(e) Phenylalanine, C₆H₅CH₂CH(NH₂)COOH, is an amino acid with an isoelectric point of 5.5.
(i) State what is meant by isoelectric point.

(ii) Draw the structure of C₆H₅CH₂CH(NH₂)COOH at pH 10.
(f) C₆H₅CH₂CH(NH₂)COOH and alanine, \(CH_3CH(NH_2)COOH\), react to form a dipeptide containing both amino acid residues. Draw the structure of this dipeptide. The peptide functional group formed should be displayed.

▶️ Answer/Explanation

Solution

(a) Benzylamine > Ammonia > Phenylamine

Explanation:

Benzylamine’s alkyl group (\(CH_2\)) donates electrons via the +I effect, increasing basicity.
Ammonia has no electron-withdrawing or donating groups.
Phenylamine’s lone pair delocalizes into the benzene ring, reducing availability for protonation.

(b)(i) Observations: White precipitate (2,4,6-tribromoaniline) forms; brown bromine color decolorizes.

Explanation: The \(-NH_2\) group activates the benzene ring, enabling electrophilic substitution at the 2, 4, and 6 positions.

(b)(ii) \(C_6H_5NH_2\) reacts because the \(-NH_2\) group donates electron density to the ring, polarizing \(Br_2\). Benzene lacks an activating group and requires a catalyst (e.g., FeBr₃) for bromination.

Explanation: The lone pair on \(-NH_2\) increases the ring’s electron density, facilitating electrophilic attack.

(c) Benzamide’s lone pair on nitrogen is delocalized into the carbonyl (\(C=O\)) group, reducing its availability to accept protons.

Explanation: The adjacent carbonyl withdraws electron density via resonance, making the nitrogen less basic than in NH₃.

(d) Nucleophilic acyl substitution mechanism:

Explanation: NH₃ attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate. Chloride leaves, yielding benzamide.

(e)(i) Isoelectric point (pI): The pH at which an amino acid has no net charge (zwitterion form).

Explanation: At pI (5.5 for phenylalanine), the \(-NH_3^+\) and \(-COO^-\) groups are balanced.

(e)(ii) At pH 10, phenylalanine is deprotonated: \(C_6H_5CH_2CH(NH_2)COO^-\).

Explanation: The \(-COOH\) group loses \(H^+\) (pKa ~2.2), and \(-NH_3^+\) deprotonates to \(-NH_2\) (pKa ~9.1).

(f) Dipeptide structure (phenylalanine-alanine or alanine-phenylalanine):

Explanation: Condensation reaction forms a peptide bond (\(-CONH-\)) between the carboxyl and amino groups of the two amino acids.

Question 9

Topic: 32.2

(a) Explain why trichloroethanoic acid, CCl₃COOH, is more acidic than ethanoic acid, CH₃COOH.
(b) Acyl chlorides are formed by reacting carboxylic acids with thionyl chloride, SOCl₂.
(i) Ethanedioyl chloride, (COCl)₂, can be prepared by reacting ethanedioic acid, (COOH)₂, with an excess of SOCl₂. Write an equation for this reaction.
(ii) Samples of (COCl)₂ are reacted separately with an excess of warm acidified KMnO₄ (aq) and with H₂NCH₂CH₂NH₂. The carbon‑containing product from the reaction with H₂NCH₂CH₂NH₂ has the molecular formula \(C_4H_6N_2O_2\). Complete the boxes in Fig. 9.1 to suggest the structure of the carbon‑containing product in each reaction.

(iii) A polyester can be synthesised from the reaction of (COCl)₂ with ethane‑1,2‑diol, HOCH₂CH₂OH. Draw two repeat units of the polymer formed. Any functional groups should be displayed.

(c) Compound H, C₆H₁₀O₃, reacts with alkaline I₂ (aq) to form yellow precipitate J but does not react with Na₂CO₃ (aq). The proton (1H) NMR spectrum of H in CDCl₃ is shown in Fig. 9.2.

(i) Identify yellow precipitate J.
(ii) Complete Table 9.2 for the proton \((^1H)\) NMR spectrum of H, C₆H₁₀O₃.

(iii) Suggest a structure for H, \(C_6H_{10}O_3\).

▶️ Answer/Explanation

Solution

(a) Due to the electron-withdrawing effect (–I effect) of chlorine atoms, which stabilizes the carboxylate anion (CCl₃COO⁻) and weakens the O–H bond.

Explanation: The three chlorine atoms in trichloroethanoic acid withdraw electron density via induction, making the conjugate base more stable and the acid more acidic compared to ethanoic acid.

(b)(i) \((COOH)_2 + 2SOCl_2 \to (COCl)_2 + 2HCl + 2SO_2\)

Explanation: Thionyl chloride (SOCl₂) converts both carboxylic acid groups into acyl chlorides, releasing HCl and SO₂ as byproducts.

(b)(ii)

Explanation:

With KMnO₄: Oxidative cleavage yields CO₂ (carbon-containing product).
With H₂NCH₂CH₂NH₂: Nucleophilic substitution forms a diamide (\(C_4H_6N_2O_2\)).

(b)(iii) Two repeat units of the polyester:

Explanation: The reaction between (COCl)₂ and ethane-1,2-diol forms a polyester with repeating units of \(-CO-CO-O-CH_2-CH_2-O-\).

(c)(i) \(CHI_3\) (triiodomethane / iodoform)

Explanation: The yellow precipitate indicates a positive iodoform test, confirming the presence of a methyl ketone (\(CH_3CO-\)) group.

(c)(ii)

Explanation: The NMR data suggests:

Singlet at 2.1 ppm (3H): Methyl group adjacent to carbonyl.
Multiplet at 2.5 ppm (2H): Methylene near carbonyl.
Triplet at 3.7 ppm (2H): Methylene next to oxygen.
Singlet at 12.2 ppm (1H): Carboxylic acid proton.

(c)(iii) Structure of H: \(CH_3COCH_2CH_2COOH\) (4-oxopentanoic acid)

Explanation: The NMR spectrum and reactions confirm a methyl ketone (\(CH_3CO-\)) and a carboxylic acid (\(-COOH\)) separated by a \(CH_2\) group.

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