9701_w24_qp

Question 1

Topic: 24.2

(a) Disodium phosphate, \((Na^+)_2(HPO_4^{2–})\), reacts with an acid to form monosodium phosphate, \(Na^+(H_2PO_4^–)\).
(i) Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved.
(ii) Define buffer solution.
(iii) Write two equations to show how a mixture of \((Na^+)_2(HPO_4^{2–})\) and \(Na^+(H_2PO_4^–)\) can act as a buffer solution.
(iv) Identify one inorganic ion that acts as a buffer in blood.

(b) Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63g of E is dissolved in water to make 250 cm³ of solution F. Solution F has a pH of 13.09 at 298K.
(i) Show that the concentration of hydroxide ions in solution F is 0.123 moldm⁻³.
(ii) Explain why the concentration of compound E in solution F is 0.0615moldm⁻³.
(iii) Use the concentration given in (ii) to identify compound E.

(c) Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10⁻⁴ moldm⁻³ at 298K. Calculate the solubility product, \(K_{sp}\), of magnesium hydroxide. Include units.

(d) Explain why compound E is much more soluble than magnesium hydroxide.

▶️ Answer/Explanation

Solution

(a)(i) Conjugate acid: \(H_2PO_4^–\), Conjugate base: \(HPO_4^{2–}\).

Explanation: The pair \(HPO_4^{2–}/H_2PO_4^–\) are conjugate base and acid respectively, as they differ by one proton.

(a)(ii) A buffer solution resists changes in pH when small amounts of acid or base are added.

Explanation: It maintains pH stability through equilibrium shifts between its weak acid and conjugate base components.

(a)(iii)
Equation 1: \(HPO_4^{2–} + H^+ \to H_2PO_4^–\) (neutralizes added acid)
Equation 2: \(H_2PO_4^– + OH^– \to HPO_4^{2–} + H_2O\) (neutralizes added base)

Explanation: The buffer system uses both reactions to maintain pH by consuming added \(H^+\) or \(OH^–\).

(a)(iv) \(HCO_3^–\) (hydrogen carbonate).

Explanation: In blood, \(HCO_3^–/H_2CO_3\) acts as the primary buffer system to regulate pH.

(b)(i) \([OH^–] = 0.123 \, \text{moldm}^{-3}\).

Explanation: Given pH = 13.09, pOH = 14 – 13.09 = 0.91. Thus, \([OH^–] = 10^{-0.91} = 0.123 \, \text{moldm}^{-3}\).

(b)(ii) Concentration of E is 0.0615 moldm⁻³.

Explanation: Since E is \(X(OH)_2\), it dissociates completely to give 2 \(OH^–\) ions per formula unit. Thus, \([E] = \frac{[OH^–]}{2} = \frac{0.123}{2} = 0.0615 \, \text{moldm}^{-3}\).

(b)(iii) Barium hydroxide, \(Ba(OH)_2\).

Explanation:
– Moles of E in 250 cm³ = 0.0615 × 0.25 = 0.0154 mol
– Molar mass = \(\frac{2.63}{0.0154} ≈ 171 \, \text{g/mol}\)
– This matches \(Ba(OH)_2\) (Ba = 137, O = 16, H = 1 → 137 + 2×(16+1) = 171).

(c) \(K_{sp} = 1.10 \times 10^{-11} \, \text{mol}^3\text{dm}^{-9}\).

Explanation:
For \(Mg(OH)_2\): \(K_{sp} = [Mg^{2+}][OH^–]^2\)
Given \([Mg^{2+}] = 1.4 \times 10^{-4} \, \text{moldm}^{-3}\), \([OH^–] = 2.8 \times 10^{-4} \, \text{moldm}^{-3}\)
Thus, \(K_{sp} = (1.4 \times 10^{-4}) \times (2.8 \times 10^{-4})^2 = 1.10 \times 10^{-11}\).

(d) Three key reasons:

Explanation:
1. Lattice energy (\(\Delta H_{latt}\)) decreases more than hydration energy (\(\Delta H_{hyd}\)) down Group 2.
2. The change in \(\Delta H_{latt}\) dominates, making \(\Delta H_{sol}\) more exothermic for larger ions like Ba²⁺.
3. \(Ba(OH)_2\) has weaker ionic bonds (due to larger ion size) compared to \(Mg(OH)_2\), enhancing solubility.

Question 2

Topic: 23.3

(a) Predict and explain the variation in the enthalpy change of hydration for the ions \(F^–\), \(Cl^–\), \(Br^–\), and \(I^–\).

(b) Fig. 2.1 shows an incomplete energy cycle involving calcium fluoride, CaF₂.

(i) Complete line D. Include state symbols.
(ii) Identify the five enthalpy changes needed to calculate process 1.
(iii) Define lattice energy, \(\Delta H_{latt}\).
(iv) Relate \(\Delta H_{latt}\) to processes 1 and 3.

(d) Define entropy.

(e) Calculate ΔS for the solution of compound T at 298K.

(f) Predict solubility change of T when heated from 298K to 360K.

▶️ Answer/Explanation

Solution

(a) The enthalpy change of hydration becomes less exothermic from \(F^–\) to \(I^–\). This occurs because larger ions (increasing ionic radius) have weaker ion-dipole interactions with water molecules.

(b)(i) Line D represents the aqueous state: \(CaF_2(aq)\) or \(Ca^{2+}(aq) + 2F^-(aq)\).

(b)(ii) The five required enthalpy changes are:
1. Enthalpy of atomization of Ca (\(\Delta H_{at}(Ca)\))
2. Enthalpy of atomization of F₂ (\(\Delta H_{at}(F_2)\))
3. First ionization energy of Ca (\(IE_1(Ca)\))
4. Second ionization energy of Ca (\(IE_2(Ca)\))
5. Electron affinity of F (\(\Delta H_{ea}(F)\))

(b)(iii) Lattice energy (\(\Delta H_{latt}\)) is the energy released when one mole of ionic solid forms from its gaseous ions.

(b)(iv) \(\Delta H_{latt} = \Delta H_3 – \Delta H_1\)

(c) Using Hess’s Law with Table 2.1 data:
\(1395 – 1650 + 2\Delta H_{hyd}(F^-) = -1214 + 13\)
Solving gives \(\Delta H_{hyd}(F^-) = -473 \text{ kJ mol}^{-1}\)

(d) Entropy (ΔS) measures the number of possible microscopic arrangements of a system.

(e) Using \(\Delta G = \Delta H – T\Delta S\):
\(6.00 = 30.0 – 298\Delta S\)
\(\Delta S = 80.5 \text{ J mol}^{-1} K^{-1}\)

(f) Compound T becomes more soluble as temperature increases because ΔS is positive. The \(-T\Delta S\) term becomes more negative, making ΔG more negative.

Question 3

Topic: 25.1

(a) A and B react together to give product AB.

\(A + B \to AB\)
When the concentrations of A and B are both 0.0100 moldm⁻³, the rate of formation of AB is 7.62 × 10⁻⁴ moldm⁻³ s⁻¹. When the concentrations of A and B are both 0.0200 moldm⁻³,
the rate of formation of AB is 3.05 × 10⁻³ moldm⁻³ s⁻¹.
(i) Complete the three possible rate equations that are consistent with these data.
(ii) Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, k. Include the units of k.

(iii) Explain why it is not possible to calculate a value for the half-life, \(t_{\frac{1}{2}}\), of this reaction using the value of the rate constant k calculated in (ii) and the equation k = 0.693/\(t_{\frac{1}{2}}\).

(b) Catalysts may be homogeneous or heterogeneous.
(i) Identify two metals that act as heterogeneous catalysts in the removal of \(NO_2\) from the exhaust gases of car engines.

(ii) Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst.

(iii) Fe²⁺ ions act as a homogeneous catalyst in the reaction between I⁻(aq) and S₂O₈²⁻(aq). Write equations for the two reactions that occur when Fe²⁺(aq) is added to a mixture of I⁻(aq) and S₂O₈²⁻(aq).

(iv) Explain the difference between a homogeneous catalyst and a heterogeneous catalyst.

(c) \(Fe^{2+}\) ions can be oxidised to \(Fe^{3+}\) ions under alkaline conditions by suitable oxidising agents.
(i) Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states.
(ii) The half-equation for the reduction of Fe³⁺ under alkaline conditions, and its \(E^o\) value, are shown.

Four more half-equations for reactions under alkaline conditions, and their \(E^o\) values, are shown.

Select two oxidising agents that can oxidise Fe²⁺ ions to Fe³⁺ ions under alkaline conditions. Write an equation, and give the \(E^0_{cell}\) value, for each of the two reactions that occur

▶️ Answer/Explanation

Solution

(a)(i) Three possible rate equations:

1. rate = k[A][B]

2. rate = k[A]²

3. rate = k[B]²

Explanation: When both concentrations double, the rate quadruples (from 7.62×10⁻⁴ to 3.05×10⁻³), indicating a second-order reaction. The rate could depend on either [A]², [B]², or [A][B].

(a)(ii) Calculation of rate constant k:

Using rate = k[A][B]:

k = rate / ([A][B]) = 7.62 × 10⁻⁴ / (0.01 × 0.01) = 7.62 mol⁻¹ dm³ s⁻¹

Explanation: The units of k are derived from the second-order rate equation (mol⁻¹ dm³ s⁻¹).

(a)(iii) Half-life explanation:

The equation \(k = 0.693/t_{1/2}\) applies only to first-order reactions. Since this reaction is second-order, the half-life depends on initial concentrations and cannot be calculated using this formula.

(b)(i) Heterogeneous catalysts: Platinum (Pt) and Rhodium (Rh).

Explanation: These metals are used in catalytic converters to reduce NO₂ emissions from car exhausts.

(b)(ii) Mode of action of iron catalyst:

1. Adsorption of N₂ and H₂ onto the iron surface.

2. Weakening of N≡N and H-H bonds.

3. Formation of NH₃, which then desorbs from the surface.

Explanation: The iron catalyst provides a surface for reactant molecules to interact, lowering the activation energy.

(b)(iii) Homogeneous catalysis equations:

1. \(S_2O_8^{2-} + 2Fe^{2+} → 2SO_4^{2-} + 2Fe^{3+}\)

2. \(2Fe^{3+} + 2I^- → 2Fe^{2+} + I_2\)

Explanation: Fe²⁺ is oxidised to Fe³⁺ by S₂O₈²⁻, then Fe³⁺ oxidises I⁻ to I₂ while being reduced back to Fe²⁺.

(b)(iv) Catalyst difference:

• Homogeneous: Catalyst and reactants in the same phase (e.g., all in solution).

• Heterogeneous: Catalyst and reactants in different phases (e.g., solid catalyst with gaseous reactants).

(c)(i) Stable oxidation states:

Iron exhibits stable +2 and +3 states because its 3d and 4s orbitals have similar energies, allowing variable electron loss.

(c)(ii) Selected oxidising agents:

1. \(H_2O_2 + 2Fe^{2+} → 2Fe^{3+} + 2OH^-\) (E°cell = +1.71 V)

2. \(ClO^- + 2Fe^{2+} + H_2O → 2Fe^{3+} + Cl^- + 2OH^-\) (E°cell = +1.70 V)

Explanation: Both reactions have positive E°cell values, indicating spontaneous oxidation of Fe²⁺ under alkaline conditions.

Question 4

Topic: 28.1

Transition metal atoms and transition metal ions form complexes by combining with ligands.
(a) Explain why transition elements form complex ions.
(b) \(Co^{2+}\) ions form complex ion G. Each G ion contains two \(Co^{2+}\) ions, both of which are octahedrally coordinated. Each G ion contains one \(O_2\) molecule, which donates one pair of electrons to each \(Co^{2+}\) ion, and one \(NH_2^–\) ion, which donates one pair of electrons to each \(Co^{2+}\) ion. The remaining ligands are NH₃ molecules.
(i) Deduce the formula of complex ion G. Include its overall charge.
(ii) The d-orbitals of the \(Co^{2+}\) ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each \(Co^{2+}\) ion.

(iii) \(Co^{2+}\) ions form a different complex ion, M. Each M ion contains two \(Co^{2+}\) ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours.

(c) Cadmium forms complex ion X, \([Cd(NH_3)_4]^{2+}\). When a solution containing CN⁻ ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no \(NH_3\) ligands and no \(H_2O\) ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction.
(i) Write an ionic equation for this reaction, using the formulae of the complex ions.
(ii) Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl⁻ ions. When NaCl(aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, \(K_{stab}\).

(d) Ethanedioate ions, \(C_2O_4^{2–}\), form complexes with transition element ions. The concentration of \(C_2O_4^{2–}\) ions can be found by reaction with acidified \(Cr_2O_7^{2–}\) ions. \(C_2O_4^{2–}\) ions are protonated and form HOOCCOOH molecules which are oxidised by \(Cr_2O_7^{2–}\). The half-equations are shown.

(i) Construct an equation for the reaction between acidified Cr₂O₇²⁻ and HOOCCOOH

(ii) A 25.0 cm³ sample of a solution of Na₂C₂O₄ reacts with exactly 16.20 cm³ of an acidified solution of 0.0500 moldm⁻³ K₂Cr₂O₇. Calculate the concentration of the solution of Na₂C₂O₄.

▶️ Answer/Explanation

Solution

(a) Transition elements form complex ions because they have energetically accessible empty d-orbitals that can accept lone pairs of electrons from ligands, forming coordinate (dative) bonds.

Explanation: The partially filled d-subshell in transition metals allows them to form stable complexes with ligands by accepting electron pairs into their d-orbitals.

(b)(i) The formula of complex ion G is \([Co_2O_2NH_2(NH_3)_8]^{3+}\).

Explanation: Each \(Co^{2+}\) is octahedrally coordinated (6 ligands). The \(O_2\) and \(NH_2^-\) each donate one pair to both Co ions (total 4 ligands), and the remaining 8 NH₃ ligands complete the coordination (4 NH₃ per Co). The overall charge is calculated as: \(2(Co^{2+}) + 1(O_2) + 1(NH_2^-) + 8(NH_3) = +3\).

(b)(ii) Higher energy d-orbitals: 2 (\(e_g\) orbitals). Lower energy d-orbitals: 3 (\(t_{2g}\) orbitals).

Explanation: In octahedral complexes, d-orbitals split into two higher energy \(e_g\) orbitals (d_z² and d_x²-y²) and three lower energy \(t_{2g}\) orbitals (d_xy, d_xz, d_yz).

(b)(iii) G and M have different colors because the energy gap (\(\Delta E\)) between split d-orbitals differs due to different ligand field strengths.

Explanation: The color depends on the energy gap between d-orbitals. Different ligands (in G vs M) cause different splitting patterns, absorbing different wavelengths of light.

(c)(i) \([Cd(NH_3)_4]^{2+} + 4CN^- \rightarrow [Cd(CN)_4]^{2-} + 4NH_3\)

Explanation: CN⁻ ligands replace NH₃ in the complex, forming the more stable \([Cd(CN)_4]^{2-}\) complex.

(c)(ii) Order of increasing stability constant: \([CdCl_4]^{2-} < [Cd(NH_3)_4]^{2+} < [Cd(CN)_4]^{2-}\)

Explanation: CN⁻ forms the most stable complex (highest \(K_{stab}\)) due to strong field strength, while Cl⁻ forms the least stable complex.

(d)(i) \(Cr_2O_7^{2-} + 3HOOCCOOH + 8H^+ \rightarrow 2Cr^{3+} + 6CO_2 + 7H_2O\)

Explanation: The balanced equation combines the half-reactions: Cr₂O₇²⁻ is reduced to Cr³⁺ (6 electrons gained), while HOOCCOOH is oxidized to CO₂ (2 electrons lost per molecule).

(d)(ii) The concentration of Na₂C₂O₄ is 0.0972 moldm⁻³.

Calculation:
• Moles of Cr₂O₇²⁻ = 0.0500 × 0.01620 = 0.00081 mol
• Moles of C₂O₄²⁻ = 3 × 0.00081 = 0.00243 mol (1:3 stoichiometry)
• Concentration = 0.00243 mol / 0.0250 dm³ = 0.0972 moldm⁻³

Question 5

Topic: 28.3

The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1. The symbol J represents an atom or ion of a transition element. The symbol L is used to represent a monodentate ligand.

(a) Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees.

(b) Identify the shapes of complexes P, Q, R and S.

(c) Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R and S. Identify all the complexes which form new complexes that show geometrical isomerism.

(d) Three L ligands are exchanged with three different monodentate ligands X, Y and Z in each of complexes P, Q and R. Identify all the complexes which form new complexes that show optical isomerism.

▶️ Answer/Explanation

Solution

(a) Bond Angles:

P: 109.5° (tetrahedral angle between any two bonds)
Q: 90° (square planar angle between adjacent ligands)
R: 90° (octahedral angle between adjacent ligands)
S: 180° (linear angle between the two ligands)

Explanation: The bond angles are determined by the geometry of each complex. Tetrahedral complexes have 109.5° angles, square planar and octahedral complexes have 90° angles between adjacent ligands, and linear complexes have 180° angles.

(b) Shapes:

P: Tetrahedral (4 ligands around central atom J)
Q: Square planar (4 ligands in a plane around J)
R: Octahedral (6 ligands around J)
S: Linear (2 ligands around J)

Explanation: The shapes are identified based on the coordination number and arrangement of ligands. P has 4 ligands (tetrahedral), Q has 4 ligands in a plane (square planar), R has 6 ligands (octahedral), and S has 2 ligands (linear).

(c) Complexes showing geometrical isomerism: Q and R.

Explanation: Geometrical isomerism (cis/trans) occurs in square planar (Q) and octahedral (R) complexes when two different ligands (X and Y) replace two L ligands. Tetrahedral (P) and linear (S) complexes cannot exhibit geometrical isomerism due to their symmetrical arrangements.

(d) Complexes showing optical isomerism: P and R.

Explanation: Optical isomerism arises when a complex is chiral (non-superimposable on its mirror image). Tetrahedral (P) and octahedral (R) complexes with three different ligands (X, Y, Z) can form chiral structures. Square planar (Q) complexes do not typically show optical isomerism.

Question 6

Topic: 34.1

Benzene, \(C_6H_6\), reacts with chloroethane, \(C_2H_5Cl\), in the presence of a suitable catalyst to form ethylbenzene, \(C_6H_5C_2H_5\). In the presence of the catalyst, the ion \(C_2H_5^+\) is formed. This ion reacts with benzene.
(a) Complete the equation for the reaction of \(C_2H_5Cl\), with this catalyst to form \(C_2H_5^+\) as one product.

(b) Ethylbenzene reacts with more \(C_2H_5Cl\), forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene.
(i) Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene.

(ii) Explain why there is very little 1,3-diethylbenzene in the product mixture.

(i) State the reagent and conditions used for this reaction.
(ii) Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as [O]. All of the atoms in the two ethyl groups are fully oxidised in this reaction

(iii) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid.

(d) The proton (1H) NMR spectra of ethylbenzene, \(C_6H_5C_2H_5\), in \(CDCl_3\), and of benzene-1,2-dioic acid, \(C_6H_4(COOH)_2\), in \(CDCl_3\) are shown. They have not been identified.

(i) Explain the use of CDCl₃, instead of CHCl₃, as the solvent when obtaining these spectra.

(ii) Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1.

(iii) Identify the substance shown by the spectrum in Fig. 6.2, and complete Table 6.2.

(iv) When \(D_2O\) is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2. Describe this difference and explain your answer.

(e) Benzene-1,2-dioic acid can be used to produce K.

Suggest the name of this type of reaction.

▶️ Answer/Explanation

Solution

(a)

Explanation: The reaction involves the formation of the ethyl cation (\(C_2H_5^+\)) from chloroethane (\(C_2H_5Cl\)) in the presence of a catalyst (e.g., AlCl₃). The catalyst facilitates the heterolytic cleavage of the C-Cl bond, releasing \(Cl^-\) and forming \(C_2H_5^+\).

(b)(i)

Explanation: 1,2-diethylbenzene has ethyl groups on adjacent carbons, while 1,4-diethylbenzene has them on opposite sides of the benzene ring. The structures are drawn accordingly.

(b)(ii) Alkyl/ethyl group is 2,4-directing OR ethyl group is electron-donating (positive inductive effect).

Explanation: The ethyl group directs incoming electrophiles to the ortho (2) and para (4) positions due to its electron-donating nature, making 1,3-diethylbenzene a minor product.

(c)(i) Hot alkaline KMnO₄ / MnO₄⁻.

Explanation: The strong oxidizing agent (alkaline KMnO₄) under heating converts the ethyl groups into carboxyl groups (-COOH).

(c)(ii)

Explanation: The two ethyl groups are fully oxidized to carboxyl groups, consuming 10 [O] atoms in the process.

(c)(iii) 4 peaks.

Explanation: The carbon-13 NMR spectrum shows peaks for the four distinct carbon environments: two carboxyl carbons and two pairs of equivalent aromatic carbons.

(d)(i) CDCl₃ does not cause a peak OR does not interfere with the spectrum.

Explanation: CDCl₃ is used because deuterium (D) does not produce a signal in proton NMR, unlike CHCl₃, which would interfere.

(d)(ii) Ethylbenzene / \(C_6H_5C_2H_5\).

Explanation: The spectrum shows a triplet (CH₃) and quartet (CH₂) due to spin-spin coupling, along with aromatic protons.

(d)(iii) Benzene-1,2-dioic acid / \(C_6H_4(COOH)_2\).

Explanation: The spectrum includes a downfield peak for the COOH proton (~13 ppm) and aromatic protons (~7-8 ppm).

(d)(iv) COOH peak disappears OR peak at 13.1 ppm is removed.

Explanation: In \(D_2O\), the COOH proton exchanges with deuterium, which does not produce a signal in proton NMR.

(e) Dehydration / elimination / (auto)condensation.

Explanation: The reaction involves the loss of water molecules to form a cyclic anhydride, characteristic of a dehydration or condensation reaction.

Question 7

Topic: 35.2

A reaction scheme is shown in Fig. 7.1. The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when \(C_2H_5NH_2\) is mixed with compound V. No special conditions are required.

(a) Identify compound U which contains only three elements.
(b) Describe the reagents and conditions for reaction 1.
(c) Identify compound V.
(d) Complete the equation for reaction 3.
\(CH_3COOH + SOCl_2 \to …………..\)
(e) Identify compound W.
(f) Describe the conditions for reaction 4.

(g) Suggest the reagent needed for reaction 6.

(h) Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in Fig. 7.1. Each of the numbers 1, 2, 3, 4, 5 and 6 should be used once only.

(i) Compare the basicities of \(C_2H_5NHCOCH_3\), \(C_2H_5NHC_2H_5\), and \(NH_3\). Explain your answer.

▶️ Answer/Explanation

Solution

(a) CH₃CN / ethanenitrile

Explanation: Compound U must contain only C, H, and N. The nitrile group (–CN) in the reaction scheme confirms it as ethanenitrile (CH₃CN).

(b) dilute HCl (aq) and heat

Explanation: Hydrolysis of nitriles to carboxylic acids requires acidic conditions (dilute HCl) and heating to break the strong C≡N bond.

(c) CH₃COCl / ethanoyl chloride

Explanation: Compound V reacts with ethylamine to form an amide, indicating it is an acyl chloride (ethanoyl chloride).

(d) \(CH_3COCl + SO_2 + HCl\)

Explanation: Thionyl chloride (SOCl₂) converts carboxylic acids to acyl chlorides, with byproducts of SO₂ and HCl gas.

(e) \(C_2H_5Br\) / bromoethane

Explanation: Compound W is a haloalkane (bromoethane) used to alkylate amines in nucleophilic substitution reactions.

(f) Heat in ethanol under pressure (sealed tube)

Explanation: Nucleophilic substitution of amines with haloalkanes requires heating in a solvent (ethanol) and pressurized conditions.

(g) LiAlH₄ / lithium aluminium hydride

Explanation: Reduction of amides to amines is achieved using the strong reducing agent LiAlH₄.

(h)

Explanation: The reactions are matched as: 1 (nitrile hydrolysis), 2 (reduction), 3 (acyl chloride formation), 4 (alkylation), 5 (amide formation), and 6 (amide reduction).

(i) Basicity order: \(C_2H_5NHC_2H_5 > NH_3 > C_2H_5NHCOCH_3\)

Explanation: – \(C_2H_5NHC_2H_5\) is most basic due to two electron-donating ethyl groups. – \(NH_3\) has intermediate basicity. – \(C_2H_5NHCOCH_3\) is least basic due to the electron-withdrawing carbonyl group.

Question 8

Topic: 34.3

(a) An aqueous solution of phenol, \(C_6H_5OH\), is acidic at 298K. Explain why phenol is more acidic than water.

(b) (i) Name the two products formed when phenol reacts with an excess of \(Br_2\) (aq).
(ii) Draw the structures of the two isomeric organic products, with \(M_r = 139\), that are formed when phenol reacts with HNO₃(aq) at room temperature.

(iii) Write the equation for the reaction between phenol, C₆H₅OH, and sodium metal.
(c) Phenol can be produced from phenylamine in a two-step synthesis.

Describe the reagents and conditions needed in each step.
step one:
reagents
conditions
step two:
reagents
conditions

▶️ Answer/Explanation

Solution

(a)

Explanation: Phenol (\(C_6H_5OH\)) is more acidic than water because the lone pair of electrons on the oxygen atom is delocalized into the aromatic ring. This delocalization stabilizes the phenoxide ion (\(C_6H_5O^-\)) formed after proton loss, making the O-H bond weaker and easier to break compared to water.

(b)(i) 2,4,6-tribromophenol and hydrogen bromide (HBr).

Explanation: When phenol reacts with excess bromine water, electrophilic aromatic substitution occurs at the 2, 4, and 6 positions (ortho and para to the -OH group) forming 2,4,6-tribromophenol (a white precipitate) and HBr as a byproduct.

(b)(ii)

Explanation: Nitration of phenol at room temperature produces two structural isomers: 2-nitrophenol (ortho-) and 4-nitrophenol (para-), both with molecular mass 139. The meta- product is not favored due to the directing effect of the -OH group.

(b)(iii) \(2C_6H_5OH + 2Na \rightarrow 2C_6H_5ONa + H_2\)

Explanation: Phenol reacts with sodium metal similarly to alcohols, forming sodium phenoxide (\(C_6H_5ONa\)) and hydrogen gas. The -OH group’s hydrogen is replaced by sodium.

(c)

Step 1: Reagents – Sodium nitrite (NaNO₂) and dilute hydrochloric acid (HCl). Conditions – Temperature maintained between 0-5°C.
Step 2: Reagents – Water (H₂O). Conditions – Heating under reflux.
Explanation: In step 1, phenylamine undergoes diazotization with NaNO₂/HCl at low temperatures to form a benzene diazonium salt. In step 2, this salt is hydrolyzed by heating with water to produce phenol, with nitrogen gas evolved as a byproduct.

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