9701_w23_qp

Question 1

Topic: 23.1

Propanone, CH₃COCH₃, reacts with iodine, I₂, in the presence of an acid catalyst.

CH₃COCH₃ + I₂ \(\to\) CH₃COCH₂I + H⁺ + I⁻
The rate equation for this reaction is shown.
\(rate = k[CH_3COCH_3][H^+]\)
(a) Complete Table 1.1 to describe the order of the reaction.

(b) An experiment is performed using a large excess of CH₃COCH₃ and a large excess of H⁺(aq). The initial concentration of I₂ is 1.00 × 10⁻⁵ moldm⁻³. The initial rate of decrease in the \(I_2\) concentration is 2.27 × 10⁻⁷ moldm⁻³ s⁻¹.
(i) Use the axes to draw a graph of [\(I_2\)] against time for the first 10 seconds of the reaction.

(ii) State whether it is possible to calculate the numerical value of the rate constant, k, for this reaction from your graph. Explain your answer.
(c) The experiment is repeated at a different temperature. The initial concentrations of \(H^+\) ions, I₂, and CH₃COCH₃ are all 0.200 mol/dm³. The value of k at this temperature is 2.31 × 10⁻⁵ mol⁻¹ dm³ s⁻¹. Calculate the initial rate of this reaction.

(d) The experiment is repeated using an excess of H+(aq). The new rate equation is shown.
rate = k₁[CH₃COCH₃]
(i) The value of k₁ is 1.1 × 10⁻³ s⁻¹. Calculate the value of the half-life, \(t_\frac{1}{2}\).
(ii) Use your answer to (i) to draw a graph of [CH₃COCH₃] against time for this reaction. The initial value of [CH₃COCH₃] on your graph should be 0.200 mol/dm³. The final value of [CH₃COCH₃] on your graph should be 0.0250 moldm⁻³.

(e) A four-step mechanism is suggested for the overall reaction.
CH₃COCH₃ + I₂ \(\to\) CH₃COCH₂I + H⁺ + I⁻ rate = k [CH₃COCH₃][H⁺]
Part of this mechanism is shown.

(i) Write an equation for step 3.
(ii) Suggest the slowest step of the mechanism. Explain your answer.
(iii) Identify one conjugate acid-conjugate base pair in the mechanism.

▶️ Answer/Explanation

Solution

(a)

Final Answer: 1, 0, 1, 2

Explanation: The rate equation \(rate = k[CH_3COCH_3][H^+]\) shows the reaction is first-order with respect to CH₃COCH₃ and H⁺, and zero-order with respect to I₂ (as it does not appear in the rate equation).

(b)(i)

Final Answer: The graph should be a straight line decreasing from \(1.00 \times 10^{-5}\) to \(0.773 \times 10^{-5}\) over 10 seconds.

Explanation: Since [CH₃COCH₃] and [H⁺] are in excess, the reaction is pseudo-zero-order with respect to I₂, leading to a linear decrease in [I₂] over time.

(b)(ii)

Final Answer: No, because the concentrations of CH₃COCH₃ and H⁺ in the rate equation are not known.

Explanation: The rate constant \(k\) requires knowledge of all concentrations in the rate equation, which are not provided in the graph data.

(c)

Final Answer: \(9.24 \times 10^{-7} \, \text{mol dm}^{-3} \, \text{s}^{-1}\)

Explanation: Using the rate equation, \(rate = k[CH_3COCH_3][H^+] = (2.31 \times 10^{-5})(0.200)(0.200) = 9.24 \times 10^{-7}\).

(d)(i)

Final Answer: \(t_{1/2} = 630 \, \text{s}\)

Explanation: For a first-order reaction, \(t_{1/2} = \frac{\ln 2}{k_1} = \frac{0.693}{1.1 \times 10^{-3}} = 630 \, \text{s}\).

(d)(ii)

Final Answer: The graph should show three equal half-life intervals, decreasing from 0.200 to 0.100, 0.050, and 0.025 mol/dm³.

Explanation: The concentration halves every 630 seconds, as calculated in (d)(i).

(e)(i)

Final Answer: CH₃COH=CH₂ + I₂ → CH₃C⁺(OH)CH₂I + I⁻

Explanation: Step 3 involves the reaction of the enol intermediate with I₂ to form the iodinated product.

(e)(ii)

Final Answer: Step 1 is the slowest step because it involves both reactants in the rate equation.

Explanation: The rate-determining step is typically the slowest step and matches the rate law.

(e)(iii)

Final Answer: Conjugate acid: CH₃C⁺(OH)CH₃, Conjugate base: CH₃COCH₃

Explanation: The pair CH₃C⁺(OH)CH₃ (acid) and CH₃COCH₃ (base) is an example of a conjugate acid-base pair in the mechanism.

Question 2

Topic: 25.2

Benzoic acid, C₆H₅COOH, is a weak acid. The Ka of benzoic acid is 6.31 × 10⁻⁵ moldm⁻³ at 298K. A 1.00 dm³ buffer solution is made at 298 K containing 1.00 g of \(C_6H_5COOH\) and a slightly greater mass of sodium benzoate, C₆H₅COO⁻Na⁺. This buffer solution has a pH of 4.15.
(a) Define buffer solution.
(b) Write equations to show how this solution acts as a buffer solution when the named substances are added to it:
(i) dilute aqueous sodium hydroxide
(ii) dilute aqueous nitric acid.
(c) Calculate the H⁺ concentration and the C₆H₅COOH concentration in the buffer solution described. Use the expression for the \(K_a\) of C₆H₅COOH to calculate the concentration of C₆H₅COO⁻Na⁺ in the buffer solution. Show your working and give each answer to a minimum of three significant figures.

(d) A 10.0 cm³ sample of the buffer solution is mixed with 10.0 cm³ of \(1.00moldm^{–3}\) KOH. Both solutions are at 298K. A reaction is allowed to occur without stirring. Two observations are recorded:
● the temperature, after the reaction is complete, is fractionally above 298K
● the pH, after the reaction, is greater than 13.
Explain these two observations.

(e) Magnesium benzoate, Mg(C₆H₅COO)₂, has a solubility in water of less than 1.00 g/dm³ at 298K.
\(K_{sp} = [Mg^{2+}][C_6H_5COO^–]^2 = 1.76 × 10^{–7}\) at 298K
(i) Calculate the solubility of Mg(C₆H₅COO)₂ in water at 298K. Give your answer in \(gdm^{–3}\). Show your working.
\([M_r: Mg(C_6H_5COO)_2, 266.3]\)

(ii) An excess of Mg(C₆H₅COO)₂ is added to a sample of 0.50 mol/dm³ MgSO₄ at 298 K. State whether the equilibrium concentration of Mg(C₆H₅COO)₂ is higher than, the same as, or lower than your answer to (i). Explain your answer. The concentration is ……………………………………………………… the concentration in (i).

▶️ Answer/Explanation

Solution

(a) A buffer solution resists pH change when small amounts of acid or alkali are added.

Explanation: Buffer solutions maintain a stable pH due to the presence of a weak acid and its conjugate base (or weak base and its conjugate acid), which neutralize added acids or bases.

(b)
(i) \(C_6H_5COOH + OH^- \rightarrow C_6H_5COO^- + H_2O\)
(ii) \(C_6H_5COO^- + H^+ \rightarrow C_6H_5COOH\)

Explanation: When NaOH is added, the weak acid (\(C_6H_5COOH\)) neutralizes the \(OH^-\). When \(HNO_3\) is added, the conjugate base (\(C_6H_5COO^-\)) neutralizes the \(H^+\).

(c)
\([H^+] = 10^{-4.15} = 7.08 \times 10^{-5} \, \text{mol dm}^{-3}\)
\([C_6H_5COOH] = \frac{1.00 \, \text{g}}{122.12 \, \text{g/mol}} = 8.20 \times 10^{-3} \, \text{mol dm}^{-3}\)
Using \(K_a = \frac{[H^+][C_6H_5COO^-]}{[C_6H_5COOH]}\), we get \([C_6H_5COO^-] = \frac{(6.31 \times 10^{-5})(8.20 \times 10^{-3})}{7.08 \times 10^{-5}} = 7.31 \times 10^{-3} \, \text{mol dm}^{-3}\).

Explanation: The \(H^+\) concentration is derived from the pH, the weak acid concentration from its mass and molar mass, and the conjugate base concentration from the \(K_a\) expression.

(d) The temperature rise is due to the exothermic neutralization reaction. The high pH (>13) indicates all \(C_6H_5COOH\) has reacted, leaving excess \(OH^-\) from KOH.

Explanation: Neutralization releases heat, and the large excess of KOH drives the pH to a very basic value.

(e)(i) Let solubility \(= s\). Then \(K_{sp} = s \times (2s)^2 = 4s^3 = 1.76 \times 10^{-7}\). Solving gives \(s = 3.53 \times 10^{-3} \, \text{mol dm}^{-3}\). Converting to grams: \(3.53 \times 10^{-3} \times 266.3 = 0.940 \, \text{g dm}^{-3}\).

Explanation: The solubility is calculated from the \(K_{sp}\) expression, accounting for the stoichiometry of dissociation.

(e)(ii) The concentration is lower than the concentration in (i).
Explanation: The common ion effect (\(Mg^{2+}\) from \(MgSO_4\)) suppresses the dissociation of \(Mg(C_6H_5COO)_2\), reducing its solubility.

Question 3

Topic: 24.2

Some electrode potentials are shown in Table 3.1.

(a) (i) Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions.

(ii) An electrochemical cell is set up using an Fe³⁺/Fe²⁺ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit.
positive electrode ……………………………………………….
Electrons flow from the ………………………… electrode to the ………………………… electrode

(b) The vanadium-containing species in the electrode reactions given in Table 3.1 are \(V, V^{2+}, V^{3+}, VO^{2+}\) and \(VO_2^{+}\).
(i) Identify one vanadium-containing species that does not react with Fe²⁺ ions under standard conditions. Use data from Table 3.1 to explain your answer.

(ii) Identify all the vanadium-containing species that will react with Fe²⁺ ions under standard conditions.

(iii) Write an equation for one of the possible reactions identified in (ii)

(c) Another electrochemical cell is set up using an Fe³⁺/Fe²⁺ electrode and an alkaline ClO⁻/Cl⁻ electrode. The concentration of Fe³⁺ is 1000 times greater than the concentration of Fe²⁺ in the Fe³⁺/Fe²⁺ electrode. All other conditions are standard.
(i) Use the Nernst equation to calculate the E value of the Fe³⁺/Fe²⁺ electrode. Show your working.

(ii) Write an equation for the reaction that occurs in the cell, under these conditions.

(d) Another electrochemical cell is set up using an Fe²⁺/Fe electrode and an alkaline ClO⁻/Cl⁻ electrode under standard conditions. Calculate the value of \(\Delta G^0\) for the cell.

(e) A solution of iron(II) sulfate, FeSO₄(aq), is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working.

(f) Iron(II) chloride, FeCl₂, is oxidised by chlorine to form iron(III) chloride, FeCl₃, under standard conditions.
\(2FeCl_2(s) + Cl_2(g) \to 2FeCl_3(s)\) \(\Delta H^0 = –128kJmol^{–1}\)

(i) Use Table 3.2 and other data to calculate the Gibbs free energy change, \(\Delta G^0\), for this reaction. Show your working.

(ii) Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer.
The reaction becomes ………………………… feasible.

▶️ Answer/Explanation

Solution

(a)(i) The standard hydrogen electrode consists of:
– A platinum electrode coated with platinum black.
– Immersed in 1.0 mol/dm³ HCl solution.
– Hydrogen gas bubbled at 1 atm pressure.
– Label: Pt electrode, H₂ gas inlet, H⁺(aq) solution.

(a)(ii) The positive electrode is the Fe³⁺/Fe²⁺ electrode (since \(E^\circ = +0.77V > 0V\)).
Electrons flow from the hydrogen electrode to the iron electrode.

(b)(i) \(V^{3+}\) does not react with Fe²⁺ because its \(E^\circ = -0.26V\) is between Fe²⁺/Fe (\(-0.44V\)) and Fe³⁺/Fe²⁺ (\(+0.77V\)), making the reaction non-spontaneous.

(b)(ii) Vanadium species that react with Fe²⁺: \(V\) (\(E^\circ = -1.18V\)) and \(VO_2^+\) (\(E^\circ = +1.00V\)).

(b)(iii) Example reaction:
\(VO_2^+ + 2H^+ + Fe^{2+} \rightarrow VO^{2+} + H_2O + Fe^{3+}\)

(c)(i) Using the Nernst equation:
\(E = E^\circ + \frac{0.059}{1} \log \left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right) = 0.77 + 0.059 \log(1000) = 0.947V\).

(c)(ii) Cell reaction:
\(2Fe^{3+} + Cl^- + 2OH^- \rightarrow 2Fe^{2+} + ClO^- + H_2O\).

(d) Cell potential: \(E^\circ_{cell} = 1.33V\).
\(\Delta G^\circ = -nFE^\circ = -2 \times 96500 \times 1.33 = -257 \text{kJ/mol}\).

(e) Calculations:
Charge passed: \(0.640A \times 1020s = 652.8C\).
Moles of Fe deposited: \(\frac{0.185g}{55.8g/mol} = 3.31 \times 10^{-3}mol\).
Avogadro constant: \(L = \frac{652.8C}{1.6 \times 10^{-19}C \times 3.31 \times 10^{-3}mol} = 6.15 \times 10^{23}mol^{-1}\).

(f)(i) \(\Delta S^\circ = -179J/mol·K\).
\(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ = -128 – (298 \times -0.179) = -74.7kJ/mol\).

(f)(ii) The reaction becomes less feasible at higher temperatures because \(\Delta G^\circ\) becomes more positive (since \(\Delta S^\circ < 0\)).

Question 4

Topic: 28.4

The structure of the polydentate ligand, \(EDTA^{4–}\), is shown in Fig. 4.1.

The stability constants, at 298K, of five octahedral complexes are given in Table 4.1.

(a) Define stability constant.

(b) Calculate the oxidation states of Cu in \([Cu(EDTA)]^{2–}\) and Cr in \([Cr(EDTA)]^–\).
(c) Deduce the number of lone pairs donated by each \(EDTA^{4–}\) ligand in a single \([Fe(EDTA)]^{2–}\) complex ion.
(d) Identify the most stable complex in Table 4.1. Explain your choice.

(e) In a solution at equilibrium at 298K, [[Cu(H₂O)₆]²⁺] = 3.00 × 10⁻¹⁰ moldm⁻³ and [EDTA⁴⁻] = 5.00 × 10⁻¹² moldm⁻³. Use the expression for \(K_{stab}\) to calculate the concentration of \([Cu(EDTA)]^{2–}\) in this solution. Show your working.

(f) A solution of \([Cu(EDTA)]^{2–}\) ions is pale blue while a solution of \([Cu(NH_3)_4(H_2O)_2]^{2+}\) ions is deep blue. Explain this difference in colour

▶️ Answer/Explanation

Solution

(a) Stability constant (\(K_{stab}\)) is the equilibrium constant for the formation of a complex ion from its constituent metal ion and ligands in solution.

Explanation: It quantifies the stability of the complex, with higher values indicating greater stability. For \([M(EDTA)]^{n±}\), \(K_{stab} = \frac{[[M(EDTA)]^{n±}]}{[M^{m+}][EDTA^{4-}]}\).

(b) Cu: +2, Cr: +3.

Explanation: – In \([Cu(EDTA)]^{2-}\), EDTA has a \(-4\) charge. Thus, \(x + (-4) = -2 \Rightarrow x = +2\) (Cu). – In \([Cr(EDTA)]^-\), \(-4 + y = -1 \Rightarrow y = +3\) (Cr).

(c) Six lone pairs.

Explanation: EDTA is hexadentate, donating 6 lone pairs (2 from N atoms and 4 from O atoms) to form an octahedral complex with Fe²⁺.

(d) \([Fe(EDTA)]^-\) is the most stable complex.

Explanation: It has the highest \(K_{stab}\) (\(1.26 \times 10^{25}\)), indicating the strongest metal-ligand bonds and greatest thermodynamic stability.

(e) \([Cu(EDTA)^{2-}] = 0.095 \, \text{mol/dm}^3\).

Explanation: \[ K_{stab} = \frac{[[Cu(EDTA)^{2-}]]}{[[Cu(H_2O)_6^{2+}]][EDTA^{4-}]} = 6.31 \times 10^{19} \] Substituting given concentrations: \[ [[Cu(EDTA)^{2-}]] = 6.31 \times 10^{19} \times 3.00 \times 10^{-10} \times 5.00 \times 10^{-12} = 0.095 \, \text{mol/dm}^3. \]

(f) The difference in color arises from different ligand field splitting energies (\(\Delta E\)).

Explanation: – EDTA (strong field ligand) causes larger \(\Delta E\), absorbing higher energy (shorter wavelength) light, resulting in pale blue. – NH₃ (weaker field) gives smaller \(\Delta E\), absorbing lower energy (longer wavelength) light, yielding deep blue.

Question 5

Topic: 27.1

Some of the ionic compounds of Group 2 elements undergo thermal decomposition. Thermal decomposition of solid anhydrous magnesium ethanedioate, MgC₂O₄, occurs above 650°C. The products are magnesium oxide and a mixture of two different gases, one of which gives a white precipitate with saturated calcium hydroxide solution.
(a) Complete the equation for the thermal decomposition of MgC₂O₄.
(b) Suggest which of MgC₂O₄ or CaC₂O₄ undergoes thermal decomposition at a lower temperature. Explain your answer.
(c) The ethanedioate ion is oxidised by acidified KMnO₄.
\(5C_2O_4^{2–} + 2MnO4^– + 16H^+ \to 10CO_2 + 2Mn^{2+} + 8H_2O\)
An experiment is performed to find the solubility of MgC₂O₄ in water. A 40.0 cm³ sample of saturated aqueous MgC₂O₄ requires 27.05 cm³ of 0.00200 mol/dm³ acidified KMnO₄ to oxidise all the C₂O₄²⁻ ions. Calculate the solubility, in mol/dm³, of MgC₂O₄ in water. Show your working.

▶️ Answer/Explanation

Solution

(a) The balanced equation for the thermal decomposition of MgC₂O₄ is:

\[ \text{MgC}_2\text{O}_4 \rightarrow \text{MgO} + \text{CO}_2 + \text{CO} \]

Explanation: The decomposition produces magnesium oxide (MgO) and two gases: carbon dioxide (CO₂) and carbon monoxide (CO). CO₂ reacts with calcium hydroxide to form a white precipitate (CaCO₃).

(b) MgC₂O₄ decomposes at a lower temperature than CaC₂O₄.

Explanation: The Mg²⁺ ion has a smaller ionic radius than Ca²⁺, leading to greater polarisation of the C₂O₄²⁻ anion. This weakens the C-O bonds, making the compound less thermally stable.

(c) The solubility of MgC₂O₄ is calculated as follows:

1. Moles of MnO₄⁻ used:

\[ \text{Moles of MnO}_4^- = 27.05 \, \text{cm}^3 \times 0.00200 \, \text{mol/dm}^3 \times \frac{1}{1000} = 5.41 \times 10^{-5} \, \text{mol} \]

2. Moles of C₂O₄²⁻ reacted (from stoichiometry):

\[ \text{Moles of C}_2\text{O}_4^{2-} = 5.41 \times 10^{-5} \times \frac{5}{2} = 1.3525 \times 10^{-4} \, \text{mol} \]

3. Solubility in mol/dm³:

\[ \text{Solubility} = \frac{1.3525 \times 10^{-4} \, \text{mol}}{0.0400 \, \text{dm}^3} = 0.00338 \, \text{mol/dm}^3 \]

Final Answer: The solubility of MgC₂O₄ is 0.00338 mol/dm³.

Question 6

Topic: 28.3

(a) Phosphine, :PH₃, and carbon monoxide, :CO, are monodentate ligands found in some transition element complexes.
(i) Define monodentate ligand.
(ii) Define transition element complex.
(iii) Explain why transition elements form complexes.
(b) The formulae of six complexes are given in Table 6.1. The abbreviation en is used for 1,2-diaminoethane. The abbreviation dien is used for the tridentate ligand H₂NCH₂CH₂NHCH₂CH₂NH₂. The dien ligand forms three bonds to the gold ion in [Au(dien)(H₂O)₂Cl]²⁺ and Au(dien)Cl₃. These three bonds all lie in the same plane. The CO ligand coordinates through the carbon atom in [Rh(CO)₂Cl₂]⁺.

(i) Complete Table 6.1 to state the geometry of the first three complexes. Each complex is either square planar, tetrahedral or octahedral. [1]
(ii) Use complexes [Au(dien)(H₂O)₂Cl]₂⁺ and Au(dien)Cl₃ to write an equation showing ligand exchange.

(iii) Draw the three-dimensional structure of Au(dien)Cl₃ in the box. The dien ligand can be drawn as Box for Au(dien)Cl₃

(iv) Draw the three-dimensional structure of Ni(PH₃)₂Cl₂ in the box.

Box for Ni(PH₃)₂Cl₂

(v) One of the complexes, [Rh(en)₂Cl₂]⁺ or [Rh(CO)₂Cl₂]⁺, can exist in three isomeric forms. Identify this complex and the types of isomerism shown.

(vi) Draw the three-dimensional structures of the two isomers of [Ni(H₂O)₂(NH₃)₄]²⁺ in the boxes and identify the type of isomerism shown.

▶️ Answer/Explanation

Solution

(a)(i)

Answer: A ligand that donates one lone pair to a metal atom/ion.

Explanation: Monodentate ligands (e.g., PH₃, CO) bind to the central metal ion through a single donor atom.

(a)(ii)

Answer: A metal atom/ion bonded to one or more ligands.

Explanation: Transition element complexes consist of a central metal ion surrounded by ligands coordinated via dative bonds.

(a)(iii)

Answer: Transition elements have vacant d-orbitals that are energetically accessible for bonding.

Explanation: The partially filled d-subshell allows transition metals to accept lone pairs from ligands, forming stable complexes.

(b)(i)

Answer: Octahedral, Square planar, Octahedral.

Explanation: [Ni(H₂O)₆]²⁺ and [Ni(en)₃]²⁺ are octahedral (6-coordinate), while [Pt(NH₃)₂Cl₂] is square planar (4-coordinate).

(b)(ii)

Answer: \([Au(dien)(H_2O)_2Cl]^{2+} + 2Cl^– \rightarrow [Au(dien)Cl_3] + 2H_2O\)

Explanation: The equation shows the replacement of H₂O ligands by Cl⁻ ions in the coordination sphere.

(b)(iii)-(iv)

Explanation: Au(dien)Cl₃ has a planar dien ligand with three Cl⁻ completing the coordination. Ni(PH₃)₂Cl₂ is tetrahedral.

(b)(v)

Answer: [Rh(en)₂Cl₂]⁺ shows optical and cis/trans (geometric) isomerism.

Explanation: The bidentate ‘en’ ligands allow for multiple spatial arrangements, creating three isomers.

(b)(vi)

Answer: Cis-trans (geometric) isomerism.

Explanation: The H₂O ligands can be adjacent (cis) or opposite (trans) in the octahedral complex.

Question 7

Topic: 34.3

Benzene can be used to make benzoic acid in the two-step process shown in Fig. 7.1.

(a) Give the reagents and conditions for step 1 and step 2.

(b) Methylbenzene and benzoic acid each have five different peaks in the carbon \((^{13}C)\) NMR spectrum.

Use Table 7.1 to complete the two sentences to suggest descriptions of these two spectra.
The carbon \((^{13}C)\) NMR spectrum of methylbenzene:
● has …………….. peak(s) in the chemical shift range of ………………… and
● has …………….. peak(s) in the chemical shift range of ………………… .

The carbon \((^{13}C)\) NMR spectrum of benzoic acid:
● has …………….. peak(s) in the chemical shift range of ………………… and
● has …………….. peak(s) in the chemical shift range of ………………… .
(c) (i) When treated with Cl₂ under suitable conditions, methylbenzene forms compound J. When treated with Cl₂ under different conditions with different reagents, methylbenzene forms compound K. Suggest and draw structures of compounds J and K in the boxes. The molecular formulam of each compound is given.

State the reagents and conditions required to form each product.
to form compound J ……………………..
to form compound K ……………………..
(ii) When treated with a chlorine-containing reagent under suitable conditions, benzoic acid forms compound L. When treated with a different chlorine-containing reagent under different conditions, benzoic acid forms compound M. Suggest and draw structures of compounds L and M in the boxes. The molecular formula of each product is given.

State the reagents and conditions to form compound M from benzoic acid.

▶️ Answer/Explanation

Solution

(a)

Step 1: Reagents: CH₃Cl + AlCl₃ (Friedel-Crafts alkylation).
Step 2: Reagents: Hot alkaline KMnO₄ (oxidation of methylbenzene to benzoic acid).

Explanation: In step 1, benzene undergoes Friedel-Crafts alkylation to form methylbenzene. In step 2, the methyl group is oxidized to a carboxyl group using KMnO₄ under hot alkaline conditions.

(b)

The carbon \((^{13}C)\) NMR spectrum of methylbenzene:
● has 4 peak(s) in the chemical shift range of 110-160 ppm (aromatic carbons) and
● has 1 peak(s) in the chemical shift range of 25-50 ppm (methyl carbon).

The carbon \((^{13}C)\) NMR spectrum of benzoic acid:
● has 4 peak(s) in the chemical shift range of 110-160 ppm (aromatic carbons) and
● has 1 peak(s) in the chemical shift range of 160-185 ppm (carboxyl carbon).

Explanation: Methylbenzene has 4 aromatic carbons (110-160 ppm) and 1 methyl carbon (25-50 ppm). Benzoic acid has 4 aromatic carbons (110-160 ppm) and 1 carboxyl carbon (160-185 ppm).

(c)(i)

Compound J: Chloromethylbenzene (side-chain chlorination).
Reagents/Conditions: Cl₂ + UV light (free radical substitution).

Compound K: 4-chloromethylbenzene (ring chlorination).
Reagents/Conditions: Cl₂ + FeCl₃ (electrophilic substitution).

Explanation: Under UV light, Cl₂ substitutes the methyl group (J). With FeCl₃, Cl₂ substitutes the aromatic ring (K).

(c)(ii)

Compound L: 3-chlorobenzoic acid (ring chlorination).
Reagents/Conditions: Cl₂ + FeCl₃ (electrophilic substitution).

Compound M: Benzoyl chloride (carboxyl group substitution).
Reagents/Conditions: SOCl₂ (thionyl chloride).

Explanation: Cl₂ + FeCl₃ chlorinates the ring (L). SOCl₂ converts the carboxyl group to –COCl (M).

Question 8

Topic: 37.3

Lactic acid, CH₃CH(OH)COOH, is the only monomer needed to form the polymer polylactic acid, PLA.

(a) (i) Draw a short length of the PLA polymer chain, including a minimum of two monomer residues. The methyl groups may be written as –CH₃, but all other bonds should be shown fully displayed. Label one repeat unit of polylactic acid on your diagram.

(ii) Give the name of the type of polymerisation involved in the formation of PLA and the name of the functional group that forms between the monomers.
(iii) Predict whether PLA is readily biodegradable. Explain your answer.

(b) The proton (1H) NMR spectrum of CH₃CH(OH)COOH in CDCl₃ is shown in Fig. 8.1. The proton NMR chemical shift ranges are shown in Table 8.1.

Table 8.1: Proton NMR Chemical Shift Ranges

(i) Use Fig. 8.1 and Table 8.1 to complete Table 8.2.

(ii) Name the substance responsible for the peak at δ = 0.0.
(iii) Explain why CDCl₃ is a better solvent than CHCl₃ for use in proton NMR.

(c) An impure sample of CH₃CH(OH)COOH contains pentan-3-one as the only contaminant. The mixture is analysed using gas/liquid chromatography. The pentan-3-one is found to have a longer retention time than the lactic acid.
(i) Explain what is meant by retention time.
(ii) Suggest suitable substances, or types of substances, that could be used as the mobile and stationary phases.
(iii) Describe how the percentage composition of the mixture can be determined from the gas /liquid chromatogram.

▶️ Answer/Explanation

Solution

(a)(i)

Explanation: The diagram shows two repeating units of PLA, formed via condensation polymerisation of lactic acid. The repeat unit is highlighted, with ester (–COO–) linkages between monomers.

(a)(ii) Type of polymerisation: condensation
Functional group: ester

Explanation: PLA forms through condensation polymerisation, releasing water. The ester functional group (–COO–) links the monomers.

(a)(iii) PLA is biodegradable because ester groups can hydrolyse under environmental conditions.

Explanation: The ester bonds in PLA are susceptible to hydrolysis, allowing microbial breakdown into smaller, non-toxic molecules.

(b)(i) Table 8.2 completed:

Chemical shifts (δ):
-COOH: ~12 ppm (singlet)
-OH: ~5.0 ppm (singlet)
-CH: ~4.4 ppm (quartet)
-CH₃: ~1.6 ppm (doublet)

Explanation: The NMR peaks correspond to the proton environments in lactic acid. Splitting patterns arise from adjacent protons (e.g., –CH splits –CH₃ into a doublet).

(b)(ii) TMS (tetramethylsilane).

Explanation: TMS is the reference compound (δ = 0 ppm) in NMR due to its inertness and sharp singlet peak.

(b)(iii) CDCl₃ has no protons, avoiding interference in the spectrum, whereas CHCl₃ would produce a peak.

Explanation: Deuterated solvents (e.g., CDCl₃) are used in proton NMR to prevent solvent peaks from obscuring sample signals.

(c)(i) Retention time is the duration a compound takes to travel through the chromatography column.

Explanation: It reflects the affinity of the compound for the stationary phase—longer retention indicates stronger interaction.

(c)(ii)
Mobile phase: Unreactive gas (e.g., helium)
Stationary phase: Non-polar liquid (e.g., silicone oil)

Explanation: The mobile phase carries the sample, while the stationary phase separates components based on polarity.

(c)(iii) The percentage composition is calculated by dividing each peak area by the total area of all peaks and multiplying by 100%.

Explanation: Peak areas in the chromatogram are proportional to the amount of each component in the mixture.

Question 9

Topic: 35.2

(a) State the reactants and conditions for two different types of reactions that both produce diethylamine, CH₃CH₂NHCH₂CH₃.

(b) Describe the relative basicities of diethylamine, phenylamine and ammonia in aqueous solution. Explain your answer in terms of structure.

(c) Phenylamine reacts with HNO₂(aq) at 4°C to form compound P. Compound P reacts with phenol under alkaline conditions at 4°C. The product of this reaction is acidified, forming azo compound Q. Draw the structure of compound Q. Circle the azo group on your structure. State one use of an azo compound such as Q.
(d) CH₃CH₂NHCH₂CH₃ reacts with ethanoyl chloride, CH₃COCl, to give the amide N,N-diethylethanamide, CH₃CON(C₂H₅)₂. An incomplete description of the mechanism of this reaction is shown in Fig. 9.1.

(i) Complete the mechanism in Fig. 9.1. You should include:
● all relevant dipoles (δ+ and δ–) and full electric charges (+ and –) on the species in box one and in box two
● all relevant lone pairs on the species in box one and in box two
● all relevant curly arrows to show the movement of electron pairs in box one and in box two
● the formula of the second product in box three.
(ii) Name this mechanism.

▶️ Answer/Explanation

Solution

(a)

Method 1: Chloroethane + ethylamine, heat/pressure in ethanol.
Method 2: N-ethyl ethanamide (CH₃CONHC₂H₅) reduced with LiAlH₄.

Explanation: Diethylamine can be synthesized via nucleophilic substitution (alkylation) or reduction of an amide. Both methods require specific conditions to ensure high yield.

(b) Basicity order: Diethylamine > Ammonia > Phenylamine

Explanation: Diethylamine is most basic due to electron-donating ethyl groups increasing N lone pair availability. Phenylamine is least basic as the lone pair delocalizes into the aromatic ring, reducing proton affinity.

(c)

Structure of Q

Use: Azo compounds like Q are used as dyes/pigments (e.g., in textiles or food coloring).

Explanation: Phenylamine forms a diazonium salt (P) at 4°C, which couples with phenol to form an azo dye (Q) upon acidification. The azo group (–N=N–) is circled.

(d)(i)

Mechanism

Key steps:
1. Nucleophilic attack by diethylamine on the carbonyl carbon (δ+).
2. Curly arrows show lone pair donation and C=O bond breaking.
3. Intermediate collapses to form N,N-diethylethanamide + HCl.

(d)(ii) Addition-elimination mechanism.

Explanation: The reaction proceeds via nucleophilic addition to the carbonyl followed by elimination of HCl, characteristic of acyl chloride-amine reactions.

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