MetadataPastPaper.sampleTitle

Let the percentage abundance of \(^{25}X\) be \(x\%\). Since the ratio of \(^{24}X\) to \(^{25}X\) is \(3:1\), the abundance of \(^{24}X\) is \(3x\%\). The remaining abundance belongs to \(^{26}X\), which is \((100 - 4x)\%\). Using the definition of relative atomic mass: \(A_r = \frac{24(3x) + 25x + 26(100 - 4x)}{100} = 24.60\). Simplifying this: \(72x + 25x + 2600 - 104x = 2460 \Rightarrow -7x + 2600 = 2460 \Rightarrow 7x = 140 \Rightarrow x = 20\). Therefore, the abundance of \(^{26}X\) is \(100 - 4(20) = 20\%\).

1 mark for the correct option (B).

Using the hydration energy cycle: \(\Delta_{\text{sol}}H^{\theta}(\text{MgCl}_2) = \Delta_{\text{hyd}}H^{\theta}(\text{Mg}^{2+}) + 2\Delta_{\text{hyd}}H^{\theta}(\text{Cl}^-) - \Delta_{\text{LE}}H^{\theta}(\text{MgCl}_2)\). Note that lattice enthalpy is defined as the exothermic formation of the lattice from gaseous ions. Substituting the values: \(-155 = -1920 + 2\Delta_{\text{hyd}}H^{\theta}(\text{Cl}^-) - (-2526) \Rightarrow -155 = -1920 + 2\Delta_{\text{hyd}}H^{\theta}(\text{Cl}^-) + 2526 \Rightarrow -155 = 606 + 2\Delta_{\text{hyd}}H^{\theta}(\text{Cl}^-) \Rightarrow 2\Delta_{\text{hyd}}H^{\theta}(\text{Cl}^-) = -761 \Rightarrow \Delta_{\text{hyd}}H^{\theta}(\text{Cl}^-) = -380.5\text{ kJ mol}^{-1}\), which rounds to \(-381\text{ kJ mol}^{-1}\).

1 mark for the correct option (B).

First, calculate initial moles: \(n(\text{HA}) = 0.0500 \times 0.200 = 0.0100\text{ mol}\) and \(n(\text{OH}^-) = 0.0250 \times 0.160 = 0.00400\text{ mol}\). The sodium hydroxide reacts completely with the propanoic acid: \(n(\text{HA})_{\text{remaining}} = 0.0100 - 0.0040 = 0.00600\text{ mol}\) and \(n(\text{A}^-)_{\text{formed}} = 0.00400\text{ mol}\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = -\log_{10}(1.35 \times 10^{-5}) + \log_{10}\left(\frac{0.00400}{0.00600}\right) = 4.870 + (-0.176) = 4.69\).

1 mark for the correct option (A).

The system with the more positive standard electrode potential undergoes reduction: \(\text{IO}_3^-(aq) + 6\text{H}^+(aq) + 5e^- \rightarrow \frac{1}{2}\text{I}_2(aq) + 3\text{H}_2\text{O}(l)\). The system with the less positive potential is reversed and undergoes oxidation: \(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\). The standard cell potential is: \(E^{\theta}_{\text{cell}} = E^{\theta}_{\text{red}} - E^{\theta}_{\text{ox}} = +1.19 - (+0.77) = +0.42\text{ V}\). To construct the overall equation, the oxidation half-equation must be multiplied by 5 to balance the electron flow: \(\text{IO}_3^-(aq) + 6\text{H}^+(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \frac{1}{2}\text{I}_2(aq) + 3\text{H}_2\text{O}(l) + 5\text{Fe}^{3+}(aq)\). Thus, the coefficient of \(\text{Fe}^{2+}(aq)\) is 5.

1 mark for the correct option (A).

The complex \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) exhibits geometric (cis-trans) isomerism. The trans-isomer has a plane of symmetry and is optically inactive. The cis-isomer is chiral and has no plane of symmetry, so it exists as a pair of non-superimposable mirror images (optical isomers or enantiomers). Therefore, there are 3 stereoisomers in total: 1 trans-isomer and 2 cis-enantiomers.

1 mark for the correct option (B).

Comparing Experiments 1 and 2: doubling \([\text{A}]\) while keeping \([\text{B}]\) constant quadruples the rate (\(2.0 \times 10^{-4} \rightarrow 8.0 \times 10^{-4}\)), indicating a second-order dependency on \([\text{A}]\). Comparing Experiments 2 and 3: doubling \([\text{B}]\) while keeping \([\text{A}]\) constant doubles the rate (\(8.0 \times 10^{-4} \rightarrow 1.6 \times 10^{-3}\)), indicating a first-order dependency on \([\text{B}]\). The rate equation is \(\text{Rate} = k[\text{A}]^2[\text{B}]\). Substituting data from Experiment 1: \(2.0 \times 10^{-4} = k(0.10)^2(0.10) \Rightarrow 2.0 \times 10^{-4} = k(1.0 \times 10^{-3}) \Rightarrow k = 0.20\). Units: \(\text{units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for the correct option (A).

The methoxy group (\(-\text{OCH}_3\)) in methoxybenzene is highly activating due to the resonance donation (+M effect) of the lone pair on the oxygen atom directly bonded to the ring. This greatly increases electron density on the aromatic ring, making it much more reactive towards electrophiles than benzene. It is ortho- (2-) and para- (4-) directing. Nitrobenzene and benzoic acid contain electron-withdrawing deactivating groups that are meta-directing. Chlorobenzene contains a deactivating group.

1 mark for the correct option (D).

Calculate the moles of \(\text{CO}_2\) using the ideal gas equation: \(pV = nRT \Rightarrow n = \frac{pV}{RT}\). Convert values to SI units: \(p = 101 \times 10^3\text{ Pa}\), \(V = 94.0 \times 10^{-6}\text{ m}^3\), \(T = 22.0 + 273.15 = 295.15\text{ K}\). Thus: \(n = \frac{101000 \times 94.0 \times 10^{-6}}{8.314 \times 295.15} = 0.00387\text{ mol}\). Since \(\text{XCO}_3 + 2\text{HCl} \rightarrow \text{XCl}_2 + \text{H}_2\text{O} + \text{CO}_2\), the molar ratio is \(1:1\), so \(n(\text{XCO}_3) = 0.00387\text{ mol}\). The molar mass of \(\text{XCO}_3\) is: \(M_r = \frac{0.385}{0.00387} = 99.5\text{ g mol}^{-1}\). The atomic mass of \(\text{X}\) is: \(A_r(\text{X}) = 99.5 - 12.0 - 3(16.0) = 39.5\text{ g mol}^{-1}\). This is closest to Calcium (\(A_r = 40.1\text{ g mol}^{-1}\)).

1 mark for the correct option (B).

Using the energy cycle for dissolving an ionic solid: \( \Delta_{sol}H = \Delta_{latt}H\text{ (dissociation)} + \Delta_{hyd}H(\text{Ca}^{2+}) + 2\Delta_{hyd}H(\text{Cl}^-) \). Substituting the given values: \( -83 = +2237 + \Delta_{hyd}H(\text{Ca}^{2+}) + 2(-378) \). Simplify the equation: \( -83 = 2237 + \Delta_{hyd}H(\text{Ca}^{2+}) - 756 \) which gives \( -83 = 1481 + \Delta_{hyd}H(\text{Ca}^{2+}) \). Therefore, \( \Delta_{hyd}H(\text{Ca}^{2+}) = -83 - 1481 = -1564\text{ kJ mol}^{-1} \).

1 mark for the correct calculation showing \( -1564\text{ kJ mol}^{-1} \). Reject positive value (D) or calculation omitting the stoichiometry of two chloride ions (B).

First, calculate initial moles: \( n(\text{HA}) = 0.250 \times 0.200 = 0.0500\text{ mol} \), \( n(\text{OH}^-) = 0.150 \times 0.150 = 0.0225\text{ mol} \). The neutralization reaction is: \( \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \). After reaction, moles are: \( n(\text{HA}) = 0.0500 - 0.0225 = 0.0275\text{ mol} \) and \( n(\text{A}^-) = 0.0225\text{ mol} \). Using the Henderson-Hasselbalch equation: \( \text{pH} = \text{p}K_a + \log_{10}(\frac{[\text{A}^-]}{[\text{HA}]}) = -\log_{10}(1.35 \times 10^{-5}) + \log_{10}(\frac{0.0225}{0.0275}) = 4.87 - 0.087 = 4.78 \).

1 mark for the correct calculation resulting in pH 4.78. Correctly accounts for neutralization stoichiometry before applying the buffer equation.

The complex has both cis and trans geometric isomers. The trans isomer is symmetrical and has no optical isomers (1 stereoisomer). The cis isomer is chiral and exists as a pair of non-superimposable optical isomers (2 enantiomers). Therefore, the total number of stereoisomers is 3.

1 mark for identifying 3 total stereoisomers (one trans-isomer and a pair of cis-enantiomers).

The reaction described is electrophilic aromatic substitution (nitration). Methylbenzene reacts most rapidly because the methyl group is an electron-donating group (+I inductive effect), which increases the electron density of the benzene ring and makes it a stronger nucleophile. Chlorobenzene and nitrobenzene contain deactivating groups, while benzene is the standard baseline.

1 mark for identifying methylbenzene as the most reactive compound due to the activating effect of the alkyl substituent.

The half-cell with the more positive standard electrode potential (dichromate, +1.33 V) undergoes reduction, while the other half-cell (tin, +0.15 V) undergoes oxidation. To balance the electrons transferred, the tin half-equation must be multiplied by 3. The overall equation is: \( \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 3\text{Sn}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 3\text{Sn}^{4+}(aq) \). The cell potential is \( E^{\theta}_{cell} = E^{\theta}_{red} - E^{\theta}_{ox} = +1.33\text{ V} - (+0.15\text{ V}) = +1.18\text{ V} \).

1 mark for identifying both the correctly balanced stoichiometric equation and the cell potential of +1.18 V.

First calculate the moles of \( \text{CO}_2 \) gas using \( pV = nRT \). Convert units: \( p = 101000\text{ Pa} \), \( V = 3.64 \times 10^{-4}\text{ m}^3 \), \( T = 298\text{ K} \). So, \( n = \frac{101000 \times 3.64 \times 10^{-4}}{8.314 \times 298} = 0.01484\text{ mol} \). Since \( 1\text{ mol} \) of \( \text{MCO}_3 \) produces \( 1\text{ mol} \) of \( \text{CO}_2 \), the molar mass of \( \text{MCO}_3 \) is \( \frac{1.25\text{ g}}{0.01484\text{ mol}} = 84.2\text{ g mol}^{-1} \). Subtracting the mass of the carbonate group (\( 12.0 + 3 \times 16.0 = 60.0 \)), we get \( M_r(\text{M}) = 84.2 - 60.0 = 24.2\text{ g mol}^{-1} \), which corresponds to magnesium.

1 mark for determining the correct group 2 metal by converting units, calculating gas moles, finding molar mass, and identifying magnesium.

At a highly alkaline pH of 12, both the acidic carboxylic acid group and the basic amine group exist in their conjugate base forms. The carboxylic acid group (\( -\text{COOH} \)) is deprotonated to form the carboxylate anion (\( -\text{COO}^- \)), and the amine group (\( -\text{NH}_3^+ \)) is deprotonated to form the neutral amine (\( -\text{NH}_2 \)). This results in the anion \( \text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^- \).

1 mark for identifying the deprotonated anionic form of the amino acid at high pH.

(a) The electronic configuration of a chromium atom is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 4\text{s}^1 \). For the \( \text{Cr}^{3+} \) ion, three electrons are removed (from the 4s orbital first, then two from the 3d orbitals), giving \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3 \). The configuration of the Cr atom is unusual because the 4s subshell is only half-filled with 1 electron to allow the 3d subshell to have 5 electrons, which provides greater stability due to a half-filled d-subshell.

(b) (i) The coordination number of \( [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ \) is 6, because there are six dative covalent (co-ordinate) bonds formed with the central chromium ion (four from water molecules and two from chloride ions). The shape is octahedral.
(ii) The stereoisomers are *cis* and *trans* isomers. In the *trans* isomer, the two chloride ligands are directly opposite each other (bond angle 180°). In the *cis* isomer, the two chloride ligands are adjacent to each other (bond angle 90°).

(c) (i) The initial precipitate formed is grey-green in colour, with the formula \( \text{Cr}(\text{OH})_3 \) (or \( [\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3] \)).
(ii) The ionic equation for the reaction with excess ammonia is:
\( \text{Cr}(\text{OH})_3(\text{s}) + 6\text{NH}_3(\text{aq}) \rightarrow [\text{Cr}(\text{NH}_3)_6]^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \)
(or \( [\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3](\text{s}) + 6\text{NH}_3(\text{aq}) \rightarrow [\text{Cr}(\text{NH}_3)_6]^{3+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) + 3\text{OH}^-(\text{aq}) \))

(d) A homogeneous catalyst is in the same physical state (or phase) as the reactants, whereas a heterogeneous catalyst is in a different physical state (or phase) from the reactants.

*(a)*
• \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 4\text{s}^1 \) (or noble gas core equivalent) [1]
• \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3 \) [1]
• Explanation: Half-filled d-subshell is more stable (or lower energy state) [1]

*(b)(i)*
• Coordination number: 6 AND Shape: octahedral [1]
• Explanation: Six dative bonds to central metal ion [1]

*(b)(ii)*
• 3D diagram of cis-isomer (Cl ligands at 90 degrees) [1]
• 3D diagram of trans-isomer (Cl ligands at 180 degrees) [1]
• Correctly labeled as cis-trans / geometric isomerism [1]

*(c)(i)*
• Colour: grey-green (accept green) precipitate [1]
• Formula: \( \text{Cr}(\text{OH})_3 \) OR \( [\text{Cr}(\text{H}_2\text{O})_3(\text{OH})_3] \) [1]

*(c)(ii)*
• Correct reactants and products: \( \text{Cr}(\text{OH})_3 + 6\text{NH}_3 \rightarrow [\text{Cr}(\text{NH}_3)_6]^{3+} + 3\text{OH}^- \) [1]
• State symbols and fully balanced equation [1]

*(d)*
• Homogeneous catalyst in same physical state AND heterogeneous catalyst in different physical state as reactants [1]

(a) A buffer solution is a solution that minimises pH changes when small amounts of an acid or a base are added to it.

(b) The dissociation of propanoic acid is: \( \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) \).
Therefore, \( K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \).

(c) First, calculate the required concentration of hydrogen ions:
\( [\text{H}^+] = 10^{-\text{pH}} = 10^{-4.85} = 1.4125 \times 10^{-5}\ \text{mol}\ \text{dm}^{-3} \).

Rearrange the \( K_a \) expression to find the concentration of propanoate ions, \( [\text{A}^-] \):
\( [\text{A}^-] = \frac{K_a \times [\text{HA}]}{[\text{H}^+]} = \frac{1.35 \times 10^{-5} \times 0.250}{1.4125 \times 10^{-5}} = 0.2389\ \text{mol}\ \text{dm}^{-3} \).

Calculate the moles of sodium propanoate needed in \( 500\ \text{cm}^3 \) (\( 0.500\ \text{dm}^3 \)):
\( n = c \times V = 0.2389\ \text{mol}\ \text{dm}^{-3} \times 0.500\ \text{dm}^3 = 0.1195\ \text{mol} \).

Calculate the mass of sodium propanoate required:
\( \text{mass} = n \times M_r = 0.1195\ \text{mol} \times 96.0\ \text{g}\ \text{mol}^{-1} = 11.47\ \text{g} \approx 11.5\ \text{g} \).

(d) (i) When acid (\( \text{H}^+ \)) is added, it reacts with the propanoate conjugate base:
\( \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) \)

(ii) Find the initial moles of components in \( 100\ \text{cm}^3 \) of buffer:
\( n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.100\ \text{dm}^3 \times 0.250\ \text{mol}\ \text{dm}^{-3} = 0.0250\ \text{mol} \).
\( n(\text{CH}_3\text{CH}_2\text{COO}^-) = 0.100\ \text{dm}^3 \times 0.2389\ \text{mol}\ \text{dm}^{-3} = 0.02389\ \text{mol} \).

Moles of \( \text{H}^+ \) added:
\( n(\text{H}^+) = 1.00 \times 10^{-3}\ \text{dm}^3 \times 1.00\ \text{mol}\ \text{dm}^{-3} = 0.00100\ \text{mol} \).

Calculate new moles after added \( \text{H}^+ \) reacts:
\( n(\text{CH}_3\text{CH}_2\text{COOH})_{\text{new}} = 0.0250 + 0.00100 = 0.0260\ \text{mol} \).
\( n(\text{CH}_3\text{CH}_2\text{COO}^-)_{\text{new}} = 0.02389 - 0.00100 = 0.02289\ \text{mol} \).

Calculate the new pH using the rearranged \( K_a \) expression (volumes cancel out):
\( [\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.35 \times 10^{-5} \times \frac{0.0260}{0.02289} = 1.533 \times 10^{-5}\ \text{mol}\ \text{dm}^{-3} \).
\( \text{pH} = -\log_{10}(1.533 \times 10^{-5}) = 4.81 \).

*(a)*
• Minimises changes in pH [1]
• When small amounts of acid or base are added [1]

*(b)*
• \( K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \) (square brackets required, do not accept H2O in expression) [1]

*(c)*
• \( [\text{H}^+] = 10^{-4.85} = 1.41 \times 10^{-5}\ \text{mol}\ \text{dm}^{-3} \) [1]
• Rearranging expression to make \( [\text{A}^-] = \frac{K_a \times [\text{HA}]}{[\text{H}^+]} \) [1]
• \( [\text{A}^-] = 0.239\ \text{mol}\ \text{dm}^{-3} \) [1]
• \( n(\text{A}^-) = 0.239 \times 0.500 = 0.1195\ \text{mol} \) [1]
• \( \text{mass} = 0.1195 \times 96.0 = 11.5\ \text{g} \) (accept range 11.4 to 11.5 g) [1]

*(d)(i)*
• \( \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH} \) (state symbols not required) [1]

*(d)(ii)*
• Calculates moles of added \( \text{H}^+ \) (0.00100 mol) AND initial moles of HA (0.0250 mol) and \( \text{A}^- \) (0.0239 mol) [1]
• Calculates new moles of HA = 0.0260 mol AND new moles of \( \text{A}^- \) = 0.0229 mol [1]
• Calculates new \( [\text{H}^+] = 1.53 \times 10^{-5}\ \text{mol}\ \text{dm}^{-3} \) [1]
• pH = 4.81 (accept 4.80 - 4.82) [1]

(a) (i) "Heating to constant mass" means heating the sample, allowing it to cool, weighing it, and repeating this sequence until two consecutive mass measurements are identical. It is carried out to ensure that all water of crystallisation has been completely driven off.
(ii) Mass of water lost = \( 4.704\ \text{g} - 3.408\ \text{g} = 1.296\ \text{g} \).
Percentage by mass of water = \( \frac{1.296}{4.704} \times 100 = 27.55\% \approx 27.6\% \).

(b) (i) The end point colour change is from colourless to a permanent pale pink (as the first excess of purple manganate(VII) ions is not reduced).
(ii) Moles of \( \text{MnO}_4^- \) used in the titration:
\( n(\text{MnO}_4^-) = 0.0200\ \text{mol}\ \text{dm}^{-3} \times 0.01200\ \text{dm}^3 = 2.40 \times 10^{-4}\ \text{mol} \).

Using the 1:5 stoichiometry from the equation:
\( n(\text{Fe}^{2+}) \text{ in } 25.0\ \text{cm}^3 = 5 \times 2.40 \times 10^{-4} = 1.20 \times 10^{-3}\ \text{mol} \).

In the original \( 250.0\ \text{cm}^3 \) solution:
\( n(\text{Fe}^{2+}) = 1.20 \times 10^{-3}\ \text{mol} \times \frac{250.0}{25.0} = 0.0120\ \text{mol} \).

(iii) Since 1 mole of the hydrated double salt contains 1 mole of \( \text{Fe}^{2+} \) (ratio of \( b = 1 \)):
Moles of hydrated double salt in the \( 4.704\ \text{g} \) sample = \( 0.0120\ \text{mol} \).

Molar mass of the hydrated salt, \( M_r \):
\( M_r = \frac{\text{mass}}{\text{moles}} = \frac{4.704\ \text{g}}{0.0120\ \text{mol}} = 392.0\ \text{g}\ \text{mol}^{-1} \).

Let's check the moles of water in the sample:
\( n(\text{H}_2\text{O}) = \frac{1.296\ \text{g}}{18.0\ \text{g}\ \text{mol}^{-1}} = 0.0720\ \text{mol} \).

Ratio of \( n(\text{H}_2\text{O}) \) to \( n(\text{Fe}^{2+}) \):
\( x = \frac{0.0720}{0.0120} = 6 \).

The molar mass of the anhydrous part is:
\( M_r(\text{anhydrous}) = M_r(\text{hydrated}) - (6 \times M_r(\text{H}_2\text{O})) = 392.0 - (6 \times 18.0) = 284.0\ \text{g}\ \text{mol}^{-1} \).
This matches the \( M_r \) of \( (\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \) given (284.0).
Hence, the full formula of the hydrated double salt is \( (\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \cdot 6\text{H}_2\text{O} \).

*(a)(i)*
• Heating, weighing, and heating again until mass does not change [1]
• To ensure all water of crystallisation is evaporated/lost [1]

*(a)(ii)*
• \( 4.704 - 3.408 = 1.296\ \text{g} \) water lost [1]
• \( \frac{1.296}{4.704} \times 100 = 27.6\% \) [1]

*(b)(i)*
• Colourless to permanent pale pink (reject purple) [1]

*(b)(ii)*
• \( n(\text{MnO}_4^-) = 0.01200 \times 0.0200 = 2.40 \times 10^{-4}\ \text{mol} \) [1]
• \( n(\text{Fe}^{2+}) \text{ in } 25\ \text{cm}^3 = 1.20 \times 10^{-3}\ \text{mol} \) [1]
• \( n(\text{Fe}^{2+}) \text{ in } 250\ \text{cm}^3 = 0.0120\ \text{mol} \) [1]

*(b)(iii)*
• Molar mass of hydrated salt = \( \frac{4.704}{0.0120} = 392.0\ \text{g}\ \text{mol}^{-1} \) [1]
• Moles of water in sample = \( \frac{1.296}{18.0} = 0.0720\ \text{mol} \) [1]
• Ratio of water to salt = \( \frac{0.0720}{0.0120} = 6 \) [1]
• Verification of anhydrous mass = \( 392.0 - 108.0 = 284.0\ \text{g}\ \text{mol}^{-1} \) [1]
• Showing that 284.0 matches the formula mass of \( (\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \) [1]
• Correct formula concluding as \( (\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \cdot 6\text{H}_2\text{O} \) [1]

(a) The strong, sharp peak at \( 1740\ \text{cm}^{-1} \) indicates the presence of a carbonyl group, \( \text{C}=\text{O} \). The lack of a broad absorption in the region \( 3200 - 3600\ \text{cm}^{-1} \) confirms the absence of an \( \text{O}-\text{H} \) bond, meaning **Y** is neither an alcohol nor a carboxylic acid. Together with the molecular formula \( \text{C}_4\text{H}_8\text{O}_2 \), this points to an ester.

(b) The presence of 4 peaks in the \( ^{13}\text{C} \) NMR spectrum indicates that the molecule contains exactly 4 different carbon environments.

(c) Analysis of \( ^1\text{H} \) NMR peaks:
- The singlet at \( \delta = 2.05\ \text{ppm} \) with relative area 3 is a methyl group, \( -\text{CH}_3 \), adjacent to a carbon with no hydrogen atoms. Its shift indicates it is attached to the carbonyl group, i.e., \( \text{CH}_3\text{C}=\text{O} \).
- The triplet at \( \delta = 1.25\ \text{ppm} \) with relative area 3 is a \( -\text{CH}_3 \) group with 2 adjacent protons (coupled to a \( -\text{CH}_2- \) group).
- The quartet at \( \delta = 4.12\ \text{ppm} \) with relative area 2 is a \( -\text{CH}_2- \) group with 3 adjacent protons (coupled to a \( -\text{CH}_3 \) group). Its downfield shift shows it is attached directly to the single-bonded oxygen of the ester group: \( -\text{O}-\text{CH}_2- \).

(d) The coupling between the triplet (area 3) and quartet (area 2) confirms the presence of an ethyl group, \( -\text{CH}_2\text{CH}_3 \). Because the quartet is shifted to \( 4.12\ \text{ppm} \), this ethyl group must be bonded directly to the single-bonded oxygen: \( -\text{O}-\text{CH}_2\text{CH}_3 \).

(e) Combining \( \text{CH}_3\text{CO}- \) and \( -\text{OCH}_2\text{CH}_3 \) yields **ethyl ethanoate**, \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \).
Its displayed formula has all atoms and bonds explicitly shown, including C-H, C-C, C=O, C-O, and O-C bonds.

(f) The peak at \( m/z = 43 \) corresponds to the acylium fragment ion \( [\text{CH}_3\text{CO}]^+ \) (or the propyl cation \( [\text{C}_3\text{H}_7]^+ \), though the former is much more characteristic of ethanoate esters).

*(a)*
• Esters/carbonyl (\( \text{C}=\text{O} \)) identified by peak at \( 1740\ \text{cm}^{-1} \) [1]
• Alcohol/acid (\( \text{O}-\text{H} \)) ruled out by absence of broad peak at \( 3200-3600\ \text{cm}^{-1} \) [1]

*(b)*
• 4 peaks = 4 different carbon environments [1]

*(c)*
• \( \delta = 2.05\ \text{ppm} \) singlet (area 3) indicates \( \text{CH}_3\text{CO}- \) environment [1]
• \( \delta = 1.25\ \text{ppm} \) triplet (area 3) indicates \( -\text{CH}_3 \) next to \( -\text{CH}_2- \) [1]
• \( \delta = 4.12\ \text{ppm} \) quartet (area 2) indicates \( -\text{CH}_2- \) next to \( -\text{CH}_3 \) AND bonded to oxygen [1]

*(d)*
• Triplet-quartet splitting pattern confirms ethyl group [1]
• Position of quartet at 4.12 ppm indicates ethyl group is attached to ester oxygen: \( -\text{OCH}_2\text{CH}_3 \) [1]

*(e)*
• Name: ethyl ethanoate [1]
• Structural formula: \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \) [1]
• Correctly drawn displayed formula with all bonds and atoms visible [1]

*(f)*
• \( [\text{CH}_3\text{CO}]^+ \) (charge must be present; accept \( \text{C}_2\text{H}_3\text{O}^+ \) or \( [\text{C}_3\text{H}_7]^+ \)) [1]

(a) (i) Step 1 is electrophilic aromatic substitution (nitration). The reagents are concentrated nitric acid (\( \text{HNO}_3 \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). The mixture is maintained at a temperature of around \( 50-60\ ^\circ\text{C} \).
(ii) The reaction to form the electrophile is:
\( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+ \)
(or \( \text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O} \)).

(b) The oxidation of the methyl group on a benzene ring requires a strong oxidising agent, typically acidified potassium dichromate(VI) (\( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+ \)) or potassium manganate(VII) (\( \text{KMnO}_4 \)) heated under reflux.

(c) (i) The catalyst is concentrated sulfuric acid, \( \text{H}_2\text{SO}_4 \).
(ii) Equation:
\( \text{O}_2\text{NC}_6\text{H}_4\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{O}_2\text{NC}_6\text{H}_4\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O} \)

(d) (i) Reduction of a nitroarene to an arylamine requires tin (\( \text{Sn} \)) and concentrated hydrochloric acid (\( \text{HCl} \)), heated under reflux, followed by treatment with aqueous sodium hydroxide (\( \text{NaOH} \)) to release the free amine from its salt.
(ii) The balanced equation is:
\( \text{O}_2\text{NC}_6\text{H}_4\text{COOCH}_2\text{CH}_3 + 6[\text{H}] \rightarrow \text{H}_2\text{NC}_6\text{H}_4\text{COOCH}_2\text{CH}_3 + 2\text{H}_2\text{O} \)

(e) Recrystallisation process:
1. Dissolve the crude benzocaine in the minimum volume of a hot suitable solvent.
2. Allow the hot solution to cool slowly to room temperature and then in ice to recrystallise, filter the pure crystals under reduced pressure (using a Buchner funnel), wash them with a small amount of ice-cold solvent, and leave them to dry.

*(a)(i)*
• Concentrated nitric acid AND concentrated sulfuric acid [1]
• Temperature of 50-60 °C (heated under reflux) [1]

*(a)(ii)*
• \( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+ \) (or equivalent) [1]

*(b)*
• Acidified potassium dichromate(VI) OR alkaline potassium manganate(VII) heated under reflux [1]

*(c)(i)*
• Concentrated sulfuric acid [1]

*(c)(ii)*
• Correct organic structures: \( \text{O}_2\text{NC}_6\text{H}_4\text{COOH} \) and \( \text{O}_2\text{NC}_6\text{H}_4\text{COOCH}_2\text{CH}_3 \) [1]
• Balanced equation with \( \text{H}_2\text{O} \) product [1]

*(d)(i)*
• Tin (Sn) and concentrated hydrochloric acid [1]
• Addition of sodium hydroxide (NaOH) to liberate the free amine [1]

*(d)(ii)*
• Reactant and product: \( \text{O}_2\text{NC}_6\text{H}_4\text{COOCH}_2\text{CH}_3 + [\text{H}] \rightarrow \text{H}_2\text{NC}_6\text{H}_4\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O} \) [1]
• Correct balancing: \( 6[\text{H}] \) and \( 2\text{H}_2\text{O} \) [1]

*(e)*
• Dissolve in minimum volume of hot solvent [1]
• Cool in ice, filter under suction/reduced pressure, wash with cold solvent, and dry [1]

(a) Standard electrode potential is the electromotive force (e.m.f.) of a half-cell compared to a standard hydrogen half-cell, measured under standard conditions: a temperature of \( 298\ \text{K} \), solution concentrations of \( 1.00\ \text{mol}\ \text{dm}^{-3} \), and gas pressure of \( 1\ \text{atm} \) (or \( 100\ \text{kPa} \)).

(b) The diagram must show:
- Left half-cell: copper metal strip in \( 1.0\ \text{mol}\ \text{dm}^{-3}\ \text{Cu}^{2+}(\text{aq}) \).
- Right half-cell: platinum wire/foil electrode in a solution containing both \( 1.0\ \text{mol}\ \text{dm}^{-3}\ \text{Fe}^{2+}(\text{aq}) \) and \( 1.0\ \text{mol}\ \text{dm}^{-3}\ \text{Fe}^{3+}(\text{aq}) \).
- Voltmeter connecting the two electrodes.
- Salt bridge (labeled) dipped into both solutions.
- Direction of electron flow: from the Cu electrode (negative electrode, since it is oxidised) to the Pt electrode (positive electrode, where reduction occurs).

(c) \( E^\theta_{\text{cell}} = E^\theta_{\text{positive electrode}} - E^\theta_{\text{negative electrode}} = +0.77\ \text{V} - (+0.34\ \text{V}) = +0.43\ \text{V} \).

(d) The half-equations are:
Oxidation: \( \text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \)
Reduction: \( \text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq}) \)
Multiplying the reduction half-equation by 2 and combining:
\( \text{Cu}(\text{s}) + 2\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \).

(e) (i) When \( [\text{Cu}^{2+}] \) is decreased, the position of the equilibrium \( \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \) shifts to the left to replace the lost copper(II) ions. This releases more electrons, causing the electrode potential of the copper half-cell to decrease (become less positive / more negative).
(ii) The cell potential is the difference between the electrode potentials: \( E_{\text{cell}} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_{\text{Cu}^{2+}/\text{Cu}} \).
Since \( E_{\text{Cu}^{2+}/\text{Cu}} \) decreases while \( E_{\text{Fe}^{3+}/\text{Fe}^{2+}} \) remains constant, the difference between them increases. Therefore, the overall cell potential, \( E_{\text{cell}} \), increases.

*(a)*
• Electromotive force / potential difference of a half-cell compared to a standard hydrogen electrode [1]
• Measured at \( 298\ \text{K} \), \( 1.00\ \text{mol}\ \text{dm}^{-3} \) and \( 100\ \text{kPa} \) [1]

*(b)*
• Correctly drawn copper electrode in \( \text{Cu}^{2+} \) solution AND platinum electrode in mixed \( \text{Fe}^{2+}/\text{Fe}^{3+} \) solution [1]
• Salt bridge connecting the two solutions AND voltmeter in the circuit [1]
• Correct labels including concentrations of all solutions (\( 1.0\ \text{mol}\ \text{dm}^{-3} \)) [1]
• Arrow showing electron flow from copper to platinum electrode through the wire [1]

*(c)*
• \( E^\theta_{\text{cell}} = +0.43\ \text{V} \) (unit required) [1]

*(d)*
• Correct species: \( \text{Cu} + \text{Fe}^{3+} \rightarrow \text{Cu}^{2+} + \text{Fe}^{2+} \) [1]
• Correct balancing: \( \text{Cu}(\text{s}) + 2\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \) [1]

*(e)(i)*
• The potential of the copper half-cell decreases / becomes less positive [1]
• Equilibrium shifts left (producing more electrons on the electrode) [1]

*(e)(ii)*
• The cell potential \( E_{\text{cell}} \) increases [1]
• Because the difference between the two half-cell potentials is larger [1]

(a) Calculate heat energy released, \( q \):
Total mass of solution: \( m = 50.0\ \text{cm}^3 + 50.0\ \text{cm}^3 = 100.0\ \text{cm}^3 \rightarrow 100.0\ \text{g} \).
Temperature change: \( \Delta T = 29.3 - 19.5 = 9.8\ ^\circ\text{C} \) (or 9.8 K).
\( q = m c \Delta T = 100.0\ \text{g} \times 4.18\ \text{J}\ \text{g}^{-1}\ \text{K}^{-1} \times 9.8\ \text{K} = 4096.4\ \text{J} \).

(b) Calculate moles of \( \text{H}_2\text{O} \) formed:
\( n(\text{HCl}) = c \times V = 1.50\ \text{mol}\ \text{dm}^{-3} \times 0.0500\ \text{dm}^3 = 0.0750\ \text{mol} \).
\( n(\text{NaOH}) = c \times V = 1.50\ \text{mol}\ \text{dm}^{-3} \times 0.0500\ \text{dm}^3 = 0.0750\ \text{mol} \).
Since they react in a 1:1 ratio, \( 0.0750\ \text{mol} \) of \( \text{H}_2\text{O} \) is formed.

(c) Calculate \( \Delta H_{\text{neut}} \):
\( q = 4.0964\ \text{kJ} \).
\( \Delta H_{\text{neut}} = -\frac{q}{n(\text{H}_2\text{O})} = -\frac{4.0964\ \text{kJ}}{0.0750\ \text{mol}} = -54.618\ \text{kJ}\ \text{mol}^{-1} \).
To 3 significant figures, \( \Delta H_{\text{neut}} = -54.6\ \text{kJ}\ \text{mol}^{-1} \).

(d) (i) Two reasons for difference:
1. Heat is lost to the surroundings / air.
2. Heat is absorbed by the polystyrene cup itself (or the thermometer/stirrer).
(Also accept: specific heat capacity or density of the salt solution is not exactly the same as pure water).
(ii) Modification: Place a lid on the polystyrene cup to reduce heat loss through evaporation and convection (or place the cup inside another polystyrene cup to improve insulation).

(e) Ethanoic acid is a weak acid, meaning it is only partially dissociated in solution. When it reacts, some energy is required to dissociate the undissociated molecules of ethanoic acid (i.e. break the \( \text{O}-\text{H} \) bonds). Because this bond-breaking process is endothermic, the overall enthalpy change of neutralisation, \( \Delta H_{\text{neut}} \), will be less exothermic (less negative) than for strong acids.

*(a)*
• Calculates \( m = 100.0\ \text{g} \) AND \( \Delta T = 9.8\ \text{K} \) [1]
• Calculates \( q = 100.0 \times 4.18 \times 9.8 = 4096.4\ \text{J} \) (accept 4100 J) [1]

*(b)*
• Calculates moles of reactants: \( 0.0500 \times 1.50 = 0.0750\ \text{mol} \) [1]
• States moles of water formed is \( 0.0750\ \text{mol} \) [1]

*(c)*
• Division of energy by moles: \( \frac{4.0964}{0.0750} = 54.6 \) [1]
• Negative sign included [1]
• Correct value to 3 significant figures: \( -54.6\ \text{kJ}\ \text{mol}^{-1} \) (allow ECF from part a and b) [1]

*(d)(i)*
• Any two of: Heat lost to surroundings / heat absorbed by apparatus / specific heat capacity not equal to 4.18 / density of solution not equal to 1.00 [2]

*(d)(ii)*
• Add a lid to the cup OR double-cup (nesting cups) [1]

*(e)*
• Value is less exothermic / less negative [1]
• Explanation: Some energy is required to dissociate the weak acid (or to break O-H bonds) [1]

To find the number of chiral carbon atoms, we look for carbon atoms that are bonded to four different groups. Carbon-2 is bonded to -H, -NH2, -COOH, and -CH2CH(CH3)CH(OH)CH3 (Chiral). Carbon-3 is a -CH2- group bonded to two hydrogen atoms (Achiral). Carbon-4 is bonded to -H, -CH3, -CH2CH(NH2)COOH, and -CH(OH)CH3 (Chiral). Carbon-5 is bonded to -H, -OH, -CH3, and -CH(CH3)CH2CH(NH2)COOH (Chiral). Therefore, there are exactly 3 chiral carbon atoms in the molecule.

1 mark for correct option C. Reject other options based on incorrect count of chiral centres.

In 4-methylphenol, there are two directing groups on the benzene ring. The -OH group is strongly activating and directs ortho- (positions 2 and 6) and para- (position 4). The -CH3 group is weakly activating and directs ortho- and para-. Since the -OH group is a much stronger activating group than the methyl group, its directing effect dominates. Since position 4 (para to the -OH group) is already occupied by the methyl group, substitution will occur at position 2 (or 6), which is ortho to the -OH group. The product of mono-bromination is therefore 2-bromo-4-methylphenol.

1 mark for correct option A. Reject B (incorrect directing position), C (side-chain substitution instead of electrophilic ring substitution), and D (di-substituted product).

Let us determine the number of carbon environments (peaks in 13C NMR) for each isomer. Pentan-2-one (CH3COCH2CH2CH3) has 5 different carbon environments (5 peaks). Pentan-3-one (CH3CH2COCH2CH3) is a symmetrical molecule with a plane of symmetry through the carbonyl group. It has 3 carbon environments: the two equivalent methyl carbons, the two equivalent methylene carbons, and the carbonyl carbon (3 peaks). 3-methylbutan-2-one has 4 carbon environments (4 peaks). 2-methylbutanal has 5 different carbon environments (5 peaks).

1 mark for correct option B. Reject options with 4 or 5 peaks.

The ester HCOOCH2CH2CH2OOCH (propane-1,3-diyl dimethanoate) has the molecular formula C5H8O4. Complete alkaline hydrolysis with NaOH yields HOCH2CH2CH2OH (propane-1,3-diol, a diol) and 2 moles of HCOONa (sodium methanoate, a sodium salt of a carboxylic acid). Let us analyze the other choices: Option A (dimethyl malonate) yields methanol (a mono-ol) and disodium malonate (a dicarboxylate salt). Option C yields methanol, sodium glycolate, and sodium acetate. Option D yields ethanol, methanol, and disodium oxalate.

1 mark for correct option B. Reject options that do not yield a diol and a single monocarboxylate sodium salt with molecular formula C5H8O4.

To find the monomers, we split the ester links in the repeating unit. The dialcohol portion is -O-CH2-CH2-O-, which comes from the diol HO-CH2-CH2-OH (ethane-1,2-diol). The dicarboxylic acid portion is -CO-CH(CH3)-CO-, which comes from the dicarboxylic acid HOOC-CH(CH3)-COOH (methylpropanedioic acid). Therefore, the correct monomers are ethane-1,2-diol and methylpropanedioic acid.

1 mark for correct option A. Reject options with incorrect diol or dicarboxylic acid structures.

X reacts with 2,4-DNPH but not Tollens' reagent, indicating X is a ketone. Reduction of a ketone with NaBH4 yields a secondary alcohol. Let us check the reduction products of the candidate ketones: Butanone reduces to butan-2-ol, which contains a chiral carbon at C2 bonded to -H, -OH, -CH3, and -CH2CH3 (chiral). Pentan-3-one reduces to pentan-3-ol, which is achiral as C3 is bonded to two identical ethyl groups. Propanone reduces to propan-2-ol, which is achiral as C2 is bonded to two identical methyl groups. Therefore, X must be butanone.

1 mark for correct option B. Reject A (aldehyde), C (achiral product), and D (achiral product).

To convert 2-chlorobutane (C4) into 2-methylbutanoic acid (C5), we must increase the carbon chain length by one carbon. Step 1: Reaction of 2-chlorobutane with potassium cyanide (KCN) in ethanol under reflux undergoes nucleophilic substitution to form 2-methylbutanenitrile. Step 2: Acid hydrolysis of the nitrile group using dilute hydrochloric acid under reflux converts the -CN group into a carboxylic acid (-COOH) group, forming 2-methylbutanoic acid. Other routes do not add a carbon atom or use inappropriate reactants.

1 mark for correct option A. Reject B (does not increase chain length), C (HCN/NaCN is for carbonyls, not haloalkanes), and D (forms an amine).

To name this compound using IUPAC rules, we first find the longest carbon chain containing the C=C double bond. Carbon-1 is CH2=, Carbon-2 is C(CH2CH3)=, Carbon-3 is CH(CH3), Carbon-4 is CH2, and Carbon-5 is CH3. This longest chain has 5 carbons containing the double bond, making it a pent-1-ene. Next, identify the substituents on this main chain: on Carbon-2, there is an ethyl group (-CH2CH3); on Carbon-3, there is a methyl group (-CH3). Combining these alphabetically gives the name: 2-ethyl-3-methylpent-1-ene.

1 mark for correct option A. Reject B (incorrect isomer), C (incorrect numbering/chain), and D (incorrectly selects a shorter 4-carbon chain as the main chain).

The infrared spectrum shows a broad absorption at \(3300\text{ cm}^{-1}\) (O-H stretch of a carboxylic acid) and a strong carbonyl absorption at \(1715\text{ cm}^{-1}\) (C=O stretch), indicating a carboxylic acid. This rules out ethyl propanoate. The singlet at \(\delta = 11.5\text{ ppm}\) (1H) corresponds to the carboxylic acid proton. The doublet at \(\delta = 1.0\text{ ppm}\) (6H) shows six equivalent protons coupled to a single proton, indicating a \(-\text{CH}(\text{CH}_3)_2\) group. The doublet at \(\delta = 2.2\text{ ppm}\) (2H) indicates a \(-\text{CH}_2-\ell\) group adjacent to a CH proton and the carboxylic acid group. This matches 3-methylbutanoic acid, \((\text{CH}_3)_2\text{CHCH}_2\text{COOH}\).

1 mark for identifying 3-methylbutanoic acid as the correct structure based on the IR and NMR splitting patterns.

The starting material is 3-methylbenzoic acid. The methyl group is an electron-donating group and is ortho/para-directing (positions 2, 4, 6 relative to the methyl group, which correspond to positions 2, 4, 6 relative to the carboxylic acid). The carboxylic acid group is electron-withdrawing and meta-directing (position 5 relative to the carboxylic acid). Activating groups (methyl) dominate over deactivating groups (carboxylic acid) in directing effects. The 2-position is sterically hindered because it lies between both substituents. Therefore, the electrophilic substitution occurs primarily at position 4, yielding 3-methyl-4-nitrobenzoic acid.

1 mark for identifying 3-methyl-4-nitrobenzoic acid as the major organic product based on directing group priorities.

The reaction of an aldehyde with hydrogen cyanide (generated from NaCN and H2SO4) is a nucleophilic addition. The cyanide nucleophile attacks the planar carbonyl carbon equally from above and below, generating a racemic mixture of the product. The reactant butanal has 4 carbons, so the addition of the nitrile group produces a 5-carbon chain: 2-hydroxypentanenitrile, which contains a chiral carbon at C2. Therefore, a racemic mixture of 2-hydroxypentanenitrile is formed.

1 mark for identifying that the reaction is a nucleophilic addition resulting in a racemic mixture of 2-hydroxypentanenitrile.

The repeating unit of the polyester is derived from a single hydroxycarboxylic acid monomer. By adding a hydrogen atom to the oxygen atom at one end and a hydroxyl group to the carbonyl carbon at the other end, we obtain the monomer structure: \(\text{HO}-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{COOH}\). The longest carbon chain containing the principal functional group (carboxylic acid) has 4 carbon atoms: C1 is the carboxylic acid carbon, C2 is the methylene carbon, C3 is the carbon with the hydroxyl group, and C4 is the methyl group. Thus, the IUPAC name is 3-hydroxybutanoic acid.

1 mark for correctly identifying 3-hydroxybutanoic acid as the IUPAC name of the monomer.

Lysine contains two basic amine groups (one on the side chain and one alpha-amine) and one acidic carboxylic acid group. When reacted with an excess of a strong acid like HCl, both basic amine groups will be protonated to form ammonium salts (\(-\text{NH}_3^+\text{Cl}^-\)), while the carboxylic acid group remains unreacted as \(-\text{COOH}\). This forms the dicationic salt represented in option C.

1 mark for identifying the correct dicationic salt formed from the protonation of both amine groups by excess acid.

Let's find the stereogenic centers in 3-methylhex-4-en-2-ol, \(\text{CH}_3-\text{CH(OH)}-\text{CH}(\text{CH}_3)-\text{CH}=\text{CH}-\text{CH}_3\). 1) Carbon-2 has four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and the rest of the chain, making it chiral. 2) Carbon-3 has four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH(OH)}\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\), making it chiral. 3) The C=C double bond (Carbon-4 and Carbon-5) exhibits E/Z stereoisomerism as both carbons have different groups attached. Total stereogenic units = 3. Because the molecule is asymmetrical, the total number of stereoisomers is \(2^3 = 8\).

1 mark for identifying two chiral centers and one E/Z double bond, calculating \(2^3 = 8\) stereoisomers.

The stationary phase (silica) is highly polar, and the mobile phase is non-polar. The rate of travel depends on the partition between the phases: more polar amino acids interact more strongly with the silica stationary phase and have lower \(R_f\) values, while less polar/more hydrophobic amino acids dissolve better in the mobile phase and travel further (higher \(R_f\)). Lysine has a polar/charged basic side chain, making it the most polar. Alanine has a small methyl side chain. Valine has a larger, more hydrophobic isopropyl side chain. Thus, polarity order is Lysine > Alanine > Valine, which corresponds to the increasing \(R_f\) order: Lysine, Alanine, Valine.

1 mark for predicting the correct order of Rf values based on the relative polarities of the amino acid side chains.

(a) Mechanism:
1. Dipole shown on carbonyl group: \(C^{\delta+}=O^{\delta-}\).
2. Curly arrow from the lone pair on the carbon of the \(:\text{CN}^-\). ion to the carbonyl carbon.
3. Curly arrow from the \(C=O\) double bond to the oxygen atom.
4. Intermediate formation of an alkoxide: \(C_6H_5C(O^-)(CH_3)CN\).
5. Curly arrow from the negative oxygen lone pair to the proton (\(\text{H}^+\)) to form the final hydroxynitrile.

(b) Reagent: Sodium tetrahydridoborate(III), \(\text{NaBH}_4\).
Conditions: Aqueous or ethanolic solution at room temperature.
Product functional group: Alcohol (secondary alcohol / hydroxyl group).

(c) Hydrolysis equation:
\(C_9H_9NO + 2H_2O + HCl \rightarrow C_9H_{10}O_3 + NH_4Cl\)
(or ionic equivalent: \(C_9H_9NO + 2H_2O + H^+ \rightarrow C_9H_{10}O_3 + NH_4^+\))

(d) Chiral carbon is the central carbon attached to: \(-C_6H_5\), \(-CH_3\), \(-OH\), and \(-COOH\).
3D drawings must show tetrahedral geometry around the chiral carbon with wedge, dash, and two flat lines, representing mirror images.
Explanation of optical activity:
- The carbonyl carbon in phenylethanone is planar.
- The nucleophile (\(\text{CN}^-\)) can attack this planar carbon atom with equal probability from either above or below the plane.
- This produces an equimolar (50:50) mixture of the two optical isomers (a racemic mixture).
- The optical rotation from each enantiomer cancels out, so the mixture is optically inactive.
- Natural sources usually produce a single enantiomer because enzyme active sites are stereoselective.

Part (a): [Total: 4 marks]
- 1 mark for showing \(C^{\delta+}=O^{\delta-}\) dipole and arrow from \(:\text{CN}^-\). lone pair to carbon.
- 1 mark for arrow from double bond to oxygen.
- 1 mark for correct intermediate structure with negative charge on oxygen.
- 1 mark for arrow from oxygen lone pair to \(\text{H}^+\).

Part (b): [Total: 3 marks]
- 1 mark for \(\text{NaBH}_4\) (accept \(\text{LiAlH}_4\) in dry ether).
- 1 mark for warm / aqueous or ethanolic conditions (or dry ether for \(\text{LiAlH}_4\)).
- 1 mark for alcohol.

Part (c): [Total: 3 marks]
- 1 mark for correct formula of reactant \(C_9H_9NO\).
- 1 mark for correct formulas of products: \(C_9H_{10}O_3\) and \(NH_4^+\) / \(NH_4Cl\).
- 1 mark for balanced equation overall.

Part (d): [Total: 4.17 marks]
- 1 mark for correct 3D tetrahedral structures of both mirror-image enantiomers.
- 1 mark for stating the carbonyl group / reactant is planar.
- 1 mark for explaining that attack of \(\text{CN}^-\). is equally likely from either side.
- 1.17 marks for stating a racemic mixture (inactive) is formed, whereas natural pathways use stereospecific enzymes.

(a) Identification and analysis of **X**:
1. **IR Spectrum**: Peak at \(1740\text{ cm}^{-1}\) indicates a \(C=O\) carbonyl group. No broad peak at \(3200-3600\text{ cm}^{-1}\) (no \(O-H\) alcohol) or \(2500-3300\text{ cm}^{-1}\) (no \(O-H\) carboxylic acid). Since the molecular formula is \(C_5H_{10}O_2\), the compound must be an ester.
2. **\(^{1}\text{H}\) NMR Spectrum** analysis:
- Singlet at \(\delta = 2.0\text{ ppm}\) (3H) indicates a \(CH_3\) adjacent to a carbonyl group: \(CH_3-C(=O)-\).
- Doublet at \(\delta = 1.2\text{ ppm}\) (6H) and septet at \(\delta = 5.0\text{ ppm}\) (1H) indicate an isopropyl group, \(-CH(CH_3)_2\).
- The splitting: The doublet arises because the two equivalent \(CH_3\) groups are adjacent to 1 proton (n+1 rule = 2). The septet arises because the single \(-CH-\). proton is adjacent to 6 equivalent protons (n+1 rule = 7).
- The chemical shift of the septet at \(\delta = 5.0\text{ ppm}\) shows this carbon is directly bonded to the ester oxygen: \(-COOCH(CH_3)_2\).
3. **\(^{13}\text{C}\) NMR Spectrum** analysis:
- 4 peaks correspond to 4 distinct carbon environments:
- Ester carbonyl carbon (\(C=O\))
- Protonated carbon of the isopropyl group (\(-CH-\))
- Methyl carbon of the acetyl group (\(CH_3-CO-\))
- Two equivalent methyl carbons of the isopropyl group (\(-(CH_3)_2\))
4. **Structure of X**: Isopropyl ethanoate, \(CH_3COOCH(CH_3)_2\).

(b) Structural isomer with only 3 peaks in \(^{13}\text{C}\) NMR:
- Structure: tert-butyl methanoate, \(HCOOC(CH_3)_3\).
- Explanation:
- Carbon 1: Formyl carbon (\(H-\underline{C}O-\))
- Carbon 2: Quaternary carbon of the t-butyl group (\(-\underline{C}(CH_3)_3\))
- Carbon 3: Three equivalent methyl carbons (\(-\underline{C}H_3\) x 3)
- This gives exactly 3 peaks in the \(^{13}\text{C}\) spectrum.

Part (a): [Total: 10.17 marks]
- 1 mark: Identify ester functional group from IR (\(C=O\) present, \(O-H\) absent).
- 1 mark: Assign 2.0 ppm singlet to \(CH_3-C=O\).
- 1 mark: Assign 1.2 ppm doublet to \(-CH(CH_3)_2\).
- 1 mark: Assign 5.0 ppm septet to \(-O-CH-\).
- 1 mark: Explain doublet splitting (adjacent to 1 H).
- 1 mark: Explain septet splitting (adjacent to 6 H).
- 1 mark: Correct structure of **X** drawn or clearly written.
- 1 mark: Explain \(^{13}\text{C}\) peak count (4 environments) matching the structure.
- 1.17 marks: Clearly linking chemical shift values to environment (especially oxygen-shielding shift at 5.0 ppm).

Part (b): [Total: 4 marks]
- 2 marks: Correct structure of tert-butyl methanoate, \(HCOOC(CH_3)_3\).
- 2 marks: Detailed explanation of the 3 carbon environments in this isomer.

(a)(i) Alcohol: Methanol, \(CH_3OH\).
Hydrolysis equation of the ester group:
\(R-COOCH_3 + H_2O \rightarrow R-COOH + CH_3OH\) (or using full structures).

(a)(ii) In highly acidic conditions (pH 1), the amine groups of the amino acids are fully protonated to form ammonium ions (\(-NH_3^+\)), while the carboxylic acid groups remain un-ionised (\(-COOH\)):
- Aspartic acid product: \(H_3N^+-CH(CH_2COOH)-COOH\)
- Phenylalanine product: \(H_3N^+-CH(CH_2C_6H_5)-COOH\)

(b) Alkaline hydrolysis with \(\text{NaOH(aq)}\) hydrolyses both the amide bond and the ester bond. Under alkaline conditions, carboxylic acid groups are deprotonated to form carboxylate salts (\(-COO^-\)), while amine groups remain uncharged (\(-NH_2\)):
1. Sodium aspartate salt: \(H_2N-CH(CH_2COO^-Na^+) - COO^-Na^+\)
2. Sodium phenylalanine salt: \(H_2N-CH(CH_2C_6H_5) - COO^-Na^+\)
3. Methanol: \(CH_3OH\)

(c) Number of chiral carbons: 2.
- The first is the CH carbon of the aspartic acid residue (bonded to \(-NH_2\), \(-CH_2COOH\), \(-H\), \(-CONH-\)).
- The second is the CH carbon of the phenylalanine residue (bonded to \(-NH-\), \(-CH_2C_6H_5\), \(-H\), \(-COOCH_3\)).

Why a single optical isomer is desirable:
- Enantiomers have different physiological/biological properties (different receptors).
- One enantiomer of a sweetener might taste sweet, while the other might be bitter, tasteless, or potentially toxic.
- Producing a single isomer reduces waste, improves efficiency, and avoids potential side effects.

Part (a)(i): [Total: 2 marks]
- 1 mark for identifying methanol / \(CH_3OH\).
- 1 mark for correct partial/full ester hydrolysis equation.

Part (a)(ii): [Total: 4 marks]
- 2 marks for structure of protonated aspartic acid (both carboxylic acid groups intact, protonated amine).
- 2 marks for structure of protonated phenylalanine (carboxylic acid group intact, protonated amine).

Part (b): [Total: 4.17 marks]
- 1.5 marks for aspartate carboxylate salt structure.
- 1.5 marks for phenylalanine carboxylate salt structure.
- 1.17 marks for methanol structure.

Part (c): [Total: 4 marks]
- 1 mark for stating "2" chiral carbons.
- 1 mark for correctly identifying/circling both chiral centres.
- 2 marks for explanation regarding distinct biological activities of enantiomers (receptors, side-effects/taste).

(a) Nitration Mechanism:
1. **Generation of the electrophile**:
\(HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-\)
(Accept: \(HNO_3 + H_2SO_4 \rightarrow NO_2^+ + H_2O + HSO_4^-\))

2. **Electrophilic Attack**:
- Curly arrow from the pi ring of methyl benzoate to the nitrogen of the \(NO_2^+\) electrophile.
- Structure of the intermediate: A hexadienyl-type carbocation with a positive charge distributed over 5 carbon atoms (the horseshoe representation open towards position 3, which is bonded to both \(-H\) and \(-NO_2\)).

3. **Loss of Proton**:
- Curly arrow from the \(C-H\) bond at position 3 back into the ring system to restore aromaticity.

4. **Regeneration of catalyst**:
- \(H^+ + HSO_4^- \rightarrow H_2SO_4\)

(b) Directing effect of \(-COOCH_3\):
- The \(-COOCH_3\) group is strongly electron-withdrawing due to the highly electronegative oxygen atoms on the carbonyl group.
- It withdraws electron density from the benzene ring inductive and resonance effects.
- This deactivates the ring toward electrophilic attack.
- The deactivating effect is more pronounced at the ortho (2-) and para (4-) positions, making the meta (3-) position relatively more electron-rich (or less deactivated) and thus directing substitution to the 3-position.

(c) Alkaline hydrolysis and purification:
- **Reagent and condition**: Heat the methyl benzoate under reflux with aqueous sodium hydroxide (\(NaOH\)).
- **Product of step 1**: This forms sodium benzoate (soluble) and methanol.
- **Acidification**: Add a strong dilute acid (e.g., dilute hydrochloric acid, \(HCl\)) to the reaction mixture until it is acidic.
- **Reaction**: \(C_6H_5COO^- + H^+ \rightarrow C_6H_5COOH\).
- **Precipitation**: Benzoic acid is only sparingly soluble in cold water and precipitates out as a white solid.
- **Separation**: Filter under reduced pressure (using a Buchner funnel) to obtain the crude solid benzoic acid.

Part (a): [Total: 6 marks]
- 1 mark for equation generating \(NO_2^+\).
- 1 mark for arrow from benzene ring to \(NO_2^+\) (directed to the 3-position).
- 1 mark for correct structure of the carbocation intermediate (with correct charge and horseshoe orientation).
- 1 mark for arrow from \(C-H\) bond to the ring.
- 1 mark for final product structure (methyl 3-nitrobenzoate).
- 1 mark for equation showing catalyst regeneration.

Part (b): [Total: 3.17 marks]
- 1 mark for identifying the ester group as electron-withdrawing.
- 1 mark for stating that it deactivates the benzene ring.
- 1.17 marks for explaining that deactivation is greatest at 2- and 4-positions, leaving the 3-position as the preferred pathway.

Part (c): [Total: 5 marks]
- 1 mark for refluxing with aqueous \(NaOH\).
- 1 mark for identifying that sodium benzoate is formed in solution.
- 1 mark for acidification using dilute acid (e.g., \(HCl\) or \(H_2SO_4\)).
- 1 mark for stating that benzoic acid precipitates.
- 1 mark for filtration to isolate the solid.

(a)(i) PLA is formed from 2-hydroxypropanoic acid, \(CH_3CH(OH)COOH\).
Repeating unit structure:
\(-[O - CH(CH_3) - CO] - \)
(The open bonds must extend from the ether oxygen and the carbonyl carbon).

(a)(ii) Type of reaction: Condensation polymerisation.
Small molecule eliminated: Water, \(H_2O\).

(b) Biodegradability:
- PLA contains ester linkages (\(C-O\) polar bonds).
- These ester linkages can be broken down by chemical hydrolysis or by microorganisms (enzymes) in the environment.
- In contrast, poly(ethene) is an addition polymer consisting of a continuous non-polar, saturated carbon-carbon (\(C-C\)) backbone.
- \(C-C\) bonds are very strong, non-polar, and inert, meaning they are not susceptible to nucleophilic attack, hydrolysis, or biological degradation.

(c)(i) Monomer structures:
- Benzene-1,4-diamine: A benzene ring with two amino groups (\(-NH_2\)) opposite each other (at positions 1 and 4).
- Benzene-1,4-dicarboxylic acid: A benzene ring with two carboxylic acid groups (\(-COOH\)) opposite each other (at positions 1 and 4).

(c)(ii) Repeating unit of Kevlar:
\(-[HN - C_6H_4 - NH - CO - C_6H_4 - CO] - \)
(Showing the amide link, \(-NH-CO-\), with extension bonds on both ends).

(c)(iii) Kevlar's high tensile strength:
- The linear chains of Kevlar can align closely parallel to each other.
- This allows extensive **hydrogen bonding** to form between the partially positive hydrogen of the \(N-H\) group on one chain and the lone pair on the partially negative oxygen of the \(C=O\) group on an adjacent chain.

Part (a)(i): [Total: 2 marks]
- 2 marks for the correct structure of the monomer repeating unit of PLA showing correct linkages.

Part (a)(ii): [Total: 2 marks]
- 1 mark for condensation.
- 1 mark for water / \(H_2O\).

Part (b): [Total: 4 marks]
- 1 mark for identifying the presence of polar ester links (\(C-O\)) in PLA vs non-polar carbon-carbon bonds (\(C-C\)) in poly(ethene).
- 1 mark for stating that ester bonds can be hydrolysed.
- 1 mark for stating that microorganisms/enzymes can break down ester links.
- 1 mark for stating that \(C-C\) bonds are resistant to biological or chemical attack.

Part (c)(i): [Total: 2 marks]
- 1 mark for the correct structural formula of benzene-1,4-diamine.
- 1 mark for the correct structural formula of benzene-1,4-dicarboxylic acid.

Part (c)(ii): [Total: 2.17 marks]
- 2.17 marks for the correct polyamide repeating unit showing a complete amide linkage and correct aromatic connectivity.

Part (c)(iii): [Total: 2 marks]
- 1 mark for identifying hydrogen bonding between chains.
- 1 mark for explaining that linear chains align closely to maximise these intermolecular forces.

(a) Step 2 Reduction:
- **Reagents**: Tin (Sn) and concentrated hydrochloric acid (\(HCl\)).
- **Conditions**: Heated under reflux, followed by addition of aqueous sodium hydroxide (\(NaOH\)) to liberate the free amine.
- **Equation**:
\(C_6H_5(OH)NO_2 + 6[H] \rightarrow C_6H_5(OH)NH_2 + 2H_2O\)
(Accept molecular formulae: \(C_6H_5NO_3 + 6[H] \rightarrow C_6H_7NO + 2H_2O\))

(b) Paracetamol formation:
- Structure of paracetamol: \(HO-C_6H_4-NH-CO-CH_3\) (amide group at position 4 relative to the phenol OH).
- Equation:
\(H_2N-C_6H_4-OH + (CH_3CO)_2O \rightarrow CH_3CONH-C_6H_4-OH + CH_3COOH\)
(The products are paracetamol and ethanoic acid).

(c) Recrystallisation procedure:
1. Dissolve the crude paracetamol in the **minimum volume** of a **hot solvent** (e.g., water or ethanol).
2. Filter the hot solution through fluted filter paper (to remove insoluble impurities).
3. Allow the filtrate to **cool slowly** to room temperature (and then in ice) to allow crystals of pure paracetamol to reform.
4. Filter the crystals under reduced pressure (using a Buchner funnel).
5. Wash the crystals with a small amount of **cold solvent** and leave to dry.

(d) Chemical test to distinguish phenol and cyclohexanol:
- **Reagent**: Bromine water (or aqueous bromine, \(Br_2\)).
- **Observation with phenol**: Bromine water decolourises (orange to colourless) and a **white precipitate** of 2,4,6-tribromophenol forms.
- **Observation with cyclohexanol**: No reaction/no change (the solution remains orange).

Part (a): [Total: 4 marks]
- 1 mark for tin (Sn) and concentrated \(HCl\).
- 1 mark for heating/reflux and subsequent addition of alkali (\(NaOH\)).
- 2 marks for balanced equation (1 mark for correct formulas of reactants/products, 1 mark for \(6[H]\) and \(2H_2O\)).

Part (b): [Total: 4.17 marks]
- 2 marks for drawing the correct skeletal/structural formula of paracetamol.
- 1.17 marks for the correct structural formula of ethanoic anhydride and ethanoic acid.
- 1 mark for a fully balanced equation.

Part (c): [Total: 4 marks]
- 1 mark for dissolving in the minimum volume of hot solvent.
- 1 mark for cooling to recrystallise.
- 1 mark for filtering (and washing with cold solvent).
- 1 mark for drying the crystals.

Part (d): [Total: 2 marks]
- 1 mark for reagent: Bromine water / \(Br_2(aq)\).
- 1 mark for observation: Phenol decolourises bromine and forms a white precipitate, while cyclohexanol shows no change.

(a)
- The reaction is slow because both reactants (S2O8^2- and I^-) are negatively charged ions.
- They repel each other, resulting in a high activation energy (Ea) for the reaction.
- Catalyst equations:
1) S2O8^2-(aq) + 2Fe^2+(aq) -> 2SO4^2-(aq) + 2Fe^3+(aq)
2) 2Fe^3+(aq) + 2I^-(aq) -> 2Fe^2+(aq) + I2(aq)

(b)
- Half-cell setup: Use an aqueous mixture containing both S2O8^2-(aq) and SO4^2-(aq) ions, each at a standard concentration of 1.0 mol dm^-3.
- An inert platinum (Pt) electrode is used to make electrical contact.
- Standard temperature of 298 K.
- Standard cell potential: E_cell = E_reduction - E_oxidation = +2.01 - (+0.54) = +1.47 V.

(c)
- Rate equation: Rate = k [S2O8^2-][I^-]
- Substituting the values: 3.84 x 10^-5 = k (0.040)(0.080)
- 3.84 x 10^-5 = k (0.0032)
- k = (3.84 x 10^-5) / 0.0032 = 0.012
- Units: rate / (conc x conc) = mol dm^-3 s^-1 / (mol^2 dm^-6) = dm^3 mol^-1 s^-1.

(a) [4 marks]
- Mention of repulsion between negatively charged ions / anions (1)
- Leads to high activation energy / few successful collisions (1)
- Equation 1: S2O8^2- + 2Fe^2+ -> 2SO4^2- + 2Fe^3+ (1)
- Equation 2: 2Fe^3+ + 2I^- -> 2Fe^2+ + I2 (1)

(b) [4.67 marks]
- Electrolyte containing both S2O8^2- and SO4^2- both at 1.0 mol dm^-3 (1)
- Platinum electrode (Pt) used (1)
- Standard temperature of 298 K / 25 °C (1)
- E_cell calculation: +1.47 V (1.67)

(c) [3 marks]
- Correct rearrangement: k = Rate / ([S2O8^2-][I^-]) (1)
- Calculation: k = 0.012 (1)
- Units: dm^3 mol^-1 s^-1 (1)

(a)
- Reaction equation: HCOOH(aq) + NaOH(aq) -> HCOONa(aq) + H2O(l) (or ionic: HCOOH + OH^- -> HCOO^- + H2O).
- When H^+ is added, the large reservoir of conjugate base (HCOO^-) reacts with the added H^+ to form weak acid molecules: HCOO^-(aq) + H^+(aq) -> HCOOH(aq).
- Thus, the ratio of [HCOOH]/[HCOO^-] and the concentration of H^+ remain relatively constant, keeping the pH stable.

(b)
- Initial moles of HCOOH = (150.0 / 1000) * 0.200 = 0.0300 mol.
- Initial moles of NaOH added = (100.0 / 1000) * 0.100 = 0.0100 mol.
- Since HCOOH is in excess, all NaOH reacts:
- Moles of HCOO^- formed = 0.0100 mol.
- Moles of HCOOH remaining = 0.0300 - 0.0100 = 0.0200 mol.
- Using the buffer equation: [H^+] = Ka * (n_acid / n_salt)
- [H^+] = 1.80 x 10^-4 * (0.0200 / 0.0100) = 3.60 x 10^-4 mol dm^-3.
- pH = -log10(3.60 x 10^-4) = 3.4437... which rounds to 3.44.

(c)
- The pH remains virtually unchanged.
- The added HCl introduces H^+ ions, which react with the conjugate base HCOO^- to form HCOOH.
- Because the buffer contains large reservoirs of both HCOOH and HCOO^-, the addition of a small amount of strong acid has only a very minor effect on the ratio [acid]/[conjugate base], so the pH changes by a negligible amount.

(a) [3 marks]
- Equation: HCOOH + OH^- -> HCOO^- + H2O (or molecular equivalent) (1)
- Explains that added H^+ reacts with HCOO^- / conjugate base (1)
- Equation: HCOO^- + H^+ -> HCOOH OR states that the [acid]/[base] ratio remains nearly constant (1)

(b) [5.67 marks]
- Moles of HCOOH = 0.0300 mol (1)
- Moles of NaOH = 0.0100 mol (1)
- Moles of remaining HCOOH = 0.0200 mol and moles of HCOO^- = 0.0100 mol (1)
- Calculation of [H^+] = 3.60 x 10^-4 mol dm^-3 (1)
- Correct pH to 2 dp = 3.44 (1.67)

(c) [3 marks]
- States that pH remains virtually unchanged (or decreases by a tiny/insignificant amount) (1)
- Explains that H^+ from HCl reacts with HCOO^- (1)
- Explains that the ratio of [HCOOH]/[HCOO^-] remains almost constant (1)

(a)
- Reagents: concentrated HNO3 and concentrated H2SO4.
- Conditions: Warm/heat at a temperature below 55 °C.
- Electrophile generation:
HNO3 + 2H2SO4 -> NO2^+ + 2HSO4^- + H3O^+
(or HNO3 + H2SO4 -> NO2^+ + HSO4^- + H2O)
- Mechanism:
- Arrow from benzene pi-ring of methylbenzene to the NO2^+ electrophile.
- Structure of the intermediate with a broken ring (opening towards the substituted carbon) and a positive charge in the ring.
- Arrow from the C-H bond of the intermediate to the ring to restore aromaticity, releasing H^+.

(b)
- Reagent: Acidified potassium manganate(VII), KMnO4/H2SO4 (or alkaline KMnO4 followed by acid).
- Conditions: Heat under reflux.

(c)
- Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl).
- Conditions: Heat under reflux.
- Isolation explanation: Under acidic conditions, the amine group gets protonated to form an ammonium salt (-NH3^+ Cl^-). To isolate the free neutral amine, a strong alkali (such as aqueous NaOH) must be added to deprotonate the ammonium salt back to the neutral amine (-NH2).

(a) [5 marks]
- Reagents: Conc HNO3 and Conc H2SO4 (1)
- Equation for electrophile: NO2^+ formation (1)
- Curly arrow from benzene ring to NO2^+ (1)
- Correct structure of carbocation intermediate (1)
- Curly arrow from C-H bond back into the ring to restore aromatic system (1)

(b) [2 marks]
- KMnO4 / potassium manganate(VII) and acid / H2SO4 (1)
- Heat under reflux (1)

(c) [4.67 marks]
- Reagents: Tin / Sn and concentrated HCl (1)
- Conditions: Heat under reflux (1)
- Explains that the product is initially formed as a protonated salt / -NH3^+ (1)
- Explains addition of NaOH / alkali is needed to release/liberate the free neutral amine (1.67)

(a)
- Energy released, q = m * c * dT
- q = 150.0 g * 4.18 J g^-1 K^-1 * 32.0 K = 20064 J = 20.064 kJ.
- Moles of propan-1-ol, n = mass / Mr = 0.900 / 60.0 = 0.0150 mol.
- Enthalpy of combustion = -q / n = -20.064 / 0.0150 = -1337.6 kJ mol^-1.
- To 3 significant figures, this is -1340 kJ mol^-1.

(b)
- Incomplete combustion of propan-1-ol (forming carbon/soot or carbon monoxide instead of carbon dioxide).
- Evaporation of propan-1-ol from the wick before/after burning.
- Non-standard conditions (as the experiment is not carried out under standard states and pressures).

(c)
- Combustion reaction: C3H7OH(l) + 4.5O2(g) -> 3CO2(g) + 4H2O(l)
- Hess's law cycle formula: H_combustion = Sum(n * H_f(products)) - Sum(m * H_f(reactants))
- H_combustion = [3 * (-393.5) + 4 * (-285.8)] - [-303.0]
- H_combustion = [-1180.5 - 1143.2] - [-303.0]
- H_combustion = -2323.7 + 303.0 = -2020.7 kJ mol^-1.

(a) [5 marks]
- Calculation of heat energy change q = 20064 J or 20.064 kJ (1)
- Calculation of moles of propan-1-ol = 0.0150 mol (1)
- Division of q by n (1)
- Negative sign included (1)
- Final value to 3 significant figures: -1340 kJ mol^-1 (1)

(b) [2 marks]
- Any two from:
- Incomplete combustion (1)
- Evaporation of alcohol from wick (1)
- Non-standard conditions (1)

(c) [4.67 marks]
- Correctly balanced equation or cycle written out (1)
- Correct calculation of products sum: 3 * (-393.5) + 4 * (-285.8) = -2323.7 kJ mol^-1 (1)
- Hess's calculation: -2323.7 - (-303.0) (1)
- Final answer: -2020.7 (or -2021) kJ mol^-1 (1.67)

(a)
- The very broad absorption band at 2500-3300 cm^-1 corresponds to the O-H stretch in a carboxylic acid.
- The strong absorption at 1715 cm^-1 corresponds to a C=O stretch.
- Together, these confirm the presence of a carboxylic acid functional group (-COOH).

(b)
- Since the molecular formula is C4H8O2 and it is a carboxylic acid, the compound must be butanoic acid: CH3CH2CH2COOH.
- NMR Assignment:
- Singlet at delta = 11.5 ppm (area 1) corresponds to the -COOH proton (labile proton, highly deshielded).
- Triplet at delta = 2.30 ppm (area 2) corresponds to the -CH2- protons adjacent to the carbonyl group (-CH2-COOH). It is split into a triplet because it is adjacent to the 2 protons of the central -CH2- group (n+1 rule: 2+1=3).
- Multiplet at delta = 1.65 ppm (area 2) corresponds to the central -CH2- protons (-CH2-CH2-CH3). It is split into a multiplet (specifically a sextet) because it is adjacent to 5 protons (2 on one side, 3 on the other).
- Triplet at delta = 0.95 ppm (area 3) corresponds to the terminal -CH3 protons. It is split into a triplet because it is adjacent to the 2 protons of the central -CH2- group.
- This structure is also supported by the 13C NMR which shows 4 distinct carbon environments.

(c)
- The molecular mass of butanoic acid is 88.
- A peak at m/z = 73 corresponds to the loss of a methyl group (M - 15) from the molecular ion.
- Therefore, the fragment ion is [CH2CH2COOH]^+ (or [C3H5O2]^+).

(a) [3 marks]
- Identifies functional group as carboxylic acid (1)
- Links 2500-3300 cm^-1 to O-H (acid) stretch (1)
- Links 1715 cm^-1 to C=O stretch (1)

(b) [6.67 marks]
- Identifies structure as butanoic acid / CH3CH2CH2COOH (1)
- Assigns delta 11.5 singlet to -COOH proton (1)
- Assigns delta 2.30 triplet to -CH2- adjacent to C=O AND explains splitting (1)
- Assigns delta 1.65 multiplet to central -CH2- AND explains splitting (1)
- Assigns delta 0.95 triplet to -CH3 AND explains splitting (1)
- Relates integration values (3:2:2:1) correctly to the number of protons (1.67)

(c) [2 marks]
- Identifies loss of CH3 group (15) from molecular ion (1)
- Writes correct formula for fragment: [CH2CH2COOH]^+ (positive charge must be included) (1)

(a)
- Kc expression: Kc = ([CH3COOC2H5][H2O]) / ([CH3COOH][C2H5OH])
- Units explanation: There are equal numbers of concentration terms in the numerator and denominator (2 on top, 2 on bottom), so the concentration units (mol dm^-3) cancel out completely.

(b)
- Set up an ICE table (Initial, Change, Equilibrium):
- Ethanoic Acid (CH3COOH):
- Initial: 1.00 mol
- Change: -0.800 mol
- Equilibrium: 0.200 mol
- Ethanol (C2H5OH):
- Initial: 1.50 mol
- Change: -0.800 mol
- Equilibrium: 0.700 mol
- Ethyl Ethanoate (CH3COOC2H5):
- Initial: 0 mol
- Change: +0.800 mol
- Equilibrium: 0.800 mol
- Water (H2O):
- Initial: 0 mol
- Change: +0.800 mol
- Equilibrium: 0.800 mol
- Volume, V, cancels out because total moles of reactants = total moles of products.
- Kc = (0.800 * 0.800) / (0.200 * 0.700) = 0.640 / 0.140 = 4.5714...
- To 3 significant figures, Kc = 4.57.

(c)
- Since the forward reaction is exothermic, increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the added heat.
- Therefore, the equilibrium yield of ethyl ethanoate will decrease.
- Since the equilibrium shifts to the left, the ratio of [products]/[reactants] decreases, so the value of Kc decreases.

(a) [2.67 marks]
- Correct Kc expression (1)
- Explanation that units cancel out because there are the same number of moles / concentration terms on both sides of the equation (1.67)

(b) [5 marks]
- Identifies that moles of water formed is also 0.800 mol (1)
- Correct calculation of equilibrium moles of CH3COOH = 0.200 mol (1)
- Correct calculation of equilibrium moles of C2H5OH = 0.700 mol (1)
- Correct substitution of values into Kc expression (1)
- Final answer to 3 significant figures: 4.57 (1)

(c) [4 marks]
- Increasing temperature shifts equilibrium to the left / in endothermic direction (1)
- Yield of ethyl ethanoate decreases (1)
- Kc value decreases (1)
- Explanation connecting shift of equilibrium to the change in the ratio of products/reactants (1)