Unit 4: Organic 2 & Physical 2 Exam Practice

The rate of reaction between peroxodisulfate(VI) ions, S₂O₈^2-, and iodide ions, I^-, was investigated at 298 K:
S₂O₈^2-(aq) + 2I^-(aq) → 2SO₄^2-(aq) + I₂(aq)

Initial rates data were obtained from a series of experiments, as shown in Table 1.

Experiment	Initial [S₂O₈^2-] / mol dm^-3	Initial [I^-] / mol dm^-3	Initial Rate / mol dm^-3 s^-1
1	0.010	0.015	1.20 × 10^-4
2	0.020	0.015	2.40 × 10^-4
3	0.010	0.045	3.60 × 10^-4

(a) State the meaning of the term order of reaction with respect to a given reactant. [1]

(b) Using the data in Table 1, deduce the order of reaction with respect to S₂O₈^2- and I^-. Show your reasoning clearly. [4]

(c) Write the overall rate equation for this reaction. [1]

(d) Calculate the rate constant, k, for this reaction at 298 K, using Experiment 1 data. State its units. [3]

(e) The rate constant, k, was measured at several different temperatures. A plot of ln k against 1/T yielded a straight line with a gradient of -6250 K. The Arrhenius equation is: k = A e^-Ea/RT. Calculate the activation energy, E_a, for this reaction in kJ mol^-1. (Gas constant R = 8.31 J K^-1 mol^-1). [3]

(f) Explain why the rate of reaction increases significantly when the temperature is raised. Refer to collision theory and the distribution of molecular energies. [2]

Show Mark Scheme

(a)

The power to which the concentration of a reactant is raised in the rate equation. [1] (Accept: "The power of concentration term in rate equation").

(b)

Order wrt S₂O₈^2- = 1 [1]
Comparing Exp 1 and Exp 2: [S₂O₈^2-] doubles while [I^-] remains constant; initial rate doubles (1.20 × 10^-4 to 2.40 × 10^-4), hence first order. [1]
Order wrt I^- = 1 [1]
Comparing Exp 1 and Exp 3: [I^-] triples while [S₂O₈^2-] remains constant; initial rate triples (1.20 × 10^-4 to 3.60 × 10^-4), hence first order. [1]

(c)

Rate = k [S₂O₈^2-][I^-] [1] (Allow error carried forward from part b).

(d)

k = Rate / ([S₂O₈^2-][I^-]) = (1.20 × 10^-4) / (0.010 × 0.015) [1]
k = 0.80 (or 0.8) [1]
Units = dm³ mol^-1 s^-1 [1] (Allow ECF for units matching candidate rate equation).

(e)

ln k = (-E_a/R)(1/T) + ln A ⇒ gradient = -E_a / R [1]
-6250 = -E_a / 8.31 ⇒ E_a = 6250 × 8.31 = 51,937.5 J mol^-1 [1]
E_a = +51.9 kJ mol^-1 (allow +52 or range 51.9 to 52.0) [1] (Must divide by 1000 for kJ mol^-1; + sign preferred but do not penalise if omitted).

(f)

Many more molecules possess kinetic energy greater than or equal to the activation energy (E ≥ E_a). [1]
A much greater proportion / fraction of collisions are successful per second (collision frequency increases). [1]

Examiner tip: Always convert Joules to kilojoules in Arrhenius calculations by dividing by 1000. For question (f), never state merely that "molecules move faster"; the crucial requirement is that a far greater fraction of particles exceed the activation energy barrier.

Methanol is manufactured industrially by the reversible gas-phase reaction of carbon monoxide with hydrogen:
CO(g) + 2H₂(g) ↔ CH₃OH(g)

A mixture of 1.50 mol of carbon monoxide and 3.00 mol of hydrogen was allowed to reach equilibrium in a sealed reactor. At equilibrium, the mixture contained 0.930 mol of methanol. The total pressure of the equilibrium system was 190 kPa.

(a) Calculate the amounts, in moles, of carbon monoxide and hydrogen present in the equilibrium mixture. [2]

(b) Calculate the mole fraction of hydrogen, x(H₂), and the partial pressure of hydrogen, p(H₂), in kPa, in the equilibrium mixture. [2]

(c) Write the expression for the equilibrium constant, K_p, for this reaction. [1]

(d) Calculate the value of the equilibrium constant, K_p, for this reaction under these conditions. State its units. [5]

(e) In a separate run, the equilibrium constant K_p was measured at two different temperatures. At temperature T₁, K_p = 2.45 × 10^-4 kPa^-2, and at temperature T₂ (where T₂ > T₁), K_p = 1.10 × 10^-5 kPa^-2. Explain what this shows about the enthalpy change (ΔH) of the forward reaction. [3]

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(a)

Moles of CO = 1.50 - 0.930 = 0.570 mol [1]
Moles of H₂ = 3.00 - 2(0.930) = 3.00 - 1.860 = 1.14 mol [1] (Must account for 1:2 stoichiometry).

(b)

Total moles = 0.570 + 1.14 + 0.930 = 2.64 mol. Mole fraction x(H₂) = 1.14 / 2.64 = 0.432 (allow 0.43 or 0.4318) [1]
Partial pressure p(H₂) = 0.4318 × 190 = 82.0 kPa (allow 82.0 to 82.1 kPa) [1] (Allow ECF from part a).

(c)

K_p = p(CH₃OH) / [p(CO) × p(H₂)²] [1] (Do NOT accept square brackets; must use partial pressure p(X) notation).

(d)

p(CO) = (0.570 / 2.64) × 190 = 41.0 kPa; p(CH₃OH) = (0.930 / 2.64) × 190 = 66.9 to 67.0 kPa [1]
Substitution into expression: K_p = 67.0 / [41.0 × (82.0)²] [1]
Denominator = 41.0 × 6724 = 275,684 [1]
Numerical value: K_p = 2.43 × 10^-4 (allow 2.4 × 10^-4) [1]
Units: kPa^-2 [1]

(e)

As temperature increases, the numerical value of K_p decreases. [1]
By Le Chatelier's principle, an increase in temperature favours the endothermic direction, shifting equilibrium to the left (reactants). [1]
Therefore, the forward reaction must be exothermic (ΔH is negative). [1]

Examiner tip: Never use square concentration brackets [ ] in a Kp expression; examiners award 0 marks if square brackets are used. Make sure to deduct 2 × moles of product formed when calculating equilibrium moles of H2 due to the 2:1 stoichiometric ratio.

Butanone, CH₃COCH₂CH₃, reacts with sodium tetrahydridoborate(III), NaBH₄, in aqueous solution to produce butan-2-ol.

(a) Name this type of organic reaction and specify the nucleophile involved. [2]

(b) Draw the complete reaction mechanism for the reaction of butanone with hydride ions (:H^-). Include all necessary curly arrows, lone pairs, and relevant dipoles. Show the structure of the intermediate and the final product. [4]

(c) (i) Define the term optical isomerism and explain what is meant by a chiral carbon. [2]

(c) (ii) Explain why the product butan-2-ol formed in this reaction is optically inactive (forms a racemic mixture). [3]

(d) State how Infrared (IR) Spectroscopy could be used to confirm that the reaction of butanone to form butan-2-ol was complete. Refer to characteristic wavenumbers from standard data. [3]

Show Mark Scheme

(a)

Type of reaction: Nucleophilic addition (also accept reduction). [1]
Nucleophile: Hydride ion / :H^- (reject "H" or "hydrogen"). [1]

(b)

M1: Curly arrow from the lone pair on :H^- to the carbonyl carbon (C^δ+). [1]
M2: Curly arrow from the C=O double bond to the oxygen atom (O^δ-). [1]
M3: Correct structure of the tetrahedral intermediate alkoxide ion CH₃CH₂C(H)(O^-)CH₃ showing a lone pair and negative charge on oxygen. [1]
M4: Curly arrow from a lone pair on the intermediate oxygen to H⁺ (or H on H₂O) to form butan-2-ol. [1]

(c) (i)

Optical isomerism: Molecules that exist as non-superimposable mirror images of each other. [1]
Chiral carbon: A carbon atom bonded to four different atoms or groups of atoms. [1] (Reject "four different bonds").

(c) (ii)

The carbonyl group (C=O carbon) is planar. [1]
The hydride nucleophile (:H^-) has an equal probability of attacking the planar carbonyl from above or below the plane. [1]
An equimolar (50:50) mixture of both enantiomers is produced; their equal and opposite rotations of plane-polarised light cancel out completely. [1]

(d)

The carbonyl (C=O) peak at 1680 - 1750 cm^-1 will completely disappear. [1]
A broad alcohol (O-H) peak will appear at 3230 - 3550 cm^-1. [1]
Comparison of the fingerprint region (below 1500 cm^-1) against a reference spectrum of pure butan-2-ol confirms purity. [1]

Examiner tip: To score full marks for optical inactivity, you must state that the carbonyl group is planar (not the whole molecule) and that attack occurs with equal probability from above or below, forming an equimolar mixture whose rotations cancel.

Methylbenzene, C₆H₅CH₃, reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid to produce 4-nitromethylbenzene.

(a) Concentrated sulfuric acid acts as a catalyst in this reaction. Write equations to show how the electrophile, NO₂⁺, is generated from concentrated nitric and sulfuric acids. [2]

(b) Draw the complete mechanism for the reaction of methylbenzene with the NO₂⁺ electrophile to form 4-nitromethylbenzene. Include curly arrows showing electron movement and draw the structure of the intermediate carbocation. [4]

(c) State the name of this type of reaction mechanism. [1]

(d) Explain why methylbenzene reacts faster than benzene in this substitution reaction. Refer to the inductive effect of the methyl group. [3]

(e) 4-Nitromethylbenzene can be reduced to 4-methylphenylamine. Identify the reagent(s) needed for this reduction and write an equation for the reaction, using [H] to represent the reducing agent. [3]

Show Mark Scheme

(a)

HNO₃ + H₂SO₄ ↔ H₂NO₃⁺ + HSO₄^- [1]
H₂NO₃⁺ ↔ NO₂⁺ + H₂O [1]
(Or accept overall single equation: HNO₃ + 2H₂SO₄ ↔ NO₂⁺ + H₃O⁺ + 2HSO₄^- for 2 marks).

(b)

M1: Curly arrow from the delocalised pi ring of methylbenzene to the nitrogen atom of NO₂⁺. [1]
M2: Correct structure of the arenium intermediate carbocation showing a horseshoe open at carbon 4 with the positive charge located inside the cavity. [1]
M3: Curly arrow from the C-H bond at carbon 4 back into the ring to restore the aromatic pi sextet. [1]
M4: Correct structure of 4-nitromethylbenzene and regenerated H⁺ (or HSO₄^- + H⁺ → H₂SO₄). [1]

(c)

Electrophilic substitution. [1] (Reject "nucleophilic" or "addition").

(d)

The methyl group is an electron-donating group. [1]
It releases electron density into the benzene pi electron cloud through a positive inductive effect (+I). [1]
The higher electron density makes the ring more attractive / more susceptible to attack by the electrophile (NO₂⁺). [1]

(e)

Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl), followed by NaOH. [1] (Reject LiAlH₄).
Equation: C₇H₇NO₂ + 6[H] → C₇H₇NH₂ + 2H₂O (or CH₃C₆H₄NO₂ + 6[H] → CH₃C₆H₄NH₂ + 2H₂O) [2] (1 mark for correct organic formulae, 1 mark for balanced 6[H] and 2H₂O).

Examiner tip: The intermediate arenium horseshoe must extend over 5 carbons (from C2 to C6, open at C4) and must not be completely closed. For reducing aromatic nitro compounds, always specify Sn and concentrated HCl; LiAlH4 is not accepted by examiners for this reaction.

Polymers and nitrogen-containing compounds form a key part of organic material systems.

(a) (i) A polyamide can be prepared by reacting butanedioic acid with hexane-1,6-diamine. Draw the repeating unit of the polyamide formed in this reaction. Show all atoms in the amide linkage clearly. [2]

(a) (ii) Identify the other small molecule produced during this polymerisation reaction. [1]

(b) In terms of the intermolecular forces between the polymer chains, explain why polyamides (such as nylon) can be spun into strong fibres suitable for weaving, whereas polyalkenes produce weak fibres. [3]

(c) Polyamides are biodegradable, whereas polyalkenes are not. Explain why polyamides can be broken down in the environment, referencing the chemical bonds in the polymer chains. [2]

(d) (i) Draw the structure of the zwitterion form of alanine, CH₃CH(NH₂)COOH. [2]

(d) (ii) Draw the structure of the species formed when this zwitterion reacts with excess concentrated hydrochloric acid. [2]

(d) (iii) Draw the structure of the species formed when this zwitterion reacts with excess sodium hydroxide solution. [2]

Show Mark Scheme

(a) (i)

Repeating unit: [-CO-CH₂-CH₂-CO-NH-(CH₂)₆-NH-] [2]
1 mark for correct carbon backbone from butanedioic acid and hexane-1,6-diamine; 1 mark for correct amide linkage (-CO-NH-) with open bonds at both ends.

(a) (ii)

Water / H₂O. [1]

(b)

Polyamides contain hydrogen bonds between adjacent polymer chains. [1]
Hydrogen bonds form between the partially positive H on N-H and the lone pair on the carbonyl oxygen (C=O) of adjacent chains. [1]
Polyalkenes only have much weaker Van der Waals / temporary dipole-dipole forces between chains, which require less energy to overcome. [1]

(c)

Polyamides contain polar carbon-nitrogen (C-N) amide bonds. [1]
These polar bonds are susceptible to nucleophilic attack and can undergo hydrolysis by water, acid, or alkalis. [1] (Polyalkenes have non-polar, inert C-C backbones).

(d) (i)

Zwitterion: CH₃CH(NH₃⁺)COO^- [2] (Must show positive charge on N and negative charge on O).

(d) (ii)

In acid: CH₃CH(NH₃⁺)COOH [2] (The carboxylate group accepts a proton; the amine remains protonated).

(d) (iii)

In base: CH₃CH(NH₂)COO^- [2] (The ammonium group loses a proton to form -NH2; the carboxylate remains -COO-).

Examiner tip: Ensure polymer repeating units have open extension bonds extending beyond the brackets. When drawing amino acid ions, remember that in acidic conditions only the carboxylate is protonated, and in alkaline conditions only the ammonium group is deprotonated.

A student synthesised a sample of the ester methyl 2-methylpropanoate, (CH₃)₂CH-COOCH₃, and purified the product.

(a) Suggest a suitable two-step synthetic pathway to prepare methyl 2-methylpropanoate starting from 2-methylpropan-1-ol. Name the reagents and conditions for each step. [4]

(b) (i) Explain the purpose of each of the following steps in a recrystallisation purification procedure:
• Filtering the hot solution through fluted filter paper. [1]
• Washing the collected crystals with a small volume of ice-cold solvent. [1]

(b) (ii) How would the student experimentally determine if the purified solid product was pure? [2]

(c) Deduce the relative integration values in the ¹H NMR spectrum of methyl 2-methylpropanoate, (CH₃)₂CH-COOCH₃. Complete Table 2. [4]

Proton Environment	Chemical Shift / ppm	Splitting Pattern	Integration Value
-CH(CH₃)₂ (methyl protons)	1.2	Doublet	[ Deduce ]
-CH- (methine proton)	2.6	Septet	[ Deduce ]
-OCH₃ (ester methyl protons)	3.7	Singlet	[ Deduce ]

Show Mark Scheme

(a)

Step 1: Oxidation
Reagent: Potassium dichromate(VI) (K₂Cr₂O₇) and dilute sulfuric acid (H₂SO₄). [1]
Conditions: Heat under reflux (to ensure complete oxidation to 2-methylpropanoic acid). [1]
Step 2: Esterification
Reagents: Methanol (CH₃OH) and concentrated sulfuric acid catalyst (conc. H₂SO₄). [1]
Conditions: Heat / warm under reflux. [1]

(b) (i)

Hot filtration: To remove insoluble impurities while keeping the desired compound dissolved in the hot solvent. [1]
Ice-cold wash: To remove soluble surface impurities without dissolving the pure crystals. [1]

(b) (ii)

Measure the melting point using a capillary tube in a melting point apparatus. [1]
A pure sample has a sharp melting point matching the literature / data value, whereas an impure sample melts over a broad range at a lower temperature. [1]

(c)

Peak 1 (-CH(CH₃)₂): Integration = 6 [1]
Peak 2 (-CH-): Integration = 1 [1]
Peak 3 (-OCH₃): Integration = 3 [1]
Correct ratio 6 : 1 : 3 correctly linked to the three proton environments. [1]

Examiner tip: In 1H NMR, the integration value corresponds strictly to the number of equivalent protons in that chemical environment (6 protons in the two equivalent methyl groups, 1 in the methine CH, and 3 in the methoxy ester methyl).

Section B: Structured Written Questions

Question 1: Kinetics, Rate Equations & Arrhenius

Question 2: Homogeneous Gas Equilibria & Kp

Question 3: Carbonyl Chemistry, Reaction Mechanisms & Spectroscopy

Question 4: Aromatic Chemistry & Electrophilic Substitution

Question 5: Amines, Condensation Polymers & Amino Acids

Question 6: Organic Synthesis & NMR Spectroscopy