A Level Chemistry: Organic Mechanisms and Arenes — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Organic Mechanisms revision notes

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Section A

1. State what a curly arrow represents, and where it must start. [2]

2. Define an electrophile and a nucleophile. [2]

3. Give three pieces of evidence that benzene does not contain three ordinary C=C double bonds. [3]

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Section B

4. 2-bromo-2-methylpropane and 1-bromobutane are both hydrolysed by aqueous sodium hydroxide.

(a) Name the mechanism each follows and explain the difference. [4]

(b) Explain why the tertiary halogenoalkane reacts by this mechanism. [3]

(c) Predict and explain which of 1-bromobutane, 1-chlorobutane and 1-iodobutane hydrolyses fastest. [3]

5. Benzene reacts with a mixture of concentrated nitric and sulfuric acids at 55 °C.

(a) Name the mechanism and the electrophile. [2]

(b) Write an equation showing how the electrophile is generated. [1]

(c) Describe the final step of the mechanism and explain why it occurs. [3]

(d) Explain why alkenes undergo addition but benzene undergoes substitution. [3]

(e) State what would happen if the temperature rose to 80 °C. [1]

6. Propene reacts with hydrogen bromide.

(a) Name the mechanism, and state which carbon the bromine ends up bonded to (the major product). [2] (b) Explain this outcome in terms of carbocation stability, not by stating Markovnikov’s rule as a fact about hydrogen. [3]

7. Methane reacts with chlorine in the presence of ultraviolet light.

(a) Name the mechanism. [1] (b) Write the initiation step, and explain why UV light is needed for it specifically. [2]

8. Benzene reacts with CH₃COCl in the presence of an AlCl₃ catalyst.

(a) Name this reaction and state the electrophile generated. [2] (b) Explain why AlCl₃ is described as a catalyst in this reaction, given that it appears to be consumed generating the electrophile. [2]

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Answers

1. The movement of a pair of electrons [1]; it must start from a bond or a lone pair — never from an atom or a positive charge [1].

2. Electrophile — an electron-pair acceptor, attracted to a region of high electron density [1]. Nucleophile — an electron-pair donor, attracted to a δ+ centre [1].

3. All C–C bond lengths are equal at 139 pm, between a single and a double bond [1]; the enthalpy of hydrogenation is about 152 kJ mol⁻¹ less exothermic than three times cyclohexene [1]; benzene does not decolourise bromine water without a catalyst [1].

4. (a) 2-bromo-2-methylpropane: Sₙ1 [1] — two steps, via a carbocation intermediate, rate depends only on the halogenoalkane [1]. 1-bromobutane: Sₙ2 [1] — one concerted step through a five-coordinate transition state, rate depends on both reactants [1].

(b) The tertiary carbocation is stabilised by three electron-donating alkyl groups (positive inductive effect) [1], so it forms readily [1]. The tertiary carbon is also sterically hindered, blocking backside attack by the nucleophile [1].

(c) 1-iodobutane [1], because the C–I bond has the lowest bond enthalpy [1] and so breaks most readily [1]. Note this follows bond enthalpy, not electronegativity — C–F is the most polar bond and the least reactive.

5. (a) Electrophilic substitution [1]; electrophile is NO₂⁺ (the nitronium ion) [1].

(b) HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ [1].

(c) The unstable intermediate loses H⁺ [1], which is taken up by HSO₄⁻ to regenerate H₂SO₄ [1]. This restores the delocalised π system, which is energetically favourable [1]. Omitting this final step is the most common error in arene mechanisms.

(d) Benzene’s delocalised π system confers extra stability [1], which addition would destroy [1]. Substitution preserves the delocalised ring, whereas an alkene’s localised π bond is expendable and addition gains two σ bonds [1].

(e) Dinitration would occur, giving dinitrobenzene [1].

6. (a) Electrophilic addition [1]; the major product is 2-bromopropane, with bromine bonded to the middle (secondary) carbon [1]. (b) The electrophile (H⁺, generated as HBr polarises) can add to either carbon of the double bond, giving either a secondary or a primary carbocation intermediate [1]. The secondary carbocation is more stable, since it is stabilised by two electron-donating alkyl groups compared with the primary carbocation’s one [1], so it forms preferentially and leads to the major product [1].

7. (a) Free radical substitution [1]. (b) Cl₂ → 2Cl• [1]. UV light provides enough energy to break the weak Cl–Cl bond homolytically, generating two chlorine radicals; this bond is weak enough to be broken by UV photons specifically, unlike the stronger C–H and C–Cl bonds involved in later steps [1].

8. (a) Friedel-Crafts acylation [1]; electrophile is CH₃CO⁺ (the acylium ion) [1]. (b) AlCl₃ generates the electrophile by accepting a lone pair from the chlorine of CH₃COCl [1], but it is regenerated at the end of the mechanism when the AlCl₄⁻ formed loses Cl⁻ back to reform AlCl₃ as H⁺ is lost from the ring — so it is not consumed overall, which is what makes it a catalyst rather than a reagent [1].

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Where marks are usually lost

• Curly arrows starting at an atom or a charge.
• Explaining C–X reactivity by electronegativity instead of bond enthalpy.
• Omitting the loss of H⁺ that regenerates aromaticity.
• Drawing benzene with alternating double bonds in a mechanism.
• Saying Sₙ1 is faster “because it is tertiary” without mentioning carbocation stability.
• Stating Markovnikov’s rule as if it were a fact about hydrogen rather than explaining it through the relative stability of the possible carbocation intermediates.
• Forgetting that a Friedel-Crafts catalyst such as AlCl₃ is regenerated at the end of the mechanism, and so is not consumed overall despite appearing to react when the electrophile is generated.