AS Chemistry: Organic Synthesis Routes — 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 Synthesis Routes revision notes

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Questions

1. Give the reagents and conditions for each conversion:

(a) alkene → halogenoalkane [1] (b) halogenoalkane → alcohol [2] (c) halogenoalkane → alkene [2] (d) alcohol → aldehyde [2] (e) alcohol → carboxylic acid [2]

2. Devise a two-step synthesis of propan-2-ol from propene, giving reagents and conditions for each step. [4]

3. Devise a synthesis of propanoic acid from bromoethane.

(a) State how many carbons each contains and what this implies. [2] (b) Give the two steps with reagents and conditions. [4]

4. Explain why aqueous NaOH and ethanolic NaOH give different products from the same halogenoalkane. [3]

5. A student proposes making butan-2-ol from but-1-ene by adding steam. Explain why the major product would not be butan-2-ol’s isomer butan-1-ol, and name the rule involved. [3]

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Answers

1. (a) HBr (or HCl) at room temperature [1]. (b) NaOH(aq) [1], heat under reflux [1]. (c) Ethanolic NaOH [1], heat under reflux [1]. (d) K₂Cr₂O₇/H₂SO₄ [1], warm and distil off the product [1]. (e) K₂Cr₂O₇/H₂SO₄ [1], heat under reflux [1].

2. Step 1: HBr, room temperature [1] → 2-bromopropane [1] (Markovnikov addition: H adds to the carbon that already has more hydrogens, giving the more stable secondary carbocation intermediate, so Br ends up on the middle carbon). Step 2: NaOH(aq), heat under reflux [1] → propan-2-ol [1]. (Direct hydration with steam and H₃PO₄ also gives propan-2-ol by the same Markovnikov rule, and is acceptable as a one-step route, but the question asks for two steps.)

3. (a) Bromoethane has 2 carbons, propanoic acid has 3 [1]. A carbon must be added, so a nitrile step is required [1].

(b) Step 1: KCN in ethanol, heat under reflux [1] → propanenitrile [1]. Step 2: dilute HCl(aq), heat under reflux [1] → propanoic acid [1].

4. In aqueous solution the hydroxide ion acts as a nucleophile, attacking the δ+ carbon and giving substitution to an alcohol [1] [1]. In ethanolic solution it acts as a base, removing a hydrogen from the adjacent carbon and giving elimination to an alkene [1].

5. Addition proceeds via a carbocation intermediate [1]. The secondary carbocation is more stable than the primary, because the electron-donating alkyl groups spread the positive charge [1], so the major product is butan-2-ol — this is Markovnikov’s rule [1].

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Additional questions

6. Devise a synthesis of ethanoic acid from bromomethane, giving reagents and conditions for each step, and explain why a direct one-step conversion is not possible. [6]

7. Propanal is reduced to propan-1-ol. Name a suitable reducing agent and give the product, then explain one difference between NaBH4 and LiAlH4 as reducing agents. [3]

8. Ethanol is converted to ethyl ethanoate. Give the second reagent needed and the conditions, and name the type of reaction occurring. [3]

Answers to additional questions

6. Bromomethane has 1 carbon, ethanoic acid has 2, so a carbon must be added via a nitrile step [1]: Step 1, KCN in ethanol, heat under reflux, giving ethanenitrile [1] [1]; Step 2, dilute HCl(aq), heat under reflux, giving ethanoic acid [1] [1]. A one-step conversion is not possible because no single AS reaction both changes the halogenoalkane’s functional group and adds a carbon at the same time [1].

7. NaBH4 (sodium borohydride) in aqueous or alcoholic solution, or LiAlH4 in dry ether, either is acceptable [1], giving propan-1-ol, CH₃CH₂CH₂OH [1]. The real distinction between them: LiAlH4 also reduces carboxylic acids to primary alcohols, but NaBH4 does not — NaBH4 is too weak a reducing agent to reduce the C=O of a carboxylic acid, so LiAlH4 is needed whenever the substrate is a carboxylic acid rather than an aldehyde or ketone [1].

8. Ethanoic acid, with a small amount of concentrated H2SO4 as catalyst, heated [1] [1]; this is esterification (a condensation reaction), producing ethyl ethanoate and water [1].

A worked approach to planning any route

Facing an unfamiliar synthesis question, first identify the functional group in the starting material and in the target, then compare carbon counts between the two – a different carbon count means a nitrile step is compulsory somewhere in the route, since it is the only AS reaction that changes chain length. From there, work backwards from the target: ask what reaction produces that specific functional group, then chain the necessary steps together, giving the reagent and conditions for each one explicitly rather than naming only the reagent. A final check worth making before answering is whether any intermediate step would destroy a functional group needed later in the route – a route that is chemically sound step-by-step can still fail if an earlier step inadvertently removes something a later step depends on.

Why “aqueous vs ethanolic” and “distil vs reflux” carry so many marks

A large share of the marks across this whole practice set come down to three recurring distinctions: aqueous versus ethanolic conditions for a halogenoalkane (giving substitution to an alcohol or elimination to an alkene respectively), distillation versus reflux when oxidising a primary alcohol (giving an aldehyde or a carboxylic acid respectively), and correctly identifying when a carbon-chain-lengthening nitrile step is required. Examiners consistently withhold marks for an answer that names the correct reagent but omits or gets wrong the specific condition that determines which product actually forms, so revision time spent drilling these three distinctions specifically pays off across almost every question in this set, not just one or two of them.

Where marks are usually lost

• Writing “oxidise” without giving both the reagent and the conditions.
• Confusing reflux with distillation.
• Using aqueous rather than ethanolic conditions for a nitrile or elimination.
• Routes that change the number of carbons with no reaction accounting for it.