Question 3 of 5: Ester & Disulfide Isomers; Acid-Catalysed Transesterification

Question 3: Ester & Disulfide Isomers; Acid-Catalysed Transesterification (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a(i) Two isomeric esters of C₅H₁₀O₂. The formula's single degree of unsaturation is the ester carbonyl; the acyl-chain/alkoxy-chain split can be varied freely as long as the two sides sum to five carbons:

Methyl butanoate (4-carbon acyl + 1-carbon alkoxy) and ethyl propanoate (3-carbon acyl + 2-carbon alkoxy) are both C₅H₁₀O₂ — constitutional isomers of each other and, in fact, of pentanoic acid from Question 1a(v).

a(ii) Two isomeric disulfides of C₄H₁₀S₂. A disulfide R–S–S–R′ carries no unsaturation at all (S–S and both C–S bonds are single bonds), so all four carbons are split between the two alkyl groups. Splitting them symmetrically (2+2) or asymmetrically (1+3) gives two genuinely different isomers:

1. Diethyl disulfide (symmetric, 2+2 split): $$\mathrm{CH_3CH_2\text{-}S\text{-}S\text{-}CH_2CH_3}$$
2. Methyl propyl disulfide (asymmetric, 1+3 split): $$\mathrm{CH_3\text{-}S\text{-}S\text{-}CH_2CH_2CH_3}$$

Both are C₄H₁₀S₂; a disulfide's formula alone does not pin down whether the two alkyl groups are the same, so both a symmetric and an asymmetric answer are valid "isomeric disulfides."

b) Acid-catalysed transesterification mechanism. HCl provides a catalytic proton; the reaction is an addition–elimination at the carbonyl carbon and is reversible (the drawn double-headed arrow), so the product distribution is governed by which alcohol is in excess, not by thermodynamics of the mechanism itself.

1. Step 1 — protonate the carbonyl oxygen. H⁺ adds to the carbonyl O, making the carbonyl carbon substantially more electrophilic (the positive charge is delocalised onto that carbon by resonance): $$\mathrm{CH_3CH_2C(=O)OCH_3 + H^+ \longrightarrow CH_3CH_2C(OH^+)OCH_3\ (\text{resonance-stabilised})}$$
2. Step 2 — nucleophilic addition of 1-propanol. A lone pair on 1-propanol's oxygen attacks the activated carbonyl carbon, forming a tetrahedral intermediate now bearing three oxygen substituents (the original carbonyl O, now –OH; the original methoxy O; and the new propoxy O, still protonated): $$\mathrm{CH_3CH_2C(OH^+)OCH_3 + CH_3CH_2CH_2OH \longrightarrow \text{tetrahedral intermediate}}$$
3. Step 3 — proton transfer, then expel methanol. An intramolecular/ intermolecular proton transfer moves the positive charge from the new propoxy oxygen onto the original methoxy oxygen, turning –OCH₃ into the neutral leaving group CH₃OH; collapse of the tetrahedral intermediate re-forms the C=O and expels methanol: $$\mathrm{\text{tetrahedral intermediate} \longrightarrow CH_3CH_2C(=O^+H)OCH_2CH_2CH_3 + CH_3OH}$$
4. Step 4 — deprotonate to regenerate the catalyst. Loss of the final proton from the protonated new ester restores the neutral carbonyl and regenerates the H⁺ catalyst (net: HCl is not consumed): $$\mathrm{CH_3CH_2C(=O^+H)OCH_2CH_2CH_3 \longrightarrow CH_3CH_2C(=O)OCH_2CH_2CH_3 + H^+}$$

$$\mathrm{\boxed{CH_3CH_2C(=O)OCH_3 + CH_3CH_2CH_2OH \xrightleftharpoons{HCl} CH_3CH_2C(=O)OCH_2CH_2CH_3 + CH_3OH}}$$

Products: propyl propanoate + methanol. The reaction is an equilibrium (shown by the double-headed arrow in the source); driving it to completion in practice requires using 1-propanol in large excess or continuously removing methanol (e.g. by distillation), per Le Chatelier's principle.