This is a good example of a problem where you should work back from the end. Since no formulae are given, you have to look for hints elsewhere.

Benzene is the product from the action of soda-lime on D; this substance could be benzoic acid, decarboxylation of which yields benzene. The acid has only one carboxyl group since it reacts in a 1:1 ratio with sodium hydrogen carbonate - this eliminates the possibility of benzenedicarboxylic acids such as phthalic acid which would also react with soda-lime to give benzene. Benzoic acid it is.

Benzoic acid is precipitated from an alkaline solution by means of sulphuric acid, so the alkaline solution contains sodium benzoate.

Both substances B and C give the iodoform reaction, and C is an aldehyde. The only aldehyde to do this is ethanal, CH₃CHO, and since this is produced by careful oxidation of B, B must be ethanol which also gives the iodoform reaction since it is oxidised to ethanal under the conditions of the iodoform test.

The products from the reaction of sodium hydroxide with A are therefore ethanol and sodium benzoate; A must be an ester, so is ethyl benzoate.

C₆H₅COOCH₂CH₃ + NaOH    à   C₆H₅COONa + CH₃CH₂OH

CH₃CH₂OH + [O] à   CH₃CHO + H₂O

CH₃CHO + 3I₂ + 4NaOH à     CH₃COONa + CHI₃ + 3NaI + 3H₂O

C₆H₅COOH + 2NaOH (as soda-lime)  à   C₆H₆ + Na₂CO₃ + H₂O

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