Question

Write the product(s) when : (I) One mol of ethanal is treated with 1 mol of \(CH_3OH\) in the presence of dry HCl gas. (II) Benzaldehyde is treated with conc. NaOH. (III) Ethanoic acid is heated in the presence of \(P_2O_5\).

Hint:

One mole of alcohol reacts with an aldehyde to form a hemiacetal. Benzaldehyde undergoes Cannizzaro reaction because it lacks \(\alpha\)-hydrogen. \(P_2O_5\) converts carboxylic acids into acid anhydrides by dehydration.

Answer 1

Step 1: Ethanal + 1 mol $CH_3OH$ + dry HCl.
Only 1 mol of alcohol undergoes partial addition to the carbonyl group, forming a hemiacetal: \[ CH_3CHO + CH_3OH \xrightarrow{\text{dry HCl}} CH_3CH(OH)(OCH_3) \]
Step 2: Benzaldehyde + conc. NaOH (Cannizzaro reaction).
Benzaldehyde has no alpha-H, so aldol condensation cannot occur. It undergoes Cannizzaro reaction: one molecule is oxidized to benzoate and another is reduced to benzyl alcohol: \[ 2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]
Step 3: Ethanoic acid + $P_2O_5$.
$P_2O_5$ is a powerful dehydrating agent that removes water from two acetic acid molecules to form acetic anhydride: \[ 2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O + H_2O \] \[ \boxed{(CH_3CO)_2O} \]