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Which one of the following on treatment with 50% aqueous sodium hydroxide yields the corresponding alcohol and acid?
A
$C_6H_5CHO$
B
$CH_3CH_2CH_2CHO$
C
$CH_3-CO-CH_3$
D
$C_6H_5CH_2CHO$
Detailed Solution
Aldehydes containing no $\alpha$-hydrogen atom, on warming with 50% NaOH or KOH, undergo disproportionation, i.e. self oxidation-reduction, known as Cannizzaro's reaction.
One molecule of the aldehyde is reduced to the alcohol and another is oxidised to the salt of the carboxylic acid.
Benzaldehyde, $C_6H_5CHO$, has no $\alpha$-hydrogen (the CHO group is attached directly to the ring).
$2C_6H_5CHO + NaOH \xrightarrow{conc.\ NaOH} C_6H_5COONa + C_6H_5CH_2OH$
It gives sodium benzoate (the acid salt) and benzyl alcohol.
$CH_3CH_2CH_2CHO$, $CH_3COCH_3$ and $C_6H_5CH_2CHO$ all have $\alpha$-hydrogen atoms, so they undergo aldol condensation with a base instead.
So the compound is $C_6H_5CHO$.
One molecule of the aldehyde is reduced to the alcohol and another is oxidised to the salt of the carboxylic acid.
Benzaldehyde, $C_6H_5CHO$, has no $\alpha$-hydrogen (the CHO group is attached directly to the ring).
$2C_6H_5CHO + NaOH \xrightarrow{conc.\ NaOH} C_6H_5COONa + C_6H_5CH_2OH$
It gives sodium benzoate (the acid salt) and benzyl alcohol.
$CH_3CH_2CH_2CHO$, $CH_3COCH_3$ and $C_6H_5CH_2CHO$ all have $\alpha$-hydrogen atoms, so they undergo aldol condensation with a base instead.
So the compound is $C_6H_5CHO$.