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A carbonyl compound reacts with hydrogen cyanide to form cyanohydrin which on hydrolysis forms a racemic mixture of $\alpha$-hydroxy acid. The carbonyl compound is :
A
Acetaldehyde
B
Acetone
C
Diethyl ketone
D
Formaldehyde
Detailed Solution
A racemic mixture can form only if the $\alpha$-hydroxy acid has a chiral carbon, i.e. the carbon carrying OH must have four different groups.
Acetaldehyde: $CH_3CHO + HCN \rightarrow CH_3-CH(OH)-CN$
On hydrolysis: $CH_3-CH(OH)-CN \xrightarrow{H^+/H_2O} CH_3-\overset{*}{C}H(OH)-COOH$ (lactic acid)
The starred carbon has four different groups ($H$, $OH$, $CH_3$ and $COOH$), so it is chiral.
The carbonyl group is planar and $CN^-$ attacks it from both faces with equal ease, so both enantiomers are formed in equal amounts, giving a racemic mixture.
Formaldehyde gives $HOCH_2COOH$, acetone gives $(CH_3)_2C(OH)COOH$ and diethyl ketone gives $(C_2H_5)_2C(OH)COOH$; in each of these the carbon has two identical groups, so it is not chiral.
So the carbonyl compound is acetaldehyde.
Acetaldehyde: $CH_3CHO + HCN \rightarrow CH_3-CH(OH)-CN$
On hydrolysis: $CH_3-CH(OH)-CN \xrightarrow{H^+/H_2O} CH_3-\overset{*}{C}H(OH)-COOH$ (lactic acid)
The starred carbon has four different groups ($H$, $OH$, $CH_3$ and $COOH$), so it is chiral.
The carbonyl group is planar and $CN^-$ attacks it from both faces with equal ease, so both enantiomers are formed in equal amounts, giving a racemic mixture.
Formaldehyde gives $HOCH_2COOH$, acetone gives $(CH_3)_2C(OH)COOH$ and diethyl ketone gives $(C_2H_5)_2C(OH)COOH$; in each of these the carbon has two identical groups, so it is not chiral.
So the carbonyl compound is acetaldehyde.