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Step-by-Step Solution
Step 1: Identify the Reaction
A carbonyl compound (either an aldehyde or a ketone) reacts with hydrogen cyanide (HCN) to form a cyanohydrin. Upon subsequent hydrolysis, the product is an α-hydroxy acid. The problem specifies that this α-hydroxy acid is optically active, existing as a racemic mixture.
Step 2: Understand the Formation of Cyanohydrin
When a carbonyl compound, denoted as R−C(=O)−H for an aldehyde or R2C=O for a ketone, reacts with HCN, the nucleophile (CN−) attacks the electrophilic carbon of the C=O group. This produces a cyanohydrin with the generic structure R−C(OH)(CN)−H for an aldehyde, or R2C(OH)(CN) for a ketone.
Step 3: Hydrolysis of Cyanohydrin to α-Hydroxy Acid
Upon acidic hydrolysis, the −CN group of the cyanohydrin is converted into a −COOH group, forming an α-hydroxy acid:
$ \text{R−CH(OH)−CN} \xrightarrow[\text{acidic}]{\text{hydrolysis}} \text{R−CH(OH)−COOH} $
If this α-hydroxy acid is chiral (i.e., it has four different groups attached to the α-carbon), it will exist in two enantiomeric forms (a pair of non-superimposable mirror images). Under typical reaction conditions, these two enantiomers form in equal amounts, resulting in a racemic mixture.
Step 4: Determining the Specific Carbonyl Compound
1. Formaldehyde ($\text{HCHO}$) upon reaction with HCN and hydrolysis would yield formic acid derivatives, which are not chiral.
2. Acetone ($\text{CH}_3\text{COCH}_3$) or more substituted ketones typically yield tertiary α-hydroxy acids if they form a chiral center, but certain steric and structural requirements are different, and often the carbon next to a ketone can be less straightforwardly chiral.
3. Diethyl ketone ($\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$) would form an α-hydroxy acid upon hydrolysis of its cyanohydrin, but the α-carbon may or may not be chiral depending on substitution.
4. Acetaldehyde ($\text{CH}_3\text{CHO}$) reacts with HCN to give $\text{CH}_3\text{CH(OH)(CN)}$, which upon hydrolysis yields lactic acid ($\text{CH}_3\text{CH(OH)COOH}$). Lactic acid is a well-known chiral molecule that exists in two enantiomeric forms. Therefore, under these conditions, one obtains a racemic mixture of lactic acid.
Step 5: Conclusion: Acetaldehyde is the Correct Carbonyl Compound
The key point is that the α-carbon in the product (lactic acid) must be chiral, leading to a racemic mixture under typical laboratory conditions. This scenario specifically corresponds to acetaldehyde.
Correct Answer: acetaldehyde
