Major Product in Reaction of Amino Acid Derivative with Nitrous Acid

To determine the major organic product of the reaction, let us analyze the reaction step-by-step:

1. Nature of the Reaction

The reactant is an $\alpha$-amino acid derivative (specifically, $(2S, 4R)$-2-amino-4-methylpentanoic acid, a leucine derivative) containing a primary aliphatic amino group ($-\text{NH}_2$) at the $\alpha$-carbon. Treating a primary amine with nitrous acid ($\text{HNO}_2$, generated in situ from $\text{NaNO}_2$ and aqueous $\text{HCl}$ at $0^\circ\text{C}$) leads to diazotization.

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2. Reaction Mechanism and Stereochemistry

1. Diazotization: The primary amino group ($-\text{NH}_2$) at the $\alpha$-position reacts with $\text{HNO}_2$ to form an unstable diazonium ion: $\text{R-CH(NH}_2\text{)-COOH} \xrightarrow{\text{NaNO}_2\text{, HCl}} \text{R-CH(N}_2^+\text{)-COOH}$

2. Neighboring Group Participation (NGP): The reaction of $\alpha$-amino acids with nitrous acid proceeds with retention of configuration at the $\alpha$-chiral center via a double inversion mechanism due to Neighboring Group Participation (NGP) by the carboxyl group ($-\text{COOH}$):

   • First Inversion: The carbonyl oxygen of the carboxylic acid group acts as an internal nucleophile and attacks the $\alpha$-carbon from the back side, expelling $\text{N}_2$ gas as a leaving group. This forms a transient 3-membered $\alpha$-lactone intermediate with inversion of configuration.
   • Second Inversion: A water molecule ($\text{H}_2\text{O}$) from the aqueous solvent subsequently attacks the $\alpha$-carbon of the $\alpha$-lactone intermediate from the back side to open the ring. This second nucleophilic substitution ($S_N2$) results in a second inversion of configuration.

3. Net Result (Double Inversion = Retention): The two successive inversions ($S_N2$ steps) yield a net retention of configuration at the $\alpha$-carbon. Thus, the incoming hydroxyl group ($-\text{OH}$) occupies the exact same spatial position (wedge) previously occupied by the $-\text{NH}_2$ group.

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3. Conclusion

• The stereocenter at C-4 bearing the methyl group ($-\text{CH}_3$, on a wedge) remains unaffected throughout the reaction.
• The stereocenter at C-2 retains its configuration, so the newly formed hydroxyl group ($-\text{OH}$) is on a wedge.

Therefore, the major product is the $\alpha$-hydroxy acid represented in Option (C).

Correct Answer: (C)