Major Product in Reduction and Hydrolysis of Amide

To determine the major product $(\text{P})$ formed in the given reaction sequence, we analyze each step sequentially:

Step 1: Selective Reduction using $\text{NaBH}_4/\text{MeOH}$

• Starting Material: The reactant contains two functional groups: an aldehyde group ($-\text{CHO}$) in the side chain and a cyclic amide ($\delta$-lactam) group in the ring.
• Reagent: Sodium borohydride ($\text{NaBH}_4$) in methanol ($\text{MeOH}$) is a mild reducing agent.
• Reaction: $\text{NaBH}_4$ selectively reduces aldehydes and ketones to primary and secondary alcohols, respectively, while leaving amides unreacted.
• Intermediate: The aldehyde group ($-\text{CHO}$) is reduced to a primary alcohol ($-\text{CH}_2\text{OH}$): $\text{R}-\text{CHO} \xrightarrow{\text{NaBH}_4/\text{MeOH}} \text{R}-\text{CH}_2\text{OH}$

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Step 2 & 3: Hydrolysis of the Lactam ($\text{NaOH(aq.)}, \Delta$ followed by $\text{H}_3\text{O}^+$)

• Reagent: Aqueous sodium hydroxide ($\text{NaOH}$) with heating ($\Delta$) hydrolyzes the amide bond (lactam ring).
• Mechanism:
  1. Base-catalyzed cleavage of the cyclic amide bond occurs via nucleophilic acyl substitution on the ring carbonyl carbon, causing the 6-membered ring to open.
  2. This process yields a carboxylate ion at one end and a secondary amine at the other end: $\text{O}^--\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CH}(\text{CH}_3)-\text{CH}_2\text{OH}$
  3. Acidification with $\text{H}_3\text{O}^+$ protonates the carboxylate group to form a carboxylic acid ($-\text{COOH}$).

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Final Product Structure:

Combining both transformations, the final product $(\text{P})$ is 5-((1-hydroxypropan-2-yl)amino)pentanoic acid: $\text{HO}-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CH}(\text{CH}_3)-\text{CH}_2\text{OH}$

This matches the chemical structure shown in Option C.