Consider the following reaction. The major pr | vedclass

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(A) Step $(i)$: $NaBH_{4}$ in $MeOH$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_{2}OH)$. The cyclic amide (lactam) rem

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(A) Step $(i)$: $NaBH_{4}$ in $MeOH$ selectively reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_{2}OH)$. The cyclic amide (lactam) remains unaffected.Step (ii): $NaOH(aq.), \Delta$ causes the alkaline hydrolysis of the cyclic amide (lactam). The ring opens to form the corresponding amino acid salt (carboxylate).Step (iii): $H_{3}O^{+}$ protonates the carboxylate to form a carboxylic acid $(-COOH)$ and the amine to form an ammonium salt $(-NH_{2}^{+}-)$.The final product is a linear amino acid derivative with a primary alcohol group: $HOOC-(CH_{2})_{4}-NH-CH(CH_{3})CH_{2}OH$.

Consider the following reaction. The major product $(P)$ formed is:

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