Consider the following reaction. [figure] The major product (P) formed is :
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Answer
(3) [figure]
Solution
(i) NaBH4/MeOH reduces only the aldehyde group: -CHO -CH2OH; the amide (lactam) C=O is not reduced by NaBH4.
(ii) NaOH(aq.), Δ hydrolyses the cyclic amide: the ring C(=O)–N bond breaks ⇒ OOC-(CH2)4-NH-CH(CH3)-CH2OH
(iii) H3O+ converts -COO into –COOH ⇒ P:
[figure]
Answer: (3)