JEE Main 2026 · 8 Apr · Shift 2 · Chemistry Q67 · Dronacharya Exams

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ChemistrySingle correct+4 / -1Question 67 of 75

Consider the following reaction.

The major product (P) formed is :

2-(2-Oxopiperidin-1-yl)propanal (a piperidin-2-one whose N carries CH(CH₃)CHO) treated with (i) NaBH₄/MeOH, (ii) NaOH(aq), Δ, (iii) H₃O⁺ gives P.

A

Option A: 2-(piperidin-1-yl)propan-1-ol.

B

Option B: 2-(2-hydroxypiperidin-1-yl)propanal.

C

Option C: 5-[(1-hydroxypropan-2-yl)amino]pentanoic acid.

D

Option D: 5-[(1-hydroxypropan-2-yl)amino]pentan-1-ol.

Answer and worked solution

Official answer: C

Step by step.

$\ce{NaBH4}$ reduces the aldehyde to $\ce{-CH2OH}$ (the lactam is untouched).
Hot aqueous NaOH hydrolyses the lactam (cyclic amide), opening the ring to the carboxylate and the secondary amine.
Acid gives the free acid.

P: $\ce{HOOC(CH2)4-NH-CH(CH3)CH2OH}$ .

(C)

Topics:Acid derivatives Oxidation and reduction

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