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Saturday, 31 October

JEE Main 2026 · ChemistryMultiple choiceDiagram basedHardMulti-step

JEE Main 8 April 2026, Shift 2, Chemistry Q67

Question 67 of 75 in this shift, Chemistry question 17 of 25, Section A.

Consider the following reaction. [figure: N-substituted 2-piperidone (six-membered lactam, C=O adjacent to ring N) with N bonded to CH(CH3)-CHO] →(iii) H3O+(i) NaBH4/MeOH (ii) NaOH(aq.), Δ\xrightarrow[\text{(iii) } \mathrm{H_3O^+}]{\text{(i) } \mathrm{NaBH_4/MeOH}\ \text{(ii) NaOH(aq.), }\Delta} (P) The major product (P) formed is :

The figure, in words

Reactant: six-membered lactam (2-piperidone) whose nitrogen carries a CH(CH3)-CHO substituent. Reagents: (i) NaBH4/MeOH, (ii) NaOH(aq.), heat, (iii) H3O+.
  1. (1)[figure: piperidine ring with N attached to CH(CH3)-CH2OH]
  2. (2)[figure: 2-hydroxypiperidine (OH on ring carbon next to N) with N attached to CH(CH3)-CHO]
  3. (3)[figure: HOOC-(CH2)4-NH-CH(CH3)-CH2OH]Official answer
  4. (4)[figure: HO-(CH2)5-NH-CH(CH3)-CH2OH]

Official answer

Option 3

NTA final key.