JEE Main 2023 · 24 January · morning · Single correct · +4 / −1
Compound (X) undergoes following sequence of reactions to give the Lactone (Y).
- (a) CH₃; |; H, -, C, -, CHO; |; CH₃
- (b) H₂C - CH₂ - CH₂ - CH₂ - CHO
- (c)
- (d) H₂C, —, CH₂, —, CHO; |; CH₃
Answer and solution
Answer (a) CH₃; |; H, -, C, -, CHO; |; CH₃
Solution
Chemistry ma’am · 4 steps
Read the lactone backwards: open the ring, undo the cyanohydrin, then undo the aldol with HCHO.
- Open the lactone. Break the ring ester bond in Y. You get the hydroxy acid HOCH₂-C(CH₃)₂-CH(OH)-COOH. Its -COOH and -CH₂OH sit in a 1,4 relation, so they close a five-membered (γ) lactone on their own in acid.
- Undo the cyanohydrin. A -CH(OH)-COOH end is the signature of a cyanohydrin. KCN adds to a -CHO to give -CH(OH)CN, and H₃O⁺ hydrolyses -CN to -COOH. So before step (ii) the compound was HOCH₂-C(CH₃)₂-CHO.
- Undo step (i). With HCHO and KOH, the α-carbon of X attacks formaldehyde in a crossed aldol and picks up a -CH₂OH group. Remove that -CH₂OH and put back an H. X must be (CH₃)₂CH-CHO, with exactly one α-hydrogen.
(CH₃)₂CH-CHO →(HCHO, KOH) HOCH₂-C(CH₃)₂-CHO →(KCN) HOCH₂-C(CH₃)₂-CH(OH)CN →(H₃O⁺) Y
- So. X is 2-methylpropanal, (CH₃)₂CH-CHO. The carbon count checks out: 4 from X, 1 from HCHO and 1 from CN⁻ give the 6 carbons of Y. The answer is (a).
The idea to keep. In multistep problems, work backwards one reagent at a time. A -CH(OH)COOH end means CHO plus HCN, and a new -CH₂OH on an α-carbon means a crossed aldol with HCHO.
Answer from the NTA final answer key. Solution by PrepWiser.