JEE Main 2022 · 30 Jul · Shift 2 · Chemistry Q44

← JEE Main 2022 · 30 Jul · Shift 2

ChemistrySingle correct+4 / -1Question 44 of 90

Major product of the following reaction is

3,3-Dimethylpent-4-enoic acid is treated with I $_2$ , NaHCO $_3$ .

Scheme: CH2=CH-C(CH3)2-CH2-COOH treated with I2 and NaHCO3.

A six-membered lactone: 5-iodo-4,4-dimethyloxan-2-one

Option A: six-membered lactone with iodine on the ring carbon next to the gem-dimethyl carbon.

A five-membered lactone: 5-(iodomethyl)-4,4-dimethyloxolan-2-one

Option B: five-membered lactone bearing CH2I on the ring carbon attached to O, next to the gem-dimethyl carbon.

4-Iodo-3,3-dimethylpentanoic acid

5-Iodo-3,3-dimethylpentanoic acid

Answer and worked solution

Official answer: B

Iodolactonisation. I₂ forms an iodonium ion on the C=C; the carboxylate (from NaHCO₃) attacks the internal carbon to close the five-membered ring (5-exo), leaving CH₂I outside: a γ-lactone with gem-dimethyl next to the ring O-bearing carbon.

(B)

Topics:Alkenes Carboxylic acids

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JEE Main 2022 · 30 Jul · Shift 2, Chemistry question 44