JEE Main 2016 · 3 Apr · Shift 1 · Chemistry Q52

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ChemistrySingle correct+4 / -1Question 52 of 90

The product of the reaction given below is :

4-tert-Butylcyclohexene

5-tert-Butylcyclohex-2-en-1-ol

Cyclohexene ring with an OH on the allylic carbon and a tert-butyl group across the ring.

5-tert-Butylcyclohex-2-en-1-one

Cyclohexenone: the allylic carbon is a C=O, with a tert-butyl group across the ring.

5-tert-Butylcyclohex-2-ene-1-carboxylic acid

Cyclohexene ring with a CO₂H on the allylic carbon and a tert-butyl group across the ring.

Answer and worked solution

Official answer: B

NBS/hν brominates the allylic position, and aqueous $\ce{K2CO3}$ then replaces Br by OH (by $S_N$ , possibly through the allylic cation, which also moves the double bond). The product is the allylic alcohol, structure (2).

Topics:Alkenes SN1 and SN2

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JEE Main 2016 · 3 Apr · Shift 1, Chemistry question 52