JEE Advanced 2021 Paper 1 · Q17 · Isomerism & Stereochemistry

Correct answer:13

Solution

Br$_2$ + hν is a free-radical bromination, which abstracts H atoms at allylic and other ring carbons. 1-Methylcyclohex-1-ene has the following replaceable H positions:

(i) The methyl group (CH$_3$ on C1): three equivalent H atoms → bromination gives 1-(bromomethyl)cyclohex-1-ene → 1 constitutional product (no stereocentre) = 1 isomer.

(ii) C3 (allylic, sp$^3$): generates a stereocentre at C3 → R and S = 2 stereoisomers. The allyl radical also delocalises to C1 so 3-bromo-1-methylcyclohex-1-ene and the rearranged isomer with the double bond shifted (from C1=C2 to C2=C3) giving 3-bromo-2-methylcyclohex-1-ene (the regio-isomer from radical migration) → +2 stereoisomers.

(iii) C6 (other allylic position): gives 6-bromo-1-methylcyclohex-1-ene with a stereocentre at C6 → 2 stereoisomers; allyl radical resonance shifts double bond → 2-bromo-1-methylenecyclohexane / 2-bromo-6-methylcyclohex-1-ene with another stereocentre → +2 stereoisomers (but 6-bromo-1-methylcyclohex-1-ene = same as a counted form).

(iv) C4 and C5 (homoallylic, sp$^3$): each gives a constitutional product with a stereocentre → 4-bromo-1-methylcyclohex-1-ene (R, S = 2) and 5-bromo-1-methylcyclohex … [truncated]

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