The compound representing a vinylic bromide is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1-bromocyclohex-1-ene
Halogen compounds with a C=C bond are classified by the position of the halogen. In a vinylic halide the halogen is bonded directly to an \(sp^2\) carbon of the C=C double bond, as in \( \ce{CH2=CH-Br} \). In an allylic halide the halogen is on the \(sp^3\) carbon next to the double bond, as in \( \ce{CH2=CH-CH2-Br} \).
In Option (3), 1-bromocyclohex-1-ene, bromine sits on C1, which is one of the two carbons of the ring double bond. So this compound is a vinylic bromide.
Why the other options are incorrect
Option (1) \(CH_3CH=CHCH_2Br\) — Incorrect. 1-Bromobut-2-ene: Br is on C1, an \(sp^3\) carbon next to the double bond (C2=C3), so it is an allylic bromide.
Option (2) 3-bromocyclohex-1-ene — Incorrect. Br is on the ring carbon next to the double bond, an \(sp^3\) carbon, so it is an allylic bromide.
Option (4) \(BrCH_2CH=CHCH_2Br\) — Incorrect. 1,4-Dibromobut-2-ene: both Br atoms are on \( \ce{CH2} \) carbons next to the double bond, so it is a diallylic bromide.
NCERT (2026-27 edition):
Vinylic Halides
