A will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → H3C – CH2 – CH2 – CH2 – CHBr – CH3
When HBr adds to an unsymmetrical alkene, the addition follows Markovnikov’s rule: the hydrogen goes to the double-bond carbon that already has more hydrogens, and the bromine goes to the other carbon. The reason is that \( \ce{H^{+}} \) adds first to give the more stable carbocation, which Br– then attacks.
For hex-1-ene, \( \ce{H^{+}} \) adds to the terminal \( \ce{CH2} \), giving a secondary carbocation at C2 (more stable than the primary one at C1), and Br– bonds to C2:
\( \ce{CH3CH2CH2CH2CH=CH2 + HBr -> CH3CH2CH2CH2CHBrCH3} \)
The product A is 2-bromohexane.
Why the other options are incorrect
Option (1) 1-bromohexane — Incorrect. This is the anti-Markovnikov product, formed only when a peroxide is present (peroxide or Kharasch effect, a free-radical addition); no peroxide is shown here.
Option (3) 1-bromo-1-methylcyclohexane — Incorrect. An open-chain alkene cannot form a ring by simple addition of HBr.
Option (4) Bromobenzene — Incorrect. An aromatic compound cannot form from an alkene and HBr; the carbons and hydrogens do not match.
NCERT (2026-27 edition):
From Hydrocarbons – “… (i) Addition of hydrogen halides: An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide. Propene yields two products, however only one predominates as per Markovnikov’s rule.”