Identify the product [B] in the following reaction:

The correct answer is Option (2) → CH₃–CH₂–C(Cl)(I)–CH₃ (2-chloro-2-iodobutane)

Both steps are additions of H–X to an unsymmetrical multiple bond, so both follow Markovnikov’s rule: the H goes to the carbon that already has more hydrogens and X goes to the carbon that gives the more stable carbocation.

First addition — but-1-yne + HCl: H adds to the terminal CH carbon; Cl goes to C-2.

    \( \ce{CH3CH2C#CH + HCl -> CH3CH2C(Cl)=CH2} \)  (A, 2-chlorobut-1-ene)

Second addition — A + HI: \( \ce{H+} \) adds to the =\( \ce{CH2} \) carbon, giving a carbocation on C-2. This cation is secondary and is further stabilised by the lone pair of the attached Cl (resonance), so \( \ce{I-} \) attacks C-2 and both halogens end up on the same carbon (a gem-dihalide):

    \( \ce{CH3CH2C(Cl)=CH2 + HI -> CH3CH2C(Cl)(I)CH3} \)  (B)

Why the other options are incorrect

• Option (1) CH₃CH₂CH(I)CH₂Cl — Incorrect. \( \ce{CH3CH2CH(I)CH2Cl} \) puts Cl on the terminal carbon, against Markovnikov’s rule in the first addition.

• Option (3) CH₃CH₂CH₂CH(Cl)(I) — Incorrect. \( \ce{CH3CH2CH2CH(Cl)(I)} \) has both halogens on C-1, which would need anti-Markovnikov addition in both steps.

• Option (4) CH₃CH(Cl)CH₂CH₂I — Incorrect. \( \ce{CH3CH(Cl)CH2CH2I} \) has the halogens on C-2 and C-4, positions that are not part of the original triple bond (C-1 and C-2).