Q20 / 160 · TS EAMCET 2016 · Chemistry

The product $Z$ of the following reaction is $$ \mathrm{H}_3 \mathrm{CC} \equiv \mathrm{CH} \stackrel{2 \mathrm{HBr}}{\longrightarrow} \mathrm{Z} $$

  • A
    $\mathrm{H}_3 \mathrm{CCH}_2 \mathrm{CHBr}_2$
  • Correct answer
    B
    $\mathrm{H}_3 \mathrm{CCHBrCH} \mathrm{Br}_2$
  • C
    $\mathrm{H}_3 \mathrm{CCBr}_2 \mathrm{CH}_3$
  • D
    $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$

Solution

$\mathrm{H}_3 \mathrm{CC} \equiv \mathrm{CH} \stackrel{2 \mathrm{HBr}}{\longrightarrow} \mathrm{H}_3 \mathrm{CCBr}_2 \mathrm{CH}_3$ The reaction follows Markovnikov's addition.

Hence, the correct answer is option: 'B'.

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