• # question_answer The product obtained by treating: $C{{H}_{3}}-CH=C{{H}_{2}}+HBr\to$ A) $C{{H}_{3}}-CH=C{{H}_{2}}Br$B) $C{{H}_{2}}BrC{{H}_{2}}=C{{H}_{2}}$C) $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br$D) $C{{H}_{3}}-\underset{Br}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{3}}$

$C{{H}_{3}}CH=C{{H}_{2}}+HBr\xrightarrow{{}}C{{H}_{3}}-\underset{Br}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}$ This is according to Markovnikov's rule. Hence, the correct option is [d].