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