A) Isopropyl bromide
B) Propyl bromide
C) 1, 2-dibromoethane
D) None of the above
Correct Answer: A
Solution :
\[C{{H}_{3}}-CH=C{{H}_{2}}+HBr\xrightarrow[\begin{smallmatrix} \text{Marko wnikoff }\!\!'\!\!\text{ s} \\ \text{ rule} \end{smallmatrix}]{}C{{H}_{3}}-\underset{Br\,}{\mathop{\underset{|\,\,\,\,}{\mathop{CH}}\,}}\,-C{{H}_{3}}\]
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