A) \[C{{H}_{3}}-\underset{\,\,}{\mathop{\underset{\underset{\,Br}{\mathop{|}}\,}{\mathop{C}}\,=C{{H}_{2}}\ \ }}\,\]
B) \[BrC{{H}_{2}}-CH=C{{H}_{2}}\]
C) \[BrC{{H}_{2}}-CH=CHBr\]
D) \[BrC{{H}_{2}}-\underset{\underset{Br}{\mathop{|\,\,\,\,}}\,}{\mathop{CH}}\,-C{{H}_{2}}Br\]
Correct Answer: B
Solution :
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