A) \[{{F}_{3}}C-CH=C{{H}_{2}}\]
B) \[Cl-CH=C{{H}_{2}}\]
C) \[C{{H}_{3}}O-CH=C{{H}_{2}}\]
D) \[{{H}_{2}}N-CH=C{{H}_{2}}\]
Correct Answer: A
Solution :
Due to higher e- withdrawing nature of \[C{{F}_{3}}\]group. It follow anti markovnikoff productYou need to login to perform this action.
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