A) \[C{{H}_{3}}C{{H}_{2}}OH\]
B) \[C{{H}_{3}}CHClBr\]
C) \[CC{{l}_{2}}BrF\]
D) \[CC{{l}_{2}}{{F}_{2}}\]
Correct Answer: B
Solution :
\[C{{H}_{3}}-\underset{Br}{\mathop{\overset{H}{\mathop{\underset{*|}{\overset{|}{\mathop{C}}}\,}}\,-}}\,Cl\], it is a chiral compound.
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