A) 2-Bromobutane
B) 1-Bromobutane
C) 2-Bromopropane
D) 2-Bromopropan-2-ol
Correct Answer: A
Solution :
\[C{{H}_{3}}-\overset{\overset{Br}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}\] 2nd Carbon of 2-Bromobutane is chiral so molecule is optically active.You need to login to perform this action.
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