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