A) 2
B) 4
C) 0
D) 3
Correct Answer: A
Solution :
\[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{2}}C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,=\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,+{{H}_{2}}\xrightarrow[{}]{{}}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{2}}C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{3}}\] \[\because \]This product is found in d and I form. \[\therefore \]Two optical isomers are possible.You need to login to perform this action.
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