Question
Which of the following will give enantiomeric pair on reaction with water due to presence of asymmetric carbon atom?
A.
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
{{C}_{2}}{{H}_{5}}\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
{{C}_{2}}{{H}_{5}}\, \\
|\,\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-\,\,C-Br}}}\,\]
B.
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
{{C}_{2}}{{H}_{5}} \\
|\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-\,\,C-Cl}}}\,\]
C.
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
| \\
\,\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
H \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,I\]
D.
\[C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
{{C}_{2}}{{H}_{5}}\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
C{{H}_{3}} \\
|\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-\,\,C-Br}}}\,\]
Answer :
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
| \\
\,\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
H \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,I\]
Solution :
Option (C) has asymmetric carbon atom which is joined to four different groups hence it will give optically active products or a pair of enantiomers.