A) \[\underset{Cl\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,=C<\begin{matrix} C{{H}_{3}} \\ C{{H}_{3}} \\ \end{matrix}\]
B) \[\underset{\,\,\,}{\mathop{\underset{\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,-CH<-\begin{matrix} Cl \\ C{{H}_{3}} \\ C{{H}_{3}} \\ \end{matrix}\]
C) \[\]\[\underset{\,\,\,}{\mathop{\underset{\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,=C<\begin{matrix} C{{H}_{3}} \\ C{{H}_{2}}Cl \\ \end{matrix}\]
D) None of these
Correct Answer: B
Solution :
\[C{{H}_{2}}=\underset{C{{H}_{3}}\,}{\mathop{\underset{|\,\,\,\,\,\,\,}{\mathop{C\,-}}\,C}}\,{{H}_{3}}+\overset{\,\,\,\,}{\mathop{HCl}}\,\to \underset{\begin{smallmatrix} \text{2-chloro-2-methyl} \\ \,\,\,\,\,\,\,\,\,\,\,\,\text{propane} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{Cl\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,}{\overset{|\,\,\,\,\,}{\mathop{C\,\,-\,}}}\,}}}\,C{{H}_{3}}}}\,\]
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