A) \[\underset{C{{H}_{3}}\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-CH=C{{H}_{2}}Cl}}\,}}\,\] and \[C{{H}_{2}}Cl-C{{H}_{2}}Cl\]
B) \[\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-CCl=C{{H}_{3}}}}\,}}\,\] and \[C{{H}_{3}}-CHC{{l}_{2}}\]
C) \[\underset{C{{H}_{3}}\,\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-CH=C{{H}_{2}}Cl}}\,}}\,\] and \[C{{H}_{3}}-CHC{{l}_{2}}\]
D) \[\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-CH=C{{H}_{3}}}}\,}}\,\] and \[C{{H}_{2}}Cl-C{{H}_{2}}Cl\]
Correct Answer: B
Solution :
\[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,}{\mathop{C\,}}\,}}\,=C{{H}_{2}}+HCl\to C{{H}_{3}}-\underset{\underset{C{{H}_{3}}}{\mathop{|\,\,\,\,\,\,}}\,}{\overset{\overset{Cl\,\,}{\mathop{|\,\,\,\,\,}}\,}{\mathop{C\,-\,}}}\,C{{H}_{3}}\] \[CH\equiv CH+HCl\to C{{H}_{2}}=CH-Cl\xrightarrow{HCl}C{{H}_{3}}-CHC{{l}_{2}}\]
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