A) \[\begin{align} & C{{H}_{3}}-\underset{|}{\mathop{C}}\,H-C{{H}_{2}}C{{H}_{2}}I \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}\]
B) \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I \\ & C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-\underset{|}{\mathop{\overset{|}{\mathop{C}}\,}}\,-H \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}\]
C) \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I \\ & C{{H}_{3}}-C{{H}_{2}}-\overset{|}{\mathop{C}}\,H-C{{H}_{2}}Cl \\ \end{align}\]
D) \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I \\ & C{{H}_{3}}C{{H}_{2}}-\underset{|}{\mathop{\overset{|}{\mathop{C}}\,}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}\]
Correct Answer: D
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-C\equiv CH+HCl\xrightarrow[{}]{{}}\] \[\begin{align} & C{{H}_{3}}-C{{H}_{2}}-\underset{|}{\mathop{C}}\,=C{{H}_{2}}\xrightarrow[{}]{HI} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}\] \[\underset{2-chloro,\,2-ido\,bu\tan e}{\mathop{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I \\ & C{{H}_{3}}-C{{H}_{2}}-\underset{|}{\mathop{\overset{|}{\mathop{C}}\,}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}}}\,\]
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