A) \[CHC{{l}_{3}}\]
B) \[C{{H}_{3}}Cl\]
C) \[C{{H}_{3}}C{{H}_{3}}\]
D) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}\]
Correct Answer: D
Solution :
\[C{{H}_{4}}\xrightarrow{C{{l}_{2}}}C{{H}_{3}}Cl\xrightarrow{C{{l}_{2}}}C{{H}_{2}}C{{l}_{2}}\xrightarrow{C{{l}_{2}}}\] \[CHC{{l}_{3}}\xrightarrow{C{{l}_{2}}}CC{{l}_{4}}\]. Since this reaction takes place by free radical mechanism. Hence, there is a possibility of formation of ethane. \[\underset{\text{Methyl free radicals}}{\mathop{\overset{.}{\mathop{C}}\,{{H}_{3}}+\overset{.}{\mathop{C}}\,{{H}_{3}}}}\,\to \underset{\text{Ethane}}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\] \[C{{H}_{4}}\xrightarrow{C{{l}_{2}}}C{{H}_{3}}Cl\xrightarrow{C{{l}_{2}}}C{{H}_{2}}C{{l}_{2}}\xrightarrow{C{{l}_{2}}}\] \[CHC{{l}_{3}}\xrightarrow{C{{l}_{2}}}CC{{l}_{4}}\]. Since this reaction takes place by free radical mechanism. Hence, there is a possibility of formation of ethane. \[\underset{\text{Methyl free radicals}}{\mathop{\overset{.}{\mathop{C}}\,{{H}_{3}}+\overset{.}{\mathop{C}}\,{{H}_{3}}}}\,\to \underset{\text{Ethane}}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\]
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