A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
B) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\]
C) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H\]
D) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{3}}\]
Correct Answer: D
Solution :
Key Idea: Write chlorination reaction for all of them to find which gives of the maximum number of monochlorination product (a) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}+C{{l}_{2}}\xrightarrow{UV}\] \[ClC{{H}_{2}}-{{(C{{H}_{2}})}_{3}}C{{H}_{3}}+C{{H}_{3}}-\overset{Cl}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-{{(C{{H}_{2}})}_{2}}C{{H}_{3}}\] \[+\,C{{H}_{3}}-C{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\] \[\therefore \] Total 3 monochlorinated products are formed. (b) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}+C{{l}_{2}}\xrightarrow{UV}\] \[ClC{{H}_{2}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\] \[+\,\,C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{Cl}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{2}}-C{{H}_{3}}\] \[+\,\,C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{2}}Cl\] \[\therefore \] Total 3-monochlorinated products are (c) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H+C{{l}_{2}}\xrightarrow{UV}\] \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-Cl+C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{2}}Cl}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-H\] \[\therefore \]Total 2- monochlorinated products are formed (d) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{3}}+C{{l}_{2}}\xrightarrow{UV}\] \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{2}}Cl\] \[\therefore \] Only one monochlorinated product formed. \[\therefore \](d) is correct answer.
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