A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
B) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}\]
C) \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-H\]
D) \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{3}}\]
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] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}+C{{l}_{2}}\xrightarrow{UV}\] \[ClC{{H}_{2}}-{{(C{{H}_{2}})}_{3}}C{{H}_{3}}+C{{H}_{3}}-\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}\,H-{{(C{{H}_{2}})}_{2}}C{{H}_{3}}\] \[+\,C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}\] \[\therefore \] Total 3 monochlorinated products are formed. [b] \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}+C{{l}_{2}}\xrightarrow{UV}\] \[ClC{{H}_{2}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}\] \[+\,C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}\,}}\,H-C{{H}_{2}}-C{{H}_{3}}\] \[+\,C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{2}}Cl\] \[\therefore \] Total 3-monochlormated products are formed. [c] \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-H+C{{l}_{2}}\xrightarrow{UV}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-Cl+C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{2}}Cl \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-H\] \[\therefore \] Total 2- monochlorinated products are formed [d] \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{3}}+C{{l}_{2}}\xrightarrow{UV}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{2}}C{{l}_{2}}\] \[\therefore \] Only one monochlorinated product formed. \[\therefore \] [d] is correct answer.
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