A) 3-methylhexane
B) 2, 2-dimethylpentane
C) 2-methylhexane
D) 2, 3-dimethylpentane
Correct Answer: A
Solution :
In 3-methylhexane, none out of the seven carbon atoms are equivalent and it gives seven different monochlorinated products upon free radical chlorination. \[{{H}_{3}}\overset{1}{\mathop{C}}\,-\overset{2}{\mathop{C}}\,{{H}_{2}}-\overset{\begin{smallmatrix} ^{7}C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\overset{3}{\mathop{C}}\,H}}\,-\overset{4}{\mathop{C}}\,{{H}_{2}}-\overset{5}{\mathop{C}}\,{{H}_{2}}-\overset{6}{\mathop{C}}\,{{H}_{3}}\]
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