A) \[\begin{align} & C{{H}_{3}}-C{{H}_{2}}-CH-C{{H}_{3}} \\ & | \\ & C{{H}_{2}}-C{{H}_{3}} \\ \end{align}\]
B) \[\begin{align} & C{{H}_{3}}-CH-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}} \\ & | \\ & C{{H}_{3}} \\ \end{align}\]
C) \[\begin{align} & C{{H}_{3}}-CH-CH-C{{H}_{3}} \\ & || \\ & C{{H}_{3}}C{{H}_{3}} \\ \end{align}\]
D) \[\begin{align} & C{{H}_{3}} \\ & | \\ & C{{H}_{3}}-C{{H}_{2}}-C-C{{H}_{3}} \\ & | \\ & C{{H}_{3}} \\ \end{align}\]
Correct Answer: A
Solution :
\[Alkane\underrightarrow{Monochlorination}\underbrace{I+II}_{Isomers}\] Mol. Wt. \[=86\Rightarrow {{C}_{6}}{{H}_{14}}\] \[Y(alkene)\xrightarrow[reduction]{Catalytic}X(isomer\,\,of\,\,A)\] \[\xrightarrow{Monochlorination}4\,\,monochloro\,\,derivatives\] As X is isomer of A thus it should have molecular formula \[{{C}_{6}}{{H}_{14}}\] \[\begin{align} & \overset{1}{\mathop{C{{H}_{3}}}}\,-\overset{{}}{\mathop{\overset{2}{\mathop{C{{H}_{2}}}}\,-\overset{3}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}}}\, \\ & | \\ & \underset{4}{\mathop{C{{H}_{3}}}}\, \\ \end{align}\] This isomer of hexane gives four monochloro products on chlorination.
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