Answer:
\[C{{H}_{3}}-\underset{\underset{\underset{2-Methylpropane}{\mathop{\text{C}{{\text{H}}_{\text{3}}}}}\,}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}\xrightarrow[(Monochlorination)]{C{{l}_{2}},hv}\]
\[C{{H}_{3}}-\underset{\underset{\underset{I({{3}^{o}}radical)}{\mathop{\text{C}{{\text{H}}_{\text{3}}}}}\,}{\mathop{|}}\,}{\mathop{\overset{\centerdot
}{\mathop{C}}\,}}\,-C{{H}_{3}}+C{{H}_{3}}-\underset{\begin{smallmatrix}
| \\
\underset{II({{1}^{o}}radical)}{\mathop{C{{H}_{3}}}}\,
\end{smallmatrix}}{\mathop{CH}}\,-\overset{\centerdot
}{\mathop{C}}\,{{H}_{2}}\]
Intermediate hydrocarbon radical \[I\]
(having nine \[\alpha \]-hydrogen atoms) is tertiary whereas radical \[II\]
(having only one \[\alpha \]-hydrogen atom) is primary. Radical \[I\] is more
stable due to hyper conjugation.
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