Answer:
The given organic compound is :
\[\underset{2-\text{methylbutane}}{\mathop{C{{H}_{3}}-\overset{\overset{\text{C}{{\text{H}}_{\text{3}}}}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}}}\,\]
This compound has 9 primary
hydrogen; 2 secondary and one tertiary hydrogen atoms. The relative reactivity
of \[1{}^\circ ,\text{ }2{}^\circ \]and \[3{}^\circ \]hydrogen atoms towards
chlorination is 1: 3.8:5.
Relative amount of product after
chlorination
= Number of hydrogen x relative
reactivity
\[\begin{array}{*{35}{l}}
Relative~~~{{1}^{o}}halide~~~\,\,\,{{2}^{o}}\text{
}halide~~~\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}^{o}}halide \\
amount~~~9\times 1=9\text{
}\,\,\,\,\,\,\,\,2\text{ }\times 3.8=\,7.6\text{ }\,\,\,\,\,\,\,\,\,1\times
5=5 \\
\end{array}\]
Total amount of mono chloro
product
\[=9+7.6+5=21.6\]
Percentage of \[1{}^\circ \]
mono =\[\frac{9}{21.6}\times 100=41.7%\]
chloro product
Percentage of \[2{}^\circ \]
mono = \[\frac{7.6}{21.6}\times 100=35.2%\]
chloro product
Percentage of \[3{}^\circ \] mono
\[=\frac{5}{21.6}\times 100=23.1%\]
chloro product
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