question_answer 1)

  The relative reactivity of \[1{}^\circ ,\text{ }2{}^\circ ,\text{ }3{}^\circ \]hydrogen's towards chlorination is 1: 3.8: 5. Calculate the percentages of all mono chlorinated products obtained from 2-methylbutane.

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