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Question
The relative reactivity of 1°, 2°, 3° hydrogen’s towards chlorination is 1 : 3.8 : 5. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.
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Solution
2-Methyl butane is \[\begin{array}{cc}
\ce{CH3}\phantom{......}\\
|\phantom{.........}\\
\ce{CH3 - CH - CH2 - CH3}
\end{array}\].
Possible compounds are A, B and C given below:
| \[\begin{array}{cc} \ce{CH3}\phantom{..}\\ |\phantom{....}\\ \ce{\underset{A(1°)}{ClCH2 - CH - CH2CH3}} \end{array}\] |
Nine possibilities for compound ‘A’ because nine methyl hydrogens are present in 2-methylbutane. |
| \[\begin{array}{cc} \ce{H3C}\phantom{........................}\\ \backslash\phantom{...................}\\ \phantom{}\ce{CH - CH - CH3}\\ \phantom{}/\phantom{.......}|\phantom{...........}\\ \phantom{}\ce{H3C}\phantom{........}\ce{\underset{A(2°)}{Cl}}\phantom{..............} \end{array}\] |
Two possibilities for ‘B’ compound because two CH hydrogens are present in 2-methylbutane. |
| \[\begin{array}{cc} \ce{H3C}\phantom{........................}\\ \backslash\phantom{...................}\\ \phantom{}\ce{C - CH2 - CH3}\\ \phantom{}/\phantom{.}|\phantom{................}\\ \phantom{}\ce{H3C}\phantom{.}\ce{\underset{A(2°)}{Cl}}\phantom{..................} \end{array}\] |
Only one possibility for ‘C’ compound because one CH hydrogen is present in 2- methylbutane. |
Relative amounts of A, B and C compounds = number of hydrogen × relative reactivity
| A(1°) | B(2°) | C(3°) | |
| Relative amount | 9 × 1 = 9 | 2 × 3.8 = 7.6 | 1 × 5 = 5 |
Total Amount of monohaloginated compounds = 9 + 7.6 + 5 = 21.6
Percentage of A = `9/21.6 xx 100` = 41.7%
Percentage of B = `7.6/21.6 xx 100` = 35.2%
Percentage of C = `5/21.6 xx 100` = 23.1%
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