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.

Answer 1

The structure of 2-methylbutane is as follows:

Relative amounts of A, B, and C compounds = $Number of hydrogen \times Relative reactivity$

$For\quad molecule\quad A =9\times1=9$

$For\quad molecule\quad B=2\times3.8=7.6$

$For\quad molecule\quad C=1\times5=5$

Total amount of mono-halogenated compounds = 9 + 7.6 + 5 = 21.6

$Percentage\quad of\quad A=\frac { 9 }{ 21.6 }\times 100=41.7$%

$Percentage\quad of\quad B=\frac { 7.6 }{ 21.6 }\times 100=35.2$%

$Percentage\quad of\quad C=\frac { 5 }{ 21.6 }\times 100= 23.1$%

Answer 2

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%of A= 41.7
%of B = 35.2
%of C = 23.1


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