Question

Calculate the percentage of all the monochlorinated products obtained from the chlorination of $2$ - methyl butane. The relative reactivity of $1^{{}^{\circ }}$ , $2^{{}^{\circ }}$ , and $3^{{}^{\circ }}$ hydrogen to chlorination is $(1:3.8:5)$.

• A. 15%, 45%, 20%, 25%
• B. 14%, 35%, 23%, 28%
• C. 12%, 30%, 22%, 29%
• D. 17%, 36%, 20%, 23%

Answer 1

Answer: (B)

There are four different (A, B, C and D) monochloro products.

			Relative amount
(A)		(A) is obtained from the reaction of three equivalent $1^{{}^{\circ }}$ H atoms.	$3\times 1=3.0$
(B)		(B) is obtained from the reaction of two equivalent $2^{{}^{\circ }}$ H atoms.	$2\times 3.8=7.6$
(C)		(C) is obtained from the reaction of one equivalent $3^{{}^{\circ }}$ H atoms.	$1\times 5=5.0$
(D)		(D) is obtained from the reaction of six equivalent $1^{{}^{\circ }}$ H atoms.	$6\times 1=6.0$ Total = 21.6

Therefore, percentages of the A, B, C and D monochloro products are as follows :
$A=\frac{3}{21.6}\times 100=13.9\approx 14\%$
$B=\frac{7.6}{21.6}\times 100=35.18\approx 35\%$
$C=\frac{5.0}{21.6}\times 100=23.15\approx 23\%$
$D=\frac{6.0}{21.6}\times 100=27.8\approx 28\%$