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

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