If relative rates of substitution of 1^(@) and 2^(@) H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane.

If relative rates of substitution of 1^(@) and 2^(@) H are in the rati

1.	If relative rates of substitution of 1^(@) and 2^(@) H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane.
Answer» Solution :According to the reaction : `underset("n-Butane")(CH_(3)CH_(2)CH_(2)CH_(3))underset("light")overset(Cl_(2), 298K)to underset("2-Chlorobutane")(CH_(3)-underset(Cl)underset(\|)CH-CH_(2)CH_(3)) + underset("1-Chlorobutane")(CH_(3)CH_(2)CH_(2)CH_(2)Cl)` The RELATIVE rates of two isomeric chlorobutanes will be equal to their number of types of H.s (`1^(@), 2^(@)` or `3^(@)`) and their relative rates of substitution. `("1-chlorobutane")/("2-chlorobutane")= ("No. of" 1^(@)H)/("No. of" 2^(@)H) xx ("Reactivity of" 1^(@) H)/("Reactivity of" 2^(@)H)` `=(6)/(4) xx (1)/(3.8)=(6)/(15.2)` Now, if x is the percentage of 1-chlorobutane, then Percentage of 2-chlorobutane = 100 - X `:. (x)/(100-x)=(6)/(15.2)` 15.2 x = 600 - 6X 21.2 x = 600 x = 28 % `:.` 1-Chlorobutanee = 28% and 2-chlorobutane = 72%