State whether each of the following is true or false.
a. Photobromination of 2-methylpropane gives a mixture of 1-bromo-2-methylpropane gives a mixture of 1-bromo-2-methylpropane and 2-bromo-2-methylpropa ne in the ratio `9 : 1`
b. The percentage of n-propyl chloride obtained in the chlorination of propane is about `56%`.
c. The percentage of 1-chloro-2-methylpropane obtained in the chlorination of isobutane is about `64%`.
d. The percentage of n-propyl bromide in the bromination of propane is `44%` . The relative reactives of `3^(@)` , `2^(@)` and `1^(@)` H atoms are `1600 : 82 : 1` .

Explanation

`Br` radical, being less reactive is less infuenced by the probability factor. The bromination primarly depends on the reactivity of `H` atom which is `3^(@), 2^(@), 1^(@)`. Relative rate of bromination of `3^(@), 2^(@), 1^(@)` is `1600: 82: 1`, respecetively.
`H_(3)C - overset(CH_(3))overset(|)(CH) - CH_(3) overset(Br_(2)//hv)(rarr)`
`Br-H_(2)C - overset(CH_(3))overset(|)underset((I))(CH) CH_(3) + H_(3) C - underset((II)Br)underset(|)overset(CH_(3))overset(|)(C) - CH_(3)`
`1^(@)`H = 9, 3^(@), H = 1`, Ratio of `(II)` and `(I)`:
`((II))/(.(I)) = ("No. of" 3^(@) H)/("No. of" 1^(@)H) xx ("Reactivity of" 3^(@)H)/("Reactivity of 1^(@)H)`
`= (1xx1600)/(9xx1) = 1600//9`
Percentage of `(II) = (1600)/(1600+9) xx100 = 99.4%`
So it is clear in case of bromination percentage of `3^(@) RX (II)` will always predominate.