Calculate the percentage of compounds obtained by monobromination of isobutane. The relative reactivity of $1, 2, 3$ H atoms to bromination is $1 : 82 : 1600.$

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Solution

Reactivity order of bromination = $3 : 2 : 1 = 1600 : 82 : 1$

(A) is obtained from the reaction of nine equivalent 1 H atoms.

(B) is obtained from the reaction of one 3 H atom.

Percentage of (A) (isobutyl bromide) = $\frac{9100}{1609}$ = $0.56 %$
Percentage of (B) (tert-butyl bromide) = $\frac{1600100}{1609}$ = $99.4$
Ratio = $177.7 : 1$

Formation of $t - b u t y l$ bromide is $177.7$ times to that of Isobutyl bromide.

Relative amount

$91 = 9.0$

$16001 = 1600.0$

Total = 1609.0

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