Question

Photobromination of 2-methyl propane gives a mixture of 1-bromo-2-methyl propane and 2-bromo-2-methyl propane in the ratio:

• A. 99:1
• B. 1:99
• C. 1:1
• D. 3:1

Answer 1

The correct option is B 1:99

$\stackrel{\bullet }{Br}$ radical, being reactive is less influenced by the probability factor. The bromination primarily depends on the reactivity of H atoms which is $3^o>2^o>1^o$ Relative rate of bromination of $3^o,2^o$ and $1^o$ is $1600:82:1$ respectively
$1^{o}H=9,3^{o}H=1$, Ration of $\left(II\right)$ and $\left(I\right)$
$\frac{\left(II\right)}{\left(I\right)}=\frac{No.of{3}^{o}H}{No.of{1}^{o}H}\times \frac{Reactivityof{3}^{o}H}{Reactivityof{1}^{o}H}$
$=\frac{1\times 1600}{9\times 1}=1600/9$
Percentage of $\left(II\right)=\frac{1600}{1600+9}\times 100=99.4$
So it is clear in case of bromination percentage of $3^{o}RX\left(II\right)$ will always predominate.