Excess of isobutane on reaction with Br$$_2$$ in presence of light at 125°C gives which one of the following, as the major product?

We have isobutane, whose condensed formula is $$(CH_3)_3CH$$ and whose structural skeleton looks like a central carbon attached to three methyl groups and one hydrogen.

The reagent is molecular bromine, $$Br_2$$, and the reaction conditions are light and a temperature of about $$125^\circ{\rm C}$$. Under these conditions bromine undergoes homolytic fission giving two bromine radicals: $$Br_2 \;\xrightarrow{h\nu,\,125^\circ{\rm C}}\; 2\,Br^{\bullet}$$. Thus the reaction proceeds by the free-radical halogenation mechanism.

Radical bromination shows a very strong preference for abstracting a hydrogen from the carbon that forms the most stable radical; the order of stability (and hence of ease of abstraction) is

$$3^{\circ}\,H \;>\; 2^{\circ}\,H \;>\; 1^{\circ}\,H$$

In isobutane there are:

• one tertiary hydrogen on the central carbon, and
• nine primary hydrogens on the three methyl groups.

Although there are many more primary hydrogens, the rate of abstraction from the tertiary position is so much higher that the tertiary radical is formed predominantly. We write the propagation step producing that radical:

$$(CH_3)_3CH \;+\; Br^{\bullet}\;\longrightarrow\;(CH_3)_3C^{\bullet}\;+\;HBr$$

The tertiary carbon radical then rapidly combines with another bromine molecule:

$$(CH_3)_3C^{\bullet}\;+\;Br_2\;\longrightarrow\;(CH_3)_3CBr\;+\;Br^{\bullet}$$

Because an excess of isobutane is present, the concentration of $$Br_2$$ is kept low, so further substitution (di- or tri-bromination) is statistically suppressed. Therefore the chief product is tert-butyl bromide, whose condensed structural formula is:

$$CH_3-C(Br)(CH_3)_2$$

This structure corresponds exactly to Option D.

Hence, the correct answer is Option D.