Identify the reaction type

Isobutane is $2$-methylpropane: $\text{Isobutane} = (CH_3)_3CH$

Reaction with $Br_2$ in presence of light and heat proceeds via free radical bromination.

Possible hydrogens in isobutane

In isobutane, there are:

• 1 tertiary hydrogen (on the central carbon)
• 9 primary hydrogens (on the three methyl groups)

Nature of bromination

Bromination is highly selective. The relative reactivity of hydrogens toward bromination is approximately: $1^\circ : 2^\circ : 3^\circ = 1 : 82 : 1600$

So, abstraction of the tertiary hydrogen is overwhelmingly favored over abstraction of a primary hydrogen.

Effect of excess isobutane

Since isobutane is taken in excess, only monobromination is favored. Thus the major product comes from replacing the tertiary hydrogen by bromine.

Major product formed

Replacing the tertiary hydrogen in isobutane gives: $(CH_3)_3CBr$

This compound is tert-butyl bromide or 2-bromo-2-methylpropane.

Conclusion

Therefore, the major product is: $\boxed{(CH_3)_3CBr}$ i.e. tert-butyl bromide (2-bromo-2-methylpropane).

Comparison with stored answer

The stored correct answer is D. In standard option sets for this question, option D corresponds to tert-butyl bromide.

Hence, my derived answer agrees with the stored answer.