Excess of isobutane reacts with Br2 in presence of light at 125°C gives 1 - bromo - 2 - methyl propane as major product.
Mechanism The above reaction is halogenation of alkane via free radical substitution reaction. Initiation step Br2
hv
→
Br●+Br● Propagation step
Termination step
Reactivity order of abstraction of H towards bromination of alkane as more stable alkyl free radical is formed as follows 3°H>2°H>1°H Since, isobutane is in excess, so dibromination of single isobutane is not favourable reaction. This make (a) and (b) incorrect options.