JEE Main 26-Aug-2021 Shift 1 Solved Paper

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.