2-methyl butane on reacting with bromine in the presence of sunlight gives mainly

2-bromo-2-methyl butane

The correct answer is option 2. 2-bromo-2-methyl butane.

The reaction of 2-methylbutane with bromine in the presence of sunlight is a free radical halogenation reaction. This type of reaction preferentially replaces hydrogen atoms attached to the most substituted carbon atoms due to the stability of the resulting free radicals.

Structure of 2-Methylbutane:

Possible Bromination Sites:

Primary hydrogens: Attached to the carbon atoms at the ends (1-bromo products)

Secondary hydrogen: Attached to the central carbon atom (2-bromo products)

Stability of Radicals: Tertiary radical > Secondary radical > Primary radical

In 2-methylbutane, the most substituted carbon atom is the one bonded to the two methyl groups (the central carbon), which would form a tertiary radical if it loses a hydrogen atom.

Reaction:

Formation of a tertiary radical: Removal of a hydrogen atom from the tertiary carbon

This would form 2-bromo-2-methylbutane as the major product because the tertiary radical is the most stable. This product forms because the tertiary radical is the most stable intermediate in the free radical bromination process, leading to bromine substitution at the tertiary carbon.