The major product obtained in the photo catalysed bromination of 2-methylbutane is:

The photo-catalysed bromination of 2-methylbutane is a free radical substitution reaction. Under UV light, bromine molecules ($$Br_2$$) undergo homolytic cleavage to form bromine radicals ($$Br^\bullet$$). These radicals abstract hydrogen atoms from the alkane, leading to alkyl radicals. The stability of the alkyl radical determines the major product, as bromination is selective for the most stable radical.

The structure of 2-methylbutane is $$CH_3-CH(CH_3)-CH_2-CH_3$$. We identify the types of hydrogen atoms:

• The hydrogen on the carbon at position 2 (the carbon with the methyl group) is tertiary (3°), as this carbon is bonded to three other carbons. There is one such hydrogen.
• The hydrogens on the carbon at position 3 ($$CH_2$$) are secondary (2°). There are two such hydrogens.
• The hydrogens on the terminal methyl groups (positions 1, 4, and the methyl branch on carbon 2) are primary (1°). Specifically:
  • Position 1 ($$CH_3$$- attached to carbon 2): 3 hydrogens.
  • The methyl branch on carbon 2: 3 hydrogens.
  • Position 4 ($$CH_3$$): 3 hydrogens.
  Total primary hydrogens: 9.

The stability order of alkyl radicals is tertiary > secondary > primary. Bromine radicals preferentially abstract the tertiary hydrogen due to its higher stability, even though there is only one tertiary hydrogen compared to multiple primary and secondary hydrogens.

Abstraction of the tertiary hydrogen at carbon 2:

This forms a tertiary alkyl radical at carbon 2. This radical then reacts with a bromine molecule:

The product is 2-bromo-2-methylbutane ($$CH_3-CBr(CH_3)-CH_2-CH_3$$).

Comparing with the options:

• Option A: 1-bromo-2-methylbutane ($$BrCH_2-CH(CH_3)-CH_2-CH_3$$) is a primary bromide.
• Option B: 1-bromo-3-methylbutane likely refers to $$CH_3CH(CH_3)CH_2CH_2Br$$, which is also a primary bromide.
• Option C: 2-bromo-3-methylbutane ($$CH_3CHBrCH(CH_3)CH_3$$) is a secondary bromide.
• Option D: 2-bromo-2-methylbutane ($$CH_3CBr(CH_3)CH_2CH_3$$) is a tertiary bromide.

Due to the high selectivity of bromine for tertiary hydrogens (relative rate approximately 1600 times that of primary), the major product is the tertiary bromide, 2-bromo-2-methylbutane.

Hence, the correct answer is Option D.