Identify the reaction type

Bromine in the presence of sunlight ($h\nu$) reacts with alkanes by free radical substitution.

So, in $2$-methylbutane, one of the hydrogen atoms is replaced by bromine.

Write the structure of $2$-methylbutane

$\text{2-methylbutane} = CH_3-CH(CH_3)-CH_2-CH_3$

Numbering the carbon chain:

$CH_3-CH(CH_3)-CH_2-CH_3$ $\ \,1\quad 2\qquad\, 3\qquad 4$

Carbon $2$ is attached to three carbons, so it is a tertiary carbon and has one tertiary hydrogen.

Find the different types of hydrogens

In $2$-methylbutane, we have:

• 1 tertiary hydrogen at carbon $2$
• 2 secondary hydrogens at carbon $3$
• 9 primary hydrogens on the three methyl groups

Use selectivity of bromination

Bromination is highly selective. The reactivity order is:

$3^\circ > 2^\circ > 1^\circ$

Therefore, bromine substitutes mainly at the tertiary hydrogen position.

Form the major product

Replacing the tertiary hydrogen at carbon $2$ by bromine gives:

$CH_3-C(Br)(CH_3)-CH_2-CH_3$

This is named:

$\text{2-bromo-2-methylbutane}$

Check options

• A: $1$-bromo-$2$-methylbutane $\rightarrow$ primary substitution, not major
• B: $2$-bromo-$2$-methylbutane $\rightarrow$ tertiary substitution, major product
• C: $2$-bromo-$3$-methylbutane $\rightarrow$ same type not corresponding to the major tertiary substitution here
• D: $1$-bromo-$3$-methylbutane $\rightarrow$ not the major product

Final answer

The main product is:

$\boxed{\text{2-bromo-2-methylbutane}}$

So the correct option is B.