The isomeric deuterated bromide with molecular formula C$$_4$$H$$_8$$DBr having two chiral carbon atoms is

We need to find the isomeric deuterated bromide with molecular formula $$C_4H_8DBr$$ that has two chiral carbon atoms.

We start by noting that

Option 1: 2-Bromo-1-deuterobutane ($$CH_2D-CHBr-CH_2-CH_3$$)

C-1 is bonded to: H, H, D, and $$CHBr-CH_2CH_3$$. Since C-1 has two identical H atoms, it is not a chiral centre.

C-2 is bonded to: $$CH_2D$$, Br, H, and $$CH_2CH_3$$. These are four different groups, so C-2 is chiral.

This compound has only 1 chiral centre. Not the answer.

Option 2: 2-Bromo-2-deuterobutane ($$CH_3-CBrD-CH_2-CH_3$$)

C-2 is bonded to: $$CH_3$$, Br, D, and $$CH_2CH_3$$. These are four different groups, so C-2 is chiral.

No other carbon has four different groups attached.

This compound has only 1 chiral centre. Not the answer.

Option 3: 2-Bromo-3-deuterobutane ($$CH_3-CHBr-CHD-CH_3$$)

C-2 is bonded to: $$CH_3$$, H, Br, and $$CHD-CH_3$$. These are four different groups, so C-2 is chiral.

C-3 is bonded to: $$CH_3$$, H, D, and $$CHBr-CH_3$$. Since D and H are different atoms (deuterium is an isotope of hydrogen), these are four different groups, so C-3 is chiral.

This compound has 2 chiral centres. This is our answer.

Option 4: 2-Bromo-1-deutero-2-methylpropane ($$CH_2D-CBr(CH_3)-CH_3$$)

C-2 is bonded to: $$CH_2D$$, Br, $$CH_3$$, and $$CH_3$$. Since two of the groups ($$CH_3$$) are identical, C-2 is not chiral.

This compound has 0 chiral centres. Not the answer.

Deuterium (D) and hydrogen (H) are different isotopes and are treated as different substituents for chirality analysis. This makes C-3 in 2-bromo-3-deuterobutane a chiral centre.

The correct answer is Option (3): 2-Bromo-3-deuterobutane.