3-Methylhex-2-ene on reaction with HBr in presence of peroxide forms an addition product (A). The number of possible stereoisomers for 'A' is _______.

3-Methylhex-2-ene: $$CH_3CH=C(CH_3)CH_2CH_2CH_3$$

In the presence of peroxides, HBr adds via anti-Markovnikov (free radical) mechanism. The Br adds to the less substituted carbon of the double bond.

The product is: $$CH_3CHBr-CH(CH_3)-CH_2CH_2CH_3$$ (Br on C-2).

Let's check for stereocenters:

C-2: $$CHBr$$ bonded to $$CH_3$$, $$H$$, $$Br$$, and $$CH(CH_3)CH_2CH_2CH_3$$ — this is a chiral center.

C-3: $$CH(CH_3)$$ bonded to $$CH_3$$, $$H$$, $$CHBrCH_3$$, and $$CH_2CH_2CH_3$$ — this is a chiral center.

With 2 chiral centers, the maximum number of stereoisomers = $$2^2 = 4$$.

Since the two chiral centers have different substituents (they are not equivalent), all 4 stereoisomers are distinct (no meso form).

The number of possible stereoisomers is $$\boxed{4}$$.