1. Write the structure of the alkene

3-Methyl-pent-2-ene has the structure:

$\mathrm{CH_3-CH=C(CH_3)-CH_2-CH_3}$

Numbering:

• C-2 and C-3 are the double-bond carbons.
• A methyl group is attached at C-3.

2. Reaction with HBr in presence of peroxide

In the presence of peroxide, HBr adds by the free-radical (anti-Markovnikov) mechanism.

So, in general:

• $\mathrm{Br}$ adds to the less substituted double-bond carbon.
• $\mathrm{H}$ adds to the more substituted double-bond carbon.

For

$\mathrm{CH_3-CH=C(CH_3)-CH_2-CH_3}$

$\mathrm{Br}$ goes to C-2 and $\mathrm{H}$ goes to C-3.

Hence the product is:

$\mathrm{CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3}$

This is 2-bromo-3-methylpentane.

3. Check for stereocentres

Now examine the product:

$\mathrm{CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3}$

• C-2 is attached to:

  • $\mathrm{Br}$
  • $\mathrm{H}$
  • $\mathrm{CH_3}$
  • $\mathrm{CH(CH_3)CH_2CH_3}$

  These are four different groups, so C-2 is chiral.

• C-3 is attached to:

  • $\mathrm{H}$
  • $\mathrm{CH_3}$
  • $\mathrm{CH(Br)CH_3}$
  • $\mathrm{CH_2CH_3}$

  These are also four different groups, so C-3 is chiral.

Thus, the product has 2 stereogenic centres.

4. Count the possible stereoisomers

Maximum possible stereoisomers for 2 chiral centres:

$2^2 = 4$

Now check for meso form.

A meso form is possible only if there is an internal plane of symmetry. Here the molecule is:

$\mathrm{CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3}$

The two stereocentres are not symmetrically equivalent because the ends/groups are different, so no meso form exists.

Therefore total stereoisomers $=4$.

5. Evaluate the options

• A: Zero — incorrect
• B: Two — incorrect
• C: Four — correct
• D: Six — incorrect

Therefore, the correct answer is:

$\boxed{4}$

So the correct option is C.