3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

We have the alkene 3-methyl-pent-2-ene, whose condensed form may be written as $$CH_3-CH=C(CH_3)-CH_2-CH_3$$. The double bond is between carbon-2 and carbon-3.

In the presence of peroxide, the addition of $$HBr$$ follows the anti-Markovnikov (peroxide) rule. Stated in words, this rule says that in a radical addition of $$HBr$$, the bromine radical adds first to the double-bond carbon that will give the more stable carbon radical; afterwards that radical abstracts a hydrogen from another $$HBr$$ molecule. Hence the final product places bromine on the less substituted alkene carbon.

Applying the rule, let us compare the two possibilities:

• If $$Br\cdot$$ added to C-3, the radical would reside on C-2, a secondary radical.
• If $$Br\cdot$$ added to C-2, the radical would reside on C-3, a tertiary radical (more stable).

Because the tertiary radical is preferred, $$Br\cdot$$ adds to C-2 and the radical forms on C-3; this radical then captures $$H\cdot$$ from another $$HBr$$. So the net result is

$$CH_3-CH=C(CH_3)-CH_2-CH_3 \;\xrightarrow[\text{peroxide}]{HBr}\; CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3$$

The IUPAC name of the product is 2-bromo-3-methylpentane.

Now we examine stereochemistry. In the product the only π bond has been converted to σ bonds, so any stereoisomerism will arise solely from chiral (asymmetric) carbon atoms.

Checking each carbon:

• Carbon-2 bears $$Br$$, $$H$$, $$CH_3$$ (left side), and $$CH(CH_3)-CH_2-CH_3$$ (right side). The four substituents are all different, therefore carbon-2 is a chiral centre.

• Carbon-3 bears $$H$$, $$CH_3$$ (its methyl substituent), $$CH_2-CH_3$$ (to the right), and $$CH(Br)-CH_3$$ (to the left). Again the four groups are different, so carbon-3 is also a chiral centre.

• All other carbons possess at least two identical substituents and are achiral.

Thus the molecule contains $$n = 2$$ stereogenic centres. The maximum number of stereoisomers obtainable from $$n$$ independent chiral centres is given by the formula $$2^{\,n}$$.

Applying the formula, $$2^{\,2} = 4$$ possible stereoisomers are predicted: the pairs $$(R,R)\;,\;(S,S)\;,\;(R,S)\;,\;(S,R)$$. No internal mirror plane exists in the molecule, so none of these coincide or collapse into a meso form, and all four are distinct.

Hence, the correct answer is Option C.