Out of the following, the alkene that exhibits optical isomerism is

To recognise optical (enantiomeric) isomerism in an alkene, look for at least one $$sp^3$$‐hybridised carbon atom that is attached to four different groups. Such a carbon is called a chiral (asymmetric) centre. The $$sp^2$$ carbons in the $$C=C$$ bond can never be chiral, so the requisite chiral centre must lie elsewhere in the molecule.

Option A : 3-methyl-2-pentene
Structure: $$CH_3CH=C(CH_3)CH_2CH_3$$.
Every $$sp^3$$ carbon here has at least two identical groups attached (either two hydrogens or two alkyl fragments that are identical when traced out), so no carbon is chiral. Hence no optical isomerism.

Option B : 4-methyl-1-pentene
Structure: $$CH_2=CHCH_2CH(CH_3)CH_3$$.
The only $$sp^3$$ carbon with four substituents is the fourth carbon, $$CH(CH_3)$$. Its four attachments are H, $$CH_3$$, $$CH_3$$ (through the straight chain) and $$CH_2CH=CH_2$$. Two of them are identical (both $$CH_3$$ fragments), so the carbon is achiral. No optical activity.

Option C : 3-methyl-1-pentene
Structure: $$CH_2=CHCH(CH_3)CH_2CH_3$$.
Examine the third carbon (marked with *):
$$CH_2=CH\text{-}C^*(H)(CH_3)\text{-}CH_2CH_3$$.
Its four different groups are:
1. H
2. $$CH_3$$ (methyl side chain)
3. $$CH_2CH_2CH_3$$ (propyl fragment towards the chain end)
4. $$CH_2=CH$$ (allyl fragment towards the double bond)
Because all four groups differ, this carbon is chiral, giving rise to a pair of non-superimposable mirror images. Therefore 3-methyl-1-pentene exhibits optical isomerism.

Option D : 2-methyl-2-pentene
Structure: $$CH_3C(CH_3)=C(CH_3)CH_3$$.
Every $$sp^3$$ carbon again has at least two identical substituents (two hydrogens or two methyl/ethyl fragments). No chiral centre, hence no optical isomerism.

Thus, the only alkene among the four that contains a chiral carbon is 3-methyl-1-pentene.

Final answer: Option C which is: 3-methyl-1-pentene.