1. Condition for optical isomerism
   An organic compound shows optical isomerism if it has a chiral carbon (an asymmetric carbon attached to four different groups) and lacks an internal plane of symmetry.

2. Check each alkene option

Option A: 3-methyl-2-pentene

Structure:

At carbon-3, the double-bonded carbon is $sp^2$ hybridized, so it cannot be chiral.
No other carbon has four different substituents.

So, A is not optically active.

Option B: 4-methyl-1-pentene

Structure:

Now check carbon-4: It is attached to:

• $\text{H}$
• $\text{CH}_3$
• another $\text{CH}_3$
• $\text{CH}_2-\text{CH}=\text{CH}_2$

Since two substituents are identical ($\text{CH}_3$ and $\text{CH}_3$), this carbon is not chiral.

So, B is not optically active.

Option C: 3-methyl-1-pentene

Structure:

Check carbon-3: It is attached to:

• $\text{H}$
• $\text{CH}_3$
• $\text{CH}_2\text{CH}_3$
• $\text{CH}=\text{CH}_2$

These are four different groups, so carbon-3 is a chiral center.

Hence, C exhibits optical isomerism.

Option D: 2-methyl-2-pentene

Structure:

The double-bond carbon is $sp^2$ hybridized and cannot be chiral.
No other carbon has four different substituents.

So, D is not optically active.

3. Final conclusion Only 3-methyl-1-pentene has a chiral carbon and therefore shows optical isomerism.

4. Comparison with stored correct answer
   Stored correct answer = C
   Derived answer = C
   So, they agree.