1. Idea: A molecule can show stereoisomerism if it has either:
   • a geometrical isomerism element (like $E/Z$ or cis/trans around a double bond), or
   • an optical isomerism element (a chiral center).

We check each option.

2. Option A: 3,4-Dimethylhex-3-ene

Structure around the double bond: $\text{CH}_3\text{CH}_2\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3$

At carbon-3 of the double bond, the two groups are:

• $\text{CH}_3$
• $\text{C}_2\text{H}_5$

At carbon-4 of the double bond, the two groups are:

• $\text{CH}_3$
• $\text{C}_2\text{H}_5$

Since each double-bond carbon has two different groups, geometrical isomerism ($E/Z$) is possible.

So, A shows stereoisomerism.

3. Option B: 3-Methylhex-1-ene

Structure: $\text{CH}_2=\text{CH}-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3$

Check geometrical isomerism:

• At carbon-1 of the double bond, it is $\text{CH}_2=$, so one carbon has two identical H atoms.
• Therefore, no geometrical isomerism.

Check chirality at carbon-3: Carbon-3 is attached to

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

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

Hence, B shows optical isomerism, so it shows stereoisomerism.

4. Option C: 3-Ethylhex-3-ene

Structure: $\text{CH}_3\text{CH}_2-\text{C}(\text{C}_2\text{H}_5)=\text{CH}-\text{CH}_2\text{CH}_3$

More clearly, at carbon-3 of the double bond, the attached groups are:

• left side chain: $\text{C}_2\text{H}_5$
• substituent: $\text{C}_2\text{H}_5$

So carbon-3 has two identical ethyl groups attached.

For geometrical isomerism, each double-bond carbon must have two different groups. That fails here. Thus, no geometrical isomerism.

Now check optical isomerism:

• There is no chiral center, because no tetrahedral carbon has four different groups.

Hence, C does not show stereoisomerism.

5. Option D: 4-Methylhex-1-ene

Structure: $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$

Check geometrical isomerism:

• Double bond is terminal ($\text{CH}_2=\text{CH}-$), so one alkene carbon has two H atoms.
• Therefore, no geometrical isomerism.

Check chirality at carbon-4: Carbon-4 is attached to

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

These are four different groups, so carbon-4 is a chiral center.

Hence, D shows optical isomerism.

6. Conclusion

Only option C does not show any stereoisomerism.

Therefore, the correct answer is: $\boxed{\text{C}}$

7. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They agree.