1. Find degree of unsaturation

For $\mathrm{C_6H_8}$,

So, $X$ has total 3 units of unsaturation.

2. Use hydrogenation data

One mole of $X$ absorbs 2 moles of $\mathrm{H_2}$ on catalytic hydrogenation.

That means $X$ contains two reducible pi bonds. Since total DBE is 3, the remaining 1 unit must be due to a ring.

So $X$ should be a cyclic diene.

This immediately rules out:

• A: hexa-1,3,5-triene because it is acyclic and would absorb 3 moles of $\mathrm{H_2}$.

3. Now check ozonolysis condition

Ozonolysis of $X$ gives two moles of methane dicarbaldehyde.

Methane dicarbaldehyde is:

which is propanedial (malondialdehyde).

So on cleavage of the two double bonds, the molecule must split into two identical 3-carbon dialdehyde fragments.

4. Test the cyclic diene options

Option B: cyclohexa-1,4-diene

Structure has a 6-membered ring with double bonds at $1,2$ and $4,5$.

On ozonolysis, each double bond is cleaved. Breaking both double bonds splits the ring into two identical 3-carbon fragments:

Thus, it gives two moles of methane dicarbaldehyde.

Also, it has exactly two double bonds, so it uses 2 moles of $\mathrm{H_2}$.

So B satisfies both conditions.

Option C: cyclohexa-1,3-diene

Here the double bonds are conjugated at $1,2$ and $3,4$.

Ozonolysis of this structure does not produce two identical molecules of $\mathrm{OHC-CH_2-CHO}$; cleavage pattern gives different fragments because the double bonds are adjacent.

So C is incorrect.

Option D: 1-methylcyclopenta-1,4-diene

It has two double bonds, so hydrogenation part fits. But on ozonolysis, because of the methyl substituent and 5-membered ring framework, cleavage will not yield two identical molecules of methane dicarbaldehyde.

So D is incorrect.

5. Conclusion

The only structure satisfying both:

• absorption of 2 moles of $\mathrm{H_2}$,
• ozonolysis giving 2 moles of methane dicarbaldehyde,

is

So the correct option is B.