'A' and 'B' respectively are
A $$\xrightarrow[(2) Zn-H_2O]{(1) O_3 \atop}$$ Ethane-1,2-dicarbaldehyde + Glyoxal/Oxaldehyde
B $$\xrightarrow[(2) Zn-H_2O]{(1) O_3 \atop}$$ 5-oxohexanal

We need to find compounds A and B based on the ozonolysis products.

For compound A:

A on ozonolysis gives ethane-1,2-dicarbaldehyde ($$OHC-CH_2-CH_2-CHO$$) and glyoxal ($$OHC-CHO$$).

In ozonolysis of cyclic dienes, each double bond is cleaved. For a six-membered ring with two double bonds:

Cyclohex-1,3-diene has double bonds at C1=C2 and C3=C4 positions. On ozonolysis followed by reductive workup ($$Zn/H_2O$$):

- Cleavage at C1=C2 and C3=C4 gives two fragments.

- The fragment from C2-C3 (single bond between the two double bonds) gives glyoxal: $$OHC-CHO$$

- The fragment from C4-C5-C6-C1 gives $$OHC-CH_2-CH_2-CHO$$ (ethane-1,2-dicarbaldehyde)

So compound A is cyclohex-1,3-diene.

For compound B:

B on ozonolysis gives 5-oxohexanal ($$CH_3-CO-CH_2-CH_2-CH_2-CHO$$).

Since only one product is formed, B must be a cyclic compound with one double bond. On ozonolysis, the ring opens at the double bond to give a single dicarbonyl compound.

5-oxohexanal has the structure: $$CH_3-CO-CH_2-CH_2-CH_2-CHO$$

The two carbonyl carbons (C1 aldehyde and C2 ketone at C5 position counted from the aldehyde end) were originally connected by a double bond in the ring.

Reconnecting these carbons: we get a 5-membered ring with a methyl substituent. The compound is 1-methylcyclopent-1-ene.

Verification: 1-methylcyclopent-1-ene has the double bond between C1 (bearing $$CH_3$$) and C2. Ozonolysis cleaves this to give a ketone at C1 ($$-CO-CH_3$$) and an aldehyde at C2 ($$-CHO$$), with the remaining chain $$-CH_2-CH_2-CH_2-$$ connecting them, giving $$CH_3CO-CH_2CH_2CH_2-CHO$$ = 5-oxohexanal. This is correct.

Therefore, A is cyclohex-1,3-diene and B is 1-methylcyclopent-1-ene.

Hence, the correct answer is Option C.