a. Write all the possible structures of `(A)` and `(B)`.
b. Write all the possible strutucres of `(A)` and `(B)`, if both are resolvable.

Explanation

`a.` Degree of unsaturation in `(A):`
`implies ((2n_(C)+2)-n_(H))/(2)=(22-16)/(2)=3^(@)`
Degree of unsaturation in `(B):`
Discard oxygen and find degree
`implies ((2n_(C)+2)-n_(H))/(2)=(12-8)/(2)=2^(@)`
Number of `C` atome in ` (A)=10`
Number of `C` atoms `(B)=5`.
Therefore, `2 mol` of `(B)` have been formed.
`2^(@)` of unsaturation in `(B)` means one keto group and one ring. If aldehyde group is obtained in reductive ozonolysis, then in oxidative ozonolysis a different product. i.e., acid will be obtained.
However, if a keto group is obtained in reductive ozonolysis, then same keto group will be obtained in oxidative ozonolysis. Accordingly, the possible structures of `(B)` and, hence, `(A)` may be `:`

Since on hydrogenation only `1 mol` of `H_(2)` will be required, this structure is not possible because cyclopentane ring is stable and does not undergo reduction at `120^(@)C`. Similarly, four `-` membered ring with one methyl group is also not possible because cyclobutane ring is also stable and does not undergo reduction at `120^(@)C`.
Other possiblities`:`

All the structures `B_(1),B_(2)B_(3),A_(1),A_(2),A_(3),C_(1),C_(2)`and `C_(3)` are not possible.
Only possibility left is alkene with cyclopropane ring.
The alkene of the type
Compound `(A)` can be `:`

`(C) ` is obtained when `(C_(1)-C_(3))` bond breaks
`(C_(1))` is obtained when `(C_(2)-C_(3))` bond breaks during hydrogenation of `A`.
Both `(A)` and `(B)` are non`-` resolvable `(` optically inactive `)`. Other possible structures for `(A)` can be as follows `:`
Four structures of `(A)` are possible if both `(A)` and `(B)` are non`-` resolvable.

`b.` If both `(A)` and `(B)` are resolvable `(` optically active `)` then,