Hydrocarbon (X), `C_(7)H_(12)`, on reaction with boron hydride followed by treatment with `CH_(3)COOH` yields (A). On reductive ozonolysis (A) yields a mixture of two aldehydes, (B) and (C ). Of these, only (B) can undergo Cannizzaro reaction. (A) exists in two geometrical isomer, `(A-1)` and `(A-2)`, of which `(A-2)` is more stable. Gives structures of (X), (A), (B), (C ), `(A-1)`, and `(A-2)` with proper reasoning.

Explanation

Form formula `C_(7)H_(12)`. It seems to be an alkyne, which forms alkene on hydroboration. Since, alkene forms two aldehydes B and C on reductive ozonolysis. Out of which B can undergo Cannizzaro's reactive. It means that aldehyde B has no `alpha`- h atom and the -CHO group is linked to tertiary butyl group. Hence hydrocarbon X has to be
`CH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-C-=C-CH_(3) overset((i) B_(2)H_(6))underset((ii) CH_(3)COOH) to CH_(3) -underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH=CH-CH_(3)overset("Reductive")underset("ozonolysis") to CH_(3) - overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-CHO+O= CH-CH_(3)`
Compound B has no `alpha`- H atom hence it undergoes Cannizzaro's reaction.
The alkene A gives two geometrical isomers, these are cis and trans.