Identify `(A)` to `(F)`.
i.
ii. What would be the structure of (A) if (F) is resolvable?

Explanation

1. DU in `(A)=((2n_C+2)-n_H)/(2)=((2xx10+2)-14)/(2)=4^@`
(A) does not give precipitate with `[Cu(NH_3)_2]^(o+)` initially (shows internal alkyne), but gives precipitate with `[Cu(NH_3)_2]^(o+)` after treatment of `(A)` with `NaNH_2`. This shows shifting of `(C-=C)` bond from internal alkyne to terminal alkyne. Skeleton of `(A)` must be same with the skeleton of `(B)`. Therefore, in `(A)` there are 2 DU for one `(C-=C)` (internal alkyne), 1 DU for ring, and 1 DU for `(C-=C)` in the ring.
Location of one `(C-=C)` bond in the ring:
(F) is optically inactive `8C`-tribasic acid. Also, `2C`-fragment is lost on oxidative ozonolysis of `(E)`. `(F)` must contain at least two same groups so that `(F)` is optically inative.
Structure of `(F)` can be:

So it is optically inactive.
Three possibilities of structure `(A)` may be:

Structure (I) is possible for (A) because on oxidative cleavage of (I), the structure (IV) is same as that of (F). Structure (V) is optically active and is not same as structure (F). So structure (II) is not possible. Structure (III) is also not possible, since it is a keto diabasic acid and is not same as the structure of (F).
So the structure of `Aimplies(I)`
Reactions:

ii. If (F) is optically active, the structure of (A) would be the structure (II) as discussed above.