Identify (A) and (D).
a.
b. What would be the structure of (A), if on reaction with `(Na+EtOH)` or `(H_2+Ni_2B)`, it gives an optically active compound?

Explanation

a. DU in (A)`=((2xx8+2)-12)/(2)=3^@`
DU in (B)`=((2xx8+2)-18)/(2)=0^@`
Reaction of (A) with `(H_2+Ni+B)` shows that it has `(-C-=C-)` bond, and it is internal alkyne because (A) does not react with `[Ag(NH_3)_2]^(o+)` (shows the absence of terminal alkyne).
Zero degree of unsaturation in (B) suggests that there is no ring in (A) or (B).
Therefore in (A), there is one `(C-=C)` bond (2 DU) and one `(C=C)` (1 DU).
(A) is optically active, containing one internal alkyne group, one `(C=C)` group, one H and one Me group, also, it is a compound with total eight C atoms.
Possibility of (A):


Therefore, (A) is `2Z-4`-Methylhept-2-en-5-yne.
Reactions:

b. In order to get an optically active compound, regardless of the method of reducing `(C-=C)` bond (syn- or anti-addition, either by `Na+EtOH` or by `H_2+Ni_2B`), position of `(C=C)` should be the terminal one.
The possibility of A: