An organic compound (A) of molecular formula `C_5H_8` when treated with sodium in liquid ammonia followed by reaction with n-propyl bromide yields (B), `C_8H_14`. (A) gives a ketone ©, `C_5H_10O` when treated with dilute `H_2SO_4` and `HgSO_4`. (B) on oxidation with alkaline `KMnO_4` gives two isomeric acids (D) and (E) `C_4H_8O_2`. Give structures of compound (A) to (E) with proper reasoning.

Explanation

(i) (A) reacts with sodium in liquid ammonia and thus it is terminal alkyne i.e. `C_3H_7C-=C-H`
`C_3H_7C-=CHoverset(Na)rarrC_3H_7C-Naoverset(CH_3=CH_2-CH_2-Br)rarrC_3H_7C-=C-underset((B))(CH_2)-CH_2-CH_2-CH_3`
(iii) `C_3H_7C-=CHoverset(HgSO_4)underset(H_2SO_4)rarrC_3H_7CO-CH_3`
(iv) `C_3H_7C-=C-CH_2-CH_2-CH_3overset([O])underset(KMnO_4)rarrCH_3-CH_2underset((D))-CH_2-COOH+underset((E))(C_3H_7COOH)`
Since (D) and (E) are isomers, thus the structure of (E) is