An organic compound (A) of molecular formula `C_(5)H_(8)` when treated with Na in liquid ammonia followed by reaction with -Propyl iodide yeidls (B) `C_(8)H_(12)` (A) gives a ketone `C_(5)H_(10)O` (e) when treated with dil `H_(2)SO_(4)` and `HgSO_(4)` (B) on oxidation with alkaline `KMnO_(4)` gives two isomeric acids (D) and (E) `C_(4)H_(8)O_(2)`. Give structures of compounds (A) to (E) with proper reasoning .

Explanation

To solve the problem step by step, we will identify the compounds A to E based on the given information.

### Step 1: Determine the structure of compound (A) with the molecular formula `C5H8`.

Given that the molecular formula is `C5H8`, we can deduce that it is an alkyne because it satisfies the formula `C_nH_(2n-2)` for alkynes. The simplest alkyne with five carbons is 1-pentyne or 2-pentyne. However, since the reaction with sodium in liquid ammonia suggests a terminal alkyne, we will consider **1-pentyne** (CH≡C-CH2-CH2-CH3) as compound (A).

**Structure of (A):**
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