An alkyl halide `C_(5)H_(11)` (A) reacts with ethanolic KOH to give an alkene 'B' which reacts with `Br_(2)` to give a compound 'C' which on dehydromination gives an alkyne 'D' . On treatment with sodium metal in liquid ammonia one mole of 'D' give one mole of the sodium salt of 'D' and half a mole of hydrogen gas . Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D . Give the the reactions involved.

Explanation

To solve the problem, we need to identify the compounds A, B, C, and D based on the given reactions and molecular formulas. Let's break it down step by step.

### Step 1: Identify Compound A
- **Given:** An alkyl halide with the formula `C5H11X`.
- **Structure of A:** This compound can be a straight-chain or branched alkyl halide. For simplicity, let's consider it as 1-bromopentane (C5H11Br).

### Step 2: Reaction of A with Ethanolic KOH
- **Reaction:** `C5H11Br + KOH (ethanolic) → C5H10 + HBr`
...