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 will identify the compounds A, B, C, and D step by step, along with the reactions involved.

### Step 1: Identify Compound A
Given that A is an alkyl halide with the formula C5H11Br, we can deduce that it is a bromoalkane. A common structure for this compound could be 1-bromopentane (Br-CH2-CH2-CH2-CH3).

### Step 2: Reaction of A with Ethanolic KOH
When 1-bromopentane (A) reacts with ethanolic KOH, an elimination reaction occurs, leading to the formation of an alkene (B) through dehydrohalogenation. The reaction can be represented as follows:

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