An alkyl halide with molecular formula `C_6H_13Cl` when treated with alcoholic KOH gave two isomeric alkenes, A and B. Reductive ozonolysis of the mixture gave the following compounds:
(i) `CH_3COCH_3` (ii) `CH_3CHO`
(iii) `CH_3 CH_2CHO and (CH_3)_2 CHCHO`.
Write the structural formula of alkyl halide.

Explanation

To solve the problem, we need to determine the structural formula of the alkyl halide with the molecular formula `C6H13Cl` that, when treated with alcoholic KOH, yields two isomeric alkenes A and B. The reductive ozonolysis of the mixture produces specific compounds. Let's break down the solution step by step.

### Step 1: Identify the Alkyl Halide
The molecular formula `C6H13Cl` indicates that we have a six-carbon alkyl halide. The presence of chlorine suggests that it is a primary or secondary alkyl halide, as tertiary halides are less likely to yield two distinct alkenes upon elimination.

### Step 2: Determine Possible Alkenes
When treated with alcoholic KOH, the alkyl halide undergoes elimination to form alkenes. The two isomeric alkenes A and B must have different structures but the same molecular formula, `C6H12`.

...