`0.450 gm` of an aromatic organic compound `(A)` on ignitin gives `0.905 gm CO_(2)` and `0.15gm H_(2)O. 0.350gm` of `(A)` on boiling with HNO_(3) and adding AgCl. The vapour density of (A) is 87.5. On hydrolysis with `Ca(OH)_(2)` yields `(B)` which on mild reduction gives an optically active compound `(C)`. On heating `(C)` with `I_(2)` and `NaOH`, iodofrom is produced along with `(D)`. With `HCl, (D)` gives a solid, which is more soluble in hot water than in cold. Idenify `(A)` to `(D)` with proper explanaion.

Explanation

To solve the problem step by step, we need to identify the compounds (A) to (D) based on the provided information. Here’s a structured solution:

### Step 1: Calculate the percentage composition of compound (A)

1. **Calculate the percentage of Carbon (C)**:
- From the combustion data, we know that 0.905 g of CO₂ is produced.
- Molar mass of CO₂ = 44 g/mol (12 g/mol for C + 32 g/mol for O).
- Moles of CO₂ = 0.905 g / 44 g/mol = 0.02059 mol.
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