A colours substance 'A' `(C_(6)H_(7))` is sparingly soluble in water and gives a water soluble comound 'B' on treating with mineral acid. On reaction wih `CHCl_(3)` and alcoholic potash 'A' produces an obnoxious smell due to the formation of compound 'C'. Reaction of 'A' with benzensulphonyl chloride gives compound 'D' which is soluble in alkali. WIth `NaNO_(2)` and HCl, 'A' forms compound 'E' which reacts with phenol in alkaline medium to give an orange dye 'F'. Identify compounds 'A' to'F'.

Explanation

To solve the problem step by step, we will identify each compound based on the reactions described in the question.

### Step 1: Identify Compound 'A'
The question states that compound 'A' is a colored substance with the formula \( C_6H_7 \) that is sparingly soluble in water. Given the formula and the context, compound 'A' is likely **aniline** (C6H5NH2), which is a primary aromatic amine.

### Step 2: Reaction of 'A' with Mineral Acid
When aniline (A) is treated with a mineral acid (like HCl), it forms a water-soluble salt. The reaction is:
\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^- \]
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