Question

The dibromo compound \(P\) (molecular formula: \(C_9H_{10}Br_2\)) when heated with excess sodamide followed by treatment with dilute HCl gives \(Q\). On warming \(Q\) with mercuric sulphate and dilute sulphuric acid yields \(R\), which gives a positive iodoform test but a negative Tollens' test. The compound \(P\) is:

• A. A
• B. B
• C. C
• D. D

Hint:

Alkynes give ketones with \(HgSO_4/H_2SO_4\); a positive iodoform test confirms a methyl ketone.

Answer 1

The Correct Option is C

Step 1: Effect of excess sodamide.
Heating a vicinal or geminal dibromide with excess sodamide leads to double dehydrohalogenation, forming an alkyne. Hence, compound \(P\) must be such that it can form an alkyne \(Q\).

Step 2: Reaction with \(HgSO_4/H_2SO_4\).
Hydration of an alkyne in the presence of mercuric sulphate and dilute sulphuric acid gives a ketone via enol–keto tautomerism.

Step 3: Analysis of tests on product \(R\).

Positive iodoform test indicates the presence of a \(\ce{CH3CO-}\) group.
Negative Tollens' test confirms the absence of an aldehyde group. Thus, \(R\) must be a methyl ketone.

Step 4: Identify the correct structure.
Among the given options, only Option (C) on treatment with sodamide forms an alkyne that, upon mercuric sulphate hydration, yields a methyl ketone satisfying both test conditions.

Final Answer: \[ \boxed{\text{Option (C)}} \]