The dibromo compound (molecular formula: ) when heated with excess sodamide followed by treatment with dilute HCl gives . On warming with mercuric sulphate and dilute sulphuric acid yields , which gives a positive iodoform test but a negative Tollens' test. The compound is:
Correct answer:C
Standard Method
Given: A dibromo compound with molecular formula gives on heating with excess sodamide followed by dilute HCl. Then on treatment with gives . Compound gives a positive iodoform test and a negative Tollens' test.
Find: The correct structure of .
Step 1: Effect of excess sodamide.
Heating a vicinal or geminal dibromide with excess sodamide leads to double dehydrohalogenation, forming an alkyne. Hence, compound must be such that it can form an alkyne .
Step 2: Reaction with .
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 .
Positive iodoform test indicates the presence of a group. Negative Tollens' test confirms the absence of an aldehyde group.
Thus, 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.
Therefore, the correct option is C.
Assuming any dibromide will give the required product is incorrect because only a structure capable of double dehydrohalogenation to the right alkyne works. First check whether the dibromide can form an alkyne under excess sodamide.
Treating the positive iodoform test as evidence for an aldehyde is wrong. A positive iodoform test indicates a methyl ketone or a related precursor, while the negative Tollens' test rules out an aldehyde.
Ignoring the hydration outcome of the alkyne is a conceptual error. Under , the alkyne must hydrate to a ketone, so choose the precursor whose alkyne gives a methyl ketone after enol–keto tautomerism.
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