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:
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Alkynes give ketones with \(HgSO_4/H_2SO_4\); a positive iodoform test confirms a methyl ketone.
Let's analyze the given information step-by-step to identify the compound \(P\).
The dibromo compound \(P\) (molecular formula: \(C_9H_{10}Br_2\)) is heated with excess sodamide, leading to the formation of an alkyne from a vicinal dihalide through an elimination reaction.
The product (\(Q\)) of this reaction is an alkyne. When treated with mercuric sulfate and dilute sulfuric acid, \(Q\) undergoes hydration to form a ketone (\(R\)).
The compound \(R\) gives a positive iodoform test but a negative Tollens' test. This indicates that \(R\) is a methyl ketone. The iodoform test is positive for compounds with the structure \(CH_3C=O\), while the Tollens' test, which is negative, implies \(R\) is not an aldehyde.
From the options, compound \(C\) is a 1,2-dibromo compound (bromines are on adjacent carbons) with a benzene ring, suitable for forming an alkyne when treated with excess sodamide.
The alkyne formed can be hydrated to produce a methyl ketone, which explains the positive iodoform test.
Thus, compound \(C\) is the correct answer as it correctly fulfills the condition of being a 1,2-dibromo compound, leading to a methyl ketone capable of a positive iodoform test.
