The dibromo compound [P] (molecular formula: C₉H₁₀Br₂ ) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yield [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is :

Solution

Chemistry ma’am · 3 steps

Work backwards: methyl ketone R points to a terminal alkyne Q, and Q points to a geminal dibromide.

1. Read the tests. A positive iodoform test with a negative Tollens' test means R is a methyl ketone, Ar-CO-CH₃, not an aldehyde. Warming an alkyne with HgSO₄ and dilute H₂SO₄ adds water the Markovnikov way, so a terminal alkyne Ar-C≡CH gives exactly this methyl ketone. So Q is Ar-C≡CH.
2. Work back to P. Excess NaNH₂ removes two HBr from a dibromide to make a triple bond. This works only when both Br atoms are on the same carbon (geminal) or on neighbouring carbons (vicinal), with H atoms there to remove. The excess NaNH₂ also turns the terminal alkyne into its sodium salt, and dilute HCl gives back Ar-C≡CH.

   Option	Where the Br atoms sit	Alkyne with excess NaNH₂?
   (a)	Vicinal, but the carbon holding the ring has no H	No, it stops at a vinyl bromide
   (b)	Two separate CH₂Br groups on the ring	No, the two Br atoms are not on neighbouring carbons of a chain
   (c)	Already on a C=C, attached to a non-aromatic ring	No clean terminal alkyne
   (d)	Geminal, CBr₂ next to a CH₃	Yes, gives Ar-C≡CH

   ArCBr₂CH₃ →(excess NaNH₂) ArC≡C⁻Na⁺ →(dil. HCl) ArC≡CH →(HgSO₄, H₂SO₄) ArCOCH₃

3. So. With Ar as 4-methylphenyl, P in (d) (C₉H₁₀Br₂) gives 4-methylphenylethyne as Q and 4-methylacetophenone as R. R is a methyl ketone, so it gives a positive iodoform test and a negative Tollens' test. The answer is (d).

The idea to keep. Iodoform positive but Tollens' negative means a methyl ketone. Mercury-catalysed hydration of a terminal alkyne gives one, and that alkyne comes from a geminal or vicinal dibromide losing two HBr.

Answer from the NTA final answer key. Solution by PrepWiser.