The dibromo compound [P] (molecular formula : $$C_{9}H_{10}Br_{2}$$) when heated with excess sodamide followed by treatment with dilute HO gives [Q]. On wanning [Q] with mercuric sulphate and
dilute sulphuric acid yield [R] which gives positive Iodofonn test but negative Tollen's test. The compound [P] is:

We need to determine the correct structure of the dibromo compound [P] ($$C_9H_{10}Br_2$$) based on its reaction sequence leading to compound [R].

Step-by-Step Reaction Analysis:

• Step 1: Conversion of [P] to [Q]

  When the geminal dihalide [P] is heated with excess sodamide ($$\text{NaNH}_2$$), it undergoes a double dehydrohalogenation (elimination of two molecules of $$\text{HBr}$$) to yield a terminal alkyne [Q]. Treatment with dilute acid protonates the acetylide ion back to the alkyne form.

• Step 2: Hydration of [Q] to [R]

  Warming the terminal alkyne [Q] with mercuric sulfate ($$\text{HgSO}_4$$) and dilute sulfuric acid ($$\text{H}_2\text{SO}_4$$) causes Kucherov hydration (Markovnikov addition of water). The resulting enol tautomerizes into a stable methyl ketone [R].

• Step 3: Characterization of [R]

  • Positive Iodoform Test: Confirms that [R] contains a methyl ketone group ($$\text{-COCH}_3$$).
  • Negative Tollen's Test: Confirms that [R] is a ketone rather than an aldehyde.

Structural Deduction:

From the reaction scheme shown, the structure of [R] is explicitly identified as 1-(3-methylphenyl)ethan-1-one (a methyl ketone with a meta-methyl group on the benzene ring).

Tracing back to the starting material, the geminal dibromo group must be located on the side chain alpha-carbon atom where the terminal alkyne and carbonyl group are formed.

Therefore, the original compound [P] is 1,1-dibromo-1-(3-methylphenyl)ethane.

Answer: Option B