The reaction proceeds in two steps. Step 1: Dehydrohalogenation to form Propyne (A) 1,2-dibromopropane reacts with alcoholic KOH and heat, followed by \(NaNH_2\), to undergo double dehydrohalogenation, resulting in the formation of propyne (compound A). The reaction is as follows: \[ CH_3{-CHBr-CH_2Br} \xrightarrow[{(ii) \, NaNH_2}]{{(i) \, \text{Alc. KOH}, \, \Delta}} CH_3{-C}\equiv{CH} \quad (\text{Propyne, A}) \] Thus, compound A is propyne. Step 2: Hydration of Propyne to form Acetone (X) Propyne (A) undergoes hydration in the presence of \(Hg^{2+}/H^{+}\) and water at 333 K to form acetone (compound X). The reaction follows Markovnikov's rule. \[ CH_3{-C}\equiv{CH} \xrightarrow[{333 \, K}]{{H_2O, \, Hg^{2+}/H^{+}}} CH_3{-CO-CH_3} \quad (\text{Acetone, X}) \] Thus, compound X is acetone (propan-2-one). Step 3: Calculating Percentage of Carbon in Acetone (X) The molecular formula of acetone is \(C_3H_6O\). Molar mass of Acetone (\(C_3H_6O\)) is calculated as: \[ M(C_3H_6O) = 3 \times M(C) + 6 \times M(H) + 1 \times M(O) \] \[ = 3 \times 12 \, \text{g/mol} + 6 \times 1 \, \text{g/mol} + 1 \times 16 \, \text{g/mol} \] \[ = 36 + 6 + 16 = 58 \, \text{g/mol} \] Mass of carbon in 1 mole of Acetone = \(3 \times 12 = 36\) g. Percentage of carbon in Acetone (X): \[ \text{Percentage of Carbon} = \frac{\text{Mass of Carbon in Acetone}}{\text{Molar mass of Acetone}} \times 100 \] \[ = \frac{36 \, \text{g/mol}}{58 \, \text{g/mol}} \times 100 \] \[ = \frac{36}{58} \times 100 \approx 62.07 \% \] The percentage of carbon in compound X is approximately 62%. Correct Answer: (2) 62
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