Identification of Organic Compound Y from Multistep Synthesis Involving Perkin Reaction

To determine the structure of compound $\mathbf{Y}$, we analyze the reaction sequence step-by-step:

1. Formation of Compound $\mathbf{X}$:

   • Step 1 (Gattermann-Koch reaction): Benzene is treated with $\text{CO}$ and $\text{HCl}$ in the presence of anhydrous $\text{AlCl}_3/\text{CuCl}$ to form benzaldehyde: $\text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{AlCl}_3/\text{CuCl}} \text{C}_6\text{H}_5\text{CHO} + \text{HCl}$

   • Step 2 (Perkin reaction): Benzaldehyde is subsequently reacted with acetic anhydride ($\text{Ac}_2\text{O}$) in the presence of sodium acetate ($\text{NaOAc}$) to furnish cinnamic acid as the predominant product $\mathbf{X}$: $\text{C}_6\text{H}_5\text{CHO} + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{CH}_3\text{COONa}, \, \Delta} \text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{COOH} \quad (\mathbf{X})$

2. Formation of Compound $\mathbf{Y}$:

   • Step 1 (Bromination): Cinnamic acid ($\mathbf{X}$) undergoes electrophilic addition of bromine ($\text{Br}_2/\text{Na}_2\text{CO}_3$) across the carbon-carbon double bond to yield 2,3-dibromo-3-phenylpropanoic acid (cinnamic acid dibromide): $\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{COOH} \xrightarrow{\text{Br}_2/\text{Na}_2\text{CO}_3} \text{C}_6\text{H}_5-\text{CH(Br)}-\text{CH(Br)}-\text{COOH}$

   • Step 2 (Dehydrohalogenation and Decarboxylation): Heating the dibromo derivative at $473\text{ K}$ ($200^\circ\text{C}$) with moist $\text{KOH}$ induces simultaneous double dehydrobromination and decarboxylation to give phenylacetylene ($\mathbf{Y}$) as the major product: $\text{C}_6\text{H}_5-\text{CH(Br)}-\text{CH(Br)}-\text{COOH} \xrightarrow{\text{moist KOH}, \, 473\text{ K}} \text{C}_6\text{H}_5-\text{C}\equiv\text{CH} + \text{CO}_2 + 2\,\text{HBr}$

Thus, compound $\mathbf{Y}$ is phenylacetylene, which corresponds to option (C).