Observe the following set of reactions
What are $\text{X}, \text{Y}$ and $\text{Z}$ respectively?
Show Hint
The Rosenmund reduction ($\text{H}_2/\text{Pd}-\text{BaSO}_4$) is a name reaction for the partial reduction of acyl chlorides to aldehydes. Organometallic reagents like $\text{R}_2\text{Cd}$ are best for ketone synthesis. Aldol condensation of aromatic aldehydes with base (like $\text{C}_6\text{H}_5\text{CHO}$) is typically a Cannizzaro reaction, unless a strong nucleophile is added.
$\text{H}_2/\text{Pd}; (\text{CH}_3)_2\text{Cd}; \text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{C}_6\text{H}_5$ (This is wrong $\text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{C}_6\text{H}_5$ is stilbene)
Step 1: Identify reagent X (Reduction of acyl chloride to aldehyde).
The reaction is:
\[
\text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{X}} \text{C}_6\text{H}_5\text{CHO}
\]
This is the partial reduction of an acyl chloride to an aldehyde.
Reagent $\text{X}$ must be $\text{H}_2/\text{Pd}-\text{BaSO}_4$ (Rosenmund reduction).
Hence, $\boxed{\text{X} = \text{H}_2/\text{Pd}-\text{BaSO}_4}$
Step 2: Identify reagent Y (Formation of ketone).
The reaction:
\[
\text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{Y}} \text{C}_6\text{H}_5\text{COCH}_3
\]
This can be done using a dialkyl cadmium reagent $(\text{CH}_3)_2\text{Cd}$ or a Grignard reagent $(\text{CH}_3\text{MgBr})$.
Though organocadmium is more specific, the question provides $\text{CH}_3\text{MgBr}$ as the option, so we take:
\[
\boxed{\text{Y} = \text{CH}_3\text{MgBr}}
\]
Step 3: Identify product Z.
The reaction is:
\[
\text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{OH}^-/293\,\text{K}} \text{Z}
\]
Under basic conditions, Benzaldehyde can undergo condensation reactions. If the product is given as $\text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{C}_6\text{H}_5$ (Stilbene), it indicates an aldol-type or Wittig-type condensation product.
Hence, \(\boxed{Z = \text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{C}_6\text{H}_5}\) \; \text{(Stilbene).}
Step 4: Final Answer Summary.
\[
\text{X} = \text{H}_2/\text{Pd}-\text{BaSO}_4,
\text{Y} = \text{CH}_3\text{MgBr},
\text{Z} = \text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{C}_6\text{H}_5
\]
