Question

Consider the following set of reactions.
$\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{B}} \text{Y} (\text{reacts with } 2,4\text{-DNP}) \xrightarrow{\text{A}} \text{X} (\text{dissolves in dil. } \text{HCl})$
What are A and B respectively?

• A. $\text{LiAlH}_4, \text{H}_2\text{O}; \text{H}_2/\text{Ni}$
• B. $\text{Na}/\text{Hg}, \text{C}_2\text{H}_5\text{OH}; \text{DIBAL-H}, \text{H}_2\text{O}$
• C. $\text{DIBAL-H}, \text{H}_2\text{O}; \text{LiAlH}_4, \text{H}_2\text{O}$
• D. $\text{Na}/\text{Hg}, \text{C}_2\text{H}_5\text{OH}; \text{H}_2/\text{Ni}$

Hint:

Nitriles can give different products depending on the reducing agent: DIBAL-H (controlled conditions) → Aldehyde Strong reduction (Na/Hg, LiAlH$_4$) → Primary amine

Answer 1

The Correct Option is B

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.
\[ \text{X} = \text{H}_2/\text{Pd}-\text{BaSO}_4, \quad \text{Y} = \text{CH}_3\text{MgBr}, \quad \text{Z} = \text{C}_6\text{H}_5\text{CH}=\text{CH}-\text{C}_6\text{H}_5 \]