Question

Benzonitrile (A) + X \(\rightarrow\) B; A + Y \(\rightarrow\) C. B + C \(\xrightarrow{dil.\ NaOH}\) \(\alpha,\beta\)-unsaturated carbonyl compound. What are X and Y?

• A. DIBAL-H, \(H_2O\); \((CH_3)_2Cd\)
• B. \(SnCl_2 + HCl,\ H_3O^+\); \(CH_3MgBr,\ H_2O\)
• C. DIBAL-H, \(H_2,\ Ni\)
• D. \(SnCl_2 + HCl,\ H_2O\); \((CH_3)_2Cd\)

Hint:

Stephen reduction converts nitriles into aldehydes without affecting the aromatic ring.

Answer 1

The Correct Option is D

Step 1: Read the target.
The final product is an alpha,beta-unsaturated carbonyl compound made by an aldol type (Claisen-Schmidt) condensation. That needs an aldehyde (B) and a ketone (C).
Step 2: Make B from benzonitrile.
The Stephen reduction with $SnCl_2/HCl$ then water turns a nitrile into an aldehyde: $C_6H_5CN\rightarrow C_6H_5CHO$. So $B$ is benzaldehyde and $X=SnCl_2+HCl,\ H_2O$.
Step 3: Make C from benzonitrile.
Dimethyl cadmium $(CH_3)_2Cd$ adds a methyl group to give a ketone: $C_6H_5CN\rightarrow C_6H_5COCH_3$. So $C$ is acetophenone and $Y=(CH_3)_2Cd$.
Step 4: Combine B and C.
Benzaldehyde and acetophenone undergo Claisen-Schmidt condensation with dilute NaOH.
Step 5: Get the product.
This gives the alpha,beta-unsaturated ketone (a chalcone), matching the requirement.
Step 6: Pick the reagents.
So $X$ and $Y$ are $SnCl_2+HCl,\ H_2O$ and $(CH_3)_2Cd$. \[ \boxed{SnCl_2+HCl,\ H_2O;\ (CH_3)_2Cd} \]