Question

What are 'X' and 'Y' respectively in the following reactions? \[ \text{C}_6\text{H}_5\text{CHO} + \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{OH}^-,\ 293\text{K}} \text{X} \xrightarrow{\text{NaBH}_4} \text{Y} \]

• A. $\text{C}_6\text{H}_5\text{COCH}=\text{C}(\text{CH}_3)\text{C}_6\text{H}_5$, $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COC}_6\text{H}_5$
• B. $\text{C}_6\text{H}_5\text{COCH}=\text{C}(\text{CH}_3)\text{C}_6\text{H}_5$, $\text{C}_6\text{H}_5\text{CH}=\text{CH}(\text{OH})\text{C}_6\text{H}_5$
• C. $\text{C}_6\text{H}_5\text{CH}=\text{CHCOC}_6\text{H}_5$, $\text{C}_6\text{H}_5\text{CH}=\text{CHCH}(\text{OH})\text{C}_6\text{H}_5$ \correctanswer
• D. $\text{C}_6\text{H}_5\text{CH}=\text{CHCOC}_6\text{H}_5$, $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COC}_6\text{H}_5$

Hint:

Key Mechanisms:
1. Aldol condensation requires \alpha-hydrogens
2. NaBH$_4$ reduces only carbonyls, not C=C bonds
3. Chalcones are important flavonoid precursors

Answer 1

The Correct Option is C

Step 1: Aldol Condensation (Formation of X)
\[ \text{C}_6\text{H}_5\text{CHO} + \text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5\text{CH}=\text{CHCOC}_6\text{H}_5 + \text{H}_2\text{O} \]
1. Benzaldehyde + acetophenone gives chalcone (X)
2. α,β-unsaturated ketone formed via dehydration 
Step 2: Reduction (Formation of Y)
\[ \text{C}_6\text{H}_5\text{CH}=\text{CHCOC}_6\text{H}_5 \xrightarrow{\text{NaBH}_4} \text{C}_6\text{H}_5\text{CH}=\text{CHCH}(\text{OH})\text{C}_6\text{H}_5 \]
1. NaBH$_4$ selectively reduces carbonyl group
2. Double bond remains intact
3. Forms allylic alcohol (Y) Thus, the correct pair is $\boxed{3}$.

Answer 2

Step 1: Identify the reaction type.
The reaction between benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)) and acetophenone (\(\text{C}_6\text{H}_5\text{COCH}_3\)) in the presence of a base (\(\text{OH}^-\)) at 293 K is an example of an aldol condensation.

Step 2: Mechanism of aldol condensation.
- Under basic conditions, the alpha hydrogen of acetophenone is abstracted to form an enolate ion.
- This enolate ion attacks the carbonyl carbon of benzaldehyde.
- The intermediate formed undergoes dehydration to give an \(\alpha, \beta\)-unsaturated ketone.

Step 3: Identify compound X.
The condensation product \(X\) is chalcone, which has the structure:
\[ \text{C}_6\text{H}_5\text{CH} = \text{CHCOC}_6\text{H}_5, \] an \(\alpha, \beta\)-unsaturated ketone formed by the aldol condensation of benzaldehyde and acetophenone.

Step 4: Reduction by \(\text{NaBH}_4\).
Sodium borohydride (\(\text{NaBH}_4\)) is a mild reducing agent that selectively reduces the carbonyl group (C=O) to an alcohol (C–OH) without affecting the C=C double bond.

Step 5: Identify compound Y.
Reduction of chalcone (\(X\)) leads to:
\[ \text{C}_6\text{H}_5\text{CH} = \text{CHCH}(\text{OH})\text{C}_6\text{H}_5, \] which is the corresponding allylic alcohol.

Step 6: Summary.
Thus,
- \(X = \text{C}_6\text{H}_5\text{CH} = \text{CHCOC}_6\text{H}_5\) (chalcone),
- \(Y = \text{C}_6\text{H}_5\text{CH} = \text{CHCH}(\text{OH})\text{C}_6\text{H}_5\) (reduced allylic alcohol).