Question

'x' is the product from propenenitrile + $SnCl_2/HCl$ followed by hydrolysis. 'y' is the product from but-2-ene by ozonolysis. Which product is not obtained when 'x' and 'y' react in alkali with heating?

Hint:

Count the carbons! Pent-2-enal has 5 carbons ($3+2$), while 3-methylbut-2-enal also has 5, but its branching pattern doesn't match the possible enolates from propenal.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
Identify 'x' and 'y' first, then perform all possible aldol condensation reactions (self and cross).
Step 2: Detailed Explanation:
1. Identify x: $CH_2=CH-CN \xrightarrow{Stephen's Reduction} CH_2=CH-CHO$ (Acrolein). 2. Identify y: $CH_3-CH=CH-CH_3 \xrightarrow{O_3} 2 CH_3CHO$ (Acetaldehyde). 3. Aldol combinations: - Self-Aldol of y: $CH_3CHO + CH_3CHO \rightarrow$ But-2-enal. - Cross-Aldol (y enolate + x): $CH_3CHO + CH_2=CH-CHO \rightarrow$ Pent-2-enal. - Cross-Aldol (x enolate + y): $CH_2=CH-CHO + CH_3CHO \rightarrow$ 2-Methylbut-2-enal. - Self-Aldol of x: $x + x \rightarrow$ 2-Methylpent-2-enal. 4. Conclusion: 3-Methylbut-2-enal would require a branched precursor like acetone, which is not present.
Step 3: Final Answer:
The product not obtained is 3-Methylbut-2-enal.