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

To solve the problem, we need to understand the chemical transformations involved:

Step 1: Reaction of propenenitrile with $SnCl_2/HCl$ and hydrolysis:

• Propenenitrile (acrylonitrile) undergoes reduction with $SnCl_2$ and HCl to form propionaldehyde.
• Upon hydrolysis, propionaldehyde is converted to propionic acid.

Step 2: Reaction of but-2-ene by ozonolysis:

• Ozonolysis of but-2-ene cleaves the double bond, yielding two molecules of ethanal (acetaldehyde).

Step 3: Reaction of 'x' (propionic acid) with 'y' (ethanal) in alkali with heating:

• Propionic acid and ethanal can undergo an Aldol condensation in the presence of a base.
• This reaction typically forms β-hydroxyaldehydes or their dehydration products (α,β-unsaturated carbonyl compounds).
• However, when propionic acid reacts with ethanal, it's unlikely to form such compounds due to steric and electronic factors.

Range Verification:

• The expected range 3,3 suggests a numerical confirmation but chemically equates here to a conceptual understanding.
• No stable $β$-hydroxyaldehyde or α,β-unsaturated carbonyl compound of this system that fits this numbering scheme is well-known.

Conclusion:

• The product not obtained when 'x' and 'y' react in alkali with heating is a stable β-hydroxyaldehyde or α,β-unsaturated carbonyl compound.