A mixture of ethanal and propanal is subjected to aldol condensation by treating with dilute NaOH. How many aldols (including stereo isomers) are expected to be formed in this reaction

When a mixture of ethanal (acetaldehyde) and propanal (propionaldehyde) undergo aldol condensation with dilute NaOH, multiple possible aldols (including stereoisomers) can form due to cross and self-condensations.

Stepwise Reasoning:

1. Possible pairs for condensation:
   • Ethanal + Ethanal (self-condensation of ethanal)
   • Propanal + Propanal (self-condensation of propanal)
   • Ethanal + Propanal (cross-condensation, two different ways depending on which is enolate and which is electrophile)
2. Products from each pair:
   • Ethanal self-condensation:
     Forms 3-hydroxybutanal (aldol), which can dehydrate to crotonaldehyde.
     1 aldol product.
   • Propanal self-condensation:
     Forms 4-hydroxy-2-methylpentanal, can dehydrate to a conjugated enal.
     1 aldol product.
   • Cross-condensation:
     Two possibilities:
     • Enolate from ethanal attacks propanal
     • Enolate from propanal attacks ethanal
       Thus, 2 aldol products from cross-condensation.
3. Stereoisomers:
   • Aldols commonly form stereoisomers (E/Z or cis/trans), especially in the double bonds after dehydration.
   • Each aldol can have 2 stereoisomers.
4. Total aldol products including stereoisomers:
   • Self-condensations:
     Each 1 product × 2 stereoisomers = 2
     Total self = 4
   • Cross-condensations:
     2 different aldols × 2 stereoisomers each = 4
5. Grand total:
   4+4=84 + 4 = 84+4=8 aldol products including stereoisomers.

Final answer

• 8 aldols (including stereoisomers) are expected from aldol condensation of ethanal and propanal mixture with dilute NaOH.