Question
(JEE Main 2023 PYQ) Identify the products A and B in the following sequence:
Name the reactions involved.
Solution — Step by Step
Acetaldehyde with -H undergoes aldol addition with dilute NaOH. Mechanism: NaOH abstracts an -H, generating an enolate ion, which attacks the carbonyl carbon of another acetaldehyde molecule.
So A = 3-hydroxybutanal (aldol product).
On heating, the -hydroxy aldehyde loses water (E1cb mechanism — dehydration of aldol):
So B = but-2-enal (crotonaldehyde), an -unsaturated aldehyde.
The combination — aldol addition followed by dehydration — is called aldol condensation. The complete sequence converts two molecules of acetaldehyde into crotonaldehyde plus water.
A = 3-hydroxybutanal, B = but-2-enal (crotonaldehyde). Reaction = aldol condensation.
Why This Works
Aldol reactions need an aldehyde or ketone with at least one -hydrogen. Acetaldehyde has three -Hs (the CH₃ next to the carbonyl), so it readily undergoes aldol.
The aldol step gives a -hydroxy carbonyl. On heating, this dehydrates to give an -unsaturated carbonyl (conjugated, more stable than the aldol).
Aldehydes with no -H (formaldehyde, benzaldehyde) cannot undergo aldol — instead, they undergo Cannizzaro reaction in concentrated alkali, giving an alcohol and a carboxylate.
Quick test for aldol vs Cannizzaro: count -Hs. ≥1 -H + dilute NaOH → aldol. No -H + conc. NaOH → Cannizzaro.
Alternative Method
Acid-catalyzed aldol: replace NaOH with H⁺. The mechanism uses an enol (not enolate), but the products are the same. Less common in JEE/NEET, more in advanced organic chemistry.
Common Mistake
Students draw the aldol product as a 1,3-diol (CH₃CH(OH)CH(OH)CH₃) — wrong. The aldol product retains one carbonyl (the attacking C=O is unchanged) and gains a hydroxyl (from the attacked C=O). So it’s a -hydroxy aldehyde, not a diol.