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Question
Explain the aldol condensation of ethanal.
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Solution 1
Aldehydes containing at least one α-hydrogen atom undergo a reaction in the presence of dilute alkali (dilute NaOH, KOH or Na2CO3) as a catalyst to form β-hydroxy aldehydes (aldol). This reaction is known as the aldol reaction.
Ethanal contains an α-hydrogen atom, so it undergoes the aldol condensation reaction.
\[\ce{\underset{Ethanal}{2CH3 - CHO} ->[dil{.} NaOH] \underset{3-Hydroxybutanal}{CH3 - CHOH - CH2 - CHO}}\]
\[\ce{\underset{3-Hydroxybutanal}{CH3 - CHOH - CH2 - CHO} ->[H+] \underset{But-2-enal}{CH3 - CH = CH - CHO} + H2O}\]
When ethanal is treated with dilute base NaOH, KOH or sodium carbonate, two molecules of ethanal add together to form 3-Hydroxybutanal. When 3-Hydroxybutanal is heated in the presence of an acid, a molecule of water is lost, and But-2-enal is formed.
Solution 2
The carbon atom adjacent to the carbonyl carbon atom is called the α-carbon atom (α − C).
\[\begin{array}{cc}
\phantom{.........}\ce{O}\\
\phantom{.........}||\\
\ce{- \overset{\beta}{C} - \overset{\alpha}{C} - C - \overset{\alpha}{C} -}\\
|\phantom{.....}\\
\ce{H^\alpha}\phantom{...}\\
\end{array}\]
Aldol condensation reaction is a characteristic reaction of aldehydes and ketones containing an active α-hydrogen atom.
When aldehydes or ketones having α-H atoms are heated with a dilute base or dilute acid, two molecules undergo self-condensation to yield β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol), respectively. The reaction is referred to as the aldol addition reaction.
For aldehyde:

Acetaldol, on heating, undergoes subsequent elimination of water, giving rise to α, β-unsaturated aldehyde.

The overall reaction is called aldol condensation. It is a nucleophilic addition-elimination reaction.
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(i) \[\ce{CH3 - CHO}\]
(ii) 
(iii) \[\begin{array}{cc}
\phantom{}\ce{O}\\
\phantom{}||\\
\ce{CH3 - C - CH3}
\end{array}\]
(iv) \[\begin{array}{cc}
\phantom{}\ce{CH3}\\
|\phantom{...}\\
\ce{CH3 - C - CHO}\phantom{..}\\
|\phantom{...}\\
\phantom{}\ce{CH3}\\
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{....................}\\
\phantom{}\ce{2CH3 - C = O ->[Ba(OH)2] A ->[Δ] B + H2O}
\end{array}\]
\[\ce{CH3-CH2-CHO ->[dil][alkali] Product}\]
The product in the above reaction is:
The major product of the following reaction is:

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\[\begin{array}{cc}
\ce{2CH3 - CHO ->[dil.NaOH] CH3 - CH - CH2 - CHO}\\
\phantom{...............}|\\
\phantom{.................}\ce{OH}
\end{array}\]
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