Question

What is the action of dilute \[NaOH\] on ethanal (acetaldehyde)? Name the reaction.

Answer 1

Hint: Two molecules of aldehyde combine together in the presence of a base to form $\beta - $hydroxyaldehyde. When sodium hydroxide is added to ethanal to give or form $3 - $hydroxybutanal.

Complete step by step answer:
Ethanal reacts with sodium hydroxide to form beta-hydroxy butyraldehyde. The product is known as aldol and this is the example of the aldol condensation reaction. It is given by ketones or aldehydes having an alpha hydrogen atom. Aldehyde and alcohol are represented by “Aldol”. When an aldehyde reacts with carbon at the $\alpha $(alpha) site to the carbonyl of another molecule under basic or acidic conditions to obtain beta-hydroxy aldehyde. The molecular formula of ethanal is $C{H_3}CHO$.
Aldol condensation is an organic reaction in which enolate ion reacts with a carbonyl compound to form $\beta - $ hydroxy aldehyde. It is followed by dehydration to give a conjugated enone and it plays a vital role in organic synthesis. It is creating a path to form carbon-carbon bonds.
The reaction is:
$C{H_3}CHO + C{H_3}CHO\xrightarrow{{dilNaOH}}C{H_3}CH\left( {OH} \right)C{H_2}CHO\xrightarrow[\Delta ]{{{H^ + }}}C{H_3} - CH = CH - CHO + {H_2}O$

Additional information:
The Aldol reaction is an important organic reaction of ketones and aldehydes and is shown by these compounds because they contain alpha – hydrogen atoms. They will undergo self-condensation in the presence of dilute alkalies like dilute sodium hydroxide. In the reaction, the aldol or ketol are formed by eliminating the water molecule or they undergo dehydration.

Note: Aldol condensation reaction is also given by ketone compounds. Ketone compounds undergo self- condensation to form $\beta $(beta) - hydroxyketone (ketol) in the presence of dilute $Ba{\left( {OH} \right)_2}$.