Draw the structure of the product formed in the following reaction.
The mechanism of an aldol condensation contains two steps. The first part is an aldol reaction; the second part is a dehydration elimination reaction. First, the base abstracts the proton to form a carbanion, and then it attacks the carbonyl group of another compound to form an oxy anion. This compound undergoes hydrolysis and forms an aldol, followed by dehydration which will give an aldol product. Mechanism:-
Step 1 of 3
Sodium hydroxide abstracts the acidic proton from an α-carbon atom of 2-phenylacetaldehyde to form the enolate and water as a byproduct. Hydroxide is not a good base to enolize an aldehyde, so the amount of enolate ion formed is less.
Step 2 of 3
The enolate ion formed in the first step reacts with the 2-phenylacetaldehyde that is not enolized. Each enolate ion attacks the 2-phenylacetaldehyde to form 4-oxo-1,3-diphenylbutan-2-olate (alkoxide ion). This is then protonated by the water molecules formed in the first step. The product 3-hydroxyl-2,4-diphenylbutanal is called the aldol.
Step 3 of 3
Hydroxide formed from the sodium hydroxide again removes an acidic α-hydrogen bond to form the enolate ion of an aldol product. The negative charge on the carbon of an enolate ion is used to form the carbon-carbon double bond. This causes the displacement of the leaving group and regenerates hydroxide, giving 2, 4-diphenylbut-2-enal.