Write a mechanism for the aldol reaction of 3,3-dimethyl-2-butanone.
The aldol reaction is a type of organic reaction that involves the formation of a new carbon-carbon bond. It typically occurs between an enol or enolate (a molecule with a double bond adjacent to a carbonyl group) and an aldehyde or ketone. In the case of the aldol reaction of 3,3-dimethyl-2-butanone, also known as pentane-2,4-dione, we will need another aldehyde or ketone as the second reactant.
Here is the mechanism for the aldol reaction of 3,3-dimethyl-2-butanone using formaldehyde (methanal) as the second reactant:
1. Step 1: Formation of enolate ion:
- Treat 3,3-dimethyl-2-butanone with a strong base such as sodium hydroxide (NaOH) to deprotonate the alpha carbon and form the enolate ion. The alpha carbon is the carbon next to the carbonyl group.
- The reaction can be represented as:
3,3-dimethyl-2-butanone + NaOH → 3,3-dimethyl-2-butanol + Na+ + CH3C(O)CH2-
2. Step 2: Nucleophilic addition:
- Formaldehyde (methanal) acts as the carbonyl compound that will undergo nucleophilic addition.
- The enolate ion attacks the carbonyl carbon of formaldehyde, resulting in the formation of a new carbon-carbon bond.
- The reaction can be represented as:
CH3C(O)CH2- + HCHO → CH3C(O)CH(OH)CH3
3. Step 3: Acid-catalyzed dehydration:
- To form the final aldol product, water molecules are eliminated from the aldol intermediate.
- This step requires an acid catalyst, typically a mineral acid like hydrochloric acid (HCl).
- The reaction can be represented as:
CH3C(O)CH(OH)CH3 → CH3C(O)CH=CH2 + H2O
Overall, the aldol reaction of 3,3-dimethyl-2-butanone with formaldehyde proceeds through the formation of an enolate ion, followed by nucleophilic addition and acid-catalyzed dehydration to produce the final aldol product, which is 3,3-dimethyl-2-penten-4-one.