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There are four aldol condensation products possible when acetone and propanal are mixed in the presence of base. Use mechanistic sequences to illustrate what are they and how are they formed?
Accepted Answer
The aldol condensation reaction between acetone and propanal can indeed yield four different products. Here's a breakdown of the mechanism and the products:
Mechanism:
1. Enolate Formation: The base (e.g., NaOH) deprotonates the alpha-hydrogen of both acetone and propanal, forming their respective enolates.
2. Nucleophilic Attack: The enolate of acetone, being more reactive due to steric factors, acts as a nucleophile and attacks the carbonyl carbon of propanal.
3. Protonation: The resulting alkoxide intermediate gets protonated by water, yielding the aldol product.
Products:
Product 1: 4-Hydroxy-4-methyl-2-pentanone (formed by the reaction of acetone enolate with propanal)
Product 2: 3-Hydroxy-2-hexanone (formed by the reaction of propanal enolate with acetone)
Product 3: 2,6-dimethyl-2,6-heptanediol (formed by self-condensation of acetone)
Product 4: 3-hydroxyhexanal (formed by self-condensation of propanal)
Illustrative Mechanism:
(Note: This is a simplified representation, and the actual reaction mechanism involves multiple steps with proton transfers.)
Product 1 Formation:
[Image of the reaction mechanism for product 1]
Product 2 Formation:
[Image of the reaction mechanism for product 2]
Product 3 Formation:
[Image of the reaction mechanism for product 3]
Product 4 Formation:
[Image of the reaction mechanism for product 4]
Note: The actual product distribution will depend on the reaction conditions and the relative reactivity of the reactants.
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