Class Summary and Notes from 10/24/00 - CHEM3312
Topic: Alkylations of Carbonyl Compounds at the alpha position (cont'd)
Related Reading
(23.4-23.6: we didn't get to this material last time), 23.7-23.11, 23.13
Related Problems
23.9-23.18, 23.23, 23.26, 23.27, 23.30, 23.34-23.36, 23.40-23.43, 23.48-23.50, 23.53
Summary
The aldol reaction (reaction of two molecules of an aldehyde, one in its enolate form) was reviewed. It was pointed out that heating the b-hydroxy carbonyl product of an aldol reaction results in an a,b-unsaturated carbonyl compound by the loss of water. This reaction is quite favorable as the resulting pi system is conjugated.
The aldol reaction between two different molecules was discussed (called a mixed aldol). After the first in-class problem, the following points were summarized:
- Mixed aldols tend to give lots of products if both reactants can form enolates (enol1 + carbonyl1, enol1 + carbonyl2, enol2 + carbonyl1, enol2 + carbonyl2)
- Mixed aldols give better yields when one reactant has no a-hydrogens (and can't form an enolate)
- If one reactant has a b-carbonyl group, the a-hydrogens are unusually acidic and the other compound will not form a competetive amount of enolate - this will also be productive.
Note: Considerations applied to mixed aldols will also apply to mixed Claisen reactions
Enolates are capable of reacting with electrophilic sites in carbonyl compounds. In addition to the carbonyl carbon, a,b-unsaturated carbonyl compounds have a second electrophilic site. It is easiest to see this second electrophilic site in a second resonance structure:
Enolates have a strong tendency to react with these other electrophilic sites in conjugated compounds. Such a reaction is termed the Michael Reaction. A general scheme for this reaction is:
Problems Discussed In Class
What products will be generated in the following reactions:
- Ethanal + propanal reacted with sodium hydroxide, followed by water.
- Benzaldehyde + ethanal reacted with sodium hydroxide, followed by water.
- Ethanal + propanedial reacted with sodium hydroxide, followed by water.
Results of Class Discussion
- 3-hydroxybutanal, 3-hydroxy-2-methylpentanal, 3-hydroxypentanal, 3-hydroxy-2-methylbutanal
- 3-hydroxy-3-phenylpropanal, 3-hydroxybutanal
- 2-(1-hydroxyethyl)-propanedial
Take-home problem: Use the compounds in Table 23.1 for the following
exercises:
- Identify the electrophilic sites (both direct and through conjugation) in
the acceptor
- Identify the most acidic hydrogen (this will be in the donor molecule)
- Use NaOCH3 to remove the most acidic hydrogen
- Draw the product (and practice naming it!)
Note: if you aren't sure if you've done these correctly, stop by my office and I'll be happy to look over your products!
Last modified 10/24/00
Dr. Abby Parrill
Department of Chemistry
University of Memphis
These pages may be downloaded and linked from other pages freely for academic and educational purposes. Questions, problems, and errors should be sent to
aparrill@memphis.edu.