EvidenceBasedReason Posted July 24, 2012 Share Posted July 24, 2012 My question is; If using the same two aldehyde's (CH3CH2CHO) for both alpha substitution and condensation reactions I always get the same end product, basically a Beta-hydroxy aldehyde. Is the only difference in these reactions simply the base used to deprotonate? I get that you need a full equivalent of something fairly strong such as Na+OEt- for the alpha and a catalytic amount of something fairly week such as -OH for the condensation. Is that all it is, or is there some fundamental flaw in my reaction mechanism? Link to comment Share on other sites More sharing options...
Dr. Lennox Posted August 6, 2012 Share Posted August 6, 2012 You will certainly proceed through a beta-hydroxy carbonyl intermediate, however if heat is specified for the reaction, the result is beta-elimination affording the enal. Additionally, OH- or EtO- is not always used as there are advantages to using LDA to obtain kinetic control. Please read the articles below: Aldol Condensation Part I: The Basics Aldol Condensation Part II: Kinetic Control Aldol Condensation Part III: Stereochemistry Good luck! Link to comment Share on other sites More sharing options...
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