Question

la) Consider the mixed aldol condensation reaction of 1-methylcyclopentane-1-carbaldehyde (shown below) and 3,3-dimethyl-2-butanone (shown below) in the presence of sodium hydroxide Provide the expected major MIXED aldol condensation product for this reaction (can exist as E or Z isomer so you can draw either) b) Would you have to heat the reaction mixture in order for the reaction to form the aldol condensation product? Why or wh y not? 2) Consider the mixed aldol condensation reaction of 3,3-dimethyl-2-butanone (shown in question 1) and propiophenone (shown below). 0 Two major MIXED aldol condensation products arise from this reaction (shown below, labeled A and B (each will exist as E and Z isomers) If you wish to form product A, to which starting carbonyl would you add a strong base such as LDA in order to form the enolate? Which carbonyl would you then add slowly to the enolate of the first carbonyl?

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Answer #1

1a)In the mixed aldol condensation reaction between 1-methylcyclopentane-1-carbaldehyde and 3,3-dimethyl-2-butanone there is an -carbon in 3,3-dimethyl-2-butanone.Therefore the OH-ion from the NaOH will attack on the -carbon to form the major aldol condensation product i.e.1-methylcyclopentane-4 4-dimethylpent-1-en-3-one.

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