Which of these two pathways is better for Robinson cycloaddition?....

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The discussion centers around a question from an organic chemistry exam regarding the synthesis of 4-phenyl-2-cyclohexenone. The main focus is on the retrosynthetic analysis, which identifies one aldol condensation reaction and two potential Michael addition reactions. The preferred pathway involves using 2-phenyl-2-propenal and acetone, as the alternative option reduces the nucleophilicity of the a-carbon due to charge delocalization in the benzene ring. A key point raised is the necessity of a base to generate the active enolate, with emphasis on the stability and ease of generation of the enolates. The discussion concludes with insights from the teacher, highlighting that aldehydes undergo aldol condensation more rapidly than ketones, and the potential side reactions with 2-phenylacetaldehyde complicate the synthesis.
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Target molecule: 4-phenyl-2-cyclohexenone. 2-phenylacetaldehyde + 3-buten-2-one or 2-phenyl-2-propenal + acetone?
So this was a question on my orgII exam yesterday. The target molecule is 4-phenyl-2-cyclohexenone. By retrosynthetic analysis there is only one possible aldol condensation reaction, but two possible michael addition reactions afterwards. My "guess" was that the best one is 2-phenyl-2-propenal + acetone because in the other case the negative charge is delocalized throughout the benzene ring reducing the nucleophilic character of the a-carbon and also activating the ring in which case EAS could take place with the ring acting as the "Michael Donor". Please correct if I was supposed to post this question in another forum. Thank you a lot in advance.
 
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You need a base to generate the active enolate. Which enolate is easier to generate? Which one is more stable?
 
TeethWhitener said:
You need a base to generate the active enolate. Which enolate is easier to generate? Which one is more stable?
Well, there's an obvious answer for that. But anyway, I went and asked my teacher and it turns out the same exact exercise was solved in the textbook. The answer was what I said, but for very different reasons. Aldehydes undergo aldol condensation in base much faster than simple ketones. So, the issue with using the 2 - phenylacetaldehyde is aldol condensation with another same molecule as a side reaction.
 
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Very interesting. Good to know.
 
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