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

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Discussion Overview

The discussion revolves around the optimal pathway for the Robinson cycloaddition reaction involving the target molecule 4-phenyl-2-cyclohexenone. Participants explore the retrosynthetic analysis, specifically focusing on the aldol condensation and subsequent Michael addition reactions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the best pathway involves using 2-phenyl-2-propenal and acetone, arguing that the negative charge in the alternative pathway is delocalized through the benzene ring, which may reduce nucleophilicity and activate the ring for electrophilic aromatic substitution (EAS).
  • Another participant emphasizes the necessity of a base to generate the active enolate and questions which enolate is easier to generate and more stable.
  • A later reply mentions that the same exercise was solved in a textbook, indicating that aldehydes undergo aldol condensation in base faster than simple ketones, which could complicate the use of 2-phenylacetaldehyde due to potential side reactions.

Areas of Agreement / Disagreement

Participants express differing views on the optimal pathway and the stability of enolates, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

There are unresolved considerations regarding the stability and generation of enolates, as well as the potential side reactions that may arise from the use of certain aldehydes versus ketones.

Aristotelis1999
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TL;DR
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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