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I realize that one of the methyl groups on either carbonyl compound will become deprotonated and then it will attack the other carbonyl to form a pentane ring. However, with the methyl group in the picture below there should be an OH group. There is no heat in the reaction at all that would make this reaction an aldol condensation, thus how does the double bond form?

  • 1
    $\begingroup$ Why do you think you need heat? The initial ring formation, an aldol REACTION, forms the beta-hydroxycyclopentanone. Base again deprotonates the ketone and elimination of hydroxide occurs with net loss of water---an aldol CONDENSATION. Some reactions do occur at ambient temperature. $\endgroup$ – user55119 Aug 21 at 19:55

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