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The following question was given in an exam.Which of the following undergo aldol condensation ?

Given answer : a,b,c

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My answer: b,c

My Logic :

Cyclopenta-2,4-dienecarbaldehyde(1) on reaction with a base gives carbanion whose resonance structures are 1a to 1e as shown below.This anion is an aromatic anion .Being stable ,tis aromatic anion now in oresence of base would undergo Cannizaro Reaction to give 3 and 4.

enter image description here

My question is, will cyclopenta-2,4-dienecarbaldehyde undergo aldol condensation?

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  • $\begingroup$ Well the CH group between 2 double bonds and a CHO group seems very "deprotonable" to me. But this is just to make your imagination "bubbling" !! $\endgroup$ – SteffX Jul 2 at 16:13
  • $\begingroup$ Hi, please don't make new tags - although this is something the system allows you to do with a handful of reputation, on Chemistry.SE we don't usually do this without some sort of consensus on Meta. In this case, there were already appropriate tags to describe your question well. If you're unsure you can always ask somebody what is a good tag. Also, please avoid the all caps. $\endgroup$ – orthocresol Jul 9 at 17:24
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    $\begingroup$ Somewhat ambiguous in the wording of the question. Presumably, we're talking about self aldol? $\endgroup$ – Zhe Jul 9 at 17:43
  • $\begingroup$ Anyway, the only legitimate reference I can find to the starting material in the literature is this onlinelibrary.wiley.com/doi/abs/10.1002/jlac.19646780107, where it's made as a mixture of isomers/tautomers, so it seems that (as has been the case before) this is a made-up question which cannot be seriously answered. (Reaxys turns up a few more hits for the compound, which upon closer inspection are revealed to be typos.) $\endgroup$ – orthocresol Jul 9 at 18:34
  • $\begingroup$ @orthocresol ,this question was given to students preparing for IIT engineering exam in my city during a weekend test.My point is ,since carbanion is involved in aromaticity, it would remain stable,and therfore may not participate as enolate. And there may no be an aldol. $\endgroup$ – Chakravarthy Kalyan Jul 9 at 20:21

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