Will cinnamaldehyde undergo self-aldol condensation?
My book says it won't, but since it has an α-hydrogen, I'm confused.
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2 Answers
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Not every α-hydrogen automatically means that a ketone is enoliseable. It would need to be an α-hydrogen capable of aligning with the π system of the $\ce{C=O}$ bond such that removing the hydrogen automatically generates the required enolate anion system. In the case of cinnamaldehyde — your compound — there is conjugation across the phenyl ring, the double bond and the aldehyde meaning that this compound is entirely planar. Thus, there is no way for the α-hydrogen to align favourably for deprotonation to occur. If it did, it would generate an allene-olate $\ce{Ph-CH=C=CH-O-}$ which is not particularly favourable.
In general, α,β-unsaturated ketones and aldehydes are deprotonated vinylogously, meaning that the double bond shifts away upon deprotonation. For example crotonaldehyde $\ce{H3C-CH=CH-CHO}$ is deprotonated in the following way:
$$\ce{H3C-CH=CH-CHO ->[][- H+] H2C=CH-CH=CH-O-}\tag{1}$$
I.e. the γ-hydrogen is abstracted in place of the α-hydrogen.
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No aldol condensation is not possible here, as you would have to extricate an alpha hydrogen(alpha to the carbonyl group), but if you do extricate a hydrogen atom from the alpha position it would result in the formation of Vinylic Carbanion which is very very unstable.
Thus the molecule would not want the alpha hydrogen to get extricated, thus aldol reaction would not proceed as you need to remove the alpha hydrogen in order to proceed forward with the reaction.
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$\begingroup$ You might want to read the question you linked; it actually does not say that vinylic carbanions are rather unstable. They are actually more stable than alkyl carbanions. $\endgroup$– JanMar 1, 2017 at 18:31
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1$\begingroup$ (Side note: if you remove that wrong bit from your answer I’ll upvote it because the remainder is pretty good.) $\endgroup$– JanMar 1, 2017 at 18:34
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$\begingroup$ Everything that calls the vinylic carbanion very unstable. And probably everything that talks about vinylic at all because the anion wouldn’t be a vinylic carbanion. $\endgroup$– JanMar 1, 2017 at 18:42