We know trihaloacetaldehyde is formed during the haloform reaction of acetaldehyde with a halogen in strong base. Since it has no alpha hydrogens, I expected that it might undergo a Cannizzaro reaction, as we are using a strong base. Why not?
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$\begingroup$ en.wikipedia.org/wiki/Haloform_reaction is for ketones not aldehydes. $\endgroup$– MithoronFeb 12, 2016 at 18:09
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$\begingroup$ @Mithoron Please go through the same link, you will find: The only primary alcohol and aldehyde to undergo this reaction are ethanol and ethanal, $\endgroup$– adianadiadiFeb 12, 2016 at 18:20
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2$\begingroup$ Och, didn't thought about them in this context... Well then it happens for the same reason all haloform reactions happen: CX3 anion is leaving group good enough to be substituted by OH-. Much better then H- for that matter. $\endgroup$– MithoronFeb 12, 2016 at 18:41
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Albeit haloform reaction is primarily for methyloketons, aldehydes which have relatively good leaving groups connected with carbonyl (like chloral), or get it during halogenation, indeed also react.
Here we have another factor limiting usage of Cannizzaro reactions, as although intermediate tetrahedral anion (6) could in principle undergo this reaction if $\ce{R = H-}$, it doesn't if there's a better leaving group then $\ce{H-}$. The point of haloform reactions is to make them if they aren't present already.
