Cannizzaro reaction is a disproportionation reaction which is given by aldehydes and ketones having no α-H, forming alcohol (reduction) and carboxylic acid (oxidation) as its products. The compound, Propynal has no α-H and it should be giving Cannizzaro reaction but, my textbook says that the reaction is not given by Propynal. Can someone please explain to me why Propynal doesn't give Cannizzaro reaction?


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    $\begingroup$ My suspicion is that while it has no alpha hydrogens, it does contain an acidic alkyne C-H proton that can be removed by base to produce an anionic conjugate base. This may thus hinder the Cannizzaro reaction. $\endgroup$ Aug 9, 2020 at 7:11
  • $\begingroup$ @Tan Yong Boon: You are correct. That was my first instinct as well. Acidities of enolizable hydrogens and terminal alkknyl hydrogens are compatible. $\endgroup$ Aug 9, 2020 at 7:29
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    $\begingroup$ I think this is assisted by the electron-withdrawing carbonyl group making the alkyne-H more acidic $\endgroup$
    – Waylander
    Aug 9, 2020 at 7:39
  • $\begingroup$ The anion is not resonance stabllized. Is propynal recovered or does base destroy it? E.g., formation of malonaldehyde anion. The lack of formation of propargyl alcohol or propargylic acid may just mean other events occur. $\endgroup$
    – user55119
    Aug 9, 2020 at 14:04

1 Answer 1


It is known that terminal alkynes reacts$1$ with $\ce{NaNH2}$ and produces salt, much like an acid-base reaction, hence it is a fast reaction.

Furthermore, the aldehyde group is an EWG, which makes it easy to lose a proton, hence drives the reaction!

Whereas, cannizzaro reaction happens through multiple reversible steps as seen in the following image,

enter image description here

Therefore, acid-base reaction in this case has much higher probability than Cannizzaro reaction.


$1$: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch09/ch9-2.html

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    $\begingroup$ Do you have a reference for the reaction of alkynes with NaOH? $\endgroup$
    – Waylander
    Aug 9, 2020 at 7:38
  • $\begingroup$ @Waylander, I've added a reference for that $\endgroup$ Aug 9, 2020 at 7:48
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    $\begingroup$ @Rahul Verma, NaOH doesn't react with terminal alkynes (your answer's first line), it is too weak to react with strong bases. Even the Reference you mentioned, has mentioned alkynes won't react with strong bases. Reference: chemistry.stackexchange.com/questions/8448/… $\endgroup$ Aug 9, 2020 at 14:26
  • $\begingroup$ I confused NaNH2 with NaOH, @Waylander. This happens with me, often. $\endgroup$ Aug 9, 2020 at 14:30
  • $\begingroup$ Yes @HariSriram, I didn't even noticed that. $\endgroup$ Aug 9, 2020 at 14:31

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