My teacher told me that the $K_\mathrm{eq}$ of acetone is greater than that of benzaldehyde in nucleophilic addition reaction.

But I am not able to understand how is this possible. Acetone has a +I group attached to it, so it will decrease the reactivity of carbonyl. If we talk about benzaldehyde, it has a −M group. Can someone tell what is the correct reason behind it?


The information given by your teacher is incorrect. The $K_\mathrm{eq}$ value of acetone for cyanohydrin formation (a nucleophilic addition reaction) is less than the $K_\mathrm{eq}$ shown by benzaldehyde

According to Organic Chemistry: Structure, Mechanism, Synthesis [1 , p. 595], the values of $K_\mathrm{eq}$ for cyanohydrin formation are as follows:

$$ \begin{array}{lr} \hline \mathrm{Compound} & K_\mathrm{eq} \\ \hline \text{Acetone} & 20 \\ \text{Ethanal} & 10^5 \\ \text{Benzaldehyde} & 210 \\ \textit{p}\text{-Methoxybenzaldehyde} & 30 \\ \text{Acetophenone} & 0.8 \\ \hline \end{array} $$

Hence the statement that the $K_\mathrm{eq}$ is greater for acetone is incorrect. This seems accurate considering that acetone is more sterically hindered than benzaldehyde, which is a dominating factor for nucleophilic addition reaction since they are kinetically stable reactions.

This document has a much larger cases of compounds that are compared for multiple reactions, in all of which benzaldehyde is more reactive than acetone.

However, there is one minor issue with your reasoning given in the question.

Resonance in Benzaldehyde

Benzaldehyde does not have a −M group in it. However if you add another functional group into the ring, then we say that the $\ce{-CHO}$ group acts as a −M group. Actually, $\ce{-CHO}$ has a +R effect from the phenyl ring attached to it which makes the ring in benzaldehyde deactivated. So, the carbon in benzaldehyde is less electrophilic than formaldehyde which has no such effects.


  1. Ouellette, R. J.; Rawn, J. D. Organic Chemistry: Structure, Mechanism, and Synthesis, 1st ed.; Elsevier, 2014. ISBN: 978-0-12-800780-8 (Google Books Preview)

A quantitative justification using tables of actual values has already been provided in the other answer, where it is evident that the $K_\mathrm{eq}$ for nucleophilic addition for acetone will be lesser than the $K_\mathrm{eq}$ for benzaldehyde.

I will be talking about how on a qualitative basis this can be predicted. Acetone is an example of a ketone, while benzaldehyde has the aldehydic functional group. In a very general sense, ketones are more sterically hindered than aldehydes. It's true that conjugation with the phenyl ring can reduce electrophilicity of the carbonyl group for benzaldehyde, but that will simply lead to nucleophilic addition to benzaldehyde becoming slower with respect to other aldehydes like formaldehyde(nucleophilicity being a kinetic property) as the activation energy barrier for addition will be slightly increased.

Comparing $K_\mathrm{eq}$ is a question of the relative product distribution in the equilibrium. It is basically a thermodynamic consideration, which will largely be governed by judging the relative stability of the tetrahedral intermediates formed after the addition(the more stable counterpart will shift the reaction towards the intermediate's side in accordance with Le Chatelier's principle). Due to steric clashes caused by the two bulky alkyl side chains, the tetrahedral intermediate formed from the ketone will always be relatively less stable than that formed from the aldehyde(which has only one alkyl side chain, while the other substituent is a small hydrogen atom). Hence you get the $K_\mathrm{eq}$ for benzaldehyde to be greater than that of acetone


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