Why does $\ce {OH^-}$ not attack the carbonyl carbon in an aldehyde during the aldol reaction and instead, deprotonate the $\alpha$ hydrogen to form an enolate intermediate?

Also, why does the enolate intermediate attack the carbonyl carbon of another aldehyde once it is formed?

  • 3
    $\begingroup$ It does, but the addition is reversible! It is an unproductive step. Welcome to ChemSE! $\endgroup$
    – user55119
    Dec 30, 2018 at 4:09
  • $\begingroup$ Thank you.Many reactions are reversible i.e in SN1 reaction carbocation formation is reversible but the reaction still proceeds. could you please provide an alternate explination $\endgroup$
    – user72730
    Dec 30, 2018 at 5:19
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/64425/… $\endgroup$ Dec 30, 2018 at 5:29
  • $\begingroup$ Could you clarify your question? It isn't clear at the moment. The phrase "But instead enolate intermediate attacks the carbonyl carbon" seems to be rather out of place. $\endgroup$ Dec 30, 2018 at 5:34

1 Answer 1


As noted by Jan here, both pathways do proceed, i.e. the hydroxide attacks the carbonyl carbon and also deprotonates the $\ce {\alpha-H}$. To give you a rough gauge of the extent of each reaction, I will now provide some data from Carey & Sundberg (2007).

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In general, the hydration of carbonyl compounds (aldehydes, ketones) is unfavourable, i.e. the equilibrium constant has a value that is less than 1. Only in cases where there are inductive (e.g. presence of halogen substituents) or mesomeric effects (e.g. presence of phenyl substituents) operating will the value be much greater than 1. Also, it is noted that the hydration proceeds very quickly, i.e. the rate of reaction is pretty fast. Note that this is reversible.

Having a greater number of steps, amongst other factors, results in the aldol reaction being slower in general. The equilibriums of each individual step in the aldol reaction mechanism are likely to not establish themselves as quickly as that for the hydration of the carbonyl.

From Carey & Sundberg (2007), p. 684:

In general, the reactions in the addition phase of both the base- and acid-catalyzed mechanisms are reversible. The equilibrium constant for addition is usually unfavorable for ketones. The equilibrium constant for the dehydration phase is usually favorable because of the conjugated $\alpha$, $\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall reaction can go to completion, even if the equilibrium constant for the addition step is unfavorable.

So what happens? In the aldol condensation reaction, the dehydration step is what essentially moves the whole equilibrium forward. As the carbonyls react in this aldol pathway and are depleted, the hydrates would convert back to this carbonyl form to react in this aldol pathway. Eventually, if the reaction were to go to completion, the amount of hydrates present would go to zero and all the carbonyls would have been converted to the $\alpha$, $\beta$-unsaturated carbonyl.


Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.


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