The title speaks for itself really.

I know that if we have a conventional ketone (with 3+ α-hydrogens) and an aldehyde with less than or equal to 3 α-hydrogens, the ketone acts as the nucleophile and attacks the aldehyde, which consequently forms the major product.

Let us take an example of a case where we have acetophenone and acetaldehyde. Both have 3 α-hydrogens. Also, the acetophenone is sterically hindered due to the phenyl group. Which is the major product here?

  • 1
    $\begingroup$ Depends on the reaction conditions. Are you setting up to get the kinetic product or the thermodynamic product? $\endgroup$
    – Waylander
    Commented Apr 1, 2019 at 14:48
  • 1
    $\begingroup$ @Waylander it wasn't mentioned in the question. Actually this was a question asked in an exam. What will be the product in each case? And why so? $\endgroup$
    – Arka Seth
    Commented Apr 1, 2019 at 16:13

1 Answer 1


There are two main steps in any aldol condensation:-
1. Formation of enolate ion
2. Using the enolate ion as a nucleophile.

The 1st step is a reversible step. If you think that the kinetically favoured enolate ion will be formed fast then you are wrong as it also disappears faster. So, in any case the one which is more thermally stable will form in more amount except in the cases where the formation of stable enolate is too slow or the backward reaction of kinetic favoured enolate is blocked or not enough time has been given to let the reaction reach equilibrium.

Now, I will tell you about above strategies to block reverse reaction or make formation of stable enolate too slow:-

1. Use a strong electron withdrawing group at alpha position to the newly formed CH2(-)COR so that the electron can spend most of time away from the original carbon.
-This blocks reverse reaction

2. Use a bulky base or somehow hinder the alpha carbon to make the extraction of proton hard which will in turn make the reaction slow.

3. Using Non-Protic Solvent stops the protonation of the enolate ion which in turn blocks the reverse reaction.

Now, the answer for your question, the conjugate base of acetaldehyde is more stable when compared to the former but the attack of conjugate base of acetophenone is more easier as compared to latter because of steric reasons and the more electrophilic character of aldehyde and when you make cannonical structures, the number of structures which has negative charge on oxygen is greater than being away which makes it a good nucleophile when you take weighted average of all. So, the major product should be 1-phenyl-but2en1one.

  • $\begingroup$ I think that a mixture should get formed as the conditions are not so much sharp. Though i provided what i think $\endgroup$ Commented Apr 3, 2019 at 6:49
  • $\begingroup$ the product you've mentioned is the one we get after heating. In the answer, they've kept the alcohol intact (the one formed after the enolate attacks the aldehyde). By the way, which exam are you preparing for? And I'll think about it and get back to you when possible. $\endgroup$
    – Arka Seth
    Commented Apr 3, 2019 at 13:19
  • $\begingroup$ I'm preparing for JEE as well. My exam is on the 8th (second half). All the best to you! $\endgroup$
    – Arka Seth
    Commented Apr 3, 2019 at 17:35

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