# Acidity of alpha hydrogens in enolate formation

I am interested in why the most right alpha hydrogen is removed to form the carbanion prior to intermolecular aldol condensation (2nd row first molecule).

Why is it that the alpha hydrogen from the secondary and tertiary alkyl substituted carbon is not removed instead?

I would think that in deciding which alpha hydrogen is removed, the most acidic is removed. Considering that a more stable conjugate base would result in a more acidic alpha hydrogen, wouldn’t the tertiary and secondary substituted alpha carbons be more acidic then? Since the carbanion formed has better charge dispersal than the non substituted alpha carbon on the right (I carried over the idea of charge dispersal from acidity in halogens where a bigger volume can stabilise a charge). also since the carbanion is not electronegative, the alkyl groups would not donate electrons towards the alpha carbanion to destabilise it.

• Welcome to ChemSE! All of the alpha protons are removed continuously under reversible conditions. Under the conditions of the reaction every step should have reversible arrows except for the last one that gives structure A. There is a protic solvent here, presumably water. – user55119 Jun 1 at 17:09
• I'm not posting it as an answer since I'm not completely sure, but I think that abstracting the least sterically hindered H-atom is favourable in this case, so that might be another reason besides the answers given by Waylander and Mathew – TRC Jun 2 at 3:44

I'm interested in the OP's curiosity. I'd agree with Waylander about disfavoring to make 4-membered rings. Yet, it is possible to have two six-membered ring products. The first $$\alpha$$-proton abstraction followed by Michael addition resulted the first intermediate as OP suggested:

Now, this newly formed intermediate contain 4 types of eligible $$\alpha$$-protons, namely, $$\ce{H_a, H_b, H_c,}$$ and $$\ce{H_d}$$. Amongst them, only $$\ce{H_a}$$ is primary, while $$\ce{H_c}$$ is tertiary and the rest, $$\ce{H_b}$$ and $$\ce{H_d}$$, are secondary. The base would abstract each proton in rates according to the dissociation energy of each (e.g., the fastest rate with tertiary and the slowest with primary) but all abstractions are in fast equilibrium as shown in the scheme below:

$$\ce{H_a}$$ and $$\ce{H_d}$$ proton abstractions resulted the formation of intermediates $$\bf{I}$$ and $$\bf{II}$$, which would abstract from a proton from water to give the corresponding aldols. Intermediate $$\bf{I}$$ is the most stable among others and resulted aldol would undergo further dehydration to give the corresponding $$\alpha,\beta$$-unsaturated ketone as the major product (as correctly suggested by the OP). Meanwhile, formation of the intermediate $$\bf{II}$$ is slow compared to that of $$\bf{I}$$ due to greater ring strain (even though it is a six-membered ring formation, the resulting intermediate is a bicyclic product with a bridged-head). Nevertheless, some would form to give the corresponding aldol as a minor product, which would not undergo dehydration because $$\alpha$$-hydrogen is locked in a position that restrict the elimination.

On the other hand, although abstractions of $$\ce{H_b}$$ and $$\ce{H_c}$$ protons are possible and in fast equilibrium, the formation of intermediates $$\bf{III}$$ and $$\bf{IV}$$ are very minute depending on the corresponding activation energies. The rates of these formations are very slow or not exist compared to other two, mostly due to very unfavorable 4-membered ring formation.

• to clarify, for intermediates III and IV would it be the activation energy leading to their formation or their stability relative to the other intermediates that hinders their formation? (i am leaning towards stability to be the main reason) Is the reason why intermediate II cannot undergo elimination because the carbons containing the α-hydrogen and soon to be alcohol functional group do not have corresponding p-orbitals to form a pi bond? for confirmation, do alkyl substituents aid in the deprotonation via charge dispersal? (my teacher insists that they have no effect so i am unsure) – ImJustKawaii Jun 3 at 13:57
• @ImJustKawaii: It's true intermediate III and IV are unstable compared to other two intermediates, and hence contain higher energy. Therefore they need high energy to form and have higher activation energy to form. When formed, the reversed reaction need less energy to go back so eventually more stable intermediates win out. You may need to look at energy profile with reaction progress to fully understand what's I'm saying. – Mathew Mahindaratne Jun 3 at 15:35

The two deprotonated species are in equilibrium (along with the neutral parent molecule). The deprotonation you suggest will occur, but the forward reaction cannot because it would be forming a [4] ring which is dis-favoured. The reaction moves forward through the other anion even though it may be the minor component of the equilibrium mixture.

• what about the secondary substituted alpha carbon on the most left? if it was deprotonated there, wouldn't it form a [6] ring making its forward reaction favourable? – ImJustKawaii Jun 1 at 17:36
• it could, but the species it creates is subject to greater ring strain so will be disfavoured. – Waylander Jun 1 at 17:44
• sorry but i do not see why the species created using the secondary substituted alpha carbon would experience a greater ring strain than the structure A. since both of them form 6 membered rings, wouldn’t they experience similar ring strain? – ImJustKawaii Jun 1 at 19:38