# Which Alpha-H will be extracted in intramolecular condensation rxn between ester and ketone

In this problem, which hydrogen will be extracted - the one near the ketone group (Marked 2) or the one near the ester (Marked 1)?

My approach:- In intramolecular reactions between aldehydes and ketones, the base extracts $\ce{H}$ from the carbon adjacent to ketone (making it a nucleophile) as the carbonyl carbon of aldehyde has more positive character, thereby rendering attack by the ketonic nucleophile easy. But here case is between ester and ketone.

One can argue that esteric $\ce{H}$ (Marked 1) should be extracted by the same logic as the ketonic carbon has more plus character. Another factor is that if the $\ce{H}$ near ketone is extracted, then it can remove the $\ce{C2H5O^-}$ group(as it is a better leaving group).

So, which effect dominates? I am unable to decide.

Your immediate concern seems to be which hydrogen is abstracted, 1 or 2, in structure 1 (see below). Examination of pKa’s is useful. pKa’s: EtOH, 16; alkyl ketone, ~20; aliphatic ester, ~25; and β-diketone, ~9; aliphatic carboxylic acid, ~5. Clearly, alkyl ketones are more acidic than aliphatic esters. The methyl ketone hydrogen 3 in structure 1 is slightly more acidic than the methylene hydrogen 2 as evidenced by the haloform reaction. But this subtle point is immaterial here. More importantly, the question is what is the structure of A and how is it formed? [A appears to be an intermediate but you have not shown the whole question.] For the sake of discussion, let’s assume that we are dealing with concentrated NaOEt/EtOH. All of these enolates are undergoing proton exchange under the reaction conditions and the concentrations of the enolates are relatively low.

There are two possible reaction pathways to consider. The enolate 2 adds to the ester carbonyl as in intermediate 3. This step is reversible (blue arrows) but elimination of ethoxide leads to β-diketone 4, which, given its low pKa, is irreversibly deprotonated to enolate 5. Acidification of enolate 5 provides β-diketone 4, aka A.) The ring closure (23) occurs through a 5-membered ring transition state that is more favorable than a 7-membered ring transition state resulting from the enolate of the methyl ketone of 1. This reaction must be conducted with at least stoichiometric base because the alkoxide is consumed in the reaction 45.

The alternative pathway (16789) invokes the ester enolate adding to the ketone. All of these steps are reversible except perhaps 98. The irreversible formation of β-diketone channels the reaction irreversibly in the direction of 5. If the formation of unsaturated ester 9 is irreversible, then pathway 14 is faster than 19. Can one argue that the pathway 14 is preferred because there is more ketone enolate than ester enolate? Not necessarily!

Consider the reaction of cyclohexanone (10) with diethylsuccinate (11) under a variety of basic conditions (Stobbe condensation). The product of this reaction is the carboxylate salt 15. Here it is the ester enolate that adds to the ketone! The favorable 5-membered ring lactone formation (1213) leads to the irreversible elimination 1315. Alternatively, the addition of the cyclohexanone enolate to the ester group of diethyl succinate (11) would lead to the irreversible formation of β-diketone enolate 17. In spite of the fact that there is expected to be a preponderance of ketone enolate in this route, it does not prevail!

• Another query if you may.If we replace the ester by an aldehyde group,then will aldehyde enolate be favoured or ketonic enolate?Although Aldehydic adjacent carbon H should be more acidic(due to fact that it has 1 less +Inductive effect of methyl as compared to ketone),My book says attack will take place by Ketone enolate on Aldehydic carbon since Aldehydic carbonyl has more positive character(hence rxn faster).Here are we to ignore acidity since there is not much difference between pka of aldehyde and ketone?Thanks in advance – Yashasvi Grover Mar 22 '18 at 22:41
• You are still too focused on the which hydrogen is more acidic. This is an issue when making kinetic enolates (LDA, aprotic solvent). If the ester is replaced with an aldehydic group, you are looking at an intramolecular aldol condensation. If the aldehyde is the electrophile, you would get 1-(cyclopent-1-en-1-yl)ethan-1-one (1-cyclopentenyl methyl ketone). If the ketone is the electrophile, then the product is 2-methylcyclopent-1-ene-1-carbaldehyde. These products are under thermodynamic control and are generally conducted in aqueous base. – user55119 Mar 23 '18 at 1:00

TL;DR

The hydrogen near the ketone will be plucked.

Comparing positive charge characters

Ketones and esters both have carbonyls. The oxygen of the carbonyl pulls electrons away from the carbonyl carbon. Thus, both carbons (in the ester and the ketone) have partial positive charge.

Ketones have 2 alkyl groups bonded to the carbonyl. The partial positive charge is partially releived by the +I effect due to these groups.

Esters have an alkyl group and an alkoxy group attached to the carbonyl. The alkyl group is responsible for a +I effect and the alkoxy group provides a +M (positive mesomeric) effect, which is stronger than the +I effect.

Thus, there is more partial positive charge on the carbonyl carbon of the ketone.

Comparing enolates

The enolate formed by the ketone is more stable than that formed by the ester.

The lone pairs on the oxygen of the alkoxy group further destabilise the enolate.