# What does the intramolecular aldol condensation of 6-oxoheptanal form?

I know that when 6-oxoheptanal is treated with NaOH, an intramolecular aldol condensation will take place. But will a five-membered ring or seven-membered ring be formed?

I know that the five-membered ring is more stable in comparison to the seven- membered ring, but as per the mechanism of aldol condensation the first step is removal of acidic hydrogen (which is also the slowest step i.e. the rate determining step {if I am not wrong}).

So if I remove the acidic hydrogen on C-7 (the terminal methyl group), then the (primary) carbanion formed will be more stable than if I were to deprotonate on C-5.

So I think that the major product should be the seven-membered ring.

Am I right? Or is my reasoning incorrect?

• You need to remember that every step in this process is reversible. – Waylander Apr 11 '20 at 8:19

B.Anshuman has given a short answer, but I felt more explanation is needed to understand this situation. This reaction can be done in kineticlly controlled conditions (e.g., using $$\ce{LDA/THF}$$ as a base at $$\pu{-78 ^\circ C}$$) or in thermodynamically controlled conditions as in this case. Each case may give totally different major products. For example, alkylation of 2-methylcyclohexanone gives 6-alkyl-2-methylcyclohexanone as the predominanet product under kinetic conditions (1. $$\ce{LDA/THF}/\pu{-78 ^\circ C}$$; 2. $$\ce{R-Cl/THF}/\pu{-78 ^\circ C}$$, alkyl chloride is $$1^\circ$$ or $$2^\circ$$). However the same starting material would gives 2-alkyl-2-methylcyclohexanone as the amjor product under thermodynamic condition such as $$\ce{NaOEt/EtOH}/\gt \pu{25 ^\circ C}$$.

In above example shows that the thermodynamic conditions allow equilibration of the two possible enolates (which are derived from $$2^\circ$$ and $$3^\circ$$ carbanions); the one enolate with greater alkyl substitution on $$\ce{C=C}$$ would form in greater concentration (Zaitsev-like). The question in hand can also be treated like the same because the situation is similar ($$1^\circ$$ vs $$2^\circ$$ carbanions).

The literature has shown that the evidence for carbanion formation might not be the rate determining step (Ref.1). The abstract if the reference is given below for your benefit for reasoning:

Rate and equilibrium constants have been determined for both the aldol addition and the elimination steps in the intramolecular condensation reactions of 2,5-hexanedione, 2,6-heptanedione, 1-phenyl-1,5-hexanedione, and 5-oxohexanal. The overall thermodynamics are similar for cyclization of 2,5-hexanedione and 2,6-heptanedione; conversion of 2,5-hexanedione to the corresponding enone is actually more favorable, but the cyclization of 2,5-hexanedione is 2400 times slower than that of 2,6-heptanedione. As expected on the basis of intermolecular analogs, the addition step is less favorable and slower for 1-phenyl-1,5-hexanedione, and the addition step for 5-oxohexanal is more favorable though similar in rate to that for heptanedione. Detailed analysis of the kinetics and equilibrium for all of these compounds, as well as 2-(2-oxopropyl)benzaldehyde, in terms of Marcus theory, leads to the same intrinsic barriers for the intramolecular reactions as were seen previously for the intermolecular reactions. This means that rate constants for intramolecular aldol reactions should be predictable from the energetics of the reactions and that the effective molarity can be calculated. Methods for estimating thermodynamic quantities for reactants and products of these reactions have been examined.

Thus, my conclusion is formation of 5-membered ring is predominated here to give 1-acetylcyclohexene as the major product among three theoretically possible products.

Reference:

1. J. Peter Guthrie, Junan Guo, "Intramolecular Aldol Condensations:  Rate and Equilibrium Constants," J. Am. Chem. Soc. 1996, 118(46), 11472-11487 (https://doi.org/10.1021/ja954247l).

Intramolecular-Aldol reactions involves heat, as you have mentioned, and the base used (NaOH) isn't much affected by steric factors. Therefore, the preferred product should be thermodynamically controlled product. Moreover, as you commented about the formed carbanion being less stable in case of 2° carbon, actually the carbanion is converted to an enolate then the reaction proceeds further. So, the 5-membered cyclic ring should be the major product.

• The carbanion isn't converted to an enolate; it is an enolate. – orthocresol Apr 11 '20 at 11:01
• The base extracts the hydrogen from the carbon, carbon accepting the electron. So though temporarily, the carbon must have an excess negative charge, i.e a carbanion, which becomes an enolate. Please explain, if otherwise. – B.Anshuman Apr 11 '20 at 11:03
• Nope; you deprotonate a carbonyl, you get an enolate. The carbanion that you're thinking of is a resonance form. It doesn't exist as an independent entity. Not even transiently. – orthocresol Apr 11 '20 at 11:06
• Enolate can also be considered as a one of the resonating structure, so actually, none of them should be an individual identity. Sorry for taking your time. – B.Anshuman Apr 11 '20 at 11:10
• Well, okay; let's call it a "resonance hybrid which looks almost like an enolate", or RHWLALAE for short. When you deprotonate a carbonyl compound, you get a RHWLALAE; there's no carbanion intermediate or transition state. So we can rephrase my original comment: the carbanion isn't converted to an enolate; it is a RHWLALAE. If you accept that the RHWLALAE is close enough to an enolate, then we can call it an enolate, but if not we can agree to call it a RHWLALAE. But that doesn't affect my point about using the word "converted" - it's not right to say that. – orthocresol Apr 11 '20 at 11:29