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?
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.