Product of intramolecular aldol condensation of 3-methyloctane-2,7-dione
For the enolates that can be formed, the others have already given comprehensive answers: All four enolates are possible and since we are under thermodynamic control (weak base), we expect o observe them all. Numbering the chain carbons according to IUPAC, and noting the non-carbonyl end of the enolate, I expect the enolate ratio to be $3&gt;6&gt;1\approx 8$.

But your question asked for the reaction’s major product. First of all, the two terminal enolates (carbons 1 and 8) are to be dismissed: The seven-membered ring is much less stable. Then, the reaction mechanism shows us which enolate leads to a thermodynamically much more stable product than the other:



The more stable enolate 2 unfortunately cyclises to give a β-hydroxyketone 5 which has a quarternary carbon between the ketone and the newly-formed hydroxy group. No elimination is possible here. The other enolate aldol product 3, however, can eliminate to the corresponding cyclopentene 4. This should likely be given as the reaction’s major product since all steps up to the elimination are reversible. (The elimination is, too, but it is still most stable.)

Indeed, Stork and Williard report that the cyclopentene 4 is obtained when 3-methyloctane-2,7-dione is refluxed with ethanolic KOH for 2 hours, although no yields are given as it is only mentioned in a footnote.[1]


Stork, G.; Williard, P. G. Five- and six-membered-ring formation from olefinic α,β-epoxy ketones and hydrazine. J. Am. Chem. Soc. 1977, 99 (21), 7067–7068. DOI: 10.1021/ja00463a053.