1,3-Dicarbonyl compounds such as A are usually mostly enolized. Why is this? Draw the enols available to compounds B-E and explain why B is 100% enol but C, D and E are 100% ketone.

Compound A is mostly enol because only the enol is delocalized over five atoms. A minor reason for this particular compound is the intramolecular hydrogen bond in the enol.

There is another equally good enol that has the other carbonyl group enolized. The two enols are tautomers of each other and of the keto-ester.

That compound B is completely enolized shows that conjugation is much more important than hydrogen bonding, which is impossible with B. However, B has extra conjugation from the lone pairs on the extra oxygen atoms.

The remaining compounds have problems with enolization. Compound C can form an enol on the side away from the other carbonyl group, but cannot form an enol between the two ketones as it would be a 'bridgehead alkene'. These do not generally exist as the four substituents around the alkene cannot become planar.

Compound D seems to have a perfectly reasonable enol. But the very large tert-butyl group, which would be out of the plane in the diketone, would have to lie in the plane if the enol were formed. The four-membered ring in E is already strained enough with two $s{p}^2$ atoms having $90°$ bond angles. The enol would have three such atoms and this is too much strain.