Reaction of cyclohexane-1,2,4-triol with acetone

Question

What is the product formed when cyclohexane-1,2,4-triol reacts with 1 equivalent of acetone in an acidic medium?

There are 3 possible hemiacetal products. The major one is formed via nucleophilic attack by a less hindered hydroxyl group:

There is one possible cyclic acetal formed by the two adjacent $\ce{- OH}$ groups:

But which of the two products will be formed in a larger quantity? I have heard that cyclic acetals are more stable than acyclic ones but don't know the reason (because of entropy change or something?).

Answer 1

As long as you have a proton source in your solution (hint: all the time) these acetals will equilibrate back and forth until they find the thermodynamic hole. So you will never observe a hemiketal like the one you drew; rather, it will always move forward with the elimination of water (there is only little water around if acetone is the solvent, so liberation of water is entropically favourable) to the formation of the $\ce{O{,}O}$-ketal.

Note that you can form two different ketals if the stereochemistry of the hydroxy groups is correct: If they are 1,3-syn, the six-membered cyclic ketal can also be formed.

Which one you end up with will likely depend solely on the reaction conditions. Kinetically, the five-membered ring is favoured. Thermodynamically, I assume the six-membered ring to be more stable, if it is able to find a conformation with little steric stress.