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What happens to 1,5-dimethyl-6,8-dioxabicyclo[3.2.1]octane when it is hydrolyzed under acidic conditions?

I know that under acidic conditions the first thing to likely happen is protonation. Does it matter which oxygen is protonated first?

Once that oxygen has been protonated it is going to want to return to neutrality.

Now we have our carbonyl restored (sort of). However, we still want to return to neutrality. Water can attack.

Then we deprotonated the attached water and regenerate the acid-catalyst and we have our original starting compound.

Is this the correct mechanism and is the product correct?

  • 1
    $\begingroup$ In your middle structure, you show water attacking the methylated carbon on the left side of the drawing, yet it is the methylated carbon on the right side of the structure that carries some positive charge due to resonance. Examine nucleophilic attack by water at this positive carbon and see what happens $\endgroup$
    – ron
    Oct 18, 2014 at 23:45

1 Answer 1


Your product is correct, but the mechanism isn't quite right. In the first step, one of the oxygens will be protonated (either is possible). The other oxygen "kicks out" the one activated in the first step. So far so good. However, in the next step, water will attack the "carbonyl" carbon to give a hemi-acetal and a hydroxyl group. Repeating a similar set of steps reveals the ketone and second hydroxyl.

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The key difference here is that by adding to the pi system, water reacts with the C-O pi-antibonding orbital, which is the LUMO of the system

  • $\begingroup$ Thanks! I had water initially attack on the side you suggested, but I just couldn't see what would happen next! This helps a ton! $\endgroup$
    – Dissenter
    Oct 18, 2014 at 23:46

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