What is the structure of the hemiacetal formed by reaction of butanone with methanol?

1 Answer
Jan 8, 2016

Butanone is more specifically 2-butanone, due to symmetry and the number of carbons on the main chain. This reaction also requires acid if you are using methanol as-written.

Furthermore, it is a hemiketal that forms, since butanone is a ketone, not an aldehyde. This reaction might then keep going if you have enough methanol and acid, and you would then get a regular ketal.

The mechanism for this is just like any other nucleophilic addition-elimination reaction for a nucleophile attacking a carbonyl carbon.

It follows the same pattern:

  1. Protonate the carbonyl because the oxygen is electron-rich (electron donor / proton acceptor).
  2. The carbonyl carbon becomes electrophilic enough due to the electron-withdrawing nature of the carbonyl oxygen, so methanol can act as a good enough nucleophile.
  3. Proton transfer part 1. Formation of hemiketal.
  4. Proton transfer part 2.
  5. Tetrahedral collapse to eliminate the leaving group (water). (The equilibrium between the hydroxide getting protonated and the methoxide getting protonated is very balanced, so it's easy to tip it either way. But if we want to see how the reaction would go forward, so we draw it that way.)
  6. Water has a slightly higher pKa than methanol (#15.7# vs. #~15.5#), so it steals a proton from the precursor to the ketal.