Ring closure to a cyclic hemiacetal: practice and mechanism
The OH group attacks the aldehyde group in the same molecule. A five- or six-membered ring forms with O in the ring and a new OH group on the former carbonyl carbon (cyclic hemiacetal).
Example
If the OH and aldehyde groups in one molecule can form a five- or six-membered ring, the OH group attacks its own carbonyl group. A cyclic hemiacetal forms. This is exactly why glucose and other sugars exist mainly as rings.
Mechanism step by step
- 1
The carbonyl group is protonated (catalytically) and becomes more electrophilic.
- 2
The OH group at the chain end attacks the carbonyl carbon in the same molecule. A low-strain five- or six-membered ring is favoured.
- 3
After proton transfer the cyclic hemiacetal is present: the ring O plus a new OH group on the former carbonyl carbon.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That would be a lactone (cyclic ester). It would need the aldehyde to be oxidised first; without an oxidant the hemiacetal forms, with an OH group on the ring carbon.
- That would be a lactone (cyclic ester). It would need the aldehyde to be oxidised first; without an oxidant the hemiacetal forms, with an OH group on the ring carbon.
- No oxygen is lost on ring closure: the carbonyl oxygen becomes the OH group on the ring. A plain cyclic ether would need a reduction.
Often confused with: Acetal formation (protecting group) · Reduction with sodium borohydride
Frequently asked questions
Why is the cyclic hemiacetal stable but the open-chain one not?
Intramolecularly the system loses hardly any entropy (no two molecules need to meet). Five- and six-membered rings are practically strain-free, so the equilibrium lies on the ring side.
What is the anomeric carbon?
The former carbonyl carbon (C1 in aldoses). In the ring it carries two O atoms (ring O and OH) and is a new stereocentre: α (OH axial, down in D sugars) or β (equatorial, up).