Reduction to the sugar alcohol: practice and mechanism
NaBH₄ reduces the open-chain aldehyde form: C1 becomes a CH₂OH group and the ring is then permanently open. A sugar alcohol (alditol) forms, with OH on every carbon.
Example
The ring opens in equilibrium to the open-chain aldehyde form. NaBH₄ reduces this aldehyde to a CH₂OH group and the equilibrium keeps supplying it until everything is reduced. Glucose gives glucitol (sorbitol), an open-chain polyol.
Mechanism step by step
- 1
The cyclic hemiacetal is in equilibrium with the open-chain aldehyde form (below 1 % for glucose).
- 2
The hydride from NaBH₄ attacks the aldehyde carbon of the open form, giving an alkoxide.
- 3
Water protonates; C1 has become a CH₂OH group. The equilibrium keeps supplying the open form until all the sugar is reduced.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- NaBH₄ reduces C=O to CH–OH, not further to a CH₃ group. The OH group on C1 is kept.
- It is not just the anomeric OH group that is removed. NaBH₄ attacks the open aldehyde form, so the ring opens and the former ring O becomes an OH group.
Often confused with: Oxidation to the aldonic acid · Glycoside formation (Fischer glycosylation)
Frequently asked questions
Why does the open form react although there is so little of it?
The equilibrium is established quickly. As the open form is consumed, more rings open (Le Chatelier).
Where do I encounter sugar alcohols?
Sorbitol (from glucose) and xylitol (from xylose) are sugar substitutes in chewing gum and diet products.