# Which conformer of beta D-glucose is more stable?

I drew the two chair conformers of $\beta$-D-glucose:

Which one is more stable?

I think it's the one with all of the OH groups in the equatorial position because of less steric hindrance, but I also know that the anomeric effect makes the axial one favored as well. So I'm unsure which is more stable.

I also know that the anomeric effect makes the axial one favored as well.

The anomeric effect does stabilize an axial group, but only at the anomeric position. Other axial groups are not similarly stabilized. Thus $\alpha$-D-glucose is more stable than you think it should be based on steric considerations alone.

Knowing this, which is the more stable?

PS - very pretty chairs.

• isn't the beta d-glucose the one with the equitorial bonds? – user176105 Oct 28 '12 at 22:23
• Both of your structures are beta-D-glucose, just different conformations. alpha-D-glucose is the anomer, and would have the OH at the anomeric position different. Otherwise, the two are the same. – Ben Norris Oct 28 '12 at 22:41
• right. so i'm going to say the one on the left is more stable because the anomeric effic is still higher in energy than having all groups equitorial – user176105 Oct 28 '12 at 22:55
• And I would agree. – Ben Norris Oct 29 '12 at 12:33

First, The anomeric effect of hydroxy group in polar protic solvent is small, becasue the dielectric constant of solvent is high. So basically you can not consider anomeric effect as the main stability of beta-D-Glucose in water.

For my understanding, an equatorial anomeric hydroxy group fits into tridymite structure of water than axial position. which make it more stable than an axial anomeric hydroxy group.

• Simply wrong. The anomeric effect is the only explanation as to why D-mannose favours the alpha configuration (you read that correctly). – Jan Nov 5 '17 at 13:29

i think in alpha d glucose the position of OH group and OH of CH2OH are trans so having trans is more stable than having cis.