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I drew the two chair conformers of $ \beta $-D-glucose: enter image description here

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.

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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.

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  • $\begingroup$ isn't the beta d-glucose the one with the equitorial bonds? $\endgroup$
    – user176105
    Commented Oct 28, 2012 at 22:23
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    $\begingroup$ 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. $\endgroup$
    – Ben Norris
    Commented Oct 28, 2012 at 22:41
  • $\begingroup$ 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 $\endgroup$
    – user176105
    Commented Oct 28, 2012 at 22:55
  • $\begingroup$ And I would agree. $\endgroup$
    – Ben Norris
    Commented Oct 29, 2012 at 12:33

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