I drew the conformer (below) and I'm wondering if this is the most stable chair conformation.enter image description here

Three of the bonds are equatorial so it looks like the most stable to me.

  • $\begingroup$ See this related question chemistry.stackexchange.com/questions/2462/… $\endgroup$ – Ben Norris Nov 1 '12 at 10:45
  • $\begingroup$ that's glucose. this is galactose $\endgroup$ – user176105 Nov 1 '12 at 14:33
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    $\begingroup$ I know that; I am not suggesting the questions are duplicates. However, the issue of the anomeric effect raised in that question is relevant to this question. Someone interested in one question will be interested in the other. $\endgroup$ – Ben Norris Nov 1 '12 at 21:13

The chair conformation that you have drawn (4C1) is likely to be the most stable one, as it minimizes the number of heavy axial groups.

To determine the chair conformation of a hexose, it is generally easiest to draw it and compare it with β-D-glucose, where all heavy groups are equatorial and the conformation is 4C1. If the number of heavy axial groups becomes smaller when the conformation is changed to 1C4 (all equatorial groups in 4C1 become axial and vice versa), then it is likely that the conformation is 1C4.


The conformation shown loses the anomeric effect, worth roughly 17 kcal/mol, and about 0.5 kcamol for the axial OH. This is opposed by the penalty for two axial OH and an axial CH2OH. The penalty for these interaction sis harder to estimate, but it probably amounts to 5 kcal/mol. Thus, on the basis of sterics alone, I expect you have drawn the least stable isomer.

Hydrogen bonding is probably not a factor because in both conformations, the same number of hydrogen bonds is available.


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