Beta-D-Glucopyranose
By Nathan E. Schultz and Matthew J. Stockinger
A Spartan calculation (HF/STO3G) showed the ground state energy of the beta form to be -447426.64 kcal/mol.
The Beta Form is 4.88 kcal/mol more unstable!!
note - HF/STO3G is not quantitatively accurate, but it is qualitatively accurate.
Unlike the alpha form, the dipole on the ring oxygen is not maximally seperated from the anomeric substituent. Hence, it is more unstable. The distance between the oxygens of interest is 2.29 Angstroms (0.05 Angstroms less than alpha)
The Beta Form a dipole of 2.560 Debye and is oriented in the opposite direction of alpha's dipole.
A map of the electrostatic potential helps to visualize why the beta form is actually more unstable.
The high population of beta is a solvent effect and not because it is sterically favored.
Spartan is also able to perform qualitative calculations on molecules. The PM3-SM3 model (developed by Chris Cramer and Donald Truhlar at the University of Minnesota) was used.
Unfortuntely, the beta form is still higher in energy (-246.25 kcal/mol).
But, the solvation energy {beta --> beta(aq)} is actually higher than for the alpha form.
beta --> beta(aq) = -15.981
That difference is not huge and the beta form is still higher in energy than the alpha form. spooky action at a distance...
Another difference is that the beta form now has a higher dipole (2.77 Debye) and it is oriented in a different direction. Also, pictorial representations of the dipole show that they are roughly aligned in the same direction.
It turns out that this problem was a lot more complicated than we had thought. A great reference (that we found after doing the project) is:
Quantum Chemical Conformational Analysis of Glucose in
Aqueous Solution, C. J. Cramer and D. G. Truhlar, Journal of the American Chemical
Society 115, 5745-5753 (1993).
Conclusions are that the alpha form is more stable. However, the solvent plays a big part in the beta stabilization. The beta form is actually "easier" to put into solution, but the alpha form is still lower in energy. The answer could be in one of the effects that PM3-SM3 does not calculate. The effects that it does calculate are:
-accurate account of the hydrophobic effect for hydrocarbons
-electric polarization of the solvent
-free energy of cavitation
-dispersion interactions
The stabilizing effect is not one of the above.