[shayne]

Urgh.. i've been trying to solve this for a whole day now but still can't figure it out.. anyone has an idea how?

A freshly prepared solution of a-D-glucose shows a specific rotation of +112 deg. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to [a]D= +52.5. In contrast, a freshly prepared solution of B-D-Glucose has a specific rotation of +19 degrees. The rotation of this solution increases over time to the same equilibrium value as that shown by the a-anomer. Temperature=25 degrees.

Q: calculate the % of each two forms of D-glucose present in equilibrium.

[Yggdrasil]

Start with this simpler problem to start.  If the solution contains 50% β-D-glucose and 50% α-D-glucose, what will the optical rotation be?  If the solution contains x % β-D-glucose and (100-x)% α-D-glucose, what will the optical rotation be in terms of x?

[shayne]

here's what i did:

52.5 x 100 = 112 x (100-x) + 19x
x = 63.98% of B

100-63.98 = 36.02% of a

correct?

[Yggdrasil]

I didn't double check the math, but the approach looks correct to me.  Now, here's something to consider.  Does the answer make sense?  Based on the structures of alpha- and beta- D-glucose, which one would you expect to be more stable (remember that 6-membered rings adopt chair conformations)?