[john2520]

Alpha-D-Mannose has a specific rotation of 20.3 while the B-D-mannose has a specific rotation of -17.0. A mixture of 85% of Alpha-D-Mannose and 15% Beta-D-Mannose is dissolved in distilled water.

a) What is the initial specific rotation of D-Mannose
b) After sitting for several hours the specific rotations was measured to be 0 degrees. What is the percentage of alpha and beta D-Mannose at equilibrium?

Can someone help me with these, for part A i got 14.705 and for part B I got Beta to be 54% and Alpha to be 46%. Am I right?

[Schrödinger]

Yeah, you're numbers seem to be fine. But next time, post your working here so that people can see if you've arrived at the answer in the right way.

[john2520]

oh ok thank you. I was just wondering for part b, does it make sense that the majority is the B anomer even though we started with a lot more of the alpha anomer?

[Schrödinger]

You can't really say anything about that since you don't have any data on the equilibrium constant for the reaction (equilibrium) that inter-converts the alpha form to beta form.

[Dan]

does it make sense that the majority is the B anomer even though we started with a lot more of the alpha anomer?

It could, as has been said it depends on the equilibrium constant for mutarotation.

In reality, the equilibrium composition of mannopyranose in water is ~2:1 α:β, so a 46:54 α:β mixture is not practically realistic. But it is a hypothetical question - If the optical rotation measured 0, then it would imply that ratio based on the data you have been given. You have answered the question correctly, but the question its self does not give a real life scenario.

The actual equilibrium rotation of D-mannose is about +14; the actual rotation of α-D-mannopyranose is +29 (not +20); the actual rotation of β-D-mannopyranose is -17.

A good reference for the equilibrium composition of sugars in water: ACIE 1969, 8, 157-166