Calculate the experimental specific rotation (°) of the equilibrium mixture of α- and β-lactose

Question

Calculate the experimental specific rotation (°) of lactose at time equals zero (t = 0)=87
Enter the observed rotation of the equilibrium mixture of α- and β-lactose (°)=5.2


Calculate the experimental specific rotation (°) of the equilibrium mixture of α- and β-lactose and Calculate the percentage of α-lactose in the equilibrium mixture (use ° and the specific rotations of α- and β-lactose found in the Table of Physical Constants in your Lab Guide.) (%) and Calculate the percentage of β-lactose in the equilibrium mixture (%)??

Answer 1

Without knowledge of the concentration used and the pathlength we cannot calculate the specific rotation of the mixture. Also, it is not entirely clear if at time zero you started with alpha or beta lactose or another mixture. Since the known specific rotations for the anomers are 84° (α) and 34° (β) I'm assuming you are starting with alpha lactose.

 

The issue I'm having here is that your observed rotation should not be below that of the beta anomer, which leads me to think that the value you gave above is not the calculated observed rotation, but rather the number you wrote down from the polarimeter. Since the value for the observed rotation needs to be calculated taking into account the concentration of the sample and the pathlenght, which we are not given, we cannot help you solve your problem.

 

Nevertheless, below are the important equations that will allow you to determine the percentages of each anomer:

 

[observed rotation] = [(polarimeter rotation)/(concentration)(pathlength)] x 100%

 

anomer excess = [(observed rotation)/(specific rotation)] x 100%

 

percent anomer = (anomer excess) + [(100 - anomer excess)/2]