OCOL

Problem #1:

The (+) enantiomer of a compound has an observed optical rotation of 1.72^o when measured in a one dm tube at a concentration of 0.3 g/15 mL. Calculate the specific rotation for this molecule. Click on the "Equation" button to review the equation and the parameters for this problem. Click on the "Answer" button to view the complete solution.

Solution:

The observed rotation, a = 1.72^o

The concentration of 0.3 g/15 mL is equivalent to 0.02 g/mL; c = 0.02 g/mL

The length of the analysis tube is one dm; l = 1.0 dm

The specific rotation is therefore:

[a] = ^{1.72^o_{/(0.02 g/mL x 1.0
dm)}}

[a] = 86^o g^-1 mL^-1 dm^-1

_{[a] = ^{a_{/(c x l)}}}

_{^{The observed
rotation, a =
1.72^o}}

_{^{The concentration of 0.3 g/15
mL is equivalent to 0.02 g/mL; c = 0.02
g/mL}}

_{^{The length of the analysis
tube is one dm; l = 1.0
dm}}