Stereochemistry: Specific Rotation and
Related
Calculations
Exercise Solutions
a. 0.856 g in 10.0 mL of solution = 0.0856 g ml-1.
[a]lT
= a/lc
So: [a]D20.0
= +1.06o
/
(1.00 dm) (0.0856 g ml-1) = +12.38o
b. [a]lT
= a/lc
c = 0.0001 M = 0.0001 moles L-1 =
(0.0001) (853.93 g mole-1)
(0.001 L) = 0.000085 g ml-1
Assume l = 1.0 dm (standard polarimeter tube)
Solving for a: a
= [a]lTlc
= (-49o) (1.0 dm) (0.000085 g ml-1) = -0.0042o.
c. Solutions of enantiomers of equal concentrations
rotate
plane polarized light to an equal extent but in opposite
directions.
If a certain compound has a specific rotation of -43.2o (c =
5, toluene) then the enantiomer has a specific rotation of +43.2o
(c = 5, toluene). From the previous question,
a = [a]lTlc
= (+43.2o) (1.00 dm) (1.00 g ml-1) = +43.2o.
We could have anticipated this result without doing a calculation, as
the
rotation is observed under the "standard conditions" of 1.00 dm tube
and
1.00 g ml-1 concentration.
d. i) If the sample is levorotatory, then the
concentration
of the (-) enantiomer is greater than the concentration of the (+)
enantiomer. We cannot calculate the exact concentration, as we do not
know how much
(-) enantiomer is being canceled by the (+) enantiomer. For
example,
the sample might contain 1.1 g ml-1 of the (-) enantiomer
and
1.0 g ml-1 of the (+) enantiomer. This solution would
have the same rotation as a 0.1 g ml-1 solution of the (-)
enantiomer
alone.
ii) Enantiomers have equal but opposite
rotations. An observed
rotation of zero implies the enantiomers are present in equal
concentrations
(a racemic mixture).
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