A sample of 2-methyl-1-butanol has a specific rotation of -3.25°. Pure (R)-2-methyl-1-butanol has a specific rotation of +5.75°. Determine the %ee and molecular
composition of the sample.
To determine the percent enantiomeric excess (%ee) and molecular composition of the sample, we need to use the formula:
%ee = (Observed specific rotation - Specific rotation of pure enantiomer) / Specific rotation of pure enantiomer * 100
1. Calculate the %ee:
%ee = (-3.25° - (+5.75°)) / (+5.75°) * 100
%ee = -9 / 5.75 * 100
%ee ≈ -1.57%
The negative sign indicates that there is an excess of the (S)-enantiomer in the sample.
2. To determine the molecular composition, we need to assume that the two enantiomers are present in equal amounts when the %ee is 0%. However, since we have a negative %ee, it means there is more of the (S)-enantiomer present than the (R)-enantiomer.
Assuming the total mass of the sample is 100 g, we need to find the masses of each enantiomer.
Let x be the mass of (R)-2-methyl-1-butanol
Then (100 - x) will be the mass of (S)-2-methyl-1-butanol
Using the specific rotation values, we can relate the observed rotation of the mixture to the amount of (R)-2-methyl-1-butanol (x) present:
(x * specific rotation of (R)-2-methyl-1-butanol) + ((100 - x) * specific rotation of (S)-2-methyl-1-butanol) = observed rotation
In this case, the observed rotation is -3.25°.
Substituting the specific rotation values (-3.25° and +5.75°) into the equation:
(x * +5.75°) + ((100 - x) * (-5.75°)) = -3.25°
Solving this equation will give us the value of x, which represents the mass of the (R)-2-methyl-1-butanol enantiomer.
Using this value, we can then calculate the mass of the (S)-2-methyl-1-butanol enantiomer (100 - x).
With the masses of the two enantiomers known, we can determine the mole fractions and molecular composition in the sample.