Question

(R)-2-methyl-1-butanol has a specific rotation of +13.5°. The specific rotation of 2-methyl-1-butanol containing 40% of the (S)-enantiomer is ................. (Round off to one decimal place)

Hint:

The specific rotation of a mixture of enantiomers is the weighted average of the specific rotations of the individual enantiomers, depending on their concentrations in the mixture.

Answer 1

Correct Answer: 2.7

Step 1: Formula for specific rotation.
The specific rotation of a racemic mixture is calculated as the weighted average of the specific rotations of the individual enantiomers.
For a mixture, the specific rotation \([ \alpha ]_{\text{mix}}\) is given by: \[ [ \alpha ]_{\text{mix}} = f_1 [ \alpha ]_{(R)} + f_2 [ \alpha ]_{(S)} \] where \(f_1\) and \(f_2\) are the fractions of the (R)- and (S)-enantiomers, respectively, and \([ \alpha ]_{(R)}\) and \([ \alpha ]_{(S)}\) are the specific rotations of the (R) and (S) enantiomers. Step 2: Given data.
- Specific rotation of (R)-2-methyl-1-butanol = +13.5°.
- The sample contains 40% (S)-enantiomer, and therefore 60% (R)-enantiomer.
- Specific rotation of (S)-2-methyl-1-butanol = -13.5° (since it is the mirror image of the (R)-enantiomer).
Step 3: Calculation.
\[ [ \alpha ]_{\text{mix}} = 0.60 \times 13.5^\circ + 0.40 \times (-13.5^\circ) = 8.1^\circ - 5.4^\circ = +2.7^\circ \] Step 4: Conclusion.
Thus, the specific rotation of 2-methyl-1-butanol containing 40% of the (S)-enantiomer is +7.2°.