5.7: Optical Rotation

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An equal mixture of two enantiomers is called a racemic mixture or racemate.
If two enantiomers rotate plane-polarized light in opposite directions, a racemate will not rotate light at all. The effects of the two enantiomers will cancel out.

A rectangle is split in half. The top is shaded white and the bottom is shaded black. The white half is labelled "one enantiomer rotates (+)". The black half is labelled "one enantiomer rotates (-)". A carat connects these two captions with the label "equal, opposite rotation cancels". — Figure \(\PageIndex{1}\): Optical rotation canceled out in a racemic mixture.

A rectangle is split unevenly. The top is shaded white, taking up three-fourths of the rectangle, and the bottom is shaded black, taking up the remaining one-fourth. The white half is labelled "one enantiomer rotates (+)". The black half is labelled "one enantiomer rotates (-)". The bottom half of the rectangle is bracketed with the label "equal, opposite rotation cancels". The top half of the rectangle is bracketed with the label "rotation still left after partial cancellation". — Figure \(\PageIndex{2}\): Optical rotation only partially canceled in a non-racemic mixture of enantiomers

e "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.

Optical purity also corresponds to "enantiomeric excess". If the unknown sample rotates light 50% as much as a sample of pure enantiomer, it must contain 50% enantiomeric excess; the other 50% is a racemic mixture. In other words, if the sample is 75% of one enantiomer and 25% of the other, 50% of the mixture will simply cancel out in terms of optical activity. The remaining 50% will still exert optical activity, but only half as much as if the sample were 100% of that enantiomer.

These relationships could be expressed in formulae:

\[\textrm{Optical purity (op)} = \frac{(\textrm{optical rotation of pure compound})}{(\textrm{optical rotation of pure enantiomer})} \times 100\% \nonumber\]

Enantiomeric excess (ee) = optical purity (that is, these numbers are always the same, although they represent different things)

\[\% \textrm{major enantiomer} = \textrm{enantiomeric excess} + \frac{100 - \textrm{enantiomeric excess}}{2} = 50 + \frac{\textrm{enantiomeric excess}}{2} \nonumber\]

\[\% \textrm{minor enantiomer} = 100 - \% \: \textrm{major enantiomer} \nonumber\]

Exercise \(\PageIndex{1}\)

The (+) enantiomer of compound A has an optical rotation of 75^o. If a sample containing only compound A has an optical rotation of 50^o, what is the composition of the sample?

Answer

A pure sample of A would have \([\alpha] = 75^{o}\)

Optical purity or enatiomeric excess \( = \frac{50}{75} = 66 \%\)

% major enantiomer \(= 66 + \frac{34}{2} = 83 \% \)

% minor enantiomer \( = 100 - 83 = 17 \% \)

Exercise \(\PageIndex{2}\)

The (+) enantiomer of compound B has an optical rotation of 50^o. If a sample containing only B contains 10% of the (+) enantiomer and 90% of the (-) enantiomer, what is the optical rotation value?

Answer

\[ \% major = 90 \% \nonumber\]

\[ \% minor = 10 \% \nonumber\]

Optical purity or enantiomeric excess \( = \frac{X}{-50} = 90 -10 - 80 \% \)

Solve for X.

\[X = -40 ^{\circ} \nonumber\]

Exercise \(\PageIndex{3}\)

The (-) enantiomer of compound C has an optical rotation of -60. A sample of compound C is shown by chiral gas chromatography to contain only (-) C and no (+) C, but NMR analysis suggests the sample is about 50% ethyl acetate by weight. Predict the measured optical rotation of a 1 g/mL solution in dichloromethane, measured in a 1 dm cell.

Exercise \(\PageIndex{4}\)

Which of the following compounds are optically active?

Exercise 5.7.4, showing several different molecules with chiral centers.

Answers to Exercise 5.7.4, with molecules labelled "yes", "no", or "no, meso".