Question

The specific rotation of (S)-phenylalanine is +16.4°. A sample of phenylalanine has a specific rotation of -10.4°. What is the predominant isomer in the mixture? What is the ratio of isomers in the solution?

Answer 1

Answer 2

Understand the Given Data:

• Pure (S)-phenylalanine has a specific rotation of +16.4°.

• The observed rotation of the mixture is -10.4°, meaning it’s levorotatory (rotates light to the left).

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1. Key Insight:

• The (R)-isomer is the mirror image of (S), so its pure rotation would be -16.4° (equal magnitude, opposite sign).

• The mixture’s rotation (-10.4°) is closer to (R)’s value, so (R) must dominate.

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2. Calculate the Enantiomeric Excess (ee):

• ee = (Observed rotation / Pure rotation) × 100%

• Here, we use the magnitude of the pure (S)-isomer’s rotation (16.4°) as the reference:

  $$\text{ee}=\left(\frac{\mathrm{\mid }-10.4\mathrm{\mid }}{16.4}\right)\times 100\mathrm{\%}\approx 63.4\mathrm{\%}$$

• The negative sign of the observed rotation tells us the excess is of the (R)-isomer.

3. Determine the Isomer Ratio:

• (R)-isomer ≈ 81.7%

• (S)-isomer ≈ 18.3%

• ee = % (R) - % (S)

• Let x = fraction of (R), so (1 - x) = fraction of (S).

• Solve:

  $$x-(1-x)=0.634\Rightarrow 2x-1=0.634\Rightarrow x\approx 0.817(81.7\mathrm{\%}R)$$

• Thus:

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4. Adjust for Racemic Contribution:

• The mixture is 63.4% enriched in (R) compared to a racemate (50:50).

• So:

• (R) = 50% + (63.4%/2) ≈ 81.7%

• (S) = 50% - (63.4%/2) ≈ 18.3%

• Wait! The above assumes only (R) and (S) are present, but the ee calculation already accounts for the net excess.

• A simpler (and more accurate) approach:

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5. Final Ratio (S:R):

• Simplify 18.3 : 81.7 ≈ 18:82 or roughly 1:4.5.