Understanding Chirality and Specific Rotation

This question involves calculating the composition of an enantiomeric mixture using specific rotation data. Key concepts are specific rotation and enantiomeric excess (ee).

Calculating Enantiomeric Excess (ee)

The enantiomeric excess (ee) measures the difference between the amounts of the two enantiomers in a mixture. It can be calculated using the specific rotations:

• Specific rotation of pure (S)-2-butanol: $[α]_{pure} = +14^\circ$
• Specific rotation of the mixture: $[α]_{mixture} = +7^\circ$

The formula for ee based on specific rotation is:

$ ee = \frac{[α]_{mixture}}{[α]_{pure}} \times 100\% $

Substituting the given values:

$ ee = \frac{+7^\circ}{+14^\circ} \times 100\% = 0.5 \times 100\% = 50\% $

Since the mixture's rotation ($+7^\circ$) is positive, the (S)-enantiomer (with rotation $+14^\circ$) is the major component.

Determining Percentage of (R)-2-butanol

An ee of 50% means the major enantiomer is in excess by 50% over the minor enantiomer. Let $X_S$ be the percentage of (S)-2-butanol and $X_R$ be the percentage of (R)-2-butanol.

• We know $X_S + X_R = 100\%$ (Total percentage).
• We also know $ee = X_S - X_R = 50\%$ (Excess of the major, S, enantiomer).

Solving these two equations:

1. Add the two equations: $(X_S + X_R) + (X_S - X_R) = 100\% + 50\%$
2. This simplifies to $2X_S = 150\%$, so $X_S = 75\%$.
3. Substitute $X_S$ back into the first equation: $75\% + X_R = 100\%$.
4. Therefore, $X_R = 100\% - 75\% = 25\%$.

The percentage of (R)-2-butanol in the mixture is 25%.