Question

The specific rotation of enantiomerically pure (S)-2-butanol is \( +14^\circ \). The specific rotation of the enantiomeric mixture of 2-butanol obtained from an asymmetric reduction of 2-butanone is found to be \( +7^\circ \).

The percentage of (R)-2-butanol present in the reaction mixture is _________ (in integer).

Hint:

When dealing with enantiomeric mixtures, use the observed specific rotation and known specific rotation of pure enantiomer to find enantiomeric excess (ee). Then calculate the percentage of each enantiomer accordingly.

Answer 1

Let the mole fraction of (S)-2-butanol = \( x \)
Then the mole fraction of (R)-2-butanol = \( 1 - x \)

Since (S)-2-butanol has a specific rotation of \( +14^\circ \) and (R)-2-butanol has \( -14^\circ \), the observed specific rotation is:

\[ [\alpha]_{\text{obs}} = x(+14) + (1 - x)(-14) = 14x - 14(1 - x) = 14(2x - 1) \]

Given that \( [\alpha]_{\text{obs}} = +7^\circ \), we solve:

\[ 14(2x - 1) = 7 \Rightarrow 2x - 1 = \frac{1}{2} \Rightarrow x = \frac{3}{4} \]

So, the percentage of (S)-2-butanol is \( 75\% \), and that of (R)-2-butanol is:

\[ 100\% - 75\% = \boxed{25\%} \]