Chapter 17, Problem 1Q

Interpretation Introduction

Interpretation:

$\text{ee}\%$ and molecular composition of a sample of $2-\text{butanol}$ is to be determined.

Concept introduction:

Optical activity is the ability of substances to rotate plane-polarized light when it falls on it.

A molecule is called chiral if it possesses no plane of symmetry and is not superimposable on its mirror image. A molecule that has a tetrahedral atom with four different substituents known as stereocenter is chiral. A stereocenter is also called a chiral or asymmetric center. Chiral compounds exhibit the property of enantiomerism. Enantiomers are stereoisomers that have non-superimposable mirror images.

The magnitude of the optical rotation relies on the concentration of the optically active compound, the length of the path of light that passes through the solution, the wavelength of the light, the nature of the solvent, and the temperature. The optical rotation is shown by the symbol ${\text{[α]}}_{\text{λ}}^{\text{T°}}$ known as specific rotation and is an inherent property of a pure optically active compound. $\text{T°}$ is the temperature of the measurement in degrees Celsius, and $\text{λ}$ is the wavelength of light used. The purity of optically active compounds is expressed as an enantiomeric excess.

Answer to Problem 1Q

The value of $\text{\% ee}$ of $\text{2-butanol}$ is $25\%$ and the molecular composition is $62.5\text{\%}(+)\text{-2-butanol}$ and $37.5\text{\%}(-)\text{-2-butanol}$ .

Explanation of Solution

Given Information:

A sample of $\text{2-butanol}$ has a specific rotation is $+3.25°$ .

The specific rotation of pure $\left(\text{+}\right)\text{-2-butanol}$ is $+13.0°$ .

The purity of optically active compounds can be determined by enantiomeric excess. The formula to calculate the enantiomeric excess $\left(\text{ee}\%\right)$ is as follows:

  $\text{\% ee}=\left(\frac{\text{[α]observed}}{\text{[α]pure}}\right)\times 100\%$

Where,

• $\text{[α]observed}$is the specific rotation of the sample actually observed.
• $\text{[α]pure}$ is the theoretical specific rotation value for the pure sample
• $\text{\% ee}$ is the enantiomeric excess.

The value for $\text{[α]observed}$ is $+3.25°$ .

The value for $\text{[α]pure}$ is $+13.0°$ .

Substitute these values in the above equation to calculate the $\text{ee}\%$ and molecular composition of a sample of $2-\text{butanol}$ .

  $\begin{array}{c}\text{\% ee}=\left(\frac{\text{[α]observed}}{\text{[α]pure}}\right)\times 100\%\\ =\left(\frac{+3.25}{+13.0}\right)100\%\\ =\left(0.25\right)100\%\\ =25\%\end{array}$

The value of $\text{\% ee}$

  $\text{2-butanol}$ is $25\%$ and that is responsible for the optical activity of the compound.

The remaining $75\%$ is optically inactive due to the cancellation of optical activity between $\text{37}\text{.5}\text{\%}(+)\text{-2-butanol}$ and $\text{37}\text{.5}\text{\%}(-)\text{-2-butanol}$ .Thus the molecular composition of $(+)\text{-2-butanol}$ is calculated as follows:

  $\begin{array}{c}\text{Molecular composition of}(+)\text{-2-butanol}=\left(\begin{array}{c}\text{37}\text{.5}\text{\%}(+)\text{-2-butanol}\\ +25(+)\text{-2-butanol}\end{array}\right)\\ =62.5\text{\%}(+)\text{-2-butanol}\end{array}$

The molecular composition is $62.5\text{\%}(+)\text{-2-butanol}$ isand $37.5\text{\%}(-)\text{-2-butanol}$ .

Conclusion

The value of $\text{\% ee}$ of $\text{2-butanol}$ is $25\%$ and the molecular composition is $62.5\text{\%}(+)\text{-2-butanol}$ and $37.5\text{\%}(-)\text{-2-butanol}$ .