Chapter 4, Problem 43P

(a) An aqueous solution containing $10$ g of optically pure fructose was diluted to $500$ mL with water and placed in a polarimeter tube $20$ cm long. The measured rotation was $-5.20°$. Calculate the specific rotation of fructose.

(b) If this solution were mixed with $500$ mL of a solution containing $5$ g of racemic fructose, what would be the specific rotation of the resulting fructose mixture? What would be its optical purity?

Interpretation Introduction

Interpretation:

The specific rotation of fructose is to be calculated and the specific rotation of the resulting fructose mixture and its optical rotation is to be calculated.

Concept introduction:

The specific rotation, denoted by $[\alpha ]$ can be calculated from the observed rotation using the following expression:

$[\alpha ]=\frac{\alpha }{cl}$

Here, $\alpha$ is the observed rotation, $c$ is the concentration of the sample in grams per mL of solution, and $l$ is the length of the polarimeter tube in decimeters.

The optical purity is numerically equal to the percentage enantiomeric excess, which can be determined from the following expression:

$\text{Optical purity}\text{=}\frac{{\text{[}\alpha ]}_{\text{sample}}}{{\text{[}\alpha ]}_{\text{pure enantiomer}}}\times 100$

The percent enantiomeric excess is the difference between the percentages of each enantiomer.

$\%\text{Enantiomeric excess}=\left(\text{\% Major enantiomer}\right)-\left(\%\text{ Minor enantiomer}\right)$

It can also be calculated according to the following equation:

$\text{(e}\text{.e}\text{.)}\text{=}\frac{\text{[R]-[S]}}{\text{[R]+[S]}}\text{×100}$

Where $\text{[R]}$ is the concentration of the major enantiomer and $\text{[S]}$ is the concentration of the minor enantiomer.

Mixtures containing equal quantities of enantiomers are called racemic mixtures.

Answer to Problem 43P

Solution:

a) The specific rotation of optically pure fructose is ${\text{-130}}^{\text{o}}$.

b) The specific rotation of resulting fructose mixture is $\text{-}\text{86}{\text{.67}}^{\text{o}}$, and its optical purity is $\text{66}\text{.67\%}$.

Explanation of Solution

a) Calculation for specific rotation of fructose.

Calculating the concentration of the sample solution as follows:

$\begin{array}{c}c=\frac{10\text{g}}{500\text{mL}}\\ =0.02\text{g/mL}\end{array}$

Now, calculate the length of the polarimeter tube as follows:

$\begin{array}{c}l=20\text{cm}\times \frac{1\text{dm}}{10\text{cm}}\\ =2\text{dm}\end{array}$

The specific rotation for the optically pure enantiomer can be calculated below:

$[\alpha ]=\frac{\alpha }{cl}$

Substitute the required values in the above equation.

$\begin{array}{l}[\alpha ]=\frac{(-{5.20}^o)}{0.02\text{g/mL}\times 2\text{ dm}}\\ [\alpha ]=-{130}^o\end{array}$

Hence the specific rotation of optically pure fructose is $-{130}^o$.

b) The amount of racemic mixture added is $5\text{g}$.

The amount of fructose present as pure enantiomer is $10\text{g}$.

The racemic mixture contains equal amounts of both the enantiomers. The total amount of pure enantiomer is calculated below.

Combined amount of the pure enantiomer is $10\text{g}+2.5\text{ g}=12.5\text{g}$.

Thus, the amount of the other enantiomer is $\text{5}-\text{2}\text{.5}=\text{2}\text{.5}\text{g}$.

Total volume of solution is $500\text{ mL}+500\text{ mL}=1000\text{ mL}$.

The concentration of the pure enantiomer is calculated as follows:

$\begin{array}{c}c=\frac{\text{12}\text{.5 g}}{\text{1000 mL}}\\ =\text{0}\text{.0125}\text{g/mL}\end{array}$

The concentration of the other enantiomer is calculated as follows:

$\begin{array}{c}c=\frac{\text{2}\text{.5 g}}{\text{1000 mL}}\\ =\text{0}\text{.0025g/mL}\end{array}$

The enantiomeric excess is given by the expression as follows:

$\text{(e}\text{.e}\text{.)}=\frac{\text{[R]}-\text{[S]}}{\text{[R]}+\text{[S]}}\text{×100}$

Substitute the required values in the above equation.

$\begin{array}{c}\text{(e}\text{.e}\text{.)}=\frac{\text{(}0.0125\text{g/mL}-0.0025\text{g/mL)}}{(0.0125\text{g/mL}+0.0025\text{g/mL)}}\times 100\\ =66.67\%\end{array}$

Optical purity is equal to the enantiomeric excess. Thus, the specific rotation of the sample is calculated using the following formula given below.

$\text{Optical purity}\text{=}\frac{{\text{[}\alpha ]}_{\text{sample}}}{{\text{[}\alpha ]}_{\text{pure enantiomer}}}\times 100$

Here, the specific rotation of pure enantiomer of fructose is ${\text{-130}}^{\text{o}}$. So, the specific rotation of sample is calculated as follows:

$\begin{array}{c}\text{66}\text{.67\%}=\frac{{\text{[}\alpha ]}_{\text{sample}}}{{\text{-130}}^{\text{o}}}\text{×100}\\ {\text{[}}_{\alpha }=-{86.67}^o\end{array}$

Hence the specific rotation of resulting fructose sample is $-{86.67}^o$.