Chapter 6, Problem 6.41AP

Interpretation Introduction

(a)

Interpretation:

The observed rotation of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is to be calculated for a $0.5M$ solution of the compound in a $5\text{-cm}$ sample path.

Concept introduction:

A carbon atom that has four nonequivalent atoms or groups attached to it is known as the chiral carbon atom. Chiral carbon centers are also called asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The specific optical rotation of a compound is given by the expression as shown below.

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

Answer to Problem 6.41AP

The observed rotation of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $2.8875\text{degrees}$.

Explanation of Solution

The molarity of the solution of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $0.5M$.

The specific rotation of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $+79{\text{drgrees mL g}}^{-1}{\text{ dm}}^{-1}$.

The path length is $5\text{ cm}$.

The molecular mass of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $146.2$.

Therefore, the molar mass of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $146.2{\text{g mol}}^{-1}$.

The specific optical rotation of a compound is given by the expression as shown below.

$\left[\text{α}\right]=\frac{\text{α}}{cl}$ …(1)

Where,

• $\alpha$ is the observed rotation.

• $c$ is the concentration of the solution in $\text{g}\cdot {\text{mL}}^{-1}$.

• $l$ is the path length in $\text{dm}$.

The molarity of the solution can be converted into the concentration by multiplying the molar mass of $\text{(3-methylbut-3-en-2-yl)benzene}$ with molarity.

$\begin{array}{rll}c& =& \left(0.5M\right)\left(\frac{1{\text{mol L}}^{-1}}{1M}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(146.2{\text{g mol}}^{-1}\right)\\ & =& 0.0731{\text{g mL}}^{-1}\end{array}$

Rearrange the equation (1) for the value of $\alpha$.

$\text{α}=\left[\text{α}\right]cl$ …(2)

Substitute the value of $\left[\text{α}\right]$, $c$ and $l$ in the equation (1).

$\begin{array}{rll}\text{α}& =& \left(+79{\text{degrees mL g}}^{-1}{\text{ dm}}^{-1}\right)\left(0.0731{\text{g mL}}^{-1}\right)\left(5\text{ cm}\right)\left(\frac{1\text{dm}}{10\text{cm}}\right)\\ & =& 2.8875\text{degrees}\end{array}$

Therefore, the observed rotation of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $2.8875\text{degrees}$.

Conclusion

The observed rotation of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $2.8875\text{degrees}$.

Interpretation Introduction

(b)

Interpretation:

The observed rotation of the resultant solution formed by mixture of $0.5M$ solution of $R$ enantiomer with equal volume of $0.25M$ solution of $S$ enantiomer is to be calculated in a $5\text{-cm}$ sample path.

Concept introduction:

A carbon atom that has four nonequivalent atoms or groups attached to it is known as the chiral carbon atom. Chiral carbon centers are also called asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The specific optical rotation of a compound is given by the expression as shown below.

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

Answer to Problem 6.41AP

The observed rotation of the resultant solution is $1.4437\text{degrees}$.

Explanation of Solution

The molarity of the solution of the $S$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $0.25M$. The molarity of the solution of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $0.5M$.

When $0.5M$ solution of $R$ enantiomer is mixed with equal volume of $0.25M$ solution of $S$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$, then a racemic mixture will be form of which do not have any rotation. The observed rotation will be only due to the $0.25M$ solution of $R$ enantiomer.

The specific rotation of the $R$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $+79{\text{drgrees mL g}}^{-1}{\text{ dm}}^{-1}$.

The specific rotations of two enantiomer are same in magnitude and opposite in sign. Therefore, the specific rotation of the $S$ enantiomer of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $-79{\text{drgrees mL g}}^{-1}{\text{ dm}}^{-1}$.

The path length is $5\text{ cm}$.

The molecular mass of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $146.2$.

Therefore, the molar mass of $\text{(3-methylbut-3-en-2-yl)benzene}$ is $146.2{\text{g mol}}^{-1}$.

The specific optical rotation of a compound is given as,

$\left[\text{α}\right]=\frac{\text{α}}{cl}$ … (1)

Where,

• $\alpha$ is the observed rotation.

• $c$ is the concentration of the solution in $\text{g}\cdot {\text{mL}}^{-1}$.

• $l$ is the path length in $\text{dm}$.

The molarity of the solution can be converted into the concentration by multiplying the molar mass of $\text{(3-methylbut-3-en-2-yl)benzene}$ with molarity.

$\begin{array}{rll}c& =& \left(0.25M\right)\left(\frac{1{\text{mol L}}^{-1}}{1M}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(146.2{\text{g mol}}^{-1}\right)\\ & =& 0.03655{\text{g mL}}^{-1}\end{array}$

Rearrange the equation (1) for the value of $\alpha$.

$\text{α}=\left[\text{α}\right]cl$ …(2)

Substitute the value of $\left[\text{α}\right]$, $c$ and $l$ in the equation (1).

$\begin{array}{rll}\text{α}& =& \left(79{\text{degrees mL g}}^{-1}{\text{ dm}}^{-1}\right)\left(0.03655{\text{g mL}}^{-1}\right)\left(5\text{ cm}\right)\left(\frac{1\text{dm}}{10\text{cm}}\right)\\ & =& 1.4437\text{degrees}\end{array}$

Therefore, the observed rotation of the resultant solution is $1.4437\text{degrees}$.

Conclusion

The observed rotation of the resultant solution is $1.4437\text{degrees}$.

Interpretation Introduction

(c)

Interpretation:

The enantiomeric excess of the major enantiomer in the corresponding solution is to be calculated.

Concept introduction:

A carbon atom that has four nonequivalent atoms or groups attached to it is known as the chiral carbon atom. Chiral carbon centers are also called asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The specific optical rotation of a compound is given by the expression as shown below.

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

Answer to Problem 6.41AP

The enantiomeric excess $\left(\text{EE}\right)$ of $R$ in the given solution is $33.3\%$.

Explanation of Solution

The solution formed by mixture of $0.5M$ solution of $R$ enantiomer with equal volume of $0.25M$ solution of $S$ enantiomer. Therefore, the solutions have $0.5\text{ mmol}$ of $R$ enantiomer and $0.25\text{ mmol}$ of $S$ enantiomer

The percentage of $D$ in the solution is calculated as,

$\%\text{major}\text{enantiomer}=\frac{n_R}{n_S+n_R}\times 100\%$

Where,

• $n_R$ is the number of mole of $R$.

• $n_S$ is the number of mole of $S$.

Substitute the value of $n_R$ and $n_S$ in the above equation.

$\begin{array}{rll}\%\text{major}\text{enantiomer}& =& \frac{0.5\text{ mmol}}{0.5\text{ mmol}+0.25\text{ mmol}}\times 100\%\\ & =& 66.66\%\end{array}$

The enantiomeric excess of a sample is given as,

$\text{EE}=2\left(\%\text{major}\text{enantiomer}\right)-100\%$

Substitute the value of percentage of major enantiomer in the above equation.

$\begin{array}{rll}\text{EE}& =& 2\left(66.66\%\right)-100\%\\ & =& 33.3\%\end{array}$

Therefore, the enantiomeric excess $\left(\text{EE}\right)$ of $R$ in the given solution is $33.3\%$.

Conclusion

The enantiomeric excess $\left(\text{EE}\right)$ of $R$ in the given solution is $33.3\%$.