Chapter 6, Problem 6.10P

Interpretation Introduction

(a)

Interpretation:

The specific rotation of pure chiral compound $D$ 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 as 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 as,

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

Answer to Problem 6.10P

The specific rotation of pure chiral compound $D$ is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Explanation of Solution

The molarity of the solution of $D$ is $0.1M$.

The observed rotation of the solution of $D$ is $+0.20°$.

The path length is $1\text{dm}$.

The molecular mass of the compound $D$ is $150$.

Therefore, the molar mass of the compound $D$ is $150\text{g}\cdot {\text{mol}}^{-1}$.

The specific optical rotation of a compound is given as,

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

Where,

• $\alpha$ represents the observed rotation.

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

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

The molarity of the solution can be converted into the concentration by multiplying the molar mass of the compound $D$ with molarity.

$\begin{array}{rll}C& =& \left(0.1M\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1M}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(150\text{g}\cdot {\text{mol}}^{-1}\right)\\ & =& 0.015\text{g}\cdot {\text{mL}}^{-1}\end{array}$

Substitute the value of $\alpha$, $c$ and $l$ in the equation (1).

$\begin{array}{rll}\left[\alpha \right]& =& \frac{+0.20°}{\left(0.015\text{g}\cdot {\text{mL}}^{-1}\right)\left(1\text{dm}\right)}\\ & =& 13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}\end{array}$

Therefore, the specific rotation of pure chiral compound $D$ is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Conclusion

The specific rotation of pure chiral compound $D$ is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The observed rotation of the given solution, that is formed with an equal volume of $0.1M$ solution of $L$ and $D$, 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 as asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$

Answer to Problem 6.10P

The observed rotation of the given solution is zero.

Explanation of Solution

It is given that two enantiomeric solutions of same molarity are mixed with same volume. The resulting solution is a racemic mixture. The specific rotation of two enantiomers is the same in magnitude but different in sign. The specific rotation of both enantiomers will cancel each other. Therefore, the observed rotation of the given solution is zero.

Conclusion

The observed rotation of the given solution is zero.

Interpretation Introduction

(c)

Interpretation:

The observed rotation of the given solution that is formed by dilution of corresponding solution of $D$ with an equal volume of solvent 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 as asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$

Answer to Problem 6.10P

The observed optical rotation of the given solution is $0.1°$.

Explanation of Solution

The molarity of the solution of $D$ is $0.1M$.

The observed rotation of the solution of $D$ is $+0.20°$.

The path length is $1\text{dm}$.

The molecular mass of the compound $D$ is $150$.

Therefore, the molar mass of the compound $D$ is $150\text{g}\cdot {\text{mol}}^{-1}$.

The specific rotation of pure chiral compound $D$ is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

It is given that the solution of $D$ is diluted with an equal volume of solvent.

The solutions are mixed in equal volume. Therefore, the molarity of new solution is calculated as

$M_{\text{f}}=\frac{M_{\text{i}}}{2}$ …(2)

Where,

• $M_{\text{f}}$ represents the final molarity.

• $M_{\text{i}}$ represents the initial molarity.

Substitute the value of $M_{\text{i}}$ in the equation (2).

$\begin{array}{rll}{M}_{\text{f}}& =& \frac{0.1M}{2}\\ & =& 0.05M\end{array}$

The molarity of the solution can be converted into the concentration by multiplying the molar mass of the compound $D$ with molarity.

$\begin{array}{rll}C& =& \left(0.05M\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1M}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(150\text{g}\cdot {\text{mol}}^{-1}\right)\\ & =& 0.0075\text{g}\cdot {\text{mL}}^{-1}\end{array}$

The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$ …(3)

Where,

• $\left[\alpha \right]$ represents the specific rotation.

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

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

Substitute the value of $\left[\alpha \right]$, $c$ and $l$ in the equation (3).

$\begin{array}{rll}\alpha & =& \left(13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}\right)\left(0.0075\text{g}\cdot {\text{mL}}^{-1}\right)\left(1\text{dm}\right)\\ & =& 0.09999975°\\ & \approx & 0.1°\end{array}$

Therefore, the observed optical rotation of the given solution of $D$ is $0.1°$.

Conclusion

The observed optical rotation of the given solution is $0.1°$.

Interpretation Introduction

(d)

Interpretation:

The specific rotation of $D$ after the dilution of 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 as asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$

Answer to Problem 6.10P

The specific rotation of $D$ after the given dilution is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Explanation of Solution

The path length is $1\text{dm}$.

The concentration of the given solution of $D$ is $0.0075\text{g}\cdot {\text{mL}}^{-1}$.

The observed optical rotation of the given solution of $D$ is $0.1°$.

The specific optical rotation of a compound is given as,

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

Where,

• $\alpha$ represents the observed rotation.

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

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

Substitute the value of $\alpha$, $c$ and $l$ in the equation (1).

$\begin{array}{rll}\left[\alpha \right]& =& \frac{+0.10°}{\left(0.0075\text{g}\cdot {\text{mL}}^{-1}\right)\left(1\text{dm}\right)}\\ & =& 13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}\end{array}$

Therefore, the specific rotation of $D$ after the given dilution is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Conclusion

The specific rotation of $D$ after the given dilution is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Interpretation Introduction

(e)

Interpretation:

The specific rotation of $L$ after the dilution of the solution given in part (c) 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 as asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$

Answer to Problem 6.10P

The specific rotation of $L$ after the given dilution is $-13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Explanation of Solution

Enantiomers are the non-superimposable mirror images of each other. Therefore, the magnitude of the specific rotation of enantiomers of the given compound is same but the sign of specific rotation is different. The specific rotation of $D$ after the given dilution is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Therefore, the specific rotation of $L$ after the given dilution is $-13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Conclusion

The specific rotation of $L$ after the given dilution is $-13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

Interpretation Introduction

(f)

Interpretation:

The observed rotation of $100\text{mL}$ of a solution that contains $0.01\text{mole}$ of $D$ and $0.005\text{mole}$ of $L$ 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 as asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$

Answer to Problem 6.10P

The observed optical rotation of the given solution is $0.1°$.

Explanation of Solution

The molecular mass of the compound $D$ is $150$.

Therefore, the molar mass of the compound $D$ is $150\text{g}\cdot {\text{mol}}^{-1}$.

The specific rotation of pure chiral compound $D$ is $13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}$.

The given solution contains $0.01\text{mole}$ of $D$ and $0.005\text{mole}$ of $L$.

The volume of the given solution is $100\text{mL}$.

The solution formed by $0.005\text{mole}$ of $L$ and $0.005\text{mole}$ of $D$ becomes a racemic mixture and result in zero rotation. The extra $0.005\text{mole}$ of $D$ will give the rotation to the solution.

The molarity of the solution is represented as,

$M=\frac{n}{V}$

Where,

• $n$ represents the number of moles of the solute.

• $V$ represents the volume of the solution.

Substitute the value of $m$ and $V$ in the above equation.

$\begin{array}{rll}c& =& \frac{0.005\text{mole}}{100\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(\frac{1M}{1\text{mol}\cdot {\text{L}}^{-1}}\right)\\ & =& 0.05M\end{array}$

The molarity of the solution can be converted into the concentration by multiplying the molar mass of the compound $D$ with molarity.

$\begin{array}{rll}C& =& \left(0.05M\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1M}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(150\text{g}\cdot {\text{mol}}^{-1}\right)\\ & =& 0.0075\text{g}\cdot {\text{mL}}^{-1}\end{array}$

The observed optical rotation of a compound is given as,

$\alpha =\left[\alpha \right]cl$ …(3)

Where,

• $\left[\alpha \right]$ represents the specific rotation.

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

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

Substitute the value of $\left[\alpha \right]$, $c$ and $l$ in the equation (3).

$\begin{array}{rll}\alpha & =& \left(13.3333°\text{mL}\cdot {\text{g}}^{-1}\cdot {\text{dm}}^{-1}\right)\left(0.0075\text{g}\cdot {\text{mL}}^{-1}\right)\left(1\text{dm}\right)\\ & =& 0.09999975°\\ & \approx & 0.1°\end{array}$

Therefore, the observed optical rotation of the given solution is $0.1°$.

Conclusion

Therefore, the observed optical rotation of the given solution is $0.1°$.

Interpretation Introduction

(g)

Interpretation:

The enantiomeric excess $\left(\text{EE}\right)$ of $D$ 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 as asymmetric or stereogenic centers. A chiral molecule is an optically active molecule. It rotates the plane of a plane polarized light. The enantiomeric excess of a sample is given as,

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

Answer to Problem 6.10P

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

Explanation of Solution

The given solution contains $0.01\text{mole}$ of $D$ and $0.005\text{mole}$ of $L$.

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

$\%\text{major}\text{enantiomer}=\frac{n_D}{n_D+n_L}\times 100\%$

Where,

• $n_D$ represents the number of moles of $D$.

• $n_L$ represents the number of moles of $L$.

Substitute the value of $n_D$ and $n_L$ in the above equation.

$\begin{array}{rll}\%\text{major}\text{enantiomer}& =& \frac{0.01\text{mole}}{0.01\text{mole}+0.005\text{mole}}\times 100\%\\ & =& 66.66\%\end{array}$

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 $D$ in the given solution is $33.3\%$.

Conclusion

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