Chapter 6, Problem 6.43AP

Interpretation Introduction

(a)

Interpretation:

The age of the skull sample 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 6.43AP

The age of the skull sample is $205.128\text{yr}$.

Explanation of Solution

The rate of racemization in skull samples follows the equation

$\ln \left(\frac{1+r}{1-r}\right)=kt$ …(1)

Where,

• $k$ is the rate constant $\left(6.24\times {10}^{-4}{\text{yr}}^{-1}\right)$.
• $t$ is the age of the sample in years.
• $r$ is the ratio of $\left(R\right)\text{-}(-)\text{-}$ and $\left(S\right)\text{-}(+)\text{-}$ aspartic acid.

The percentage of $\left(R\right)\text{-}(-)\text{-}$ aspartic acid in the skull sample is $6\%$.

The percentage of $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the skull sample is $94\%$.

The expression for $r$ is shown below,

$r=\frac{P\%\left(R\right)}{P\%\left(S\right)}$ …(2)

Where,

• $P\%\left(R\right)$ is the percentage of $\left(R\right)\text{-}(-)\text{-}$ aspartic acid.
• $P\%\left(S\right)$ is the percentage of $\left(S\right)\text{-}(+)\text{-}$ aspartic acid.

Substitute the value of $P\%\left(R\right)$ and $P\%\left(S\right)$ in the equation (2).

$\begin{array}{rll}r& =& \frac{6\%}{94\%}\\ & =& 0.064\end{array}$

Substitute the values of $r$ and $k$ in the equation (1).

$\begin{array}{rll}\ln \left(\frac{1+0.064}{1-0.064}\right)& =& \left(6.24\times {10}^{-4}{\text{yr}}^{-1}\right)t\\ \ln \left(\frac{1.064}{0.936}\right)& =& \left(6.24\times {10}^{-4}{\text{yr}}^{-1}\right)t\\ \ln \left(\frac{1.064}{0.936}\right)& =& \left(6.24\times {10}^{-4}{\text{yr}}^{-1}\right)t\\ 0.128& =& \left(6.24\times {10}^{-4}{\text{yr}}^{-1}\right)t\end{array}$

Rearrange the above expression for the value of $t$.

$\begin{array}{rll}t& =& \frac{0.128}{6.24\times {10}^{-4}{\text{yr}}^{-1}}\\ & =& 205.128\text{yr}\end{array}$

Therefore, the age of the skull sample is $205.128\text{yr}$.

Conclusion

The age of the skull sample is $205.128\text{yr}$.

Interpretation Introduction

(b)

Interpretation:

The enantiomeric excess of the $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the sample 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 6.43AP

The enantiomeric excess of the $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the sample is $88\%$.

Explanation of Solution

The percentage of $\left(R\right)\text{-}(-)\text{-}$ aspartic acid in the skull sample is $6\%$.

The percentage of $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the skull sample is $94\%$.

The enantiomeric excess of a sample is given by the formula as shown below.

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

Substitute the value of percentage of $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the skull sample in the above equation.

$\begin{array}{rll}\text{EE}& =& 2\left(94\%\right)-100\%\\ & =& 88\%\end{array}$

Therefore, The enantiomeric excess of the $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the sample is $88\%$.

Conclusion

The enantiomeric excess of the $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the sample is $88\%$.

Interpretation Introduction

(c)

Interpretation:

The observed rotation of $100\text{mg}$ of the forensic sample in $10\text{mL}$ of $6M$ aqueous $\text{HCl}$ 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 6.43AP

The observed rotation of $100\text{mg}$ of the forensic sample in $10\text{mL}$ of $6M$ aqueous $\text{HCl}$ is $0.2156\text{degrees}$.

Explanation of Solution

The specific rotation of $\left(S\right)\text{-}(+)\text{-aspartic}$ acid, ${\left[\text{α}\right]}_{\text{D}}^{20}=+24.5\text{degrees}\text{mL}{\text{g}}^{-1}{\text{dm}}^{-1}$.

The mass of the forensic sample is $100\text{mg}$.

The volume of $6M$ aqueous $\text{HCl}$ is $10\text{mL}$.

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

The enantiomeric excess of the $\left(S\right)\text{-}(+)\text{-}$ aspartic acid in the sample is $88\%$.

The specific rotation of the mixture is given by the expression as shown below.

$EE=100\%\times \frac{{\left[\text{α}\right]}_{\text{mixture}}}{{\left[\text{α}\right]}_{\text{pure}}}$

Where,

• ${\left[\text{α}\right]}_{\text{mixture}}$ is the specific rotation of the mixture.
• ${\left[\text{α}\right]}_{\text{pure}}$ is the specific rotation of the pure enantiomer.

Rearrange the above equation for the value of ${\left[\text{α}\right]}_{\text{mixture}}$.

Substitute the value of ${\left[\text{α}\right]}_{\text{mixture}}$ and ${\left[\text{α}\right]}_{\text{pure}}$ in the above equation.

${\left[\text{α}\right]}_{\text{mixture}}=\frac{EE\times {\left[\text{α}\right]}_{\text{pure}}}{100\%}$

Substitute the value of $EE$ and ${\left[\text{α}\right]}_{\text{pure}}$ in the above equation.

$\begin{array}{rll}{\left[\text{α}\right]}_{\text{mixture}}& =& \frac{\left(88\%\right)\left(+24.5\text{degrees}\text{mL}{\text{g}}^{-1}{\text{dm}}^{-1}\right)}{100\%}\\ & =& +21.56\text{degrees}\text{mL}{\text{g}}^{-1}{\text{dm}}^{-1}\end{array}$

The concentration of solution is given by the formula as shown below.

$c=\frac{m}{V}$

Where,

• $m$ is the mass of the solute.
• $V$ is the volume of the solution.

Substitute the value of and in the above equation.

$\begin{array}{rll}c& =& \left(\frac{100\text{mg}}{10\text{mL}}\right)\left(\frac{1\text{g}}{1000\text{mg}}\right)\\ & =& 0.01\text{g}{\text{mL}}^{-1}\end{array}$

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

$\text{α}=cl\left[\text{α}\right]$ …(3)

Where,

• $\text{α}$ is the observed rotation.
• $\left[\text{α}\right]$ is the specific rotation.
• $c$ is the concentration of the solution in $\text{g}{\text{mL}}^{-1}$.
• $l$ is the path length in $\text{dm}$.

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

$\begin{array}{rll}\text{α}& =& \left(0.01\text{g}{\text{mL}}^{-1}\right)\left(1\text{dm}\right)\left(+21.56\text{degrees}\text{mL}{\text{g}}^{-1}{\text{dm}}^{-1}\right)\\ & =& 0.2156\text{degrees}\end{array}$

Therefore, the observed rotation of $100\text{mg}$ of the forensic sample in $10\text{mL}$ of $6M$ aqueous $\text{HCl}$ is $0.2156\text{degrees}$.

Conclusion

The observed rotation of $100\text{mg}$ of the forensic sample in $10\text{mL}$ of $6M$ aqueous $\text{HCl}$ is $0.2156\text{degrees}$.