Chapter 24, Problem 24.AE

a)

Interpretation Introduction

Interpretation:

The name of two straight chain hydrocarbons that is responsible for the elution of 2-pentanone has to be identified.

Concept Introduction:

Kovats retention index:

Retention index, I for linear alkane equals 100 times the number of Carbon atoms.

For a molecule of Octane, $\text{I=800}$

For a molecule of Nonane, $\text{I=900}$

The elution of compound between Octane and Nonane will have retention index between 800 and 900 is calculated using the formula,

$\text{I=100}\left[\text{n+}\left(\text{N-n}\right)\frac{\text{log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{unknown}\right)\text{-log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{n}\right)}{\text{log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{N}\right)\text{-log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{n}\right)}\right]$

Where, $\text{n}$= number of Carbon atoms in smaller alkane

$\text{N}$ = number of Carbon atoms in larger alkane

${\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{n}\right)$ = adjusted retention time of smaller alkane

${\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{N}\right)$ = adjusted retention time of larger alkane

To give the name of two straight chain hydrocarbons that is responsible for the elution of 2-pentanone

Answer to Problem 24.AE

The elution of 2-pentanone takes place between Nonane and Decane, but very close elution takes places through Decane.

Explanation of Solution

The retention index of the compounds is based on the overall polarity phases and the selectivity that depends on the types of polar interaction offered by each column. For poly (ethylene glycol) column, the elution of 2-pentanone takes place between Nonane and Decane, but very close elution takes places through Decane.

b)

Interpretation Introduction

Interpretation:

The retention time of the compound has to be calculated.

Concept Introduction:

Kovats retention index:

Retention index, I for linear alkane equals 100 times the number of Carbon atoms.

For a molecule of Octane, $\text{I=800}$

For a molecule of Nonane, $\text{I=900}$

The elution of compound between Octane and Nonane will have retention index between 800 and 900 is calculated using the formula,

$\text{I=100}\left[\text{n+}\left(\text{N-n}\right)\frac{\text{log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{unknown}\right)\text{-log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{n}\right)}{\text{log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{N}\right)\text{-log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{n}\right)}\right]$

Where, $\text{n}$= number of Carbon atoms in smaller alkane

$\text{N}$ = number of Carbon atoms in larger alkane

${\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{n}\right)$ = adjusted retention time of smaller alkane

${\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{N}\right)$ = adjusted retention time of larger alkane

To calculate the retention time of the unknown compound

Answer to Problem 24.AE

The retention time of the compound is $\text{16}\text{.40}\text{min}$.

Explanation of Solution

Given,

$\begin{array}{l}{\text{t}}_{\text{r}}{}^{\text{'}}\text{of}{\text{C}}_{\text{10}}\text{=13}\text{.83}\\ \\ {\text{t}}_{\text{r}}{}^{\text{'}}\text{of}{\text{C}}_{\text{11}}\text{=15}\text{.42}\\ \\ {\text{t}}_{\text{m}}\text{=1}\text{.80}\text{min}\end{array}$

The adjusted retention time of ${\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{unknown}\right)$ is calculated as,

$\begin{array}{l}\text{1050=100}\left[\text{10+}\left(\text{11-10}\right)\frac{\text{log}{\text{t}}_{\text{r}}{}^{\text{'}}\left(\text{unknown}\right)\text{-log13}\text{.83}}{\text{log15}\text{.42-log13}\text{.83}}\right]\\ \\ {\text{t}}_{\text{r}}{}^{\text{'}}\text{(unknown)=14}\text{.60}\text{min}\end{array}$

The adjusted retention time of unknown along with ${\text{t}}_{\text{m}}$ gives the retention time of the compound.

$\begin{array}{l}{\text{t}}_{\text{r}}{\text{=t}}_{\text{r}}{}^{\text{'}}\text{of}{\text{unknown+t}}_{\text{m}}\\ \\ {\text{t}}_{\text{r}}\text{=14}\text{.60+1}\text{.80}\\ \\ {\text{t}}_{\text{r}}\text{=16}\text{.40}\text{min}\end{array}$

The retention time of the compound= $\text{16}\text{.40}\text{min}$