Question

Use the retention data given in the following table to calculate the retention index of $1-$$hexene.$

$\begin{array}{lc}\text{Sample}& \begin{array}{c}\text{Retention}\\ \text{Time, min}\end{array}\\ \text{Air}& 0.571\\ n\text{-pentane}& 2.16\\ n\text{-hexane}& 4.23\\ \text{1-hexene}& 3.15\end{array}$

Short answer

$\text{Theretention index for 1-hexene is}558\text{.}$

Step-by-step solution

Definition of Retention time:

Retention time is the amount of time a compound spends on the column after it has been injected. If a sample contains numerous compounds, each will spend a varied length of time on the column depending on its chemical composition, resulting in a variable retention time for each.

Retention time can be calculated from the formula (1) for the non-alkane:

$I=100\times \left[\mathrm{n}+(\mathrm{N}-\mathrm{n})\times \frac{{\mathrm{logt}}_{R(\text{unknown}}-{\mathrm{logt}}_{R\left(n\right)}}{{\mathrm{logt}}_{R\left(N\right)}-{\mathrm{logt}}_{R\left(n\right)}}\right]$

Where $t_R\left(\text{unknown}\right)$is adjusted retention time for the unknown compound ($(1-hexene)$in this case, $n$refers to smaller alkane's number of carbons $(n-pen\tan e)$and $N$to larger one $(n-hexane)$.

First calculate adjusted retention time tR' for each of the samples by following formula:

$t_R^{\text{'}}=t_R-t_M$

Table $1$shows that $t_M$is the time for the air. As a result, for each of them, it will be:

$t_R^{\text{'}}\left(\mathrm{n}\text{-pentane}\right)=t_R\left(\mathrm{n}\text{-pentane}\right)-t_R\left(\text{air}\right)=(2.16-0.571)\min =1.589\min$

$t_R^{\text{'}}\left(\mathrm{n}\text{-hexane}\right)=t_R\left(\text{n-hexane}\right)-t_R\left(\text{air}\right)=(4.23-0.571)\min =3.659\min$

$t_R^{\text{'}}\left(1\text{-hexane}\right)=t_R\left(\text{1-hexane}\right)-t_R\left(\text{air}\right)=(3.15-0.571)\min =2.579\min$

Applying the equation:

From row $1,$we can now return to equation $\left(1\right)$to $\left(2\right)$. In the formula, $n$will correspond to $n-pen\tan e$and will be $5$, whereas $N$will correspond to $n-hexane$and will be $6$. Then, in relation $1$, we may insert data from row $\left(3\right)$into retention times:

$I=100\times \left[5+(6-5)\frac{\log (2.58)-\log (1.59)}{\log (3.66)-\log (1.59)}\right]=\mathbf{558}$