# Question 186e8

Sep 2, 2016

Good question. The boiling points of the alkanes should increase in the order, $\text{2,2,4-trimethylheptane}$, $\text{2-methylheptane}$, $\text{octane}$.

#### Explanation:

The longer, and straighter the chain (i.e. the less unbranched) the greater should be the interaction between chains, and thus the greater the boiling point.

Now of course I did not know these boiling points off the top of my head, but I can account for their order:

$\text{2,2,4-trimethylheptane,}$ $99$ ""^@C

$\text{2-methylheptane,}$ $116$ ""^@C

$\text{octane} , 125$ ""^@C

Longer, less branched chains have greater opportunity for intermolecular dispersion forces, and this is certainly reflected in the boiling point. That the longest chain alkane, $\text{octane}$, is the most involatile supports our analysis. And even though $\text{2,2,4-trimethylheptane}$ is the most massive molecule, branching prevents as effective intermolecular interaction as exists in the other pair.

The boiling point of $n - \text{heptane}$ is $98.5$ ""^@C# (this I did know off the top of my head, because I found it to be a highly useful laboratory solvent), which is comparable to the boiling point of
$\text{2,2,4-trimethylheptane}$. Once again, branching serves to decrease the boiling point.