Question

Question #186e8

Answer 1

Good question. The boiling points of the alkanes should increase in the order, `"2,2,4-trimethylheptane"`, `"2-methylheptane"`, `"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:

`"2,2,4-trimethylheptane,"` `99` `""^@C`

`"2-methylheptane,"` `116` `""^@C`

`"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, `"octane"`, is the most involatile supports our analysis. And even though `"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-"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
`"2,2,4-trimethylheptane"`. Once again, branching serves to decrease the boiling point.