# What are the intermolecular forces that operate in "butane", "butyraldehyde", "tert-butyl alcohol", "isobutyl alcohol", "n-butyl alcohol", "glycerol", and "sorbitol"?

Apr 27, 2017

A scientist interrogates data. Quite properly, you should have given the normal boiling points, and asked how to rationalize them.......

#### Explanation:

Anyway, here is the order, and we insert the data appropriately. From MOST to LEAST volatile, we gets:

$\left(i\right)$ $\text{n-butane,}$ $- 1$ ""^@C

$\left(i i\right)$ $\text{n-butyraldeyde,}$ $74.8$ ""^@C

$\left(i i i\right)$ $\text{tert-butyl alcohol,}$ $82.2$ ""^@C

$\left(i v\right)$ $\text{isobutyl alcohol,}$ $108$ ""^@C

$\left(v\right)$ $\text{n-butyl alcohol,}$ $117.4$ ""^@C

$\left(v i\right)$ $\text{glycerol,}$ $290$ ""^@C

$\left(v i i\right)$ $\text{sorbitol,}$ $296$ ""^@C

Now, I have inserted the normal boiling points (after I looked them up of course). However, this is a reasonable order of prediction, because we can reasonably assess the order of intermolecular force between molecules, and this assessment should (and does!) correlate with boiling point.

Most volatile is butane, with only dispersion forces between molecules. Next is the $C 4$ aldehyde, which has some polarity by virtue of the terminal carbonyl group.

And now we have to assess the boiling point on the basis of hydrogen-bonding, because the remaining species are ALL alcohols, and thus have hydrogen bonding available as a potent intermolecular force. Isobutyl alcohol is next, and this is MORE volatile than n-butyl alcohol, because the latter has dispersion forces between molecules, which are always greater for LINEAR alkyl chains as opposed to BRANCHED ones.

And the last 2 cabs in the rank are glycerol, a triol, and sorbitol, a linear polyol with 6 hydroxyl groups available for hydrogen bonding. Thus sorbitol should be the most involatile, and the given order of boiling points represents the degree of intermolecular interaction; greatest for alcohols, and least for alkanes.