Question

Give the order of volatility for the following molecules...and can you account for the proposed order?

`underbrace(H_3C-CH_2OCH_2CH_3)_"A".`
`underbrace(H_3C-CH(OH)CH_3)_"B";`
`underbrace(H_3C-CH_2CH_2OH)_"C";`
`underbrace(H_3C-CH_2CH_2OCH_3)_"D";`

Answer 1

A tentative order....`underbrace(C>B>A>D)_("decreasing boiling point"rarr)`

Explanation:

So why so? And note, as physical scientists, that we really should examine what data exist, but this is up to you.

Boiling point is a function of the intermolecular force between molecules, and of such forces, hydrogen bonding, where hydrogen is bound to a strongly electronegative atom such as oxygen, or fluorine, or nitrogen, is the most potent.

And thus for the given molecules, which are comparable size, `C` and `B` should BE THE MOST involatile because hydrogen bonding can occur thru the alcoholic function. As for the volatility of `C` versus `B`, we plump for `C`, the more linear molecule, in that long hydrocarbyl chains can more efficiently interact thru `"dispersion forces"` than cyclic or branched hydrocarbons. For a macroscopic example of this, is it easier to untangle short threads of string or long threads when they are tangled together?

`A` has some degree of polarity by virtue of the ether linkage. And while dipole-dipole interaction is an order of magnitude LESS significant than hydrogen bonding, this should be the next cab off the rank.

And finally, `D` HAS ONLY DISPERSION FORCE available as an intermolecular force. `D` should be the LEAST INVOLATILE or the MOST VOLATILE.

Please note that as a chemist, as a physical scientist, you rely on data. Can you find the data? They really should have been supplied with the question.