How do we explain the normal boiling points of #"ethanol"# (#78.4# #""^@C#), #"methanol"# (#64.7# #""^@C#), #"ethane"# (#-89# #""^@C#), #"ethyl acetate"#, (#77.1# #""^@C#), and #"methyl acetate"# (#56.9# #""^@C#)?

1 Answer
anor277
Mar 9, 2017

Because the esters are not capable of effective hydrogen bonding.

Explanation:

Look at the details. Ethanol and methanol are infinitely miscible in water, due to the fact that both alcohols can engage in intermolecular hydrogen bonding. And both formic, and acetic acids, their oxidation products, retain the solubility of their parent alcohols in aqueous solution, inasmuch as they contain a polar carboxyl group, that is certainly capable of interaction with a hydrogen-bonding solvent. The elevated boiling points of both acids (#118# #""^@C#, #"H"_3"CCO"_2"H"#, #100.8# #""^@C#, #"HCO"_2"H"#, ) with respect to their parent alcohols (and even more so with respect to the boiling points of their parent alkanes), again illustrates the intermolecular force present for each molecule.

On the other hand, methanol is INSOLUBLE in petroleum ether, whereas ETHANOL is SOLUBLE. Why should this be so?

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