Question 07e2d

Sep 8, 2017

Because a greater degree of intermolecular interaction occurs in ethanol.....

Explanation:

The diethyl ether molecule, ${\text{H"_3"CCH"_2"OCH"_2"CH}}_{3}$ has a normal boiling point of $34.5$ ""^@C; that of ethanol is $78.5$ ""^@C. Clearly a greater degree of intermolecular interaction operates in ethanol, even tho this is a lighter molecule than $E {t}_{2} O$.

And the dominant intermolecular interaction is hydrogen bonding, which occurs when hydrogen is bound to a strongly electronegative element, such as fluorine, or oxygen, or nitrogen. If you look at the simple hydrides of these elements, water, ammonia, and hydrogen fluoride, their respective normal boiling points of $100$ ""^@C, $- 33.5$ ""^@C, and $19.5$ ""^@C#, are disproportionately high.

For ethanol there is intermolecular interaction between the hydrogen dipoles on adjacent molecules, which elevates the boiling point.

For ether, only dispersion forces operate between the hydrocarbyl chains, however, here hydrogen IS BOUND to CARBON, which is not strongly electronegative, and there is no resultant bond dipole. And thus ether is more volatile than ethanol.