3F.5 part b (7th Ed.) Melting point

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Julia Lindner 1I
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Joined: Fri Sep 28, 2018 12:17 am

3F.5 part b (7th Ed.) Melting point

Postby Julia Lindner 1I » Wed Nov 28, 2018 12:55 pm

This problem asks which has the higher melting point, C2H5OC2H5 (diethyl ether) or C4H9OH (butanol). The answer is butanol, but I don't understand why. Anyone know?

Patrick Cai 1L
Posts: 93
Joined: Fri Sep 28, 2018 12:25 am

Re: 3F.5 part b (7th Ed.) Melting point

Postby Patrick Cai 1L » Wed Nov 28, 2018 1:17 pm

Butanol is a somewhat polar molecule, as evident by the presence of its alcohol functional group (-OH), whereas diethyl ether is a non-polar molecule. The stronger intermolecular forces that come along with butanol as a result of its polar nature make it have a higher melting point than diethyl ether.

Searra Harding 4I
Posts: 68
Joined: Fri Sep 28, 2018 12:29 am

Re: 3F.5 part b (7th Ed.) Melting point

Postby Searra Harding 4I » Wed Nov 28, 2018 1:20 pm

Butanol has stronger interactions therefore the melting point is higher. H bonding can usually explain a higher melting point.

Milena Aragon 2B
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Re: 3F.5 part b (7th Ed.) Melting point

Postby Milena Aragon 2B » Sun Dec 02, 2018 11:36 am

It's mainly because Butanol has the ability to form hydrogen bonds whereas diethyl ether cannot. This additional bond that butanol can form means it has a higher melting point. Just think of it as more/stronger bonds=harder to break apart and thus higher melting/boiling point.

Tony Chung 2I
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Re: 3F.5 part b (7th Ed.) Melting point

Postby Tony Chung 2I » Sun Dec 02, 2018 7:58 pm

it has a stronger IMF, resulting in a higher melting point


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