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Topic: Boiling point Hydrocarbon  (Read 3189 times)

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Offline JNW2

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Boiling point Hydrocarbon
« on: August 03, 2016, 08:06:51 AM »
I know it's a well known question but still don't have a clear explanation.

Why boiling point of alkyne > alkane > alkene?

I have tried some searches.

These including factors:
1) polarizability from London dispersion force
2) Pi electron

and from http://forums.studentdoctor.net/threads/boiling-point-of-alkanes-vs-alkenes-vs-alkynes.649390/

 I wonder a clear explanation.
Could you help me?
« Last Edit: August 03, 2016, 08:21:58 AM by JNW2 »

Offline Corribus

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Re: Boiling point Hydrocarbon
« Reply #1 on: August 03, 2016, 06:52:33 PM »
It's an interesting question. For hydrocarbons really the only forces involved are induced dipoles, the strength of which largely comes down to the amount of surface area available (a somewhat abstract concept for atoms and molecules) and how easy it is for electrons to be pulled away (so-called polarizability). A knee-jerk guess might have been that because pi-electrons are generally more polarizable than sigma electrons, alkenes and alkynes should both have stronger intermolecular forces and thus higher boiling points than alkanes. But a look at the data shows that's clearly not the case.  For kicks, I plotted some of it out - the difference in boiling points between the alkene or alkyne and the corresponding alkane). In fact, alkynes have significantly higher boiling points than alkanes with the same number of carbons, and contrarily alkenes are moderately lower. As the number of carbons increases, the differences shrink (not surprisingly). The magnitude of the effect also depends pretty significantly on where the double or triple bond is, and even that doesn't necessarily follow a logical progression - e.g., 2-ynes have higher boiling points than either 1-ynes or 3-ynes. It's probably a pretty complex algorithm of how the molecules pack together in the liquid state and the relative polarizability of the electrons involved. I'm not sure there's an easily generalized answer unfortunately but I'll continue to dwell on it. :)

(There are also some very small differences between the various E- or Z- isomers for the alkenes, but in many cases the measurement errors are larger than the absolute difference in boiling points. All the data were taken from the NIST webbook and correspond to linear carbon chains.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline JNW2

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Re: Boiling point Hydrocarbon
« Reply #2 on: August 05, 2016, 08:54:21 AM »
It's an interesting question. For hydrocarbons really the only forces involved are induced dipoles, the strength of which largely comes down to the amount of surface area available (a somewhat abstract concept for atoms and molecules) and how easy it is for electrons to be pulled away (so-called polarizability). A knee-jerk guess might have been that because pi-electrons are generally more polarizable than sigma electrons, alkenes and alkynes should both have stronger intermolecular forces and thus higher boiling points than alkanes. But a look at the data shows that's clearly not the case.  For kicks, I plotted some of it out - the difference in boiling points between the alkene or alkyne and the corresponding alkane). In fact, alkynes have significantly higher boiling points than alkanes with the same number of carbons, and contrarily alkenes are moderately lower. As the number of carbons increases, the differences shrink (not surprisingly). The magnitude of the effect also depends pretty significantly on where the double or triple bond is, and even that doesn't necessarily follow a logical progression - e.g., 2-ynes have higher boiling points than either 1-ynes or 3-ynes. It's probably a pretty complex algorithm of how the molecules pack together in the liquid state and the relative polarizability of the electrons involved. I'm not sure there's an easily generalized answer unfortunately but I'll continue to dwell on it. :)

(There are also some very small differences between the various E- or Z- isomers for the alkenes, but in many cases the measurement errors are larger than the absolute difference in boiling points. All the data were taken from the NIST webbook and correspond to linear carbon chains.)

Thank you very much for the data.
Obviously , 1-ene always have the less boiling point.
Why? Because the property that alkene have the lock chain?
1-ene have the longest lock chain then the less polarizability of electron?

Offline JNW2

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Re: Boiling point Hydrocarbon
« Reply #3 on: August 05, 2016, 07:38:42 PM »
Any help is much appreciated.

Offline Corribus

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Re: Boiling point Hydrocarbon
« Reply #4 on: August 06, 2016, 12:10:32 PM »
I'm not sure what you mean by "lock chain".
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Babcock_Hall

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Re: Boiling point Hydrocarbon
« Reply #5 on: August 08, 2016, 09:14:52 AM »
We had a thread a few months ago on branched versus unbranched alkanes.  IIRC both melting and boiling points were mentioned, but the former were emphasized.  I came away thinking that simple models could not explain the data that we were examining.

Offline Enthalpy

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Re: Boiling point Hydrocarbon
« Reply #6 on: August 09, 2016, 03:35:58 PM »
Pity, I hoped you'd find a simple model for the melting point. I didn't, after a few days of collecting, formatting and observing the data. I keep suggesting it as a research project - true research, whose result would be useful, but where one ignores if he'll find something, nor which direction to explore.

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