Sciencemadness Discussion Board

Intermolecular Forces contradiction?

SunriseSunset - 29-12-2015 at 17:19

Inter-molecular forces.

The stronger the inter-molecular forces, the higher the boiling point.

Reference from Organic Chemistry 2nd Edition textbook Janice Smith
Quote:

VDW aka London Forces (weakest inter-molecular force)
Surface area of a molecule determines the strength of VDW, and polarity.

Dipol-Dipol interactions (stronger than VDW)
Dipol-Dipol interactions are forces between the permanent dipols of two polar-molecules. Acetone for example.

Hydrogen Bonding (stronger than D-D & VDW)
Typically occurs when a hydrogen atom is bonded to O, N, or F is electro-statically attracted to a lone pair of electrons on an O, N, or F atom from another molecule. Water for example.


So I was comparing some random molecules practicing this concept out some and ran into a problem on the 3rd comparison.

The 1st one given to me was Propylene Glycol vses Vegetable Glycerin. Which has the higher BP? Vegetable glycerin. I solved this quiet easily from just learning the basic concepts of.

The 2nd question was compare 25I-NBOMe vses 2C-I.
I chose 25I-NBOMe. Not because it was a larger molecule (surface area only strengthens London Forces which are the weakest inter-molecular force to begin with), but because it seemed to have the same amount of hydrogen bonding locations, and an overall greater dipole dipole than 2C-I. And it turned out that my slightly-educated guess was correct. 25I-NBOMe has the greater boiling point.

Now for the 3rd question, was to compare 4-methylmethcathinone vses Methylenedioxypyrovalerone.
It seems that even though Methylenedioxypyrovalerone may have more dipole-dipole situations? and and more surface area to it, that it still seemed 4-methylmethcathinone has more of a hydrogen bonding potential to display. So I chose 4-methylmethcathinone as being the one with the higher boiling point. And I think I might be incorrect.

The data on finding out the actual boiling point comparisons of these two molecules is sparse. Nothing exactly specifies whether the detailed boiling point is for the freebase oil, or the hydrochloride salt. Am I wrong for assuming 4-methylmethcathinone has the higher boiling point, and if so, please explain.
Thanks

[edit]The only assumption I can make as to why MDPV would have more inter-molecular forces than Mephedrone has is because there's more dipol-dipol going on and it's a larger molecule that has more surface area. But would that really overpower the fact that Mephedrone has a location for hydrogen-bonding to occur, where as MDPV does not?

[Edited on 30-12-2015 by SunriseSunset]

DraconicAcid - 29-12-2015 at 17:46

Quote: Originally posted by SunriseSunset  

Now for the 3rd question, was to compare 4-methylmethcathinone vses Methylenedioxypyrovalerone.
It seems that even though Methylenedioxypyrovalerone may have more dipole-dipole situations? and and more surface area to it, that it still seemed 4-methylmethcathinone has more of a hydrogen bonding potential to display. So I chose 4-methylmethcathinone as being the one with the higher boiling point. And I think I might be incorrect.


Mephedrone has a carbonyl group and is an amine. MDPV is similar, but also has the methylenedioxy group (more polarity, more polaraizable, so more London forces), but it doesn't have the N-H bond, so there won't be any hydrogen bonding in the pure substance. Tough to tell.

SunriseSunset - 29-12-2015 at 17:56

I always take basic concepts a bit too far, and find a contradiction point to stop at. I think someone else with a lot more knowledge on determining boiling points might be able to explain what I haven't learned yet though.

blogfast25 - 29-12-2015 at 19:01

Quote: Originally posted by SunriseSunset  
I always take basic concepts a bit too far, and find a contradiction point to stop at. I think someone else with a lot more knowledge on determining boiling points might be able to explain what I haven't learned yet though.


Determining BPs from pure theory is almost impossible. The general principles you cited stand firm but they are qualitative and don't really allow quantitative calculations of BP.

SunriseSunset - 30-12-2015 at 01:45

Oh wow, I just realized I didn't put into account that the two oxygens on the methylenedioxy ring each have two lone pairs of electrons and just because the oxygen aren't bonded to hydrogen atoms, doesn't mean the lone pairs wouldn't cause hydrogen bonding force between the N-H2's on the other sides of the molecule. This literally should explain why MDPV has a higher boiling point. It does in fact have more intermolecular forces due to more H.B. potential. Edit: I apologize, this theory is invalid because I realized afterwards there are no N-H bonds in MDPV. So I'm incorrect.

[Edited on 30-12-2015 by SunriseSunset]

j_sum1 - 30-12-2015 at 02:28

I have attached a spreadsheet that I have used when showing students trends on the periodic table. It mostly looks at melting point and boiling point.

A couple of obvious trends:
1. Higher molecular mass leads to higher MP ad BP. (Most sheets)
2. Noble gases exhibit very narrow temperature ranges where they exist as a liquid. Only very weak dispersion forces in all cases. (Sheet 1)
3. Other factors can throw off all trends easily. (Sheet 3)
4. The relationship between molecular mass and MP/BP is only relevant for molecules. Does not hold for ionic lattices. (Other trends however may be present for quantities such as bond length, bond strength etc.) (Sheet 4)
5. Hydrogen bonding -- molecules with N-H, O-H or F-H bonds leave the H essentially without any shielding electrons. Therefore dipole-dipole interactions are significantly stronger leading to much higher MP and BP. (Sheets 6, 7 & 8) Carbon does not show this (Sheet 5).

IMO, investigating these kinds of relationships is best done with simpler molecules than the monsters you chose. In the more complex molecule there is more going on and other factors may influence the parameter that you are interested in.

Attachment: Molecular trends in periodic table.xlsx (60kB)
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