Follow-up Question Regarding Attractive Forces: London Forces, Dipole-Dipole, Hydrogen Bonding, etc.

[Zellmer, Robert]

I got the following question from someone about the attractive forces (AF) e-mail I sent previously

and about a related homework problem.  I get these questions every semester in 1210 and 1220.



"How can we tell what attractive forces are stronger in two different molecules if one has hydrogen

bonding but the other is bigger? In other words, which aspect, size or hydrogen bonding, should we

consider first when determining which molecule has the stronger intermolecular force"



Here's my answer:



There's no hard and fast rule for this (sorry).  For instance, lets look at the following molecules



            H2O        C8H18        CCl4

MW       18            114            154

b.p.      100            125            77    (all in degC)



In Ch 11, look at the Clausius-Clapeyron Eqn. in the book and my notes (Ch 11) and the associated

graph, H2O has a bigger slope than CCl4 and thus a higher H_vap (proportional to slope).  H2O has

a higher b.p.  This means it has stronger attractive forces (AF) than CCl4, even though CCl4 is much

larger.  The LF for CCl4 are much larger than the LF between H2O molecules.  However, H2O is polar

and can form HB between H2O molecules as well.  Water's higher b.p. is mainly due to the HB between

H2O molecules.  It forms lots of HB, up to 4 per H2O molecule, averaging between 2-4 HB between

molecules in the liquid state.  Pretty much all of them must be broken to boiling the water (go from liquid

to vapor).



What about octane (C8H18, more specifically, n-octane, the straight-chain isomer)?  It's nonpolar and

has only LF, like CCl4.  It's smaller in size (MW) than CCl4.  Yet, it has a higher b.p., indicating stronger

AF (LF) between the molecules in the liquid state.  Why?  CCl4 has a rather spherical charge distribution

while C8H18 is long and cylindrical.  That means C8H18 has more points of contact (more surface area

contact) with another C8H18.  This maximizes the LF.  The CCl4 molecules have less surface contact

and can't maximize the LF which it would have based on its size if it wasn't sort of spherical.



In fact, the C8H18 is big enough with lots of points of contact that it's AF are stronger than the AF

between the H2O molecules, even though water has DD and HB in addition to it's small LF.  n-Heptane

(C7H16) which is only slightly smaller than octane has a b.p. of 98 degC, slightly below that of H2O.



There in lies the problem.  There's no hard and fast rule for how much larger a molecule must be than

H2O to have stronger AF than H2O.  Remember, as discussed above, shape also plays a role.  Even so,

water has a remarkably high b.p. relative to its size due to HB.  Nonpolar molecules have to be

"significantly" larger (The MW of 118 for octane is 6.3 times bigger than the MW of 18 for H2O).



Propanol (CH3CH2CH2-OH) and butanol (CH3CH2CH2-OH) have LF, DD and HB, like H2O.  Propanol has

a b.p. of 97 degC and butanol has a b.p. of 118 degC.   Propanol, butanol and H2O all have LF, DD and HB.

H2O is more polar and can form more HB per molecule. This fact is enough to outweigh the extra size and

larger LF of C3H7-OH (which can form only up to 3 HB per molecule).  However, adding another CH2 group

to get butanol (C4H9-OH) makes it big enough so its overall AF consisting of DD (weaker than H2O) and HB

(fewer and weaker than H2O) and larger LF are enough to outweigh the stronger and more frequent HB in

H2O.  Butanol's molar mass is about 4 times larger than that of water.



Furthermore, keep in mind, the strength of a single HB between two molecules increases in the following

order, N < O < F.  This is because F is the most electronegative of the three atoms and a single HB between

two HF molecules is stronger than a single HB between two H2O molecules. Water has a higher b.p. than

HF because water can form more HB per molecule (up to 4 for H2O compared to up to 2 for HF).  More HB

AF must be broken to separate the H2O molecules than to separate HF molecules so more energy has to

be added to separate (boil) the H2O molecules.  Thus, the number of HB a molecule can form plays a role.

NH3 is similar to HF in that it forms only up to 2 HB per molecule (HF has 3 lone pair electrons and 1 H and

NH3 has 1 lone pair electrons and 3 H atoms, H2O has a balance of 2 H atoms and 2 lone pairs).



Generally, when we give questions about AF and how they affect properties of different substances we try

to give molecules which are relatively close in size or very different in size.  Certainly, if I gave you H2O and

C20H42 you would expect C20H42 to have much stronger AF than H2O even though C20H42 has only LF.

It's really large.  A molecule that large is most likely to be a solid at room temp.  In fact, this is icosane (or

eicosane) and the m.p. of its straight-chain isomer is ~ 37 degC and it's b.p. is 343 degC.



I hope this answers the question.  Perhaps not as satisfying as it would be if there was a hard and fast cutoff

but that's the way it is sometimes when you're speaking in generalities.  We must often speak generally

because there's a whole lot of different substances and sometimes there's exceptions to these general

statements.  The general statements help and when we find those interesting exceptions we try to explain

them.  Like the fact ice (solid H2O) floats on liquid water.  That's very unusual.  It's due to the fact the water

molecules move further apart as they form the many HB between the H2O molecules (each H2O in

the interior of ice has 4 HB between it and 4 other H2O molecules).



Dr. Zellmer
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