<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8">
</head>
<body text="#000000" bgcolor="#FFFFFF">
I got the following question from someone today about the AF e-mail
I sent earlier<br>
today.<br>
<br>
"How could we tell what attractive forces were stronger in two
different molecules<br>
if one has hydrogen bonding but the other is bigger? In other words,
which aspect,<br>
size or hydrogen bonding, should we consider first when
determining which molecule<br>
has the stronger intermolecular force"<br>
<br>
Here's my answer:<br>
<br>
There's no hard and fast rule for this. For instance, lets look at
the<br>
following molecules<br>
<br>
H2O C8H18 CCl4<br>
MW 18 114 154<br>
b.p. 100 125 77 (all in degC)<br>
<br>
As I pointed out in class when discussing the Clausius-Clapeyron
Eqn.<br>
and the associated graph, H2O has a bigger slope than CCl4 and thus<br>
a higher H_vap. It has a higher b.p. This means it has stronger<br>
attractive forces (AF) than CCl4, even though CCl4 is much larger.<br>
The LF for CCl4 are much larger than the LF between H2O molecules.<br>
However, H2O is polar and can form HB between H2O molecules as<br>
well. Water's higher b.p. is mainly due to the HB between H2O
molecules.<br>
<br>
What about octane (C8H18, more specifically, n-octane the
straight-chain<br>
isomer)? It's nonpolar and has only LF, like CCl4. It's smaller in
size than<br>
CCl4. Yet, it has a higher b.p., indicating stronger AF (LF)
between the<br>
molecules in the liquid state. Why? CCl4 has a rather spherical
charge<br>
distribution while C8H18 is long and cylindrical. That means C8H18
has<br>
more points of contact (more surface contact) with another C8H18.
This<br>
maximizes the LF. The CCl4 molecules have less surface contact and<br>
can't maximize the LF which it would have based on its size if it
wasn't<br>
spherical.<br>
<br>
In fact the C8H18 is big enough that it's AF are stronger than the
AF<br>
between the H2O molecules, even though water has DD and HB in<br>
addition to it's small LF. n-Heptane (C7H16) which is only slightly<br>
smaller than octane has a b.p. of 98 degC, slightly below that of
H2O.<br>
<br>
There in lies the problem. There's no hard and fast rule for how
much<br>
larger a molecule has to be than H2O to have stronger AF than H2O.<br>
Remember, shape also plays a rule. Even so, water has a remarkably<br>
high b.p. relative to its size due to HB. Nonpolar molecules have
to be<br>
"significantly" larger (The MW of 118 for octane is 6.3 times bigger<br>
than 18 for H2O).<br>
<br>
Propanol (CH3CH2CH2-OH) and butanol (CH3CH2CH2-OH) have<br>
LF, DD and HB, like H2O. Propanol has a b.p. of 97 degC and butanol<br>
has a b.p. of 118 degC. Propanol, butanol and H2O all have LF, DD<br>
and HB. H2O is more polar and can form more HB per molecule.<br>
This fact is enough to outweigh the extra size and larger LF of<br>
C3H7-OH (which can form only up to 3 HB per molecule). However,<br>
adding another CH2 group to get butanol (C4H9-OH) makes it<br>
big enough so its overall AF consisting of DD (weaker than H2O) and<br>
HB (fewer and weaker than H2O) and larger LF are enough to outweigh<br>
the stronger and more frequent HB in H2O. Butanol's molar mass is<br>
about 4 times larger than that of water.<br>
<br>
Furthermore, keep in mind, the strength of a single HB between two<br>
molecules increases in the following order, N < O < F. This
is because<br>
F is the most electronegative of the three atoms and a single HB
between<br>
two HF molecules is stronger than a single HB between two H2O
molecules.<br>
Water has a higher b.p. than HF because water can form more HB per<br>
molecules (up to 4 for H2O compared to up to 2 for HF). More HB AF
have<br>
to be broken to separate the H2O molecules than to separate HF
molecules<br>
so more energy has to be added to separate (boil) the H2O molecules.<br>
So the number of HB a molecule can form plays a role. NH3 is
similar to<br>
HF in that it forms only up to 2 HB per molecule (HF has 3 lone pair
electrons<br>
and 1 H and NH3 has 1 lone pair electrons and 3 H atoms, H2O has a<br>
balance of 2 H atoms and 2 lone pairs).<br>
<br>
Generally, when we give questions about AF and how they affect
properties<br>
of different substances we try to give molecules which are
relatively close<br>
in size or very different in size. Certainly if I gave you H2O and
C20H42<br>
you would expect C20H42 to have much stronger AF than H2O even
though<br>
the former has only LF. It's really large. A molecule that large
is most likely<br>
to be a solid at room temp (in fact, this is icosane or eicosane and
the m.p.<br>
of its straight-chain isomer is ~ 37 degC and it's b.p. is 343
degC).<br>
<br>
I hope this answers the question. Perhaps not as satisfying as it
would be<br>
if there was a hard and fast cutoff but that's the way it is
sometimes when<br>
you're speaking in generalities. We have to often speak generally
because<br>
there's a whole lot of different substances and sometimes exceptions
to<br>
these general statements. The general statements help and when we
find<br>
those interesting exceptions we try to explain them. Like the fact
ice (solid<br>
H2O) floats on liquid water. That's very unusual.<br>
<br>
Dr. Zellmer <br>
<br>
<br>
<p style="margin-top:0;margin-bottom:0"><br>
</p>
</body>
</html>