Name ______________________________________
In class today we'll start talking about aromatic compounds. These are discussed in Ch 16. Of
course we haven't covered the specific reactions mentioned in section 16.1, but the point of that
section should still be clear. The discussion of molecular orbitals in sections 16.3 and .4 is
beyond where we need to go right now, and it doesn't really give us conceptual insight into
aromaticity anyway. Skip sections 16.9 and .10 on NMR.
1 Draw benzene.
2 What are the CC bond lengths in benzene?
3 Benzene's aromatic stabilization, relative to a hypothetical version of benzene with three
isolated double bonds, is said to be 36 kcal/mol. Hydrogenation of benzene is exothermic, i.e.
heat is released. Does hydrogenation of benzene release 36 kcal/mol more heat or 36 kcal/mol
less heat than it would if it had three isolated double bonds (like the double bond of
cyclohexene)?
4 Draw cyclobutadiene.
5 (a) Is cyclobutadiene stabilized or destabilized by resonance?
(b) By how much (in kcal/mol)? If you can't find a number, make an educated guess.
Lecture outline
The relative stabilities of alkenes and dienes can be determined by hydrogenation calorimetry.
This involves combining the alkene and hydrogen gas in the presence of a catalyst and very
precisely measuring the amount of heat given off.
Here are the results of hydrogenation calorimetry on three isomeric dienes...
!H°hydrog
H2
Pd
H2
Pd
H2
Pd
pentane
– 60.2 kcal/mol
pentane
– 54.1
pentane
– 70.5
Which is the most stable based on those numbers? That question is much easier to answer if we
draw an enthalpy diagram showing the energies of the dienes relative to their common
hydrogenation product, pentane.
Can we draw resonance structures for a conjugated diene? Yes. Should we? Probably not. It's
normally sufficient to recognize that conjugation is slightly stabilizing (by about 5 kcal/mol on
average) because of favorable interaction of the π-bonds when the p-orbitals are lined up and
leave it at that. We don't really gain any additional insight by drawing the minor resonance
contributors.
Double bonds can be conjugated in two different ways — some dienes can adopt both
conformations; others are force to have one or the other. Each of these has the alignment of porbitals that results in conjugative stabilization.
C
C
C
C
C
C
C
s-cis
("cisoid")
C
s-trans
("transoid")
(lines imply unspecified
substits — Hs or other stuff)
We can see the alignment of the p-orbitals in "side views" of these structures
(Add the p-orbitals, and note the overlaps that make the formal π-bonds.)
Certain cyclic conjugated π-systems benefit from even more stabilization. These are typified by
benzene, and are classified as aromatic.
Benzene has a cyclic, planar,
fully conjugated π-system —
containing 6 electrons
Aromaticity => special stability
The special aromatic stability of benzene makes it much less reactive than simple alkenes or
conjugated dienes.
This special stability is not limited to benzene and its derivatives. More generally, any cyclic,
fully conjugated (therefore planar or nearly planar) array of p-orbitals containing a "magic
number" of electrons, i.e. 2, 6, 10, 14, 18, etc, will display special stability, i.e. aromaticity.
This is the Hückel "4n + 2 rule" — plugging in integers n = 0, 1, 2, 3, ...
gives the magic numbers of electrons for aromatic cycles
(the n itself has no physical meaning)
Thermochemistry —
How do we go about quantifying something like aromaticity?
The classic experiment (by Kistiakowski in 1936) is a comparison of heats of
hydrogenation —
H2
cat.
! !, pressure
H2
!H°hydrog = –49.8 kcal/mol
!H°hydrog = –28.6 kcal/mol
cat.
Q: What does this mean? A: Nothing by itself, but we cna use the result for cyclohexene
to get an idea of what benzene might be like if it had three simple non-interacting double bonds.
If benzene were nothing special, i.e. if it had 3 isolated, non-interacting, π-bonds, what would we
expect its ΔH°hydrog to be?
ΔH°hydrog (for hypothetical "benzene" with isolated π-bonds) =
Let's call this "model 1"
Q: So how do we compare this to real benzene??? Could we please draw an enthalpy
diagram? Because these numbers are making my brain swell. Ow. A: Mine too...
E
Q: But, aren't the π-bonds conjugated? Can't the extra stability of benzene
be explained just by the fact that it has 3 conjugated pairs of π-bonds?
If we used a model that had conjugated π-bonds, wouldn't this have
the same energy as real benzene?
A: Let's see — We can improve the localized π-bond model (model 1) by
considering the 3 π-bonds to be conjugated instead of isolated, and
assuming, based on what we know about dienes, that each pair of
conjugated π-bonds stabilizes the molecule by about 5 kcal/mol, how
would the energy of our new and improved hypothetical benzene
change?
Q: Could we please call this "model 2" and add it to the diagram above?
A: We would be happy to .... let's do that now...
We now have 2 measures of benzene's resonance energy or aromatic stabilization energy —
Benzene is more stable than the localized π-bond model (#1) by...
(the "Kistiakowsky resonance energy")
Benzene is more stable than the conjugated π-bond model (#2) by...
(the "Dewar resonance energy")
Q: Isn't there an easier way to get these numbers? A: Why yes, there is.
I'm glad you asked. We can use the enthalpy of formation of benzene and compare this
with that of a hypothetical benzene by using Benson's group increments.
— From measurements of H°fs of many conjugated polyenes, it can be determined that
each conjugated =CH– unit contributes 6.78 kcal/mol to the compound's H°f.
That’s called a “group increment”.
— A hypothetical "benzene" with 3 conjugated polyene-type π-bonds (i.e. "model 2"),
and thus 6 of these =CH– units, would have H°f = 6 x 6.78 = 40.7 kcal/mol.
— Benzene's enthalpy of formation, H°f, is 19.7 kcal/mol.
(Note that these really these should really be written as "ΔH°f", but because we
normally treat them as energies of molecules, I like to drop the Δ — the change is
going from elements in standard states to the molecule of interest, but this in
effect just sets all the enthalpies relative to the same "zero" point.)
— So based on these numbers, real benzene is stabilized by:
In contrast to the aromaticity of benzene and other molecules with cycles of 4n+2 π-electrons,
molecules with 4n π-electrons (i.e., 4, 8, 12, 16, etc) are surprisingly unstable. These have been
called anti-aromatic.
Cyclobutadiene (cbd) is the prototypical example. When cbd is created in solution, it instantly
(within microseconds) dimerizes, or it finds something else to react with. It's impossible to
isolate and purify cbd using ordinary techniques. This means that it is impossible to do the sort
of calorimetric measurement described above for benzene.
Cbd can be made photochemically and observed spectroscopically only at very low temperatures
(e.g. 10K in a matrix of solid argon), but calorimetric measurements can't be done under such
conditions.
H°f of cbd has been measured by photoacoustic calorimetry. This technique uses a short
laser pulse to initiate a fast photochemical reaction and measures the enthalpy change via the
sound pulse produced.
h!
O
P
O
+
266 nm
(= 107.5 kcal/mol)
CBD
fast rxn — ns timescale
H°f = 81.0 kcal/mol
– 8.4 kcal/mol
?
slower rxn —
µs timescale
dimer
Measuring the ΔH°rxn provides the one "missing" H°f value for cbd.
The photoacoustic experiment —
soln of P
laser pulse
1 ns, 266 nm
heat released in
irradiated region
of sample
sudden expansion
microphone
sound pulse —
detected by microphone
amplitude proportional
to heat released
Q: Please draw another one of those enthalpy diagrams. Those things are great. A: Why
certainly.
measured heat release
= 83 kcal/mol
E (kcal/mol)
h!
+ 107.5
H°f =
O +
H°f = 81
P
!H°rxn =
The diagram shows photoexcitation of P by a 107.5 kcal/mol photon, followed by an 83 kcal/mol
heat release. (a) Based on these two events, what is ∆H°rxn?
(b) Based on ∆H°rxn, and the H°f of P, what is the H°f of the two products (i.e. together)?
(c) Since we know that phthalan (the aromatic ether) has a H°f of –8.4 kcal/mol, what must the
H°f of cbd be?
H°f (cbd) =
So what does this tell us about the antiaromaticity? We can use the same group increments
approach we used to evaluate the aromaticity of benzene —
In a conjugated polyene, each =CH– unit contributes 6.78 kcal/mol to the compound's H°f, so
H°f for a hypothetical "cbd" with simple conjugated π-bonds is 4 x 6.78 = 27 kcal/mol.
Real cbd is less stable than this by...
But we have to account for the fact that cbd is destabilized by a purely geometrical phenomenon
called ring strain. The fact that the C-C-C angles are too small causes some destabilization. We
can confidently estimate the ring strain as about 32 kcal/mol.
This means that the destabilization due to antiaromaticity is
(for comparison, recall that benzene is stabilized by about 21 kcal/mol)
Ashok A. Deniz, Kevin S. Peters, Gary J. Snyder Science 1999, 286, 1119-22.