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## Conformations of cycloalkanes

Current time:0:00Total duration:8:09

# Stability of cycloalkanes

## Video transcript

- [Voiceover] At one
time it was thought that the cycloalkanes were all planar, so cyclopropaner was thought
to be a flat triangle, cyclobutane was thought
to be a flat square, cyclopentane was thought
to be a flat pentagon and cyclohexan was thought
to be a flat hexagon. And in terms of analyzing
them, the idea of angle strain was introduced, and angle
strain is the increase in energy that's associated with a
bond angle that deviates from the ideal bond angle of 109.5 degrees, and this number should sound
familiar to you, this was the bond angle for a carbon
of tetrahedral geometry. So if you go through and you
analyze these, the bond angle in here for this triangle
must be 60 degrees, and 60 degrees is a long
ways off from 109.5 degrees meaning cyclopropane has
significant angle strain. For cyclobutane, this
angle would be 90 degrees, and 90 degrees is still a
ways off from 109.5 degrees so cyclobutane also has a
large amount of angle strain, although not as much as cyclopropane. For cyclopentane, this
bond angle is 108 degrees, and 108 degrees is pretty
close to 109.5 degrees, closer than it would be for cyclohexane; This bond angle is 120 degrees. And so the theory was,
cyclopentane is the most stable out of the cycloalkanes,
because this bond angle is closest to 109.5 degrees. However that conclusion doesn't hold up if you look at the heat of
combustion of the cycloalkanes. And first we'll start with cyclopropane. So if you define the heat of combustion as the negative change in the enthalpy, cyclopropane gives off 2,091
kilojoules for every one mole of cyclopropane that is combusted, and if you count the number
of CH2 groups on cyclopropane, let's go back here and
let's count them up. So here's one, two, and
three on the drawings; That's why there's a three here. Now you can't analyze the cycloalkanes in terms of just the heats of combustion. So if we look at those we
can see that they increase. 2,091 to 2,721, to 3,291 and then 3,920. But that's what we expect to
happen because as we go from cyclopropane to cyclobutane,
-pentane, and -hexane, we're increasing in number of
carbons and we already know from the earlier video
on heats of combustion, if you increase the amount
of carbons that you have, you'd expect an increase
in the heats of combustion. So you can't really compare
the cycloalkanes directly in terms of just the heats
of combustion, you have to compare them in terms of
their heats of combustion divided by the number of CH2
groups, and that gives you a better idea of the stability. So if you take 2,091, which
is the heat of combustion of cyclopropane, and divide that
by the number of CH2 groups, which is three, you get approximately 697. So again, this is the heat
of combustion divided by the number of CH2 groups
in kilojoules per mole. And this is a much better way to compare the stability
of the cycloalkanes. Notice cyclopropane has the
highest value here, 697. Cyclobutane goes down to
680, cyclopentane is 658 and cyclohexane is approximately 653. If you remember back to the
video on heats of combustion, this number here, 653 kilojoules
per mole, is approximately the same value we got for
a straight chain alkane when you add it on a CH2 group,
so each additional CH2 group increased the heat of
combustion by approximately 653, or 654 kilojoules per mole,
and that tells us that cyclohexane is pretty much strain-free, cyclohexane is about as stable
as an open chain alkane, and so we know that this idea
of cyclohexane being flat must not be true, so
cyclohexane isn't flat as we'll see in later videos. So cyclohexane is the most
stable out of these cycloalkanes. Cyclopentane is a little
bit higher in energy and therefore a little bit more unstable, and cyclobutane even higher than that, and finally cyclopropane at
697 for a heat of combustion per CH2 group, this is the most unstable, this is the highest heat of combustion, this is the highest in
energy, and so let's analyze why cyclopropane has
such a relatively high heat of combustion per CH2 group. Here we have a model of
the cyclopropane molecule. If I turn it to the side you can see that all three carbon atoms
are in the same plane. So cyclopropane is planar. You can also see that
these bonds are bent. The significant angle strains means the orbitals don't overlap very well, which leads to these bent bonds. You can see the plastic is
even bending in the model set. There's another source
of strain associated with cyclopropane and we can
see it if we look down one of the carbon-carbon
bonds, so the front hydrogens are eclipsing the hydrogens in the back, and cyclopropane is locked
into this eclipse confirmation. All this increased strain means
that a three-membered ring is very reactive, and highly susceptible to ring-opening reactions. So I'm gonna take the ring
here, I'm gonna break it and open up the ring so we can
see that decreased the strain, those bonds even look straight now. Here we have the cyclobutane molecule, and you can see there is
some angle strain here although not as much as in cyclopropane. If we turn it to the side you also see some torsional strain,
the hydrogens in the front are eclipsing the hydrogens in the back. To relieve this torsional strain, cyclobutane can adopt a
non-planar confirmation, this is called the puckered confirmation. And if you turn it to the side here so you're staring down one
of the carbon-carbon bonds, you can see how that's relieved
some of the torsional strain so the hydrogens in front
are no longer eclipsing the hydrogens in the back. Finally we have the
cyclopentane molecule, which has much less angle strain than
cyclopropane or cyclobutane but if you turn it to
the side you can see that the planar confirmation is
destabilized by torsional strain, so we have some eclipsed hydrogens there. Some of that torsional
strain can be relieved in a non-planar
confirmation, so one of the non-planar confirmations
would be to rotate the carbon up like that, and that's called
the envelope confirmation. And you can see that four
carbons are in the same plane. So this carbon right here,
this one, the one in the back and this one in the
back are the same plane is this fifth one here
is up out of the plane. So this looks a little
bit like an envelope, and so that's why it's called
the envelope confirmation. Now that we understand the
ability of cycloalkanes, let's do a quick problem. On the left we have ethylcyclopropane, on the right we have methylcyclobutane. They're isomers of each
other, they both have the molecular formula C5H10. The first question is which
isomer is more stable. Well, we're comparing
a three-membered ring to a four-membered ring, and
we know that cyclopropane is higher energy, there's more
strain associated with it. So ethylcyclopropane must
be the less-stable isomer. So this one is less stable,
which makes methylcyclobutane the more stable isomer, so
there's not as much strain in methylcyclobutane. So I've answered our first question. Our second question is which one has the higher heat of combustion. We know that the more stable compound has a lower heat of combustion. We know that from the
heat of combustion video. That must mean
methylcyclobutane has the lower heat of combustion because
this one is more stable. Which means that ethylcyclopropane must have the higher heat of combustion. So the higher heat of combustion. The higher heat of combustion
is the one that's less stable.