Stability of cycloalkanes

- [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.