STEREOISOMERISM - GEOMETRIC

ISOMERISM

Geometric isomerism (also known as cis-trans isomerism or E-Z

isomerism) is a form of stereoisomerism. This page explains
what stereoisomers are and how you recognise the possibility of
geometric isomers in a molecule.

Further down the page, you will find a link to a second page
which describes the E-Z notation for naming geometric isomers.
You shouldn't move on to that page (even if the E-Z notation is
what your syllabus is asking for) until you are really confident
about how geometric isomers arise and how they are named on
the cis-trans system.

The E-Z system is better for naming more complicated

structures but is more difficult to understand than cis-trans. The
cis-trans system of naming is still widely used - especially for the
sort of simple molecules you will meet at this level. That means
that irrespective of what your syllabus might say, you will have to
be familiar with both systems. Get the easier one sorted out
before you go on to the more sophisticated one!

What is stereoisomerism?

What are isomers?

Isomers are molecules that have the same molecular formula,

but have a different arrangement of the atoms in space. That
excludes any different arrangements which are simply due to the
molecule rotating as a whole, or rotating about particular bonds.

Where the atoms making up the various isomers are joined up in

a different order, this is known as structural isomerism.
Structural isomerism is not a form of stereoisomerism, and is
dealt with on a separate page.

Note: If you aren't sure about structural isomerism, it might

be worth reading about it before you go on with this page.

What are stereoisomers?

In stereoisomerism, the atoms making up the isomers are joined

up in the same order, but still manage to have a different spatial
arrangement. Geometric isomerism is one form of
stereoisomerism.

Geometric (cis / trans) isomerism

How geometric isomers arise

These isomers occur where you have restricted rotation

somewhere in a molecule. At an introductory level in organic
chemistry, examples usually just involve the carbon-carbon
double bond - and that's what this page will concentrate on.

Think about what happens in molecules where there is

unrestricted rotation about carbon bonds - in other words where
the carbon-carbon bonds are all single. The next diagram shows
two possible configurations of 1,2-dichloroethane.

These two models represent exactly the same molecule. You

can get from one to the other just by twisting around the carbon-
carbon single bond. These molecules are not isomers.

If you draw a structural formula instead of using models, you

have to bear in mind the possibility of this free rotation about
single bonds. You must accept that these two structures
represent the same molecule:

But what happens if you have a carbon-carbon double bond - as

in 1,2-dichloroethene?

These two molecules aren't the same. The carbon-carbon

double bond won't rotate and so you would have to take the
models to pieces in order to convert one structure into the other
one. That is a simple test for isomers. If you have to take a
model to pieces to convert it into another one, then you've got
isomers. If you merely have to twist it a bit, then you haven't!

Note: In the model, the reason that you can't rotate a

carbon-carbon double bond is that there are two links joining
the carbons together. In reality, the reason is that you would
have to break the pi bond. Pi bonds are formed by the
sideways overlap between p orbitals. If you tried to rotate the
carbon-carbon bond, the p orbitals won't line up any more
and so the pi bond is disrupted. This costs energy and only
happens if the compound is heated strongly.

If you are interested in the bonding in carbon-carbon double

bonds, follow this link. Be warned, though, that you might
have to read several pages of background material and it
could all take a long time. It isn't necessary for understanding
the rest of this page.

Drawing structural formulae for the last pair of models gives two
possible isomers.

In one, the two chlorine atoms are locked on opposite sides of

the double bond. This is known as the trans isomer. (trans :
from latin meaning "across" - as in transatlantic).

In the other, the two chlorine atoms are locked on the same side
of the double bond. This is know as the cis isomer. (cis : from
latin meaning "on this side")

The most likely example of geometric isomerism you will meet at

an introductory level is but-2-ene. In one case, the CH3 groups
are on opposite sides of the double bond, and in the other case
they are on the same side.

Geometric isomerism in cyclic compounds

Geometric isomers can only occur where there is restricted

rotation about a bond. So far we have looked at the simplest
example of this where there is a double bond between two
carbon atoms, but there are other possibilities as well.

If you have a ring of carbon atoms there will also be no

possibility of rotation about any of the carbon-carbon bonds.
Cyclohexane is a simple example:

Now suppose you replaced two of the hydrogens in the

cyclohexane molecule by two bromine atoms:

The shape around each carbon atom is tetrahedral, and there

are two different ways the bromine atoms can arrange
themselves. They can both lie above the ring, or one can be
above the ring and the other below.

The next diagram is taken from PubChem and shows the

molecule where one bromine is above and the other below the
ring. This would be a trans form.

If you swapped the hydrogen and bromine atoms around on one

of the carbon atoms, then both bromines would be on the same
side - a cis form.

Note: If you followed the link to PubChem, you will find that
this diagram can be rotated in space so that you can see it
more clearly. You will find it in the section titled "1.2 3D
Conformer".

The importance of drawing geometric isomers properly

It's very easy to miss geometric isomers in exams if you take

short-cuts in drawing the structural formulae. For example, it is
very tempting to draw but-2-ene as

CH3CH=CHCH3

If you write it like this, you will almost certainly miss the fact that
there are geometric isomers. If there is even the slightest hint in
a question that isomers might be involved, always draw
compounds containing carbon-carbon double bonds showing
the correct bond angles (120°) around the carbon atoms at the
ends of the bond. In other words, use the format shown in the
last diagrams above.

How to recognise the possibility of geometric isomerism

You obviously need to have restricted rotation somewhere in the

molecule. Compounds containing a carbon-carbon double bond
have this restricted rotation. (As we have seen, other sorts of
compounds may have restricted rotation as well, but we are
concentrating on the case you are most likely to meet when you
first come across geometric isomers.) If you have a carbon-
carbon double bond, you need to think carefully about the
possibility of geometric isomers.

What needs to be attached to the carbon-carbon double

bond?

Note: This is much easier to understand if you have actually

got some models to play with. If your school or college hasn't
given you the opportunity to play around with molecular
models in the early stages of your organic chemistry course,
you might consider getting hold of a cheap set. The models
made by Molymod are both cheap and easy to use. An
introductory organic set is more than adequate. Google
molymod to find a supplier and more about them, or have a
look at this set or this set or something similar from Amazon.

Share the cost with some friends, keep it in good condition

and don't lose any bits, and resell it via eBay or Amazon at
the end of your course.

Alternatively, get hold of some coloured Plasticene (or other

children's modelling clay) and some used matches and make
your own. It's cheaper, but more difficult to get the bond
angles right.

Think about this case:

Although we've swapped the right-hand groups around, these

are still the same molecule. To get from one to the other, all you
would have to do is to turn the whole model over.

You won't have geometric isomers if there are two groups the
same on one end of the bond - in this case, the two pink groups
on the left-hand end.

So . . . there must be two different groups on the left-hand

carbon and two different groups on the right-hand one. The
cases we've been exploring earlier are like this:

But you could make things even more different and still have
geometric isomers:

Here, the blue and green groups are either on the same side of
the bond or the opposite side.

Or you could go the whole hog and make everything different.

You still get geometric isomers, but by now the words cis and
trans are meaningless. This is where the more sophisticated E-Z
notation comes in.

Summary

To get geometric isomers you must have:

restricted rotation (often involving a carbon-carbon double

bond for introductory purposes);

two different groups on the left-hand end of the bond and

two different groups on the right-hand end. It doesn't
matter whether the left-hand groups are the same as the
right-hand ones or not.

Note: The rest of this page looks at how geometric

isomerism affects the melting and boiling points of
compounds. If you are meeting geometric isomerism for the
first time, you may not need this at the moment.

If you need to know about E-Z notation, you could follow this
link at once to the next page. (But be sure that you
understand what you have already read on this page first!)

Alternatively, read to the bottom of this page where you will

find this link repeated.

The effect of geometric isomerism on physical properties

The table shows the melting point and boiling point of the cis
and trans isomers of 1,2-dichloroethene.

isomer melting point (°C) boiling point (°C)

cis -80 60

trans -50 48

In each case, the higher melting or boiling point is shown in red.

You will notice that:

the trans isomer has the higher melting point;

the cis isomer has the higher boiling point.

This is common. You can see the same effect with the cis and
trans isomers of but-2-ene:

isomer melting point (°C) boiling point (°C)

cis-but-2-ene -139 4

trans-but-2-ene -106 1

Why is the boiling point of the cis isomers higher?

There must be stronger intermolecular forces between the

molecules of the cis isomers than between trans isomers.

Taking 1,2-dichloroethene as an example:

Both of the isomers have exactly the same atoms joined up in

exactly the same order. That means that the van der Waals
dispersion forces between the molecules will be identical in both
cases.

The difference between the two is that the cis isomer is a polar
molecule whereas the trans isomer is non-polar.

Note: If you aren't sure about intermolecular forces (and also

about bond polarity), it is essential that you follow this link
before you go on. You need to know about van der Waals
dispersion forces and dipole-dipole interactions, and to follow
the link on that page to another about bond polarity if you
need to.

Use the BACK button on your browser to return to this page.

Both molecules contain polar chlorine-carbon bonds, but in the

cis isomer they are both on the same side of the molecule. That
means that one side of the molecule will have a slight negative
charge while the other is slightly positive. The molecule is
therefore polar.

Because of this, there will be dipole-dipole interactions as well

as dispersion forces - needing extra energy to break. That will
raise the boiling point.

A similar thing happens where there are CH3 groups attached to

the carbon-carbon double bond, as in cis-but-2-ene.

Alkyl groups like methyl groups tend to "push" electrons away

from themselves. You again get a polar molecule, although with
a reversed polarity from the first example.

Note: The term "electron pushing" is only to help remember

what happens. The alkyl group doesn't literally "push" the
electrons away - the other end of the bond attracts them more
strongly. The arrows with the cross on (representing the more
positive end of the bond) are a conventional way of showing
this electron pushing effect.

By contrast, although there will still be polar bonds in the trans

isomers, overall the molecules are non-polar.

The slight charge on the top of the molecule (as drawn) is

exactly balanced by an equivalent charge on the bottom. The
slight charge on the left of the molecule is exactly balanced by
the same charge on the right.

This lack of overall polarity means that the only intermolecular

attractions these molecules experience are van der Waals
dispersion forces. Less energy is needed to separate them, and
so their boiling points are lower.

Why is the melting point of the cis isomers lower?

You might have thought that the same argument would lead to a
higher melting point for cis isomers as well, but there is another
important factor operating.

In order for the intermolecular forces to work well, the molecules

must be able to pack together efficiently in the solid.

Trans isomers pack better than cis isomers. The "U" shape of
the cis isomer doesn't pack as well as the straighter shape of the
trans isomer.

The poorer packing in the cis isomers means that the

intermolecular forces aren't as effective as they should be and
so less energy is needed to melt the molecule - a lower melting
point.

Questions to test your understanding

If this is the first set of questions you have done, please read the
introductory page before you start. You will need to use the BACK BUTTON
on your browser to come back here afterwards.

questions on geometric isomerism

answers

Where would you like to go now?

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© Jim Clark 2000 (last modified February 2020)