GARETH J ROWLANDS

ORGANIC & BIOLOGICAL CHEMISTRY

Welcome to a little more revision...

Last lecture we looked at orbitals and bonds...now we need to address a subject that seems to strike fear in to undergraduates...

LECTURE TWO 1

I’m sorry to do this but it’s quite important!

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...well kind of simple...

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This is the carbonyl moiety. A double bond connecting a carbon and an oxygen.

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The double bond is made up of the sideto-side overlap of two p orbitals...

...and this gives us a pi bond.

Lets now look at two adjacent pi bonds.

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We have two double bonds or two pi bonds in this molecule (called butadiene for those that are interested).

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These double bonds are, of course, made up of p orbitals.

As we have seen...any two p orbitals can overlap to make a pi bond (as long as they can overlap side-to-side; this adds a geometric consideration to things).

The electrons are not confined to the alkenes but can spread out over all four atoms. The two resonance structures shown above are extremes; reality is in the middle.

So we can shuffle our electrons to get the molecule above.

The movement of the electrons is represented by our friend the curly arrow. We don’t have unpaired electrons as radicals are too reactive.

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Reality spreads the electrons out over the four atoms in a resonance hybrid.

Another way to view resonance structures and hybrids is as follows:

In the complicated terms of molecular orbitals this is represented as: four p orbitals combine to give us four new orbitals (the two bonding orbitals are shown. The anti-bonding are not). We only have four electrons so only the energetically lowest two orbitals are filled (shown above)

We are going to combine two extremes; one is a man...

...the other is a spider.

...no, all the time he is Spiderman, combining the best qualities of both (along with a strong moral code).

Now the combination does not flip back and forth between the two...it is not a case that at one moment the new entity is a man and then suddenly he turns into a spider...

Simplistic, but effectively the difference between resonances structures (the man & the spider) and resonance hybrids (the real thing, a mixture of the extremes). ..and yes...I’m using my first year analogy, sorry for the repetition.

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It turns out that any adjacent p orbitals that can overlap side-to-side can form a pi bond and this leads to resonance...

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Probably the most famous example of resonance is benzene...

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Benzene is C6H6 with each carbon attached to three different atoms...

If its attached to three groups then we need to combine three orbitals to make three new orbitals to form the bonds.

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This gives the sigma bonds (in black) and leaves each carbon with an unused p orbital. All these can combine to give us pi bonds.

...this forms a symmetrical cloud of electrons above and below the plane of the benzene ring. The six pi electrons are evenly spread around the whole ring and are not localised in discreet double bonds.

They can combine in any direction (either with the p orbital on their left or the p orbital on their right), and they do...

All bond angles are 120˚; all bond lengths are 139 Å (C–C = 147Å; C=C = 133Å). Above are the resonance structures, the extremes.

...resonance hybrid, halfway between the two.

But remember, benzene is not flicking back and forth between these extremes. It is a...

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We normally draw benzene like this, with localised bonds (Kekulé structure) as it makes electron pushing / curly arrows / reaction mechanisms easier to depict. But...

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...this is more correct (but it makes electron-pushing mechanisms more complicated).

Now lets have a look at stereochemistry and the shape of molecules.

This is revision and builds on material you have seen before but might go in to a little more detail occasionally.

An unbroken circle in the middle of the sixmembered ring is WRONG. Do not draw it!

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This section is going to cover quite a bit of material as we look at conformation, shape, ring systems and stereochemistry.

All of these are important to reactions...well ring systems aren’t always important to reactions but they probably are more useful than you think...

So why is shape important?

On one level its all about life and how it occurs. If we have lots of different molecules in our bodies (or any other living system) it is important that the correct ones react at the correct time in the correct place... This selectivity is achieved by enzymes...

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Enzymes have a specific shape and they will only react with molecules that have a complimentary shape.

Once the molecule is in the enzyme then a reaction can take place (which is often shape specific as well).

This allows them to be highly selective. It’s like a lock and a key; only the correct key has the correct shape to operate the lock mechanism.

Of course, this is a hideous simplification! But it starts to give an idea why shape is important.

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Shape is also of vital importance in some reactions. As you shall see, some reactions only occur if a molecule can adopt the correct shape.

Lets first look at the conformations of molecules. Sorry but movies often don’t come out in picture form...hence the big black square...

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Conformational analysis simply looks at the rotation of single bonds (as double and triple bonds can’t rotate).

A single bond is free to rotate meaning there are an infinite number of conformations. But it turns out that some are more preferable (favoured) than others...this is is due to energy differences. Electrons in adjacent bonds (as shown) repel each other and this results in torsional strain.

Lets look at a simple molecule...ethane.

As we rotate the red and and the blue hydrogen we see that there are two extreme conformations that differ by the amount of torsional strain.

(again this is a simplification but it’ll do for now...) 21

The first is the staggered conformation. All the hydrogen are as far apart as possible and the molecule feels no torsional strain. It is the favoured conformation.

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The other conformation is the eclipsed conformation. All C–H bonds are aligned so it feels the maximum torsional strain of 12 kJmol–1 (or 3 x 4 kJmol–1).

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You will notice I’m using a wonderful molecular representation (and yes, this is my sarcastic typing), called the Newman projection. What do these mean and how do you draw them...

So what is a Newman projection? Well if you look at our standard representation and then...

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...place our eye in the plane of paper and look along the axis connecting the two carbon atoms...

...the first thing we get to is a carbon atom and this is represented as a dot...

...connected to this carbon are three hydrogens draw as above...

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...we get to the second carbon atom. We cannot see the dot representing this as it is hidden by the bond (the big circle)...

...if we then continue along the bond, which is represented as the big circle...

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...but the hydrogen atoms on it do peek out and so we draw them on...

Now lets turn our attention to a slightly more complex situation and lets have a look at butane...

Instead, we are interested in the C2–C3 bond. With this we see we have four important conformations of varying amounts of strain.

...thus we have our Newman projection! Now we could look along a number of different C–C bonds; if we look along the C1–C2 (or C3–C4) bond then we get the same situation as ethane (although the energy is slightly different; 14 kJmol–1 in the eclipsed if my memory serves me well). 27

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The anti-periplanar conformation is the lowest in energy; it is staggered with the two methyl groups as far apart as possible.

The other extreme is the syn-periplanar conformation. This is eclipsed with the methyl groups as close to each other as possible.

It is the most favourable and most important conformation (so try and remember it).

8 kJmol–1 is accounted for by torsional strain...

It is the least favourable and has a strain energy of 19 kJmol–1.

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...where is other 11 kJmol–1...

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It is steric strain and is caused by two groups trying to occupy the same space.

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Effectively, the two methyl groups are just too close.

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There are more conformations but they are not as exciting (or important)

There is a second high energy (but not quite as high as the syn-periplanar), eclipsed conformation. Its strain is simply the result of electron repulsion (torsional strain).

And there is another staggered conformation. This has no torsional strain as no bonds are aligned but it does suffer from steric strain as the two methyl groups are getting close...

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Please note conformations are just the result of bond rotation...they are not new compounds or different compounds etc.

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...you all remember the next molecule and the fun we can have with it...

We can change between different conformations without breaking any bonds (unlike constitutional isomers or stereoisomers)

If you only remember one conformation of butane...

Cyclohexane...six carbons in a cyclic molecule with no double bonds. It is C6H12, with each carbon attached to four atoms so they are sp3 hybridised and thus tetrahedral in shape.

Cyclohexane can exist in a number of different conformations. The lowest energy and hence favoured conformation is the chair. I cannot stress how important this conformation is to organic chemistry...many reactions occur through a transition state that looks shockingly similar to this...

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Six positions on the cyclohexane are called equatorial and these stick out into space.

Six positions on the ring are called axial and these rise (or fall) vertically above and below the ring.

If we look along one of the C–C bonds of the chair conformation of cyclohexane we see that it is staggered. Therefore, it has...

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Now here is a common question, “How do you draw the chair conformation?”

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Do not try to draw it as a continuous line, instead, start at one end with a simple V (that has fallen over)

Now add two parallel lines of equal length. Ideally, the bottom line should be in line with the corner of the molecule.

Next, add another parallel line as shown.

Or perhaps it’s not really a question but an invaluable skill...

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Finally, close the circle with yet another parallel line.

This gives you the carbon skeleton.

Next come the substituents and this is where most people make mistakes...

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If you remember that carbon is tetrahedral then you want your diagram to look like a chain of tetrahedrons...

Axial substituents are vertical lines.

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...so add the vertical lines to the carbons on the top of the ring. These lines also go upwards...

On alternate carbons the vertical lines go down. So far so good?

The equatorial lines are a little more complicated but not much. Just need to remember two things...

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...the carbons are tetrahedral...

...and they are parallel to the C–C bonds (which kind of makes them a little easier to draw)

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Whilst the last are parallel to the final two C–C bonds.

...and two are parallel to these two C–C bonds...

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Practice, practice, practice...it makes so much organic chemistry easy to do (and almost fun to work out) Somewhere on Stream Trevor has a video showing you how to draw the chair conformation; watch it.

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So two C–H bonds are parallel to the central C–C bonds...