the Clayden et al. book; it is well written and very ... pdf... 2. For those of you that
don't like organic chemistry (what's wrong with you?) and don't want to buy a ...
GARETH J ROWLANDS
ORGANIC & BIOLOGICAL CHEMISTRY LECTURE ONE
Welcome to gjr’s 6 lectures for 123.202. Much of it should be revision but hopefully it will be enjoyable!
The lectures are based on Trevor Kitson’s old notes (for those of you that can get a copy) and material from the following four books. For those of you interested in organic chemistry I thoroughly recommend the Clayden et al. book; it is well written and very approachable. Its index is sh*t but you can download a new version as a pdf...
2
For those of you that don’t like organic chemistry (what’s wrong with you?) and don’t want to buy a book, I recommend that you get any organic text from the library as they should all cover the material.
If you are reading these, I have either been kind enough to give them to you or you have already found the website above!
3
The section of the course will revise bonding, stereochemistry and then go in to a little more detail of the mechanism of substitution and elimination reactions.
4
So...without further ado...
...and we will start with revision of nature of bonds within organic molecules.
This was originally introduced in your first semester here...
So, which orbitals are we interested in?
I give you lecture 1...
5
Lets first look at the orbitals of an atom...
6
...the first one we a really interested in the s orbitals. These are simple spheres. The 1s is small with no nodes whilst the 2s is shown above and is considerably bigger with one node (the inner sphere’s surface).
7
The three 2p orbitals are dumbbell shaped. They have the same energy and only differ by the direction they are orientated.
So how do these orbitals permit atoms to bond?
For most of organic chemistry, these four orbitals are the only orbitals we are interested in. Occasionally, pesky atoms like sulfur, phosphorus or silicon will require us to consider d orbitals but this is rare. 8
If we combine two s orbitals we can form new molecular orbitals; one is known as the bonding orbital (constructive overlap of atomic orbitals). Funnily enough, it contributes to joining the two molecules together. This is a sigma bond and it has a circular cross-section containing the electrons that attract and bind the two atoms nuclei.
We also form the sigma(star) orbital or anti-bonding orbital (destructive overlap). Again it has a circular cross-section and hence is a sigma orbital but it has a node between the atoms so does not bind them. If this orbital gets filled with electrons then the bond breaks. Its importance is due to the fact it governs the direction from which reagents will approach.
Note: if we start with 2 orbitals we will finish with 2 orbitals (either 2 molecular orbitals or 2 bonding orbitals). It makes the maths easy... We’re happy to just deal with this simplistic representation. A more rigorous treatment is known as the Linear Combination of Atomic Orbital Theory.
9
Destructive overlap gives the sigma(star) or anti-bonding orbital...
We can do the same with p orbitals but now there are two different orientations the overlap can occur in. The first is ‘head-tohead’ and this gives a sigma bond (it has a circular cross-section).
10
Again, two atomic orbitals give two molecular orbitals. It’s not that difficult really... (yeah right! as the Tui ads might say...)
p orbitals can also overlap side-to-side and this gives rise to a pi bond. The electrons are found above and below the plane of containing the two atoms but are not found in the plane (the cross-section is no longer cylindrical). It may look like two orbitals but it is only one orbital (it just has a node running through the centre).
Once again we also get destructive overlap that results in the high energy anti-bonding orbital of pi(star). It is only one orbital (but has two nodes). To be honest, I don’t think it looks like the diagram above but should have each lobe vertical.
Best not to think about it too much...it’ll make you cross-eyed... 11
12
Guess you should start to see a pattern here...start with two atomic orbitals get two molecular (or bonding) orbitals.
Now things start to get a bit tricky...hybridisation allows us to mix atomic orbitals prior to forming bonds.
So lets take a look at the atom of organic chemistry...
...our friend carbon...
It’s a useful model but it is just a model...
13
Carbon has six electrons. These electrons are arranged in orbitals (which are arranged in shells). Carbon’s ground state is shown above. Electrons fill the lowest energy orbitals first (Aufbau principle); each orbital can only have 2 spin-paired electrons (Pauli exclusion principle); orbitals of equal energy (p orbitals) accept a single electron until all are half full (Hund’s rule)
14
Basically, it looks something like this! The 1s orbital is so low in energy (or buried deep inside) that we tend to ignore it for our simple organic models (as opposed to complex physical chemistry models).
If all the important electrons are in the 2p orbitals why does carbon form four bonds and why are they pointing to the corners of a tetrahedron (and not at right angles to each other)?
we need to bond the carbon to four groups so we need four bonding orbitals...
The answer to this can be found in hybridisation...
15
16
To get four bonding orbitals we must combine four atomic orbitals. This is done by mixing the 2s and three 2p orbitals...
By combining the orbitals we are effectively mixing the four orbitals to get four new, energetically identical orbitals...
These are known as sp3 orbitals as they are made of 1 x s and 3 x p orbitals.
17
...so here’s our tetrahedron...with the four orbitals pointing to the corners. Each orbital contains one electron. By combining these new sp3 orbitals with another orbital with a single electron we form a bond and hence a new molecule.
They all have the same shape and only differ by their orientation in space. They are directed towards the corners of a tetrahedron. (those of you who can do vector additions can see how the mathematics leads to this...the rest of us will just accept it)!
18
Thus by combining with the four electrons of four hydrogens (using the 1s orbital of H) we have methane.
The overlap of any orbital with a s orbital gives a sigma bond...
So, what about ethene? What shape is it and what orbitals are involved?
Understanding these orbitals allows us to understand the Lewis structures and the octet rule we met earlier (as in 123.101)... 19
20
The carbon atom is attached to three other atoms (or groups) so we need to form three bonds. This means we will use three atomic orbitals on carbon...
Thus we mix the s orbital with two p orbitals. This gives us three new orbitals and leaves one p orbital as it was (nonhybridised / un-hybridised...none of these words are probably English).
The new orbitals are called sp2 orbitals as they are made from 1 x s and 2 x p...
They have the same shape as the sp3 orbitals, a kind of distorted dumbbell. They differ by the direction they are orientated (as we shall see) But please remember that we also have a simple 2p orbital...
21
The three sp2 orbitals are all in the same plane and are roughly 120˚ apart. This means they point to the corners of a triangle and are thus called trigonal planar. Each still has one electron so they will form sigma bonds by head-to-head overlap with other singly occupied orbitals.
22
We still have the non-hybridised p orbital perpendicular (at 90˚) to the three sp2 orbitals...this can participate in side-toside overlap and formation of pi bonds.
23
So here is our sp2 hybridised carbon atom...
...and here are all the atoms that make up ethene. Sigma bonds are formed along the axis linking the atoms (head-to-head). The pi bond does not lie on this axis but above and below (side-to-side).
24
So this is the sigma framework...
...and this is the pi bond...
Guess what’s next?
So a double bond is comprised of one sigma and one pi bond.
That’s right...ethyne.
25
So we only mix the s orbital with one p orbital to give us two new bonding orbitals.
26
Funnily enough, this is called an sp orbital as it is made from one s and one p orbital!
We are left with two unused p orbitals.
Each carbon is attached to two other atoms (or groups), so we need to use two orbitals to form the bonds...
So looking at our orbitals again...we have two new sp orbitals that are the same shape as the sp2 and sp3 orbitals but differ in their orientation.
The sp orbitals are linear...180˚ apart pointing in opposite directions (how many times can I say the same thing?)
We also have two unused p orbitals.
27
28
They are effectively a cylinder...
Remember: we have two unused p orbitals; these are perpendicular to both the sp orbitals and each other.
Combined this gives the four orbitals above (two black, one green and one blue. For those of you in black and white...)...
29
One sigma bond...
30
...and two pi bonds.
31
These allow the sigma bonds to form as a straight line along the axis connecting all the atoms. The p orbitals then overlap above and below this axis (2 x p = 1 pi) and to each side of the axis. Thus a triple bond comprises of...
Can we apply the same principles to heteroatoms?
Yes...but we must know their Lewis structures. The reason for this is...
32
...we need to know how many lone pairs the atom has...
...because lone pairs count as a group when we are working out the hybridisation. (or you can imagine that we have to join the lone pair to the molecule so it needs an orbital)
Therefore, ammonia has four groups around the nitrogen.
33
Nitrogen is sp3 as we have combined 1 x s and 3 x p orbitals. As a result, its shape is based on a tetrahedron.
This means we need to form four new orbitals so we must hybridise four atomic orbitals on nitrogen.
34
Ammonia is actually said to be pyramidal as there is no atom on the top corner of the tetrahedron...but you can see that the shape is due to the fact we have a tetrahedral arrangement of orbitals.
What about water?
sp3 hybridised. Its shape will be based on a tetrahedron. As two corners of the tetrahedron do not contain atoms (only lone pairs of electrons) it is called bent but it is clear where this came from...
Well, the oxygen has two bonds and two lone pairs. So we need four new orbitals... Thus the oxygen will be...
Well, I hope you can see that!
35
36
The whole system works for all atoms and all forms of hybridisation (well you’d hope so or it would be a bit pointless trying to teach it)!
Lewis structure reveals that the nitrogen is joined to two atoms and has one lone pair so, overall, it is attached to three groups so we need three new orbitals...
So it is sp2 (1 x s and 2 x p). Thus it is trigonal planar with a spare p orbital that is used to form the pi bond (and hence the double bond) with the carbon.
The oxygen atom in a carbonyl group is the same...it is attached to one carbon and has two lone pairs. Therefore, it is attached to three groups and so we need three new orbitals.
So lets look at the nitrogen in this imine (hopefully you are happy that the carbon is sp2 hybridised and thus trigonal planar)
37
We use three orbitals so the oxygen is sp2 (1 x s and 2 x p). It is trigonal planar...flat as a pancake.
38
Remember, the p orbitals that make up the pi bond are perpendicular (90˚) to the plane of the atoms (or the axes joining the atoms).
39
Note: the p orbitals are different sizes...in the words of a famous song, “there maybe trouble ahead...” but not in my lectures so I don’t care! If you are interested in this crazy (but important part of organic chemistry) I recommend “Molecular Orbitals and Organic Chemical Reactions” by Ian Fleming (1st edition is called “Frontier Orbitals and...”); it is a fantastic book.
A tricky one?
We only combine 1 x s and 2 x p to get three sp2 orbitals.
Not really, boron is attached to three atoms. That’s it. Nada. Zip. Diddly-squat. Nothing else.
As a result borane is trigonal planar with an empty p orbital running through the centre...
NO LONE PAIR So we only need three new orbitals...
40
...and the last one... Looking at the nitrogen in a nitrile.
41
If we looked at the Lewis structure we would see that it has one lone pair. So the nitrogen is attached to two groups. Therefore, we need two new orbitals.
sp hybridised. It is linear with 2 unused p orbitals that go on to make the two pi bonds (and overall it is part of a triple bond).
So the nitrogen is...
The end...for today!
42
