Showing posts with label Molecular Orbital Theory. Show all posts
Showing posts with label Molecular Orbital Theory. Show all posts

Monday, September 7, 2009

Ass'N Tool (Pun Intended!)

Well, so we all know (or do we? Haha!) that the famous SN2 reaction proceeds via a 'backside' attack.

Yep, you got that right - when the nucleophile attacks an alkyl halide, it hits it right on its gluteus maximus (actually, I see it from another angle, but never mind).

And of course the age-old question: why? Why in particular, the 'backside'? Well, everything will be made clear with Molecular Orbital theory, so let us consider what the LUMO (Lowest Unoccupied Molecular Orbital) of a typical alkyl halide (in this case, chloromethane) looks like:


Notice the huge lobe on the 'backside' of the carbon atom? That's where the nucleophile attacks if if wants to bond to the carbon atom - by donating electrons into this antibonding orbital, the carbon-chlorine sigma bond effectively breaks, and a SN2 reaction ensues.

Ah well. I probably haven't explained in detail enough. But hey, I'm tired. :)

Monday, August 24, 2009

Water Me Down!

You know, I’ve always wanted to construct a Walsh Diagram for a simple molecule? So what exactly is a Walsh diagram? Well, let me illustrate using water as an example; one should know that water has an equilibrium geometry that is bent, with a H-O-H angle of 104.5 degrees as such:


And of course, the other limiting (non-equilibrium) geometry would be a linear geometry with a H-O-H angle of 180 degrees as follows:


So now the question is, why is the former geometry adopted by water molecule? Phrasing this question another way: is it such that the energy of the water molecule is lower when it has a H-O-H bond angle of 104.5 degrees?

Well, the easiest way to find out whether this is true or not (not so much why it is true!) is to plot a graph of the energy of the water molecule against the H-O-H bond angle.

Now usually you wouldn’t be able to calculate the total electronic energy of the water molecule by hand, but since I’m bundled up with this molecular calculation suite called Gaussian03 (yes, I know the latest version is G09!), I’ve performed the calculations using my trusty laptop.

So here’s the graph, or like how they call it, the Walsh Diagram for the water molecule (click to enlarge):


This graph actually shows that the 104.5 degrees H-O-H bond angle gives rise to a lower total electronic energy of water than the linear geometry! So be convinced! That the Singaporean education system isn't wrong! :)

So the next step comes in explaining why that's the case.

Another time. :p

And all calculations were clumsily performed with a minimal STO-3G basis set using the Hartree Fock level of theory, which merely took 5 seconds or so for each single point energy calculation.

Yay! Time to sleep. :)

Tuesday, February 5, 2008

Pre-Eminent Molecule: Water (I)

God uses water in the bible, in very powerful ways - He used water to flood the Earth in the Noah's Ark saga; water is used for baptism, even to baptise Jesus; Jesus turned water to wine; Jesus and Peter both walked on water itself; Moses knocked his staff on a rock and water came out, springing forth life, that quenched the dying Israelites' thirst and the list is endless! The bible makes so many referenes to the humble water molecule, and it's a really incredible molecule!

I bet you the only reason why it's such an unsung hero in our lives is because it seems to have an ubiquitous existence, and that's why we tend to take the precious water we have for granted. I feel that Singaporeans have themselves taken the potable water flowing out of their taps for granted, myself inclusive.

Therefore, inspired by my friend, Shaun Lin Darong, I shall now start posting a series of posts on the water molecule, and the first post is a simple one, to whet all of your appetites: have you ever wondered why the water molecule can only accept a single proton when it has two lone pairs of electrons?

What I'm trying to say is, why can't the following reaction occur:



After all, water has two lone pairs of electrons right? So why not? A seasoned Chemistry student would simply retort: Of course it can't! The first protonation results in a positive charge on the water molecule, which naturally repels the addition of another positively charged proton!

However, another could simply argue - then if I just increase the concentration of acid in water, eventually this will result in both lone pairs of electrons on the oxygen being protonated right?

And the argument could go on endlessly in fact, and we'd have no idea who's right! The problem with such arguments is that all of them are purely qualitative explanations, and there is no rigour to back them up at all. So how should we go about solving this problem?

Easy, we fall back to first principles: we do a simple molecular orbital calculation! All will come to light when we analyse the water molecule from its very fundamental electronic structure, and here goes:


Water is a simple triatomic bent molecule, with a bond angle of 109.5 degrees - upon closer examination, we see that its point group is C2v, such that we say it has C2v symmetry:





I've included the symmetry operations above for reference, and here is the associated character table for C2v molecules:


So let's start assigning each irreducible representation in the table to each orbital in the three atoms of water! Let's however, not forget one postulate in molecular orbital theory, that in order to form bonds, orbitals must go into overlap effectively, both in a spatial and energetical sense. However, this means that the only orbitals in effective overlap will be the valence orbitals, since the inner core orbitals are buried deep within the atoms. Therefore in our analysis, we shall only consider the valence orbitals of both oxygen and hydrogen atoms.

If so, let us then consider the symmetry labels of the oxygen valence atomic orbitals first, where we concern ourselves with the 2s and three 2p orbitals. Taking a look at the 2s atomic orbital of the central oxygen atom, we see that:



Indeed, it has a symmetry label of a1, corresponding to the totally symmetric irreducible representation of A1, since all operations result in the same 2s orbital. If we take a look at the 2pz orbital, we find that it also has the a1 symmetry label:


The 2px and 2py orbitals however, have symmetry labels of b1 and b2 respectively, because they correspond to the irreducible representations of B1 and B2 respectively (please verify this yourself by looking at the diagrams below and performing the symmetry operations yourself!):



If so, then we can summarise everything in a single table, on the oxygen atomic orbitals:


It is now time to take a look at the hydrogen atomic orbitals, namely the 1s atomic orbitals; however, we require not the 1s atomic orbitals but rather, a symmetry-adapted linear combination of the 2 1s atomic orbitals, which means that we could either have an additive linear combination, or a subtractive linear combination (note that this is only the case for simple additions of two atomic orbitals, since these are the only two combinations that can result). The SALCs are illustrated below:



I coloured them differently to show whether they're in phase or not, and leaving the verification to the reader, these two SALCs have symmetry labels of a1 (additive linear combination) and b2 (subtractive linear combination), and thus we summarise this in a table again:



So let's plot the molecular orbital energy level diagram:


One now easily sees that the b1 molecular orbital is non-bonding, indicating a lone pair of electrons centred on the oxygen atom a la 2px fashion. Where is the other lone pair of electrons? Easy, it's the lowest a1 molecular orbital, which is also non-bonding (notice it isn't lowered very much in energy), which has the the form of a 2s atomic orbital centred on the oxygen as well.

But what does this mean? Notice that the b1 molecular orbital has an energy of -13.6 eV, which is just nice for the bonding to a proton, whose energy is also at -13.6 eV! Howevr, the lowest a1 molecular orbital is simply too low in energy (i.e. too unreactive) to even bond with a proton, and therefore we expect only one proton to be able to bind, for there is only one reactive binding site.

Wow. This post has been heavy on graphics. :p