Week 4 of 52: Unimolecular elimination

Apparently January will have been the month of reactions on alkyl halides by bases and nucleophiles. I didn't initially set out to do it that way, but in retrospect it does make sense. I have no idea what reaction I'll start February with. Oh well, it's not time for that yet. It's time for unimolecular elimination.

If you've been paying close attention (which you haven't), you'll probably have deduced that this has the "unimolecular" feature of the S_N1 reaction and the "elimination" feature of the E1 reaction. So you practically know what will happen just from the name, you clever reader, you. But just in case you aren't that bright, which let's face it, is pretty likely, I'll offer a brief explanation...

• This is a two-step reaction.
• In the first step, the bond between the α-carbon and the leaving group breaks. The leaving group leaves.
• The departure of the leaving group results in the formation of a carbocation.
• In the second step, a base removes a proton from a β-carbon.
• As in the E2 reaction, the electron pair from the broken C-H bond forms a π-bond between the α-carbon and the β-carbon.

Still confused? Let's try some pictures. One note: with the S_N2 and E2 reactions, I just assumed that we were dealing with primary alkyl halides. With the unimolecular reactions, a carbocation is formed, so these reactions will be fastest with tertiary alkyl halides. So I'll draw some more groups, just to show that. So, here's the first step...
That's the same first step for both the substitution reaction and the elimination reaction. It's the second steps that are completely different. Here's S_N1...
And you already know exactly what the E2 reaction will look like now, but I'm showing you anyway, just on the off-chance that you really are that incompetent...

Week 3 of 52: Bimolecular elimination

The bimolecular elimination reaction is typically just referred to as the "E2" reaction. It has some things in common with the S_N2 reaction featured in the first week. Both are bimolecular: the reaction is driven by a collision of two molecules. Both involve the concept of a leaving group: an atom or group of atoms that can accept electron density. A classic example is an alkyl halide. In both the E2 reaction and the S_N2, something attacks the alkyl halide, and the halogen is broken away as an anion. I was being lazy for the first two weeks of this and didn't use any illustrations, so let's throw in a picture for this...
I say that's totally a step up from my previous use of pictures in this blog. Anyway, in the S_N2 reaction, which I covered in the first week, not this week, the "something" would be a nucleophile and it would attack the carbon that the halogen is bonded to, the α-carbon. The bond between the α-carbon and the halogen would break and a new bond would be formed between the nucleophile and the α-carbon. Here's a reaction mechanism. Oh, I'm leaving out the hydrogens bonded to carbon in these pictures because I want to. But realize that the α-carbon has two hydrogens attached to it that are just sitting there, not doing anything...
This week's reaction has some important differences. Firstly, the "something" is a base. Basicity and nucleophilicity are similar concepts and the same thing can behave in both ways. A nucleophile attacks a relatively exposed area of positive charge, the nucleus of the α-carbon. A base participates in a traditional acid/base reaction, reacting with a proton. No bond is formed between the base and the alkyl halide. Instead, a bond is broken. Here, I'll show you, but this time, I need to draw some hydrogens...
All of the electron density in the C-H bond at that β-carbon (it's called a β-carbon because it's adjacent to an α-carbon) is dumped onto the carbon. This simultaneously forms a double bond between the α-carbon and the β-carbon and drives the leaving group (the halogen) away. More importantly, I think making this image has placed me officially beyond all redemption.

Emphasizing this yet again, rather than providing new information about this reaction, because I know you can only handle so many new things at once: in nucleophilic substitution, the nucleophile replaces the leaving group, hence the name. In elimination, a π-bond (double bond) is formed between two carbons while the leaving group and a proton are removed, hence the name. Easy, right?

Week 2 of 52: Unimolecular nucleophilic substitution

Because you read the previous post and totally didn't forget everything I said there, you already deduced that this nucleophilic substitution reaction (S_N1) occurs in two steps. Instead of the nucleophile attaching at the same time that the leaving group is removed, first the leaving group leaves, then the nucleophile attacks the "intermediate" and together they form the product of this reaction. In order for this to happen, the bond to the leaving group has to actually break on its own. If the intermediate would not be stable, this won't happen.

I started writing this post too late in the week for me to cover relative carbocation stability, so you'll just have to believe me when I tell you that it's—important. Yeah, that wasn't very convincing. Whatever. Shut up. Carbocations in which the carbon attached to the leaving group have more bonds to hydrogen atoms are less stable. If the carbon attached to the leaving group is attached to more carbons, the carbocation will be more stable. The greater the stability, the faster the S_N1 reaction.

Also note that unlike the backside attack of the S_N2 reaction, the leaving group in this case is already out of the way, so stereochemistry (if the carbon in question is a chiral center) is randomly split between both possible configurations. That means there will be an even mixture of both possible products, not that the each individual molecule will somehow be halfway between both possible products, obviously.

Week 1 of 52: Bimolecular nucleophilic substitution

I think this was the first reaction I learned in an organic chemistry class. Regardless of whether I'm right, it will be the first reaction in this series. Now, I know the name seems intimidating to you, because you're so pathetic. But I'll confess something: I didn't remember that name when I set out to write this post. At least I don't think so. It didn't really cross my mind. I'm used to just calling this reaction by the same name everyone else calls it. The more common name for this reaction is S_N2.

"Bimolecular" in this case refers to the fact that the reaction involves a collision of two molecules. Unlike some other reactions I'll be dazzling you with, this entire reaction happens in one step. One bond breaks at the same time as another bond is formed. Consequently, the thermodynamics of this follow the "second-order rate equation." I'll be gleefully ignoring that for now, so you can too, if you want. The important thing about it is that the rate at which this reaction occurs depends on the concentrations of both molecules involved (increase the amount of either in a system, and the rate of reaction speeds up).

"Nucleophilic" refers to the fact that one of the two reactants is, well, a nucleophile. Nucleophiles are attracted to positive charge. Remember: nucleii of atoms are positively charged. Here's the part where I could elaborate on the intricacies of nucleophilicity as a property, which molecules make good nucleophiles and which ones do not and why, but it turns out that I've been procrastinating on writing this post, so we're pretty much skipping that. Anyway, I will tell you that nucleophiles are often negatively charged particles, which should be obvious anyway.

"Substitution" means that the nucleophile replaces another group. The other group is aptly known as a leaving group. You know, because it leaves. To be a leaving group, an atom or group of atoms must be able to accept electron density. This leaves less electron density on the other side of the bond (which is to a carbon atom) and more exposed nucleus for the nucleophile to do its thing. The most popular leaving groups are halogen atoms, especially bromine and iodine (they're bigger, so the electron density is spread over a larger space). One thing that I didn't remember, but that my textbook deemed noteworthy is that "all good leaving groups are weak bases with strong conjugate acids having low pK_a values."

Another fun fact that can sometimes matter is that this reaction happens by "backside attack." As you may have noted, the nucleophile reacts with the carbon atom, not the leaving group and it wouldn't make sense for it to form a bond in the same spot where the bond to the leaving group is simultaneously breaking. This means that the stereochemistry of the carbon can be completely changed. It also allows for the pickup line: "Baby, if I were a reaction, I'd be S_N2, so I could attack your backside."