Welcome to part three, of the well…okay four. I originally intended to make this summary three parts, but in the end it looks like it’s going to be four. However, this entry will pretty much perform what I had intended, that is to wrap up the Sn1/Sn2 & E1/E2 reactions. I needed another entry to cover the infamous competing reactions. However, I didn’t want to hold things up for people with looming tests. So, here’s the plan:
- Wrap up substitutions, and finish with mechanisms of eliminations (E1 & E2)
- Fourth article will cover competing reactions
- If you have any questions, or things you’d like to see in fourth entry just message me here on at twitter https://twitter.com/masterofsleep
We have spent all of our time looking at 2-chloro-3-methylbutane. So far, we’ve seen that this secondary alkyl halide undergoes Sn1 reactions quite willingly, and less enthusiastically Sn2. There’s one more thing we can do with our secondary alkyl halide, and it’s called an elimination. There are two ways to form an elimination product, those are E1 & E2.
Although there are two new reactions to think about, because the Sn1 mechanism pathway is related to E1, and the Sn2 mechanism pathway is related to E2, if you’ve mastered Sn1/Sn2 then these next two aren’t that bad — I will refer back often to the mechanism of Sn1 and Sn2 as a foundation. Therefore when we use NaOH and 2-chloro-3-methylbutane, we should expect not only the formation of the nucleophilic substitution product i.e. 3-methylbutan-3-ol, but we should also expect the formation elimination products.
It is interesting to note that we could use the very same reagent NaOH, with the same reactant 2-chloro-3-methylbutane , and still form two disparate products i.e. a substitution & elimination product. This is also why you need to work hard in lab to purify your sample usually — fortunately Organic Chemistry labs usually don’t grade you on percent yield. The reason why this happen is because the hydroxide ion in our reaction can act as a nucleophile or a base, in this case a Bronsted base (you’ll never escape acid-base chemistry).
So, referring back to 2-chloro-3-methylbutane let’s see what happens when the elimination product forms….
Now, our first decision to make is, what mechanism does it follow? E1 or E2? If you’ve read the last two blog entries you probably saw that I went line for line on a chart of reaction properties to justify my answer. I could do that here, to justify how I would select for an E1 and E2 reaction. But, I want to do this in a more fluid and quicker method, relying on some basic Organic Chemistry ideas. As I already mentioned earlier, Sn2 shares similar mechanism traits with E2, and the same goes for Sn1 and E1. So, I just need to ask myself, does this problem look more like an Sn1 or Sn2 type of problem? Begrudgingly, I remember that this secondary alkyl halide can undergo both Sn1 & Sn2, so that means that this problem can actually undergo both E1 & E2. But, in this specific problem, the reagents favor Sn2 (NaOH is a strong base, and the solvent is an ether) so they should also favor E2. Therefore, the problem above is an E2 reaction.
Let’s not forget the minor product, this time I won’t show the mechanism:
Elimination Bimolecular (E2)
As a rule of thumb, when you see a reactant, and a double bond (or any pi bond) suddenly appears on your product then it was likely an elimination. There are three things to carefully annotate in the mechanism of E2:
The confirmational isomer of our reactant should be in anti-coplanar orientation. The easiest way to pull this off is to orientate the hydrogen that’s going to be plucked off by the base (the acidic hydrogen) and the leaving group into the planes I left them in using the 3D orientation drawing I used in the mechanism.
Besides the usual arrow pushing mechanism routine be sure to clearly label the transition state demonstrating:
- Three partial bonds, 2 sigma bonds and 1 pi bond (see E2 mechanism drawing)
- Two delta negatives, one on the leaving group and another on the base. If you don’t there’s a good chance you’ll only receive partial credit. (see E2 mechanism drawing)
- The rate limiting step of the reaction is the formation of the transition state. (see E2 mechanism drawing)
Later in Organic Chemistry you likely won’t be asked to show the mechanism of E2, but at some point you will be tested directly on it.
Elimination Molecular (E1)
For our next reaction we shall use our good friend 2-chloro-3-methylbutane as a reactant, and leave it boiling water for same time. You should already know that these conditions favor substitutions along the pathway of Sn1. If this doesn’t make sense, look back at the Sn1 versus Sn2 Chart. Now, being that we know that these conditions favor Sn1 we can extrapolate that this reaction can also proceed only the pathway of E1.
If you are already familiar with Sn1, then E1 shouldn’t be a big change. Like Sn1, the rate limiting step of the reaction is the formation of the carbocation intermediate. It also should be noted that carbocation intermediates when formed can be isolated, whereas transition states never can be isolated in practice — for this reason a transition state should never be reformed to as an intermediate. There are several things you should be sure to capture in the mechanism:
- Identify the ionization step of the substrate
- Label the intermediate carbocation, watch for hydrid shifts etc.
- Double check to make sure you’ve accounted for all of the products
As you can see from our answer in E1, we end up with multiple products, just like E2. In my example we ended up with the same result whether using E1 or E2, this won’t always be the case. For example, we could of started with a trickier example reactant, and done a number of hydrid shifts, this would of given us a product likely only obtainable via E1.
Within the world of eliminations the most thermodynamic stable product is the major elimination product, fancy talk for pick the product that is the most substituted. Then be able to use factors such as Zaitev’s Rule to discriminate further if there’s a close call. Predicting product stability tends to be a pretty big testing point, so you should feel comfortable ranking the products of E1/E2 reactions [collectively called alkenes] in order of stability.
That’s pretty much it for E1 and E2, they actually aren’t that bad if you already have a solid hold onto Sn1 and Sn2 because they’re closely related. The only difference really is that the reagent acts as a nucleophile, attacking a electron deficient center on our substrate and under going a substitution for a leaving group. In eliminations our reagents act as bases, abstracting a slightly acidic protein away from our substrate, and the loss of our leaving group, ultimately this results in the formation of a double bond.
That’s pretty much it. Now, there’s nothing left to do but practice. If you felt sort of lost about the whole explanation I’d encourage you to go read over Sn1 and Sn2 again, or perhaps rework some problems etc. If you are having problems with the whole guide as a whole then my best suggestion for you is to go review acid base chemistry in the Organic Chemistry text. This may sound like a strange suggestion, but this is because at it’s heart most of Organic Chemistry is just about an electron-deficient area motivating an electron-rich area to share the wealth, i.e. acid base chemistry.
The next, and last entry will cover competing reactions. A lot of people start to get frustrated around competing reactions, but it’s not that bad if you the logic of the reactions straight.
So, here’s a table summary of everything we’ve covered so far with eliminations in a fancy compiled table:
Things to Consider
|Substrate||3 >2||3> 2>1 (notice that methyl-X is impossible)|
|Kinetics||1st order kR[substrate]; kinetics depends on only the concentration of the substrate||2nd order kR[substrate][nucleophile], i.e. the kinetics depend on both the concentrations of the substrate and|
|Base (not nucleophile!)||Weak bases are a-ok||Strong bases only – bring a strong base or else loom in lab forever|
|Leaving Group||Good Leaving Group Needed|
|Solvents||Solvents should encourage ionization||A lot of solvents work*, not too picky|
|Stereochemistry||No special orientation needed to start||Must be in coplanar transition state to proceed.|
|Rate Limiting Step||Formation of carbocation, this also means hydrid shifts are possible so expect rearrangements!||Formation of the coplanar transition state|
Well folks, that’s it for today. Good luck in your premed/MCAT days ahead of you. It’s all worth it, seriously.
Will be back soon with the last installment on competing reactions, then switching gears back to AMCAS entries.
As always, just follower or say what’s up on twitter, I love hearing from you guys: https://twitter.com/masterofsleep