So here we are at part two of the Sn1 vs Sn2: Quick-Guide. Here’s a reminder of the layout of these articles:
- The chief goal of this article [the second article] will be show how the chart is one of the few things you’d have to memorize to have general intuition about Organic Chemistry. This is especially true when weighing competing reactions.
- Review Mechanisms of Sn1 versus Sn2, referring to the chart.
The tentative third article would then exist:
- Wrap up substitutions, and finish with mechanisms of eliminations (E1 & E2).
A fourth article might come to add information.
Depending on how comfortable you feel about Orgo, some things will be intuitive. I’m not sure how convincing I’ll be, but Orgo isn’t that bad if you’re methodical. Although this guide will only cover Sn1 and Sn2 my methodology for learning other core mechanisms is similar.
So, let’s just get straight to it. I decided to use 2-chloro-3-methylbutane to form an alcohol, nefariously I have chosen to use a secondary chiral carbon. If we refer to the chart, then it’s blatant that if I had chosen a tertiary alkyl halide, or a Me-X or a primary alkyl halide, then the unequivocal mechanisms would of been Sn1 and Sn2 respectively. However, by picking a secondary halide we are put into a tough situation where we need to rely on our understanding of Orgo because it’s possible to form products with both Sn1 and Sn2, albeit in practice one reaction pathway might be more productive.
So, feast your eyes on the majestic 2-chloro-3-methylbutane [substrate] below. Here’s the typical question stem: In lab you start with 1M of S-2-chloro-3-methylbutane, and .2M of NaOH. Your partner who is in a hurry dumps an additional .5M of NaOH into the solution with your reactants and original concentration of .2M of NaOH. You both note that the reaction precedes more quickly. After purifying your product, find that all of it is now in the R configuration. Predict the products that would form given the reactants and the reagents, also provide the appropriate mechanism.
Depending on how fluent you are in Orgo, this question may seem to have an obvious answer or you’re perhaps experiencing traumatizing flashbacks of the last time you were tested on this type of thing. Now, is the time where I go get coffee while I wait for you to try the first problem. But, before your start, let’s use the following approach to solving all Orgo problems:
- Calm down, it should be solvable, otherwise it wouldn’t of been assigned –unlike in the real world.
- Take the known variables into account, and perform a differential diagnosis to narrow your options before diving deep into the problem.
- Get ‘er down, just start working the problem, staring never gets you anywhere.
- Compare you answer with the things you expected to see in step two, if your experimental data agrees with the theoretical path you originally presumed then there’s a good chance you’ve go the answer.
- Confirm, then do some accounting.
So, you’ve tried the problem yourself first right?
Okay, let’s discuss how I quickly run through the things in the table provided to quickly discriminate between Sn1 and Sn2 to take care of the problem. All I need to do is weigh the evidence for either mechanism, the mechanism with the most evidence is the winner.
Sn1 versus Sn2: Which Exhibits the Greater Preponderance of Evidence?
I could of chosen to start to weigh the evidence at any point of the chart, but I have purposely chosen to start with focusing on the substrate (reactants).
After verifying it as a worthwhile leaving group, I’d classify the substrate first (3, 2, 1, Me-X etc.), it may help narrow down if it’s going to be Sn1 or Sn2. Recalling that I chose a secondary alkyl halide, I won’t get off the hook that easily because the reaction can be either Sn1 or Sn2 — in practice it’s always worth checking. Next, let’s look at the conditions: the reactants are immersed in an ether solvent (ROR), and the temperature of the solution is at 0 degrees Celsius. Ethers promote Sn2 reactions, while doing nothing to help Sn1 reactions because they do not help encourage stable ionization of the alkyl halide. Furthermore, the low temperature does little to provide the activation energy needed to ionize the alkyl halide necessary to initiate a Sn1 reaction. Looking at the nucleophile candidate, the hydroxide ion, all we need to do is recognize that it’s a moderately strong nucleophile (and a moderate base later when we discuss eliminations). Typically strong nucleophiles are seen in Sn2 reactions while weak nucleophiles are used in Sn1, even more evidence for a Sn2 reaction. This should make sense if we consider that the first step in Sn1 is to encourage ionizing of the reactant, a strong nucleophile would try to steal the glory before the alkyl halide had a chance to ionize, i.e. try to act as an Sn2 nucleophile if conditions permitted, even more Sn2 evidence. Next, there’s that bit about the varying concentrations of sodium hydroxide and the reaction rate. If we recall that Sn2 reactions can proceed more quickly if we vary the substrate or the reagent (aka Sn2 = bimolecular nucleophilic substitution), this sort of puts the nail in the coffin, giving us more than enough evidence to assert that it’s going to provide via Sn2. And, for the overkill note that all of the product had a 100% configuration switch, and only Sn2 does that.
So, the correct mechanism should show a Sn2 mechanism, and all the pertinent details hallmark to a Sn2 mechanism e.g. showing the transition state. The general answer of the Sn2 pathway is below, however please note that I didn’t include the best answer because I didn’t show the mechanism in proper stereochemistry terms (that’s for you to enjoy).
Now, let’s use the same substrate again, 2-chloro-3-methylbutane to do an Sn1 pathway. Here’s another fake question stem: Your lab partner puts 2-chloro-3-methylbutane in a test-tube with water and prepares for distillation. After 45 minutes you’re finished and purify the product, you partner then runs the sample through a NMR machine. You find that the product solution has both a secondary alcohol and a tertiary alcohol. Provide the mechanism to explain this solution, and predict the products. Assume there is no contamination of the sample.
[It’s your turn to draw 2-chloro-3-methylbutane with water]
To be consistent, I’ll run down the list on the table to show that all of those little things you learned in Orgo still apply. First off, the leaving group is a proper one, so it’s not a waste of time to attempt this problem. We know that it’s secondary, so classifying the substrate only confirms to us that Sn1 is a viable idea. What about the solvent? While the solvent is this case is the water. We all know that water allows for salts to dissolve in it, so by nature water encourages ionization, sounds like a Sn1. Let’s also note that the temperature is boiling, this is likely done to input energy to overcome the activation energy, smells like a Sn1 reaction. And if we consider the weak nucleophile, water, we just simply need to recall that Sn2 requires a good nucleophile, so process of elimination leads us to Sn1; it sure tastes like a Sn1 reaction doesn’t it? Lastly, the mixture of products should of been a dead give away, because Sn1 mechanisms allow for hydrogen/methyl shifts.
I’d like to take this time to remind you that the same product is formed using Sn1 and Sn2. Sn2 formed the product, but only the inverted form. So if you wanted to have retention of the configuration you’d have to use Sn1, but then you’d have to deal with having multiple products form, complicating the purification of the final product.
Anyways, I hope this Sn1 and Sn2 guide wasn’t too confusing, and hopefully now you feel better when it comes time to discriminate between Sn1 and Sn2 reactions. Look out for the next entry (part 3 of 3, hopefully) in this series, where we will quickly go over E1 and E2, and truly discuss competing reactions.
As always, you can find me on twitter https://twitter.com/masterofsleep
Future changes to this post:
Will go back and redo examples with a scanner, I had to use my whiteboard in my room — yes, I have a whiteboard in my room.
I will probably append this article with the chart I keep talking about from part 1 to make this post easier to read.
Search and destroy for errata.