Class 2 Carbocations Last time we talked about neighboring group participation in substitution reactions. I want to start today by talking about a few more real life examples. Real life example 1 Let s look at this series of chloroalcohols, and how fast the chloride gets displaced by an external nucleophile. If the chloride is the same in each case, and that s the part of the molecule that reacts, why is there such a difference in the reaction rate? It must have to do with the length of the alkyl chain. The OH participates in the reaction mechanism to generate a cyclic ether. This step happens before the incoming nucleophile attack. This ring is most favored for the third compound in the series because that forms a 5 membered ring: Why are five membered rings so much more favored than any other ring size? This is really a two part question: Why are five membered rings more favored than small rings? and why are they more favored than large rings? The first part should be obvious to most people the smaller rings (3 and 4 membered) have substantially more ring strain that makes them less favorable. But why should it be easier to form a five membered ring compared to a six membered ring? The answer to this has to do with entropy. The free energy of any reaction relates to the enthalpy of the reaction and its entropy this is the chemical equation: delta G= delta H T delta S S = entropy. Things in nature in general prefer to move towards a system of more disorder, which corresponds to an increase in entropy. When you have an open chain molecule in solution, that molecule has lots of potential rotations and positions that it can assume. This makes the molecule very happy because it is more disordered/ less constrained. When you want to form a ring, you are effectively freezing out some of the motions available to that chain. The bigger the chain, the more it
was moving freely before ring formation, and the more difficult it is to form the target ring. This is why it is that much slower to form a six membered ring compared to a five membered ring, even though the stability of six membered rings (once formed) is roughly equivalent to that of a five membered ring. A lot of these cases generated cations as reaction intermediates. I want to talk today in some more detail about carbocations. Carbocations are generated from SN1 reactions, but also from other reactions such as elimination reactions (E1): SN1 reaction: E1 reaction: The stability of carbocations depends on their substitution the more substituted, the more stable. tertiary > secondary > primary > methyl Why does more substitution make carbocations more stable? Because the alkyl groups in general can delocalize the positive charge via a hyperconjugation effect. I don t want to spend too much time talking about hyperconjugation, but I do think I would be remiss as an instructor if we didn t talk about it at all. The Wikipedia definition of hyperconjugation is: In organic chemistry, hyperconjugation is the interaction of the electrons in a sigma bond (usually C H or C C) with an adjacent empty (or partially filled) non bonding p orbital or antibonding π orbital or filled π orbital, to give an extended molecular orbital that increases the stability of the system. NOT THAT ANYBODY HERE SHOULD EVER USE WIKIPEDIA AS A RELIABLE SOURCE. BUT STILL IT CAN BE USEFUL.
Basically, this means that sigma bonds can participate in conjugation in the same way that we typically think of double (pi) bonds participating in conjugation. Pi bond conjugation looks something like this: and the positive charge is delocalized over carbons 2 and 4 (numbering the carbons from left to right) Sigma bond conjugation is harder to draw, but basically involves orbital overlap of the bond: What is depicted here are the orbitals on the adjacent carbons being used to delocalize the positive charge of the carbocation. This hyperconjugation is the reason why substituted carbocations are more stable. Primary and methyl cations almost never form because they are so unstable. Those sorts of precursors would undergo SN2or E2 reactions instead. Tertiary halides will almost always form the corresponding tertiary cations. Secondary halides can go either way a secondary carbocation may form, or they may undergo the SN2/E2 reaction pathway. Secondary cations are also highly prone to rearrangements to generate more stable carbocations. In these kinds of rearrangements, some group moves (usually a hydride or an alkyl group) together with its corresponding electrons. We can divide this discussion of carbocation rearrangements into a few sub topics: (1) What things move in the rearrangement? (2) Where can they be in the molecule (i.e. how far away from the positive charge, before they start moving)? (3) What are actual examples of rearrangements? (1) What moves during a rearrangement? Let s start with a simple example of a secondary carbocation: This secondary carbocation can rearrange easily if the hydrogen on the neighboring carbon moves over:
The resulting carbocation is more stable because it is tertiary. The hydrogen migrates together with its two electrons which is why it is called a hydride shift, and the charge moves to the carbon from which the hydrogen originated. In theory the alkyl group could also have moved. This however would give a different secondary carbocation, and so we would be no better off than if the rearrangement had not occurred. In the case above, the hydrogen atom moved. Other things can also move in carbocation rearrangements including alkyl groups, vinyl groups, or aryl groups. We can create an ordered list of how apt things are to migrate, called their migratory aptitude, but a lot will depend on the particular reaction conditions. The rough migratory aptitude is: H > 3 o alkyl > 2 o alkyl > 1 o alkyl > methyl although in real life, hydrides and methyl groups are the things that most often end up moving. The reason for this order is that the migrating group develops some degree of positive charge. Certain fragments are better able to stabilize the positive charge (like a hydrogen is not particularly unhappy to be positively charged). Here s an example where the methyl group migrates: (2) Where in the molecule is the rearranged group coming from? Most rearrangements occur via a 1,2 shift which means that the migrating group only moves one carbon. (3) Real world examples: Real world example 1:
The carbon 14 labeled secondary tosylate reacts in acetic acid to give predominantly the non rearranged product, and 9% of the rearranged one. The rearrangement product arises through the mechanism shown below: If you have a hard time following this mechanism, then re draw it with all of the implied hydrogens shown, and that will probably clear up some of the confusion. Real world example #2: The first product comes from straight forward substitution of the OH group. But how do we get the second product? It looks like the phenyl group has migrated, which is probably exact what happened. However, the phenyl group likely goes through a three membered ring intermediate during the migration step: I want to move now to a discussion of non classical carbocations which roughly translated means weird carbocations and refers to things with delocalization of the positive charge over single bonds.
Norbornyl carbocation Consider the following case: If you start with either isomer of brosylate and treat it with acetic acid, you end up with the same isomer of acetate product. What is going on here? I ll give you some other information about this case: 1. The exo brosylate is 350 times more reactive than the endo isomer. 2. Chiral starting materials give racemic products. It turns out that the exo brosylate is the unusual case. What you actually form is a non classiccal carbocation, due to the assistance of the C1 C6 single bond: The mechanism is that a C C single bond that is anti periplanar to the leaving group breaks, and forms a new bond to the carbon with the leaving group. This bond formation occurs at the same time that the leaving group is being kicked off: Anti periplanar refers to two substituents that are in the same plane, but are also anti to each other. I know that doesn t make so much sense, so let s look at an example: In this particular example, the H and the LG are anti periplanar to each other. The bonds that connect them are all in the same plane, and the two groups are on opposite sides of the single bond ( anti ).
Oftentimes, two things have to be in an anti periplanar relationship in order for elimination to occur, or for their bonds to communicate with each other. This is because an anti periplanar stereochemical relationship maximizes the overlap of the orbitals that exists. So that is what you are looking for in the norbornyl case which one has a bond from the carbon to the leaving group that has an anti periplanar relationship to another bond on the molecule it is the EXO ISOMER! I ve redrawn the exo isomer with the two anti periplanar bonds highlighted. Now, as the brosylate leaves, a non classical carbocation forms: This intermediate has a positive charge delocalized over three carbon atoms. The acetate can add to either side of the ring, because they are both chemically equivalent, to give identical acetate products: Note that even though the positive charge is delocalized over three carbon atoms, you can only add to two of them i.e. nothing adds to the bridgehead position. Why not? This is your take home question for today. This intermediate is achiral (plane of symmetry). The fact that either starting material gives an exo acetate product is merely because you generate a carbocation, which gets substituted with acetate to give the most stable product. This was the first example of a non classical carbocation (first published in 1949). Since then, there have been many more examples of non classical carbocations.