Chem3x11 Lecture 8
Being constructed Sat May 19
This lecture is about how things add to alkenes, using bromine as an example.
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- The stereochemical outcome of a reaction gives clues to the mechanism of that reaction
- The addition of bromine to double bonds is via a cyclic bromonium ion that is formed in a concerted process
The Addition of Bromine to a Double Bond
Alkenes are important because they occur naturally in great quantity and we can use the reactive double bond to make more interesting molecules. The formation of epoxides is one example we have seen of how alkenes can be transformed into other molecules. The addition of bromine is another widely-used reaction. The jargon for this reaction is an electrophilic 1,2-addition. The bromine adds such that one bromine is added to one carbon and one to the other (a 1,2-addition) and the reagent we add (bromine) is seeking out the electrons of the double bond, meaning we refer to this as an electrophilic addition (unlike when a nucleophile adds to a carbonyl group, which is a nucleophilic addition).
Some Mechanistic Possibilities
How might this reaction be occurring? Perhaps a concerted mechanism like this:
...or a stepwise polar mechanism like this:
...or maybe something exotic like a stepwise radical mechanism like this:
On paper there is no easy way to say which is right, so we need to go to the lab, do the reaction carefully and see what happens. If you remember back to when we looked at the opening of epoxides, we found that we learned a lot about the reaction mechanism by using Z- and E- isomers of an alkene. Let's try that trick again.
Stereochemical Clues to the Addition Mechanism
Imagine we have some double bond with things on the end (R1 and R2). Let's start with the Z-isomer, and let's add some reagent Y2 to the double bond and let's say for the moment that the addition does not go via some carbocation or radical intermediate and think about what happens if the addition is cis or trans (i.e., if both atoms of Y2 add to the same face or to opposite faces.)
This rather complex diagram just shows that the outcomes from the different starting material isomers will be different for either cis or trans mechanisms, so we ought to be able to work out what's happened when we do the experiment. To make it simpler, imagine we go in the lab and pick the two isomers of stilbene, meaning that R1 and R2 are both phenyl groups and X = H. We do the reaction, and discover the following:
...which implies trans addition without a doubt. The question is, how? We'd need to measure more things about this reaction. We look at the same reaction in a cyclohexenyl system, and we observe selective formation of trans-diaxial products:
...which is reminiscient of what we've seen before in epoxide openings. Could it be a similar mechanism - the formation of an epoxide-like intermediate, with rear-face attack? When we measure the kinetics of the reaction we find it's first order in both alkene and bromine. We also change the alkene structure a little and find that the reaction is accelerated when there are electron-donating substituents on either carbon, and significant rate enhancements when both carbons have electron-donating substituents attached. Then you see that someone has tried the reaction with a very sterically crowded alkene shown below and trapped an intermediate bromonium ion (which was later crystallized) where one bromine is bridging the two carbons that used to form the double bond. How cool is that?
(in this case the bromonium ion, which usually reacts with bromide to give the final product, is so sterically hindered that it can't)
All this, and other bits of evidence (See Anslyn and Dougherty section 10.5.3), suggest this bromonium ion is the intermediate formed (that we do not normally see) and that it's on the productive reaction pathway (and not something irrelevant). Its existence is the reason the electrophilic addition of bromine to alkenes proceeds to give the product with trans stereochemistry.
How Do We Draw The Mechanism for Formation of the Bromonium Ion?
The reaction to give the bromonium ion involves the formation of two bonds, so our mechanism needs to show that. We can do such a thing in one step if electrons come from the bromine and add to the alkene, and at the same time come from the alkene and add to the bromine. Thus the HOMO and LUMO of each interact with each other in this way:
Though this drawing may at first site be rather odd, the lobes of the MOs match up perfectly for this to take place. Clearly in most cases the HOMOs and LUMOs are close in energy here and can interact productively. The reaction is exothermic. The curly arrows depict one of the bromines adding to one of the carbons, and the double bond adding "back" to the same bromine, with concerted cleavage of the Br-Br bond. The end result is a single bromine atom bridging the two carbons.
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