Class 3 – More carbocations IMPORTANT SYLLABUS UPDATES: In light of the fact that we haven’t had all that many classes so far, the first test is being postponed to March 1st. Your literature report will be due February 24th instead. I will post notes for how to write a good literature report on the website before class next time. I will also post the actual article for the report as well. Last time I asked you to think about why you can’t actually substitute on the bridgehead carbon in a norbornyl carbocation case: The reason for this is related to something called “Bredt's rule.” Bredt realized that you can never put a double bond at the bridgehead carbon in a bicyclic ring system, because you need a planar geometry around an sp2 hybridized carbon, and that would put too much strain on the bicyclic system. Same deal here – in order for substitution to occur at the bridgehead carbon, the carbocation would have to be localized there, and you would need to have a planar geometry. Because this puts too much strain on the bicyclic system, it doesn’t actually occur. Let’s talk about another real‐world example involving carbocation rearrangements – in particular the Pinacol rearrangement: The overall reaction is shown below: a 1,2 diol is converted into a ketone. This occurs under acidic conditions, and the mechanism is shown below: The key step here is a 1,2 migration of the methyl group. In this case, the starting carbocation and the carbocation after rearrangement are both equivalently substituted. However, the driving force for this rearrangement is that the carbocation product can be converted into a stable, neutral ketone relatively easily. And there is yet another interesting real‐world example that I would like to go through: The Tiffeneau‐
Demjanov rearrangement (TDR) The overall reaction is shown below: Let’s look at this mechanism in some detail as well. The first question is – what does the HNO2 do? It forms a diazo compound with the NH2 group – basically a way to convert NH2 into a good leaving group, N2 (nitrogen gas). And the key step for our purposes is where the N2 leaves, and a bond in the six membered ring moves at the same time to create a seven membered ring‐cation. In order for this rearrangement to occur, you need to be able to draw an intermediate where one of the bonds in the ring is anti‐periplanar to the leaving group bond. See the structure I’ve drawn below with the anti‐periplanar bonds highlighted in red. It would probably be a good idea to look at models of this compound so you can work out for yourself how this rearrangement proceeds. In particular, pay attention to why the resulting positive charge ends up on the carbon that has the OH group attached to it. A general strategy when you encounter new types of molecules would be to look for an anti‐periplanar arrangement to identify a migrating bond. FREE RADICAL CHEMISTRY This is the end of carbocation‐based rearrangements. Now we are going to move on to a discussion of free‐radicals. In general we’ve been talking about forming bonds where one species provides two electrons (nucleophile) and one species accepts the two electrons (electrophile): Another way to form a bond is if each species donates one electron to form the two electron bond. Each of these reactants (with a single, unshared electron) is referred to as a “free‐radical”: In general, a normal arrow refers to the movement of two electrons. When you want to indicate that only one electron is moving, you use half an arrow (see above). There are a couple of interesting questions we can ask about radicals: (1) A carbon‐centered radical can be considered intermediate between a carbocation (six electrons) and fully‐substitued carbon (eight electrons). What is the geometry of the 7‐electron species? These radicals usually are planar like a carbocation. The free electron is in a 2p orbital that is perpendicular to the plane of the actual bonds. A carbanion by contrast – if you had two electrons on that carbon with an overall negative charge – is tetrahedral because you consider the lone pair like a fourth substituent. (2) How do you make radicals? Usually you need some kind of radical initiator. The radical initiator is something with a weak bond that will break homolytically – meaning one electron goes to each atom. Once that first weak bond breaks, you generate two radicals. Each of these radicals can react with another neutral compound to generate a new radical and a new neutral compound. Here are two examples of initiators: 1. AIBN – this stands for azobisisobutyronitrile. It is a classic radical initiator, which undergoes decomposition to generate two radical species. The structure and decomposition pathways are shown below: In the deoxygenation reaction, the AIBN first decomposes to form a radical, that then abstracts a hydrogen from tin to generate a tin radical. 2. Another example of a radical initiator (separate from AIBN) is benzoyl peroxide: This initiator forms free‐radicals by breaking the O‐O single bond: And this particular compound happens to be used a lot as an anti‐acne medication. In general, peroxides are fairly good radical initiators, because they tend to break at the O‐O single bond. When you are working with ethers (diethyl ether, dioxane, tetrahydrofuran) in a laboratory, you are always supposed to label the container with the date you open it. Why is that? Because these kinds of ethers slowly form peroxides. Because peroxides are good radical initiators, they are highly reactive and potentially explosives. You need some kind of external energy source – usually ultraviolet light or heat – that breaks a bond on the initiator homolytically. Once that bond is broken, the initiator will react further to generate a carbon centered radical. Again, you can’t generate a carbon‐centered radical directly with heat or ultraviolet light because its bonds in general are too strong. Next time we are going to talk about the reactions that radicals undergo.