- The Sigma complex o When the double bonds from an aromatic ring donate their electrons to a strong electrophile, it forms a sigma complex because there is a new sigma bond between the aromatic ring and the electrophile
o The positive charge exists ortho and para to the site of the electrophile being added
o The sigma complex is not aromatic, but once the proton at the site of addition gets abstracted, the ring goes back to being aromatic because the electrophile donates a p orbital to the ring
- Sticking a halogen onto a benzene ring
o You start with Br-Br, but that isn’t a strong enough electrophile to pull electrons out of the double bonds in the benzene
o You mix the Br2 with FeBr3, which bonds with one of the electron pairs on one of the Br molecules. In this complex, the Br in between the other Br and the Fe has a partial positive charge. Then, the double bond can attack the leftmost Br, which then gives an electron pair to its partially positive neighbor, forming the sigma complex.
o Then, the FeBr4- acts as a base and rips out the hydrogen, leaving the final product.
o The transition state is rate-determining, and is endothermic (because it makes an aromatic nonaromatic)
o Overall, the substitution of Br into the benzene ring is exothermic, but you require the Lewis acid catalyst (FeBr3, because it sucks in the electrons from the Br-Br)
o When using chlorine, you use AlCl3 as your catalyst
o When using iodine, you need HNO3, nitric acid. This oxidizes the iodine, which removes electrons from it and makes it electrophilic. It also produces NO2 and water.
- Sticking an NO2 group onto a benzene
o You mix benzene and nitric acid in sulfuric acid.
o The sulfuric acid protonates the OH group in the nitric acid. A protonated OH group is an OH2 group, or water. So that leaves. You’re left with NO2, which is nitronium.
o The middle of the nitronium is a positively charged N. The benzene double bond attacks this N and forms the sigma complex.
o The HSO4- ion left over from the protonation acts as the base to abstract the H
- Sticking a sulfonyl group onto a benzene
o Add SO3 to benzene in sulfuric acid.
o The SO3 has resonance forms where the oxygens are – and the central S is +. This central S is a strong electrophile and gets attacked by the double bond in benzene.
o The HSO4- ion abstracts the proton from the sigma site, and the sulfuric acid present protonates the O- in the SO3 group, forming benzensulfonic acid.
§ Blast from the past: you can use this product with alcohols to stick an OTs or tosylate group where an OH group used to be… that OTs group is a good leaving group, etc.
- Taking a sulfonyl group off of a benzene
o You mix the sulfonated benzene with dilute acid and heat and water (steam)
o The acid protonates the ring and recreates the sigma complex. Then, the SO3 can leave the benzene ring and react with the water to form normal sulfuric acid (the equilibrium favors the sulfuric acid, so the SO3 gets used up).
- Putting D into a benzene ring
o If you react benzene with D3O+, you form a sigma complex with one of the carbons α to the + carbon having one H and one D. Then, either the H or the D is abstracted to give the substituted benzene
o The final equilibrium shows the relative concentrations of H and D in the solution
- The categories of groups that you can stick on benzene and what they do
o Activating groups make substitutions go faster, deactivating groups make substitutions go slower
o Ortho-para directing groups result in substitutions ortho and para to them, meta directing groups result in substitutions meta to them
o Example: toluene
§ When you try nitrating a toluene, the NO2 could come in at any carbon
§ When it sticks to the carbons ortho and para to the CH3 of toluene, one of the resonance forms has a + charge on a tertiary carbon, making the overall transition state more stable
§ With meta, it’s still a little more stable than benzene because the CH3 group is electron donating.
- Sticking different things on benzene promote different reactions
o Alkyl groups: inductive stabilization, they stabilize the transition state by donating electron density to the + charge area
o Methoxyl groups: the O back donates a pair of pi electrons into a double bond with the adjacent carbon, making the + charge more stable
§ When the substituent is added ortho or para to the methoxyl group, one of the positive charges from the resonance forms lands on the carbon next to the O. This positive charge then goes onto the O which back donates an electron pair to double bond with the C (very stable)
o Amino groups: the N does a job similar to O in the methoxy group, donating its available pi electrons
o Nitro group: strongly deactivating, slows down reaction, puts a positive charge on the N next to the carbon in the ring. This means that any attack at the ortho or para positions puts 2 + charges next to each other which is BAD. So the whole reaction is slower, of course, but of the products that you do get, you get more of the meta.
o You can apply this to other substituents with double bonded O’s (which such up electron density, leaving the atom next to the carbon with a partial positive charge) – such as carbonyls, sulfonic acid, esters, cyanos (nitrogen, but same idea), ammonia, etc.
o Halogens:
§ they withdraw electron density from the ring, which means they are deactivating, because you want the ring to have lots of electron density to draw in the electrophile
§ they also back donate electron density because of their lone pairs
§ so it’s harder for the substituents to attack, but when they do, they will be ortho-para so that the + charge can land next to the halogen in one or more of the resonance forms.
§ When that + charge gets there, it’s like a halonium ion (-X+-), and is more stable then without it (meta attack)
- When you have competing effects
o There are 3 classes of groups
§ Strong: the powerful ortho para directors
§ Medium: ortho para directors like alkyl and halogen
§ Weak: meta directors
o The group of the stronger class dominates. If they are within the same class, you get a mixture
o If you have more than one ring, pick one ring that is most activated and consider the activation of that ring only!
- Friedel-Crafts alkylation
o For tertiary halides:
§ Mix your tertiary halide with a lewis acid catalyst (sucks up electrons) such as aluminum trichloide. That takes the halide off of the alkyl group and leaves you with a carbocation
§ The aromatic ring attacks the carbocation
§ The AlCl4- is the base that abstracts the proton
o For primary halides
§ Mix the primary halide with the same lewis acid. This time, the carbocation does not break off, but there is a partial positive charge on the carbon α to the halogen.
§ The aromatic ring attacks this carbon, and the AlCl4- breaks off and then acts as the base
§ This is basically SN2
- How to prepare various carbocations for the F-C reaction
o Alkenes + HF = carbocation (remember the markovnikov orientation)
o Alcohols: the O donates a lone pair of electrons to the B in a BF3, forming a BF3OH leaving group and a carbocation, which gets attacked by the ring
o Later, the BF3OH acts as the base and HF breaks off, leaving BF2O, which is NOT what you started with (BF3), so you need a full equivalent of starting material
- Problems with F-C
o If the aromatic ring has a deactivating group, the synthesis doesn’t work
o Carbocations rearrange
o Once you put an alkyl group onto benzene, the product becomes activating, even more activating than benzene. Therefore, it will react with your halogenated substrate faster than the benzene itself, and you will only get a small amount of mono-substituted produc.
o Solution? Add a ton of benzene. Or use another system.
- Friedel-Crafts acylation
o You start with an acyl halide, and mix it with a catalyst like AlCl3
o The catalyst takes available electrons from the halogen atom and rips it off
o Now you are left with an acylium ion, which has a positive charge on the carbon in one of the resonance forms
o So the ring attacks that carbon and the Cl- acts as the base, forming HCl
o Afterwards, the O part of the carbonyl that is now attached to the aromatic ring draws in the AlCl3 to form a complex, which can be hydrolyzed off with water
o The acylium ion is usually bulky, so it mostly adds in para
o The acylium ion is stable, so it doesn’t rearrange
o The product of an acylation has a carbonyl group, which makes for a deactivated product, hindering further reaction
- Turning acylation products into alkylation products using the Clemmenson reduction
o You start off with your normal F-C acylatoin
o You then use zinc in aqueous HCl to reduce the carbonyl back to an alkane
- How to make an aldehyde out of benzene (Gatterman-Koch)
o Carbon monoxide reacts with HCl to make HC==OCl, which is unstable. Along with AlCl3 and CuCl, you get the HC==O ion, which gets attacked by benzene
