Benzene Chemistry part 2

Electrophilic Substitution

the main reaction of benzene chemistry is electrophilic substitution

✓ at no other point will you see electrophilic substitution. So, as soon as you see benzene or another similar molecule, the words electrophilic substitution must be in your head!

The mechanism is very similar, but not identical, to the electrophilic addition to alkenes in AS.

The way to tackle this section is to learn and understand the basic mechanism, which is the same for all benzene reactions.

+ E represents any electrophile:

Step 1: draw an arrow from the benzene ring to the electrophile. A new bond is formed between a carbon atom on the benzene and the electrophile.

✓ remember the arrow goes from the area of high electron density to the electron deficient electrophile.

Step 2: the structure drawn here is the carbocation intermediate. As we have used a pair of electrons to form the bond, the benzene ring has been broken.

✓ draw the intermediate as a broken circle with a ‘+’ in the middle that covers at least 3 carbon atoms.

Step 3: As was mentioned above, the key point about benzene is that it is unreactive in comparison to alkenes due to delocalization of electrons. So in step 3 we need to substitute/remove a hydrogen atom to regenerate the benzene ring for stability reasons.

A large part of this topic is knowing how to generate various electrophiles and then apply the basic mechanism.

+ + + + + Electrophiles required: NO2 , Cl or Br , R (alkyl) and RC =O (acyl)

✓ It is definitely worth knowing that there are only 4 electrophile types as shown in the box above. There’s only 4, so no excuse for not knowing them.

+ NO2 (nitro group)

You need to use a mixture of concentrated nitric acid and sulphuric acid.

+ They often ask for the equation to generate the NO2 . To help remember the equation, I recommend doing it in two steps.

1.Sulphuric acid acts as a catalyst, which simply means it supplies an H+. This is just a conjugate acid-base reaction as you do in the acid-base topic.

Nitric acid is acting as a base i.e. accepting an H from the sulphuric acid:

+ - HNO3 + H2SO4 → H2 NO3 + HSO4

2.From the equation above, water is then lost → the electrophile:

+ + H2 NO3 → NO2 + H2O

+ + To write out the equation in one step, simply replace H2 NO3 in the first equation with NO2 + H2O from step 2:

+ - HNO3 + H2SO4 → NO2 + H2O + HSO4

The mechanism to then form nitrobenzene is identical to that in part 1.

Regenerating the catalyst

Questions often ask you to show that for example H2SO4 is acting as a catalyst. There are two ways to do this:

In the mechanism

- + Instead of just losing the ‘H’ atom we can use the HSO4 formed initially to remove the H and reform the H2SO4 catalyst.

In an equation

You can also write out an equation to show this.

- + HSO4 + H → H2SO4

✓ you can use either method. If they don’t ask about the catalyst being regenerated then you can just draw the mechanism as normal.

Polysubstitution

A problem associated with this nitration reaction is polysubstitution i.e. more than one NO2 group is added.

You must keep the temperature below 55 °C to get the monosubstituted compound (one NO2 group).

If the temperature is higher than 55 °C you will get something like this:

Friedel-Crafts Reactions There are two parts to this: Friedel-Crafts alkylation and Friedel-Crafts acylation but they are soooooo similar you can almost treat it as one reaction:

Alkylation

the starting reagent that you need for alkylation is a haloalkane

To generate the electrophile, the haloalkane is reacted with a Friedel-Crafts catalyst, often referred to as a halogen carrier. AlCl3 or FeBr3 are used (both do the same thing).

The catalyst polarises the haloalkane, which results in the formation of the electrophile. The AlCl3 “pulls” the Cl away from the haloalkane:

After the electrophile is generated, the electrophilic substitution reaction can take place exactly as described in part 1. The alkylation product will look like this:

Imagine now that you were given the above product and they asked you to identify the starting reagent.

All that you need to do is break the bond between benzene and the alkyl group and put a + sign on the alkyl part.

Remember that the electrophile is not the reagent but something that is generated in the reaction mixture. To get the starting reagent, add a Cl to the electrophile to give, in this example, CH3-CH2-Cl.

✓ After alkylation the benzene ring becomes more activated/reactive (due to the presence of an electron donating alkyl group) towards electrophiles so further substitution can be a problem.

- - You can regenerate the catalyst as we did above for nitration but using AlCl4 instead of HSO4 . Either write out the mechanism

CH2CH3 CH2CH3

- AlCl4

AlCl3 + HCl

Or we can write out an equation: - + AlCl4 + H → AlCl3 + HCl

✓ This also applies to the next two electrophiles in the section below. Exactly as above so I’m not writing it all out again.

Acylation Friedel-Crafts acylation is very similar to the alkylation reaction, but with the addition of a carbonyl group.

the starting reagent that you need for acylation is an acyl chloride

We start off with an acyl chloride, for example:

As for alkylation above, we need to use AlCl3 → electrophile. But starting with an acyl chloride rather than a haloalkane (that’s the only difference):

Again the mechanism is the same as we have done already, just make sure that it is the carbonyl that attaches to the benzene to give the following product:

✓ The product above is more versatile than an alkylation product. You can now do nucleophilic reactions at the carbonyl group, which is an exam favourite in organic synthesis.

✓ The acylation product is more stable than the alkylation product making it easier to control in the laboratory, meaning further substitution is unlikely to happen. I’m sure you’re just as enthralled as I am to find this out!

Halogenation The good news….this is very very similar to the Friedel-Crafts stuff. The bad news? None!

+ A halogen carrier such as FeBr3 or AlCl3 is again used to polarise Br2 or Cl2 enough to generate the Br or Cl+ nucleophile.

- + AlCl3 + Cl2 → AlCl4 + Cl

I have also seen questions where they start from Fe to make the halogen carrier. They would do something like:

2Fe + 3Br2 → 2FeBr3 - + FeBr3 + Br2 → FeBr4 + Br

It doesn’t matter if you use FeBr3 or AlCl3 with halogenation. You can even use FeBr3 with the Cl+ electrophile.

✓ Everything else is the same as we have looked at already: same mechanism and same way to regenerate the catalyst.