Benzene Reactions - Online Tutor, Practice Problems & Exam Prep

• Benzene is an aromatic compound known for its unique stability and undergoes substitution reactions instead of addition reactions to maintain its stable aromatic structure.
• Halogenation of benzene involves substituting one hydrogen atom on the benzene ring with a halogen (bromine or chlorine) using a catalyst that matches the halogen (FeBr3 for Br2 and FeCl3 for Cl2).
• Friedel-Crafts Alkylation involves substituting a hydrogen atom on the benzene ring with an alkyl group from an alkyl halide, using a catalyst like AlBr3 or AlCl3 that matches the halogen in the alkyl halide.
• Both halogenation and Friedel-Crafts Alkylation reactions require the use of a catalyst that contains the same halogen as the reagent, and both result in the substitution of a hydrogen atom on the benzene ring.

Benzene Reactions

Now, we've seen benzene a few times within this chapter. Now benzene itself is what we call an aromatic compound. An aromatic compounds have an unusually or uniquely stable condition. Benzene possesses π bonds, but because of its aromaticity and its unique stability, it doesn't undergo additional reactions like we would expect.

Benzene undergoes substitution reactions instead, and that's unlike alkenes and alkynes, which do addition because they have π bonds, but they're not uniquely stable. Now here we're going to cover two types of substitution reactions that benzene undergo, one being halogenation reactions, where we're substituting in a halogen onto the benzene ring, and then Friedel–Crafts alkylation, we're substituting in alkyl group onto the benzene ring.

So just remember, benzene is pretty special. It's aromatic and very stable. It's gonna do substitution reactions instead of addition reactions because it wants to maintain this very highly stable form that it possesses.

Halogenation Reactions

In this video we're going to take a look at the halogenation of benzene. Here, benzene would react with either Br2 or Cl2. In this reaction, one of the hydrogen atoms on the benzene is substituted with an X, which stands in for either bromine or chlorine.

Here we'd say that a catalyst in a form of FeX3 where X3 is either Br3 or Cl3 must be used and has to contain the same halogen as the reagent. So basically what I mean by this is if X2 here is Cl2 then this has to be FeCl3. If X2 here is Br2 then this has to be FeBr3.

So if we take a look here at this benzene halogenation reaction example, we have our benzene with its six hydrogens on it. Remember, X here equals either bromine or chlorine. What happens is that one of these hydrogens gets replaced.

Hydrogens gets replaced by this halogen. Here, one of them. So here the H is now gone, and instead in its place we put one of the halogens. That halogen could be chlorine or bromine. This is a way that we can substitute in a halogen at the expense of losing a hydrogen on benzene.

Benzene Reactions Example

Which of the following is a correct product of benzene substitution reaction with BR₂? So remember here we have the bromanation of benzene where we're going to substitute out a hydrogen on benzene for a bromine.

A doesn't work because we substituted in somehow an iron, which doesn't work. B doesn't work because we only substitute in one bromine, not 2C. Doesn't work because the H is still there.

The answer here is D because there was an H here and we substituted it out to get the bromine there, creating bromobenzene as our answer. So here option D is the correct answer.

Name the product of the following benzene halogenation reaction.

• A. chlorobenzene
• B. 1,1-dichlorobenzene
• C. 1,2-dichlorobenzene
• D. 1,4-dichlorobenzene

Friedel Craft Alkylation

In this video, we're going to take a look at Friedel–Crafts alkylation. Here in this reaction, benzene reacts with an alkyl halide, and what happens is that one of the hydrogens on the benzene ring is substituted with an alkyl group. To do this, a catalyst of aluminum X₃ must be used, and this has to contain the same halogen as the reagent.

So basically what we mean by this, if we take a look at this Friedel–Crafts reaction down below, this is CH₃X. This X could represent either bromine or chlorine. If it's bromine, then this has to be AlBr₃. If this X is chlorine, then this has to be AlCl₃. So the halogens have to match between these two.

Now in this reaction, all that happens is that one of the hydrogens on benzene gets substituted out, and the alkyl portion of our alkyl halide comes in. The alkyl portion of our alkyl halide here would be methyl, so we wind up making methyl benzene as our final answer.

Benzene Reactions Example

Name the major product of the following Friedel–Crafts alkylation reaction. All right, so here we have our benzene and we know we're still going to have benzene at the end. So we can redraw it. And we know that essentially what happens is one of the hydrogens on benzene gets substituted out with the alkyl group portion of my alkyl halid.

So here we have ethyl chloride and we have aluminum chloride as the catalyst. So here we're going to attach ethyl to my benzene. This would be my major product form and its name would be ethyl benzene.

So just remember with Friedel–Crafts alkylation, which you're substituting out an H and bringing in an alkyl group.

Summary of Benzene Reactions

So here we have a summary of our benzene reactions. We have halogenation in it. We have the reagent of X₂ which could represent BR₂ or CL₂. We use our catalyst to get things going. In essence what happens is that we substitute at one of the hydrogens on benzene with one of these halogens.

Here if this is BR₂ then this has to be Fe BR₃ if this is CL₂ and this has to be Fe CL₃. At the end the H is substituted out. The halogen comes in next. We have Friedel–Crafts Alkylations here we use as an our reagent and alkyl Halid X again could be CL or BR. But now we're using the catalyst of aluminum and it could be either aluminum chloride or aluminum bromide.

Here we substitute out our hydrogen and bring in our alkyl group here. If the X is CL, then this is aluminum chloride. If the X here is bromine, then this is aluminum bromine. So remember both of these reactions are quite similar. They use similar types of catalysts and they're just substituting out an H. This one substitutes in a halogen, this one here in alkyl group.

Provide a complete reaction of benzene Friedel Crafts alkylation with 2-bromo-2-methylpropane and name the product.

• A.
• B.
• C.
• D.

Fill in the missing reagent for the provided reaction.

• A.
• B.
• C.
• D.