Why do you get isopropyl benzene on treating benzene with 1 - chloropropane instead of n - propyl benzene?

Answer 1

Your question is in regard to the "Friedel-Crafts Alkylation" reaction, which is probably the most common way to add an alkyl group to a benzene ring. The reaction is carried out with benzene or a substituted benzene, an alkyl chloride or bromide, and a small amount of a Lewis acid catalyst; usually either FeCl3 or AlCl3 when an alkyl chloride is the reagent, or FeBr3 or AlBr3 when an alkyl bromide is the reagent.

The first step in the reaction forms a carbocation when the halide on the alkyl group is removed by the Lewis acid. If 1-chloropropane is used, the first step of the reaction is:

H3C-CH2-CH2Cl + FeCl3 ''Ä¢√á H3C-CH2-C+H2 + Fe{Cl}4-

Primary carbocations are the least stable ones after the methyl carbocation, therefore they will almost always rearrange if possible to yield a secondary or sometimes even a tertiary carbocation. Thus, the next thing to occur is formation of a secondary carbocation. This happens when a hydrogen atom on the middle carbon migrates to the carbon bearing the positive charge and brings two electrons with it. This is called a hydride shift because the hydride ion is H-, and it takes place as shown:

H H

| '§µ |

H3C-CH-C+H2 ''Ä¢√á H3C-C+-CH3

(I apologize, but these were the best graphics I could do given the tools available here.) The equilibrium shown is in much greater favor of the secondary carbocation, however I would not be surprised if some propyl benzene is formed.

So, there are two parts to the answer to your question: 1) A primary carbocation is formed when a 1-chloro-n-alkane reacts with a strong Lewis acid. 2) Because the reaction of a carbocation with benzene is slower than formation of the primary carbocation, the primary carbocation has time to undergo a hydride shift, thereby creating a much higher concentration of the more stable isopropyl carbocation.

P.S.: Have you studied which groups activate a benzene ring and which groups deactivate the ring, and which groups direct subsequently added groups to the meta position and which groups direct groups later added to the ortho and para positions? If you have, what compound would you expect to be the major product of the reaction? What can one do practically speaking when performing the reaction above to minimize multiple alkylations?

I'm correcting two molecular drawings included in the original posting. They did not come out anywhere close to how they appeared on the screen when I wrote the first answer.

The first one didn't come out that badly:

+

H3C-CH2-CH3 -----> H3C-CH2-CH2 + Fe(Cl)4-.

The second graphic was a drawing of the secondary (isopropyl) carbocation:

+

H3C-CH-CH3. (The "+" symbol is normally drawn much closer to the atom bearing the formal positive charge. It was not possible to do this here as answers rarely require graphics.)

Answer 2

Isopropylbenzene forms when treating benzene with 1-chloropropane instead of n-propylbenzene because 1-chloropropane undergoes an S N 1 mechanism due to the carbocation intermediate formed. This leads to rearrangement of the carbocation to a more stable secondary carbocation, favoring the formation of isopropylbenzene over n-propylbenzene.

Answer 3

It is because the carbocation formed (primary) is less stable than the secondary carbocation. So that's why.