{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch17Su08 - Reactions of Aromatic Compounds Just like an...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Reactions of Aromatic Compounds Just like an alkene, benzene has clouds of π electrons above and below its sigma bond framework. Although the π electrons are in a stable aromatic system, they are still available for reaction with strong electrophiles. This generates a carbocation which is resonance stabilized (but not aromatic). This cation is called a sigma complex because the electrophile is joined to the benzene ring through a new sigma bond. The sigma complex (also called an arenium ion) is not aromatic since it contains an sp 3 carbon (which disrupts the required loop of p orbitals). The loss of aromaticity required to form the sigma complex explains the highly endothermic nature of the first step. (That is why we require strong electrophiles for reaction). The sigma complex wishes to regain its aromaticity, and it may do so by either by a reversal of the first step (i.e. regenerate the starting material) or by loss of the proton on the sp 3 carbon (leading to a substitution product). When a reaction proceeds this way, it is electrophilic aromatic substitution . There are a wide variety of electrophiles that can be introduced into a benzene ring in this way, and so electrophilic aromatic substitution is a very important method for the synthesis of substituted aromatic compounds. I) Substitution of Benzene A) Halogenation of Benzene 1) Bromination of Benzene 093164d6d9c46d877fee149145455383331fe4ca.doc Page1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Bromination follows the same general mechanism for the electrophilic aromatic substitution (EAS). Bromine itself is not electrophilic enough to react with benzene. But the addition of a strong Lewis acid (electron pair acceptor), such as FeBr 3 , catalyses the reaction, and leads to the substitution product. The bromine molecule reacts with FeBr 3 by donating a pair of its electrons to the Lewis acid, which creates a more polar Br-Br bond, and thus a more reactive electrophile. Benzene will now attack this electrophile to generate the sigma complex. Bromide ion from the FeBr 4 - can act as a weak base to remove the proton, thus generating the aromatic product, H-Br, and regenerating the catalyst (FeBr 3 ). The formation of the sigma complex is an endothermic and energetically unfavorable process - it is therefore the rate determining step . 093164d6d9c46d877fee149145455383331fe4ca.doc Page2
Image of page 2
The second step is exothermic since it regenerates the aromatic π system. The overall reaction is exothermic by about 11 kcal/mol. Comparison with Alkenes Alkenes react spontaneously with bromine to give addition products. E.g. This reaction is exothermic by 29kcal/mol. An analogous addition reaction between benzene and bromine would be endothermic by 2kcal. 093164d6d9c46d877fee149145455383331fe4ca.doc Page3 H H Br H Br H Br 2 H o = -29kcal H H Br H Br H Br 2 H o = +2kcal
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
The destruction of the aromatic sextet causes this endothermicity.
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern