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chemistry. catalyst in bromination of benzene

I have a doubt.
in the edescel advenced science chemistry unit 5 textbook the catalyst given is iron wire and the reagent, of course, bromine. it says that
2Fe + 3Br2 --> 2FeBr3
and that the electrophile is formed like this
Br----Br + FeBr3 ---> Br+ + Br---FeBr3.
this all makes sense. i haven't been able to draw the curly arrows. one arrow starts between the two Br and finishes betwees tje Br an the FeBr3 (near the plus sign).

what my teacher told me was to use anhydrous aluminium BROMIDE. this again works and is shown in a slide prepared by the RSC (it can be found on this link http://www.wbateman.demon.co.uk/asa2sums/sum5.3/sum5.3.htm )
The edexcel chemistry specifications talk about
"benzene with.... bromine and chloroalkane and acid chlorides in the presence of anhydrous aluminium CHLORIDE. again this works fine, but i am still in doubt as to which one to memorise for the exam.

in short which is the expected catalyst for the bromination of benzene:
iron
anhydrous aluminium bromide
or anhydrous aluminium chloride.

by the way did you all get MnSO4 and Zn(NO3)2 for the salts in the practical. i wrote Al nitrate which i fear is wrong. my precipitate wouldn't dissolve (or maybe i didn't add enough to make it an excess)

Reply 1

john !!

We are told Iron or Iron Brominde will work for Bromination of Benzene
This catalyst works by the reaction:
Br-Br + FeBr(3) ---> Br+ + [FeBr(4)]-
As the bromine's lone pair occupies one of the Iron's vacant orbitals which breaks the bond. Later on when the Br+ substitutes fr an H+, this reacts with the [FeBr(4)]- to make HBr and the original FeBr(3).

Aluminium chloride is used in Friedel-Craft's reactions like alkylation and aclyation.

Reply 2

Hawk

Are you sure it actually breaks the Br---Br bond. I was told that it polarises the bond like so,

Bromination.jpg63.2KB

Reply 3

john !!

No I'm not sure, just what I've been told. That seems possible too I guess.

Reply 4

corkskrew

Br is quite electronegative. Hold your middle three fingers up. That W shape is what the bond angles are in FeBr3.
The problem you've got is catalyst regeneration. For a catalyst to be a catalyst is must:
-Lower the activation energy
-Get regenerated.

It goes something like this....

1) Br-Br polarised by FeBr3
2) Curly arrow, from Br-Br bond to the Fe atom
3) Br is now afull blown +ive charge, FeBr4- formed
4) Br attracted to high electron density in delocalised ring
5) Two electrons move from ring to form a bond between a carbon and the Br
6) At this point you should be able to draw a benzene ring, one carob with a Br AND a H shown bonded. The ring is now "broken" (pi bonds are disrupted....if you dont recognise pi boning dont worry about it). Instead of a full ring, there is a half ring like a U shape, the open end facing the carbon with the H and Br shown bonded. This U shape thing should also have a + in the middle
7) Electron pair on the C-H bond shown is donated to the ring (curly arrow required)
8) H+ now floating around near the FeBr4-. One of the Fe-Br bonds break

Now to the explanation of the importance of using FeBr3....
Look at step 3. If AlCl3 were used, there would be AlCl3Br- and this would mean in step 8 there would be AlCl3Br as well (obviously). Consider the possibilities. To regenerate the catalyst the original molecule must be created.
With FeBr4- the ONLY bond that can break is an Fe-Br bond, so ONLY FeBr3 can be created.
With AlCl3Br- a Al-Cl OR Al-Br bond can be broken, so EITHER Al-Cl3 (the original catalyst) or AlCl2Br (a different molecule) can be created.

The important thing here is, to use the same halogen in the catalyst that you need to substitute into the benzene ring.
You can use Aluminium or Iron
It's anhydrous because Metal halides react violently with water to give metal hydroxide and a diatomic halide gas (I wouldn't fancy chlorine gas floating around in the lab).

Hope this helps :-)