Benzene and phenol
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Why reacting phenol with bromine water will yield a tri substituted product while reacting phenol with bromine in tetrachloromethane (CCl4) will yield only a mono substituted product?
Jenkin says:
There seems to be no simple answer to this one. It is worth noting that with bromine water, the reaction is with phenate anions, C6H5O-, and involves a ketone-like intermediate. Our emails don’t send diagrams, but you can perhaps try working through this:
Phenate anion + Br2 gives an intermediate with =O (in position 1, where the original –OH was), H and Br in the ‘2’ position and double bonds at 3,4 and at 5,6.
This loses the proton from position 2 to give 2-bromophenol (or rather the 2-bromophenate anion)
A further Br2 reacts to give a similar intermediate with =O, Br still in the 2 position, H and Br in the 4 position and double bonds at 2,3 and 5,6. Loss of the proton from position 4 gives 2,4-dibromophenol, again as its phenate anion.
A third Br2 reacts similarly to give the final product, 2,4,6-tribromophenol.
In CCl4 we are looking at a non-polar solvent and the reaction is with phenol itself rather than with the phenate ion and the product is mainly 4-bromophenol. A similar result is obtained if an acidic solution is used.
In both cases we are looking at an electrophilic substitution and the –OH group enhances the reactivity of the aromatic ring.
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updated: 04 April 2008
