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In phenol, the $-\ce{OH}$ group activates the benzene ring towards electrophilic substitution. But why is it that the electrophilic bromination of phenol gives trisubstitution (forming 2,4,6-tribromophenol) when aqueous bromine is used, but monosubstitution (either 2- or 4-bromophenol) when a $\ce{CCl4}$ solution of bromine is used? How does the solvent control the extent of bromination?

Related but does not explain completely: How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?

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In aqueous medium, phenol is deprotonated to a certain extent, forming the phenoxide ion in which the ortho and para positions are even more activated than in phenol itself. Hence, trisubstitution occurs here.

$$\ce{PhOH + H2O <<=> PhO- + H3O+}$$

In an organic solvent, phenol is not deprotonated and monosubstitution tends to occur.

ron also pointed out in a comment that:

  1. The concentration of $\ce{Br^+}$ is higher in water than in $\ce{CCl4}$ making the reaction faster in water.
  2. The dielectric constant of water is much greater than that of $\ce{CCl4}$, so the former solvent will better stabilize the charged intermediate, again accelerating the reaction.
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