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Consider the bromination reactions of phenol and anisole. We know that the former can be brominated by aqueous bromine, while the latter requires the catalyst $\ce{FeBr3}$ instead.

Our teacher said that $\ce{-OH}$ is a stronger ring activator. So, polarization of bromine takes place easily and, hence, phenol doesn’t require a Lewis acid for bromination.

But, this related question had the conclusion that both anisole and phenol are practically equally activating. So, does aqueous bromine react with anisole too? If not, why is there a difference in the reagents? If yes, do we get 2,4,6-tribromomethoxybenzene?

We have a similar case with chlorobenzene also. Difference in electronegativity between $\ce{C-O}$ is more than that in $\ce{C-Cl}$. So, chlorine must be able to give one of its lone pairs and activate the ring more readily than oxygen. But even bromination of chlorobenze requires a Lewis acid. Why is this so?


PS: I checked this similar question on our site but a proper reason wasn’t stated to my question.

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  • $\begingroup$ $\ce{-OH}$ is a better activator than $\ce{-OCH3}$ see this question $\endgroup$ – Avnish Kabaj Mar 30 '18 at 10:00
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    $\begingroup$ Possible duplicate of Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution? $\endgroup$ – Archer Mar 30 '18 at 10:04
  • $\begingroup$ So even anisole undergoes the same reaction? $\endgroup$ – MollyCooL Mar 30 '18 at 10:07
  • $\begingroup$ The accepted answer says they both are practically equally reactive. $\endgroup$ – MollyCooL Mar 30 '18 at 10:11
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    $\begingroup$ It might be true that $\ce{-OH}$ and $\ce{-OCH3}$ are about equally activating. But phenol in water is partially deprotonated, and $\ce{-O^-}$ is a great deal stronger than both. $\endgroup$ – Ivan Neretin Mar 30 '18 at 11:11

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