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Question:

treatment of anisole with HI(conc.)

My Attempt: In most of the ethers reaction of HI in excess yields alkane iodide products. But in case of anisole two scenarios are possible, I believe:

  1. The HI breaks bond between methyl and oxygen leading to formation of phenol and methyliodide.

  2. The phenol formed further reacts to form phenyliodide.

But would HI be strong enough to break partial double bond in phenol?

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  • $\begingroup$ Try drawing a possible mechanism: what would the first step be? $\endgroup$ – The_Vinz Apr 21 '20 at 6:09
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    $\begingroup$ Protonation of O ? $\endgroup$ – Aditya suresh Apr 21 '20 at 6:37
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    $\begingroup$ Correct, and then where is the easier site for I- to attack? Remember how big iodide is relative to carbon $\endgroup$ – Waylander Apr 21 '20 at 6:53
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    $\begingroup$ It would be easier to attack methyl group. But as we normally see in case of reaction of ethers with alkane iodides, would the product be phenol or can HI attack attack it too? Had it been dilute i wouldn't have thought that, but conc makes me think. $\endgroup$ – Aditya suresh Apr 21 '20 at 16:00
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The correct answer here should be C. $\ce{I-}$ is an excellent nucleophile, it would attack the $\ce{C-O}$ bond in the methly group because phenoxide is a good leaving group as it is resonance stabilized and quite polarizable. So, the products are phenol and methyliodide. As for the excess of conc. $\ce{HI}$, it is just there so that some other nucleophile like $\ce{OH-}$ doesn't grab the limelight. I don't think it is energetically favourable to remove hydroxide from phenol to make it iodobenzene.

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