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A question in my book (an archive for IIT JEE exam questions) asks:

In the reaction:

enter image description here

The intermediate(s) is/are?

The correct answers are given as

enter image description here

However, as per my knowledge, compound (b) does not appear in the reaction mechanism. Is this an error in the book or do I know the wrong mechanism?

Correct mechanism according to me:

enter image description here

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    $\begingroup$ Can you give the reference for the book? $\endgroup$ Mar 26, 2021 at 5:18
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    $\begingroup$ Also, (b) and (c) are not intermediates, they are products. $\endgroup$ Mar 26, 2021 at 5:55
  • $\begingroup$ @Mathew If (IV) is the product then (b) and (c) could be intermediates. $\endgroup$
    – S R Maiti
    Mar 26, 2021 at 16:20
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    $\begingroup$ @Shoubhik R Maiti: In that way, (I) and (III) are also intermediates. But they were not included in the answer. That's why my comment. In my experience, (II)-(IV) are in the product mixture, in which (IV) predominates. $\endgroup$ Mar 26, 2021 at 16:56
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    $\begingroup$ @Hridai Khurana: I understand now. Yes, your mechanism is acceptable. It's believed to be step by step addition of $\ce{Br2}$ to the ring. However, I doubt ipso addition of $\ce{Br2}$ (structure (b)) exists because of inductive effect of $\ce{Br}$ (it is not in your mechanism). $\endgroup$ Mar 26, 2021 at 19:23

2 Answers 2

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In a 1922 paper, the intermediate of panel B is shown:

enter image description here

More than a 100 years later, this looks a bit funky.

A more recent paper from 1989 shows a dienone intermediate for the reaction in neutral or acidic aqueous solution. However, there is only one and not two bromine atoms in the para-position:

enter image description here

These are both screen-grabs from the freely available first page, thus the low quality.

It is possible that the mechanism varies with solvent and pH. Panel C shows the product of the first bromination step in its deprotonated form. Not sure why this would be called an intermediate, other than that the subsequent bromination steps are faster. This means that in the presence of an excess of bromine, you probably could not isolate the singly-substituted product.

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    $\begingroup$ +1. This question was in my mind for a long time. Glad you answered it. I have linked the other question which I answered through comments but since this post didn't have a decent answer, I didn't refer this post in my own answer. $\endgroup$ Aug 28, 2023 at 14:06
  • $\begingroup$ In NaOAc/HOAc phenol forms 2,4,4,6-tetrabromocyclohexa-2,5-dien-1-one. An Organic Synthesis Preparation $\endgroup$
    – user55119
    Aug 28, 2023 at 15:47
  • $\begingroup$ @user55119 I see. The reaction posted by the OP is basic, not acidic, though. The 1922 paper does not mention pH on the first page. I wonder what they mean with "tribromophenol bromide", and whether it is related to the dienone in your comment. $\endgroup$
    – Karsten
    Aug 28, 2023 at 16:01
  • $\begingroup$ @Karsten: I suspect that the tetrabromide in strong base may form the tribromophenoxide or at least be in equilibrium. $\endgroup$
    – user55119
    Aug 28, 2023 at 20:05
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The correct answer per my checks is B, however I stand for correction. Since the solvent is a polar solvent, the OH of the solvent will abstract the H of the OH on the phenol. Making the phenol a phenolic oxide PhO. Which will activate or the electrons on the O of the phenoxide will be pushed to the Ortho and the Para position on the ring, making them more activated hence the formation of a trisubstituted ring.

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