Rearrangement Reaction Practice: OH Protonation and Phenolic Group Migration

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Discussion Overview

The discussion revolves around a rearrangement reaction involving the protonation of an OH group and the migration of a phenolic group in a specific organic compound. Participants explore the implications of this reaction, including the formation of intermediates and potential products, while addressing the clarity of the initial problem statement.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes the initial reaction where the OH group is protonated and leaves as water, suggesting that the phenolic group migrates to the left carbon, leaving a positive charge on the right carbon.
  • Another participant questions the accuracy of the drawn compound, indicating that it may not be represented correctly.
  • A subsequent post clarifies the compound as 1,1,2-triphenyl-1,2-propandiol, correcting the initial representation.
  • One participant proposes that the reaction generates a stable benzylic cation, which is further stabilized by an adjacent bromine, leading to the formation of a bromonium ion.
  • There is a suggestion that the water might re-add to the bromonium ion system, potentially regenerating the original bromohydrin through an E1 mechanism.
  • Another participant speculates on the possibility of the hydroxyl group migrating to form an epoxide, discussing the behavior of epoxides in the presence of strong acids.
  • It is noted that the first reaction involving the addition of H+ to the OH group would likely occur, leading to the regeneration of a specific product.
  • One participant concludes that the bromonium ion intermediate is the most reasonable outcome, though this is presented as a personal submission rather than a definitive answer.

Areas of Agreement / Disagreement

Participants express differing views on the accuracy of the initial compound representation and the subsequent steps in the reaction mechanism. There is no consensus on the final outcome of the reaction, with multiple competing hypotheses presented.

Contextual Notes

Participants highlight the need for clearer presentation of the chemical structure and reaction details, indicating that assumptions about the compound's representation may affect the discussion. The discussion also reflects uncertainty regarding the stability and behavior of intermediates.

chaoseverlasting
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Homework Statement




phph
| |
ph-c-c-ch3 + H+ ->
| |
OH Br

Homework Equations





The Attempt at a Solution



Here, the OH is protonated, and leaves as water. Then the phenolic group on the right carbon migrates to the left one leaving a +ive charge on the right carbon. What happens after that?

If there was another OH group instead of Bromine, the hydrogen on the OH would leave, forming a c=O. What happens here?

Im sorry if the presentation is unclear, but I don't know how to use latex for chem.
 
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Have you drawn the compound correctly? It isn't supposed to be...
Code:
ph ph
 | |
 C-C-C-CH3
 | |
HO Br
is it?
 
No. That didnt come out correctly. 1,1,2-triphenyl-1,2-propandiol, is the compound.
 
Use Chemdraw, it's free. Be sure to state the problem again, this time in a clearer form, mentioning all of the necessary information including those that were in the original post.
 
I'm assuming that you meant to write 1,1,2-triphenyl-2-bromo-1-propanol.

You are correct that water will leave and generate a very stable benzylic cation which is further stabilized by the adjacent bromine. This will form a three membered ring, bromonium ion. The positive charge will reside on both the 1 and 2 carbons. This will probably not revert back to the vinyl compound through loss of Br+ (Br+ is a poor leaving group) even if Br- is present. What may happen is that the water might re-add to the bromonium ion system and regenerate the original bromohydrin. This will undoubtedly occur by the E1 mechanism and the fleeting, isolated charge (and thus the hydroxyl) will reside on the most stable carbonium ion... on the diphenyl-substituted carbon (the "1" position), regenerating starting material.

You might consider that the H+ adds to the bromine (in the neutral compound) and then loses HBr leaving behind a stabilized cation alpha to a hydroxyl group. If this happens, the oxygen of the adjacent alcohol could migrate over to form an epoxide and regenerate a proton. What do we know about epoxides and strong acids? Strong acids add to epoxides in a Markovnikov fashion and the product would be 112-triphenyl-1-bromo-2-propanol, a new product. This is unlikely to happen however since the hydroxyl is much more basic (a likely target for H+) than bromine. Even if it did happen, the first reaction (adding H+ to OH group) would definitely occur and regenerate 1,1,2-tripheny-2-bromo-1-propanol.

Therefore, if I had to provide the answer, I would submit the bromonium ion intermediate as the only reasonable answer.
 

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