Confirming the major product...


by maverick280857
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maverick280857
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#1
Oct27-04, 02:37 AM
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Hi

The major product in the reaction of but-2-en-1-ol with HBr (no peroxide) should be 1-bromo-but-2-ene. Am I correct in saying so? I've done this before and I reason that cleavage of the C-OH bond will produce the more stable (resonance stabilized) carbocation.

Thanks for your help.
vivek
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chem_tr
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#2
Oct27-04, 05:21 AM
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Hello, I think [itex]HO-CH_2-CH=CH-CH_3[/itex] with no given cis or trans configuration will make our task harder; your thought is an applicable one though, an alkyl bromide with elimination of water is likely to be produced. But don't forget that [itex]^+CH_2-R[/itex] is a primary carbocation and very reactive. So another approach involving the saturation of the double bond is also possible. Secondly, systems favor additions rather than condensations. So I conclude that the major products should be 2- and 3-bromobutanols, with 3-isomer being slightly more dominant because of OH-Br steric interaction.

Please check my answer and give some feedback so that we may find the correct one quickly.
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#3
Oct27-04, 09:54 AM
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Maverick, I think that you're right. This is a special kind of SN1 reaction because you make an allylic carbocation which is much more stable than a primary carbocation. The bromide should favor attack at the terminal positon to give 1-bromo-2-butene. You may get some 3-bromo-1-butene as well, but that product is less stable.

As for cis and trans isomers, once you have the allylic carbocation, the olefin should be able to isomerize, so I would expect predominantly trans product since it is more thermodynamically stable.

chem_tr
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#4
Oct27-04, 12:46 PM
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Confirming the major product...


You are not wrong, HBr is a powerful brominating agent indeed. But what about the double bond? Does this remain intact? I am doubtful about this, what about bromine (Br2)? This should give the vicinal dibromide around the former double bond, right?
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#5
Oct27-04, 01:13 PM
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For bromine to react directly with a double bond you need a source of Br+, but in HBr you just have Br-. I suppose that there is a possibility of protonating the olefin (to give a carbocation at the 2 or 3 positon) and then adding Br- to the carbocation. However, the oxygen in the alcohol is much, much more basic than the olefin, so the alcohol would get protonated first, leading to the SN1 product at the 1 position.

With Br2 you would first get olefin activation (via the bromonium ion) and then attack of Br- on the activated olefin to give the 2,3-dibromide, as you predicted chem_tr.
chem_tr
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#6
Oct27-04, 01:38 PM
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Okay, but please note that OH is acidic when compared with very "short-living" resonance contributor of carbanion; it is indeed a very strong base, so the corresponding carbocation after resonance is the most acidic structure, thus bromine prefers to attack there, as hydroxyl group is partly saturated by hydrogen. This was my reasoning. Please discuss this one. Thank you for your explanation anyway.
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#7
Oct27-04, 08:22 PM
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The acidity of the OH will be negligiable in the presence of HBr, since it is far more acidic.

The OH is still a whole lot more basic than the alkene, so the OH will get protonated before the alkene.
maverick280857
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#8
Oct28-04, 12:49 AM
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Thanks for your help. What if the question says "I mol HBr" or "1 equivalent HBr". What will the major and minor products be in these cases? I understand that the major product should be the olefinic bromide but what about the minor product? Will it be bromohydrin or another olefinic bromide?
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#9
Oct28-04, 02:35 PM
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I think the next most prevalent product would be from the internal attack of the bromide with the allylic cation, so 3-bromo-1-butene (the major product being 1-bromo-2-butene).

The only way to get a bromohydrin would be by protonating the olefin, which is a possibility, but I think it would be formed considerably less than the 3-bromo-1-butene. To get this product you would have to protonate the olefin in the presence of the alcohol, which is very unlikely, but still a possibility.
maverick280857
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#10
Oct29-04, 04:47 AM
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Thanks for your help.


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