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Topic: Why the molecule does not go protonation?  (Read 2473 times)

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Offline dorratz

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Why the molecule does not go protonation?
« on: October 12, 2016, 07:25:29 PM »
Hey

I got a question about the picture here:


Why does the OH group doesn't go another protonation and leaving of a water molecule in the end, just like it can be happen during E1 reactin after protonation.

Thanks
Dor
« Last Edit: October 12, 2016, 08:40:31 PM by Arkcon »

Offline AWK

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Re: Why the molecule does not go protonation?
« Reply #1 on: October 13, 2016, 02:41:16 AM »
Why does the OH group doesn't go another protonation and leaving of a water molecule
it does, but elimination of water is easy only from tertiary alcohols (draw mechanism for 2,2,5,5-tetramethyltetrahydrofuran)
AWK

Offline dorratz

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Re: Why the molecule does not go protonation?
« Reply #2 on: October 13, 2016, 04:58:01 AM »
I got you.

Thanks a lot!

Offline Vidya

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Re: Why the molecule does not go protonation?
« Reply #3 on: October 13, 2016, 11:32:42 AM »


Why does the OH group doesn't go another protonation and leaving of a water molecule in the end, just like it can be happen during E1 reactin after protonation.

Thanks
Dor
You already has one positive charge on the chain after first protonation ..do you think molecule is ready to undergo for one more positive charge which is very unstable primary carbocation.

Offline orgopete

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Re: Why the molecule does not go protonation?
« Reply #4 on: October 13, 2016, 02:36:55 PM »
For a student, that is a good question as it you may find the same conditions listed for the dehydration of 1-butanol, a common student lab experiment. Furthermore, the reaction conditions are also a key predicting the product. I'm guessing heating will result in elimination and low temperature will favor re-cyclization to the starting material. A factor favoring elimination of the alcohol would include the amount of sulfuric acid. Conc sulfuric acid is a good dehydrating agent. Loss of water will dilute the sulfuric acid, hence higher concentration will increase elimination. This would be difficult to know simply from a listing of the reagent.
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