December 05, 2021, 08:37:41 AM
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Topic: Alcohol reactivity exercise  (Read 295 times)

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Offline luscofusco

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Alcohol reactivity exercise
« on: November 10, 2021, 10:08:57 AM »
In this exercise I have to explain the mechanisms for that transformation.
I understand that the H+ protonates the OH and then the molecule loses the H2O and forms a carbocation, but I'm stuck there.
I know there is a rearrangement but I'm not able to see it. Maybe C1 will bond with C3 somehow? There are no H in C2 for the hydride shift to happen, is it coming from one of the methy on that C2? And after that it would react with water in a elimination reaction to form the double bond. It is just a gess, I know I'm missing something.

Offline Meter

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Re: Alcohol reactivity exercise
« Reply #1 on: November 10, 2021, 12:11:35 PM »
You start with a seven-membered ring and end with a six-membered ring. Is there a way your ring can rearrange such that the carbocation is still "moved" to a stable position?

Offline luscofusco

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Re: Alcohol reactivity exercise
« Reply #2 on: November 12, 2021, 06:51:53 AM »
I've done this:

C1 bonds with C3 and the carbocation is displaced to C2 and then water takes one of the beta H of one of the methyl groups and forms the double bond. At first I thought I was missing a carbon and had to do it a couple of times. Does this make sense?

Offline Meter

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Re: Alcohol reactivity exercise
« Reply #3 on: November 12, 2021, 12:04:03 PM »
I've done this:

C1 bonds with C3 and the carbocation is displaced to C2 and then water takes one of the beta H of one of the methyl groups and forms the double bond. At first I thought I was missing a carbon and had to do it a couple of times. Does this make sense?
Can you draw and show a mechanism? Because it sounds right.

Offline luscofusco

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Re: Alcohol reactivity exercise
« Reply #4 on: November 12, 2021, 03:54:48 PM »
Yes, I did this.

Offline Meter

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Re: Alcohol reactivity exercise
« Reply #5 on: November 13, 2021, 08:44:18 AM »
Looks about right. Remember to draw your arrows from the bond, not the atom.

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