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Topic: Sulphur ylide problem  (Read 1162 times)

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Offline electrogeek

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Sulphur ylide problem
« on: March 09, 2020, 08:41:10 AM »
Hi everyone,

I'm confused about a question involving sulphur ylides which I have attached below along with my attempt. I reacted the enone with the soft ylide to form the cyclopropyl ring but now I need to remove the bromine and hydrogen from the compound.

I was wondering whether you could use the ylide to deprotonate the alcohol on the left of the molecule, and then attack the carbon and displace the bromine from the compound, which suggests a compound for G Ive attached this below as well) with the right chemical formula? I'm not sure whether this ylide would be basic enough to do this though... I know the pKa of phenol is about 10 and so the pKa of the alcohol in question is also about 10 but I don't know about the ylide.

Any help will be greatly appreciated!

Cheers

Offline chenbeier

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Re: Sulphur ylide problem
« Reply #1 on: March 09, 2020, 11:31:02 AM »
Probably Bromine and die H Form OH Form HBr and die Ether is formed.

Offline clarkstill

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Re: Sulphur ylide problem
« Reply #2 on: March 09, 2020, 12:57:33 PM »
Does it do an intramolecular 1,4-addition first, followed by bromide elimination to give an intermediate chromone, then a corey chaykovsky on that compound?

Offline pgk

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Re: Sulphur ylide problem
« Reply #3 on: March 09, 2020, 01:58:12 PM »
Just a moment. Formation of a sulfur ylide, demands basic conditions that simultaneously lead the corresponding phenolic anion (phenoxide). Does this help?

Offline electrogeek

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Re: Sulphur ylide problem
« Reply #4 on: March 10, 2020, 10:19:12 AM »
Thanks for the help everyone! The comments have helped me a lot - forgot about the implied base in the solution!

Offline pgk

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Re: Sulphur ylide problem
« Reply #5 on: March 10, 2020, 02:30:36 PM »
And a last thing!
The key point is that after methylene insertion, bromine group is tertiary and thus susceptible to SN1 nucleophilic substitution; which explains why the cyclopropane ring is not opened by the phenoxide anion.

Offline clarkstill

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Re: Sulphur ylide problem
« Reply #6 on: March 11, 2020, 05:40:52 AM »
And a last thing!
The key point is that after methylene insertion, bromine group is tertiary and thus susceptible to SN1 nucleophilic substitution; which explains why the cyclopropane ring is not opened by the phenoxide anion.

To generate a carbocation α-to a ketone?

Offline pgk

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Re: Sulphur ylide problem
« Reply #7 on: March 11, 2020, 01:09:26 PM »
Yes and so, the phenoxide immediately attacks to the formed α-carbocation that prevents the nucleophilic cyclopropane ring-opening.

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