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Topic: PBr3  (Read 24281 times)

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Offline rleung

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PBr3
« on: July 28, 2005, 07:52:38 PM »
Hi,

I am wondering why adding PBr3 to cis-3-methycyclohexanol would result in an alkyl halide product with an inversion configuration?  Don't PBr3, SOCl2/pyridine, and P/I2 reagents always give alkyl halide products with retention configurations, as opposed to using HCl, HBr, and HI.  Thank you.

Ryan

GCT

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Re:PBr3
« Reply #1 on: July 28, 2005, 10:35:50 PM »
why don't you observe the mechanism in your text

Offline rleung

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Re:PBr3
« Reply #2 on: July 29, 2005, 03:33:19 PM »
In my text, it does not say what conversions these lead to.  However, in one of the answers to a problem in the solutions manual, it shows SOCl2 to lead to a retention configuration, so I assumed PBr3 would lead to retention configuration as well.  

So now, I guess I learned that PBr3 leads to inversion configuration, while SOCl2 leads to retention.  What about P/2/I2?  Also, does PCl3 and PI3 also lead to inversion?  Lastly, do HCl, HI, and HBr all lead to inversion?  

Thanks.  I am sorry for the simplicity of these questions, but my book honestly does not discuss the configurations of the products of reactions with these reagents.

Ryan

laotree

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Re:PBr3
« Reply #3 on: August 08, 2005, 11:13:16 PM »
You have to study the mechanism in different reaction condition. Go to read Adv. org. Chemistry, Carey/Sundberg.

Offline Winga

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Re:PBr3
« Reply #4 on: August 09, 2005, 01:59:28 AM »
Read this!

Offline rleung

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Re:PBr3
« Reply #5 on: August 09, 2005, 01:34:33 PM »
Thanks for the slides Winga.  I am still not certain though what configurations result when an alcohol reacts with P/2/I2, PCl3, PI3, HCl, HBr, and HI.  My book does not clarify at all.

Ryan

Offline Winga

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Re:PBr3
« Reply #6 on: August 09, 2005, 01:44:08 PM »
Thanks for the slides Winga.  I am still not certain though what configurations result when an alcohol reacts with P/2/I2, PCl3, PI3, HCl, HBr, and HI.  My book does not clarify at all.

Ryan
It should be inversion of configuration if the carbon is stereogenic, because the halides attack from backside.

The Good Doctor

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Re:PBr3
« Reply #7 on: August 16, 2005, 02:58:29 PM »
Well that's a good remark of u! Why don't we get any inversion (in walden) of the configuration. Well, first mention that the reaction with PBr3 goes via a SN2-like mechanism and all reaction via SN2 have a inversion!!!! But! Don't forget the SN2 rival: SN1! All the factors that reduce the speed of an SN2 (ex. steric hyndrance) will compete with SN1! In your case it's first good to draw your conformation...indeed if u draw your conformation in the most stable conformer that is all the substituents equatorial...u'll see that periplanar attack of the bromine to the phosphoroxi-moiety is very steric hyndered by the axial hydrogens of the ring...in this case SN1 will play the major role here... you should have a secundary carbocation who then will attack via the most steric less hyndered side giving retention...
pfiiuw,
cheers :P

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