[rleung]

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]

why don't you observe the mechanism in your text

[rleung]

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]

You have to study the mechanism in different reaction condition. Go to read Adv. org. Chemistry, Carey/Sundberg.

[Winga]

Read this!

[rleung]

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

[Winga]

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]

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