[AlbertoA]

let's say that we have chloramphenicol


under SN1/E1 conditions with acidic conditions, so, I think that the bencylic OH will be protonated, thus, when it goes away, we will have a bencylic carbocation.
There's only one H than can be removed.
I have Organic Chemistry by Marye Anne Fox and it says that the leaving hydrogen must be ANTI to the carbocation, then only one alkene will be formed.
but I also have Organic Chemistry by Paula Yurkanis Bruice and it says that an E1 elimination can be both syn and anti, there fore there will be 2 alkenes as products.

Do anyone has an answer to this?

[discodermolide]

let's say that we have chloramphenicol


under SN1/E1 conditions with acidic conditions, so, I think that the bencylic OH will be protonated, thus, when it goes away, we will have a bencylic carbocation.
There's only one H than can be removed.
I have Organic Chemistry by Marye Anne Fox and it says that the leaving hydrogen must be ANTI to the carbocation, then only one alkene will be formed.
but I also have Organic Chemistry by Paula Yurkanis Bruice and it says that an E1 elimination can be both syn and anti, there fore there will be 2 alkenes as products.

Do anyone has an answer to this?

Carbocations are usually planar.
In eliminations the two participating groups are usually trans.
Perhaps this helps you further?

[the.khemist.ds]

I can only think that where it says that the carbocation must be anti to the leaving group, what is meant is:

The vacant p-orbital of the carbocation must be orientated so that it is aligned with the bond that will be broken as your leaving group departs.

That is to say that in a sufficiently constrained system (I can't bring any specific examples to mind immediately) it might be possible to generate a carbocation which has a p-orbital which cannot eliminate as it is at right angles to any potential leaving groups.

[mickelcorleone]

well,


anti carbocation stable than syn in this reaction but in E1 elimination can be both syn and anti

[Dan]

p orbitals have a plane of symmetry, this means that syn and anti are the same (ignoring configurational isomerism, which will have an effect, but this is to do with sterics).

I don't actually see how it is possible to label syn/anti conformations for a planar carbocation...

I think you might get NGP from the amide directing the stereochemistry of the product rather than simple sterics. I'm not 100% sure about that though.

[mickelcorleone]

anti conformations stable than syn. orbitals  symmetry is not  important in this reaction because thermodynamics, the Gibbs free energy and internal energy, importer than  symmetry  plane(plane of symmetry  )

[fledarmus]

It seems to me that you are confusing E2 elimination (which it appears the Fox reference is describing) and the E1 elimination (which Bruice is describing). E2 eliminations must have the hydrogen anti to the leaving group, because the double bond is forming at the same time that both substituents are leaving. In contrast, E1 eliminations happen in two steps - the leaving group leaves to form the cation, and later the hydrogen atom leaves to form the double bond. The bond between the two carbons (which will eventually become a double bond) is free to rotate in the meantime, and the hydrogen can be either above or below the newly forming double bond when it forms. So E1 can give either syn or anti configurations, as long as the bond is free to rotate once the carbocation has formed.

[AlbertoA]

It seems to me that you are confusing E2 elimination (which it appears the Fox reference is describing) and the E1 elimination (which Bruice is describing). E2 eliminations must have the hydrogen anti to the leaving group, because the double bond is forming at the same time that both substituents are leaving. In contrast, E1 eliminations happen in two steps - the leaving group leaves to form the cation, and later the hydrogen atom leaves to form the double bond. The bond between the two carbons (which will eventually become a double bond) is free to rotate in the meantime, and the hydrogen can be either above or below the newly forming double bond when it forms. So E1 can give either syn or anti configurations, as long as the bond is free to rotate once the carbocation has formed.

when the bond is NOT free to rotate (besides cycloalkanes)
the size of the groups interfere with the free rotation?

[AC Prabakar]

It seems to me that you are confusing E2 elimination (which it appears the Fox reference is describing) and the E1 elimination (which Bruice is describing). E2 eliminations must have the hydrogen anti to the leaving group, because the double bond is forming at the same time that both substituents are leaving. In contrast, E1 eliminations happen in two steps - the leaving group leaves to form the cation, and later the hydrogen atom leaves to form the double bond. The bond between the two carbons (which will eventually become a double bond) is free to rotate in the meantime, and the hydrogen can be either above or below the newly forming double bond when it forms. So E1 can give either syn or anti configurations, as long as the bond is free to rotate once the carbocation has formed.

This Sounds good!

[nox]

It seems to me that you are confusing E2 elimination (which it appears the Fox reference is describing) and the E1 elimination (which Bruice is describing). E2 eliminations must have the hydrogen anti to the leaving group, because the double bond is forming at the same time that both substituents are leaving. In contrast, E1 eliminations happen in two steps - the leaving group leaves to form the cation, and later the hydrogen atom leaves to form the double bond. The bond between the two carbons (which will eventually become a double bond) is free to rotate in the meantime, and the hydrogen can be either above or below the newly forming double bond when it forms. So E1 can give either syn or anti configurations, as long as the bond is free to rotate once the carbocation has formed.

Just to add to this, E2 reactions are not necessarily always anti. There is a preference for anti elimination no doubt, but when constrained enough conformationally (eg in cyclobutane systems), syn elimination is preferred instead.

[fledarmus]

It seems to me that you are confusing E2 elimination (which it appears the Fox reference is describing) and the E1 elimination (which Bruice is describing). E2 eliminations must have the hydrogen anti to the leaving group, because the double bond is forming at the same time that both substituents are leaving. In contrast, E1 eliminations happen in two steps - the leaving group leaves to form the cation, and later the hydrogen atom leaves to form the double bond. The bond between the two carbons (which will eventually become a double bond) is free to rotate in the meantime, and the hydrogen can be either above or below the newly forming double bond when it forms. So E1 can give either syn or anti configurations, as long as the bond is free to rotate once the carbocation has formed.

Just to add to this, E2 reactions are not necessarily always anti. There is a preference for anti elimination no doubt, but when constrained enough conformationally (eg in cyclobutane systems), syn elimination is preferred instead.

Good point Nox - the appropriate overlap can occur on the syn side in very constrained systems, but there aren't many systems constrained enough to overcome the gauche interactions that usually prevent syn overlap. Besides cyclobutane rings, cyclopropane rings, and constrained boat-form cyclohexane rings, I can't think of any others.