[cyrosceals]

Hi everyone. I have recently learnt from lesson that how an unsymmetrical alkene react to a molecule ('electrophilic addition') will depend on the stability of the carbocation. I can tell most of the answers using Markonikov (sp?) rule by looking at the no of H. But my teacher hates the rule and ask us to look at the carboncation intermediates. I know the basic stuff, but I am wondering about a molecule such as http://img234.imageshack.us/my.php?image=wierddf8.jpg which has no alkyl groups.

My teacher told me that the carbocation stability depends on the no of alkyl or "R" groups. And about secondary, primary, tertiary etc. But what happens if they do not have an R group or a hydrogen group? (Like above but I would replace the hydrogen with something else). How would I determine the stability of the carbocation ( How to tell whether it is P, S or tertiary)?

[spirochete]

You're correct both alkene carbons there are equally substituted, so you can't use markovnikov's rule in a straight forward way.

No matter what, the reaction goes via the carbocation which is the most stable.  I see two electronegative halogens on that molecule.  The only difference is their distance from the pi bond.  How would you expect the proximity of an E.N. atom to influence stability of a carbocation?

[cyrosceals]

Ok...Since there is a halogen, which means each halogen exhibits an electron withdrawing effect. This will destabilize the carbocation, which means that the carbon which has the halogen group furthest from it will form the most stable carbocation. Meaning that the delta positive part of the attacking molecule will form a bond with the carbon closest to the halogen group?

Sry if my concept it wrong....

[cyrosceals]

But another question. The carbocation intermediate is which one (from my first post with the two halogen grp). P, S or tertiary? It must be classified under one right???

[spirochete]

Your explanation of the mechanism is correct. 

The carbocation produced is secondary because it's attatched to two other carbons.  As you've noticed,  the other hypothetical route also would go by a secondary carbocation.

[spirochete]

Edit:  I realized I read your OP a little too fast.  A carbon with a halogen attatched can still be considered an alkyl group and be counted for determing substitution (ie primary, secondary etc). 

Typically an SP3 carbon bonded to the carbon of interest will be counted for determining substitution.   Normally we expect akyl groups to stabilize a carbocation because they are electron donating.  Of course as you add more Electro Negative atoms to the alkyl group it starts to behave less like we'd expect.

[azmanam]

But I would contend that the lone pair electrons on bromine (talking about the bromine on the right) can donate into the empty p orbital of the carbocation to stabilize the carbocation through a bromonium ion (sort of a neighboring group participation argument).  Thus I think it might be more likely that the carbocation forms on the carbon on the right.

I could be thinking about it wrong, though.  Any takers?

[spirochete]

That's a reasonable looking possibility I never considered.  Nice thinking.

However I'm not sure if that kind of stabilization would be the dominant effect here.  Using electrophilic aromatic subsitutiton as an precedent, inductive effects tend to dominate when dealing with halogens. This leads to net destabilization despite the stabilizing effect of resonance.  Going further down the periodic table you also end up with progressively worsening overlap between the P orbitals. 

Still I'm far from an authority so I'd wait for somebody else's opinion.

Sorry if that confused you cryoseals, you may not have even leaned about reactions of aromatic compounds yet.  Have you even learned about the addition of halogens to an alkene yet?  If the answer is no then this probably not the answer your teacher is looking for.

[spirochete]

On the other hand resonance donation does determine the directing effect of halogens on EAS, so I'm probably wrong in my previous statement.

[cyrosceals]

Yes I have learnt it. The question is actually something I thought of as I couldnt find the question. I just remembered that the carbocation is not attached to a normal alkyl grp (aka CH3, C2H5 etc) but is lacking a hydrogen or so and so I came out with the question.

Maybe my example is a little strange or wierd and may involve some complex stuff ^^^. Anyone got a simpler example?

But as long as it is a alkyl group, even it is substituted, it is still considered an "alkyl group"? (even it has weaker electron donating effect than normal) and can be considered in the stability of carbocation?

[azmanam]

I spoke with an organic prof in our department, and she thought the stabilization through resonance would be stronger than the destabilization by the electron withdrawing nature of the halogen.  Thus, she thought the alkene would open to put the carbocation on the right.  She also thought it was a probably outside the scope of a standard undergrad organic course.

To answer what I think is your main question, if the alkene is disubstituted, and there is no major stabilizing factor on one side (like in the example above), Markovnikov's rule will not predict one regioisomer over the other - you would probably see both regioisomers in about the same yield.

[miro_2k]

Br is having negative inductive effect but it can share its lone pair of electrone by resonance
so i think 1st-the electrophile will be added to the carbon far from bromide
             2nd- the carbon that bear positive charge (close to bomide)..and then 1,2 hydride shift occurs leads to decolization of positive charge to the carbon directly attached to the bromide ..the carbocation which is stabilized by resonace
             3rd- attack by bromide anion (if we add HBr) to yield the jeminal derivative

if that hypothesis is worng or correct ..may somebody help me