ok so i took an exam and got 0/8 for this one question.  to be honest, im quite sure i got the correct answer, although it differs with the exam answer key in one place.


so we start with 3 methyl 2 pentene.  we add bromine and water in one step, then HCl.

my first step showed the pi C=C bond attacking a bromine atom, resulting in the more substituted 3 carbocation and bromide anion.

the p electrons in the oxygen atom of water then attack the carbocation from the opposite side that the bromine was added on, forming C-OH2 with a + charge on O.  these first two steps are basically a concerted reaction, for all intents and purposes.

the solvent then accepts H+ from the C-OH2+ group, yielding the final product of the first step, 2 bromo 3 methyl 3 pentanol, with the vicinal Br and OH on opposite sides.

upon addition of HCl, the OH group is protonated by H3O+, which was formed by the HCl.  this turns the OH into a good leaving group.  finally, Cl- attacks the sigma C-O antibonding orbital yielding 2 bromo 3 chloro 3 methyl pentane, the product with Br and Cl substituents on the same plane with respect to the C2-C3 bond.  thus, the S,S enantiomer was formed, with free rotation about the C2-C3 bond.



the answer key differs with my answer only towards the end.  according to their answer, the bromine at carbon 2 performs a backside attack on the sigma C-O antibonding orbital of the vicinal carbon with the hydroxyl group.  this yields a bromonium cation, which is displaced due to yet another backside attack performed by Cl- on the sigma C-Br antibonding orbital at the more substituted (3) carbon.

i believe this is not an accurate answer for a number of reasons.  OH is not a good leaving group by any means.  TsCl, PBr3, SOCl2, etc were not used.  there is no way that Br would displace OH when already bonded to a vinincal carbon.  why form a bromonium cation when the vinical carbon is already stable?  the angle of the bond would make the bromonium cation significantly less stable than an epoxide.  yes, the Cl- nucleophilic substitution would work, but the first step is incredibly unlikely.  on the other hand, protonation of OH by an acid and displacement by the Cl- is much more likely.

i want to talk to my professor about this, but i decided i may as well ask first to make sure i'm seeing things correctly.

I agree with GCT, the first part is probably not as concerted as you think.  The bromonium ion is pretty stable.  It's best to draw the formation discretely.  So then it would be alkene --> bromonium ion + Br^- --> halo-alcohol.  If I were grading the exam I would want to see the bromonium ion as a discrete intermediate.

As for the second part, If the answer key has the bromine displacing -OH, then it is wrong.  However, if you protonate the alcohol first, as you suggested in your answer, then the displacement of OH₂ and reforming the bromonium ion is certainly reasonable.  Your argument about the stability of the tertiary carbocation is a good one, but you have to think of these things relatively; is an unstabilized tertiary carbocation better than one that is stabilized by the donation of the bromine?  

Think of it this way: the reaction of a trisubstituted alkene in the presence of bromine requires only mild conditions while the reaction with acid alone requires a strong mineral acid (e.g., H₂SO₄), and usually rather high temperatures.  Both of these reactions have a tertiary carbocation intermediate, but on is stabilized by the bromine while the other is not.

You can also explain this effect via microscopic reversibility, but that's probably beyond the scope of your course.

Also, the mechanism you proposed requires an S_N2 attack at a tertiary carbon.  Reactions like this are exceedingly slow (due to sterics) and tend to go via an S_N1 mechanism instead.  So, if you have to form the carbocation anyway it's better to form the one that is stabilized by the bromine.

Finally, I can't believe that you received no points for your answer.  Were I the one grading your exam you would have probably received 3 or 4 points for sure, but that is something you'd have to take up with your professor.

But attack on the C-O antibonding orbital by Cl^- is sterically disfavored because it is a tertiary carbon.  Since the bromine is tethered in close proximity it will react much, much faster than an external nucleophile to reform the bromonium ion.

Even if the protonated alcohol is more stable than the bromonium ion, it won't be reactive in the substitution reaction, only the bromonium species will be reactive.

There will not be an Sn2 reaction for two main reasons:

1) Carbon 3 of 2-bromo-3-methyl-3-pentanol is tertiary.  It is connected to a methyl group, an ethyl group, and a bromoethyl group in addition to the hydroxyl.

2)  Even if carbon 3 were secondary, Cl^- generally is not a strong enough nucleophile to cause an Sn2 reaction to occur.