budullewraagh - 20-7-2005 at 09:19
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.
frogfot - 20-7-2005 at 13:22
Do the Br on carbon 2 really displaces the OH?? I'd rather guess that the OH is protonated and the OH2+ is displaced by bromine.. Maybe it will
rather be displaced by Br since its intermolecular reaction.
But really, it becames clearer when you know the answer.. 
But still, I dont see
any reason why the mech proposed by you wouldn't work..
budullewraagh - 21-7-2005 at 13:10
thats what i said. the OH gets protonated to OH2+ (a good leaving group) and is displaced by the bromine, supposedly. surely it must compete with
the sN2 of the chloride anion. i actually intend to do much physics to prove this to my professor
Quibbler - 27-7-2005 at 05:51
First the question should have stated whether the starting material was cis or trans. Second the intermediate is believed to be a cyclic bromonium
ion. Which can then be attacked from the opposite side by whatever hapens to be around in this case Br and H2O.
Madandcrazy - 27-7-2005 at 06:43
Do attack the Br ion the C=C bond instead the C-C bond 
and an wich position ?
The question, started with Br2 in H20,
bromic aid or bubbling bromine thrue the liquid and it`s generally possible produce
a OH with the arangement ?
2 >CH3-CH=C(CH3)-CH2-CH3> + Br2
-->
2 >CH2Br-CH=C(CH3)-CH2-CH3> + H2
I would say the OH group can better implemented by a syntheses with for
instance 3-methyl-2-penteneone.
[Edited on 27-7-2005 by Madandcrazy]