Reaction mechanism from an exam

[budullewraagh]

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.

[Yggdrasil]

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

 

The addition of Br₂ to a pi bond does not generate a carbocation. This is shown by the fact that no rearangements associated with carbocations are seen during such reactions. Rather, a bromonium ion is formed.

 

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 that the OH itself would not act as a leaving group in the final part of the reaction. However, if the OH group were protonated by the HCl, I could see how a bromonium ion would be formed. Cl^- is not a good enough nucleophile to displace the oxonium ion in an Sn2 reaction. Rather, the oxonium will leave as water, forming a carbocation at carbon 3. Either the Br or Cl^- could form a bond with the carbocation; however, since intramolecular reactions are much faster than intermolecular reactions, the bromonium ion will form before the chloride has a chance to react with the carbocation. Remember, the dissociation of water and formation of the carbocation is the rate limiting step. Therefore, the favorability of the steps following the formation of the carbocation will be determined by kinetics (which reaction goes faster), not thermodynamics (which reaction produces a more stable product).

 

Once the bromonium ion is formed, the chloride can attack carbon 3 and displace the bromine, forming 2-bromo-3-chloro-3-methyl-pentane in which the bromine and chlorine are anti to each other [the product itself will be a racemic mixture of the (R,S) and (S,R) enantiomers].