ongoingdat

5+ Year Member
Jun 29, 2009
84
0
Status
Dental Student
Hi,

I'm confused about Ochem on Destroyer #103 choice (d).

1-butene treat with Br2, methanol ----> [product]

Is this Sn1 mechanism? I think methanol is week base..

I'm a little confused about Sn1/Sn2 substitution..
Could anyone explain this with mechanism if possible?:oops:

Thanks in advance!
 
May 22, 2009
341
0
Status
Dental Student
This is kind of long but I'll try to explain it. It's not really SN1 or SN2 solely.

Remember how Br2 adds to an alkene? (If you don't, look it up in kaplan or ochem book). First, one of the Br s gets added to the alkene and a 3-membered ring with Br having a positive charge is made (I think it was called bromidium ion? but don't quote me on that). Here if only the other half of Br2 (a Br-) is present, the Br- will attack one of the two carbons in a mechanism that's more SN2-like than SN1-like in that it the ring will break at the same time that Br- attacks. This is why when you add Br2 to an alkene, they will always add on two opposite sides.

However if there's a stronger nucleophile or a nucleophile in higher concentration (like a methanol solvent) is present, the second step, it's not Br- that attacks, but the O in CH3OH. One thing to note is that even though the Bromidium ion ring is not broken until CH3OH attacks, the Br has a positive charge in that ring and is going to be more tilted towards the carbon that is more substituted, so when methanol attacks, it's going to attack the less substituted carbon (again more SN2-like), the H of methanol then gets removed. So it results in addition of Br on the more substituted carbon and the other nucleophile on the less substituted carbon, on opposite sides of the alkene.

Damn that was long. Hope it's clear at least.
 
OP
O

ongoingdat

5+ Year Member
Jun 29, 2009
84
0
Status
Dental Student
This is kind of long but I'll try to explain it. It's not really SN1 or SN2 solely.

Remember how Br2 adds to an alkene? (If you don't, look it up in kaplan or ochem book). First, one of the Br s gets added to the alkene and a 3-membered ring with Br having a positive charge is made (I think it was called bromidium ion? but don't quote me on that). Here if only the other half of Br2 (a Br-) is present, the Br- will attack one of the two carbons in a mechanism that's more SN2-like than SN1-like in that it the ring will break at the same time that Br- attacks. This is why when you add Br2 to an alkene, they will always add on two opposite sides.

However if there's a stronger nucleophile or a nucleophile in higher concentration (like a methanol solvent) is present, the second step, it's not Br- that attacks, but the O in CH3OH. One thing to note is that even though the Bromidium ion ring is not broken until CH3OH attacks, the Br has a positive charge in that ring and is going to be more tilted towards the carbon that is more substituted, so when methanol attacks, it's going to attack the less substituted carbon (again more SN2-like), the H of methanol then gets removed. So it results in addition of Br on the more substituted carbon and the other nucleophile on the less substituted carbon, on opposite sides of the alkene.

Damn that was long. Hope it's clear at least.
I totally got it now. Thank you so much!!:D