WHy is this a Sn1 Reaction and NOT Sn2?

[DrTacoElf]

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Ok i will agree that the "conditions" favor Sn1, but i fail to see what would be protonated (if anything) does BR just decide to leave on its own, and if so i thought that only happened with tertiatry carbons?

Thanks!

 

[lissa1217]

a bromine attached to a tertiary OR secondary carbon can be considered a good leaving group in acidic solution.

 

[btowngirl]

generally, you see carbocations forming on tiertiery carbons, but they can form on secondary ones also. Since it will be planar, you can see how answer C should be correct with both products forming. feel free to PM me with other questions, i like OChem.

 

[DrTacoElf]

Thank you i was not aware of this 🙂

 

[Pinkertinkle]

Hahah I don't remember a sliver of this and its only been a semester.

 

[btowngirl]

another thing to keep in mind for the sn1 vs sn2 is that sometimes it is unclear which one it is (such as when you have a secondary carbon). for the mcat, you can help yourself out by looking at the answers and eliminating answers or determing the reaction mechanism which will then allow you to find the answer.

 

[Ragun Cajun]

I need some help. How do you do a new post? Like were is the "New Post" Button? I really need some help on this, thank you, oh and sorry for taking up some of your space on this topic, well I would really be a happy camper if you guys could help. Thanks. 😕

 

[btowngirl]

above threads in forum is a link that says new thread. click it.

 

[Ragun Cajun]

I don't see it, sorry I'm being such a pain, but I really need to find it.

 

[btowngirl]

go to the page with all the threads on the mcat forum (or which ever forum you are on). at the top where it says page 1 of 100, look all the way to the left and you will see "new thread". click that.

 

[ChemMN]

1) The question is odd because Br is a good leaving group, period. There is no need to have H+ around for this reaction to go.
2) H+ would actually slow this reaction since many water molecules would be protonated and therefore not want to attack the carbocation.
3) Answer C is correct because of the planarity of the intermediate.

 

[vinoyp]

Ours is not to wonder why, ours is but to do or die.

--Vinoy

 

[rtmcad2319]

Answer C is correct because of the planarity of the intermediate.

Could you explain wt you mean by this?

Thanx!

 

[DrTacoElf]

Sn1 reactions proceed through a planar carbocation (Sp2 hybridized carbon) therefore attack is equally likely from below or above, hence the 2 potential products.

 

[rtmcad2319]

I see said the blind man!

 

[NE_Cornhusker1]

reaction mech is this [just basic sn1]:

H+ assoc. with the Br- to get something like this H+-Br+ [+ is the hydrocarbon].

H:Br leaves and the center hydrocarbon becomes protantated.

HOH [H2O] electrons from the oxygen fill the open orbital on the cationinc carbon. half the time the open orbital is on 'top' half o bottom=> racemic mixture.

there's no way this would ever go sn2. that [realtively] huge a$$ ethyl group will put the breaks on anything trying to push off br. if there was a pair off methyl groups maybe.

 

[neda]

..but here's a shorthand list of general guidelines for deciding if an alkyl compound will have a 1st order or 2nd order rxn.


* if it's 1', you pretty much mostly get Sn2. but if not, usually the easiest way to figure out the answer is to look at the reagents:

* if you've got a strong base/nucleophile, you generally get a one-step rxn (that is, Sn2 or E2).

* protic solvents (water & ethanol) stabilize carbocations, so it favors the two-step rxn (Sn1 or E1)


why:
if it's a one-step rxn, the rate-determining step is the attack of the nucleophile or base. so the stronger it is, the more the rxn is favored. but with the two-step rxns, that first step (the leaving group popping off, making a carbocation) is the slow step. anything that favors forming the cation favors the rxn, so protic solvents that stabilize the carbocation favor Sn1 & E1 rxns. since that next step (the base or nucleophile's attack) is a fast (non-rate determining) step, the strength of the nuc/base isnt' terribly important. So in this problem, the presence of the H+ guarantees that your nucleophile is crap; you're stuck with a first-order rxn. (And, as mentioned before, it makes Br- "happier" to be a leaving group.)

 

[ih8biochem]

You guys are making it harder than the question really is. It's merely asking "is this sn1 or sn2?" Obviously it's sn1 (and therefore a racemic mixture instead of answer choice b) because the solvent is protic, which would definitely screw up the nucleophile in an sn2 rxn. Both reactions require a good leaving group to that's not the point. Just remember, protic solvents usually means sn1. OH is also a poor nucleophle (the stong base of the very weak conjugate acid, h20), which doesn't mater in sn1 reactions.