e1 or e2.......
Started by polarmolar
Good question. I would have thought E1 or Sn1 because of the polar protic solvent and the 2* substrate. But the answer is a double bond so it must be an elimination reaction. The nucleophile is really strong which makes it possible to be E2, but the solvent is polar protic. Does E2 happen in polar protic solvents? I thought polar protic solvents were only used because of the carbocation intermediate in Sn1 and E1 reactions. Anybody?
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If this is the case then what kind of solvent is used for E2 reactions?
First of all, alcohol is a weak acid. pKa ~16. You won't have much protons to stabilize that carbocation ion intermediate if this was to be an E1 mech. Second, even if the H was to deprotonate, Na+ would grab that proton pretty quick. Last and probably most important in differentiating E1 and E2 other then sterics is that E2 will occur in basic conditions while E1 will occur in acid conditions. The ethoxide is clearly a very strong base and this should be E2.
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Darn! This is a really good questions. At first, i thought it was E2 since ethoside is a strong base. But when i look at closely, the carbocation form is a allylic carbocation, which mostly preferred E1. second, the solvent is polar protic.
But.... I will go with my ochem lecture notes and choose E2 since the ethoxide is a really strong base which determines the second order reaction.
Yes please explain this!!
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I thought for E2 reactions the base must be strong and bulky like having a X shape or something!!
Also ethanol is slightly acidic. So I think since the base is not a bulky base, the carbocation is not primary or secondary, and we have slighlty acidic ethanol, as well as polar protic it must be E1. no?
Also ethanol is slightly acidic. So I think since the base is not a bulky base, the carbocation is not primary or secondary, and we have slighlty acidic ethanol, as well as polar protic it must be E1. no?
That's the problem, we don't know whether it's E1 or E2. It just gives the answer, not the mechanism.
Opps, i mean it is the benzylic carbocation, not allylic carbocation. Sorry!
The questions tells us that this is the elimination, so it is either E1 or E2. For first order reaction, the carbocation is as followed:
carbocation stability: Benzylic > tert > allylic > sec > primary.
As you can see, this carbocation formed by removing the bromide is not the primary carbocation but benzylic. Thus, it is very stable so it can undergoes first order reaction or in this question E1.
But from what i learned is that the reaction mechanism order is determined by how strong is the base or nucleophile, so in this case i guess the strong base ethoxide overule the stability of substrate (benzylic carbocation).
What is the answer by the way?
The questions tells us that this is the elimination, so it is either E1 or E2. For first order reaction, the carbocation is as followed:
carbocation stability: Benzylic > tert > allylic > sec > primary.
As you can see, this carbocation formed by removing the bromide is not the primary carbocation but benzylic. Thus, it is very stable so it can undergoes first order reaction or in this question E1.
But from what i learned is that the reaction mechanism order is determined by how strong is the base or nucleophile, so in this case i guess the strong base ethoxide overule the stability of substrate (benzylic carbocation).
What is the answer by the way?
ummm, the carbocation that would be formed would be on the carbon of the substituent not the carbon in the benzene ring if this was E1. That carbon on the benzene, although would be a very stable carbocation has no Hs to be removed for it to even have a cation. In this reaction we are looking at a secondary carbocation IF it undergoes E1. This is where the problem lies, because it is hard to tell if the reaction mechanism will undergo E1 or E2 in secondary substituted carbons.
Anyways, in a real reaction, im sure both mechs will occur, which one dominates?... I dont know although Im still sticking with E2. 😀
I don't know, thats why I posted it!!!
Answer from this site: http://www.mhhe.com/physsci/chemist...raphics/carey04oc/ref/ch11benzylicsystem.html
"[FONT=Arial, Helvetica, sans-serif]Benzylic halides undergo base catalyzed dehydrohalogenation via an E2 pathway (review) to form conjugated alkenes (review)".
"[FONT=Arial, Helvetica, sans-serif]Benzylic halides undergo base catalyzed dehydrohalogenation via an E2 pathway (review) to form conjugated alkenes (review)".
http://library.tedankara.k12.tr/carey/ch5-3.html
http://www.chem.memphis.edu/parrill/EMOC/roadmap/dehydhal.html
i woulda said E1
but its E2
http://www.chem.memphis.edu/parrill/EMOC/roadmap/dehydhal.html
i woulda said E1
but its E2
So the strong nucleophile overrides the polar protic that would cause use to think E1?
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If E2 is anti then is E1 - the carbocation is stabilized.
Answer is E2. Talked to my Ochem professor and he said that this is a classic example of E2. He told me that strong base should immediately indicate E2.
i think its E2 because
1- its bulky, strong base.
this will prevent from steric hind.
its not Sn1 because you did Elimination. the DB should be the first sign!
also, here is another tip:
Sn1: polar protic (h2o, nh3, oh)
Sn2: polar aprotic (we dont want the H to react with the C+)
E1: Polar solvents
(the way to tell if the rxn will go E1 or Sn1 is that E1 will have heat with it.)
(remember that E1 also forms the most subs. form)
E2: forms the most stable DB also and it uses Bulky base! this reaction used a Bulky base. c2 H5 its a good nucleophile used in Sn2 and E2 rxns😳.
1- its bulky, strong base.
this will prevent from steric hind.
its not Sn1 because you did Elimination. the DB should be the first sign!
also, here is another tip:
Sn1: polar protic (h2o, nh3, oh)
Sn2: polar aprotic (we dont want the H to react with the C+)
E1: Polar solvents
(the way to tell if the rxn will go E1 or Sn1 is that E1 will have heat with it.)
(remember that E1 also forms the most subs. form)
E2: forms the most stable DB also and it uses Bulky base! this reaction used a Bulky base. c2 H5 its a good nucleophile used in Sn2 and E2 rxns😳.
Thats what I was thinking!!
Yo, its a very simple case of E2 rxn. This is a 2ndary haloalkane, so still cant tell rxn, but u form double bond so u know its an elimination. The nucleophile tells u if its E1 or E2, -0C2H5 is a strongly basic unhindered nucleophile so it will undergo only E2 rxn. It is that clear, if it was E1 the nucleophile would need to be a weakly basic good nucleophile like I-
