2010 Destroyer ochem #50

[fuqele]

Hey guys,

I'm a little confused about this question. Can somebody help me?

The question gives a bunch of reactions and ask which one produces alkene. My question is for one of the choices. The reaction is a (Ch3)3-C-I reacting with Ch3OH(aq). I thought that Since Ch3OH is a weak base/nuc and the alkyl halide is tertiary, this could go E1 or SN1, so you get a mixture of both products. The answer says that it only does SN1. Can someone tell me why?

Thanks in advance.

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[Neil45]

I think in order to yield an E1 product, you need heat. This isn't shown in choice E (sometimes heat is denoted with a triangle symbol) with the (CH3)3-C-I reacting with methanol.

 

[drejto]

The above post is correct. It is easier to think about it this way though. SN1 requires both a GOOD nucleophile and weak base however and E1 only needs a weak base. This is because the E1 will have the assistance of acid and heat to help to remove the leaving group. Since the question you have says that you have a good nucleophile and weak base, then there is no need for it to go under a E1 reaction.

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch08/ch8-9.html

 

[thejamespark]

oh yeah i remember this problem.
you need heat in order to continue a elimination reaction (E1).

 

[fuqele]

Ahh, got it. Thanks for all the replies!

 

[Kittenz]

Bringing this up because I'm on the same problem, and I have a hard time believing any rxn that undergoes SN1 cannot undergo E1. I believe heat will make a rxn favor elimination, but elimination should not be prevented in the absence of heat. Anyone else agree?

 

[Thanhn]

I came across this problem and I definitely thought the answer was CH3OH. I think that ALL SN1 reactions will having E1 competing with it, however, the major product and minor product ratio can be slightly skewed depending on the type of thermo/kinetic control the reaction conditions are set. Question 50 doesn't include any extra information which kind of makes it a bad question. However, I think it is definitely reasonable to say that a tertiary alkyl halide treated with CH3OH will have an elimination product along with a substitution product.

 

[panmit]

Heat is not the only thing missing in order for the answer to be right.

I just came on this problem today, and looked over my chad video notes. In order for it to go E1, there needs to be a weak base present. CH3OH is not a base, just a nucleophile. Therefore, all it will do is SN1 because its a weak nucleophile and a tertiary electrophile.

 

[Kittenz]

I'm not sure what you mean by CH3OH not being a weak base. Throughout my Ochem, I don't believe I've seen anything that supports that.

 

[matty675]

E1 will always be competing with SN1. There are certain things you can do to promote E1, and thus increase the amount of alkene in your product such as adding heat. Though your yield will not be 100% alkene.

However, it sound as though there was a better answer choice- something that produced only alkene product.

 

[panmit]

I'm not sure what you mean by CH3OH not being a weak base. Throughout my Ochem, I don't believe I've seen anything that supports that.

CH3OH is not a weak base nor a weak acid; it's just an alcohol and a weak nucleophile. If NaOCH3 was present, it can now be a nucleophile and a base that will allow E1 to happen. For E1 to happen, CH3OH would have to deprotonate an H in order to form the alkene, which it cannot do.

 

[matty675]

CH3OH is not a weak base nor a weak acid; it's just an alcohol and a weak nucleophile. If NaOCH3 was present, it can now be a nucleophile and a base that will allow E1 to happen. For E1 to happen, CH3OH would have to deprotonate an H in order to form the alkene, which it cannot do.

I have to disagree. If NaOCH3 were present, E2 would occur. CH3OH can certainly pull off a beta Hydrogen because the tertiary carbocation is very reactive.

 

[panmit]

Sorry, my mistake you are right NaOCH3 is a strong base and E2 will occur. But CH3OH cannot pull of the hydrogen.