Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: PoetryInMotion on November 05, 2013, 05:41:51 PM
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Problem. Neopentyl bromide is treated with sodium methoxide. What happens?
My thoughts. Neopentyls cannot react through SN2 or E2. But what about E1 or SN1?
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Or is the first step to slow for this to happen at all?
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Why can't this react through SN2?
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Oups went a bit too fast there. The SN2 reaction is not impossible, but really slow, due to steric hindrance. (For instance, ethyl bromide is ~40,000 times more reactive through SN2 than neopentyl bromide.)
So, as a matter of fact, we actually have four possibilities.
1. Neither SN2 nor E1/SN1 is are enough to take place to any notable extent.
2. Both reactions take place to a notable extent.
3. Only SN2 takes place to a notable extent.
4. Only E1/SN1 takes place to a notable extent.
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Almost NEVER get an Sn2 with 3 Prime Carbon. Too much steric hinderance.
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1. Neither SN2 nor E1/SN1 is are enough to take place to any notable extent.
2. Both reactions take place to a notable extent.
3. Only SN2 takes place to a notable extent.
4. Only E1/SN1 takes place to a notable extent.
I would say that 4 is correct, I would also stick my neck out and say that methoxide is neucleophillic enough to favour the 1,2-rearrangement to give tert-pentyl methyl ether rather than the elimination product.
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Almost NEVER get an Sn2 with 3 Prime Carbon. Too much steric hinderance.
It's a primary halide with branching, not a tertiary halide.
To the OP: you cannot make a primary carbocation, as you did in your mechanism. If it were to go by ionization there would have to be a methyl group shift concerted with loss of the leaving group, similar to what happens in certain Friedel Craft's reactions.
This is an unusual compound, given the fact that it is both primary and yet very sterically hindered at the same time. I do not know if it would go by Sn2 or a weird Sn1 where the carbocation formation was concerted with an alkyl shift to give a tertiary cation as your first intermediate. Either way not a very good reaction, because it would take a long time and require high temps, etc probably.
Edit: the polarity and protic nature of the solvent will be critical in determining the mechanism. Polar protic will favor Sn1/E1, while polar aprotic will favor Sn2. Overall the higher the dielectric constant the more Sn1 will be favored, I assume also.
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My bad Spiro! I have never seen this in practice, in the lab, nor have I ever asked this on a quiz. Quite far beyond the scope of an Orgo 1/2 class I'd imagine! I would say that the Methyl-Shift is the most likely, as every piece of advice/literature/ etc. says Steric Hinderance BIGTIME disfavors Sn2 Reactions.
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If anything happens at all, there will be a rearrangement
however, MeO- isn't involved here directly in the first step, but only will act as a possible nucleophile for the 2-methylbutane-2-yl cation resulting from the rearrangement / proton catcher should elimination occur
it all depends on the side conditions "solvent, temperature"
regards
Ingo
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We did this in the lab. It does not go through Sn2, well it didn't in the 1 hour time we had. This will however prefer SN1 or E1. I believe the major product was SN1. However it is not fair to say Sn2 is impossible as it is not.
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We did this in the lab. It does not go through Sn2, well it didn't in the 1 hour time we had. This will however prefer SN1 or E1. I believe the major product was SN1. However it is not fair to say Sn2 is impossible as it is not.
What was the solvent? What was the E1/Sn1 ratio? I would guess mostly E1 with such a strong base. I usually don't think of ionization ever happening in the presence of such a strong base/nucleophile but this is such a weird case.
1. Neither SN2 nor E1/SN1 is are enough to take place to any notable extent.
2. Both reactions take place to a notable extent.
3. Only SN2 takes place to a notable extent.
4. Only E1/SN1 takes place to a notable extent.
I would say that 4 is correct, I would also stick my neck out and say that methoxide is neucleophillic enough to favour the 1,2-rearrangement to give tert-pentyl methyl ether rather than the elimination product.
Doesn't a strongly basic nucleophile suggest E1 rather than Sn1?
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We did this in the lab. It does not go through Sn2, well it didn't in the 1 hour time we had. This will however prefer SN1 or E1. I believe the major product was SN1. However it is not fair to say Sn2 is impossible as it is not.
this sounds very strange to me, as esp. the treatment of neopently-bromide as starting material is the most prominent example for neopentyl-rearrangments Wagner-Meerwein type, and in most of my literature it is stressed that this mechanism is starting with a SN1 - type breaking of the C-Br bond, yes, but that methyl rearrangement would be much faster than any nucleophile that might show up.
regards
Ingo