Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: sharbeldam on July 02, 2019, 06:43:28 AM
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If we have a tertiary alkyle halide with a strong base like OH- without heat, would SN1 OR E2 occur?
with heat it's E2 But I learned that Elimination needs heat in the lab to occur, so would SN1 occur in that case? since SN2 Can't happen on a tertiary carbon. but can SN1 occur with a strong base?
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I think you will get a mix, how much of each is hard to say.
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Hmm, well E1 needs heat also but I guessed E2 would be more probable than E1 because of the strong base...
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I agree with rolnor. Heat favors elimination on thermodynamic grounds, but that is not to say that elimination cannot happen without it. Some questions do not have simple answers.
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Thank you both
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Don't forget you need heat for SN1 also - you have to form that carbocation!
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Why do they say that elimination product would be prefferable with heat then if SN1 Needs heat as well?
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https://www.masterorganicchemistry.com/2012/11/21/deciding-sn1sn2e1e2-1-the-substrate/
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At "high" concentrations of base, I believe the typical result is E2. Technically it depends on the stability of the hypothetical carbocation, and also on polarity/hydrogen bonding ability of the solvent. A more stable carbocation and a more polar solvent would both accelerate E1/SN1 pathway.
If you are in a Sophomore organic chemistry class, probably your professor just wants you to say E2 for this.
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There are various factors that contribute.
SN1 needs high temperature and a nucleophile, but the concentration of nucleophile doesn't matter. For the SN1 to occur, the leaving group has to dissociate and form a carbocation. If the temperature is too low, the SN1 wont occur.
In competition between nucleophilic substitution and elimination, higher heat tends to favor elimination over substitution. Substitution always takes 2 particles in (nucleophile and substrate) and generates 2 particles (leaving group and product), while elimination takes in 2 particles (base and substrate) and generates 3 particles (leaving group, protonated base, product). Increasing number of particles means increased entropy, and is favored by TΔS.
Elimination also needs high temperature and a base. Many nucleophiles are also bases (like hydroxide). E2 can be increased in rate by increasing concentration of base.
So in your original statement, you have tertiary alkyl halide, hydroxide and "no heat". Both SN1 and E2 will be slow with no heat. Which one will be favored depends on whether "no heat" is 40 C, 20 C, 0 C or -20 C. The lower the temperature, the more lilley the answer will be "no reaction". If "no heat" is 30-50 C, E2 will probably be more likely. Setting fluorides aside, Chlorides will favor E2 the most while Iodides will more easily dissociate and give SN1.
If you want pure E2, high concentration of hydroxide and about 80 C with the chloride or bromide would be good. This is not quite high enough temperature to get much dissociation and SN1.
If you want pure SN1, water instead of hydroxide and about 100 C would be good, probably with the bromide - that is enough heat to dissociate, and the water is a poor base. If you want to do SN1 at low Temperature, use an iodide - although tertiary iodides dissociate so easily they are not very stable.
There is a good explanation of this in the Organic review book by Fredlos.
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Wow! that is some deep explanation, why dont professors explain it that way in the university? Nor taking o-chem 1 or 2 nor labs gave me that knowledge.
thanks a lot guys
is this the book? https://www.amazon.co.uk/Organic-Chemistry-Maximum-Success-Minimum-ebook/dp/B07RK49QC8
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If heat is strong enough you can have Ei-mecanism
https://en.m.wikipedia.org/wiki/Ei_mechanism
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That's the book - I don't know when he will publish part 2.