Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xshadow on January 14, 2018, 09:47:33 AM
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I don't understand why but my textbook says that:
a secondary alkyl halide + Br- gives a "Sn2 reaction"
a secondary alcohol + Br- gives a "Sn1 reaction"
In either the leaving group in on a secondary carbon ...so why there is this differecne!!!
Thanks for helping!
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What is the difference between the leaving groups?
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Secondary carbons can proceed by either mechanism, depending on the conditions, which include the nature of the solvent and the leaving group. mjc123 has pointed us in the right direction IMO.
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Water better leaving group??
So the water can" break the bond " and go away more easily (before the arrival of nucleophilic , i.e. Sn1) ... pherhaps??
Is it correct? !
Some help,please!!
: (
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@OP,
What sort of solvent would one use in a typical SN1 reaction? What are the common conditions used to convert an alcohol into alkyl halide?
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A protic solvent in a SN1 ...in order to stabilize the carbocation
In the alcohol reaction witch halide I use HBr,for example and the alcohol itself...that works also as a solvent I think
But I don 't understand why my textbook says that secondary alcohol gives Sn1 with halide(Br- ,for example) while secondary alkyl halide with Br- gives much more SN2
I can understand that a solvent like an alcohol can help a Sn1 mechanism but why do I have a Sn2 reaction for a seondary alkyl halide?
Why @mjc123 speaks about the difference of the leaving group?
The difference in Sn1/Sn2 in this case is due to the solvent ...or the leaving group?...I don' t understand very well
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Whether a reaction proceeds by one mechanism or another is never an easy question. With respect to SN1 versus SN2, there are several factors. In this instance, the leaving group is neutral water, which is an indication that the reaction is done under acidic conditions. One can say that SN1 reactions are sometimes performed under acidic conditions (and in hydroxylic solvents), whereas SN2 reactions are sometimes performed under basic conditions in polar, parotid solvents.
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My last sentence should read polar, aprotic solvents (it's amazing what autocorrect spelling will do). Polar aprotic solvents include DMF and DMSO. @OP What is the first step of the conversion of an alcohol to an alkyl bromide when H-Br is the source of the bromide ion?
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My last sentence should read polar, aprotic solvents (it's amazing what autocorrect spelling will do). Polar aprotic solvents include DMF and DMSO. @OP What is the first step of the conversion of an alcohol to an alkyl bromide when H-Br is the source of the bromide ion?
The first step is the protonation of the alcohol.
Then the secondary alcohol gives Sn1
Usually a protic solvent stabilizes the charged species ,much more. (In alkyl halide as substrate , they help the Sn1 mechanism)
Can you tell my why a secondary alcohol gives Sn1 while secodary alkyl halide a Sn2 ?!
Protic solvent help help the stabilization of charge species(usually help Sn1 because I have a carbocatoon and a neutral reagent),but in Sn1 with secondary alcohol , either the reagent ,the protonted alcohol, AND the intermediate (the carbocation,for a Sn1) are charged.
So the solvent stabilizes both!!
I need an answer ..I 've searched everywhere but no answer why there is this difference.
Thanks.
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So is it correct?
I need an explanation please.
Thanks.
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I was offline for a few days. I think what you have written so far about protic solvents sounds basically correct. They will stabilize the carbocation intermediate. Let us assume that the secondary alkyl halide is dissolved in a polar, aprotic solvent and that the nucleophile is a halide salt (such as NaBr) not an acid (such as HBr). Then what do you think the mechanism will be and why?
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HI!
Someone can tell me if this exercise is correct?
The nucleophile is CH3O-
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi65.tinypic.com%2Ff1dzyq.jpg&hash=48ad186ef92e3116aae00304a73b0b16faab1258)
I think 5) is wrong...because actually I've a tertiary carbon (sorry noy a secondary one) and so a Sn2 mechanism is not possible (but also a E2 is not possibile...because I have an equatorial leaving gropup)
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What made you chose E2 for question 2? With respect to cyclohexane derivatives, can they undergo ring flips?
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What made you chose E2 for question 2? With respect to cyclohexane derivatives, can they undergo ring flips?
Hi
Oh yes...I see now that for 2) the answer is wrong
Perhaps is Sn2?
Because I should have:
a) a primary carbon (Sn2 is the best mechanism for this substrate)
b) strong nucleophile /base (so E2 or Sn2)
At 25 °C I think that cyclohexane derivatives can undergo ring flips
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I was offline for a few days. I think what you have written so far about protic solvents sounds basically correct. They will stabilize the carbocation intermediate. Let us assume that the secondary alkyl halide is dissolved in a polar, aprotic solvent and that the nucleophile is a halide salt (such as NaBr) not an acid (such as HBr). Then what do you think the mechanism will be and why?
Hi..mee too
I think it will be a Sn2 mechanism...because I have an aprotic solvent ,that usually helps witch Sn2
But with secondaty alcohol my textbook says that there will be a Sn1 mechanism instead...now I assume this is because with alcohol i have a protic solvent (the alcohol itself)...is it correct?
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I will assume that the secondary alcohol is its own solvent, and the nucleophile is HX. Then I would expect the SN1 mechanism to predominate.
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So if you perform a ring-flip on the compound in 5, what do you see?
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So if you perform a ring-flip on the compound in 5, what do you see?
With a ring flip I should have the E2 mechanism...because the leaving group needs to stay in an axial position in order to give E2 ,for a cyclo....
So I have Sn2 when the leaving group is equatorial(the more stable conformer)while an elimination E2 in the other conforme...???
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Is the SN2 reaction fast or slow for tertiary alkyl halides? How fast are ring flips?