SN1/SN2 vs E1/E2

[aspiringdoc09]

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What makes a reaction prefer E1 vs SN1 or E2 vs SN2? I know this question has been asked and I did a search and found this thread: http://forums.studentdoctor.net/showthread.php?t=428844

I understand their mechanisms enough to answer a basic question, but I'm not sure why a reaction would choose E2 over SN2. In the TPRH orgo book, it states that most times the substitution reaction will compete with the corresponding elimination reaction. If this is true, how do you know how to choose one over the other if given both answer choices? The mechanisms and kinetics follow the same trend and they are most often using the same bases and solvents.

Please explain. Thanks.😕

 

[syoung]

Main difference is in SN you want a strong nucleophile to attack the carbocation rather than deprotonate as a strong base would.
Both strong bases and nucleophiles can deprotonate, but you can use bulky bases to ensure they don't do nucleophilic attack, as that would require the nucleophile to attach to the electrophilic carbon
Yes they do use same bases/solvents (all e1/sn1 like polar protic, while e2/sn2 like polar aprotic) but you will have to make the distinction between the nucleophiles, temperature

 

[aspiringdoc09]

Main difference is in SN you want a strong nucleophile to attack the carbocation rather than deprotonate as a strong base would.
Both strong bases and nucleophiles can deprotonate, but you can use bulky bases to ensure they don't do nucleophilic attack, as that would require the nucleophile to attach to the electrophilic carbon
Yes they do use same bases/solvents (all e1/sn1 like polar protic, while e2/sn2 like polar aprotic) but you will have to make the distinction between the nucleophiles, temperature

Are you saying elimination reactions use strong bases to prevent nucleophilic reactions?

 

[aspiringdoc09]

I was reviewing GS-1 and they explained nucleophilic and elimination reactions. I will have to read over it multiple times for it to sink. I hope it helps others.

ExplanationDiscuss this question


Questions 35-41

Since this is our first organic chemistry passage together we'll discuss it in detail.Afterwards, you'll begin to notice the trends then we'll speed up a bit more.

< Principle #1: Organic Chemistry is simple.

< Principle #2: It's OK to learn new words (i.e. solvolysis) during the exam!

Let's solve the problem. Read the first paragraph.Now you should want to know what tert-butyl bromide looks like.Draw it now on some scratch paper.{If you can't, see ORG 3.1 for tert-butyl and simply attach "Br" to the free bond}

< Principle #3:Think about the preceding molecule.Begin by assessing electronics: bromine is a halogen to the right of carbon on the periodic table thus it is more electronegative (inscribe d- next to Br).Since Br is drawing electrons away from the central carbon which it is attached to, that carbon becomes partially positive, d+. {see ORG 1.5; CHM 2.3}

< Principle #4:Recognize.Notice how bulky the molecule is.Compare the molecule with less hindered ones with the same number of carbons, i.e. sec-butyl which has a secondary carbon or n-butyl (normal butyl) which has a primary carbon.Steric hindrance {see ORG 6.2.3/4} means that the partially positive carbon has so many attached groups around it that if you were a nucleophile, with your negative or partially negative charge, you would have a difficult time gaining access to that carbon which attracts you.If you can't reach the carbon nucleus then you can't engage in nucleophilic substitution.

Remember, you're still a nucleophile.You want something positive but you can't get to the carbocation.However, you have your pick of hydrogens!{H is slightly more electropositive compared to carbon see ORG 1.5; CHM 2.3}Removing or eliminating H from t-butyl is the start of an elimination reaction.Note, for this reaction to occur, we would need a hydrogen-hungry strong nucleophile like OH- or C2H5O- (ethoxide, which was added to Reaction I in the passage).

KEY CONCEPT: If we started with a primary compound (i.e. n-butyl bromide) and added a nucleophile (you!), the nucleophile has easy access, and is attracted, to the partially positive carbon.Thus the nucleophile adds to the central carbon and bumps off the leaving group (Br) = nucleophilic substitution, second order = SN2."Second order" emphasizes that the rate-determining step depends on the concentration of two compounds - the nucleophile and n-butyl bromide (ORG 6.2.3).

However, this problem begins with a tertiary hindered compound which, as we described, may undergo elimination, and for the same reason as previous, it is also second order = E2 (ORG 6.2.4).Once a hydrogen is eliminated by the nucleophile, we are left with the very unstable but intermediate primary carbanion whose electrons are quickly put to use by forming a double bond and bumping off Br-.Thus the product is (CH3)2C=CH2. Its name is 2-methyl-1-propene. It contains hydrogens that are arranged similar to those of the vinyl group (ORG 4.1).Draw the E2 mechanism and the product below (ORG 6.2.4).

Is there a second possible mechanism?That question is both excellent and rhetorical!Here is an expression to memorize: tertiary carbocations are relatively stable.Why?Here is another expression to memorize: alkyl groups are somewhat electron donating.In other words, if you were a primary carbocation, you would be pretty unstable!You would try to hold on to a formal positive charge without any help - too difficult.However, if you were a tertiary carbocation surrounded by three alkyl groups - each of which will tend to reduce the burning positive charge on the central carbon with cool electrons!

Because of the stability of tertiary compounds, the second mechanism has one*** compound in its rate-determining step: t-butyl bromide in solution can simply dissociate into Br- and the stable t-butyl .Draw this reaction.

Now a nucleophile would be happy to quickly mate with the positive carbocation (nucleophilic substitution, first*** order =SN1).If the nucleophile is hydroxide, or water (Q 36), then the product would be the tertiary alcohol tert-butanol (CH3)3C-OH {note -OH substituted -Br}.If the nucleophile is ethoxide, or ethanol, then the product would be the ether (CH3)3C-OCH2CH3.{The preceding product can be named t-butyl ethyl ether, or, ethoxy t-butane; see ORG 10.1}

Now let's read the second paragraph in the problem!By analyzing the possible mechanisms we now can say with confidence that Compound A must be the SN1 product ethoxy t-butane [C6H14O = (CH3)3C-OCH2CH3].Compound B, with its vinylic protons, must be the E2 product 2-methyl-1-propene [C4H8 =(CH3)2C=CH2].The latter is the answer to Q 35.

 

[syoung]

I said bulky bases (prevents nucleophilic attack of carbocation) but can still strip a proton, resulting in elimination
Now for SN2 is favored over E2 when conditions are like this: Low temperature; Modest bases (too strong of a base results in proton stripping), good nucleophiles; Sterically small bases (so they can attack the electrophilic carbon)

 

[aspiringdoc09]

Ok. Thank you.

 

[MedPR]

What makes a reaction prefer E1 vs SN1 or E2 vs SN2? I know this question has been asked and I did a search and found this thread: http://forums.studentdoctor.net/showthread.php?t=428844

I understand their mechanisms enough to answer a basic question, but I'm not sure why a reaction would choose E2 over SN2. In the TPRH orgo book, it states that most times the substitution reaction will compete with the corresponding elimination reaction. If this is true, how do you know how to choose one over the other if given both answer choices? The mechanisms and kinetics follow the same trend and they are most often using the same bases and solvents.

Please explain. Thanks.😕

You can't call something a nucleophile or a base until you see what it reacts with. In other words, the same anion can act as a nucleophile and a base depending on the reaction.

In general, the substitution product will be the major product at low temperatures and with a small anion and the elimination product will be the major product in the presence of heat and a bulky (large) anion. For example, if you see methoxide, you are probably looking at a substitution reaction. If you see tert-butoxide, you are probably looking at an elimination.