[davidenarb]

Hi all,

Let's consider the following inputs:
-a reagent that functions as strong nucleophile and strong base.
-a tertiary substrate.

with these conditions, we expect only an E2 reaction
 -No E1 because the base is strong, and E1 is not sensitive for the type of the base
 -No Sn2 because the substrate is too sterically hindred (tertiary substrate)
- However, I don't understand why don't we have Sn1 product ( in mixture with E2)

Thank you

[orgopete]

I think an SN1 could be considered as a product if one looked very closely at every single molecule that reacts. That would leave us with a statistical question, which would give the larger amount. For this, I make a rhetorical and chemical argument. By virtue of indicating a base is present means that it is mechanistically relevant. It would not make sense to ask to predict a product for a given set of reagents with the proviso that one of the reagents doesn't react. So, it will be an E2 reaction.

[zsinger]

This was recently argued about on this forum.  Statistically (but rarely), the leaving groups can "block" the attacking nucleophile from entering on one side, leading to a NON-50/50 racemate.
                 -Zack

[davidenarb]

This was recently argued about on this forum.  Statistically (but rarely), the leaving groups can "block" the attacking nucleophile from entering on one side, leading to a NON-50/50 racemate.
                 -Zack

In contrary, Sn1 reacts perfectly with tertiary substrates because it is a stepwise process, and the rate determining step is the loss of the leaving group. In addition, a tertiatry carbocation the most stable one (alkyl groups are electron donating, and thereby stabilizing the positive charge of the carbocation).

I am still quite puzzled.

[davidenarb]

Is it because we have a strong base/nucleophile, and Sn1 is not involved with strong nucleophiles (the rate determining step is the loss of the LG, and the rate of this step is incomparable and much more faster than the attack of the nucleophile ) ?

Is this a reason for not having Sn1 ?

[orgopete]

If the reaction shows a base is one of the reagents, that MEANS it is part of the rate determining step. The rate determining step is removal of the proton, hence it is an E2 reaction.

If you were to ask whether a tertiary halide could lose a halogen to the solvent without encountering the base, that could happen, but this would be a concentration dependent reaction competing with the E2 elimination. It is the concentrations that control the rates. It will be an E2 elimination. If concentrations were given, then it might, might be possible to argue that some of the reaction could have been SN1, but that is way beyond my pay grade.

[davidenarb]

If the reaction shows a base is one of the reagents, that MEANS it is part of the rate determining step. The rate determining step is removal of the proton, hence it is an E2 reaction.

E2 is a concerted process, so it doesn't have a rate-determining-step (there is only one step)

I am still looking for an answer please

[Dan]

E2 is a concerted process, so it doesn't have a rate-determining-step (there is only one step)

The one step is the rate determining step. All reactions have a rate determining step.

Considering your system:

1. Steric hinderance precludes S_N2
2. Formation of a stabilised carbocation is possible - so S_N1/E1 may operate
3. You have a strong base - this increases the rate of E2 (see rate equation)

E2 is likely to be favoured due to the fact that a concentration of strong base is present (but without concentrations and rate constants, nobody can know without doing the experiment).

[davidenarb]

E2 is a concerted process, so it doesn't have a rate-determining-step (there is only one step)

The one step is the rate determining step. All reactions have a rate determining step.

Considering your system:

1. Steric hinderance precludes S_N2
2. Formation of a stabilised carbocation is possible - so S_N1/E1 may operate
3. You have a strong base - this increases the rate of E2 (see rate equation)

E2 is likely to be favoured due to the fact that a concentration of strong base is present (but without concentrations and rate constants, nobody can know without doing the experiment).

Thank you Dan. Why are you saying that the E2 product is favoured ?

In the system that I considered, the reagent can act not only as a strong base but also as strong nucleophile. In my textbook, only the E2 product is favoured, so I didn't understand the logic behind the fact that Sn1 product will not be formed.

[spirochete]

E2 is a concerted process, so it doesn't have a rate-determining-step (there is only one step)

The one step is the rate determining step. All reactions have a rate determining step.

Considering your system:

1. Steric hinderance precludes S_N2
2. Formation of a stabilised carbocation is possible - so S_N1/E1 may operate
3. You have a strong base - this increases the rate of E2 (see rate equation)

E2 is likely to be favoured due to the fact that a concentration of strong base is present (but without concentrations and rate constants, nobody can know without doing the experiment).

Thank you Dan. Why are you saying that the E2 product is favoured ?

In the system that I considered, the reagent can act not only as a strong base but also as strong nucleophile. In my textbook, only the E2 product is favoured, so I didn't understand the logic behind the fact that Sn1 product will not be formed.

Because with reasonable base concentrations the rate of bimolecular elimination (E2) is much faster than the rate of ionization that would lead to Sn1 or E1 products. This is the explanation that matches the empirical observation.

Like Orgopete said in reality an ionization pathway might be possible if the concentration of base was very low but there would be no practical reason I know of to run the experiment like that so the book says it's an E2 process.

The formation of an E2 product in this case is analogous to the formation of Sn2 products when a strong nucleophile/weak base is mixed with a secondary alkyl halide. Yes, a secondary halide can form a carbocation but a strong nucleophile attacks the alkyl halide faster than it can ionize.

[orgopete]

If the reaction shows a base is one of the reagents, that MEANS it is part of the rate determining step. The rate determining step is removal of the proton, hence it is an E2 reaction.

I can see my point is not being understood as it is a rhetorical point. It would be disingenuous to list a reagent for a reaction, but also say it plays no role in the formation of the product. It is like listing the drying agent. That would only make things more confusing. I therefore assume that listing a reagent literally MEANS the reagent participates in the rate determining step. In this case, I am referring to the kinetics of the reaction. If the reaction were an E1 reaction, the rate would be solely dependent on the substrate concentration and independent of the base. If it were independent of the base, this should tell you the base is not participating in the causing the reaction to occur. If that were so, that is if the base is not causing the reaction to occur, why would it be listed? Hence, I believe that by convention, listing a reagent MEANS it causes the reaction to take place as part of the reaction kinetics, namely E2.

If we were to get real picky about it, I would expect that given the large number of molecules present per mole, that in any E2 reaction, that at least some of the products could have formed via an SN2/E2 mechanism. These amounts should depend on the concentrations which will affect the rates. As the base becomes consumed, the bimolecular (and unimolecular) rate will slow. If an exact stoichiometric equivalent of base were used, I might expect some substrate could reaction unimolecularly while in search of an ever decreasing amount of base.