Why doesn't this reaction use an E2 mechanism?

In summary, the use of MeOK in MeOH would likely result in an E2 reaction due to the dissociation of MeOK, making it a strong enough base to initiate the reaction with tert-butyl bromide. The tertiary structure of the central carbon atom in tBut-Br also eliminates the possibility of Sn2 and leaves E1, E2, or Sn1 as potential reactions. However, the strong nature of MeOK makes Sn1 and E1 less likely, leaving E2 as the most probable reaction. While Sn1, E1, and Sn2 may still occur, they can be considered negligible in this scenario.
  • #1
gauss44
49
0
tert-butyl bromide + MeOK in MeOH ----->

I would think that MeOK would dissociate in solution making it MeO- which would be strong enough to yank a proton off tert-butyl bromide and initiate an E2 reaction. Where am I going wrong?

PS - I am a tutor who hasn't done organic chemistry for a while, so please leave this here AND do NOT send me to the ¨Homework¨ section. This is NOT homework.
 
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  • #2
The central carbon atom in tBut-Br is tertiary, which gives rise to high steric hinderance, which means that we can exclude Sn2, leaving E1, E2 or Sn1. Continuing, if we consider the other reactant Me-OK, it is negatively charged and thus a strong nucleophile/base. However, since it is strong Sn1 or E1 are unlikely to happen, leaving us only E2 to happen. Don't forget though that Sn1, E1 and Sn2 will happen, but it is to a such small extend that it can be approximately unconsidered.
 

1. Why is an E2 mechanism not used in this reaction?

An E2 mechanism is not used in this reaction because it involves a one-step elimination process where a proton and leaving group are removed simultaneously. This requires a strong base and a good leaving group, which may not be present in this particular reaction. Additionally, the reaction conditions and reactant molecules may favor a different mechanism.

2. What are the characteristics of a reaction that uses an E2 mechanism?

A reaction that uses an E2 mechanism typically involves a strong base, a good leaving group, and a molecule with a beta-hydrogen (such as an alkyl halide). The reaction also tends to occur in a single step, with the formation of a double bond and the elimination of a proton and leaving group.

3. How does an E2 mechanism differ from an E1 mechanism?

The main difference between an E2 mechanism and an E1 mechanism is the number of steps involved. E2 reactions occur in one step, while E1 reactions involve two steps – the formation of a carbocation intermediate before the elimination of a proton and leaving group. Additionally, E2 reactions require a strong base, while E1 reactions can occur with either a strong or weak base.

4. Can an E2 mechanism occur with a primary alkyl halide?

An E2 mechanism can occur with a primary alkyl halide, but it is less likely compared to secondary or tertiary alkyl halides. This is because primary alkyl halides have a less substituted beta-hydrogen, making it more difficult for the base to remove the proton and form a double bond. As a result, primary alkyl halides tend to undergo substitution reactions instead of elimination reactions.

5. How do reaction conditions affect the likelihood of an E2 mechanism?

The reaction conditions, such as temperature and solvent, can greatly influence the likelihood of an E2 mechanism. For example, a polar solvent may stabilize the transition state and favor an E2 mechanism, while a nonpolar solvent may hinder the formation of the transition state and result in a different mechanism. Additionally, a higher temperature may favor an E2 mechanism by providing enough energy for the reaction to occur in one step.

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