Why e1 or e2

If it is a strong base — the mechanism is E2. If a weak base is used — then the mechanism is E1. Remember also that E1 reactions cannot occur on primary substrates since primary carbocations are very unstable. Other than this, the reactivity pattern is the same for both E1 and E2 — they go faster with more substituted alkyl halides:. Take Now. Determine if the following reactions will go through E2 or E1 mechanism and draw the structure of the major product expected in each reaction:.

The reaction mechanisms of E1 reactions are known as unimolecular eliminations. E1 reactions are two-step reactions, which means, an E1 reaction occurs through two steps named as ionization and deprotonation. In the ionization process, a carbocation is formed due to the removal of a substituent. In the second step deprotonation , the carbocation is stabilized by removal of a hydrogen atom as a proton. Usually, E1 reactions take place with tertiary alkyl halides.

But sometimes, secondary alkyl halide also undergoes this type of elimination reactions. There are two reasons for this; bulky alkyl halides highly substituted are unable to undergo E2 reactions, and highly substituted carbocations are more stable than primary or secondary carbocations.

In E1 reactions, the carbocation formation is the slowest step. Therefore, it is the rate-determining step pf E1 reactions, and the reaction rate depends only on the concentration of the alkyl halide. E1 reactions usually take place in the complete absence of bases or the presence of weak bases. Acidic conditions and high temperatures are preferred for successful E1 reaction. And also, E1 reactions include carbocation rearrangement steps. E2 reactions are a type of one-step elimination reactions found in organic chemistry.

Products are trisubstituted and monosubstituted alkene. How do we determine which of these carbons will participate in the double bond? Using the Zaitsev rule! But when we draw all these products, we can see that disubstituted alkenes are the same when rotating the structure. These prefixes denote the number of carbon atoms attached to the double bond rounded red. But this is not all. We have another rule about the stereochemistry that we have to apply.

This is the case that we can see in the examples with substituted cyclohexanes. If we applied the previous rule to get a product, we would see that it is correct for just one reaction E1. But as our starting compounds are the same, the same product could be expected for another reaction, but this is not true.

In these reactions, we need to apply the following rule:. In our cyclohexane ring here, the hydrogen has to be axial. We can always do a ring flip to make this H axial, so we can form the Zaitsev product. Using the Newman projection, we can more clearly demonstrate the anti-periplanar relationship between the leaving group and the hydrogen.

This also applies to E2. Notify me via e-mail if anyone answers my comment. This site uses Akismet to reduce spam. Learn how your comment data is processed. Previous The E2 Mechanism. The first case is an E2 reaction. The leaving group must be anti to the hydrogen that is removed. The second case is an E1 reaction. In our cyclohexane ring here, the hydrogen has to be axial. We can always do a ring flip to make this H axial, so we can form the Zaitsev product.

Part XI. Kinetics of olefin elimination from tert. Dhar, E. Hughes, and C.