![]() ![]() That’s the only way we can form a π bond between these two carbons we need the p orbital of the carbocation to line up with the pair of electrons from the C-H bond that we’re breaking in the deprotonation step. In our cyclohexane ring here, the hydrogen has to be axial.The leaving group must be anti to the hydrogen that is removed. The Key Requirements Of Stereochemistry In The E2 Reaction Remember this reaction – where one elimination gave the “Zaitsev” product, whereas the other one did not. ![]() Now we’re in a position to answer a puzzle that came up when we first looked at elimination reactions. E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not? The stereochemistry of the hydrogen to be removed must be anti to that of the leaving group the pair of electrons from the breaking C-H bond donate into the antibonding orbital of the C-(leaving group) bond, leading to its loss as a leaving group. A strong base is generally required, one that will allow for displacement of a polar leaving group. The rate of the E2 reaction depends on both substrate and base, since the rate-determining step is bimolecular (concerted). ![]() Finally there is no requirement for the stereochemistry of the starting material the hydrogen can be at any orientation to the leaving group in the starting material. Forming the carbocation is the “slow step” a strong base is not required to form the alkene, since there is no leaving group that will need to be displaced (more on that in a second). Hence, the more stable that carbocation is, the faster the reaction will be. The rate of the E1 reaction depends only on the substrate, since the rate limiting step is the formation of a carbocation. Now, let’s also look at how these two mechanisms are different. How Are The E1 and E2 Reactions Different? both reactions follow Zaitsev’s rule (where possible)ģ.in both reactions, a species acts as a base to remove a proton, forming the new π bond.in both cases, we form a new C-C π bond, and break a C-H bond and a C–(leaving group) bond.Here’s what each of these two reactions has in common: What Do The E1 and E2 Reactions Have In Common? Comparing The Mechanism Of The E1 and E2 ReactionsĢ. (Advanced) References and Further Readingġ.The Key Requirements Of Stereochemistry In The E2 Reaction.E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not?.How Are The E1 and E2 Reactions Different?.What Do The E1 and E2 Reactions Have In Common?.Comparing The Mechanism Of The E1 and E2 Reactions.Now that we’ve gone through the mechanisms of the E1 and E2 reactions, let’s take a moment to look at them side by side and compare them. This only happens if the leaving group is especially poor.E1 versus E2 : Comparing The E1 and E2 Reactions Lastly, under #"E"2#, we have one special reaction called #"E1cB"#, meaning "first-order elimination, forming a conjugate base intermediate". #"E"1# has a carbocation intermediate, which allows for 1,2-hydride shifts or 1,2-alkyl shifts, but #"E"2# does not have or allow either.#"E"1# has no need for an antiperiplanar orientation, but #"E"2# does.Favorable for higher steric hindrance on the electrophile and/or nucleophile.A proton from the #beta# carbon leaves, forming a #pi# bond across a carbon-carbon bond such that the most substituted product is made (Zaitsev's Rule).Second-order would mean that the rate-limiting step involves two molecules. First-order simply means the rate-limiting step involves one molecule only. These are called #"E"1# and #"E"2#, respectively. We have a first-order and a second-order process associated with elimination. ![]() Note that they don't necessarily all happen in one step. In general, an elimination reaction (specifically, it's called #beta#-elimination) involves the elimination of a proton from the #beta# carbon, forming a #pi# bond, and ejecting a leaving group. ![]()
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