Question

Hexachloro cyclohexane has 8 positional isomer out of this one does not show bimolecular elimination in hydrohalogenation raection​

Answer 1

Answer:

Here’s what each of these two reactions has in common:

in both cases, we form a new C-C π bond, and break a C-H bond and a C–(leaving group) bond

in both reactions, a species acts as a base to remove a proton, forming the new π bond

both reactions follow Zaitsev’s rule (where possible)

both reactions are favored by heat.

Answer 2

Answer:

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Home / E1 vs E2: Comparing the E1 and E2 Reactions

Elimination Reactions

By James Ashenhurst

E1 vs E2: Comparing the E1 and E2 Reactions

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Last updated: January 16th, 2020 |

E1 versus E2 : Comparing 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.

Table of Contents

Comparing The Mechanism Of The E1 and E2 Reactions

What Do The E1 and E2 Reactions Have In Common?

How Are The E1 and E2 Reactions Different?

E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not?

The Key Requirements Of Stereochemistry In The E2 Reaction

1. Comparing The Mechanism Of The E1 and E2 Reactions

Here’s how each of them work:

comparing e1 vs e2 mechanism e1 is two step e2 is one step

2. What Do The E1 and E2 Reactions Have In Common?

Here’s what each of these two reactions has in common:

in both cases, we form a new C-C π bond, and break a C-H bond and a C–(leaving group) bond

in both reactions, a species acts as a base to remove a proton, forming the new π bond

both reactions follow Zaitsev’s rule (where possible)

both reactions are favored by heat.

3. How Are The E1 and E2 Reactions Different?

Now, let’s also look at how these two mechanisms are different. Let’s look at this handy dandy chart:

table comparing e1 vs e2 reactions rate law big barrier strong base stereochemistry

The rate of the E1 reaction depends only on the substrate, since the rate limiting step is the formation of a carbocation. Hence, the more stable that carbocation is, the faster the reaction will be. 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). 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 [although we’ll see in a sec that we do require that the C-H bond be able to rotate so that it’s in the same plane as the empty p orbital on the carbocation when the new π bond is formed].

The rate of the E2 reaction depends on both substrate and base, since the rate-determining step is bimolecular (concerted). A strong base is generally required, one that will allow for displacement of a polar leaving group. 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.

4. E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not?