Consider the following pair of reactions. predict the type of elimination mechanism

Consider the following pair of reactions. Predict

Consider the following pair of reactions. Predict the type
of elimination mechanism, predict which reaction of the pair will
occur at the fastest rate, and draw the correct organic
product.

My mechanisms are correct but I'm having an issue with the
structure.


Answer

General guidance

Concepts and reason
A pair of the reactions is given and the type of elimination mechanism needs to be given along with the major product formed. Also, the faster and the slower reaction needs to be mentioned between the pair

Decide the reaction mechanism based on the structure of the substrate (haloalkane) and the reagent (nucleophile/base) used.

Fundamentals

Elimination reaction (E1 & E2): In order to decide the elimination reaction mechanism, following factors are considered:

• Polarity of solvent: The polarity of solvent increases the rate of E1 reaction mechanism.

• Strength of base: The stronger bases favor E2 reaction mechanism.

Step-by-step

Step 1 of 3

The reaction pathway for the given pair of reactions is E1.

The reaction proceeds through E1 reaction mechanism because a polar protic solvent is used both the reactions.

The base used is also a neutral weak base that also favors E1 reaction.

Step 2 of 3

The first reaction between the given pair is the faster one as shown beblow:

Methanol
-OH
Heat
E1/Faster
Methanol
OH
Heat
E1/Slower

The first reaction is faster reaction because in an E1 reaction, the rate determining step is the first step involving the heterolytic cleavage of C-X (X=halogen) bond resulting in the formation of a carbocation as shown below:

The stability of carbocation determines the rate of reaction. In the first reaction, tertiary carbocation (carbon attached to three other carbons) is formed after the cleavage of C-Cl bond whereas in the second reaction, secondary carbocation (carbon attached to two other carbons) is formed.

The stability of carbocation is

Hence, first reaction is faster and the second one is slower.

Step 3 of 3

The product formed in the pair of reactions are as follows:

Methanol
-OH
Heat
-OH
Methanol
Heat

The products of the pair of reactions are as follows:

Methanol
-ОН
Heat

E2/same rate
E1/slower
E2/slower
El/same rate
E2/faster
El/faster

CI:
Methanol
Heat

E2/same rate
E1/slower
E2/slower
El/same rate
E2/faster
El/faster

The above products are obtained by the abstraction of the proton of the tertiary and secondary carbocations formed respectively in the previous step as shown below:

Answer

The products of the pair of reactions are as follows:

Methanol
-ОН
Heat

E2/same rate
E1/slower
E2/slower
El/same rate
E2/faster
El/faster

CI:
Methanol
Heat

E2/same rate
E1/slower
E2/slower
El/same rate
E2/faster
El/faster

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