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Question: draw the expected major elimination product and identify the mechanism–Free Chegg Question Answer

Transcribed text From Image: Draw the expected major elimination product and identify the mechanism. These conditions will under go an reaction.

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Answer:

Answer

Step-by-step

Step 1 of 2

The ionization of an alkyl halide is given below:

slow

Explanation | Hint for next step

The substrate present in the reaction is 1-bromo-1-methylcyclobutane and it goes through a {{\rm{E}}_{\rm{1}}}E1​ pathway. Initially, the reactant undergoes ionization and forms a tertiary carbocation intermediate and bromide ion \left( {{\rm{B}}{{\rm{r}}^ – }} \right).(Br−). The formation of a carbocation involves the breaking of a carbon-bromine bond and it is the slowest step (or rate limiting step) of the reaction and the bromine leaves as leaving group.

Step 2 of 2

The mechanism of the reaction is given below:

fast
+ CH3OH,
H
CH3
Zaitsavs product

The expected major elimination product of the reaction is given below:

CH3OH

These conditions will undergo an reaction,

Ο Ε1
O E2

Explanation | Common mistakes

The solvent methanol \left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}} \right)(CH3​OH) acts as a base and it can also abstract the proton from the carbon adjacent to the carbocation and form a new (C-C) π-bond. The final elimination product is formed according to Zaitsev’s rule.

Answer

The expected major elimination product of the reaction is given below:

CH3OH

These conditions will undergo an reaction,

Ο Ε1
O E2


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