Maximize elimination product and minimize substitution product So I\'m working o
ID: 841649 • Letter: M
Question
Maximize elimination product and minimize substitution product
So I'm working on the following question:
In the reaction written below, we want to get as much of product 3 as possible and minimize the production of compund 4. Which alcoholate 2a, b or c is best suited for this task?
My answer: 2C. Product 4 is created through a SN2 reaction and the speed of a SN2 reaction is directly proportional to the nucleophilicity of the nucleophile. Because the compound in 2C is a tertiary alcoholate, it's the least nucleophilic and it will therefore yield the slowest SN2 reaction. Considering how SN2 and E2 reactions are in direct competition with eachother, it can be concluded that a slow SN2 reaction will result in a higher yield of the E2 compound, product 3.
Is this correct?
Explanation / Answer
Yes the correct answer is 2c. And it will be E2 elimination. Reasons are:
1. The option 2c is both a nucleophile as well as a base.
In the case when there are two competing reactions like acid-base reactions and nucleophilic substitution reactions, Acid base reactions are favoured because acid base reactions are faster.
So if a strong base (CH3)3C-O- is present, it will favour the removal of an acidic hydrogen first, i.e removal of hydrogen adjacent to Br. Then elimination will proceed.
Secondly, (CH3)3C-O- bulky groups attached across O- reduces its nucleophilicity due to steric hindrance, thereby favouring more elimination reaction.