36. Which of the following reactions is Hofmann elimination!
CI Me
Alc. KOH
Me
Ph
Pyridine
OH
a. (1)
b.(I), (III)
d.
ca
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1
Answer:
Your options aren't able to read. But I can explain the reaction.
Explanation:
Hofmann elimination, also known as exhaustive methylation, is a process where a quaternary ammonium reacts to create a tertiary amine and an alkene by treatment with excess methyl iodide followed by treatment with silver oxide, water, and heat. After the first step, a quaternary ammonium iodide salt is created.
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