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What is the major alkene formed in the following Hofmann elimination?
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A. i
B. ii
C. iii
D. iv
E. v

Answer
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Hint: Hofmann elimination is the reaction where an aliphatic or aromatic amine that has beta-hydrogen is converted to an unsaturated compound. It takes place in the presence of a base along with high temperature (heat).

Complete answer:
Hofmann elimination is the reaction where an amine reacts with a base to form an alkene. The amine consist of a $\beta $ - hydrogen whose removal marks the formation of a double bond. This elimination is based on the Saytzeff rule that states that the major product will be that product which is more stable.
The given compound when subjected to a base and heat, have the $\beta $ - hydrogen removed, after the removal of $\beta $ - hydrogen, the most stable carbocation will form the major product. The carbocation is stable when the saturation takes place on the $2{}^\circ $ carbocation, which is the product (i), while other products contain less stable $1{}^\circ $ carbocations. Therefore the reaction will be as follows:
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Hence, the product formed is (i), so, option A is correct.

Note:
Hofmann elimination reaction and Hofmann bromamide degradation reactions are two different reactions, but both consist of a base as a reactant. Hofmann bromamide consists of the formation of aliphatic or aromatic amines from the reaction of amide, bromine and sodium hydroxide (base). Hofmann elimination is the preparation of unsaturated compounds.