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(D) Hofmann exhaustive methylation involves the following steps:$1$. The amine reacts with excess $CH_3I$ to form a quaternary ammonium salt.$2$. Trea

Vedclass · Accepted Answer

$H_2C = CH_2$

Vedclass · Answer

(D) Hofmann exhaustive methylation involves the following steps:$1$. The amine reacts with excess $CH_3I$ to form a quaternary ammonium salt.$2$. Treatment with moist $Ag_2O$ $(AgOH)$ replaces the iodide ion with a hydroxide ion.$3$. Upon heating,the quaternary ammonium hydroxide undergoes an elimination reaction (Hofmann elimination) to form an alkene and a tertiary amine.In the given compound,the nitrogen is attached to a cyclohexyl group and an ethyl group. During elimination,the $OH^-$ base abstracts a proton from the $\beta$-carbon of the ethyl group,leading to the formation of ethene $(H_2C = CH_2)$ as the alkene product and $N,N$-dimethylcyclohexanamine as the tertiary amine. The question asks for the major product,which is the alkene $H_2C = CH_2$.

What is the major product of the reaction when the given compound undergoes Hofmann exhaustive methylation?

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