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Question

(B) The given reaction is a Hofmann elimination reaction of a quaternary ammonium hydroxide.In this reaction,the base $(HO^-)$ abstracts a proton from

Vedclass · Accepted Answer

methylenecyclohexane

Vedclass · Answer

(B) The given reaction is a Hofmann elimination reaction of a quaternary ammonium hydroxide.In this reaction,the base $(HO^-)$ abstracts a proton from the $\beta$-carbon atom.According to the Hofmann rule,the less substituted alkene is formed as the major product because the base attacks the most accessible (least sterically hindered) $\beta$-hydrogen.In the given molecule,there are two types of $\beta$-hydrogens: those on the methyl group and those on the ring carbons.Removing a proton from the methyl group leads to the formation of the exocyclic double bond,resulting in methylenecyclohexane.Removing a proton from the ring carbon leads to the formation of $1-$methylcyclohexene.Since the exocyclic double bond (methylenecyclohexane) is less substituted than the endocyclic double bond ($1$-methylcyclohexene),methylenecyclohexane is the major product.

The major product $A$ is:

Similar Questions

Product $(B)$ of this reaction is

The order of basic strength of the following in aqueous solution is
$I$. $C_6H_5NH_2$
$II$. $(CH_3)_3N$
$III$. $NH_3$
$IV$. $CH_3NH_2$
$V$. $(CH_3)_2NH$

The incorrect statement regarding the reaction given below is:

Identify the correct decreasing order of the basic strength of the following amines in the aqueous phase: $CH_3NH_2$, $(CH_3)_2NH$, $C_6H_5NH_2$, and $NH_3$.

Aniline $\xrightarrow[{{0^{\circ}}C - {5^{\circ}}C}]{{NaNO_2/HCl}}A$ $\xrightarrow{{Ph-N(CH_3)_2}}B$
What is $B$?

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