Deamination (or) diazotization of n-butylamin | vedclass

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(B) The reaction of $n$-butylamine with $NaNO_2/HCl$ proceeds via the formation of a diazonium salt,which is unstable and loses $N_2$ to form a primar

Vedclass · Accepted Answer

$3$

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(B) The reaction of $n$-butylamine with $NaNO_2/HCl$ proceeds via the formation of a diazonium salt,which is unstable and loses $N_2$ to form a primary carbocation $(CH_3CH_2CH_2CH_2^+)$.This primary carbocation undergoes a $1,2$-hydride shift to form a more stable secondary carbocation $(CH_3CH_2CH^+CH_3)$.The primary carbocation eliminates a proton to form $1$-butene $(CH_3CH_2CH=CH_2)$.The secondary carbocation eliminates a proton to form a mixture of $cis$-$2$-butene and $trans$-$2$-butene.Thus,the isomeric butenes formed are $1$-butene,$cis$-$2$-butene,and $trans$-$2$-butene,totaling $3$ isomers.

Deamination (or) diazotization of $n$-butylamine $(CH_3CH_2CH_2CH_2NH_2)$ with $NaNO_2/HCl$ gives how many isomeric butenes?

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