What is formed when 1-propanamine is treated | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The reaction of $1-$propanamine $(CH_3CH_2CH_2NH_2)$ with $NaNO_2$ and $HCl$ produces a diazonium salt $(CH_3CH_2CH_2N_2^+Cl^-)$ as an intermediat

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) The reaction of $1-$propanamine $(CH_3CH_2CH_2NH_2)$ with $NaNO_2$ and $HCl$ produces a diazonium salt $(CH_3CH_2CH_2N_2^+Cl^-)$ as an intermediate. This aliphatic diazonium salt is highly unstable and decomposes to form a carbocation $(CH_3CH_2CH_2^+)$. The primary carbocation undergoes rearrangement to form a more stable secondary carbocation $(CH_3CH^+CH_3)$. These carbocations react with water,chloride ions,or undergo elimination to form a mixture of products including $1-$propanol,$2-$propanol,propene,$1-$chloropropane,and $2-$chloropropane. Therefore,all the mentioned products are formed.

What is formed when $1-$propanamine is treated with $NaNO_2$ and $HCl$?

Similar Questions

In the reaction sequence: $C_6H_5NH_2$ $\xrightarrow{CHCl_3/KOH, \Delta} (X)$ $\xrightarrow{LiAlH_4/H_2O} (Y)$,the product $(Y)$ is:

The strongest nucleophile is

The product obtained by the electrolytic reduction of nitrobenzene in the presence of $H_2SO_4$ is:

Arrange the following compounds in increasing order of their basicity:
$CH_3NH_2, (CH_3)_2NH, C_6H_5NH_2, (CH_3)_3N$

The major product $(R)$ of the following reaction is:
Aniline $\xrightarrow[{pyridine}]{{CH_3COCl}}$ $P$ $\xrightarrow[{CH_3COOH}]{{Br_2}}$ $Q$ $\xrightarrow{{{H_3O^{+}}}}$ $R$

Vedclass Test Series

Exam Paper Generator

Online Exam Module