How many amines with molecular formula C_3H_9 | vedclass

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

Question

(B) The molecular formula $C_3H_9N$ corresponds to the following isomeric amines:$1$. $CH_3CH_2CH_2NH_2$ (Propan-$1$-amine,$1^{\circ}$ amine)$2$. $CH_

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The molecular formula $C_3H_9N$ corresponds to the following isomeric amines:$1$. $CH_3CH_2CH_2NH_2$ (Propan-$1$-amine,$1^{\circ}$ amine)$2$. $CH_3CH(NH_2)CH_3$ (Propan-$2$-amine,$1^{\circ}$ amine)$3$. $CH_3CH_2NHCH_3$ ($N$-methylethanamine,$2^{\circ}$ amine)$4$. $(CH_3)_3N$ ($N,N$-dimethylmethanamine,$3^{\circ}$ amine)Benzene sulphonyl chloride (Hinsberg reagent) reacts with $1^{\circ}$ and $2^{\circ}$ amines to form sulphonamides.$1^{\circ}$ amines ($CH_3CH_2CH_2NH_2$ and $CH_3CH(NH_2)CH_3$) react to form $N$-alkylbenzenesulphonamides.$2^{\circ}$ amines $(CH_3CH_2NHCH_3)$ react to form $N,N$-dialkylbenzenesulphonamides.$3^{\circ}$ amines $((CH_3)_3N)$ do not react with benzene sulphonyl chloride.Therefore,there are $3$ amines ($2$ primary and $1$ secondary) that can react with benzene sulphonyl chloride.

How many amines with molecular formula $C_3H_9N$ can react with benzene sulphonyl chloride?

Similar Questions

Explain the basicity of aromatic amines.

The following is a confirmatory test for aromatic primary amines. Identify reagent $(A)$ and $(B)$.

$C_{6}H_{5}NO_{2}$ $\xrightarrow{Sn+HCl} A$ $\xrightarrow{C_{6}H_{5}N_{2}^{\oplus}Cl^{\ominus}} P$
(Yellow coloured compound)
Consider the above reaction,the Product "$P$" is:

$CH_3-C(CH_3)_2-NH_2 \xrightarrow{NaNO_2 + HCl}$ Product. Which of the following is not formed as a product?

Which of the following will not give Hoffmann bromamide reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module