Answer

(i) Considering the inductive effect of alkyl groups, $\mathrm{NH_3}, \mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}$, and $\left(\mathrm{C_2} \mathrm{H_5}\right)_{2} \mathrm{NH}$ can be arranged in the increasing order of their basic strengths as:

$ \mathrm{NH_3}<\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}<\left(\mathrm{C_2} \mathrm{H_5}\right)_{2} \mathrm{NH} $

Again, $\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}$ has proton acceptability less than $\mathrm{NH_3}$. Thus, we have:

$ \mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}<\mathrm{NH_3}<\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}<\left(\mathrm{C_2} \mathrm{H_5}\right)_{2} \mathrm{NH} $

Due to the - I effect of $\mathrm{C_6} \mathrm{H_5}$ group, the electron density on the $\mathrm{N}$-atom in $\mathrm{C_6} \mathrm{H_5} \mathrm{CH_2} \mathrm{NH_2}$ is lower than that on the $\mathrm{N}$-atom in $\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}$, but more than that in $\mathrm{NH_3}$. Therefore, the given compounds can be arranged in the order of their basic strengths as:

$ \mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}<\mathrm{NH_3}<\mathrm{C_6} \mathrm{H_5} \mathrm{CH_2} \mathrm{NH_2}<\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}<\left(\mathrm{C_2} \mathrm{H_5}\right)_{2} \mathrm{NH} $

(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, $\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}$, $\left(\mathrm{C_2} \mathrm{H_5}\right)_{2} \mathrm{NH_2}$ and their basic strengths as follows:

$ \mathrm{C _2} \mathrm{H _5} \mathrm{NH _2}<\left(\mathrm{C _2} \mathrm{H _5}\right) _{3} \mathrm{~N}<\left(\mathrm{C _2} \mathrm{H _5}\right) _{2} \mathrm{NH} $

Again, due to the - $\mathrm{R}$ effect of $\mathrm{C_6} \mathrm{H_5}$ group, the electron density on the $\mathrm{N}$ atom in $\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}$ is lower than that on the $\mathrm{N}$ atom in $\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}$. Therefore, the basicity of $\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}$ is lower than that of $\mathrm{C_2} \mathrm{H_5} \mathrm{NH_2}$. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:

$ \mathrm{C _6} \mathrm{H _5} \mathrm{NH _2}<\mathrm{C _2} \mathrm{H _5} \mathrm{NH _2}<\left(\mathrm{C _2} \mathrm{H _5}\right) _{3} \mathrm{~N}<\left(\mathrm{C _2} \mathrm{H _5}\right) _{2} \mathrm{NH} $

(iii) Considering the inductive effect and the steric hindrance of alkyl groups, $\mathrm{CH _3} \mathrm{NH _2}$, $\left(\mathrm{CH _3}\right) _{2} \mathrm{NH}$, and $\left(\mathrm{CH _3}\right) _{3} \mathrm{~N}$ can be arranged in the increasing order of their basic strengths as:

$ \left(\mathrm{CH _3}\right) _{3} \mathrm{~N}<\mathrm{CH _3} \mathrm{NH _2}<\left(\mathrm{CH _3}\right) _{2} \mathrm{NH} $

In $\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}, \mathrm{~N}$ is directly attached to the benzene ring. Thus, the lone pair of electrons on the $\mathrm{N}$ atom is delocalized over the benzene ring. In $\mathrm{C_6} \mathrm{H_5} \mathrm{CH_2} \mathrm{NH_2}, \mathrm{~N}$ is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the $\mathrm{N}$ atom are more easily available for protonation in $\mathrm{C_6} \mathrm{H_5} \mathrm{CH_2} \mathrm{NH_2}$ than in $\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}$ i.e., $\mathrm{C_6} \mathrm{H_5} \mathrm{CH_2} \mathrm{NH_2}$ is more basic than $\mathrm{C_6} \mathrm{H_5} \mathrm{NH_2}$.

Again, due to the - I effect of $\mathrm{C _6} \mathrm{H _5}$ group, the electron density on the $\mathrm{N}$ - atom in $\mathrm{C _6} \mathrm{H _5} \mathrm{CH _2} \mathrm{NH_2}$ is lower than that on the $\mathrm{N}$ - atom in $\left(\mathrm{CH _3}\right) _{3} \mathrm{~N}$. Therefore, $\left(\mathrm{CH _3}\right) _{3} \mathrm{~N}$ is more basic than $\mathrm{C _6} \mathrm{H _5} \mathrm{CH _2} \mathrm{NH_2}$. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows.

$ \mathrm{C _6} \mathrm{H _5} \mathrm{NH _2}<\mathrm{C _6} \mathrm{H _5} \mathrm{CH _2} \mathrm{NH _2}<\left(\mathrm{CH _3}\right) _{3} \mathrm{~N}<\mathrm{CH _3} \mathrm{NH _2}<\left(\mathrm{CH _3}\right) _{2} \mathrm{NH} $

Answer

(i) $ \underset{n-\text{Propylamine}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{NH_2}}+\mathrm{HCl} \longrightarrow \underset{n-\text{Propylammoniumchloride}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{NH_3} \stackrel{+}{\mathrm{Cl}}} $

(ii)

$\underset{\text { Triethylamine }}{\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}}+\mathrm{HCl} \longrightarrow \underset{\text { Triemethylammoniumchloride }}{\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{\stackrel{+}{N}H_3} \stackrel{-}{\mathrm{Cl}}}$

Answer

Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of $\mathrm{Na} 2 \mathrm{CO} 3$ solution, $\mathrm{N}, \mathrm{N}$-dimethylaniline produces $\mathrm{N}, \mathrm{N}, \mathrm{N}$-trimethylanilinium carbonate.

Answer

Answer

The structures of different isomers corresponding to the molecular formula, $\mathrm{C_3} \mathrm{H_9} \mathrm{~N}$ are given below:

(a) $\mathrm{CH_3}-\mathrm{CH_2}-\mathrm{CH_2}-\mathrm{NH_2}$

Propan-1-amine $\left(1^{\circ}\right)$

(b)

Propan-2-amine $\left(1^{\circ}\right)$

(c)

$\mathrm{CH_3}-\mathrm{NH}-\mathrm{C_2} \mathrm{H_5}$

$\mathrm{N}$-Methylethanamine $\left(2^{0}\right)$

(d)

N,N-Dimethylmethanamine $\left(3^{0}\right)$

$\left(1^{0}\right)$ amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

$ \underset{\text{Propan-1-amine}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{NH_2}}+\mathrm{HNO_2} \longrightarrow \underset{\text{Propan-1-ol}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{OH}}+\mathrm{N_2}+\mathrm{HCl} $