Question

Write the balanced chemical equations for the (1) reaction of each of the following amines with water and (2) neutralization with HBr:

a. ethylmethylamine

b. propylamine

c. N-methylaniline

Short answer

a.

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right)\mathrm{NH}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH} \\ {\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH}\rightleftharpoons \left[{\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2^{+}\right]+\left[{\mathrm{OH}}^{-}\right] \\ {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{Br} \end{gathered}$$

b.

role="math" localid="1653993426532" $$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{OH} \\ {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{OH}\rightleftharpoons \left[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3^{+}\right]+\left[{\mathrm{OH}}^{-}\right] \\ {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{Br} \end{gathered}$$

c.$$\begin{gathered} \begin{array}{l}{\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right)\mathrm{NH}+{\mathrm{H}}_2\mathrm{O}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH}\\ {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\rightleftharpoons \left[{\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2^{+}\right]+\left[{\mathrm{OH}}^{-}\right]\end{array} \\ {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right)\mathrm{NH}+\mathrm{HBr}\to {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{Br} \end{gathered}$$

Step-by-step solution

Part (a) Step 1: Given Information

Ethylmethylamine

Representing the reaction of ethylmethylamine with water and neutralization with HBr.

Part (a) Step 2: Explanation

Ethylmethylamine is an illustration of auxiliary amine which is water dissolvable and structure ethyl methyl ammonium hydroxide, which separates into particles.

Reaction with water is -

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right)\mathrm{NH}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH} \\ {\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH}\rightleftharpoons \left[{\mathrm{CH}}_3{\mathrm{CH}}_2\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2^{+}\right]+\left[{\mathrm{OH}}^{-}\right] \end{gathered}$$

Ethylmethylamine produces salt on reaction with Hbr as it is a base.

Neutralization with Hbr is -

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{Br}$

Part (b) Step 1: Given Information

Propylamine

The reaction of propylamine with water and neutralization with HBr.

Part (b) Step 2: Explanation

Propylamine is an illustration of primary amine which is water dissolvable and structure ethyl propyl ammonium hydroxide, which separates into particles.

Reaction with water is -

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{OH} \\ {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{OH}\rightleftharpoons \left[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3^{+}\right]+\left[{\mathrm{OH}}^{-}\right] \end{gathered}$$

Ethylmethylamine produces salt on reaction with Hbr as it is a base.

Neutralization with Hbr is -

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_3\mathrm{Br}$

Part (c) Step 1: Given Information

N-methylaniline

The reaction of N-methylaniline with water and neutralization with HBr.

Part (c) Step 2: Explanation

N-methylaniline is an illustration of auxiliary amine which is water dissolvable and structure ethyl methyl ammonium hydroxide, which separates into particles.

Reaction with water is -

$\begin{array}{l}{\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right)\mathrm{NH}+{\mathrm{H}}_2\mathrm{O}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH}\\ {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{OH}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\rightleftharpoons \left[{\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2^{+}\right]+\left[{\mathrm{OH}}^{-}\right]\end{array}$

N-methylaniline produces salt on reaction with Hbr as it is a base.

Neutralization with Hbr is -

${\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right)\mathrm{NH}+\mathrm{HBr}\to {\mathrm{C}}_6{\mathrm{H}}_5\left({\mathrm{CH}}_3\right){\mathrm{NH}}_2\mathrm{Br}$