Preparation of Amines

AMINES

Amines are derivatives of $N{H}_3.$ They are represented by general formula $C_n{H}_{2n+3}N.$

General methods of preparation:

1. Alkylation of $N{H}_3$with alkyl halides or alcohols:

$C{H}_{3}I+N{H}_3\to C{H}_{3}N{H}_2+HI$

This reaction is called Hoffmann's Method. Exhaustive methylation is as follows:

Aryl halides show low reactivity for this reaction.

2. By the action of ammonia on alcohol:

This method yields a mixture of $1^o,2^o,3^o$ amines and $4^o$ salts which are separated from each other by means of Hinsberg method, Hofmann method and fractional distillation. However, $1^o$amines can be prepared in good yield by using excess of ammonia.

$ROH+N{H}_3\underset{{300}^{o}C}{\overset{ZnC{l}_2}{\to }}RN{H}_2+H_{2}O$

Illustration 1. Which of the following compounds exists as non resolvable racemic mixture?

Solution: 3° amine contain all three alkyl groups different, giving non resolvable mixture.

Hence (D) is correct

Illustration 2. The statement not correct for piperidine is

(A) it is heterocyclic compound (B) it is homocyclic compound

(C) it is more basic than pyridine (D) it is more basic than aniline

Solution: Piperidine is heterocyclic compound.

Hence (A), (C) and (D) are correct and (B) is wrong.

Methods yielding only $1^{o}amines:$

1. By reduction of nitroalkanes:

{{C}_{2}}{{H}_{5}}N{{O}_{2}}+6[H]\xrightarrow[{{H}_{2}}/\,\,Ni\,\,or\,\,LiAl{{H}_{4}}]{Sn\,\,+\,\,HCl\,\,or}\underset{Ethyla\min e}{\mathop{{{C}_{2}}{{H}_{5}}N{{H}_{2}}}}\,+2{{H}_{2}}O

$C_6{H}_{5}N{O}_2+6[H]\underset{orFe+HCl}{\overset{Sn+HCl}{\to }}\underset{aniline}{C_6{H}_{5}N{H}_2}+2H_{2}O$

2. Mendius reduction of alkyl cyanides:

$C{H}_{3}CN+4H\underset{orLiAl{H}_4}{\overset{Na-C_2{H}_{5}OH}{\to }}C{H}_{3}C{H}_{2}N{H}_2$

Ethylamine

Reduction of alkyl isocyanides with $Na/C_2{H}_{5}OHgives{2}^o$amines eg.

$C{H}_{3}NC\stackrel{Na,C_2{H}_{5}OH}{\to }C{H}_{3}NHC{H}_3$

3. By reduction of amides and oximes:

$RCON{H}_2\stackrel{LiAl{H}_4}{\to }RC{H}_{2}N{H}_2$

4. By Hofmann bromamide reaction:

$C{H}_{3}CON{H}_2+B{r}_2+4KOH\to C{H}_{3}N{H}_2+2KBr+K_{2}C{O}_3+2H_{2}O$

$C_6{H}_{5}CON{H}_2+B{r}_2+4KOH\to C_6{H}_{5}N{H}_2+2KBr+K_{2}C{O}_3+2H_{2}O$

5. By Gabriel phthalimide reaction:

Potassium phthalimide formed after reaction of phthalimide with KOH, on heating with alkyl halide gives N – alkyl phthalimide. This on hydrolysis with 20% hydrochloric acid under pressure gives 1° amine.

Illustration 3.

The product B is

Solution: Gabrial pthalamide reaction.

Hence (B) is correct.

6. By action of chloramine on Grignard's reagent:

7. By decarboxylation of amino acids:

$\underset{Glycine}{H_{2}N.C{H}_2.COOH}\stackrel{Ba{(OH)}_2}{\to }\underset{Methyla\min e}{H_{2}NC{H}_3}$

8. By Wurtz method:

$\underset{Methylisocyanate}{C{H}_3-N=C=O}+2KOH\to C{H}_{3}N{H}_2+K_{2}C{O}_3$

9. Schmidt reaction:

\begin{align}  C{{H}_{3}}COOH+H{{N}_{3}}\xrightarrow[{{H}_{2}}S{{O}_{4}}]{conc.}C{{H}_{3}}N{{H}_{2}}+C{{O}_{2}}+{{N}_{2}} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Hydrazoic\,\,acid\,\,\,\,\,\,\,Methyla\min e \\ \end{align}

Methods giving $2^o$ amines only:

1. By reduction of alkyl isocyanide with sodium and ethanol

\begin{align}  C{{H}_{3}}NC+4H\xrightarrow[{}]{Na-{{C}_{2}}{{H}_{5}}OH}C{{H}_{3}}NHC{{H}_{3}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Dimethyla\min e \\ \end{align}

2. By heating an alcoholic solution of $1^o$ amine with alkyl halide

\begin{align}  {{C}_{2}}{{H}_{5}}N{{H}_{2}}+I{{C}_{2}}{{H}_{5}}\xrightarrow{{}}{{({{C}_{2}}{{H}_{5}})}_{2}}NH \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Dimethyla\min e \\ \end{align}

3. By hydrolysis of p – nitroso dialkyl aniline with boiling alkali

METHODS YIELDING ONLY 3⁰ AMINES

1. By heating alcoholic solution of NH₃ with excess of alkyl halide -

\begin{align} 3C{{H}_{3}}I+N{{H}_{3}}\xrightarrow{{}}{{(C{{H}_{3}})}_{3}}N+3HI \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Trimethyla\min e \\ \end{align}

2. By decomposition of tetra – alkyl ammonium hydroxide

However $(C{H}_3{)}_4{N}^{+}O{H}^{-}$ decomposes in different way.

${(C{H}_3)}_4{N}^{+}O{H}^{-}\stackrel{\Delta }{\to }{(C{H}_3)}_{3}N+C{H}_{3}OH$

Illustration 4. Consider the following sequence of reactions

Which of the following is correct?

Solution. Hoffmann's elimination.

Hence (B) is correct.

Illustration 5. Which of the following compounds on hydrolysis yield carboxylic acid and a secondary amine?

(C) CH₃CH₂NC (D) CH₃CONHCH₃

Solution: Compound having two alkyl group on N will give 2° amine.

Hence (A) is correct.