Chapter 23, Problem 23.66AP

Interpretation Introduction

(a)

Interpretation:

A synthesis for the compound $N\text{-methyl-1-hexanamine}$ from pentanoic acid is to be stated.

Concept introduction:

Carboxylic acid reacts with ammonia and its derivatives to give ammonium salts of primary secondary and tertiary amines. On heating the ammonium salts dehydrate to form amides. The amides can be reduced to amines with ${\text{LiAlH}}_4$ or sodium in the presence of ethanol.

Answer to Problem 23.66AP

The compound, $N-\text{methyl}-1-\text{hexanamine}$ can be synthesized from pentanoic acid as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOH+CH}}_3{\text{NH}}_2\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONHCH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONHCH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{NHCH}}_3\end{array}$

Explanation of Solution

Pentanoic acid combines with methyl amine to give corresponding amides. By addition reaction, nitrogen of amino group of methyl amine attaches to carbonyl carbon of pentanoic acid. Amide obtained is reduced with a suitable reducing agent like sodium in ethanol or ${\text{LiAlH}}_{\text{4}}$. The product obtained is secondary amine, $N-\text{methyl}-1-\text{hexanamine}$. The corresponding reactions are shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOH+CH}}_3{\text{NH}}_2\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONHCH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONHCH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{NHCH}}_3\end{array}$

Conclusion

The amines can be prepared by the reaction of carboxylic acid with ammonia and its derivatives. The amides obtained are reduced to required amine by a suitable reducing agent.

Interpretation Introduction

(b)

Interpretation:

A synthesis for the compound, pentylamine from pentanoic acid is to be stated.

Concept introduction:

Carboxylic acid reacts with ammonia and its derivatives to give ammonium salts of primary secondary and tertiary amines. On heating, this ammonium salts dehydrate to form amides. The amides can be reduced to amines with ${\text{LiAlH}}_4$ or sodium in the presence of ethanol.

Answer to Problem 23.66AP

The compound pentylamine can be synthesized from pentanoic acid as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONH}}_2\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONH}}_2\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{NH}}_2\end{array}$

Explanation of Solution

When pentanoic acid is treated with ammonia, the pentonamide is obtained. In this by addition reaction of ammonia to carboxylic acid amide is formed. This pentonamide gets reduced with sodium in ethanol or ${\text{LiAlH}}_{\text{4}}$.to required pentylamine. This pentylamine is a primary amine. The reactions involved are given below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONH}}_2\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONH}}_2\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{NH}}_2\end{array}$

Conclusion

Reaction product is a primary amine, pentyl amine. To prepare the given compound acid is treated with ammonia as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONH}}_2\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CONH}}_2\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{NH}}_2\end{array}$

Interpretation Introduction

(c)

Interpretation:

A synthesis for the compound, $N,N-\text{dimethyl}-1-\text{pentanamine}$ from pentanoic acid is to be stated.

Concept introduction:

Carboxylic acid reacts with ammonia and its derivatives to give ammonium salts of primary secondary and tertiary amines. On heating this ammonium salts dehydrate to form amides. The amides can be reduced to amines with ${\text{LiAlH}}_4$ or sodium in the presence of ethanol.

Answer to Problem 23.66AP

The compound, $N,N-\text{dimethyl}-1-\text{pentanamine}$ can be synthesized from pentanoic acid as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOH+CH}}_3{\text{NHCH}}_3\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{CONH}\left({\text{CH}}_2\right){\text{CH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{CONH}\left({\text{CH}}_2\right){\text{CH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2\text{NH}\left({\text{CH}}_2\right){\text{CH}}_3\end{array}$

Explanation of Solution

The compound $N,N-\text{dimethyl}-1-\text{pentanamine}$ is treated with ammonia derivative, $1,1-\text{Dimethylamine}$. This prepares the corresponding amide. The amide is reduced with suitable reducing agent to required tertiary amine. The reactions are shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOH+CH}}_3{\text{NHCH}}_3\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{CONH}\left({\text{CH}}_2\right){\text{CH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{CONH}\left({\text{CH}}_2\right){\text{CH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{$\text{Na}$}\!\left/ \!\raisebox{-1ex}{$\text{EtOH}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2\text{NH}\left({\text{CH}}_2\right){\text{CH}}_3\end{array}$

Conclusion

The tertiary amine $N,N-\text{dimethyl}-1-\text{pentanamine}$ can be prepared from penatanoic acid by first treating it with secondary amine and then reducing it to amine.

Interpretation Introduction

(d)

Interpretation:

A synthesis for the compound butylamine from pentanoic acid is to be stated.

Concept introduction:

Carboxylic acids when treated with ammonia give primary amines. This reaction is known as Schmidt reaction. Under acidic conditions ammonia is added to acid by nucleophilic addition. One molecule of carbon dioxide and nitrogen gas each is removed and amine is formed with one carbon less than parent acid.

Answer to Problem 23.66AP

The compound, butylamine can be synthesized from pentanoic acid as shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{NH}}_2+{\text{N}}_{\text{2}}+{\text{CO}}_{\text{2}}$

Explanation of Solution

The pentanoic acid, when treated with ammonia undergoes nucleophilic addition reaction. First ammonia is added to acid with elimination of a nitrogen molecule and amide is formed. Under acidic condition this amide breaks to give primary amine and a carbon dioxide molecule is evolved. The reaction is known as Schmidt reaction and is given below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{NH}}_2+{\text{N}}_{\text{2}}+{\text{CO}}_{\text{2}}$

Conclusion

The pentanoic acid can be converted to butylamine as shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{NH}}_2+{\text{N}}_{\text{2}}+{\text{CO}}_{\text{2}}$

Interpretation Introduction

(e)

Interpretation:

A synthesis for the compound hexylamine from pentanoic acid is to be stated.

Concept introduction:

The no of carbon atoms in acid can be increased by its esterification. Acid is treated with alcohol to obtain desired ester. The ester gets reduced to alkane with strong reducing agent. The alkyl halide of the alkane on treatment with ammonia produces amines of all types. This is known as Hofmann’s method.

Answer to Problem 23.66AP

The compound, hexylamine can be synthesized from pentanoic acid as shown below

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{CH}}_{\text{3}}\text{OH}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOCH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOCH}}_3\stackrel{\raisebox{1ex}{$\text{HI}$}\!\left/ \!\raisebox{-1ex}{$\text{P}$}\right.}{\to }{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_3+\text{HX}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_2\text{X}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_2\text{X}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2\end{array}$

Explanation of Solution

In this reaction, the number of carbon atom is more in the desired product than in given compound. To increase the carbons first the esterification of acid is done with methyl alcohol. As methyl group is introduced, the ester is reduced to alkane. Halogenation of this hexane is done and then with any hydrogen halide. The obtained halide is treated with ammonia to get the required amine as in Hofmann’s method.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{COOH}+{\text{CH}}_{\text{3}}\text{OH}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOCH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{COOCH}}_3\stackrel{\raisebox{1ex}{$\text{HI}$}\!\left/ \!\raisebox{-1ex}{$\text{P}$}\right.}{\to }{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_3\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_3+\text{HX}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_2\text{X}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_2\text{X}+{\text{NH}}_{\text{3}}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2\end{array}$

Conclusion

The hexylamine is having more carbon than pentanoic acid, therefore number of carbon atoms is increased first, and then it is converted to desired amine.

Interpretation Introduction

(f)

Interpretation:

A synthesis for the given quaternary ammonium salt from butyraldehyde is to be stated.

Concept introduction:

The aldehydes can easily be reduced to alcohols. The common reducing agents for the same are hydrogen with nickel or lithium aluminum hydride or sodium borohydride. The alcohols react with $\text{HBr}$, and bromoalkanes are obtained. The bromoalkanes, when react with tertiary amines, produce quaternary ammonium salts.

Answer to Problem 23.66AP

The given quaternary ammonium salt can be synthesized from butyraldehyde as given below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHO}\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{${\text{H}}_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{Ni}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHOH}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHOH}\stackrel{\text{HBr}}{\to }{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHBr}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHBr}+{\text{PhCH}}_2\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\to \to {\text{PhCH}}_2{\text{N}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CH}}_2{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}{\text{CH}}_3{\text{Br}}^{-}\end{array}$

Explanation of Solution

The compound butyraldehyde, in presence of nickel and hydrogen, gets hydrogenated and give alcohol. This butyl alcohol reacts with hydrogen bromide to give the corresponding bromobutane. Bromobutane when comes in contact with tertiary amine $N,N-\text{dimethyl}-\text{benzylamine}$, it gives the quaternary ammonium salt as desired. The reaction sequence is given below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHO}\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{${\text{H}}_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{Ni}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHOH}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHOH}\stackrel{\text{HBr}}{\to }{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHBr}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHBr}+{\text{PhCH}}_2\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\to \to {\text{PhCH}}_2{\text{N}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CH}}_2{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}{\text{CH}}_3{\text{Br}}^{-}\end{array}$

Conclusion

The desired quaternary ammonium salt can be prepared from butyraldeyd as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHO}\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\raisebox{1ex}{${\text{H}}_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{Ni}$}\right.}{\to }}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHOH}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHOH}\stackrel{\text{HBr}}{\to }{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHBr}\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CHBr}+{\text{PhCH}}_2\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\to \to {\text{PhCH}}_2{\text{N}}^{\text{+}}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CH}}_2{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}{\text{CH}}_3{\text{Br}}^{-}\end{array}$

Interpretation Introduction

(g)

Interpretation:

A synthesis for the given compound $N\text{-ethyl-3-phenyl-1-propanamine}$ from toluene is to be stated.

Concept introduction:

Wurtz reaction is used to increase the numbers of carbon atoms in an alkane. Two alkyl halides according to required compound are taken. Sodium metal in dry ether solution is used for the coupling reaction, to form the higher alkane. The halides of these alkanes can go under controlled substitution reaction with amines, to give primary, secondary and tertiary amines.

Answer to Problem 23.66AP

The compound $N\text{-ethyl-3-phenyl-1-propanamine}$, can be synthesized by the following reaction sequences.

$\begin{array}{l}{\text{PhCH}}_{\text{3}}+{\text{Br}}_{\text{2}}\to {\text{PhCH}}_{\text{2}}\text{Br}\\ {\text{PhCH}}_{\text{2}}\text{Br}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Br}\underset{\text{dry ether}}{\overset{\text{Na}}{\to }}{\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\\ {\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\underset{\text{heat, light}}{\overset{{\text{Br}}_{\text{2}}}{\to }}{\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{Br}\\ {\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{Br}+{\text{NH}}_2{\text{C}}_2{\text{H}}_5\stackrel{\text{NaOH}}{\to }{\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2{\text{NH}}_2{\text{C}}_2{\text{H}}_5\end{array}$

Explanation of Solution

The toluene is first halogenated with bromine to give bromotoluene. This bromotoluene reacts with ethyl bromide in presence of sodium metal and dry ether, to produce the phenyl propane (wurtz reaction). This is again halogenated with bromine to get the substituted bromo propane. When this substituted bromo propane reacts with ethyl amine, the desired product, $N\text{-ethyl-3-phenyl-1-propanamine}$ is obtained as shown below.

$\begin{array}{l}{\text{PhCH}}_{\text{3}}+{\text{Br}}_{\text{2}}\to {\text{PhCH}}_{\text{2}}\text{Br}\\ {\text{PhCH}}_{\text{2}}\text{Br}+{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Br}\underset{\text{dry ether}}{\overset{\text{Na}}{\to }}{\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\\ {\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\underset{\text{heat, light}}{\overset{{\text{Br}}_{\text{2}}}{\to }}{\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{Br}\\ {\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{Br}+{\text{NH}}_2{\text{C}}_2{\text{H}}_5\stackrel{\text{NaOH}}{\to }{\text{PhCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2{\text{NH}}_2{\text{C}}_2{\text{H}}_5\end{array}$

Conclusion

The compound, $N\text{-ethyl-3-phenyl-1-propanamine}$ can be obtained by halogenation and substitution of toluene.

Interpretation Introduction

(h)

Interpretation:

A synthesis for the given compound, $2\text{-pentanamine}$ from diethyl malonate is to be stated.

Concept introduction:

By substitution reaction the alky groups can be introduced to active methylene carbon of diethyl malonate. Further on its hydrolysis corresponding acid is obtained which can be converted into amide and then reduced to amine.

Answer to Problem 23.66AP

The given compound $2\text{-pentanamine}$, can be synthesized from diethyl malonate as given below.

$\begin{array}{l}{\text{CH}}_{\text{2}}{\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_2\underset{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}}{\overset{\text{NaOEt}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_2\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)\underset{{\text{H}}_{\text{2}}\text{O, heat}}{\overset{{\text{H}}^{+}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left(\text{COOH}\right)}_2\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left(\text{COOH}\right)}_{\text{2}}\stackrel{-{\text{CO}}_{\text{2}}}{\to }{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}+{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CONHCH}}_3\end{array}$

The amide is reduced to final product as shown below.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CONHCH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\text{NaOEt}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_2{\text{NHCH}}_3$

Explanation of Solution

The diethyl malonate has an active methylene group $-{\text{CH}}_{\text{2}}$. The hydrogen of this can be replaced with ethyl group by treating it with ethyl chloride in presence of dry ether. When hydrolysis of substituted product is done, it gives monocarboxylic acid, after decarboxylation. This monocarboxylic acid is treated with methyl amine to produce amide, which is reduced to the desired $2-\text{pentanamine}$. The reactions are given below.

$\begin{array}{l}{\text{CH}}_{\text{2}}{\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_2\underset{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}}{\overset{\text{NaOEt}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_2\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)\underset{{\text{H}}_{\text{2}}\text{O, heat}}{\overset{{\text{H}}^{+}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left(\text{COOH}\right)}_2\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left(\text{COOH}\right)}_{\text{2}}\stackrel{-{\text{CO}}_{\text{2}}}{\to }{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}+{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CONHCH}}_3\end{array}$

The amide is reduced to final product as shown below.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CONHCH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\text{NaOEt}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_2{\text{NHCH}}_3$

Conclusion

The substituted $2-\text{pentanamine}$ can be prepared by ethyl malonate as shown below.

$\begin{array}{l}{\text{CH}}_{\text{2}}{\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_2\underset{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}}{\overset{\text{NaOEt}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_2\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\left({\text{COOC}}_{\text{2}}{\text{H}}_{\text{5}}\right)\underset{{\text{H}}_{\text{2}}\text{O, heat}}{\overset{{\text{H}}^{+}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left(\text{COOH}\right)}_2\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\left(\text{COOH}\right)}_{\text{2}}\stackrel{-{\text{CO}}_{\text{2}}}{\to }{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{COOH}+{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CONHCH}}_3\end{array}$

The amide is reduced to final product as shown below.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CONHCH}}_3\underset{{\text{LiAlH}}_{\text{4}}}{\overset{\text{NaOEt}}{\to }}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_2{\text{NHCH}}_3$

Interpretation Introduction

(i)

Interpretation:

A synthesis for the given compound, isobutyl amine from acetone is to be stated.

Concept introduction:

When ketones are treated with Grignard reagents, the addition product, tertiary alcohol is formed. This alcohol can be converted to halide and halide can be substituted with amine group to get the required compound.

Answer to Problem 23.66AP

The given compound is butyl amine can be synthesized from acetone as given below.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{C}={\text{OCH}}_{\text{3}}+{\text{CH}}_{\text{3}}\text{MgI}\underset{{\text{H}}_{\text{2}}\text{O}}{\overset{\text{ether}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{CH}}_{\text{3}}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{CH}}_{\text{3}}+\text{HBr}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{Br}\right){\text{CH}}_{\text{3}}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{Br}\right){\text{CH}}_{\text{3}}+{\text{NH}}_{\text{3}}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left({\text{NH}}_{\text{2}}\right){\text{CH}}_{\text{3}}\end{array}$

Explanation of Solution

The given acetone is treated with Grignard’s reagent. The result is addition product. This upon hydrolysis gives the corresponding isobutyl alcohol. Obtained isobutyl alcohol is treated with hydrogen bromide to get the corresponding halide, isobutyl bromide. The isobutyl bromide gives substituted tertiary amine upon treatment with ammonia. The reactions are given below.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{C}={\text{OCH}}_{\text{3}}+{\text{CH}}_{\text{3}}\text{MgI}\underset{{\text{H}}_{\text{2}}\text{O}}{\overset{\text{ether}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{CH}}_{\text{3}}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{CH}}_{\text{3}}+\text{HBr}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{Br}\right){\text{CH}}_{\text{3}}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{Br}\right){\text{CH}}_{\text{3}}+{\text{NH}}_{\text{3}}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left({\text{NH}}_{\text{2}}\right){\text{CH}}_{\text{3}}\end{array}$

Conclusion

The isobutyl amine, a tertiary amine can be prepared from acetone by the reaction sequences as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{C}={\text{OCH}}_{\text{3}}+{\text{CH}}_{\text{3}}\text{MgI}\underset{{\text{H}}_{\text{2}}\text{O}}{\overset{\text{ether}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{CH}}_{\text{3}}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{CH}}_{\text{3}}+\text{HBr}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{Br}\right){\text{CH}}_{\text{3}}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{Br}\right){\text{CH}}_{\text{3}}+{\text{NH}}_{\text{3}}\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left({\text{NH}}_{\text{2}}\right){\text{CH}}_{\text{3}}\end{array}$

Interpretation Introduction

(j)

Interpretation:

A synthesis for the compound isopentylamine from acetone is to be stated.

Concept introduction:

When ketones are treated with Grignard reagents, the addition product, tertiary alcohol is formed. This alcohol can be reduced to corresponding alkane. Alkanes react with halogens to produce halides. This halide can be substituted with amine group to get the required compound.

Answer to Problem 23.66AP

The compound isopentylamine can be synthesized from acetone as shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{C}={\text{OCH}}_{\text{3}}+{\text{C}}_2{\text{H}}_5\text{MgI}\underset{{\text{H}}_{\text{2}}\text{O}}{\overset{\text{ether}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{C}}_2{\text{H}}_5\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{C}}_2{\text{H}}_5\underset{\raisebox{1ex}{${\text{H}}_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{Pt}$}\right.}{\overset{{\text{Al}}_{\text{2}}{\text{O}}_3}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_5\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_5+{\text{Br}}_2\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4\text{Br}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4\text{Br}+{\text{NH}}_3\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4{\text{NH}}_2\end{array}$

Explanation of Solution

The acetone gives an addition product with Grignard’s reagent. This product is hydrolyzed to get the isopentyl alcohol. The isopentyl alcohol, in presence of aluminium chloride gets reduced and then hydrogenated to get the parent alkane that is pentane. The pentane is treated with bromine, to produce the bromo pentane. This, when treated with ammonia, produces the required compound isopentyl amine. The corresponding reactions are shown below.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{C}={\text{OCH}}_{\text{3}}+{\text{C}}_2{\text{H}}_5\text{MgI}\underset{{\text{H}}_{\text{2}}\text{O}}{\overset{\text{ether}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{C}}_2{\text{H}}_5\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{C}}_2{\text{H}}_5\underset{\raisebox{1ex}{${\text{H}}_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{Pt}$}\right.}{\overset{{\text{Al}}_{\text{2}}{\text{O}}_3}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_5\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_5+{\text{Br}}_2\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4\text{Br}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4\text{Br}+{\text{NH}}_3\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4{\text{NH}}_2\end{array}$

Conclusion

Reaction product is the required primary amine, isopentyl amine. To prepare the given compound acetone is subject to treatment with Grignard’s reagent and other reactants as shown.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{C}={\text{OCH}}_{\text{3}}+{\text{C}}_2{\text{H}}_5\text{MgI}\underset{{\text{H}}_{\text{2}}\text{O}}{\overset{\text{ether}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{C}}_2{\text{H}}_5\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{OH}\right){\text{C}}_2{\text{H}}_5\underset{\raisebox{1ex}{${\text{H}}_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{Pt}$}\right.}{\overset{{\text{Al}}_{\text{2}}{\text{O}}_3}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_5\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_5+{\text{Br}}_2\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4\text{Br}\\ {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4\text{Br}+{\text{NH}}_3\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHC}}_2{\text{H}}_4{\text{NH}}_2\end{array}$

Interpretation Introduction

(k)

Interpretation:

The synthesis for the compound, $m\text{-chlorobromobenzene}$ from nitrobenzene is to be stated.

Concept introduction:

The replacement of hydrogen atom attached to a carbon atom of electron-rich benzene ring by another electrophile is known as electrophilic aromatic substitution reaction. The rate of electrophilic aromatic substitution reaction depends on the substituted group on the aromatic ring. The ring deactivating group retards the electrophilic aromatic substitution reaction and ring activating group enhances the electrophilic aromatic substitution reaction.

Answer to Problem 23.66AP

The synthesis of $m\text{-chlorobromobenzene}$ from nitrobenzene is shown below.

Explanation of Solution

The nitro group is meta directing group. The nitrobenzene undergoes electrophilic aromatic substitution reaction with ${\text{Br}}_{\text{2}}$ in the presence of catalyst ${\text{FeBr}}_{\text{3}}$ to form $\text{1-bromo-3-nitrobenzene}$. The nitro group of $\text{1-bromo-3-nitrobenzene}$ get reduced to amino group with help of reagents $\text{Sn}/\text{HCl}$. The amino group undergoes diazotization reaction with ${\text{NaNO}}_{\text{2}}$ and $\text{HCl}$. The resultant compound reacts $\text{CuCl}$ to form $\text{m-chlorobromobenzene}$. The corresponding chemical reaction sequence is shown below.

Figure 1

Conclusion

The synthesis of $m\text{-chlorobromobenzene}$ from nitrobenzene is shown in Figure 1.

Interpretation Introduction

(l)

Interpretation:

The synthesis for the compound $p\text{-chlorobromobenzene}$ from nitrobenzene is to be predicted.

Concept introduction:

In presence of mixed acids, conc. ${\text{HNO}}_{\text{3}}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, the nitro benzene behaves as a $\text{p}-$ directing group. Here the nitrogen shares its electron with the ring. So, the electron density increases at ortho and para position and the incoming group are directed towards the ortho and para position in the ring.

Answer to Problem 23.66AP

The synthesis for the compound $p\text{-chlorobromobenzene}$ from nitrobenzene is shown below.

Explanation of Solution

The nitro group of nitrobenzene get reduced to amino group with help of reagents $\text{Sn}/\text{HCl}$. The amino group undergoes diazotization reaction with ${\text{NaNO}}_{\text{2}}$ and $\text{HCl}$. The resultant compound reacts $\text{CuCl}$ to form chlorobenzene. The chlorobenzene undergoes electrophilic aromatic substitution reaction with ${\text{Br}}_{\text{2}}$ in the presence of catalyst ${\text{FeBr}}_{\text{3}}$ to form $p\text{-chlorobromobenzene}$. The corresponding chemical reaction sequence is shown below.

Figure 2

Conclusion

The synthesis for the compound $p\text{-chlorobromobenzene}$ from nitrobenzene is shown in Figure 2.

Interpretation Introduction

(m)

Interpretation:

A synthesis for the compound $p\text{-methoxybenzonitrile}$ from phenol is to be stated.

Concept introduction:

The alcohol group of phenol is ring activating and para directing group. The oxygen shares its lone pair of electron with the ring. The increased electron density makes the ring prone to electrophilic substitution reactions. In presence of a base the phenols give ether with alkyl halides.

Answer to Problem 23.66AP

The compound, $p\text{-methoxybenzonitrile}$ can be synthesized from phenol as shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH+}\stackrel{\text{NaOH}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}+{\text{CH}}_{\text{3}}\text{Br}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OCH}}_{\text{3}}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OCH}}_{\text{3}}\underset{\text{KCN}}{\overset{\text{CuCN}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\left(\text{CN}\right){\text{OCH}}_{\text{3}}\end{array}$

Explanation of Solution

The phenol is treated with a base. It gives sodium salt of phenoxide ion. This reacts with methyl bromide to produce methoxybenzene. This methoxybenzene is treated with potassium cyanide in presence of cuprous cyanide to give required $p\text{-methoxybenzonitrile}$. As the methoxy group is ring activating, it is ortho and para directing. The incoming nitrile group is directed towards the para position. The corresponding reaction sequence is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH+}\stackrel{\text{NaOH}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}+{\text{CH}}_{\text{3}}\text{Br}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OCH}}_{\text{3}}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OCH}}_{\text{3}}\underset{\text{KCN}}{\overset{\text{CuCN}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\left(\text{CN}\right){\text{OCH}}_{\text{3}}\end{array}$

Conclusion

The compound, $p\text{-methoxybenzonitrile}$ can be prepared from phenol by the reaction sequence as shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH+}\stackrel{\text{NaOH}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{ONa}+{\text{CH}}_{\text{3}}\text{Br}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OCH}}_{\text{3}}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OCH}}_{\text{3}}\underset{\text{KCN}}{\overset{\text{CuCN}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\left(\text{CN}\right){\text{OCH}}_{\text{3}}\end{array}$

Interpretation Introduction

(n)

Interpretation:

The synthesis for the given amine from given alcohol is to be stated.

Concept introduction:

When alcohols undergo substitution nucleophilic reaction with ${\text{SOCl}}_{\text{2}}$ the alkyl halides are formed and stereochemistry remains unaffected. The alkyl halide on treatment with potassium cyanide, gives cyanoethane with changed stereochemistry. This cyanoethane on reduction gives the primary aminopropanamine.

Answer to Problem 23.66AP

The synthesis for the given amine from given alcohol is shown below.

Explanation of Solution

First the alcohol is treated with ${\text{SOCl}}_{\text{2}}$, to gives chloroethylalcohol. This is treated with potassium cyanide, the stereochemistry of the molecule changes. The $\text{-CN}$ group attacks from one side and the $-\text{OH}$ group leaves from the other side (${\text{S}}_{\text{N}}2$ reaction). Obtained cyanoethane is reduced with lithium borohydride in presence of sodium and ethanol to the primary amine with changed stereochemistry. The corresponding chemical reaction sequence is shown below.

Figure 3

Conclusion

The synthesis for the given amine from given alcohol is shown in Figure 3.

Interpretation Introduction

(o)

Interpretation:

The synthesis for the given substituted pyrrole from levulinic acid is to be predicted.

Concept introduction:

The acids when react with amines they give corresponding amides. These amides can be reduced to get higher amines. The amines with ketone group in vicinity, undergo nucleophilic addition reaction to give cyclic compounds.

Answer to Problem 23.66AP

The synthesis for the given substituted pyrrole from levulinic acid is shown below.

Explanation of Solution

The given keto acid is first treated with methyl amine. The corresponding amide is obtained. This amide is now undergoes cyclisation reaction to form the pyrrole ring. The alcoholic group can be reduced by the selective reduction with sodium borohydride and cerium chloride ( luche reduction ). The reactions are given below.

Figure 4

Conclusion

The substituted pyrrole can be prepared from levulinic acid as shown in Figure 4.