Chapter 22, Problem 45P

Devise efficient syntheses of each of the following compounds from the designated startingmaterials. You may also use any necessary organic or inorganic reagents.

$\text{3,3-Dimethyl-1-butanamine}$ from $\text{1-bromo-2,2-dimethylpropane}$

Interpretation Introduction

Interpretation:

Efficient syntheses are to be designed for each of the five compounds with designated starting materials and necessary organic or inorganic compounds.

Concept introduction:

Synthesis of amines can be done in a variety of ways, based on forming a bond between a carbon and a nitrogen.

Nucleophilic substitution reactions on alkyl halides or alcohols by a nitrogen containing reagent is one of the common methods.

Primary amines can be synthesized by nucleophilic substitution of an alkyl halide with cyanide. The resulting compound, called a nitrile, yields a primary amine on catalytic reduction with hydrogen.

Secondary and particularly tertiary amines can be formed by reaction between ammonia and an excess of alkyl halide.

Another method involves nucleophilic addition of simpler amines to an aldehyde or a ketone.

Carboxylic acid derivatives such as acid halides, anhydrides, and esters undergo nucleophilic substitution by ammonia or a variety of simpler amines to give amides. The amides on reduction gives amines.

Benzyl radical is stabilized by delocalization of the unpaired electron on to the benzene ring. This helps in selective radical substitution of hydrogens and halogens in benzylic position.

Answer to Problem 45P

Solution:

$\text{3, 3-Dimethyl-1-butanamine}$ from $\text{1-bromo-2, 2-dimethylpropane}$:

from $10\text{-undecenoic acid}$ and $\text{pyrrolidine}$:

c)

from :

d) ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{N(CH}}_{\text{3}}{\text{)CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{NHCH}}_{\text{3}}$ and ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CN}$:

e)

from :

Explanation of Solution

a) $\text{3, 3-Dimethyl-1-butanamine}$ from $\text{1-bromo-2, 2-dimethylpropane}$:

The substrate is a primary alkyl halide; therefore, substitution of the bromine requires a strong nucleophile such as cyanide ion via an ${\text{S}}_{\text{N}}\text{2}$ reaction. $\text{1-Bromo-2, 2-dimethylpropane}$ is first treated with sodium cyanide in a polar aprotic solvent like DMF to yield $\text{3, 3-dimethylbutanenitrile}$.

The nitrile yields the desired amine on catalytic reduction.

b)

from $10\text{-undecenoic acid}$ and $\text{pyrrolidine}$:

Carboxylic acids and amines can react to form amides by nucleophilic substitution of the hydroxyl group by the amine. However, carboxylic acids are poor electrophiles and are relatively unreactive. The first step is then to activate the acid by converting it to a stronger electrophile such as an acid chloride. $10\text{-Undecenoic acid}$ is treated with thionyl chloride ${\text{SOCl}}_{\text{2}}$, converting it to more reactive $10\text{-undecenoyl chloride}$.

The chloride then undergoes substitution by $\text{pyrrolidine}$ in an ${\text{S}}_{\text{N}}\text{2}$ reaction. Deprotonation of the quaternary ammonium ion yields the amide.

This contains both a carbonyl group that needs to be reduced to a methylene and a double bond. Catalytic reduction will result in reduction of the carbonyl group as well as an unwanted reduction of the double bond. Therefore, reduction of the carbonyl to methylene bridge is done with lithium aluminum hydride.

The hydride ion (${\text{H}}^{-}$) can only reduce the carbonyl via nucleophilic addition. The double bond, being electro rich, is not attacked by the hydride ion.

c)

from :

The synthesis requires substitution of the hydroxyl group of the substrate. Hydroxyl group is a poor leaving group as the hydroxide ion is a strong base. It first needs to be converted to a better leaving group. This is done by reaction with tosyl chloride, to give a tosylate.

Tosylate ion is the conjugate base of a strong acid, so it can be easily replaced by a suitable nucleophile like phthalimide.

Reaction of the tosylate with phthalimide is an ${\text{S}}_{\text{N}}\text{2}$ reaction and leads to inversion. This is followed by a reaction with hydrazine to replace the phthalimide with an amine group.

d) ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{N(CH}}_{\text{3}}{\text{)CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{NHCH}}_{\text{3}}$ and ${\text{BrCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CN}$:

The first of the designated starting compounds $\text{N-methyl-phenylmethanamine}$ is a secondary amine. It can react with one mole of an alkyl halide to form mixed tertiary amine. The second starting compound contains bromine as well as cyanide. Bromide is a much weaker base than cyanide and therefore a much better leaving group. The reaction between the two reactants therefore results in nuclephilic substitution of bromide to yield $\text{4-(benzyl(methyl)amino)butanenitrile}$.

The nitrile (cyanide) function can then be reduced with hydrogen over palladium to yield the desired product.

e)

The methyl substituent on the benzene ring needs to be converted to an amine. This requires first converting it to an alkyl halide. This is a substitution of a hydrogen in a benzylic position. A convenient method is to treat it with $\text{N-bromosuccinimide}$ in combination with a peroxide, which generates bromine radicals.

The resulting bromide can be reacted directly with dimethyl amine to yield the product. However, there is a possibility of formation of a quaternary ammonium compound. Therefore, the bromide is converted to a primary alcohol by reacting it with hydroxide.

The alcohol is then oxidized to aldehyde using pyridiniumchlorochromate.

Reaction of the aldehyde can then be subjected to reductive amination, reaction with dimethyl amine in the presence of sodium borohydride, to yield the desired product. The reaction initially forms an imine, but that is directly reduced by sodium borohydrideinstead of isolating it.