Chapter 25, Problem 28P

Interpretation Introduction

Interpretation:

Using octadecanoic (stearic) acid and any necessary organic and inorganic reagents, an efficient synthesis for each compound is to be described.

Concept introduction:

The reduction of carboxylic acid gives primary alcohol by using reducing agent lithium aluminum hydride $\left({\text{LIAlH}}_{\text{4}}\right)$.

The primary alcohol on oxidation with pyridinium dichromate $\left(\text{PDC}\right)$ or pyridinium chlorochromate $\left(\text{PCC}\right)$ in dichloromethane gives aldehyde.

In Clemmensen reduction, the carbonyl group (aldehyde or ketone) is converted to methylene and the reagent used iszinc–mercury amalgam in concentrated hydrochloric acid, $\text{Zn(Hg)/HCl}$.

The esters can be synthesized by acid catalyzed condensation of carboxylic acid with alcohol.

The ester on reaction with one molar equivalent of Grignard’s reagent in diethyl ether gives ketone by carbon-carbon bond formation.

The ester on reaction with two molar equivalents of Grignard’s reagent in diethyl ether gives tertiary alcohol.

The dehydration of alcohol is the loss of ${\text{H}}_{\text{2}}\text{O}$ molecule to form a corresponding alkene. The $\text{H}$ and $\text{OH}$ eliminates from the adjacent carbons.

The alcohol on acid catalyzed dehydration gives corresponding alkene.

The alkene on hydrogenation with the catalyst undergoes addition of hydrogen across the double bond and forms an alkane.

The primary amine can be prepared by the acylation of ammonia.

The secondary amide can be prepared by the nucleophilic substitution of acyl chloride by amine. The two moles amines used with one mole of acyl chloride, because one amine molecule acts as a nucleophile and second acts as a Brønsted base.

The carboxylic acids on reaction with thionyl chloride forms acyl chloride by replacing the hydroxyl group of carboxylic acid with chlorine atom.

The primary amide on reduction with lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$ yields primary amine.

The reaction of thionyl chloride with alcohol gives alkyl halide.

The reaction of alkyl halide with sodium cyanide gives alkyl cyanide.

The cyanide (nitrile) can be reduced to primary amine using lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$.

The alkyl bromide can be prepared by the reaction of alcohol with phosphorus tribromide $\left({\text{PBr}}_{\text{3}}\right)$.

Grignard reagents are prepared by the reaction of the magnesium metal with an alkyl or aryl halide usually in diethyl ether as the solvent.

The potassium or sodium dichromate in presence of strong acid forms chromic acid which is a good oxidizing agent, in hydrous medium oxidizes primary alcohol to carboxylic acid.

The epoxide on treatment with Grignard reagent undergoes epoxide ring opening by forming a corresponding alcohol.

Answer to Problem 28P

Solution:

a)

b)

c)

d)

e)

f)

Explanation of Solution

The structure of octadecanoic (stearic) acid is shown below:

a) Octadecane

The synthesis of octadecane from octadecanoic acid can be done by following reactions sequence:

The octadecanoic acid on reaction with lithium aluminum hydride in aqueous medium reduced to octadecanol which is further on oxidation with pyridinium dichromate $\left(\text{PDC}\right)$ in dichloromethane gave octadecanal. The carbonyl group of octadecanal reduced to methylene by Clemmensen reduction using the reagents zinc–mercury amalgam in concentrated hydrochloric acid, $\text{Zn(Hg)/HCl}$ and formed octadecane.

b)$\text{1-Phenyloctadecane}$

The synthesis of $\text{1-Phenyloctadecane}$ from octadecanoic acid shown in following sequence:

The octadecanoic acid first converted to an ester by reacting it with ethanol in acidic condition. The ester formed is then reacted with a Grignard’s reagent phenylmagnesium bromide $\left(\text{PhMgBr}\right)$ to form $\text{1-phenyloctadecanone}$. The $\text{1-phenyloctadecanone}$ is subjected to Clemmensen reduction using the reagents zinc–mercury amalgam in concentrated hydrochloric acid, $\text{Zn(Hg)/HCl}$ to produced $\text{1-Phenyloctadecane}$.

c) $\text{3-Ethylicosane}$

The synthesis of $\text{3-Ethylicosane}$ from octadecanoic acid can be done by following reactions sequence:

The octadecanoic acid is first converted to an ester by reacting it with ethanol in acidic condition. The ester formed is then reacted with a Grignard’s reagent ethyl bromide bromide $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}\right)$ which produced a tertiary alcohol. This tertiary alcohol on heating in presence of strong acid dehydrated to an alkene. The alkene on hydrogenation in presence of nickel catalyst formed a desired product of $\text{3-Ethylicosane}$.

d) Icosanoic acid

The synthesis of Icosanoic acid from octadecanoic acid can be done by following reactions sequence:

In the first step, the octadecanoic acid is reduced to primary alcohol by reducing agent lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$. The primary alcohol formed then reacted with the phosphorus tribromide $\left({\text{PBr}}_{\text{3}}\right)$ which gave a primary octadecanyl bromide. The primary octadecanyl bromide is converted to a Grignard’s reagent by reacting it with magnesium metal in a solvent diethyl ether. This Grignard’s reagent is used in ring opening of epoxide in diethyl ether and acid and gave the product of primary alcohol, which is on oxidation with chromic acid (prepared by using reagent potassium dichromate in strong acid) formed a desired product Icosanoic acid having two additional carbons than that of octadecanoic acid.

e) $\text{1-Octadecanamine}$

The synthesis of $\text{1-Octadecanamine}$ from octadecanoic acid can be done by following reactions sequence:

The octadecanoic acid on reaction with thonyl chloride in presence of pyridine gave the product of acyl chloride. The acyl chloride is converted to $\text{1-Octadecanamide}$ by using two moles of ammonia. The ammonium chloride is also formed as byproduct. The $\text{1-Octadecanamide}$ is then reduced by lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$ in ditheryl ether followed by the hydrolysis formed a desired product $\text{1-Octadecanamine}$.

f) $\text{1-Nonadecanamine}$

The synthesis of $\text{1-Nonadecanamine}$ from octadecanoic acid can be done by following reactions sequence:

In the first step, the octadecanoic acid is reduced to primary alcohol by reducing agent lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$. The primary alcohol formed then reacted with thionyl chloride $\left({\text{SOCl}}_{\text{2}}\right)$ in pyridine which gave a primary octadecanyl chloride. The chlorine atom in a primary octadecanyl chloride is substituted by a cyanide group $\text{CN}$ to form octadecanyl cyanide by the treatment of sodium cyanide on primary octadecanyl chloride. This octadecanyl cyanide is reduced to $\text{1-Nonadecanamine}$ by using a reducing agent lithium aluminum hydride $\left({\text{LiAlH}}_{\text{4}}\right)$.