Chapter E, Problem 1PP

Interpretation Introduction

Interpretation:

The structural formulas involved in the sequence of given reactions for the formation of adipic acid or hexamethylene diamine are to be written, with given molecular formula of each compound and IR data of some compound.

Concept introduction:

A polymer is a large molecule or macromolecule composed of many repeated subunits.

Polymerization is a process of reacting monomer molecules together in a chemical reaction to form polymer chains or three-dimensional networks.

Nylon is a polymer that consists of adipic acid and hexamethylene diamine as monomer units.

Nylon $6,6$ is formed by the reaction of six carbon dicarboxylic acid that is adipic acid with a six carbon diamine, that is, hexamethylene diamine (hexane $1,6$ diamine).

The formation of Nylon $6,6$ involves the loss of water molecules due to the condensation reaction between $\begin{array}{l}-\text{C=O}\\ \text{|}\\ \text{O-}\end{array}$ group and $-{\text{NH}}_{\text{3}}{}^{\text{+}}$ group of the salt to give polyamide.

The addition of hydrogen to double or triple bond is a hydrogenation reaction.

Hydrogenation of alkenes or alkynes is an addition reaction, which is done in the presence of reducing agents, such as ${\text{H}}_2/Ni$.

Infrared spectroscopy is a simple, instrumental technique, which helps to determine the presence of various functional groups.

It depends on the interactions of atoms or molecules with the electromagnetic radiation.

The molecules that have dipole moment are IR active and the molecules that do not have dipole moment are IR inactive.

Cyclohexanol shows IR peak at $3300{\text{ cm}}^{\text{-1}}$

Cyclic ketone shows IR peak at ${\text{1714 cm}}^{\text{-1}}$.

Answer to Problem 1PP

Solution:

(a) The structures involved in the sequence are as follows:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$-, ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}\text{O}$, ${\text{C}}_{\text{6}}{\text{H}}_{\text{10}}\text{O}$

(b) The structures involved in the sequence are as follows:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}$ , ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}$

(c) The structures involved in the sequence are as follows:

${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{Cl}}_{\text{2}}$ , ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{N}}_{\text{2}}$,

${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}$

(d) The structures involved in the sequence are as follows:

${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{Cl}}_{\text{2}}$ , ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}$

Explanation of Solution

a)

$\begin{array}{l}\text{Benzene }\stackrel{{\text{H}}_{\text{2}}\text{, Cat}\text{.}}{\to }\text{}\text{}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}\stackrel{{\text{O}}_{\text{2}}\text{, Cat}\text{.}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O + C}}_{\text{6}}{\text{H}}_{\text{10}}\text{O}\stackrel{{\text{HNO}}_{\text{2}}\text{, Cat}\text{.}}{\to }{\text{Adipic acid + N}}_{\text{2}}\text{O}\\ \text{ IR:}\sim {\text{3300cm}}^{-\text{1}}\text{IR:}\sim {\text{1714cm}}^{-\text{1}}\\ \text{ (broad)}\\ (\text{as a mixture)}\end{array}$

For the formation of adipic acid, first, benzene undergoes hydrogenation reaction to form cyclohexane, which then undergoes oxidation in the presence of catalyst to give cyclohexanol ( I.R. peak at $3300{\text{ cm}}^{\text{-1}}$) and cyclic ketone ( I.R. peak at ${\text{1714 cm}}^{\text{-1}}$). Then, these two compounds react in the presence of nitric acid to form adipic acid. The sequence of reactions from cyclohexane can be written as

Hence, adipic acid is formed in this sequence of reactions.

b) $\text{Adipic acid}\stackrel{{\text{2NH}}_{\text{3}}}{\to }\text{}\text{}\text{a salt }\stackrel{\text{heat}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\underset{\text{catalyst}}{\overset{{\text{350}}^{\text{0}}\text{C}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}\underset{\text{catalyst}}{\overset{{\text{4H}}_{\text{2}}}{\to }}\text{hexamethylenediamine}$

Here, adipic acid reacts with two molecules of ammonia to form ammonium salt of adipic acid. Then, on heating, this salt loses two molecules of water to form ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}$(contains two $-{\text{CONH}}_{\text{2}}$), after that, this molecule is treated with a catalyst under high temperature so as to convert $-{\text{NH}}_{\text{2}}$ group to $-\text{CN}$ group resulting in the formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}$. Then, this compound (${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}$) reacts with hydrogen gas in the presence of catalyst to form hexamethylene diamine.

The whole sequence can be written as

Hence, hexamethylene diamine is formed in this sequence of reactions.

c)

$\text{1,3-Butadiene}\stackrel{{\text{Cl}}_2}{\to }\text{}\text{}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{Cl}}_{\text{2}}\stackrel{\text{2NaCN}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{N}}_{\text{2}}\underset{\text{Hi}}{\overset{{\text{H}}_{\text{2}}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}\underset{\text{catalyst}}{\overset{{\text{4H}}_{\text{2}}}{\to }}\text{hexamethylenediamine}$

Here, $1,3$ butadiene undergoes chlorination reaction to form ${\text{C}}_4{\text{H}}_6{\text{Cl}}_{\text{2}}$. Then, the compound formed reacts with two molecules of sodium cyanide, which results in the replacement of two $-\text{Cl}$ groups from two cyanide $-\text{CN}$ groups resulting in the formation of $-{\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{N}}_{\text{2}}$. This compound undergoes hydrogenation in the presence of catalyst to form hexamethylenediamine.

The whole sequence can be written as

Hence, hexamethylene diamine is formed in this sequence of reactions.

d)

$\text{Tetrahydrofuran}\stackrel{\text{2HCl}}{\to }\text{}\text{}{\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{Cl}}_{\text{2}}\stackrel{\text{2NaCN}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}\underset{\text{catalyst}}{\overset{{\text{4 H}}_{\text{2}}}{\to }}\text{hexamethylenediamine}$

Here, tetrahydrofuran undergoes chlorination in the presence of hydrochloric acid resulting in the formation of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{Cl}}_{\text{2}}$. Then, this ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{Cl}}_{\text{2}}$ compound reacts with two molecules of sodium cyanide, which results in the replacement of two $-\text{Cl}$ groups from two cyanide $-\text{CN}$ groups resulting in the formation of $-{\text{C}}_{\text{6}}{\text{H}}_8{\text{N}}_{\text{2}}$. This compound undergoes hydrogenation in the presence of catalyst to form hexamethylenediamine.

The whole sequence can be written as

Hence, hexamethylene diamine is formed in this sequence of reactions.