Chapter 11, Problem 11.74SP

a)

Interpretation Introduction

Interpretation:

A reaction that leads to the formation of triglyceride, starting with glycerol and carboxylic acids has to be suggested.

Concept introduction:

Ester formation reaction: Reaction of alcohol and carboxylic acid using acid catalyst results the ester formation with the elimination of water molecule.

Explanation of Solution

The hydroxyl group act as nucleophile and the carboxylic group act as electrophile in presence of acid catalyst; the nucleophile attack at electrophilic carbon of carboxylic acid leads to the formation of ester with the elimination of water molecule.

Mechanism of condensation reaction:

As shown above, the successive steps lead to the formation of triglycerides containing three ester group with the elimination of three water molecules.

b)

Interpretation Introduction

Interpretation:

An equation for the base hydrolysis of ester has to be written.

Concept introduction:

Ester formation reaction: Reaction of alcohol and carboxylic acid using acid catalyst results the ester formation with the elimination of water molecule.

Explanation of Solution

The hydroxyl group acts as nucleophile and the carbonyl carbon act as electrophile; the nucleophile attack at electrophilic carbon of ester leads to the formation of alcohol with the elimination of fatty acid salts (soap).

Base hydrolysis of Esters:

c)

Interpretation Introduction

Interpretation:

Difference between fats and oils has to be explained.

Concept introduction:

Melting point: At temperature begins the solid to melt.

Unsaturation bonds: The presence of double or triple bonds in the molecules.

Explanation of Solution

The presence of unsaturated bonds in the molecules tight close packing will be less due to bend of double bonds and the intermolecular attraction between them is less and less energy is required to overcome the interaction. More the double bonds lower the intermolecular interaction. Hence, the melting point decreases.

d)

Interpretation Introduction

Interpretation:

Reagent and catalyst used in hydrogenation process has to be identified.

Concept introduction:

Hydrogenation of alkene: Two hydrogen atoms are added across the double bond of an alkene resulting alkane product.

Homogeneous catalyst: Catalyst used is in same phase as the reactants.

Heterogeneous catalyst: Catalyst used is in different phase as the reactants.

Explanation of Solution

Liquid oil is obtained from plants, having double bonds the presence of reactive double bond is converted into single bonds in order to solidify. Hydrogenation of double bonds is the process in which hydrogen molecule is added across the double bond forming alkane product. The alkane is highly facilitated for close packing and solidifies the oil.

Reaction carried out is hydrogenation reaction; hydrogen molecule is the reagent used in presence of either heterogeneous or homogeneous catalyst.

e)

Interpretation Introduction

Interpretation:

Iodine number has to be calculated.

Concept introduction:

Iodine number: number of grams of Iodine that react with given quantity of oil is called Iodine number.

Number of moles = Molarity $\text{×}$ volume

Explanation of Solution

Given: molarity of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ is $\text{0}\text{.142}$ M; volume is $\text{20}\text{.6ml}$

Number of moles of ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ reacted is:

$\begin{array}{l}\text{20}\text{.6ml×}\frac{\text{1L}}{\text{1000mL}}\text{×}\frac{\text{0}\text{.142}\text{mol}{\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{1}\text{L}}\\ \\ \text{=}\text{2}\text{.9252}\text{×}{\text{10}}^{-3}\text{mol}{\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$

The mol ratio between ${\text{I}}_2$ and ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ is 1:2. The number of grams of ${\text{I}}_2$ leftover is:

$\begin{array}{l}\left(\text{2}{\text{.93×10}}^{\text{-3}}\text{mol}{\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\right)\text{×}\frac{\text{1}\text{mol}{\text{I}}_{\text{2}}}{\text{2}\text{mol}{\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{\text{253}\text{.8}\text{g}{\text{I}}_{\text{2}}}{\text{1}\text{mol}{\text{I}}_{\text{2}}}\\ \\ \text{=}\text{0}\text{.372}\text{g}{\text{I}}_{\text{2}}\end{array}$

Number of grams of ${\text{I}}_2$ reacted is: $\left(\text{43}\text{.8}\text{-}\text{0}\text{.372}\right)\text{g}\text{=}\text{43}\text{.4}\text{g}{\text{I}}_2$

The iodine number is the number of grams of iodine that reacts with 100 g of corn oil.

$\begin{array}{l}Iodinenumber=\frac{\text{43}\text{.4}\text{g}{\text{I}}_2}{35.3gcornoil}\times 100gcornoil\\ \\ =123\end{array}$

Hence, Iodine number calculated is 123