Chemistry
Chemistry
13th Edition
ISBN: 9781259911156
Author: Raymond Chang Dr., Jason Overby Professor
Publisher: McGraw-Hill Education
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Concept explainers

Question
Chapter 24, Problem 24.74QP

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.

Chemistry, Chapter 24, Problem 24.74QP , additional homework tip  1

a)

Expert Solution
Check Mark

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:

Chemistry, Chapter 24, Problem 24.74QP , additional homework tip  2

Chemistry, Chapter 24, Problem 24.74QP , additional homework tip  3Chemistry, Chapter 24, Problem 24.74QP , additional homework tip  4

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.

Chemistry, Chapter 24, Problem 24.74QP , additional homework tip  5

b)

Expert Solution
Check Mark

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:

Chemistry, Chapter 24, Problem 24.74QP , additional homework tip  6

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.

c)

Expert Solution
Check Mark

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.

d)

Expert Solution
Check Mark

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 × volume

e)

Expert Solution
Check Mark

Explanation of Solution

Given: molarity of Na2S2O3 is 0.142 M; volume is 20.6ml

Number of moles of Na2S2O3 reacted is:

20.6ml×1L1000mL×0.142molNa2S2O31L=2.9252×103molNa2S2O3

The mol ratio between I2 and Na2S2O3 is 1:2. The number of grams of I2 leftover is:

(2.93×10-3molNa2S2O3)×1molI22molNa2S2O3×253.8gI21molI2=0.372gI2

Number of grams of I2 reacted is: (43.8-0.372)g=43.4gI2

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

Iodinenumber=43.4gI235.3gcornoil×100gcornoil=123

Hence, Iodine number calculated is 123

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Chapter 24 Solutions

Chemistry

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