Organic Chemistry
Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 2, Problem 2.45AP
Interpretation Introduction

(a)

Interpretation:

The molecular formula of Y is to be stated for the given mass percent of carbon and hydrogen.

Concept introduction:

The combustion is the process of heating the compounds in the presence of oxygen. The combustion of alkanes yields carbon dioxide and water. In the process of combustion of alkanes a high amount of heat is evolved. These products are given by all kind of alkanes generally.

Expert Solution
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Answer to Problem 2.45AP

The molecular formula of Y for the given mass percent of carbon and hydrogen is C8H14

Explanation of Solution

The moles of H present in X per mole of C can be calculated by finding out the atomic ratio and simple ratio. The atomic ratio can be calculated by taking the ratio of percent composition of element to its atomic weight. The percent composition for carbon and hydrogen is 87.17% and 12.83% respectively. The atomic mass of carbon and hydrogen is 12g and 1g respectively. Thus, the atomic ratio is calculated as,

Atomicratioofelement=Percent composition ofelementAtomicmassofelement …(1)

Substitute the values in equation (1) for carbon as,

AtomicratioofC=Percent composition ofCAtomicmassofC=87.1712=7.2

Substitute the values in equation (1) for hydrogen as,

AtomicratioofH=Percent composition ofHAtomicmassofH=12.831=12.83

The simplest ratio is the ratio of atomic ratio of carbon and hydrogen by atomic ratio of carbon. It is calculated to get the number of moles of element in the compound. The simplest ratio for carbon is as follows.

SimplestratioofC=AtomicratioofCAtomicratioofC=7.27.2=1

The simplest ratio for hydrogen is as follows.

SimplestratioofH=AtomicratioofHAtomicratioofC=12.837.2=1.7

Thus, the number of moles of H present in X per mole of C is 1.7 and the formula can be expressed as C1H1.7.

The formula obtained on multiplying the number of moles in C1H1.7 by 4 is C4H6.8. In the above formula, number of moles of hydrogen is approximately equal to 7. Thus, the number of moles of carbon and hydrogen is 4 and 7 respectively in the compound Y. But it is given that the compound Y contains even number of hydrogen’s, therefore the formula of the compound Y is C8H14.

Conclusion

The molecular formula of Y for the given mass percent of carbon and hydrogen is C8H14

Interpretation Introduction

(b)

Interpretation:

The structures of the alkane obtained in (a) having two tertiary carbons and all other carbons secondary are to be drawn.

Concept introduction:

The carbons in a chemical formula may or may not be branched. The branched carbons are categorized as primary, secondary, tertiary, and quaternary carbons depending on the substituent attached to it. The branched carbons are categorized to understand the chemical reactions in which these carbons are involved.

The hydrogen’s attached to branched carbons are also categorized as primary, secondary and tertiary hydrogen’s.

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Answer to Problem 2.45AP

The structures of the alkane obtained in (a) having two tertiary carbons and all other carbons secondary are shown below as,

Organic Chemistry, Chapter 2, Problem 2.45AP , additional homework tip  1

Explanation of Solution

A primary carbon is the one which is attached to only one carbon. Similarly, the secondary, tertiary, and quaternary carbons are attached to 2,3 and 4 carbons respectively. The cycloalkane has secondary carbons and thus, contains secondary hydrogen’s. Thus, the possible compound is a cycloalkane. The formula obtained in part (a) is C8H14. Thus, the possible cycloalkanes having two tertiary carbons and all other carbons secondary with formula C8H14 can be drawn as shown in figure 1.

Organic Chemistry, Chapter 2, Problem 2.45AP , additional homework tip  2

Figure 1

Conclusion

The structures of the alkane obtained in (a) having two tertiary carbons and all other carbons secondary are shown in figure 1.

Interpretation Introduction

(c)

Interpretation:

The structure of the alkane obtained in (a) having no primary hydrogen’s, no tertiary carbon atoms, and one quaternary carbon atom is to be drawn.

Concept introduction:

The carbons in a chemical formula may or may not be branched. The branched carbons are categorized as primary, secondary, tertiary, and quaternary carbons depending on the substituent attached to it. The branched carbons are categorized to understand the chemical reactions in which these carbons are involved.

The hydrogen’s attached to branched carbons are also categorized as primary, secondary and tertiary hydrogen’s.

Expert Solution
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Answer to Problem 2.45AP

The structure of the alkane obtained in (a) having no primary hydrogen’s, no tertiary carbon atoms, and one quaternary carbon atom is shown below as,

Organic Chemistry, Chapter 2, Problem 2.45AP , additional homework tip  3

Explanation of Solution

A primary carbon is the one which is attached to only one carbon. Similarly, the secondary, tertiary, and quaternary carbons are attached to 2,3 and 4 carbons respectively. The cycloalkane has secondary carbons and thus, contains secondary hydrogen’s. Thus, the possible compound is a cycloalkane. The formula obtained in part (a) is C8H14. Thus, the possible cycloalkane having no primary hydrogen’s, no tertiary carbon atoms, and one quaternary carbon atom with formula C8H14 can be drawn as shown in figure 2.

Organic Chemistry, Chapter 2, Problem 2.45AP , additional homework tip  4

Figure 2

Conclusion

The structure of the alkane obtained in (a) having no primary hydrogen’s, no tertiary carbon atoms, and one quaternary carbon atom is shown in figure 2.

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