ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
6th Edition
ISBN: 9781319306977
Author: LOUDON
Publisher: INTER MAC
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Chapter 2, Problem 2.26AP
Interpretation Introduction

(a)

Interpretation:

The isomeric structures of octane having five carbons in their principal chains are to be drawn. The naming for all the structures is to be stated.

Concept introduction:

The naming of the chemical compound is done using the parameters given by IUPAC. IUPAC stands for International Union of Pure and Applied Chemistry. This system of nomenclature is accepted worldwide. The IUPAC system provides the set of rules in order to do the naming of the chemical compounds.

Isomers are the chemical compounds which have same chemical formula but different arrangement of molecules.

Expert Solution
Check Mark

Answer to Problem 2.26AP

The isomeric structures of octane having five carbons in their principal chains along with their names are drawn below as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  1

Explanation of Solution

The structure of octane can be drawn as shown in figure 1.

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  2

Figure 1

The isomers of the octane having five carbons in their principal chains can be drawn by rearranging the carbon atoms.

The first isomer is shown in figure 2 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  3

Figure 2

For the naming of the compound, the longest carbon chain is selected first. In the given compound the longest carbon chain has 5 carbon atoms; therefore, it is pentane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atoms are numbered as 2,3. The group attached is the methyl group at C-2 and ethyl group at C-3. Thus, the name of the compound is 3-ethyl-2-methylpentane.

The second isomer is shown in figure 3 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  4

Figure 3

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 5 carbon atoms, therefore, it is pentane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 2,4. The groups attached are the methyl groups at C-2 and C-4. Thus, the name of the compound is 2, 2, 4-trimethylpentane.

The third isomer is shown in figure 4 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  5

Figure 4

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 5 carbon atoms; therefore, it is pentane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 2,3. The groups attached are the methyl groups at C-2 and C-3. Thus, the name of the compound is 2, 3, 3-trimethylpentane.

The fourth isomer is shown in figure 5 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  6

Figure 5

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 5 carbon atoms; therefore, it is pentane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 2,3. The groups attached are the methyl groups at C-2 and C-3. Thus, the name of the compound is 2, 2, 3-trimethylpentane.

The fifth isomer is shown in figure 6 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  7

Figure 6

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 5 carbon atoms, therefore, it is pentane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 3. The group attached is the methyl group at C-3 and ethyl group at C-3. Thus, the name of the compound is 3-ethyl-3-methylpentane.

The sixth isomer is shown in figure 7 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  8

Figure 7

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 5 carbon atoms; therefore, it is pentane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 2,3,4. The groups attached are the methyl group at C-2, C-3 and C-4. Thus, the name of the compound is 2, 3, 4-trimethylpentane.

Conclusion

The isomeric structures of octane having five carbons in their principal chains along with their names are shown in figure 2 to 7.

Interpretation Introduction

(b)

Interpretation:

The isomeric structures of octane having six carbons in their principal chains are to be drawn. The naming for all the structures is to be stated.

Concept introduction:

The naming of the chemical compound is done using the parameters given by IUPAC. IUPAC stands for International Union of Pure and Applied Chemistry. This system of nomenclature is accepted worldwide. The IUPAC system provides the set of rules in order to do the naming of the chemical compounds.

Isomers are the chemical compounds which have same chemical formula but different arrangement of molecules.

Expert Solution
Check Mark

Answer to Problem 2.26AP

The isomeric structures of octane having six carbons in their principal chains along with their names are drawn below as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  9

Explanation of Solution

The isomers of the octane having six carbons in their principal chains can be drawn by rearranging the carbon atoms.

The first isomer is shown in figure 8 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  10

Figure 8

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 6 carbon atoms; therefore, it is hexane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atoms are numbered as 2,3. The groups attached are the methyl groups at C-2 and C-3. Thus, the name of the compound is 2, 3-dimethylhexane.

The second isomer is shown in figure 9 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  11

Figure 9

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 6 carbon atoms; therefore, it is hexane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atoms are numbered as 2,4. The groups attached are the methyl groups at C-2 and C-4. Thus, the name of the compound is 2, 4-dimethylhexane.

The third isomer is shown in figure 10 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  12

Figure 10

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 6 carbon atoms; therefore, it is hexane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atoms are numbered as 2,5. The groups attached are the methyl groups at C-2 and C-5. Thus, the name of the compound is 2, 5-dimethylhexane.

The fourth isomer is shown in figure 11 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  13

Figure 11

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 6 carbon atoms; therefore, it is hexane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 2. The groups attached are the methyl groups at C-2. Thus, the name of the compound is 2, 2-dimethylhexane.

The fifth isomer is shown in figure 12 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  14

Figure 12

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest carbon chain has 6 carbon atoms; therefore, the root word is hexane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 3. The groups attached are the methyl groups at C-3. Thus, the name of the compound is 3, 3-dimethylhexane.

The sixth isomer is shown in figure 13 as,

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX, Chapter 2, Problem 2.26AP , additional homework tip  15

Figure 13

For the naming of the compound the longest carbon chain is selected first. In the given compound the longest chain carbon has 6 carbon atoms, therefore, it is hexane. The branched groups are then arranged in the order that they get the lowest possible number. In the given compound the branched carbon atom is numbered as 3. The group attached is the ethyl group at C-3. Thus, the name of the compound is 3-ethylhexane.

Conclusion

The isomeric structures of octane having six carbons in their principal chains along with their names are shown in figure 8 to 13.

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