EBK ORGANIC CHEMISTRY-PRINT COMPANION (
EBK ORGANIC CHEMISTRY-PRINT COMPANION (
4th Edition
ISBN: 9781119776741
Author: Klein
Publisher: WILEY CONS
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Chapter 1, Problem 32PP

a.

Interpretation Introduction

Interpretation: All the constitutional isomers that have molecular formulas C2H5Cl should be drawn.

Concept Introduction: The molecules that possess the same molecular formula but differ in the structural arrangement of atoms in the molecule are said to be isomers of each other.

a.

Expert Solution
Check Mark

Answer to Problem 32PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  1

Explanation of Solution

For the first isomer structure, all two-carbon atoms are written in a straight chain and the chlorine atom is bonded to one of the carbon atoms as:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  2

There is no other possibility for the different structural formula C2H5Cl so, the above-mentioned structure is the only isomer.

b.

Interpretation Introduction

Interpretation: All the constitutional isomers that have molecular formulas C2H4Cl2 should be drawn.

Concept Introduction: The molecules that possess the same molecular formula but differ in the structural arrangement of atoms in the molecule are said to be isomers of each other.

b.

Expert Solution
Check Mark

Answer to Problem 32PP

Isomer I:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  3

Isomer II:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  4

Explanation of Solution

The different structural the arrangement of atoms in the molecule C2H4Cl2 are:

  1. All two-carbon atoms are written in a straight chain and two chlorine atoms are bonded to one of the carbon atoms resulting in:
  2. Isomer I:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  5

  3. Interchanging the position of one hydrogen atom at first carbon with one chlorine atom of the second carbon atom.
  4. Isomer II:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  6

c.

Interpretation Introduction

Interpretation: All the constitutional isomers that have molecular formulas C2H3Cl3 should be drawn.

Concept Introduction: The molecules that possess the same molecular formula but differ in the structural arrangements of atoms in the molecule are said to be isomers of each other.

c.

Expert Solution
Check Mark

Answer to Problem 32PP

Isomer I:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  7

Isomer II:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  8

Explanation of Solution

The different structural arrangement of atoms in the molecule C2H3Cl3 are:

  1. All two-carbon atoms are written in a straight chain and three chlorine atoms are bonded to one of the carbon atoms resulting in:
  2. Isomer I:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  9

  3. Interchanging the position of one hydrogen atom at first carbon with one chlorine atom of the second carbon atom.
  4. Isomer II:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  10

d.

Interpretation Introduction

Interpretation: All the constitutional isomers that have molecular formulas C6H14 should be drawn.

Concept Introduction: The molecules that possess the same molecular formula but differ in the structural arrangements of atoms in the molecule are said to be isomers of each other.

d.

Expert Solution
Check Mark

Answer to Problem 32PP

Isomer I:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  11

Isomer II:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  12

Isomer III:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  13

Isomer IV:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  14

Isomer V:

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  15

Explanation of Solution

The different structural arrangement of atoms in the molecule C6H14 are:

  1. All six-carbon atoms are written in a straight chain resulting in:
  2. Isomer I:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  16

  3. Five-carbon atoms are written in a straight chain and a methyl group is attached to the second carbon resulting in:
  4. Isomer II:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  17

  5. Five-carbon atoms are written in a straight chain and a methyl group is attached to the third carbon resulting in:
  6. Isomer III:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  18

  7. Four-carbon atoms are written in a straight chain and a methyl group is attached to the second and third carbon resulting in:
  8. Isomer IV:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  19

  9. Four-carbon atoms are written in a straight chain and two methyl groups are attached to the second carbon resulting in:
  10. Isomer V:

      EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 1, Problem 32PP , additional homework tip  20

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