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Chapter 7.SE, Problem 46AP
Interpretation Introduction

a)

Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!), Chapter 7.SE, Problem 46AP , additional homework tip  1

Interpretation:

To assign E or Z configuration for the compound given.

Concept introduction:

The member that ranks higher can be determined by considering the atomic number of the first atom of the two substituents separately. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms away from double bond are considered until the first difference is found. Multiple bonded atoms are considered as equivalent to the same number of single bonded atoms. The isomer that has the higher ranked groups on each carbon are on the same side of the double bond is said to have Z configuration. If the higher ranked groups are on opposite sides, the alkene is said to have E configuration.

To assign:

The configuration for the compound given as E or Z.

Interpretation Introduction

b)

Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!), Chapter 7.SE, Problem 46AP , additional homework tip  2

Interpretation:

To assign E or Z configuration for the compound given.

Concept introduction:

The member that ranks higher can be determined by considering the atomic number of the first atom of the two substituents separately. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms away from double bond are considered until the first difference is found. Multiple bonded atoms are considered as equivalent to the same number of single bonded atoms. The isomer that has the higher ranked groups on each carbon are on the same side of the double bond is said to have Z configuration. If the higher ranked groups are on opposite sides, the alkene is said to have E configuration.

To assign:

The configuration for the compound given as E or Z.

Interpretation Introduction

c)

Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!), Chapter 7.SE, Problem 46AP , additional homework tip  3

Interpretation:

To assign E or Z configuration for the compound given.

Concept introduction:

The member that ranks higher can be determined by considering the atomic number of the first atom of the two substituents separately. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms away from double bond are considered until the first difference is found. Multiple bonded atoms are considered as equivalent to the same number of single bonded atoms. The isomer that has the higher ranked groups on each carbon are on the same side of the double bond is said to have Z configuration. If the higher ranked groups are on opposite sides, the alkene is said to have E configuration.

To assign:

The configuration for the compound given as E or Z.

Interpretation Introduction

d)

Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!), Chapter 7.SE, Problem 46AP , additional homework tip  4

Interpretation:

To assign E or Z configuration for the compound given.

Concept introduction:

The member that ranks higher can be determined by considering the atomic number of the first atom of the two substituents separately. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms away from double bond are considered until the first difference is found. Multiple bonded atoms are considered as equivalent to the same number of single bonded atoms. The isomer that has the higher ranked groups on each carbon are on the same side of the double bond is said to have Z configuration. If the higher ranked groups are on opposite sides, the alkene is said to have E configuration.

To assign:

The configuration for the compound given as E or Z.

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Chapter 7 Solutions

Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!)

Ch. 7.5 - Which member in each of the following sets ranks...Ch. 7.5 - Prob. 12PCh. 7.5 - Prob. 13PCh. 7.5 - Prob. 14PCh. 7.6 - Prob. 15PCh. 7.8 - Prob. 16PCh. 7.8 - Prob. 17PCh. 7.9 - Show the structures of the carbocation...Ch. 7.9 - Draw a skeletal structure of the following...Ch. 7.10 - Prob. 20PCh. 7.11 - On treatment with HBr, vinylcyclohexane undergoes...Ch. 7.SE - Prob. 22VCCh. 7.SE - Prob. 23VCCh. 7.SE - The following carbocation is an intermediate in...Ch. 7.SE - Prob. 25VCCh. 7.SE - Predict the major product and show the complete...Ch. 7.SE - Prob. 27MPCh. 7.SE - When 1, 3-butadiene reacts with one mole of HBr,...Ch. 7.SE - When methyl vinyl ether reacts with a strong acid,...Ch. 7.SE - Addition of HCl to 1-isopropylcyclohexene yields a...Ch. 7.SE - Addition of HCl to...Ch. 7.SE - Limonene, a fragrant hydrocarbon found in lemons...Ch. 7.SE - Prob. 33MPCh. 7.SE - Calculate the degree of unsaturation in the...Ch. 7.SE - Prob. 35APCh. 7.SE - Prob. 36APCh. 7.SE - Name the following alkenes:Ch. 7.SE - Draw structures corresponding to the following...Ch. 7.SE - Prob. 39APCh. 7.SE - Prob. 40APCh. 7.SE - Prob. 41APCh. 7.SE - Prob. 42APCh. 7.SE - Prob. 43APCh. 7.SE - Draw and name the 17 alkene isomers, C6H12,...Ch. 7.SE - Prob. 45APCh. 7.SE - Prob. 46APCh. 7.SE - Which of the following E, Z designations are...Ch. 7.SE - Prob. 48APCh. 7.SE - trans-2-Butene is more stable than cis-2-butene by...Ch. 7.SE - Prob. 50APCh. 7.SE - Normally, a trans alkene is more stable than its...Ch. 7.SE - trans-Cyclooctene is less stable than...Ch. 7.SE - Prob. 53APCh. 7.SE - Prob. 54APCh. 7.SE - Use Hammond’s Postulate to determine which...Ch. 7.SE - Prob. 56APCh. 7.SE - Predict the major product in each of the following...Ch. 7.SE - Prob. 58APCh. 7.SE - Prob. 59APCh. 7.SE - Prob. 60APCh. 7.SE - Allene (1,2-propadiene), H2C=C=CH2, has two...Ch. 7.SE - The heat of hydrogenation for allene (Problem...Ch. 7.SE - Retin A, or retinoic acid, is a medication...Ch. 7.SE - Prob. 64APCh. 7.SE - tert-Butyl esters [RC02C(CH3)3] are converted into...Ch. 7.SE - Vinylcyclopropane reacts with HBr to yield a...Ch. 7.SE - Prob. 67APCh. 7.SE - Prob. 68APCh. 7.SE - Prob. 69APCh. 7.SE - Prob. 70APCh. 7.SE - Prob. 71APCh. 7.SE - Reaction of 2, 3-dimethyl-l-butene with HBr leads...
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