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Organic Chemistry I: 5.2 Geometric Isomers and E/Z Naming System

Organic Chemistry I
5.2 Geometric Isomers and E/Z Naming System
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table of contents
  1. Cover
  2. Title Page
  3. Copyright
  4. Table Of Contents
  5. Introduction
  6. Acknowledgements
  7. Chapter 1 Basic Concepts in Chemical Bonding and Organic Molecules
    1. 1.1 Chemical Bonding
    2. 1.2 Lewis Structure
    3. 1.3 Resonance Structures
    4. 1.4 Resonance structures in Organic Chemistry
    5. 1.5 Valence-Shell Electron-Pair Repulsion Theory (VSEPR)
    6. 1.6 Valence Bond Theory and Hybridization
    7. Answers to Practice Questions Chapter 1
  8. Chapter 2 Fundamental of Organic Structures
    1. 2.1 Structures of Alkenes
    2. 2.2 Nomenclature of Alkanes
    3. 2.3 Functional Groups
    4. 2.4 IUPAC Naming of Organic Compounds with Functional Groups
    5. 2.5 Degree of Unsaturation/Index of Hydrogen Deficiency
    6. 2.6 Intermolecular Force and Physical Properties of Organic Compounds
    7. Answers to Practice Questions Chapter 2
  9. Chapter 3 Acids and Bases: Organic Reaction Mechanism Introduction
    1. 3.1 Review of Acids and Bases and Ka
    2. 3.2 Organic Acids and Bases and Organic Reaction Mechanism
    3. 3.3 pKa of Organic Acids and Application of pKa to Predict Acid-Base Reaction Outcome
    4. 3.4 Structural Effects on Acidity and Basicity
    5. 3.5 Lewis Acids and Lewis Bases
    6. Answers to Practice Questions Chapter 3
  10. Chapter 4 Conformations of Alkanes and Cycloalkanes
    1. 4.1 Conformation Analysis of Alkanes
    2. 4.2 Cycloalkanes and Their Relative Stabilities
    3. 4.3 Conformation Analysis of Cyclohexane
    4. 4.4 Substituted Cyclohexanes
    5. Answers to Practice Questions Chapter 4
  11. Chapter 5 Stereochemistry
    1. 5.1 Summary of Isomers
    2. 5.2 Geometric Isomers and E/Z Naming System
    3. 5.3 Chirality and R/S Naming System
    4. 5.4 Optical Activity
    5. 5.5 Fisher Projection
    6. 5.6 Compounds with More Than One Chirality Centers
    7. Answers to Practice Questions Chapter 5
  12. Chapter 6 Structural Identification of Organic Compounds: IR and NMR Spectroscopy
    1. 6.1 Electromagnetic Radiation and Molecular Spectroscopy
    2. 6.2 Infrared (IR) Spectroscopy Theory
    3. 6.3 IR Spectrum and Characteristic Absorption Bands
    4. 6.4 IR Spectrum Interpretation Practice
    5. 6.5 NMR Theory and Experiment
    6. 6.6 ¹H NMR Spectra and Interpretation (Part I)
    7. 6.7 ¹H NMR Spectra and Interpretation (Part II)
    8. 6.8 ¹³C NMR Spectroscopy
    9. 6.9 Structure Determination Practice
    10. Answers to Practice Questions Chapter 6
  13. Chapter 7 Nucleophilic Substitution Reactions
    1. 7.1 Nucleophilic Substitution Reaction Overview
    2. 7.2 SN2 Reaction Mechanism, Energy Diagram and Stereochemistry
    3. 7.3 Other Factors that Affect SN2 Reactions
    4. 7.4 SN1 Reaction Mechanism, Energy Diagram and Stereochemistry
    5. 7.5 SN1 vs SN2
    6. 7.6 Extra Topics on Nucleophilic Substitution Reaction
    7. Answers to Practice Questions Chapter 7
  14. Chapter 8 Elimination Reactions
    1. 8.1 E2 Reaction
    2. 8.2 E1 Reaction
    3. 8.3 E1/E2 Summary
    4. 8.4 Comparison and Competition Between SN1, SN2, E1 and E2
    5. Answers to Practice Questions Chapter 8
  15. Chapter 9 Free Radical Substitution Reaction of Alkanes
    1. 9.1 Homolytic and Heterolytic Cleavage
    2. 9.2 Halogenation Reaction of Alkanes
    3. 9.3 Stability of Alkyl Radicals
    4. 9.4 Chlorination vs Bromination
    5. 9.5 Stereochemistry for Halogenation of Alkanes
    6. 9.6 Synthesis of Target Molecules: Introduction of Retrosynthetic Analysis
    7. Answers to Practice Questions Chapter 9
  16. Chapter 10 Alkenes and Alkynes
    1. 10.1 Synthesis of Alkenes
    2. 10.2 Reactions of Alkenes: Addition of Hydrogen Halide to Alkenes
    3. 10.3 Reactions of Alkenes: Addition of Water (or Alcohol) to Alkenes
    4. 10.4 Reactions of Alkenes: Addition of Bromine and Chlorine to Alkenes
    5. 10.5 Reaction of Alkenes: Hydrogenation
    6. 10.6 Two Other Hydration Reactions of Alkenes
    7. 10.7 Oxidation Reactions of Alkenes
    8. 10.8 Alkynes
    9. Answers to Practice Questions Chapter 10
  17. About the Author

5.2 Geometric Isomers and E/Z Naming System

Geometric Isomers of Alkenes

In the discussions about 1,2-dimethylcyclohexane in Chapter 4, we have learned that there are two geometric isomers possible for that compound, that are cis and trans. The restricted C-C bond rotation of cyclic structure result in the cis or trans isomer of 1,2-dimethylcyclohexane. Restricted rotation also can be caused by a double bond, so  geometric isomers apply to some alkenes as well.

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Figure 5.2a Geometric isomers of disubstituted cycloalkanes

For the example of 2-butene, the condensed structural formula CH3-CH=CH-CH3 does not really represent the trigonal planar shape of the sp2 carbons with double bonds. To show the shape explicitly, we need to draw the Kekulé structure that show all the bond angles. Then it will be noticed that there are two different shapes of 2-butene, with the CH3 groups on either the same side or opposite side of the double bond.

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Figure 5.2b Geometric isomers of some alkenes

They are geometric isomers and can be labelled as cis or trans in a similar way as disubstituted cycloalkane. Cis/trans is the common designation for geometric isomers and might be ambiguous for some structures, here we will learn the IUPAC naming system for geometric isomers of alkene, that is the E/Z naming system.

E/Z Naming System

To do the E/Z designation, at first, the groups connected on each sp2 double bond carbon will be assigned the priority based on the atomic number (see following guidelines for details), then the isomer with same priority group on the same side of double bond is assigned as “Z”, and the isomer with the same priority group on the opposite side of double  bond is called “E”. Both E and Z come from German, “Zusammen” means same side and “Entgegen” means opposite.

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The guidelines for assigning group priority in E/Z naming system

1. Priority is assigned based on the atomic number of the atoms bonded directly to the sp2 double bond carbon, the larger the atomic number, the higher the priority (isotopes with higher mass number has higher priority). For example: S > O > N > C > H.

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For the above structure of 2-penetene: on the left side sp2 carbon, methyl group CH3 is higher than hydrogen atom because C > H; on the right side sp2 carbon, ethyl group CH2CH3, is also higher than hydrogen. With higher priority group on both side of the double bond, this is the Z isomer, the complete name of the compound is (Z)-2-pentene.

The group withhigher priority is labelled as #1, and the group with lower priority is labelled as #2 in this book.

2. If the two groups bonded directly on an sp2 carbon start with the same atom, means there is a tie from step 1, then we move on to the atoms that connected to the “tied” atom, priority increases as the atomic number of the next attached atom increases.

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For the above structure, it is obvious that Cl is higher than C (C of CH2CH3 group) on the right side sp2 carbon.

On the left side sp2 carbon, we need to compare between methyl CH3 group and ethyl CH2CH3 group. Both groups has carbon atom attached directly on the sp2 carbon, that is a tie. In CH3 group, the carbon atom is bonded to H, H, H; while in CH2CH3 group, the carbon atom is bonded with H, H, C. So ethyl CH2CH3 is higher than methyl CH3 (see Note below). With higher priority group on opposite side of the double bond, this is the E isomer, the complete name of the compound is: (E)-3-chloro-4-methyl-3-hexene.

Note #1: For this round of comparison between H, H, H and H, H, C, compare the single atom with the greatest number in one group verse the single atom with the greatest number in the other group. So H in one group verse C in the other group, since C > H, therefore CH2CH3 is higher than CH3. Remember do not add the atomic numbers. For example, if one group has C, C, C, and the other group has C, O, H, then the C, O, H side is higher because O is higher than C.

Note #2: The above compound is cis-isomer if using the cis/trans naming system (both ethyl group are on the same side of double bond), but is E-isomer for E/Z system. So the cis/trans and E/Z are two different naming systems, don’t always match.

3. Repeat step 2 if necessary, until the priority is assigned.

Examples: What is the correct structural formula of (E)-2-bromo-3-chloro-2-butene?

A=CH3 (upper left), CH3 (upper right), br (lower left), & cl (lower right), B=CH3's are opposite (upper left & lower right)

The answer is B.

Examples: Draw the structure of (E)-3-methyl-2-pentene

Answer:

H (upper left), CH2CH3 (upper right), CH3 (lower right), & CH3 (lower right)

Examples: Order the following groups based on increasing priority.

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Approach:

1st round: C, C, C, C (tie);

2nd round:

A: C bonded to C, C, C; (3rd)

B: C bonded to H, Cl, Cl; (Cl is the 2nd high)

C: C bonded to H, C, C; (4th)

D: C bonded to H, H, Br (Br is the highest)

Solution: C < A < B < D

Exercises 5.1

Order the following groups based on decreasing priority for E/Z naming purpose.

A=-CH2CH2Br, B=-CH2CH3, C=-CH(CH3)2, D=-C(CH3)3

Answers to Practice Questions Chapter 5

4. When multiple bond is part of the group, the multiple bond is treated as if it was singly bonded to multiple of those atoms. Specifically:

CCH on blue C, CCCon blue C, & OOH on blue C

For these three groups involve multiple bonds, they all start with the carbon atom (the carbon atom  highlighted in blue color), and we should compare the group of atoms that connected on the blue carbon by converting the multiple bond to “multiple single bonds”, as shown above. So, if we compare the order of these three groups, it is:

CCH on blue C < CCC on blue C < OOH on blue C

Examples: Assign E/Z of the circled double bond.

HOCH2 (upper left), HC=CH2 (upper right), HC(three bonds to)C (lower left), & CH2CH3 (lower right)

Thinking:

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The answer is: Z-isomer.

Annotate

Next Chapter
5.3 Chirality and R/S Naming System
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Chemistry
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