1. Chapter 1 Lecture Introduction and Review Organic Chemistry, 8 th Edition L. G. Wade, Jr.

2. Lecture 2: Formal Charge, Resonance, Structural Formulas Formal Charges Common Bonding Patterns Resonance Forms Evaluating Resonance Structures ◦ Major versus Minor Contributors Ways to Write Organic Structures 2

3. Formal Charges Using Fractions H 3 O + NO + Formal charge = [group number ] – ½ [shared electrons] – [nonbonding electrons] 6 – ½ (6) – 2 = +1 5 – ½ (6) – 2 = 0 + + Formal charges are a way of keeping track of electrons. They may or may not correspond to actual charges in the molecule. Chapter 13

4. Formal Charges without Fractions H 3 O + NO + Formal charge = [group number ] – [bonds] – [nonbonding electrons] 6 – 3 – 2 = +1 5 – 3 – 2 = 0 + + A bond is simply a pair of electrons shared by two atoms. Note: You may use either formula, but you must specify which one and be consistent in showing calculations. Chapter 14

5. © 2013 Pearson Education, Inc. Chapter 95 Compute the formal charge (FC) on each atom in H 3 N – BH 3. Solved Problem 1 Solution Chapter 15

6. Common Bonding Patterns Chapter 16

7. Work enough problems to become familiar with these bonding patterns so you can recognize other patterns as being either unusual or wrong. Chapter 17

8. Resonance Forms The structures of some compounds are not adequately represented by a single Lewis structure. Resonance forms are Lewis structures that can be interconverted by moving electrons only. The true structure will be a hybrid between the contributing resonance forms. Chapter 18

9. Resonance Forms Resonance forms can be compared using the following criteria, beginning with the most important: 1.Has as many octets as possible. 2.Has as many bonds as possible. 3.Has the negative charge on the most electronegative atom. 4.Has as little charge separation as possible. Chapter 19

10. Resonance Forms Resonance theory states that whenever a molecule or ion can be represented by two or more Lewis structures that differ only in the positions of the electrons, two things will be true: ◦ None of these structures represent the actual bonding of the molecule or ion. ◦ None will agree with the physical or chemical properties of the substance. ◦ Key Point: The actual molecule or ion will be better represented by a hybrid (average) of these structures. 10

11. Evaluating Resonance Forms ◦ The more stable a structure is (when taken by itself), the greater is its contribution to the hybrid. ◦ The more covalent bonds a structure has, the more stable it is ◦ Charge separation decreases stability ◦ Structures with all the atoms containing complete valence shells of electrons are more stable ◦ Example: Resonance Structures for Formaldehyde 11

12. Major and Minor Contributors The major contributor is the one in which all the atoms have a complete octet of electrons. Chapter 112

13. Major and Minor Contributors (Continued) When both resonance forms obey the octet rule, the major contributor is the one with the negative charge on the most electronegative atom. MAJOR MINOR The oxygen is more electronegative, so it should have the negative charge. Chapter 113

14. Opposite charges should be on adjacent atoms. Non-Equivalent Resonance The most stable one is the one with the smallest separation of oppositely charged atoms. MAJOR MINOR Chapter 114

15. © 2013 Pearson Education, Inc. Chapter 915 Solved Problem 2 Draw the important resonance forms for [CH 3 OCH 2 ] +. Indicate which structure is the major and minor contributor or whether they would have the same energy. The first (minor) structure has a carbon atom with only six electrons around it. The second (major) structure has octets on all atoms and an additional bond. Solution Chapter 115

16. © 2013 Pearson Education, Inc. Chapter 916 Solved Problem 3 Draw the resonance structures of the compound below. Indicate which structure is the major and minor contributor or whether they would have the same energy. Both of these structures have octets on oxygen and both carbon atoms, and they have the same number of bonds. The first structure has the negative charge on carbon, the second on oxygen. Oxygen is the more electronegative element, so the second structure is the major contributor. Solution Chapter 116

17. Resonance Forms for the Acetate Ion When acetic acid loses a proton, the resulting acetate ion has a negative charge delocalized over both oxygen atoms. Each oxygen atom bears half of the negative charge, and this delocalization stabilizes the ion. Each of the carbon–oxygen bonds is halfway between a single bond and a double bond and is said to have a bond order of 1½. Chapter 117

18. Condensed Structural Formulas Lewis Condensed 1 2 Condensed forms are written without showing all the individual bonds. Atoms bonded to the central atom are listed after the central atom (CH 3 CH 3, not H 3 CCH 3 ). If there are two or more identical groups, parentheses and a subscript may be used to represent them. Chapter 118

19. Condensed Structural Formulas Lewis Condensed 1 2 3 4 Chapter 119

20. © 2013 Pearson Education, Inc. Chapter 920 Condensed Structural Formulas (Continued) Chapter 120

21. © 2013 Pearson Education, Inc. Chapter 921 Condensed Structural Formulas (Continued) Chapter 121

22. © 2013 Pearson Education, Inc. Chapter 922 Condensed Structural Formulas (Continued) Chapter 122

23. © 2013 Pearson Education, Inc. Chapter 923 Line-Angle Drawings 1 2 3 4 Sometimes called skeletal structure or stick figure. Bonds are represented by lines, and carbons are present where a line begins or ends and where two lines meet. Hydrogens attached to carbon are not shown. Nitrogen, oxygen, and halides must be shown. Chapter 123

24. © 2013 Pearson Education, Inc. Chapter 924 Line-Angle Drawings 1 2 3 4 5 6 Atoms other than carbon must be shown. Double and triple bonds must also be shown. Chapter 124

25. © 2013 Pearson Education, Inc. Chapter 925 Line-Angle Formulas (Continued) Chapter 125

26. Skill Building: Practice Problems Problems 1-6 thru 1-11 26