1. Homework DUE Friday, 5 Sept Problems in McMurry 1.24; 1.28; 1.31; 1.45; 1.46; 1.47 => (1.48—1.52 BONUS Problems) Organic Chemistry - 246A

2. Summary of 1 st Week’s Lectures Carbon atoms can be oxidized or reduced (oxidation state) Electronic structure of an atom described by wave equations  –Electrons occupy orbitals around the nucleus –1st row elements require 8 e — to achieve a “filled shell” octet –s orbitals are spherical (o) –p orbitals are dumbell-shaped (∞) Covalent bonds => the bonding e — pair is shared between atoms Sigma bonds (  ) are circular in cross-section and are formed by head- on interaction of s, sp, sp 2 or sp 3 hybridized orbitals Pi bonds (  ) are “dumbell” in shape and are formed from side-side interaction of two p orbitals (88) s and p orbitals can be combined to form multiple bonds Double bonds use sp 2 orbitals, have trigonal (120°) geometry (  +  ) Triple bonds use sp orbitals, have digonal (180°) geometry (  + 2  ) N and O have lone pairs (:) that can accept a proton

3. Molecular Orbital Theory (MOT) Robert Mulliken in the 1940’s proposed that molecules have orbitals (MOs) that are the result of combinations of atomic orbitals (AOs) A molecular orbital describes a region of space in a molecule where electrons are most likely to be found (e — s no longer associated with an individual atom) AO combinations are additive and subtractive to form MOs + combination (AO + AO = MO) forms MOs that is lower in energy (bonding) — combination (AO — AO = MO) that is higher in energy (anti-bonding)

4. MO’s for Ethylene Bonding and antibonding  molecular orbitals result from combination of two p atomic orbitals in ethylene The  bonding MO (+ combination) has no node between nuclei The  antibonding MO (— combination) has a node between nuclei Only the bonding MO is occupied (with 2 e — )

5. Electronegativity Electronegativity (EN): ability of an atom to attract the shared e — in a covalent bond F is most electronegative (EN = 4.0), Cs is least (EN = 0.7) Differences in EN produce bond polarity Greatest Least

6. Dipoles and Polar Covalent Bonds The greater the electronegativity difference between two bonded atoms, the more polarized the bond Polarization gives rise to partial charges, and a dipole,  If the electronegativity difference of the two atoms is greater than 2, the bond usually ionizes

7. Inductive Effect For a polar covalent bond C—X, where X is an electronegative atom (e.g. Cl in CH 3 CH 2 Cl) the partial negative charge (  — ) on the chlorine atom induces a partial positive charge (  + ) on the neighboring carbon atom The partial charges cause a dipole to be formed (  = Q x r) Experimentally, it is easy to measure , and if we know the bond length, r, then the partial charge Q can be calculated Chloroethane (Ethyl Chloride)

8. Dipoles Can Cancel Each Other Even though each C— Cl bond has a strong dipole, the tetrahedral symmetry causes them to oppose each other Resulting molecular dipole,  = 0

9. Charge Distribution Calculations

10. Resonance (VBT Concept) Some molecules are have structures that cannot be shown with a single valence bond representation (e.g. carbonate, CO 3 = ) In these cases resonance contributors are drawn, which differ in the position of the  bond(s) and lone pair(s), and all contribute to the real structure of the molecule The resonance forms are connected by a double-headed arrow In this case, we say that the negative charges on CO 3 = are delocalized

11. Resonance Hybrids A structure with resonance forms does not alternate between the forms Instead, it is a hybrid of the two resonance forms, so the structure is called a resonance hybrid Benzene (C 6 H 6 ) has two resonance forms with alternating double and single bonds, yet all the bonds are of equal length- intermediate in length between single and double bond length

12. Rules for Resonance Individual resonance forms are imaginary - the real structure is a hybrid (only by knowing all the contributors can you visualize the actual structure) Resonance forms differ only in the placement of their  e — and/or nonbonding e — Different resonance forms of a substance don’t have to be equivalent (e.g. amide H 2 N—C=O + H 2 N=C—O — ), and can contribute to different degrees Resonance forms must be valid Lewis structures: the octet rule has to be satisfied The resonance hybrid is more stable than any individual resonance form (resonance stablization)

13. Curved Arrows (Sir Robert Robinson) We can imagine that electrons move in pairs to convert from one resonance form to another Of course electrons are delocalized and this is only a bookkeeping device

14. Delocalization Radicals (), anions (-) and cations (+) can also be delocalized Reactions can take place at either end of the intermediate

15. Acid-Base Theories The Arrhenius concept that acids are solutions containing “H + ” and bases are solutions containing “OH — ” is not very useful in organic chemistry Brønsted–Lowry theory defines acids as H + donors, and bases as H + acceptors Lewis theory defines acids as lone pair acceptors, and bases as lone pair donors

16. Acetone Anion Displays Resonance Acetone can give up a proton to a strong base to form an anion Acetone anion can react at O or at C

17. Acetyl Acetone Is More Acidic than Acetone 2,4-Pentanedione (Acetyl Acetone, or AcAc) is deprotonated even by very weak bases It has 3 resonance forms

18. pKapKa Protonated Deprotonated