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Instructor: Prof. Duncan Wardrop Time/Day: T, 4:30-8:15 p.m. September 9, 2010

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## Instructor: Prof. Duncan Wardrop Time/Day: T, 4:30-8:15 p.m. September 9, 2010

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**Welcome to CHEM 494**Instructor: Prof. Duncan Wardrop Time/Day: T, 4:30-8:15 p.m. September 9, 2010**Course Website**http://www.chem.uic.edu/chem494 • Syllabus • Course Policies • Other handouts • Announcements (Course News) • Course Calendar**Keys to Success**• 1. Attend all lectures and discussion sections. • 2. Don’t fall behind. Organic chemistry is easy, but each topic builds on the previous. • 3. Don’t memorize. Organic chemistry is conceptual. • 4. Always ask yourself “Why” for everything you read or hear. You may not always find the answer, but just asking will help you to find connections and remember more.**Origins and Examples of Organic Chemistry**Introduction**Vitalism**Living Nonliving. vs. • posses vital force • compounds derived from are “organic” (coined by J. Berzelius, 1807) • could not be synthesized in the laboratory • termed “inorganic” • derived from nonliving matter • can be synthesized in the lab**Wohler Synthesis Debunks Vitalism**F. Wohler (Germany, 1828) AgOCN(aq) + NH4Cl(aq) AgCl(s) + NH4OCN(aq) X Expected Product Actual Product ammonium cyanate (inorganic) urea (organic)**Self Test Question**What is the minimum requirement, today, for a chemical substance to be classified as organic? A. derived from living matter B. contains carbon C. cannot be synthesized D. bought at Whole Foods E. combustion yields SO2**Vitalism Lives On. . .**terpinen-4-ol sold as tea tree oil or Melaleuca oil fructose “fruit sugar” also in HFCS**Natural Products**Order: coleoptera Family: coccinellidae Order: Tetraodontoformes**Pharmaceuticals**Morphine (analgesic) Tamiflu (influenza) Celebrex (arthritis)**Optics - Transition Lenses?**does not absorb visible light UV light heat does absorb visible light**Atomic Structure**A General Chemistry Review**Atomic Composition**- + + + + - - - Bohr Model**Atomic Number & Mass Number**X A A = protons + neutrons Z = protons Z Li 6 - + + + 3 - For neutral molecules, the number of electrons equals the number of protons. -**Self Test Question**How many neutrons are in the following atom: 14 C 6 A. 14 B. 6 C. 8 D. 20 E. cannot be determined**Tenets of Schrӧdinger Equation**• electrons have wave properties • wave equation gives energy of electron at a location • solutions to wave equation are wave functions (Ψ); a.k.a orbitals • probability distribution = Ψ2 (Heisenburg uncertainty principle)**Probability Distribution vs. Boundary Surface**1s 1s Probability distribution (Ψ2) (“electron cloud”) Boundary Surface (1s orbital) (where the probability = 90-95%)**Wave Function Values:Quantum Numbers**• Shrodinger equation → wave function (orbital, Ψ) • many solutions for Ψ, energies of electrons in atom**s-Orbitals**• spherically symmetric • possible for n≥1 • 1 s-orbitals for each value of n • 1s = no nodes, 2s = 1 nodes, 3s = 2 nodes, etc. • probability of finding s-electron at nodal surface = 0 • s-orbital energy increases with increasing nodes (as n increases)**p-Orbitals**• shaped like dumbells • not possible for n = 1 (n≥2) • 3 p-orbitals for each value of n; they are degenerate (equal in energy) • wave function changes sign at the nucleus (node) • probability of finding p-electron at nodal plane = 0 • higher in NRG than s-orbitals of the same shell**Relative Energies of Orbitals**3d 3d 3d 3d 3d 4s n = 3/4 3p 3p 3p 3s n = 2 Energy 2p 2p 2p 2s n = 1 1s**Order of Orbitals from Periodic Table**Read left to right starting at the top left; just like a typewriter. 1s 2s 2p 3s 3p 4s 3d 4p 4d 5p 5s 6p 6s 5d 7s 6d 7p 4f 5f**Electron Configuration: Filling Orbitals**3d 3d 3d 3d 3d 4s 3p 3p 3p n = 3/4 3s spin (ms = +1/2) n = 2 Energy 2p 2p 2p spin (ms = –1/2) 2s • Aufbau principle: fill lowest energy orbitals first. • Pauli exclusion principle: two electrons with same quantum numbers cannot occupy a single orbital. Electrons must have opposite spins in the same orbital. • Hund’s rule: degenerate orbitals filled singly first. n = 1 1s**Self Test Question**Which of the following orbitals is highest in energy? A. 3f B. 3p C. 2s D. 3s E. 2d**Self Test Question**Which is the correct electron configuration for carbon? A. 1s2, 2s8 B. 1s2, 2s4 C. 1s2, 2s2, 2s2 D. 1s2, 2s2, 2p2 E. none of the above**What is a Covalent Chemical Bond?**valent = bonding covalent = electrons shared between two nuclei • Forces involved: • electron-electron repulsion • nucleus-nucleus repulsion • electron-nucleus attraction electron-electron repulsion < electron-neutron attraction**Three Models of Bonding**• Lewis Model: • atoms gain, lose or share electrons in order to achieve a closed-shell electron configurations (all orbitals fully occupied) • closed-shell for row 2 = 8 electrons (octet rule) • only valence electrons (in outermost shell) are involved in bonding Molecular Lewis Structures Molecular Lewis Structures Atomic Structures H–H F–F**Three Models of Bonding**• Valence Bond Model: • in-phase overlap of two half-filled orbitals • in-phase = constructive interference • increases probability electrons between two nuclei Boundary Surfaces H–H H H Electrostatic Potential maps: red = negative blue = postive**Three Models of Bonding**• Molecular Orbital (MO) Model: • combine atomic orbitals (AO) of all atoms, then extract molecular orbitals • number of atomic orbitals in equals the number of molecular orbitals out • combination possibilites: • additive = produces bonding molecular orbitals • subtractive = produces antibonding molecular orbitals**Molecular Orbital Diagram**antibonding (σ*) MO nodal plane e- probability = 0 = no bond H H 1s AO 1s AO Rules Still Apply: 1. Aufbau 2. Pauli 3. Hund bonding (σ) MO**Self Test Question**Which is the correct Lewis dot structure for CH4O? D. A. B. E. none of the above C.**Structural, Bond-line and Condensed Formulas**CH3CH2CH2CHO CH3CH(OH)CH3 (CH3)2CHCH2CH2CO2H**Self Test Question**Which bond-line drawing correctly represents the following condensed formula? CH3CH(OCH2CH3)CH2Br D. A. B. E. C.**Memorize These Functional Groups NOW**• functional group: a defined group of atoms with a specific connectivity • responsible for properties • predictable reactivity • well-defined nomenclature Memorizing: “What is the minimum number of atoms needs to define a functional group and in what order are they bonded?” Flashcards! Table 4.1, page 140**Formal Charge - Example 1**• My logic (no memorizing equations): • Recognize that opposing charges cancel each other • Determine difference in protons vs. electrons by asking: • How many valence (outer shell) electrons does the atom have? • How many valence (outer shell) electrons does the atom “want” (groups #)? • If atom hasmore than it “wants,” negatively charged. • If atom hasless than it “wants,” positively charged. O 8 Has? = 16 2– Wants? = 6 8 Difference? = 2 extra**Formal Charge - Example 2**• My logic (no memorizing equations): • Recognize that opposing charges cancel each other • Determine difference in protons vs. electrons by asking: • How many valence (outer shell) electrons does the atom have? • How many valence (outer shell) electrons does the atom “want” (groups #)? • If atom hasmore than it “wants,” negatively charged. • If atom hasless than it “wants,” positively charged. 5 Has? = Wants? = 6 Difference? = 1 less “formal” = count one electron for each bond**Shortcut for Some 1st and 2nd Period Atoms**• For each additional bond = +1 • For each “missing” bond = –1**Self Test Question**What is the formal charge on the carbon atom in red? A. -2 B. -1 C. 0 D. +1 E. +2**Self Test Question**According to VSEPR, what is the bond angle between 2 H-atoms in methane (below)? • 105º • 109.5º • 107º • 180º • 120º**Structure and Bonding: Resonance**Section: 1.8, 1.11 You are responsible for Section 1.10.**The Contradictory Case of Ozone**Lewis Structure suggests ozone is non-symmetrical electrostatic potential map red = negative blue = positive Microwave spectroscopy shows that ozone is symmetrical (C2v)**Curious Case of Benzene**Predicted Actual resonance C–C bond length: 150 pm C=C bond length: 134 pm All bonds = 140 pm**Resonance Provides the Solution**• Problems: • Lewis structures fail to describe actual atom electron densities for molecules with more than one possible electron distribution. • Lewis formulas show electrons as localized; they either belong to a single atom (lone pair) or are shared between two atoms (covalent bond). • Electrons are not always localized; delocalization over several nuclei leads to stabilization (lower energy).**Physical Proof of Resonance in Amides**C-N (amide) Bond Length = 133 pm C-N (amine) Bond Length = 147 pm C=N (imine) Bond Length = 129 pm**A Question of Arrows**Equilibrium between distinct species Reaction from one species to another “Movement” of electrons from donor to acceptor Indicates that 2 species are contributing resonance structures**Resonance**• Solutions: • Actual molecule is considered a resonance hybrid (weighted average) of contributing Lewis structures. (Note: double headed arrow does NOT indicate interconversion.) • Dashed-line notation is sometimes used to indicate partial bonds. • Not all structures contribute equally.