Week 1 Class Notes (8-24-15 to 8-28-15)
Week 1 Class Notes (8-24-15 to 8-28-15) CHM 2210
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This 3 page Class Notes was uploaded by Amanda Weaver on Sunday August 30, 2015. The Class Notes belongs to CHM 2210 at University of Florida taught by Dr. Laura Beth Peterson in Summer 2015. Since its upload, it has received 26 views. For similar materials see Organic Chemistry 1 in Chemistry at University of Florida.
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Date Created: 08/30/15
CHM2210 EXAM 1 CLASS NOTES Did not take notes on 82415 First day of class Class Notes 82615 Lewis Structures and Covalent BondsMolecules 0 Octet Rule atoms are going to share electrons until their obtain a stable octet 8 e39 0 Valence number different from valence shell the number of bonds required to achieve and octet Ionic bonds loss or gain of electrons to acquire filled valence shells 0 Ex NaCl KF LiBr 9 practice writing the electron configurations 0 Atoms that give an electron have a positive charge Cation 0 Atoms that accept and electron have a negative charge Anion 0 Covalent bonds with an EN difference of greater than 19 are considered ionic Covalent bonds the sharing of electrons to achieve a stable octet 0 Nonpolar covalent bonds just called covalent bonds difference of EN is lt05 0 Polar covalent bonds difference of EN between 05 19 o Ionic bonds see above EN difference of gt19 this means that this is no longer covalent Electronegativity EN a measure of force of an atom s attraction for e39 o Fluorine F has the highest EN of all elements and the rest of the elements are scaled to it 0 Table 15 0 General trend for EN is up and to the right of the PT Need to know this Drawing Lewis Structures 1 Count the number of valence electrons also the Group number 2 Determine the connectivity determine central and surrounding atoms 3 Connect the atoms with single bonds first then assign lonenonbonding pairs 4 Determine if octets are complete If not move e39 around to make multiple bonds Example 1 Ammonia NH3 1 Nhas 5Hhas1 2 Since H only has potential for one bond N is the central atom 3 N has 2 free electrons after bonding 9 drawn as two dots on the top of N Example 2 Acetylene C2H2 1 Valence electrons C 4 H 1 valence electrons 24 21 10 2 Draw the atoms with single bonds There is no central atom here both C are central 3 H C3C H there is a triple bond between the Carbons Example 3 Methanol CH3OH 1 Valence 4 3 6 1 14 valence electrons 2 Draw with single bonds 3 Fill octets On the exam she will only see the final step Make sure to draw lone pairs as dots Exceptions to the Octet Rule 1 Sulfur and Phosphorus can have an expanded valence because they have d orbitals Ex SF6 Bond Polarity 0 Polar covalent bond the unequal sharing of e39 o C OC25035 o O has a higher EN and so it will receive a bond dipole vector 9 towards it 0 Can also be depicted by delta 0 Ex Formaldehyde CHzO She drew Lewis structure EN towards O 0 Molecular Dipole Moment sum of individual bond dipoles in a molecule 0 Takes into account the shape of the molecule Formal Charges FC charge on an atom in a polyatomic ion of molecule 0 Consequence of its valence number 0 Each atom within the molecule will have a formal charge 0 The sum of these individual formal charges must equal the overall charge on the molecule 0 Formal charge valence e39 of nonbonding e39 12 bonding e39 Example 1 Cyanide CN39 C 4 N 5 Addtnl 1 1 Number valence 4 5 1 10 a Molecule has an extra 639 which gives it its negative charge 2 Draw Lewis structure a Is the same as N2 but with one C and one N 3 The negative charge is on Carbon which is a consequence of the valence number N fills whereas C does not fill therefore when the molecule receives an extra 639 C can take it Molecules and Shorthand Notations Bring White Paper to Next Class 1 Structural formulas a C3Hg propane 2 Lewis Structure a Drawn on the board C C C with H single bonds surrounding them 3 Condensed formula a H3C CH2 CH3 4 Fully condensed formula a CH2CH32 Line Angle Formulas This is the style that we will primarily use 1 Propane look at her notes looks like this A C is where a line meets or ends 2 Each C is assumed to have a full valence So we can assume that the H fill the octet in a natural way Class Notes 82815 Recap Line Angle Formulas o Hydrogens and lone pairs on heteroatoms are always drawn 0 Heteroatoms nonCarbon nonHydrogen atoms I O N S P 0 Cyclohexane C6H12 0 We will learn how to name structures in Chapter 2 39 Cyclo means that the Carbons are in a ring 0 Draw the line angle versus thee Lewis structure 0 Diethyl ether C4H100 o CH3 CH2 O CH2 CH3 o 2Butene C4Hg o CH3 CHCH CH3 o Incorporate double bonds by drawing a line above the line Functional Groups 0 Atoms or groups of atoms in a molecule that shows a characteristic set of physicalchemical properties 0 Check the inside cover of the book make ashcards Resonance 0 Theory that many molecules are best described as a hybrid of several Lewis structures 0 Ex Carbonate CO3239 24 e39 I Draw Lewis Structure O392 CO I Experimentally all C O bonds are 13 0 C 0 121 A 0 C 143 A o Resonance structures structure that only differ in the position of e39 nonbondedmultibond e39 0 Ex Carbonate again moved around the double bond 0 Resonance hybrid weighted average of resonance structure 0 Draw dashed lines to represent the double bond that changes between them 0 The charge on each 0 atom is 23 because it s 2 between 3 O atoms 0 Rules of Resonance Structures 1 Only nonbondedmultibond e39 are delocalized can move a Pi bonds p orbitals 2 Every resonance structure must be a valid Lewis structure 3 A11 structures must have the same amount of valence e39 4 The position of all nuclei must be the same in each structure 0 Depicting Resonance Structures CURVED ARROWS 1 Curved arrows depict movement of e39 a Goes by e39 pair at the time Curved arrows always start at lone pair e39 or at the center of a multiple bond e39 sourcedensity Curved arrows end at regions of lower e39 density e39 sink Flow of e39 is from negative to positive Curved arrows will denote e39 delocalizationmovement EX Nitromethane CH3N02 I Starts at lone pair on 0 goes to sigma bond I There are too many e39 on N now so we take the pi bond on the other 0 and move those electrons to a lone pair on it 0 9593