CHEM 2211 Exam # 1 Study Guide Lectures: 1-8Lecture 1 (14 August 2013)I. Molecular Orbital TheoryA. Molecular Orbitals can either constructively overlap or a destructively overlap.B. Constructive overlap results in a bond. Destructive overlap results in a nodal space and Antibonding.II. Linear Conservation of Atomic OrbitalsA. The number of atomic orbitals consumed equals the number of molecular orbitals producedB. Sigma star and pi star are Antibonding orbitals that allow atoms to break bonds with each otherC. Molecules of equal electronegativity each contribute an atom to the bond.D. Atoms of different electronegativity unequally contribute electrons to the bond meaning that the more electronegative atom is more similar to the sigma bond than to the sigma star bond. Conversely, the least electronegative atom is more similar to the sigma star bond than to the sigma bond.Lecture 2 (16 August 2013)I. Brᴓnsted-Lowry Acid Base TheoryA. Bases are proton acceptors and form OH- ions. Acids are proton donors and form anions.II. Ka valueA. Ka is the likelihood a chemical will lose a proton. The higher the Ka value, the higher the likelihood the chemical will lose its proton and the stronger the acid.B. Since Ka is a very small number, the pKa is used instead. The formula for pKa is+¿H3O¿¿−¿A¿¿¿p Ka=−log ¿C. The lower the pKa, the stronger the acid.III. Factors that effect pKaA. Type of bondi. Because hydrogen bonds are a stronger attraction between molecules,molecules with them are harder to depronate and have a higher pKa value.B. Sizei. Poor orbital overlap cause molecules of missed matched size to be easier to depronate than molecules of similar size. C. Electronegativityi. The more electronegative an atom or molecule is the less likely it is to depronate.IV. Conjugate Acids and BasesA. Every base has a conjugate acid and every acid has a conjugate base. To determine the equilibrium of a system, the side with the highest pKa is where the equilibrium will shift.V. Resonance StructuresA. A resonance structure is the average of two or more structures which differ only in the placement of electrons. Resonance structures are more stable because the electrons are shared across the molecule and make the molecule less reactive.VI. IUPAC nomenclatureA. Created a standardized way of naming organic moleculesB. Each name has a standard prefix for the number of carbons and a suffix that specifies the type of molecule. KNOW the first thirteen prefixesi.ii. Meth-iii. Eth-iv. Prop-v. But-vi. Pent-vii. Hex-viii. Hept-ix. Oct-x. Non-xi. Dec-xii. Undec-xiii. Dodec-xiv. Tridec- VII. Line angle structures (aka Skeleton structures)A. Remember that lines are bonds, angles are points, and hydrogens are implicit.Functional groups and other atoms are completely drawn in, meaning their hydrogens are includedVIII. AlkanesA. Suffix –aneB. All single bonded carbons, symmetrical, nonpolar, have London dispersion forcesLecture 3 (19 August 2013)I. An Alkyl Group is when a hydrogen is removed from an alkaneA. Can have different functional groups attachedB. The carbons can be primary, secondary, or tertiaryC. For the common naming system, an alkyl group with three carbons can be either propyl or isopropyl groupi. A propyl group is a carbon bonded to one other carbonii. An isopropyl group has a secondary carbon bonded to two CH3 groupsD. For the common naming system, a four carbon alkyl substituent can be eitherbutyl, isobutyl, sec-butyl, or tert-butyli. The rules for butyl and isobutyl are the same as for propyl and isopropyl.ii. Sec-butyl is when a hydrogen is removed from a secondary carboniii. Tert-butyl is when a hydrogen is removed from a tertiary carbonII. IUPAC naming of AlkanesA. Determine the longest parent chainB. Number the chain in the direction that gives the substituent(s) the lowest number possibleC. When more than one substituent exists, the chain is numbered in the direction that yields the lowest number and the substituents are named in alphabetical orderD. Prefix “di—“, “tri—“, “tetra—“, etc. indicate the number of identical substituents on the parent chainE. If numbering both way yields the same number for different substituents, then the first named group gets the lowest numberF. In a highly branched compound where the substituent also has branched portions, the substituent’s name is in parenthesis and is accompanied by a number indicating a position on the substituent.G. In the case that two hydrocarbon chains have the number of carbons choose the one with the most substituents.Lecture 4 (21 August 2013)I. CycloalkanesA. Alkanes that form a ring and have two less hydrogens than the straight chain alkanesB. Namingi. Cycloalkanes are considered the parent chain unless a straight chain substituent has more carbons, then the straight chain is the parent chain1. If only one substituent exists, then no numbers are needed to identify its place on the cycloalkaneii. If two substituents are present on the ring, then the first one alphabetically is denoted as number one.iii. When a cycloalkane has more than one substituent, then the first substituent is the substituent that yields the lowest number for the second substituentII. AlcoholsA. Formed when OH- attaches to an alkyl group and can be primary, secondary, or tertiary depending on the carbon they’re attached toB. Namingi. First drop the “—e” at the end of the alkane name and add “—ol”ii. The parent chain is the longest chain with the OH- attachediii. OH group has the lowest possible number when numbered on the parent chainiv. The suffixes “—diol”, “—triol”, etc. is added to compounds with more than one OH attachedv. When both the functional group and the substituent are present and/or have the same number when numbered both ways, then the functional group gets the lowest numbervi. Multiple substituents on an alcohol are listed alphabeticallyLecture 5 (23 August 2013)I. Alkyl HalidesA. Can be primary, secondary, or tertiary B. The common name is Alkyl Group Name + Halide + “—ide” suffixC. IUPAC nomenclature states that the halide is named first and the alkane group secondII. EthersA. One oxygen bonded to two alky groupsB. Common name is to add names of two alkyl groups followed by etherC. IUPAC way is to drop the “—yl” and add the suffix “—oxy” and then the alkane namei. Shortest substituent gets the “—oxy” suffixIII. AminesA. Deemed primary, secondary, tertiary, or