Exam 2 Study Guide: Organic Chemistry
Exam 2 Study Guide: Organic Chemistry CHEM 330
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This 8 page Study Guide was uploaded by Maria Dion on Thursday October 8, 2015. The Study Guide belongs to CHEM 330 at Northern Illinois University taught by John G. Kodet in Fall 2015. Since its upload, it has received 359 views. For similar materials see General Organic Chemistry 1 in Chemistry at Northern Illinois University.
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Date Created: 10/08/15
Exam 2 Study Guide Organic Chemistry Chapters 35 Ring Systems Newman Projections and Double Bonds 0 Eclipsed When the bonds are aligned with one another 0 Staggered When the bonds are opposite with one another I Anticonformation in staggered is more stable than Gauche however gauche is higher in energy Most molecules are in staggered conformations Eclipsed is less stable than staggered Torsional strains is a decreased stability in eclipsed bonds are Steric Strain is the combination of torsional strain and steric hindrance which is when 2 atoms are too close together Interconverted by rotation of CC bonds 0 Newman projections I Eclipsed Hydrogens separated with torsion angle of 0 degrees I Gauche Hydrogens separated with torsion angle of 60 degrees I Anti Hydrogens separated with torsion angle of 180 degrees I Step 1 Draw structure I Step 2 Highlightthebondthatyouarewant I Step 3 Determine the molecule for what the designated and the other carbon is attached too I Step 4 Place in either Anti or gauche provided from your directions I You can identify the Newman projection by breaking down the 2 separate bonds and recreating the Lewis structure 0000 O Polycyclic Ring Systems 0 Polycyclic rings are compounds that contain more than 1 ring 0 Major types of polycyclic rings include I Bicyclic2 I Tricyclic3 I Tetrecyclic4 I And so on O Named according to minimum number of bond cleavages and the respect to which the rings are joined Heterocyclic Compounds 0 O O Heterocyclic compounds are compounds that has elements of 2 different elements as members based on a ring Heteroatom is the name of the substance on a heterocycle Some major heterocyclic compounds in the book were I Ethylene oxide Triangle With oxygen I Tetrahydrofuran Pentagon With oxygen I Pyrrolidine Pentagon With Nitrogen and Hydrogen I Piperidine Hexagon With Nitrogen and Hydrogen Sulfur heterocycles are common in 3 6 rings Lipoic acid and lenthionine contain SS bonds Speed of Reaction 0 A reaction can go faster or slower based on I temperature I concentration of reactions I dispersion I physical states I catalysts Nomenclature 0 Substitute Preferred I uoro chloro bromo or iodo substituent 01 for an alcohol I Number based on the side the substituent lies I Placement of substituent uoro chloro bromo or iodo Abbr of of subs ane I Halogen and alkyl are equal I Same thing goes When both halogen and alkyl are presented I alkylhalogen Placement uoro chloro bromo or iodo alkylhalo gen Placement Molecule name Abbr of of subs ane Functional More Common I Fluoride chloride bromide or iodide I Common alkyl group names npropyl isopropyl nbutyl secbutyl isobutyl tert butyl and neopentyl I Halide and Alkyl group Listed separately I Can be determined With the longest chain staring at carbon Where the halogen is connected Alcohols I Subsitutive 393 Longest chain With OH group 393 Number closest to OH replace ane With 01 I Functional 393 Alkyl nameAlcohol Priority 0 In nomenclature for alkyl halides the alkyl group and the halogen have the same priority 0 CahnIngoldPrelog states that higher atomic number takes priority over lower atomic number 0 Highest priority functional group 0 Determine whether it is acyclic cyclic and nonaromatic or cyclic and aromatic Direction 0 The carbon chain is numbered in the direction that gives the substituted carbon or halogen the lower number 0 In alcohols the hydroxyl group has more priority over the halogen or alkyl group and therefore numbered from the hydroxyl group Cis or Trans 0 O O Cis and Trans are stereoisomers or isomers with the same constitution and a different arrangement Cis is when the substituents are on the same side of the ring or double bond Trans is when the substituents are on opposite sides of the ring or double bond E Z Nomenclature O E Z nomenclature is used when alkenes are too complex to name with cis or trans The element with the higher atomic number has higher priority when 2 atoms are directly connected to each other When the 2 elements with the higher atomic number is on the opposite side then it is E When the 2 elements with the higher atomic number is on the same side then it is in Z Look up atomic numbers for all sides and determine which are higher and then classify them as E or Z Potential Energy Diagrams OOOOOOOOO Plots the change in potential energy that occurs during a chemical reaction Reactants in the beginning Products in the end Activation energy is space between reactant beginning and peak of curvature If reactants have lower energy than products Endothermic If products have lower energy than products Exothermic Staggered conformations are more stable than the eclipsed Torsional angle as X axis 0 60 120 180 240 300 360 More than 1 step gives more activation states and transition states to consider Substitution reactions 0 O SNl Rate determining step 1 is unimolecular Step 1 I Formation of carbocation be separation I Slow and reversible Step 2 I Carbocation reacts With nucleophile Step 3 I Removal of proton With base forming H30 and alcohol SN2 I Rate determining step is bimolecular I Occurs at sp3 carbon center I Attached to stable electronegative group I Nucleophile attacks carbon at 180 I Provides best overlap I Very common PBr3 SOC12 Pyridine O PBr3 I Reacts With alcohols to give alkyl bromides and phosphorous acid I Formation of an intermediate of the type ROPBr2 I Then reacts With the nucleophilic bromide ion 0 SOC12 O I Reacts With alcohols to give alkyl chlorides I Prepare primary and secondary alkyl chlorides I Conversion of the alcohol to a chlorosulfite Pyridine I Solvent but also acts as a weak base and catalyst Elimination reactions 0 E1 Protonation Dissociation 393 Unimolceular 393 Rate determining step Deprotonation 393 Base steals proton from a Br nsted acid carbocation 393 Forms acid catalyst 393 When adjacent substituents for alkenes 393 First order elimination reaction E2 I Primary cations unstable I Primary alcohols use different mechanism I Rate determining step is bimolecular I Second order elimination reaction O In El elimination reactions there is generally a weak base and a tertiary alkyl halide Dehydrogenation reactions 0 O O O XYH removal of hydrogen from a molecule reverse process of hydrogenation form a double bond Dehydrohalogenation reactions loss of HX O O O O Alkyl halide with alcoholic alkali gives alkene BElimination reaction Reactions that form more substituted alkenes have lower activation energy and are therefore faster Substituted transition states are more stable Step 1 I Ionization of alkyl halide I Unimolecular E1 I Rate determining step I Needs a primary alkyl halide for carbocation formation Step 2 I Deprotonation of primary and secondary hydrogen Condensation reactions 0 Two molecules or functional groups combine to form a larger molecule 0 Loss of a small molecule H20 Rearrangements O Rearrangements happen when the molecule wants to go from an unstable to a more stable state 0 Shifts within the molecule through 3 center transition 0 Connectivity of the atoms can change 0 Creates a secondary carbocation Determining Unimolecularity and Bimolecularity O O Unimolecular I When a molecule reconstructs itself to have one or more different molecules Unimolecular has first order rate law I RatekA Bimolecular I When 2 molecules collide to exchange atoms or even energy Bimolecular has second order rate law I Rate kA B This can be determined by looking at the written reaction mechanism Changing Rate of a Reaction 0000 Temperature can be changed to increase or decrease the rate of reaction Increasing Temperature I Decreases EaRT in the Arrhenius equation I Increases k Decreasing Temperature I Increases EaRT I Decreases k Tertiary alcohols react quickest and can be done to low temperatures Primary are slowest needing higher temperatures Fewer steps DO NOT translate to faster reaction rates activation energy of the slowest step determine reaction rate Determining Rate Limiting Step and Mechanisms O O O O A chemist can determine the rate determining step by finding out the step with the greatest or lowest activation energy He can also see if reactions are primary secondary or tertiary Mechanism can only experimentally determined Rate law can be deduced from equation Radical Mechanism Steps 0 O Radical reactions have 3 steps initiation propagation and termination Example of Methane is used I In the first step the bond is broken with heat providing the energy I In the second step the hydrogen is stolen by the chlorine forming HCL and the radical This then takes steals a chlorine to make a different molecule chloromethane and a chlorine radical and this cycle is repeated with the chlorine radical I In the last step it is terminated when 2 radicals react together Arrow Characteristics 0 O O O O O 0 Used to track ow of electrons Curved arrows used to convert from one resonance to the next Goes from where they begin to where they end up Singlebarbed arrow shows the movement of one electron Doublebarbed movement of a pair of electrons Double headed arrows indicates resonance The curved arrow should begin where the electrons are to begin with and ends on the final place where they are an unshared pair Types of Bonds Dashed Lines and 5 and 5 symbols 0 O A dashed line is used to indicate a bond that s only partial Used when drawing resonance hybrids OOOOOOO Can show the delocalization in the electron pattern Usually avoided 5 and 5 show partial charges Alkanes are hydrocarbons in which all the bonds are single bonds Alkenes contain at least one carbon carbon double bond Alkynes contain at least one carbon carbon triple bond Alcohol I CO bond is made by overlap of an sp3 orbital on carbon With a sp3 orbital on oxygen and can be I Have polar bonds Alkyl Halides I Has a carbonhalogen bond and is a sigma bond I Bond distances increases in order I Have polar bonds Dipole Dipole Attractive Forces I Permanent dipole have stronger intermolecular interaction Hydrogen Bonding I Dipoledipole I Partially positive proton H of one I OH group interacts With the partially negatively oxygen I Oxygen hydrogen bond acceptor I OH hydrogen bond donor I Weaker than covalent bonds by 1520 times I Imposes structural order because its strong I Organic compounds contain 0 and N Rate Limiting Step 0 Rate limiting step the slowest step of a chemical reaction that determines the speed of how the overall reaction proceeds Not all reactions have rate determining steps because it is only in cases Where one step is significantly slower than the rest of the steps in the reaction DaWing a Reaction Mechanism 00 Individual steps must equal whole equation Rate determining step is in step 2 and is unimolecular Step 1 Protonation Protonation by a strong acid Both reactants change causing it to be bimolecular Step 2 Dissociation Forms carbocation that is intermediate Carbocation has high energy but is unstable Unimolecular Classi ed as 5N2 because of the nucleophilic substitutions 0 Step 3 Lewis acidLewis base reaction Bimolecular Nucleophiles seeks the nucleus
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