Organic Chemistry Midterm 2 Teacher's Recommended Study Guide
Organic Chemistry Midterm 2 Teacher's Recommended Study Guide Chem241
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Popular in Chemistry
This 9 page Study Guide was uploaded by Colin on Sunday March 6, 2016. The Study Guide belongs to Chem241 at Drexel University taught by Ms. Susan Rutkowsky in Winter 2016. Since its upload, it has received 34 views. For similar materials see Organic Chemistry in Chemistry at Drexel University.
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Date Created: 03/06/16
Chapter 5 (Sections 5.1 – 5.16): Be able to determine whether a compound is chiral or achiral. o Chiral - Not superimposable o Achiral - Superimposable, plane of symmetry Be able to identify chiral carbons in compounds. o A carbon with 4 different groups attached Be able to assign (R) and (S) nomenclature. o Assigned based on bonded groups Highest atomic number, highest priority and vise versa o Put lowest priority in back and highest on top (3D representation) If not, reverse final answer o Arrow from high to low priority (1 to 2 to 3) o R - clockwise o S - counterclockwise Be able to explain optical activity o Active A compound that rotates the plane of polarized light o Inactive A compound that doesn’t change the plane of polarized light Achiral Mirror image superimposable Be able to explain the differences between enantiomers – rotation of plane‐polarized light and biological discrimination o Enantiomer Chiral molecules that are mirror images of another Non-superimposable o Rotation of plane-polarized light In chiral compounds, rotated with angle () measured in degrees Dextrorotary - clockwise, positive Levorotatory - counterclockwise, negative o Biological discrimination Active site of enzyme is selective for specific enantiomer Be able to explain a racemic mixture o Equal rotation of two enantiomers o Two enantiomers rotate the plane of polarized light equal amounts but in opposite directions o Rotations cancel Be able to calculate specific rotation, enantiomeric excess, and % composition of a mixture o Specific rotation Amount that a chiral compound rotates the plane of polarized light Use path length (l), concentration (c), temperature (T), and angle ( o Enantiomeric excess Optical purity Specific rotation:  pure Observed rotation: [ mix o Percent composition of a mixture Measure of how much one enantiomer is present in excess of the race mixture Be able to recognize the chirality of conformationally mobile systems o Chiral compounds do not form any achiral conformers o Consider most symmetric conformation Be able to recognize if a compound is chiral even if it does not have any asymmetric carbons o Asymmetric carbons Not conformationally locked Central carbon sp hybridized and linear 2 Outer carbons sp hybridized and trigonal planar Chiral center o No asymmetric carbons No chiral carbons Due to bulky groups on rings Conformationally locked (stuck in one of two possible structures) Conformer is sterically hindered Molecule may exhibit enantiomers Achiral Be able to draw Fischer projections o Horizontal lines are forward into the plane, vertical lines are backwards out of the plane o Prioritized based on highest to lowest oxidation Highest oxidized carbon at top Oxidation: Counting the number of electrons Highest at bottom left o Line up by priority Be able to represent the stereochemistry of compounds with one or more asymmetric carbons o Maximum number of stereoisomers is 2 n (n) is number of chiral centers o With more than one asymmetric carbon, R and S configuration must be assigned o Have enantiomers, diastereomers, and meso compounds Be able to identify pairs of enantiomers, diastereomers, and meso compounds and explain how they differ in their physical and chemical properties o Enantiomer Opposite configurations at each chiral carbon Chiral molecules that are mirror images Different in reaction with other chiral molecules Different direction polarized light is rotated o Diastereomers Some matching and some different configurations Stereoisomers that are not mirror images of one another Chiral Different B.P and M.P o Meso compounds Have internal mirror plane Same on both sides Symmetrical Achiral compound with tetrahedral asymmetric carbon n 2 Doesn’t apply Be able to explain how different types of stereoisomers can be separated. o Enantiomer Difficult to separate Resolution - separation of enantiomers If pure product is synthesized from achiral reagents, race mixture will occur Can form pair of diastereomers to separate Then hydrolysis o Diastereomer Distillation Recrystallization o Meso Compound Chapter 4 (Sections 4.1 – 4.16): Be able to propose a mechanism for the free‐radical halogenation of an alkane o 3 main steps: Free radical - reactive species with odd number of electrons Initiation Generates a reactive intermediate Halogen splits homolytically into free radicals Propagation (usually has 2 parts) Intermediate reacts with stable molecule to produce additional intermediate and product molecule Carbon radical Halogen radical collides with a methane molecule and abstracts (removes) and H atom Forms another free radical H and one product H-halogen Product formation Methyl free radical collides with another halogen Produces another product halogen radical and joins the previous product to the carbon Termination Side reactions destroy reactive intermediate Slow or stop reaction Collisions of any two free radicals produces non- radical compound Combination of free radicals with contaminant or collision with wall Concentration of radicals is very low during chain reaction Low likelihood that they will terminate with another radical As reaction proceeds, likelihood increases as there are few reactants left available Be able to predict the major halogenation products based on the stability of the intermediates and the selectivity of the halogenation o 3° free radical intermediates are more stable than 2° free radicals which are more stable than 1° free radicals which are more stable than methyl free radicals. o The more stable the intermediate, the more likely the reaction will follow that pathway o A double bond next to a free radical carbon will also help to stabilize the radical o Bromination is much more selective than chlorination Bromination will react almost entirely at the most highly substituted carbon Chlorination will react at all viable carbons, though the major product will most likely be the most substituted carbon o Increased temperature, increased rate, decreased (E ) a o Fluorine reacts explosively o Chlorine reacts at moderate rate o Bromine needs heat to react o Iodine doesn’t react o Product mixture 40% Hydrogens located in position one 60% Hydrogens located in position two o Hydrogen Reactivity Amount of product per hydrogen / number of hydrogens Smaller percentage is primary H Be able to draw a reaction‐energy diagram for a mechanism, and point to the corresponding transition states, activation energies, intermediates, heats of reaction, and rate‐limiting steps. o Drawing Y is energy X is process of reaction or reaction coordinate Start with reactants, draw activation energy to transition state at max, draw down to change of energy for product For multi-step reaction, draw separate diagrams, then one total diagram, each with own transition state o Transition state (‡) Located at energy maximum Unstable, never isolated Bonds partially broken and partially formed (represented by dashed line) Partial charge on any atom that gains or loses a charge during transition state Drawn in brackets o Activation energy (Ea) Energy difference between transition state and starting material Minimum amount of energy needed to break bonds in reactant o Intermediate A species that is produced in one step and used in another Located at energy minimum o Heats of reaction Catalyst - speeds up reaction without affecting it o Rate-limiting steps Step with the higher energy transition state Be able to use bond‐dissociation enthalpies to calculate the enthalpy change for each step of a reaction, and the overall enthalpy change for a reaction o Bonding Energy (H Change in bonding energy Heat released or absorbed in a chemical reaction in standard conditions Negative - exothermic Positive - endothermic Low enthalpy - strong bond Change in bonding energies - calculated from bond dissociation energies o Bond Dissociation Enthalpies BDE Energy needed to homolytically cleave a covalent bond BDE decrease down periodic table Size of halogens increases down Bond strength decreases down Stronger bond strength, higher BDE Positive H) More energy needed to break bonds than is released Bonds broken in starting material are stronger Negative (H More energy released in forming bonds than breaking bonds Bonds formed in product are stronger than bonds broken Lower energy more likely to happen Be able to calculate free energy changes from equilibrium constants, and calculate the position of an equilibrium from the free‐energy changes o Free Energy G) Overall energy difference between reactants and products (product-reactant) Inversely proportional to log[(Keq] o Position of equilibrium Equilibrium favors products (K eq> 1 (G < 1 Equilibrium favors reactants (K ) < 1 eq (G > 1 Be able to determine the kinetic order of a reaction based on its rate equation o Kinetic - the study of reaction rates o Rate Law Rate = k * [reactants]n K is constant [reactants] is the concentration of the reactants Order is the sum of reactant exponents (n) o Rate determining step Rate equation contains concentration in terms of reactants only for a multistep reaction Reaction can occur no faster than this step Be able to use the Hammond postulate to predict whether a transition state will be reactant‐like or product‐like, and explain how this distinction affects the selectivity of a reaction. o Hammond postulate Relates reaction rate to stability to estimate the energy of the transitions state Endothermic Closer in energy to product Transition state resembles product Stabilize product, stabilize transition state Decreased (E )a increased reaction rate, lowering energy at transition state Exothermic Closer in energy to reactant Transition state resembles reactant Product has little to no affect on transition state o Affect on Selectivity Order depends on number of H bonded to H 3s first order (1) H 2s second order (2 ) H is third order (3) Be able to understand the structures of carbocations, carbanions, free radicals, and carbenes and the structural features that stabilize them. Explain which are electrophilic and which are nucleophilic o Carbocations 6 electrons and a positive charge Vacant p orbital Strong acid Properties: electrophilic and strong acids Stability: 3° > 2° > 1° > methyl Stabilized through inductive effects or hyperconjugation or resonance o Carbanions 6 electrons, one lone pair, trivalent and negative charge Not electron deficient Strong bases Properties: nucleophilic and strong bases Stability: methyl > 1° > 2° > 3° Alkyl groups and other electron-donating slightly destabilizes carbanion Order of stability is usually opposite of carbocations o Free radicals Electron deficient Perpendicular p orbital contains odd electron Properties: electron deficient (electrophilic) Stability: 3° > 2° > 1° > methyl Stabilized by alkyl substitution and resonance Higher substitution, higher stability o Carbenes 6 electrons, 4 bonding, one lone pairs Vacant p orbital Can be both electrophilic and nucleophilic Chapter 6 (Section 6.2): Be able to correctly name alkyl halides using IUPAC and common names. o IUPAC Prefixes: Fluorine: fluoro- Chlorine: chloro- Bromine: bromo- Iodine: iodo- Steps: Find main carbon chain Number main chain Use lowest possible numbers for position Identify all prefixes and their position numbers Write the full name Prefix - Parent - Suffix o Common names o Steps Name all the C atoms of the molecule as a single alkyl group Name the halogen bonded to the alkyl group Combine the names of the alkyl group and halide, separating the words with a space o Some are different CH 2 2s methylene halide CHX 3s haloform CX 4s carbon tetrahalide Be able to name linear, cyclic, and bicylic systems that have halogens. o Linear o Cyclic o Bicylic
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