CHM 25500 Exam 2 Study Guide
CHM 25500 Exam 2 Study Guide CHM 25500 - 001
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CHM 25500 - 001
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This 4 page Study Guide was uploaded by Gayatri on Wednesday October 14, 2015. The Study Guide belongs to CHM 25500 - 001 at Purdue University taught by Christopher H Uyeda in Fall 2015. Since its upload, it has received 230 views. For similar materials see Organic Chemistry in Chemistry at Purdue University.
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Date Created: 10/14/15
CHM 25500 Exam 2 Study Guide Chapter 4 Acids and Bases BronstedLowry Acids and Bases o Acids proton H donors I A conjugate base is formed when an acid loses a proton I Strong acids deprotonation 9 weak conjugate bases 0 Bases proton H acceptors I A conjugate acid is formed when a base accepts a proton I Strong bases protonation 9 weak conjugate acids Acid Dissociation Constant Ka Keq H20 A39 H3O HA pKa used to de ne acid strength 0 pKa log Ka 0 low pKa strong acid 0 high pKa weak acid 0 Reactions go toward higher pKa formation of weaker acidsbases from stronger ones Acidity Trends evaluated by looking at stability of the conjugate base conjugate base stability increases 9 acidity increases Five factors 1 Electronegativity of atom with charge more EN element more stable anion EN increases going 9 on periodic table More EN atoms can hold charge better 2 Size of atom with charge larger element more stable anion Valence shell for large elements is more diffuse so orbitals that are delocalized occupy a larger volume at a lower energy making them more stable Opposite of EN trend Size takes precedence over EN going up and down on table 3 Hybridization of carbon center with charge more s more stable anion sp gt sp2 gt sp3 hybridization 4 Resonance delocalization of charge more delocalization more stable anion More delocalization of electrons gives rise to a more stable compound Strength increases with increasing of major contributing resonance structures 5 Inductive effects and e39 stabilization presence of withdrawing groups e39 more stable anion Electronegative elements that pull electron density stabilize the charge Chapter 3 Stereochemistry and Chirality Chiral molecules objects that are not superimposable on their mirror images 0 Exist in L and R forms shoes feet treaded screws Achiral molecules objects that are superimposable on their mirror images 0 Have planes of symmetry socks shells nails Stereocenter a tetrahedral atom with four different substituents a common cause of chirality in molecules 0 Can be either R clockwise right or S counterclockwise leftsinister 0 Steps for assigning stereocenters 1 Order substituents from 1 4 by decreasing atomic weight 2 Rearrange so that 4 is out of the plane away from you 3 Draw arrows connecting l 9 2 9 3 a If clockwise R b If counterclockwise S 4 If two substituents are the same atom go to next atom attached to it to assign priority 5 If doubletriple bonds are present atoms bound w those bonds count as twothree of those substituents Relationships between molecules 0 Enantiomers nonsuperimposable mirror images chiral have onnosite stereocenters o Diastereomers stereoisomers that are not mirror images 0 Meso compounds superimposable mirror images achiral o Identical when two compounds are identical not unique one is usually a meso compound Maximum possible stereoisomers 211 where n number of stereocenters Meso compounds arise from achiral molecules have plans of symmetry Mirror images of chiral compounds have opposite stereocenters Fischer projections use directions of lines to depict 3D structures 0 Vertical lines going back into page 0 Horizontal lines coming out of page Physical properties of enantiomers and diastereomers o Diastereomers are different compounds so they have different physical and chemical properties 0 Enantiomers have identical physical and chemical properties in an achiral environment I They interact with plane polarized light and also in different ways with other chiral molecules I In nature all biological molecules are chiral different chiral receptors recognize the molecules in different ways Eg R vs S Carvone RCarvone 9 caraway SCarvone 9 spearmint Can exist as racemic mixtures equal amounts of two enantiomers or racemization can occur inside the body Percent optical purity anglesampleanglepure enantiomer x 100 o Enantiomeric excess RS or vice versa I 0 9 11 mixture I 100 9 exclusively R or S check rotation direction Chapter 5 Alkene Structure Alkenes unsaturated hydrocarbons with CC double bonds 0 Formula CnHzn o Orbitals involved in bonding for alkenes 3 sp2 hybridized one p unhybridized one sigma bond between spz s one pi bond between p s Cistrans alkenes with 2 substituents o The lockedin double bond between CC cannot rotate giving rise to stereoisomers diastereomers with two configurations I Cis substituents on same side I Trans substituents on opposite side EZ alkenes with Z 2 substituents 0 When more than two substituents exist priorities are assigned to each and two varieties exist I Z two highest priority groups are on the same side I E two highest priority groups are on the opposite side No ZE when same substituents are repeated 9 terminal alkenes Nomenclature for molecules containing alkenes o Ending ene Parent chain is numbered so alkene has lowest Stereolabels E or Z are in Alkenes take priority over substituent groups Compounds with 1 alkene add on pref1xes diene triene etc Cyclic compounds start with 1 at start of alkene 2 at end C C vs CC OOOOO Properties C C CC Bond length 154 Angstroms 134 Angstroms Strength BDE 90 kcalmol 172 kcalmol Rotation Free 3kcalmol barrier Restricted 63 kcalmol barrier Chapter 6 Alkene Reactions Potential Energy Diagrams show change in energy as a reaction proceeds 0 Starting materials Reactants o Intermediates molecules made in between steps of a reaction are produced from reactants but consumed to make products 0 Products made from reactants 0 Transition state peak of potential energy diagrams usually unstable and shortlived 0 Activation energy energy needed to get from potential energy of reactants to transition state peak energy As this increases rate of the reaction decreases Nucleophiles and Electrophiles o Nucleophile a compound that has excess electrons and wants nuclei I Compound with an atom that has a lone pair I Compound with a pi bond usually CC Strong nucleophiles are usually strong bases 0 Electrophile a compound that wants electrons from others I Compound with less than 8 electrons I Compound with a pi bond usually CO I Compound with an electron de cient site polar covalent bond Carbocations three groups attached to an sp2 hybridized carbon with a formal charge also have an unhybridized p orbital 0 Work as good electrophiles because they are electron de cient only 6 e o Classi ed in four ways I Methyl C with 3 H s I Primary C with 1 CR3 group and 2 H s I Secondary C with 2 CR3 groups and l H I Tertiary C with 3 CR3 groups 0 Stability is acquired through various ways I Electron donating releasing groups so molecules with more CR3 groups are more stable 9 methyl lt primary lt secondary lt tertiary I Adjacent alkyl groups also stabilize carbocations through two methods 1 Inductive effects since C is slightly more electronegative than hydrogen it has a very very slight tendency to pull electrons toward it and have a negative charge so carbons can donate some electron density and spread the positive charge over more atoms 2 Hyperconjugation the adjacent CH or CC sigma bonds can overlap with unoccupied p orbitals nearby to make the moleculebond more stable Reaction mechanisms 0 Arrow pushing arrows go from areas of electron excess to de ciency N E o Bonds must be broken in some cases to follow octet rules 0 Reaction intermediates must be somewhat stable cannot have atoms with more than eight electrons or hydrogens with more than two electrons Product stereochemistry o Electrophillic addition reactions are stereoselective o Nucleophiles like to react at opposite side of C1 or Br less hindered site 0 Achiral molecules 9 achiral OR chiral racemic products 0 Bromination of cis alkene intermediate is an achiral meso compound but product is a chiral racemic mixture Bromination of trans alkene intermediates are chiral enantiomers but product is an achiral meso compound 0 Markovnikov vs antiMarkovnikov Addition 0 Markovnikov product has OH on more substituted part 39 USES H20 H2804 o antiMarkovnikov product has OH on less substituted part I Uses BH3 Oxidation and Reduction Reactions 0 Oxidation loss of H2 gain of O I Dihydroxilation using OsO4 and then NAHSO3 H20 9 OH groups on same side I Ozonolysis using O3 and CH3ZS 9 CC is broken 2 aldehydes are formed 0 Reduction gain of H2 loss of O I Alkenes can be reduced to alkanes using H2 and Pd as catalyst alkene hydrogenation I H s are added on same side of double bond
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