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Modern Organic Chemistry I

by: Marilou Hyatt Jr.

Modern Organic Chemistry I CHEM 341

Marketplace > Colorado State University > Chemistry > CHEM 341 > Modern Organic Chemistry I
Marilou Hyatt Jr.
GPA 3.81

Debbie Crans

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Debbie Crans
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This 83 page Class Notes was uploaded by Marilou Hyatt Jr. on Tuesday September 22, 2015. The Class Notes belongs to CHEM 341 at Colorado State University taught by Debbie Crans in Fall. Since its upload, it has received 37 views. For similar materials see /class/210346/chem-341-colorado-state-university in Chemistry at Colorado State University.

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Date Created: 09/22/15
9 November Mass spectrometry Reveals a molecule s exact mass Molecule gets bombarded by beam which kicks an electron out leaving behind positively charged molecular ion on may fragment further to smaller ions Can reveal exact mass of molecular ion to 4 decimals It allows determination of molecular formula accurately Fragmentation of Functional Groups Alkyl halides Alkyl bromides lose organic fragment Alky chlorides maintain Cl atom in organic fragment Ethers Heterolytic with e s going with O atom Homolytic alpha cleavage forming carbonyl cations Alcohols Between C and O heterolytic Between C and C or H hemolytic Cleavage of weakest bond Ketones McLafferty rearrangement 18 November What NMR tells us Chemical shift 6 Integration Spin spin splitting Signal splitting defining H atom neighbors NMR spectra usually complex not only singlets Number of peaks n 1 n of H attached to C next to that group Protons bound to a carbon that has a carbon with other protons are affected by the other protons spin spin splitting Practical Aspect of Signal Splitting How Many amp Sizes N 1 Rule HC signal splits into n 1 peaks n H s on adjacent C s Signal Splitting Structural Complexities If adjacent H s are not all equivalent chemically equal situation can be more complicated split signals may be split individually Recommended Order of Determination List signals to assign and 6 to match up with Begin with easiest peaks Such as CH next to O CH2 or CH3 with only one neighbor with H s is used to make sure assignment is proper Splitting is the final test The coupling constant of a trans is bigger than that of a cis 9 September Markovnikov s Rule electrophile adds to sp2 carbon bonded to the greater number of hydrogens Why does the Markovnikov rule work Tertiary more stable than secondary more stable than primary Hyperconjugation delocalization of electrons by overlap of 0 bond orbital and empty p orbital n carbocation rearrangement the driving force is the stability of tertiary carbocations over secondary carbocations 14 September Hydroboration AntiMarkovnikov Overall net addition of water across double bond hydration but opposite to acid catalyzed sense antiMarkovnikov Secondary carbon is a little less electronegative Attracts H rather than BHZ B ends up on smaller less substituted carbon Peroxyacid addition to alkene Expoxide formation Ethylene oxide simplest epoxide Addition of hydrogen Reaction does not occur because barrier is too high H2 bond is too strong In presence of catalyst Pd on carbon or PdC Pt on carbon Ni others metal facilitates H2 activation and reaction occurs Metal facilitates lower energy route Catalyst Ithat helps quot39 39 a f 39 but does not itself get consumed Concept helping govern reactions Alkene Stability Hydrogenation is exothermic negative AH CC n bond 63 kcalmol HH 0 bond 104 kcalmol C H bond 100 kcalmol x 2 Bonds broken 167 kcalmol Bonds formed 200 kcalmol AH 167 200 33 kcalmol Heat of hydrogenation Different C H bonds have different energies Alkenes with more alkyl substituents less hydrogen on alkene are more stable Synthesis Synthesis building or putting something together Organic synthesis making molecules building molecular structures 21 October Where are we going with 5N1 and SNZ reactions We want to be able to use so we need to understand how different substrates react Alkyl halide Primary 5N2 Tertiary SNl Secondary both 5N2 or SNl can occur Other factors nucleophile leaving group solvent Aryl and Vinyl Halides 5N2 backside attack unfavorable e repulsion SNl requires cation vinyl cations unfavorable SNl and 5N2 donotoccur spZC CI doesn t often leave Order of leaving 39gt Br gt OH gt F Reason Hi is strongest acid I39 is most stable among HBr HCI HF The Nucleophile Strong nucleophiles tend to be strong bases and vice versa H0 is more nucleophilic than H20 also better base CH30 gt CH30H NH2 gt NH3 Nucleophilicity HZN39gt HO gt F39 also order of basicity Nucleophiles in order of nucleophilicity RS39gt I gt CN39gt CH3039gt Br gt NH3gtCI gt F39 gt CHSOH How can I be a good leaving group m good nucleophile Easily polarizable IC bonds are weak Comparing SNl and 5N2 reactions Substrate Resonance Nucleophile Leaving group Solvent Comparing 5N2 to SNl Rate for 5N2 is proportional to alkyl halide and nucleophile Rate for SNl is prop to alkyl halide Increased concentration of nucleophile favors 5N2 Solvation favors SNl Strong nucleophiles favor 5N2 Polar solvents favor SNl when neutral reactants become charged intermediates and transition states polar solvents solvate charged species better than nonpolar solvents 30 September Alkynes Alkynes are similar to alkenes Nomenclature ending in yne position indicated with number Alkyne Reactions Alkynes undergo addition reactions Alkynes are less reactive than alkenes Cation stability on double bond is less lower reactivity Tertiary gt secondary gt secondary vinyl gt primary gt primary vinyl Why are vinyl cations less stable Structure of cations Alkyl on sp2 carbon Vinyl on sp carbon sp carbon is more electronegative than sp2 closer to nucleus Hyperconjugation less for vinyl cations than alkyl Addition of HX Regioselecitivity Markovnikov vs anti With one equivalent same moles reagent as substrate of HX addition of one occurs with excess two adds Geminaldibromide 2 bromides on same carbon Tautomerization shift of group and n electrons Addition of hydrogen Hydrogenation of alkynes is similar to alkenes Semireduction is possible Lindar s catayst H2 Pd BaSO4 quinolone Lindar s catalyst is deactivated Reduced alkynes but not alkenes 29 August Conformations different rotations around a bond Staggered bonds staggered when seen end on lower energy Eclipsed bonds overlap when seen end on higher energy Ring strain Angle strain when bond angles differ from ideal Torsional strain repulsion between bonding electrons or neighboring substituents Steric strain atoms or groups that approach each other too close 5 December How to evaluate where you are at in class for final Grades 50 65 D 70 80 C 80 90 B 90 A Grades and points what is needed on final 121 pts FD 179 C about 65 75 ish 250 B 286 A 321 A 357 Hydroxybenzene is phenol Methylbenzene is toluene Benzene as a substituent is phenyl Why are 2 6 10 etc aromatic but 4 8 are not Unpaired e result in unstable compound e don t pair Hund s rule because of degenerate orbitals 24 October Review Chapter 8 Y H3C39X 9 HCH3 X Simple reaction alkyl halide reagent Exchange of one group for another at sp3 carbon Chapter 8 is all about variations in Alkyl halide Alkyl group Leaving group Reagent Selecting which reaction SNl or SNZ To determine which reaction we need to understand each reaction w different substrates First consider substrate Primary SNZ Tertiary SNl Secondary both SNZ and SNl can occur Then consider other factors Nucleophile Leaving group Chance for resonance Solvent Secondary substrates are often challenging Reactions of multifunctional reactants Examine each functionality separately with that reagent Consequences of Chapter 8 Why reactions are illustrated by alkyl halide reagent Exchange of halide at sp3 carbon possible Because halides are great leaving groups What happens when X is replaced by OR What happens when X is replaced by NHR What are other good leaving groups Substitution not always the only product Chapter 9 Elimination Reactions Substitution 5N1 or 5N2 is not the only possible reaction quotEliminationquot is a common alternative Types E2 and E1 E2 elimination only one product Rate olt alkyl halide base Leaving Group Dependence gt Br gtCgt F 1 On the molecule to the right indicate the following N W On the molecule to the right indicate the following C341 Extra Practice a Provide the hybridization of the indicated atoms b Circle any parts of the molecule that are conjugated c Indicate any aromatic rings d Put a box around a quaternary 49 carbon e How many 39 carbons are there a Add the lone pairs to all atoms that possess them b Provide the hybridization of the indicated atoms c Circle the most polar bond d Put a box around the shortest carboncarbon bonds D e Is the alcohol 19 29 or 39 f Is the amine 19 29 39 or 49 Rank the following sets ofmolecules from most acidic to least acidic H a CH 3 H30 v N H CH3 CH3 H o b HoCH3 H2NCH3 HoCF3 JL HO 4 Rank the following sets ofmolecules from most basic to least basic SH NH2 3 NH2 OH 2 NEE CH3 lt32 5 Predict whether the equilibrium of the following reactions will sit towards the products reactants or have equal quantities of both a A CT JET gt AOH T bl OH O eEEquot ll gt 2 H ll 6 Rank the following molecules from lowest to highest boiling point Indicate the types of intermolecular forces present in each case vym 4 JVMHZ Mo 7 Rank the following molecules from most to least water soluble 8 Indicate whether each of the following molecules would be most likely to react as a nucleophile or an electrophile 20 b H c Br d 0H e S f CI ngr Br h A 9 Draw Newman projections down the C2C3 bond of the most and least stable conformers of 2 uoro4methylpentane Indicate any gauche interactions 10 a Draw the most stable chair conformation ofa nonmeso chiral diastereomer of 2 isopropyl13dimethylcyclohexane Either enantiomer is fine b Draw the chair ip of the above cyclohexane and explain why this conformer is less stable higher in energy 11 Provide the IUPAC names of the following compounds using both RS and EZ designations where appropriate a b CI OH 12 Indicate the relationship between the following pairs of molecules Choose from Identical I Diastereomers D Conformers C Constitutional or Structural Isomers S Enantiomers E Not Isomers N Br Br a b H2N CH3 and and I H 2N Br Br CH3 c 939 d F and O F H30 H and 39 CH3 B CH3 Cl C Br HO Br Br d l e NOH NNHz 3 and 3 and Br B 3 NH2 r 13 Draw the following molecules a 35453chloro4methylhexane b SB45dimethylheX3en2ol 14 How many stereoisomers do the following molecules have Draw the structures of any meso compounds a CEO b Br c 0 gt C CHCH3 W CI H OH NH2 Br Br 15 Answer the following questions the energy diagram below a How many intermediates are there in the mechanism b How many transition states are there in the mechanism c Is the overall reaction exothermic or endothermic d Label the axis of the energy diagram and the activation energy of the rate determining step 16 Rank the following alkenes from lowest to highest heat of hydrogenation QWQW JJVVWQW 17 Which of the following E2 elimination reactions would you expect to be faster Brie y explain your reasoning CH3 CH3 reaction A quot1039 CH3O quot CH3 CH3 CH3 CH3 391 reaction B O CI CH30 O quotCH3 CH3 18 Predict the products of the following reactions and answer the proposed questions Br 81 CH3SNa Br b CH3ONa gt Br C CH3OH Br I CH3ONa e JV NaCN Br gt f gtBr OH a Br H 0 g gtr 4 h Circle the best solvent to use for the reaction in part a DMSO CH30H hexane i Circle the best solvent to use for the reaction in part c DMSO CH30H hexane 19 Predict the major products of the following reactions a gt0 NaSCH3 o b H ON NaCN c CH3OH H2804 d OH 6 W POCI3 pyridine OH f OH gt g Y SOCI2 pyridine gt OH 11 PBr3 pyridine gt OH 20 What would be the best route to synthesize isopropyl pentyl ether using the Williamson Ether synthesis 21 Provide a mechanism for the reaction in question 19e 2 2 Provide a mechanism for the reaction in question 19g 23 Provide a mechanism for the following transformation accounting for both products that are formed CH30H H2304 CH3 CH3 gt H0 OCH3 OCH3 CH3 2 2 Provide a mechanism for the following transformation Brz H20 OH Br racem ic 23 Predict the major products for the following reactions 3 Br2 e b H20 H2804 Y c F AA HBr gt d 1 BH3 gt 2 H202 OH 6 H2 PdC f H20 H2804 24 Provide a mechanism for the following transformation and draw an energy diagram assuming the overall reaction is exothermic gtr HCI CI 25 How could you prepare each of the following products from acetylene H lquot a 1 b gt OH racemic 26 Predict the major products of the following reactions 1 Wm y K 1 BH3 2 H202 OH 27 Provide a mechanism for the following reaction 0H Cr03 H2804 H20 0 gt k 28 Predict the products of the following reactions a Na NH3 b H2 Lindlar39s catalyst K 20 r207 C OH gt Y H2804 H20 01 P00 YOH gt 2 SCH32 CI 0 1 0so4 2 NaH803 29 Provide the products of the following reactions 3 Brz hV bl O CI2 hv C quot39 HBr ROOR d HBr ROOR e NBS hv a f NBS hv M 30 Which of the following molecules are conjugated I j 1 l M N W O GM 31 a The following cation has 4 additional reasonable resonance structures Draw all four of these structures OCH3 N O b Which of the five resonance structures the four you drew plus the original cation is the major resonance contributor to the hybrid structure Brie y explain why 3 2 Rank the following molecules from longest Kmax to shortest Kmax 9lt lt 4 y e 4 5 33 Provide the products for the following reactions Ifmore than one product is formed indicate which is the major product a HBr high temperature HBr low temperature W b W O 0 34 Which of the following molecules are aromatic Which are antiaromatic NH 8 H 2 O DC I I OH N N l N 5 October 154 A normal single bond 133 A normal double bond 139 A bond length in benzene ring The Rules of Resonance How to show resonance structures Electrons move atoms never move n electrons and lone pair electrons move Electrons move towards 6 or n bond How does resonance affect stability Delocalization is beneficial Rule of thumb every good resonance structure increases stability and decreases overall energy Reaction Summary for the C341 Final Exam denotes you are responsible for knowing the mechanism Substitution Reactions NU SNZ R X strong nucleophile R Nu R R 5N1 Rx NuH R x gt R X weak nucleophile R Nu Elimination Reactions R H B R EZ X gt R strong base R R R E H BH R X gt R weak base R R Alcohol Ether and Epoxide Reactions Alcohol formation A OH gt V111 5N2 R X R OH Williamson 1 1 N H R th th 39 a J 9 er syn 9515 R OH R 0 AR 2 R X Epoxide formation HO 0 via halohydrl39n H L A quotI Br Dehydration of alcohols with acid Dehydration 0f Alcohols with POClg Conversion of an alcohol to an alkyl halide with HX Conversion of an alcohol to an alkyl bromide with PBr3 Conversion of an alcohol to an alkyl chloride with 50672 Conversion of an alcohol to a tosylate with tosyl chloride EpoXide opening with a strong nucleophile EpoXide opening with acid and a weak nucleophile R R R R OH OH OH OH OH OH R R H2804 or TsOH POCI3 pyridine HX PBr3 pyridine SOCI2 pyridine TsCI pyridine 1 Nu 2H20 HNU H2804 R NR X JVR R racemic Ejr RR OTs Nu Alkene Reactions Electrophilic addition of HX Hydration of Alkenes Electrophilic addition ofX2 Halohydrin formation WA K Hydroboration oxidation of alkenes EpoXidation using peroxy acids WA Hydrogenation of alkenes Y HX X 7lt H20 H2804 Brz Br gt w Br racemic Brz H20 quot Br racemic 1 BH3 2 H202 OH39 OH racemic mCPBA fgt racemic H2 PdC WA Alkyne Reactions Electrophl39lic addition of HX Electrophilic addition osz Hydration of alkynes Hydrob oration oxidation of alkynes Hydrogenation of alkynes SNZ substitution with acetylide anion Epoxide opening with acetylide anion HX X2 H 20 H2804 1 BH3 2 Hon OH H2 PdC 1 NaH V V XX X Mx XX Oxidation and Reduction Reactions Reduction ofalkyne H2 Lindlar39s catalyst t0 Cis alkene Reductl39on ofalkyne Na NH3 to trans alkene r SNZ substitution NX LiAH4 with LiAIH4 X Cl Br I Epoxide opening 0 1 LiAIH4 wzth L1AIH4 7Q 2 H20 1 mCPBA Trans dihydroxylation 2 OH Or H20 H2804 1 0804 Cis dihydroxylation 239 NaHSO3 V OR KMnO4 H20 0H 1 O3 2 SCH32 or Zn Oxidative Cleavage ofalkenes ozonolysis Y 5 H OH A 6H racemic OH H OH racem ic YOJ 0A HOJI r Oxidative Cleavage 2 HzO ofalkynes OH ozonolysis 1 O 3 gt O 002 2 H20 OH mild oxidation of OH PCC 390 primary alcohols gt OH CrO3 or Na2Cr207 or K20r207 0 strong oxidation of gt L primary alcohols H20 H2804 OH PCC OH OR 0 oxidation of gt k secondary alcohols Cro3 or Na20r207 or K20r207 H20 H2804 Radical Reactions B h Radical bromination A L gt9 of alkanes Br Radical allylic bromination of NBS hv 39 Br alkenes W W w Br Br racemic Radical addition HBr ROOR of HBr racemic Reactions of Conjugated Dienes HBI H Addition ofIIX N Na to conjugated dienes h39gh temperature thermodynamic Addition ofIIX B to conjugated dienes N HBr H N gt kmetlc low temperature racemic I 39 DielSAlder reaction gt racemlc 39I 11 November Infrared Spectroscopy Bonds display two types of motion stretch and bend The frequency of motions are dependent on some substituents Infrared IR Bond Vibrations The energy must match frequencies for absorbance in the IR 24 August Stronger acids have more electronegative atoms Stronger acids have the electronegative atom closer to the H that leaves The equilibrium favors the side of the highest pKa Curved arrows indicate where a pair of electrons starts from and where it ends up Drawing organic molecules Every point is a carbon atom unless it has a letter on it On carbon hydrogens are implied fill its required valence Alkanes CH4 Methane Csz Ethane caHg Propane C4H10 Butane C5H12 Pentane C6H14 Hexane C7H16 Heptane CBH18 Octane Cgqu Nonane CH7sz Decane Formula Cn HZM somerism Constitutional isomers molecules with same molecular formula but differing in how they are connected Nomenclature Substituents are named as the parent alkane but ending in yl One carbon group methane drop ane becomes methyl Name alkane according to longest linear chain Number longest linear chain from end closest to substituent Ex 3methyloctane 36dimethylnonane 6ethyl3methylnonane Ethyl is first because E comes before M in the alphabet 1 N C341 Study Session December 2 and 3 For each molecule below draw the most stable radical that can be formed QWM Ar The following questions surround the bromination of 3methylpentane a Provide the initiation step for radical bromination b Provide the propagation steps to form the monobromide product c Provide the possible termination steps for the radical bromination d Why would you not want to run the reaction with a large excess of bromine 3 It is widely known that radical chlorination results in low selectivity for monochlorination Chlorination reactions of certain alkanes however can be used for laboratory preparations Explain why the alkanes below can undergo radical chlorination to provide a single monochlorination product X 4 What would be the product ratio expected from chlorination of 2 methylpropane if all of the hydrogen atoms were to react at equal rates 5 If each of the following molecules were treated with 1 equivalent of Br2 and light hv which would produce a single major bromination product For each molecule drawthe major bromination products expected DWNX QM Hint benzene can stabilize a radical in a similiar way as an allyl 6 Provide the products of the following reactions 51 Gr Nth b HBr c5 c ACT SQ d Y L hv 7 What would the major and minor allylic bromination products be for the following reaction where there is more than one allylic position gt hv CO Starting with the compound or compounds indicated and using any other necessary reagents propose syntheses for each of the following compounds a Ethyl bromide from ethane b Diethyl ether from ethane c Cyclopentene from cyclopentane d 2 Bromo3methylbutane from 2 methylbutane e 2 Butyne from methane and acetylene f 2 Butanol from ethane and acetylene g propiononitrile CH3CH2CN from ethane 31 August Alkenes Hydrocarbons w at least one double bond Unsaturated Cycloalkanes and Alkenes General Formula CnHzn Each ring and each n bond removes 2 H s from alkane general formula Endless possibilities including more than one double bond Molecules based on formulas can have different structures Nomenclature Ending in ene double bond should be numbered Rules same as alkanes double bond must be in longest linear chain Alkenes are unsaturated hydrocarbons they have less than max of hydrogens Vinyl carbon sp2 carbon of alkene Vinyl hydrogen bound to vinyl carbon Allyl carbon sp3 carbon attached to alkene Allyl hydrogen bound to allyl carbon Alkene Structure Double Bond composed of 1 0 bond 1 n bond n bond requires p orbital from each participating atom C must be sp2 Recall that sp2 atom has all sp2 orbitals in same plane Cis isomer priority constituents are on same side of n bond Trans isomer priority constituents are on opposite sides of n bond Cis and trans alkene isomers are different compounds E Z Nomenclature Compare substituents on first carbon involved in double bond Assign priority based on atomic number Higher number is higher priority Repeat on other carbon Zusanmen quottogetherquot higher priority on same side 2 Entegen lloppositequot higher priority on opposite sides E You may need to go beyond first atom to determine priority Keep moving down the chain until one atom on one side has priority over the other Doesn t matter what s on the far end if you reach something else first Mechanism How Organic Molecules React Has to have a nucleophile and an electrophile Nucleophile electron density Organic reactions result after bond formation and cleavage the movement of electrons Electronrich molecules nucleophiles are attracted to electronpoor atoms or molecules electrophiles n bond nucleophile gives 2 electrons to electrophile Proton will bind to least substituted C Markovnikov rule 10 October The DielsAlder Reaction Reaction of alkenes to form rings No intermediates continuous movement of electrons Electron withdrawing substituents on dienophile accelerate reaction Anything w e drawing properties is a good dienophile electronegative Diene requirement is scis arrangement HOMO highest occupied molecular orbital LUMO lowest unoccupied molecular orbital Relating to energy Going for best HOMOLUMO gap Electron distribution governs regiochemistry 28 October Elimination Reactions Related to 5N1 E1 Reagent is a poor nucleophile or base Substrate a tertiary alkyl halide E2 more substituted alkene is major product Bulky bases change main product E2 elimination reaction is contrary to E1 reaction E1 elimination Rate dependent only on alkyl halide Weak base may remove protons Alkyl halide 5N1 5N2 E1 E2 Tertiar Yes when weak No Yes when weak Yes when strong y Nuc or weak base Nuc or weak base base Stereochemistry of Eliminations E2 requires leaving group and hydrogen beantiperiplanar E1 carbocation intermediate stereochemistry is thermodynamically determined 2 November Alcohol Substitution Recall OH and R0 are horrible leaving groups But H20 is a good leaving group Alcohol Substitution Other Methods Strategy Activate alcohol and displace in same pot SOCL2 thionyl chloride PBr3phosphorous tribromide Sulfate esters Introduce nucleophile in second step Alcohol Elimination Dehydration Reverse of alkene hydration use nonnucleophilic acid Ether Substitution Similar to alcohols acid mediated substitution HI is faster than HBr iodide better nucleophile Epoxides Epoxides as 3membered ring ethers are easy to cleave because of strain energy Reactions acids open epoxides Regioselectivity Partial positive charge is better tolerated at secondary sites Organometallics Binds between C amp other elements Li Mg Zn ampc Reverse polarity such that C is negative 30 November 13C NM R Spectroscopy 13C NMR uses 13C isotope Simple because CC coupling is not observed natural abundance of 13 C Is 1 Spinspinsplitting tells how many protons are on the C Chapter 14 Aromatic Chemistry Benzene C5H5 has a high C to H ratio special compound resonance derivatives have pleasant odor aromatic Aromatic denotes a class of compounds Aromatic criteria are Continuous cyclic cloud of electrons Cyclic Every atom has a porbital in a planar structure n cloud contains an odd number of electron pairs 4n 2 rule Antiaromatic Cyclic Continuous Even e pairs 7 December Reactions of Benzene Addition fails but substitution occurs under forcing conditions More stable than akanes and akenes Always remains aromatic Electrophilic reagents will be successful Additionelimination rxn Adding halogen Brz Clz Iz not sufficiently electrophilic Must be llactivated FeBr3 etc as catalyst 7 Octo be r Dienes Nomenclature as alkenes w di or tri in front of ene Allenes Allenes 12dienes cumulenes resemble CO2 Sp hybridized central carbon 9 n bonds are perpendicular Diene Stability Diene stability order Conjugated dienegt isolated dienegt cumulated diene Diene Reactions Isolated diene Reacts to make more stable carbocation Most nucleophilic alkene reacts faster Diene Reaction Mechanism Allylcation n bond plus adjacent empty p orbital sharing of e39 Regioselectivity in Diene Additions Most sable resonance contributer determines regioselecitivity 12 or 14 addition favored Thermodynamic and kinetic control Thermodynamic or kinetic Themodyanmic control relative stability of products Reversible process Kinetic control relative stability of transition states Irreversible process Kinetic control low temp Thermodynamic control high temp 9 December Stille Reaction 2 Pd cat 1 2 RlSnalkyl3 R X gt R R XSnalkyl3 R1 allyl alkenyl aryl R2 alkenyl aryl acyl X Cl etc Important notes John Stille was a CSU chemistry professor and would have gotten a Nobel Prize in 2010 if he had been alive One of several reactions that uses a Pd catalyst to make a CC bond Is the only one of these that uses Sn Look at the pieces that change and really matter Spectroscopy Spectroscopy is the study of the interaction of energy with matter Energy applied to matter can be absorbed emitted cause a chemical change or be transmitted Spectroscopy can be used to elucidate the structure of a molecule Examples of spectroscopy Infrared IR Infrared energy causes bonds to stretch and bend IR is useful for identifying functional groups Nuclear Magnetic Resonance NM R Energy applied in the presence of a strong magnetic field causes absorption by the nuclei of some elements NMR is used to identify connectivity of atoms in a molecule Mass Spectrometry Molecules are converted to ions Ethyl group triplet quartet in NMR Infrared Check C 0 Check OH peak Nucleophilic Substitution vs Elimination Methyl 1 2 3 quot39 39 39 reactions onl SNlEl or E2 No 5N2 rxn n solvolysis gives lees mamlySNzw SNlEl and at lower Gives mainly 5N2 except with a weak bases eg I tem eratures S 1 is Gives 5N2 reactions hindered strong base CN RCOZ and p N favored When a and then gIves maInly maInly E2 With strong strong base eg R0 2 bases eg RO Is used E2 predominates Solvolysis solvent reacts Hydrohalogenations Markovnikov s Rule Addition of HX to an alkene proceeds so that the hydrogen atom adds to the C that already has the most The product w the most stable carbocation intermediate predominates Most stable carbocation formed fastest because it has lower AGi Hydroborations AntiMarkovnikovSyn Hydration Rxn leads to syn and antiMarkovnikov addition of water to alkenes Steric Hydroboration Synthesis of Alkylboranes Hydrogen and boron are added across double bond In practice borane complex w solvent tetrahydrofuran THF often used Mechanism of hydroboration Boron hydride adds successively to three molecules of alkene Boron becomes attached to the least substituted C of double bond Bulky boron group approaches least sterically hindered C easily Orientation allows 6 in transition state most substituted C The boron and hydride add w syn stereochemistry Hydrogenations Alkynes Rxn of H using regular metal catalysts 9 alkane Lindlar s catalyst produces cisalkenes from alkynes DielsAlder 14cycloadditionrxn ofdienes Forms cyclohexane product 2 new s bonds formed at expense of 2 p bonds Conjugated diene 4p 339 system Dienophile is 2p e system Product called adduct Factors favoring Very low yield amp high temp necessary for simplest For good yieldlow temp e withdrawing groups Helps if diene has e39 releasing groups Aromaticity Huckel s rule 4n 2n e rule Planar monocyclic rings w continuous system of p orbitals and 4n 2n e n 0 1 2 3 etc Substantial resonance stabilization Undergo electrophile aromatic substitution EAS Electrophile has full or partial charge Addition of bromine 9bromionium ion 9 backside attack trans addition Prefixes alphabetical 2 chains of equal length chain w greatest number of substituents is parent Substituents equal distance from either end lower first point of difference Alcohols similar to H20 Weak acid weak base H donor and acceptor Ether weak base H bond acceptor Amines bad leaving group Ammonia great leaving group 31 October Alkyl halide 5N1 5N2 E1 E2 Good nuc weak Yes good base but anary No base No 5N2 usually better Secondar Yes when weak Yes w good nuc Yes when weak Yes w good y nuc or weak base and weak base nuc or weak base bulky base Tertiary Yes weak nuc or No Yes when weak Yes w Strong base wea k ba se nuc or weak base 22 August Homework for chapter 1 posted on ramCT Practice exam 1 posted Grade based on 2 exams out of 3 drop worst exam Homework due Friday Bonds in a tetrahedron are 1095 apart Carbon based Chemistry geometry around Carbon All single bonds o Tetrahedral sp3 All double bonds 1 o 1 nTrigonal planar sp2 All triples bonds 1 o 2 n Linear sp Lone pairs occupy as much space as bonds Radicals single electrons and empty orbitals do not Acidity How do you know which protons quotreactquot deprotonate Bronsted Acid proton donor Bronsted Base proton acceptor Lewis Acid electron acceptor Lewis Base electron donor HA 9 H A 11 A7 K 39 HA PKA 39logKA 4 November Carbonmetal bond reacts as carbon anion It is common to do an aqueous workup after the end of the reaction and the alkoxide gets protonated by water Coupling Reactions Organometaollics can be directly coupled with aryl and vinyl halides in the presence ofa Pd catalyst Ch 11 Alkanes are unreactive by heterolytic reaction Alkanes react by hemolytic cleavage Alkane Reactions Combustion w oxygen high T CnH2n2 029 CO2 C0 C H20 Radical Halogenation w F2 Cl or Br at high T or under irradiation light CH4 Clz Cchl HC Homolytic cleavage of ClCl bond by a radical chain reaction Stability of radicals same as stability of carbocations 12 September Skeletal rearrangement carbocation rearrangement more likely if carbocation is less stable primary carbocations very likely to rearrange secondary likely tertiary unlikely Ring strain drives ring expansion Addition of Halogen to Alkene Brz and Clz are highly polarizable to the point where they may act as Br and Br equivalents Note carbocations arenot intermediates in this reaction Solvent organic reactions are rarely conducted in absence of solubilizing medium Solvent must be inert to reaction Oxymercuratoin HgOAc2 induces addition of wateralcohol to alkene Fancy way of adding water to a double bond Overall alkene is hydrated adds water First step does not proceed through carbocation Second step NaBH4 reduces C Hg bond to C H bond Hydroboration AntiMarkovnikov Overall net addition of water across double bond hydration but opposite to acid catalyzed sense antiMarkovnikov 19 September Each optically active compound has its own specific rotation Two enantiomers have specific rotations with opposite signs Enantiomers A pair of molecules that are nonsuperimposable mirror images Models needed to convince yourself whether they are superimposable RS Nomenclature Naming System Enantiomers are identical in all respects except the direction they rotate plane polarized light other physical properties are the same mp bp density ampc Naming system indicates substituents in space the RS system Priority Rules Cahn ngold Prelog Greater atom number higher priority Same atoms Move to neighbors and apply rule 1 Note atomic numbers do not add up one group with higher atomic number overrules two or more with a smaller one Multiple bonds Add a neighboring atom for each extra bond Isotopes Higher mass higher priority quotRquot and quotSquot configuration assignment Prioritize substituents following the rules from highest 1 to lowest 4 Orient molecule so that lowest priority substituent 4 is pointing away from you Starting from highest priority substituent 1 move to second highest 2 and then to third highest 3 Clockwise movement right R Counterclockwise movement left S Useful tips Taking the mirror image ofa stereoisomer switches R and S Exchanging any two groups on a stereocenter switches R and S Fischer Projections Important for Sugars Carbohydrates 3D representation in a 2D drawing Unusual case Show 3D drawing Learn to translate D L nomenclature R S nomenclature Very important for biochemistry microbiology and biology classes Sometimes not covered in Ochem 2 Pages 1 213 14 October Substitution Reactions of Alkyl Halides Exchange of one group for another at sp3 carbon 5N2 and SNl 5N2 are onestep reactions CN and S are both good nuceophies SNl reactions Have carbocation intermediate Reaction rate independent of H20 concentration Primary alkyl halide 5N2 Tertiary alkyl halide 5N1 Mechanism of SNl Bond is not strong enough to hold halide carbocation forms Variations in Alkyl Halides amp Nucleophile How is SNl reaction affected by variations Order of reactivity Most reactive tertiary gt secondary gt primary halide least reactive RI gtRBrgtRCgt RF Nucleophile no effect Carbocation involved rearrangements possible 26 August Cycloalkanes Note 2 H s less than in linear alkanes Cycloalkane general formula Cn H2n Alkyl Halides Nomenclature Named as parent alkane with halide as substituent Cl chloro Br bromo iodo Alcohol Nomenclature Named as parent alkane with ol ending Ether Nomenclature Place ether at end of substituent names Amine Nomenclature Add amine to end of substituent name Primary Secondary Tertiary Carbon with only one other carbon on it primary Carbon with two other carbons on it secondary Carbon with three other carbons on it tertiary Van der Waals force induced dipole induced dipole interaction When there is hydrogen bonding boiling point is higher Molecules are harder to pull apart 26 September Application of Concepts in Chap 5 So far before exam 1 Chirality enantiomers how to recognize their structures Chirality enantiomers how to recognize its formation in reactions Before your framework was Reaction define it list SM reagent and products Reagent Starting material Product Stereochemistry of products Memorize framework new material updating Remember you will be tested on different types of applications Molecules with two asymmetric centers Maximum 4 stereoisomers are possible Enantiomers nonidentical mirror images Diastereomers stereoisomers that are not enantiomers cannot be superimposed with their mirror images Molecule with n asymmetric centers can have a maximum of 2quot stereoisomers 3 centers 23 8 max possible stereoisomers Special case only 3 stereoisomers exist when four identical groups are bonded to each of the centers Meso compound has asymmetric centers but it is achiral Reaction not at the asymmetric center Reagent and product have the same relative configuration Bond breaking at the asymmetric center Product configuration depends on the reaction mechanism Two substituents added to same side of bond syn Two substituents added to opposite sides of bond anti Alkene addition stereochemistry Creating one asymmetric center in the product Reactant with no asymmetric center 9 pair of enantiomers equal amounts R and S Reactant with asymmetric center 9 pair of diastereomers unequal amounts Creating two asymmetric centers in the product Carbocation intermediate 9synamp anti 9 mixture of four stereoisomers three of all substituents identical H2 borane and peroxyacid addition 9syn9cisalkene yields cis enantiomers erythro trans alkene yields trans enantiomer threo Cyclic bromonium ion intermediate addition of Br2 Br2 H20 Br2 ROH 9 anti 9cisalkene yields trans enantiomers threo transalkene yields cis enantiomers erythro 7 September n bond is nucleophile H is electrophile Double bond quotattacksquot H llThe rich gets richer H goes to C that already has more H Markovnikov rule Thermodynamics and Kinetics Thermodynamics relative amounts on and 2 present when the reaction has reached equilibrium Kinetics how fast Y is converted to 2 Reaction Coordinate Diagram Plot of change in energy as a function of reaction progress Transition state point on energy surface where progress in either direction leads downhill Equilibrium Stability Thermodynamics mA n3 9 sC tD products CSDt reactants Am 3quot Keq Gibbs Free Energy Change AG free energy of products free energy of reactants An exergonic reaction AG is negative Keqgt 1 An endergonic reaction AG is positive Keqlt 1 0 AG RT InKeq 3 kcal R gas constant 1986 x 10 mol K Temperature in Kelvin AG AH T AS Kinetics AH change in enthalpy Energy of bonds broken energy of bonds formed AS changed in entropy Freedom of motion in products freedom of motion in reactants Exothermic AH negative Endothermic AH positive 1 indicates transition state AGi free energy of activation AGIiI AHIiI T A51 Rate of a reaction Rate of a reaction number of collisions per unit of time x fraction with sufficient energy x fraction with proper orientation A 9 B Rate X A Rate k A K is a first order rate constant A B 9 C D Rate k A B K is a second order rate constant Rate determining step has highest energy HX hydrogen halide Usually HBr HCI or HI 16 November Using HardyWeinberg to Estimate Frequencies in SexLinked Traits Redgreen color blindness in humans is caused by a recessive Xlinked allele Frequency of affected males q Frequency of affected females q2 8 of males are redgreen color blind q 008 p 1 q 092 064 qz of females are color blind 2 pq 2 x 092 x 008 0147 or 147 of females have normal vision but are carriers for color blindness Migration Movement of individuals from one population to another causes gene flow If gene flow occurs between populations with different gene pools allele frequencies will change Change in allele frequency calculated as Aq m q qm Where q original allele frequency qm allele frequency among arriving migrants m rate of migration proportion by which migrants increases population Calculating Change in Allele Frequency Following Migration Population A 100 bears frequency of cinnamon gene q 01 Population B frequency of cinnamon gene qm 03 10 bears migrate to population A New allele frequency in next generation following migration Aq 11 3 Aq 02 new q 012 Genetic Drift Change in allele frequency due to random gamete sampling Can have major effect on gene pool in small populations lt100 Probability of allele being fixed is same as initial frequency Probability of allele being lost from gene pool through drift 1 initial frequency 16 September Nonsuperimposable mirror images Chapter 5 Stereochemistry Isomers compounds same molecular formula not identical Constitutional isomers Compounds with different connectivities Stereoisomers Cistrans isomers Restricted rotation CC or rings Isomers that contain chirality centers CisTrans Isomers Geometrical isomers cistrans isomers molecules differ as a result of restricted rotation Chiral Molecules Chirality handedness objectmolecule with nonsuperimposable mirror image is chiral Stereocenters Chiral molecule molecule whose mirror image is nonsuperimposable Chiral center or asymmetric center carbon atom bonded to four different groups Polarimeter measures the degree of optical rotation T L all 026 C T temperature in C A wavelength ot observed rotation deg 1 path length dm c concentration gmL of solution Key Concepts Chap 4 Summary of reactions Electrophilic addition reaction Addition of HX n organic solvent in water Addition of H20 and alcohols Addition of halogens n organic solvent in water Oxymercurationreduction Addition of peroxyacid Hydroboration Addition of hydrogen Addition of H20 and alcohols is cousins with oxymercurationreduction Alcohols similar to H20 Weak acid weak base H donor and acceptor Ether weak base H bond acceptor Amines bad leaving group Ammonia great leaving group 31 October Alkyl halide 5N1 5N2 E1 E2 Good nuc weak Yes good base but anary No base No 5N2 usually better Secondar Yes when weak Yes w good nuc Yes when weak Yes w good y nuc or weak base and weak base nuc or weak base bulky base Tertiary Yes weak nuc or No Yes when weak Yes w Strong base wea k ba se nuc or weak base


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