Organic Chemistry I Lectures
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Aug 21 Intro and Review I Tips for success index card 9 one side write the concept Other side write the answer Office Hours T Th 35 pm Friedrich Wohler 18001882 Heated inorganic salt produced urea shattered the theory of vitalism the belief that organic compounds only come from life Jesuits sent to Peru in the early 1800 s discovered Quinine from tree bark which was used to treat malaria Pelletier and Caventou isolated quinine in the lab Organic chemists later synthesized it Atomic Structure Hydrogen 1s1 1 electron 1 proton 1 neutron Carbon 1s22s22p2 6 electrons 6 protons 6 neutrons Aufbau principle the orbital with the lowest energy must be populated first Hunds rule electrons should first be spread over different orbitals as widely as possible Pauli Exclusion Principle electrons fill orbitals in one spin state first then fill in opposing spin state electron pair Bond Types Organic Chemistry deals primarily with the second period elements which when bonded have 8 electrons in their valence shells Octet Rule atoms bonded in a molecule are most stable and therefore most likely to be found with 8 valence electrons Ionic Bonding bond formed between metal and nonmetal Resulting material usually in the form of ionic salt with regular highly ordered and rigid crystalline lattice structure Covalent Bonding bond formed between two non metals Resulting material can have equal nonpolar or unequal polar sharing of electrons Multiple bonds ie double or triple bonds can be made between appropriate atoms via sharing of more than one electron Bond Formation a Lewis Structure 1916 electrons represented by dots bonds represented by electron pairing between atoms lines or pairs of dots b Formal Charge of an atom in a molecule FC group number of nonbonding electrons 12number of shared electrons Aug 26 Predicting relative acidity Everything we need to know about relative acidity is on the first slide of Li s lecture What makes a compound more acidic Splitting to give you more hydronium ion H30 If the conjugate base is stable there will be a high Ka tendency to dissociate giving conjugate base The smaller the pKa is the more acidic the substance is Factors that impact the stability of the conjugate base Charge electronegativity henceforth eneg size hybridization resonance inductive effects Charge A neutral molecule is more stable than a charged one A negative pKa indicates a strong likely inorganic acid Eneg We know F is more e neg than 0 which is more e neg than C The more e neg the conjugate base is the more acidic the acid is Size The larger the size of the conjugate base anion the more stable it is because it can better disperse the negative charge and thus the stronger the acid Also the larger the anion the easier it is to break off the H proton Hybridization The more s character electrons near the nucleus the conjugate base is the more stable Resonance Resonance 9 delocalization of the neg charge on the anion 9 easier dissociation 9 stronger acid Inductive effects of substituents F being very e neg tries to attract maximum electron density An anion containing very e neg atoms is more stable thus the acid is stronger In class exercise We learn that in NaNH3CN has a negative charge that actually lies on the Boron because the BH3 is a Lewis acid Also formal charge group number of non bonding electrons 12 number of bonding electrons Trick when you encounter a very esoteric molecule google the name and Lewis structure and click image Someone has done it before Ozone has two resonance structures with a single bond and a double bond on either side of the central 0 which is partially positive The geometry is bent Remember in resonance structures the major contributor is the most stable structure and it likely has the most dispersed or neutral charge We are encouraged to draw arrows to signify moving pairs of electrons between resonance structures If N is more eneg than C then it is more stable for the N in the molecule to have the lone pairs and extra negative charge In Class Exercise Two Formic acid combined with cyanide gives a formate and a cyanic acid which is the gas that kills you if you try killing yourself with cyanide Equilibrium goes to the right because the negative charge on the formate is shared between two 0 atoms rather than concentrated on one C atom which is not as eneg The equilibrium will lie to the side of the equation with the electron density concentrated on a more eneg atom Compare HNO3 to HNO2 the oxidative state is 5 on HNO3 and 3 on HNO2 N03 has three resonance structures and three O s to share the negative henceforth neg charge N02 has only two and is thus less stable therefore HNO3 is more acidic Acetonitrile is a very polar solvent CH3 C3N and it s a stronger acid than H2C CH3 because the N withdraws electrons Even though S is less eneg than 0 a neg charge spread over a S is more stable because S is a much bigger atom Aug 30 Structure and Properties II In the textbook there are some symbols you will encounter from time to time Sometimes they represent orbitals using waves a mathematical interpretation based on the function of psi Psi squared is electron density Psi symbol Orbitals 1 Wave properties of e orbitals There can be or s orbitals and lobes of p orbitals these are not charges they are mathematical interpretations 2 Linear combination of atomic orbitals Refers to the combination of and When different atoms orbitals interact these are called molecular orbitals MOS Linus Lewis at Cal Tech came up with hybridization theory to explain many of the observations in chemistry The theme of the theory is that orbitals of one atom can interact that s called hybridization Symmetry of orbitals we will learn more about this later 3 Molecular Orbitals Most important bond is a sigma 6 bond Constructive interference between the two electrons creates a bond Destructive interference creates antibonding There is a nodal plane between the two parts of the antibonding orbital at a nodal plane the density of the electrons is 0 and an antibonding orbital is higher energy than the preceding atomic orbital whereas a bonding orbital is lower energy All bonds whether single double or triple have a sigma bond Sigma bonds can be formed from two s orbitals or an s and a p orbital An antibonding o version will have a large splike orbital a nodal plane then a small spherical orbital Bonding orbitals create exactly what you would expect a big lobe and a small lobe Antibonding orbitals participate in reactions they explain a lot There are many other sigma combinations but these are the key ones In Organic Chemistry there are many multiple bonds often between Carbons Whenever there is destructive interaction there is a nodal plane plane with O electron density as with a 7t orbital Constructive interference between two p orbitals in a multiple bond creates two lobes we draw them to represent the polarity between and amplitudes Sidenote Conservation of orbitals EJ Corey in the 1960 s composed the conservation of orbitals His theory was stolen by another and Fukui and Hoffman got the prize Corey and Prof Li are friends and publish books together Multiple Bonds A double bond is one sigma and one pi and pi bonds are weaker and easier to break generally A triple bond consists of a sigma and two pi bonds 0 prefers double bonds and N prefers triple C can form them all Where do organic molecules get their shapes The answer is hybridization Bond angles cannot be explained with simple s and p orbitals we need Valenceshell electron pair repulsion theory basically the way electrons in different orbitals try to stay as far away as possible from each other Hybridized orbitals are lower in energy because electron pairs are farther apart For the tetrahedral sp3 orbital the angle between bonds is 1095 Dec 2 Last Exam Review and Chapter 4 NMR Question from Last Exam If you are not able to figure out a complete NMR structure figure out as much as possible if you get the degree of unsaturation or see that there s a hydroxyl group etc you get some points As far as IR there are many things you could have gotten but if IR tells there s OH that s all you need to know NMR OH or acid is a broad peak but this one was obviously not acid because that peak shows up Very downfield Aromatic region shows two doublets that means the substituents are para to each other The nine protons upfield that are a singlet must be tbutyl Another NMR problem from Exam 2 There are two questions the first one has an IR peak at 1659 degree of unsaturation is 1 Formula is C5H11NO At 3 ppm on the NMR there are two singlets The 1659 peak can indicate an amide can be an imine as well Actually it turns out this was a Very simple molecule the quartet at 25 is a methylene next to a methyl Why do the two methyl groups on the N have different chemical shifts This molecule has a resonance form with a double bond between NC The small double bond character splits the methyl groups we will see this next semester as well in 0 Chem II Next NMR The methylene group is really downfield showing there s a neighboring ewithdrawing group the methyl group is still at 19 unaffected The only e withdrawing group is 0 so we know the O is by the methylene Then the methylene at 51 is so downfield it must be between the O and carbonyl This final will look a lot like Exam 4 but longer and harder What are 13diaxial interactions Let s say you have a tbutyl group in the most stable chair conformation The other conformation where the tbutyl is in the axial position is less stable due to 13 diaxial interaction you can see clearly on a model that the tbutyl and axial H are encroaching on each other s space If you cannot draw chair properly you will have huge problems For Wednesday please bring questions to ask Now for a few multiplestep syntheses Cispent 2en1thiol from alkenyl halide Just do an SN2 with NaSH HeXan 25dimethyl Chop it up You could split this into an alkyl halide and a Grignard Epoxide Synthesis Show how to make this Use mCPBA in methylene chloride but you don t have to worry about solvent Remember there are multiple ways to do this Add PhMgBr to formaldehyde then PCC then you get this O 0 Once you have the double bond just throw in mCPBA and you get the epoxide Now synthesize this CH After you synthesize the alcohol add PBr3and then Mg to get a Grignard then you can add the ketone 0 After that it s SNl in the presence of CH3OH and sulfuric acid sulfuric acid protonates the hydroxyl 0 To do with alcohol of less than four carbons Chapter 4 Material all that will be on the exam We will cover in class 0 Bond dissociation energy 0 In order to appreciate BDE s you must understand two concepts 0 One is hemolytic cleavage the other is heterolytic bond cleavage 0 Fact a negative heat of reaction means exothermic 0 For the hemolytic cleavage We have a 5050 divorce of assets for example dissociating C12 0 For heterolytic cleavage We saw so much in SN1 and E1 0 For these bonds that have been broken 0 The heat of cleavage for Cl Cl is 242 and for Me3C H is 435 very hard 0 But the energy change for bond formation comes to 782 so the overall energy change is 677 782 so the overall energy change is 105 reaction is exothermic Dec 4 Chapter 4 Continued Example Radical Chlorination of Propane Reaction is exothermic happens very easily The distribution of products is 40 primary halide and 60 secondary though This seems off given there are six primary protons and two secondary protons Breaking the primary C H bond takes 410 KJmol while making the H Cl bond is 431 KJmol the change in energy is 21 KJmol so the reaction happens easily The secondary C H bond breaks more easily 397 KJmol so the primary bond is stronger harder to break For full appreciation know that the tert C H bond is weaker breaking it takes 381 KJmol The change in energy for the secondary reaction is 397431 34 so that reaction gives off more energy that is one perspective to show why the secondary reaction is more favored In additional to this experimental explanation there is the theoretical the secondary protons have six other protons available for hyperconjugation whereas the primary only have two Because this is exothermic the products are lower in energy The difference between these products is worth 13 KJmol However there is always a transition state The transition state for the secondary alkyl chloride is lower in energy than that for the primary chloride The difference between the reactant and transition state energies is called activation energy and it is smaller if the transition state energy is closer to the starting material energy Radicals bear closer resemblance to cation than to anion Therefore whatever we have learned about cation can almost always apply to the radicals The stronger force at play in any ionic stability is resonance Although the tert radical is quite stabilized if you have say an allylic radical there are two resonance forms which contributes a lot A methyl benzene radical will have more resonance structures pretend it s a cation This is an extremely stable radical this was the first radical they discovered at the University of Michigan triphenyl radical so here is the order of importance resonance gt hyperconjugation gt inductive effect you will be able to use this to solve all the problems related to the stability of ions Bromination of Propane the reaction would not go until you heat it up very high 125 deg C still get a primary and a secondary product the primary bromide you only observe 3 you needed heat so you know the reaction is endothermic the primary product yields 3 and the secondary 97 sec must be way more reactive the bond forming energy on the right has something to do with this reaction the H Cl bond is way stronger than the H Br bond 431 KJ and 368 KJ respectively the product in the HBr case is higher energy than the starting material the secondary though is below the primary in energy there are transition states in both cases but the activation energy is lower for the secondary product the transition states are similar configuration to the Cl ones in the endothermic bromide reaction the transition state is closer to the product in energy and structure as well this is the crude experimental observation For clarif1cation s sake here are the energy time diagrams for the bromination reaction left and chlorination reaction right H g 3939 4 PL4 PIN caycM Ada Hi quot quot rxCk L Hammond extracted a lot of facts from different reactions and made a postulate The First Hammond postulate says related species that are closer in energy are also closer in structure Go in the book as far as knowing the second Hammond postulate this will be important next semester Professor Li says that those of us who came to class today will be glad we did when we see the final and that he has enjoyed teaching this class and will be around for the next two days to answer questions we have before the final November 4 Chapter 8 Continued Hydration Addition of H3O H adds to the less substituted side in accordance with the Markovnikov Rule Then H20 fills the void attaches to the carbocation site Also could use oxymercurationdemercuration These reactions are good teaching tools but are not carried out anywhere They aren39t done because mercury is very toxic The alkene gets HgOAc2 added in the presence of water product follows Markovnikov Rule Addition of mercury yields a three membered carbocation properly termed a mercurium ion The bonds are not equal in length because the cation is shared by all three members of the triangle according to where the positive charge is more stable The subsequent addition of H20 looks like SN2 but can39t be due to steric hindrance Li calls it SN15 but that39s not official terminology Water then comes in and deprotonates the OH on the product yielding an alcohol If we switch to alcohol as a solvent it s alkoxymercuration This mechanism begins the same with the creation of the mercurium ion SN15 adds a molecule of the alcohol then another alcohol molecule deprotonates the product Then this is reduced with sodium borohydride to yield the Markovnikov product Hydroborationoxidation BH3 has a partial positive on the B Then basic oxidation occurs to yield an alcoholic product 9anti Markovnikov product This mechanism puts the carbocation on the more substituted side The transition state is a 4 membered transition state a ring formed of the alkene and the newly attaching substituents BH2 and H This addition has to have the BH2 and H cis to each other Build a model to clarify the stereospecificity Hydrogen peroxide is not very acidic at all you need a very strong base to deprotonate it If it is deprotonated you get a hydroperoxide ion A tri akyated boron is a strong Lewis Acid very electrophilic the hydroperoxide ion is a strong Lewis base it s like Romeo met Juliet Three substitutions of OR for R yields a borate ester An alcohol group can attach and an alkoxy group is expelled In class practice Consider the H on BH3 to be very negative it prefers to be near the positive more substituted side of the substrate Primary carbocations are very unstable barely exist The first problem forces a secondary carbocation Boron has a steric preference for the less hindered side Substitution of R the olefin for H on the boron occurs twice more by the same mechanism November 6 Addition Continued Announcements Claire Castro giving a seminar on computational organic chemistry in CSI 103 at 1145 Exam Average was 71 median was 77 max was 95 Addition of Halogens Anti addition is the exact opposite of dehalogenation The geometry should be anti copanar Pi electrons are aggressive attract the Br A triatomic carbocation is formed with Br and named bromonium With chloride it would be chloronium with iodide it s iodonium ion There is experimental data to support this mechanism Although this bromonium intermediate was proposed probably 100 years ago it was not found experimentally until 1985 when a building block of diamond was treated with bromine The bromide ion that was kicked out by that mechanism goes through what39s officially called quotloose SN2 Way we start with cyclopentene and add a bromine The intermediate looks like bromonium then the bromide that got kicked out takes revenge shifts the H into the trans position Anti addition thus gives a trans product Friday39s quiz will include the assignment to draw a perfect chair no points given for any ambiguity between axial and equatorial Haohydrins Halohydrin nomenclature chlorohydrin bromohydrin iodohydrin substituents are halogen and OH Example add Br and H20 in excess to methylcyclopentene Bromonium formed first then a bond is broken and the one that s broken is the one that gives a tertiary carbocation Because water is a stronger nucleophile it attacks the carbocation and we get the E isomer When someone says quotmechanism that means draw every step In Class Exercise A Remember that addition is from the opposite side to the carbocation When asked for mechanism you need every step but not transition state Practice the trans b at home Other Reactions of Olefins Catalytic hydrogenation Most famous homogeneous catalyst is Wilkinson39s catalyst soluble in most organic solvents Wilkinson British professor won a Nobel Prize for finding the sandwich compounds Asymmetric hydrogenation If you add chiral ligands the structure can be found in the textbook you can engineer the hydrogen a dfrom one direction o n L V quot v A 3 Attack can happen from either direction But Monsanto discovered how to do asymmetric hydrogenation with a chiral ligand so the hydrogen comes from the back of the board This is dopa when given to dopamine deficient brain cells the cells do decarboxylation this helped many Alzheimer patients Venezuela issued a stamp to commemorate In the movie Awakening there was a severe Alzheimer patient used dopa to recover well enough to start having affairs class laughs But after a few months of taking dopa relapse occurs Paul Sabatier hydrogenated organic compounds using finely disintegrated metals won Nobel Prize in 1912 Oxidation Treating an olefin with peroxy acid we end up getting oxidized product called epoxide also known as oxirane the snobby name If you oxidize the peroxy acid must be reduced November 8 Ch 8 Finish Oxidation Epoxidation This mechanism looks complicated but all that ultimately happens is addition of an O to the double bond Electrons flow from the pi bond to the O to the carbonyl and then to the carbocation Bond strength C C gt C X gt X Y The mechanism is concerted no real intermediate r J 9 II 0 0 Syn addition Eg take a cis alkene and add meta choro peroxybenzoic acid Transition state has three member ring epoxide Same with trans Acid catayzed opening of epoxide You end up with a diol The first step is when you add acid to epoxide the proton attacks the oxygen and we get a protonated epoxide Loose SN2 from backside proceeds Then it gets deprotonated Novice s dream wherever you attack you get the same product Syn dihyd roxylation Catalyzed by OsO4 or potassium permanganate Mechanism pi bonds of OsO4 attack the alkene and we get a fivemembered ring E K 0 f O quotT 40 v 1 Vl gt o quotQ 6 6 With permanganate we use a base not hydrogen peroxide Potassium permanganate is purple when you add it to the olefin the purple disappears solution becomes clear The precipitate MnO2 is almost black OsO4 is the catalytic one because it can be recycled Oxidative cleavage Cleaving the double bond via diol If we add permanganate to olefin we get a diol intermediate If we heat it up the reaction does not stop there The product is cloven in two and becomes a ketone and an aldehyde r Qf V I L W 1 Q 1k P i l u Ob r g 39v 5 t I2 j l 3 0 j lt K H c on 0H r H Aldehydes are very vulnerable to oxidation The intermediate in heat is transient gets cloven quickly Ozonolysisz Write the resonance structures of ozone E9 g lgt0zE Off Old British Professor Criegee proposed the right mechanism and had it named after him 0VlvoQ Heads up ozonide is highly explosive The mechanism is called 13 dipoarcycoaddition This formed another explosive intermediate called primary ozonide however it is not stable by itself it undergoes a cleavage immediately Er v 0nV 39 quot O Q T J T 339 M p B 5 l U 53 39 t V quot 5 ozonide 1390 1 quoty cl X I I3 al v Jolcw O 9 2 vi rl e v om L gtquot 0ho quot 09 do S I The left part becomes a ketone and the right part becomes a carbonyl oxide or zwitlerionic peroxide 2 oa These two combine again through another 13 dipoarcycoaddition and you end up with another ozonide called secondary ozonide explosive Add reducing agent dimethyl sulfide is popular the smell in rotten eggs is dimethyl sulfide We have found dozens of substitutes since dimethyl sulfide is most smelly though it works the best It is also a pollutant You end up with a nice ketone and aldehyde cloven products If we want a ketone acid pick potassium permanganate but for a ketone product ozonolysis is good Nov 11 Chapter Ten Beginning Quiz Mercurinium ion will have a longer bond between the CH3 and mercury The nucleophile is CH3OH and carries out a loose SN2 backside attack Intermediate is AcOHg ion Question B mechanism similar use double bond to attack bromine ion to give bromonium attack from the back to yield trans bromide When you switch Bromide with water then the nucleophile is water and you get a trans haohydrin If you have a mixture of bromine and water bromonium forms first then the attack is always by water Alcohols Professor Li39s favorite subject Alcohol modulates the GABA receptor gammaaminobutyric acid Has multiple domains that modulate different aminobutyric acids The hybridization of oxygen in water is sp3 but with a 1089 substituent angle in H3COH and only 1045 in water decrease due to the steric force of the two lone pairs in the O competing with the substituent sterics In organic chemistry the tricks are sterics electronics kinetics and thermodynamics most all questions can be answered this way A primary alcohol has one non hydrogen substituent If that substituent is tertiary it doesn39t matter the alcohol is still primary Isopropanol is common in the lab it is secondary r wkl 7 7 L5 I 3 quot710 Q l T quot 0 I K W 2 cu 0 39 Pampem39 T J g E N gt H 1 K V Idm39L L A phenol is an alcohol attached to a benzene ring used to be called carbonic acid by product of coal gasification This thing once saved thousands of lives from infection it kills bacteria Nomenclature No not booze and moonshine IUPAC nomenclature 1 Pick your main chain this is where the hydroxyl group and the longest chain with a hydroxyl Then change alkane to alkanol 2 your main chain starting at the end nearest to alcohol 3 Name the substituents If there are two OHs it39s a diol if they are vicinal it39s glycol if they are on a cyclopentane it39s trans cycopentane 12 dio quotglyco is a branch of the a incusive diol group Ethylene glycol was responsible for the creation of the FDA A chemist dissolved sulfa compound in ethylene glycol to give to children as an antibiotic but it turns III out ethylene glycol is toxic dozens of children died hundreds have liver failure Ethylene glycol is oxidized to a bisaldehyde very toxic government decided to reign in the chemists Ortho bromopheno is called today 2 bromopheno Most of us call metanitrophenol m nitropheno but now IUPAC calls it 3 nitropheno Similar with paraethylphenol and 4 ethypheno There are some famous bis phenos The ortho one is called catechol The 13 one is called resorcinol the antioxidant in red wine The para one is called hydroquinone a radical quencher and reducing agent A little oxygen oxidizes it to quinone hydroxyls become carbonyls and it39s important in the body That process goes back and forth in the human body What we learn in o chem is foundation in biochemistry Acidity Acid Ka HCI 7 CH3CO2H 48 H20 henol 10 Tertia alcohol 18 i ro 165 CH3CH2OH 16 CH3OH 155 To determine acidity look at the stability of the conjugate base Methyl is an electronpushing group and Oxygen already has a lot of electrons so adjacent methyls make oxygen less stable That makes the conjugate base of the tertbutanol very unstable It is a strong bulky base which weakens the acid Where inductive effect applies Chlorine can stabilize an alcoxide making the alcohol more acidic F3COH behaves like an acid not an alcohol Why is phenol so acidic Phenol used to be used as a cheap antiseptic in China This is acidic due to resonance An alcohol with a hexane has a pKa of 18 just to show the importance of resonance The negative charge will strongly want to be with oxygen making that a major contributor but the other three resonance structures make a big difference Nov 13 Making of Alcohols How to make alcohols review SN2 Simple backside attack inversion of configuration The three remaining substituents become flat in the transition state In the transition state the charge is split between the nuc and LG Acid catayzed hydration of olefins The olefin pi bond electrons attack the H3O proton The proton ends up on the less substituted side to make a tertiary carbocation the rich get richer and the poor get poorer If you think about organic chemistry you will find you don t have to remember anything OxymercurationDemercu ration Pi bond attacks the HgOAc2 mercury to give mercurinium ion Water attacks the more substituted side which has a weaker bond with mercury After deprotonation of water this is reduced to the desired alcohol HydroborationOxidation BH3 attacks less substituted side gets displaced by OH Electrons from pi bond move to the HZB of H2B H making a partial positive charge on the more substituted side which is attracted to the negative charge on the H I K CJ Opening of epoxide Bonds between heteroatoms are the weakest Electrophilic oxygen attacked by pi electrons of substrate then gets protonated by an acid like hydronium H flaw Iquot 9 O quot 39 m H 3H 2 R I u 34 397 lt QHltJ gt Water then attacks and gets deprotonated making a diol Grignard Reagents New material A Grignard Reagent is when you treat an organic halide with magnesium Organometallic reagents are made How do we distinguish covalent from ionic bonds For nonpolar covalent bonds the difference in e neg btwn the two atoms is ltO5 If it39s greater than 05 that becomes a polar covalent if the difference is under 21 If the two atoms are very different it becomes an ionic bond Their difference is greater than 21 eg sodium chloride So what is the carbon Mg bond So polar covalent that it s close to ionic It has been drawn both ways in literature so if you draw either it s acceptable Textbooks has a table of organic reagents they all fall in that range R most reactive gt R Br gt R C gt gt R F RF bond is so tight it rarely reacts Teflon is carbon covered with fluoride so it never reacts An allobromide can react with Mg and make the reagent allylmagnesium bromide Most often used are bromide and chloride compounds Cyclopentene bromide treated with magnesium metal creates a Grignard reagent If you treat that with an aldehyde for example formaldehyde you get the corresponding primary alcohol If you throw water at a Grignard reagent you kill it it gets protonated and kicks out the metal When two formaldehydes get together you get a formalin ring which decomposes to formaldehyde in heat and you can bubble formaldehyde vapor into your Grignard reagent The Grignard reagent attacks formaldehyde to create alkoxide If you quench it with water and acid you get your desired product The mechanism is similar if we use a ketone But what we get is a tertiary rather than primary alcohol Ethylene Oxide et s react ethylene oxide epoxide ring with the Grignard reagent reagent carries out backside attack relieves epoxide ring strain if we have a magnesium bromide reagent deuterium can easily replace magnesium bromide which also attaches to CD and makes a salt if you work with Grignard reagents you have to have dry equipment or you ruin the reagent We have a convention in 0 Chem to name reactions after their discoverers because IUPAC conventions would be very cumbersome Nov 15 More Chapter 10 There will be IR and NMR in the next midterm exam The quiz from today will give extra points to everyone because Prof Li feels he didn t teach the material well Organolithium Reagents Treat bromobutane with 2 Li in hexane solvent creates n BuLi and LiBr N stands for normal That was a primary bromide if you treat a secondary organohalide with lithium you end up getting secBuLi If you treat a tertiary halide with lithium you get t buty lithium t BuLi C Li can be written as a polar covalent bond or as n Bu39 Liquot Among the three organolithium bases the tert base is the most basic due to carbocation stability from high substitution The more edonating groups of which methyl is one the more stable the carbocation If you add water to n BuLi you form butane If you add water to Grignard reagent you get protonation If you treat a Grignard reagent with water it gets protonated T BuLi burns in the air remember the story about the technician who had a syringe collapse on her and the t BuLi spilled on her clothes and burned her up very dangerous Popular use of n BuLi is to make another base treat diisopropyl amine with n BuLi Yields a very bulky base and butane the bulky base is called lithium diisopropyl amide LDA 0 xh Mr 0 v x39Lquot1um quotquotquotF l 6W o1 39 unquotY0r M mLgj The butane is volatile bubbles out of solution so that39s how you know the reaction worked N BuLi is also a strong nucleophile easily attacks carbon whereas LDA is very bulky so the chance of it adding to a carbonyl is nonexistent Addition to acid chloride and esters If you replace the OH on a carboxylic acid with chloride you get an acid chloride R CO OR is an ester N BuLi carries out nucleophilic attack on acid chloride Intermediate is tetrahedral with 0 and Cl substituents then Cl leaves and CO bond reforms Q requot ml 1 M1 T quotluv I l 4130 L 0 1 acw Cl 6y 039 I quot393 5 94 39 h L 39I K T 39 39 7 0H Then there is another nucleophilic attack to the acid yielding a di tbuty and the 0 gets protonated Ester example Phenol Grignard carries out nuc attack on the carbonyl in H3C CO OEt yielding an unstable tetrahedral intermediate That collapses to give you a ketone OEt leaves Then the Grignard Ph attacks the ketone yielding a diphenol alcohol Brine means saturated NaC which absorbs moisture from solution In Class Exercise I1m Class Exercise 11 1 Provide the etaitiing materia sj teagentsI or produeiisj for the fol1ewit1g Ieae1ien5 a 1 eis12e539e1uopenta11ediuol IC 2Iu1D W t CHg CClll391 warm e s eh AL H T 39l5l39l2I l H FlaCH 39 ran 31 3 Make a diol with hot KMnO4 then that gets cloven and you get the products above However the phenol carboxylic acid gets rapidly oxidized C ozonize the ten member alkene D In cold KMnO4you get a cis diol To get the stereoisomer you could add m CPBA with acid to get a trans diol We will pick up the rest of chapter ten next week Do the rest of the in cassroom exercise over the weekend it may show up on the exam Nov 18 Chapter 10 end Chapter 11 Beginning Reduction of Carbonyl Groups Hydride Reagents Include sodium borohydride NaBH4 and lithium aluminum hydride LiAlH LAH A ketone in a solution of NaBH4EtOH or LAHether will react to form a secondary alcohol An aldehyde by the same mechanism yields a primary alcohol The pi electrons of carbonyl attack the H in BH4 yielding an alkoxide which is then protonated Note from notetaker never put LAH in water explosion Chapter 11 Definition of OxidationReduction in Organic Chemistry Oxidation Adding O O2 X2 or removing H2 Reduction Adding H2 H or removing O O2 or X2 Neither Adding or removing H quotOH H2O or HX So dibromination of an alkene is an oxidation and reversing it is a reduction Any reaction that creates new C O bonds is an oxidation including tuming a hydroxyl into a doublebonded carbonyl CO You can oxidize alcohol with bleach to a ketone OH attacks the Cl and is deprotonated then the carbinol carbon gives up a H and double bonds with O Cl is the leaving group Chromic acid Dichromate ion sodium dichromate gets hydrogenated in the presence of an acid then splits into 2 HOOCrOOH also works with chromium trioxide CrO3 You don t need the whole mechanism but the results see Nov 20 Nov 25 ROH 9 RX and ROTs tosylate Reactions with HBr or HC1 hydrohalic acids SNl Mechanism The OH lone pair attacks the H Br yielding an H20 leaving group H20 leaves and a carbocation forms Then Br attacks the carbocation Example shows a tertiary alcohol SN2 Mechanism H Br protonates the hydroxyl as before but simultaneously Br carries out a backside attack Example shows a primary alcohol Reactions with Phosphorus Halides PBr3 has an electrophilic central phosphorus The hydroxyl lone pair attacks the phosphorus of PBr3 which attaches to the hydroxyl yielding a positive 0 one Br leaves Then PBr2OH acts as a leaving group while Br carries out a backside attach on the alkyl group PBr2OH gets deprotonated and Br is a leaving group 9 HBr Mechanism 14 Kw ow 3 V0PfWwgt Aj 0 9 DY 13 ryy 39Q 7 7 9P v W VvQ GK jgtlY V quotCquot 6 I 1 6 J F H H H r y gt lr Cq 39l H3 0F IHE 141Q J21 F I V g 0quoto Z 3 WWW 1 PPM 9 77K l5v1 fH Final equation ends up being 3 R OH PBr3 9 3 R Br POH3 Reaction with SOCI2 thionyl chloride Overall equation R OH 9 R Cl in the presence of SOCI2 and a cyclic diether The hydroxyl lone pair attacks the SOCI2 back and Clquot is a leaving group The product is deprotonated then by C1 yielding a chlorosulfite ester Mechanism H I Hf HLC1 ul Ci E M Q 2 That collapses into a carbocation and dSOCl Cl attacks the carbocation leaves the S00 9 alkyl halide and sulfur dioxide O H M C ag Reaction with Tosyl Chloride A paradisubstituted methyl benzene can have a sulfur dioxide substituent that is also attached to a OH group or a Cl yielding toluene sulfonic acid TsOH or toluene sulfonyl chloride TsCl Tosylate S03quot is a good leaving group due to resonance stabilization An alcohol lone pair can attack a tosyl chloride with Clquot as the leaving group The combined product is deprotonated by the solvent pyridine then the halide ion carries out a backside attack on the alkyl group with a tosylate as the leaving group The result is an alkyl halide and a toluene sulfonic acid Note in the following mechanism the halide changes from C1 to I the notetaker thinks that either one would play the same role myoz4 a e t 9 gt campJEgtllt2 H 4 ow 0 Wt W gig quot gt T rTv 0 7 3 Nov 27 Chapter 4 The chlorination of CH4 to CCl4 and HC1 is a step by step synthesis adding one Cl at a time and gradually releasing energy Thermodynamics relates energy kinetics relates reaction rates Mechanism Initiation cleaves Cl Cl to 2C1 radicals The chlorine radical attacks methane stripping off an H atom to make HCl and an alkyl radical The alkyl radical attacks C12 creating an alkyl chloride and regenerating the Cl Radical The stability of the alkyl radicals is similar to that of carbocations Therefore the trend in bond dissociation energy BDE is methyl lt primary lt secondary lt tertiary Oct 2 NMR All atoms used for NMR have odd numbers of protons paired protons are not active in the nucleomagnetic field In the magnetic field the sample is irradiated with a magnetic wave just enough energy to make the proton jump to the beta state The change in energy is equal to hnu The radiomagnetic pulse is continuous re exed in our NMR spectrum Protons are not naked but have electrons around them That electric current generates a small magnetic field against the external magnetic field The consequence is shielding Chemical Shift For the proton adjacent to the carbon there is a current generated against the extemal field resulting in considerable shielding Oxygen however is much more eneg and that sucks edensity toward the oxygen thus the proton H is deshielded Remember deshielded protons are to the left of the NMR spectrum Shielded protons are upfield and this is where the electromagnetic field is more powerful What can we do with NMR lt s the most powerful tool for organic chemistry you can solve most organic chemistry problems with it Information given by NMR 1 Number of types of protons 2 Location of the protons as a re ection of their environment more deshielded protons downfield 3 Intensity the thing IR UV and Mass Spec cannot do gives us a quantitative result of how many protons in each type 4 Splitting pattern tells us the number of protons attached to the neighbor atom References 5060 years ago there was not very advanced computer technology They decided to use relative shift using TMS trimethyl silane as a universal 00 ppm value Chemical shift was defined in units of ppm delta 8 Chemical shift is shift downfield from TMS in Hertz total spectrometer frequency mHz Computers are able to calibrate on their own without TMS now Molecular environment This is where NMR pays the bill protons molecular environments are different and show up in different places Eneg atoms deshield protons For methane all four protons have the same chemical shift at 02 But if you put a chlorine in the proton is deshielded and the chemical shift value increases to 03 If you put two chlorines it jumps to 053 and for chloroform it is 072 The inductive effect diminishes with increased distance so with 1bromobutane the chemical shift is largest closest to the bromine 34 Then it decreases to 17 13 and 09 Standard chemical shifts will be attached to the exam We don t need to memorize Aromatic Rings In benzene every bond is the same and the six pi electrons rotate around the benzene nonstop This creates a very powerful intemal magnetic field so that everything inside the benzene ring is shielded The outside atoms are deshielded and the effect is dramatic Aromatic protons thus have a chemical shift between 7 and 8 A large compound ring will have delta values even higher whereas the inside protons are so shielded their chemical shift is in the negative The example used here is 18 annulene which we will leam more about when we leam about aromatic protons Alkenes Pi bond is split between two sides of single bond These generate an induced magnetic field themselves The protons outside the cylinder of the bond are deshielded Alkenes come in around 56 ppm that region is reserved solely for alkenes Remember regions like that for the test Aldehydes characteristically come in around 910 ppm Aldehyde proton always comes out as a singlet plus acid protons are usually farther downfield r The picture at the left is of the drawing Professor Li did of the alkene shieldingdeshielding regions if L 39velfquot Alkynes This has two pi bonds and ends up caving in to the powerful external magnetic field so instead of horizontal it becomes parallel to the magnetic field Electrical current then goes this way shielding the proton Hbonded protons For amine it generally shows up at 35 ppm and alcohol at 45 ppm Unfortunately these two scenarios never happen only if sample is very pure as well as the solvent Usually there are impurities or some moisture in the deuterated solvent 9 H bonds This has a deshielding effect so it s really unpredictable for both of these in the real world Addition of D20 takes out H bonds Carboxylic Acid This creates shielding up to around 12 ppm and it s usually a broad peak due to multiple lMF s the O H and N H also have broadened peaks due to H bonding We are unable to get to the in class exercise today so do it for Friday and be wellprepared Friday is our last day to go over NMR and we will do practice Integration Areas of peaks Only the ratio is given you must correlate with molecular formula to figure out exactly how many protons of each type you have There s an addition of chemical equivalence identical chemical environment 9 equivalent protons So if the two protons have the same shielding or deshielding environment they will have the same chemical shift delta value On a benzene ring two identical groups have identical chemical environment as do the two ortho protons and the two meta protons Remember the computer only gives us the relative ratio between types of protons so one and 3 can be 39 618 etc Spinspin splitting Define N as the number of protons on the neighbor carbon of the proton of concern the number of peaks is Nl 9 something next to a methyl will be a quartet Something with no H s on the neighboring carbon will be a singlet Oct 4 NMR Continued SpinSpin Splitting The following photos are from a c1assmate s notebook The note taker felt that the concepts could not be well explained without pictures They will be followed by captions explaining the material In the presence of single bonds Ha and Hb and He are chemically the same However in the presence of double bonds the three are chemically different The Nl rule is that if the carbon adjacent to the carbon with the protons in question has N protons attached to it the protons in question will be split into Nl peaks Thus referring to the singlebonded drawing above Ha is split into a triplet by H and He Metaortho and their relation to benzenes will not be on the test Alcohols most often do not split and are not split so the proton bonded to oxygen is a singlet this is due to moisture and H bonding hydrogen bonding When the OH peak is near 3 it s on its own whereas when the OH peak is around 11 you have a carboxylic acid Ethanol with a small amount of acidic or basic impurities will not show splitting Nitrogen protons often create broad peaks Make sure you can label each proton on a molecule as s singlet d doublet t triplet q quartet etc Elements of unsaturation One pi bond one element of unsaturation so a triple bond is two elements of unsaturation A benzene ring counts as four elements of unsaturation When calculating consider halogens as the same thing as a proton How to calculate Calculate the number of hydrogens the carbons would have if the molecule were saturated then subtract the actual number and divide by two Or add one to the number of Carbons in the molecule the subtract half the number of hydrogens Either Way you end up with the number of elements of unsaturation Example C5Hg 682 2 so there are two elements of unsaturation Ignore 0 in the formula as it does not change the hydrocarbon ratio Calculate degree of unsaturation first no matter What when it comes to resolving NMR spectra If you have four or more elements there is a good chance you have a benzene ring Oct 9 Chapter Six Intro Nomenclature read this in the textbook I Cl Br F EN 27 30 32 90 C X bond C C C C Br C F Length in 3 214 194 178 138 Dipole Moment 29 148 156 151 C bond is the longest because I is so large compared to F Nucleophilic Substitution Reactions a SN2 Nucleophilic Substitution Bimolecular That which is nucleophilic attracts nuclei and is an electron donor Lewis base That which is electrophilic conversely attracts electrons and is a Lewis acid A leaving group LG is the group that is substituted out for a nucleophilic group Elimination Reaction J 71lt z quot7 39r7rltVt H S H 6 A 1 do cw 2 S I 1 1 1 on L 4 5 I My 9 quot gt 5 7 V A 1 5 1 2 7117fAWe7 Jfa6 TI Note that the transition state pictured above has the neg charge split between the nucleophile and the outgoing LG OH I l l l The rate equation is Rate KHO39CH3 In the above reaction there is also a conversion of configuration from R to S Note from the note taker Professor Li later suggested that you use molecular models of the reactants and products to see the conversion of configuration for yourself Factors affecting SN2 reaction rates Level of substitution CH3X gt 1 gt 2 no reaction thereafter Strength of nucleophile nucleophilicity CH3O39 as a base is 1000000 times stronger than CH3OH Something that is too large to attack the backloops of the reactant molecule is not a good nucleophile Consider this example of conversion of configuration from cis to trans in an SN2 reaction Oct 11 Factors in SN2 Reactions How to do better many of us have asked Prof Li One word PRACTICE Also next time our chair drawings better be understandable axial must be vertical Factors affecting SN2 reactions Nuceophiicity negative charge always gt neutral 0 For example H0quot is a better nucleophile than H20 0 Nucleophilicity decreases from left to right 9 on the periodic table 0 So Nitrogen anion is orders of magnitude more nucleophilic than OH which is more nucleophilic than Fluoride 0 For the neutral molecules ammonia is more nucleophilic than water 0 Nucleophilicity INCREASES down the periodic table making fluoride one of the weakest nucleophiles 0 Nucleophilicity could be compared to the quothard and quotsoft concepts in gen chemistry 0 Hard likes hard and soft likes soft The reason Fluoride is bad nucleophile is that it s hard compared to Carbon 0 Example CH3 X is sp3 hybridized has four double loops with one loop on each being the large one that forms the sigma bond with the substituent 0 However each of the four orbitals has a little back loop 0 The nucleophile tries to attack the back loop and doesn t succeed at first there is a transition state where each reagent has part of the negative charge 0 Remember F is small and tight and has small valence shell 0 During the transformation the orbital becomes more symmetrical the two loops being more symmetrically sized 0 As the size of the nucleophile increases the nucleus has less control over the outer electrons so when those electrons approach the lobe it s more easy to form a covalent bond 0 When a reaction is not working throw in some KI 0 Iodide is soft with a large valence shell Living Group Effect 0 LG s are eectron withdrawing species strongly polarized 0 Basically this is a group that makes a dipole moment when bonded to Carbon 0 Sometimes a weaker LG can be transformed into a LG if you add a proton to CH3 OH 0 You get a really good LG then so the Brquot ion for example can attack the CH3 0 LG s must be stable 0 That means the product must have a lower energy than the starting material as when you add Iquot to CH3 X 0 For strong acids their conj anions must be weak such bases are good living groups 0 So is methylate or trifilate a better living group Trifilate of course Trifilate is sulfate with CF3 rather than CH3 0 There are more examples in the book 0 Hard species are usually not good in organic chemistry because they are too different from Carbon Generality of SN2 Reactions 0 Eg Halogen exchange reaction 0 Still take the Fquot and Iquot as example Iquot can readily kick Clquot off a methylene group but Fquot cannot 0 Note that this is a very cumulative subject you have to know about hybridization to do things later for example 0 We have many nucleophiles containing sulfur in us for example cysteine the amino acid 0 HSquot can replace bromine on an alkyl chain for example to make a thiol group 0 Nitriles are another example CN39 trumps Iquot 0 We don t need to always draw the transition state on the exam 0 In reactions like this the configuration changes often but not always 5 becomes R and vice versa Here s an example of an alkl halide reaction and its product Stereochemistry of SN2 reactions 0 The stereochemistry of SN2 reactions demands that the nucleophile attack from the back 0 The nucleophile must form a sigma bond with the central atom and this is the only way 0 The stereochemistry of SN2 reactions 9 inversion configuration also called Walden inversion 0 Stereospecific reactions are reactions that only go one way if you react a nucleophile with an electrophile that is S the only product you have is R a reaction that only gives you Sis also stereospecific 0 Stereoselective reactions give you two mixtures one of which is more prominent 0 quotSeective implies that there are factors that favor one over the other Examples in the real world 0 Methionine reacts with ATP to make an intermediate SAM which any nucleophile can attack to attach to methyl 0 Iodomethane is a clear and present danger in the lab as is dimethyl sulfate 0 In our body there are many nucleophiles and when you methylate those polar groups as with dimethyl sulfate as a potent methylating agent the compounds become less polar and cannot draw the water from your lungs you can drown 0 There was a provost at Dartmouth dealing with NMR of methyl mercury as NMR standard 0 Just by one drop seeping through her glove she died a painful death 0 The chemistry profession is one of the safest professions on earth however 0 The reaction mechanism for CCH33Br and OHquot has nothing to do with the concentration of OHquot and is not SN2 explaining the surprisingly high reaction rate 5N1 Substitution Nucleophilic Unimolecular 0 In the reaction below the slowest step is the breaking of C Br making it the RateDetermining Step RDS 0 Remember the hybridization of O is sp3 so you get a tet configuration 0 The nucleophilic reaction happens very fast 0 The H comes right off the oxygen associated with the methyl and it may be taken up by either CH3OH or Brquot but eventually your side product is HBr The trend in reaction rate for SN1 is exactly the opposite of SN2 with tertiary cations being more stable 9 faster reaction The most straightforward explanation is inductive effect Is methyl e donating or e withdrawing Draw the dipole moments Ifyou add the three small dipole moments of a methyl together you get a larger one in the direction of the carbon creating an e donating group The methyl sigma bonds will overlap just slightly with the empty p orbital of the cation this is called hyperconjugation Hyperconjugation is less significant effect than resonance but may be stronger than inductive effect The stability trend of alkenes comes from this hyperconjugation effect Stereochemistry of SN1 reactions The methylation of a cycloalkane halide is slow because the halide bond is hard to break The spz cation that results is flat which means it can be attacked from either side with nucleophiles This gives you two products there is a slight steric bias if the cycloalkane has another methyl substituent the trans product is favored However a deuterium substituent still gives a racemate Energetics and Rate SN1 reaction has intermediates which makes a more complicated energy diagram There is a transition state to get to the intermediates and another to get to the products Diff mechanism 9 diff reaction rate If you understand the mechanism you can predict what factors influence reaction rate SN2 dependent on both substrates but the rate of SN1 depends on only the R X substrate Carbocation rearrangement How do you explain the circled product here Remember that a secondary cation is more stable than a primary tertiary than primary The OEt alkane does a quothydride shift to make the cation on the tertiary part of the molecule so that stable cation can then grab EtOH 0 Methyl can shift as well as long as the cation becomes more stable Do the in cass exercises when you have time Oct 21 End of Chapter 6 and Chapter 7 Intro Exam 3 will be November 1 E1 Reaction unimolecular elimination Nucleophile acts as base abstracts proton Reaction conditions are about the same as SN1 reactions so mixtures of the products of the two reactions are commonly obtained Step 1 ionization formation of the carbocation the rate determining step Step 2 product is the more highly substituted double bond in accordance with Zaitchev s Rule Reasoning hyperconjugation Sigma bonds to methyl or hydrogen alpha to the double bond allow for overlap in p orbitals which stabilizes the more substituted product A carbocation has the option to react with its own leaving group or a nucleophile Primary halide always SN2 and E2 Secondary halide depends since all are possible Tertiary halide always SN1 and E1 Chapter Seven Introduction The pi bond is weaker than the sigma bond Dissociation for double bond 611 KJmol Sigma bond component 347 KJmol pi bond component 264 KJmol Allows for a lot of chemistry as the pi bond is easier to break This is further confirmed by the greater length of sigma bonds as compared to double bonds and the fact that sigma bonds allow for free rotation about the bond while pi bonds do not Degree of unsaturation 2C 2 H X N2 1 for alkenes Nomenclature Main chain chain of carbons bonded with the most double bonds Numbering start from the end closest to the double bond Substituents same naming rules as alkanes Cistrans EZ system or r rV r vvvT rv quotTWi 39p Suffixes ene diene and triene denote 1 2 or 3 double bonds EZ system of cis trans isomer nomenclature becomes less applicable as the alkenes increase in complexity For this system substituents on each side of the double bond are RANKED in accordance with the Cahn ngodPrelog system start at the left side rank then look to the right side Z Zussman same side E Entgegen opposite sides Oct 18 0 The trend of nucleophilicity from F to 1 is because the size increases and the basicity decreases 0 Hyperconjugation can be used to partially explain hydride shift 0 A tertiary carbocation is Very stable and makes a hydride or methyl shift likely In Class Exercise 2 In Classroorn Exercise 2 for Chapter E Oct 216 2D1l3v Propose ta mechani5m involving ta hy dri de shi1itcran alkyl shi t for each sclvcl sis reacti cn Esrpiain how each rearrangemeni forms ta more stable inten1nediaie n J o H V UI l wt 1 I ll ll Tl c I T U T y V heal T CH L H 2 Hy quotx ciarllJI1ia 1 uh you I lITlCquotlL P 1 T L MI I Part c The rate limiting step is taking off the Iquot to create a carbocation the carbonyl O is more nucleophilic than the hydroxyl 0 so at first it will attack the carbocation and then there is rearranging Then what happens is a hydride shift and the product is more stable because of the resonance structures M 2 l 5 T quot 39 V g Pi U Kiri 2 r 39 73 J i Practice d at home Factors affecting SNl Reactions substrate structure tertiarygtsecondarygtprimarygtCH3X leaving groups a good leaving group for SN2 is a good leaving group for SNl A good leaving group is 1 polarizable 2 Weak base con1ing from a strong conjugate acid 3 stable after leaving That completes the nuclear substitution reactions Elimination Reactions There are two kinds El and E2 E1 unimolecular E2 bimolecular Imagine a tertiary carbon with a good leaving group but a lot of steric interference an SN2 reaction will not happen even with a good nucleophile The negatively charged nucleophile behaves like a base The leaving group leaves as does a H and We get a double bond The carbon initially has sp3 hybridization attached to little protons The CCH32Br C is also sp3 and when the Br leaves the orbitals originally bound to H and Br join together to create a pi bond 0 It is stereospecific 0 The two C H and C Br are anticoplanar bonds and therefore you have an optimal transition state the least crowded possible 0 If they were on the same side of the bond they would be syn coplanar and the transition state Would be more crowded 0 Anticoplanar looks like the staggered Newman projection Whereas syncoplanar H and Br are eclipsed 0 An E2 example 0 l bromo1methylcyclohexane loses its bromide in the presence of NaOEt and EtOH 0 We end up With a CC bond and the major contributor has the double bond in the ring 0 ZaitseV s rule is that the major contributor is more substituted quot quot39lt 0 Hyperconjugation explains the trend in this case because the H s in the ring have opportunities for hyperconjugation Oct 23 Chapter 7 Continued The final will be December 6 from 101 in ED 110 our classroom Dr Bruce Roth will be in McLaren 250 at 1245 tomorrow the 24 Lipitor inventor now VP or Medical Chemistry at Genentech Stability of Alkenes Heat of Hydrogenation AH KJmol Is negative for both four carbon alkenes even more for the 23 one Hyperconjugation makes it more stable Stability also decreases along this line tetra substituted gt tri gt di gt mono Heat of hydrogenation increases as substitution increases also due to hyperconjugation Cycloalkane stability increases with decreasing ring strain Stability is higher in the cyclohexene than in the cyclopentene cyclobutene Trans is possible in cycloalkenes with 10 C or more but cis stability is greater than trans stability Brendt s Rule says that a bridged bicyclic compound cannot have a double bond at the bridgehead unless there are more than eight carbons in the ring Synthesis of Alkenes Dehydrogenation HX E2 A base abstracts the proton a halogen acts as the leaving group and the product is the alkene Concerted stereospecific and the bonds in the reaction must be anti copanar Use of a bulky group bad nucleophile ensures it s E2 not SN2 o tBuLithium tButoxide triethylamine Hofman ProductRuleElimination Antithesis of Zaitchev s Rule bulkiness of the base steric force 9 must abstract H See in cass exercise for Chapter 7 Oct 25 Clarification on Brendt s Rule it only applies to double bonds directly attached to the bridgehead Note Quiz retake will be held in Monday s class since many of us were sick and worried about the Cell Physiology exam SN1 Step by step 0 Step 1 is carbocation formation ionization 0 Step 2 is nucleophilic attack of the carbocation 0 Assume there is stereochemistry the reaction is not stereospecific since the nucleophile can easily attack from either side 0 Step 3 is deprotonation of the nucleophile Note that it does not require much basicity to displace the proton 0 The energy diagram then will show three peaks three transition states 0 Third peak is very small given deprotonation happens easily the book leaves it out but Professor Castro wants us to be aware of it SN2 review 0 The nucleophile carries out a backside attack and there is a distinct intermediate 0 Dashed lines in the intermediate diagram signify half bonds 0 The transition state is a continuous process of moving the electrons and negative charge from one atom to another 0 The transition state diagram only captures the instant in time when the charge is split evenly between nucleophile and LG 0 The reaction is stereospecific and often converts S into R or Vice Versa it may remain S depending on priorities 0 The SN2 reaction only has one transition state so the energy diagram has one peak E2 reactions 0 As long as it s elimination you prepare olefins 0 Sodium is a spectator ion in most cases 0 In this example there is too much steric hindrance to carry out SN2 0 Ethoxide does not act as a nucleophile but instead acts as a base to take the hydrogen then a double bond forms p orbitals join into pi bond and the bromide is the leaving group El reactions 0 Step 1 is the same as in SNl 0 This is a slow rate determining step 0 Ethoxide or Whatever nucleophile then opts to act as a base to absorb a hydrogen then a hydride shift leaves a double bond 0 Any E reaction is a reaction Where you lose HX to create a double bond 0 If you don t have a really strong nucleophile E reactions are often preferred Preparation of olefins 0 In the 3D diagram to be accurate the medial sigma bond must be rotated to make the proton and leaving group anticoplanar to each other on the same plane but pointing in opposite directions 0 You can take the diastereomer of the olefin and do the exact same exercise 0 For testing purposes ensure you can distinguish E from Z Oct 30 Chapter 8 Part and Quizzes Chapter 8 The New Markovnikov Rule Always draw the arrow from the electrons to the positive charge of the electrophile Say you have a cycloalkane that becomes a carbocation cycloalkane the trans is way stabler with bonds remaining in the equatorial positions Basically the rich get richer the side with hydrogen gets more hydrogen and the poor get poorer the side with less hydrogen gets a bromide The antiMarkovnikov Rule If we use the exact same substrate but add some peroxide we get the anti Markovnikov product The two oxygens have a communist system and share the bonding electrons With heat the weak O O bond breaks and you get two R0 radical HBr splits and you get an ROH and a bromine radical The stability of radicals has similar trend to carbocation The properties lean toward carbocation properties The first and second steps of this reaction are radical initiation then it s radical propagation When two bromides collide with each other you form a bromine and both radicals die that39s termination 1 That is the sequence associated with all radical reactions One thing we need to know is that because electrophilic addition and radical addition are different mechanism they have different chemistry This is all for chapter 8 so the Friday exam will not go beyond this Quizzes Quiz 6A Question 1 Count the hydrogens for hyperconjugations That which has the most hydrogens is the most stable A and c have the same hyperconjugation so to determine stability you need to use dipoledipole moment A has an overall dipole moment but in c the dipole moment goes in opposite directions and cancels out so c is more stable than a Answer bgtdgtcgta Quiz 6B Question 1 The most important factor in carbocation is resonance so c and e both have the most and the same amount of resonance E is more stable because of greater hyperconjugation so it s e and then c The next factor is hyperconjugation and d is the next most stable because it s tertiary with a methyl A would be next because it has more hyperconjugation than b You are using the knowledge from chapter 1 onwards to solve these problems 6A final question If a methyl group is ionized the hydride shift is from a vicinal H to the methyl group Remember the last step in SN1 is deprotonation assuming the nucleophile is neutral In organic chemistry we don t care as much about the peripherals the little spectator ions We just care about the substrate and the product After the ionization that creates a primary carbocation substituent the ring can open up to make a secondary carbocation How this works Same hydride shift Quizzes do very little to your grade just promote studying 6B question 2 Conformation must be anticoplanar for elimination to happen and this is the staggered conformation If it s syn it s also eclipsed So rotate the medial sigma bond such that the leaving H and the Br are 180 degrees apart You can draw it one of two ways looking at the molecule from either direction it s the same LP l JF quotL mg v s 5 PI quot quot39 N441 CH 1 L 1 A 3 I l7 Ll 39 V V 5 7 I v V F For Friday39s exam bring your molecular model it helps For bulky bases only E1 and E2 generally happen but if it39s weak it can serve as a nucleophile in SN1 6B final question The first two products will follow Zaitschev s rule as to which is a major contributor The third product is a pretty selfevident mechanism How do we form a six member ring Transition state has partial C C bonds remember the sigma bond moves toward the cation not the other way around Sep 4 Chapter Two As a general rule pi bonds are easier to break than sigma bonds Valence Shell Electron Pair Repulsion Theory VSEPR 9 used to explain molecular shapes via hybridization 9 hybridized orbitals are lower in energy therefore favorable Look at the two different structures of CH3 NO2 the one with the positive charge on the Nitrogen and no charge on the Carbon is much more favorable due to less polarization of charge and thus is the major contributor Hybridization and Molecular Shape 9 sp example BeHz the 2s2 electrons are non ideal for bonding so they are promoted to the p orbital in sp hybridization 9 regarding Carbon structure is 1s2 2s2 2p2 and again one e from the 2s2 is promoted up to the p orbital and thus all orbitals now hybridized are 12 full contain one electron each Promotion raises the energy of a molecular system so hybridization is the solution to this raise in energy Forming a bond lowers the energy increasing the molecular stability 9 sp2 Example BH3 Boron is 1s2 2s2 2p1 and the 2s2 e are promoted to 2p creating three hybrid orbitals Boron also has an unhybridized p orbital making it Very electrophilic 9 sp3 Everything becomes hybridized and four sigma bonds are formed In Class Exercise determine the hybridization for the central atom 1 H3CCHNH 2 H3C CN 3 CH33O 4 CH2CCH2 5 CH3CHNO239 6 03 Answers 1 sp2 2 sp 3 sp3 4 sp 5 sp2 6 sp2 Note Lone pairs count as groups bonded to the central atom and therefore contribute to hybridization September 6 Molecular structure and intermolecular forces Structure a Single bonds b Multiple bonds c Isomerism Constitutional isomerism same molecular formula but atoms are connected differently Example from the book ethanol Versus dimethyl ether Steroisomerism same molecular formula and connections but atoms are arranged differently in space Example cis trans isomerism has two pi bonded atoms connected to the same side groups but in the cis structure the side groups are on the same side of the atom whereas in the trans structure they aren t Classic example of cis trans isomerism H H H CH3 EC E3E C C I H I H3C CH3 H3 H c393 2 h11Ie11e mrs 2 h utene methyl gmups Iuethyl gmups on same side of an opposite sides of II hand I hand These are also an example of geometric isomers Polarity a Bond dipole Example C N where the dipole is towards the N because it is more electronegative The bigger the difference in electronegativity the more polar the bond C is more e neg than H so methyl is actually an electron donating group Remember the dipole arrow points toward the more e neg atom with the cross representing the positive charge on the other end Example 6H Cl5 41 b Molecular dipole The molecular dipole is the vector sum of all the bond dipoles and it has both magnitude and direction If the bond dipoles are in opposite directions and have the same magnitude they cancel each other out as with carbon dioxide below F lg 0 39 1 H H X H O C O II39ul39c lt39m39m39 lth 39Im39 39 mumlt39r39 c Intermolecular attraction and repulsion 9 Hydrogen bond which is the most important kind The slightly positive end is the electron donor whereas the slightly positive end of another molecule is the electron acceptor There must be an H bonded to a C N or F Lone pairs such as those on Oxygen also contribute because they charge polarities themselves 9 Dipoledipole forces intermolecular 9 London dispersion forces Example Br is a largely nonpolar molecule but random perturbation as the molecule moves through space causes temporary random electron displacement and thus temporary weak charge polaritydipole This dipole moment is maybe a fraction of a formal charge Many of these together have a net attractive effect Between several isomers the isomer with the larger surface area will have a larger London dispersion effect and thus a higher boiling point Homework in addition to regular reading read about the classes of organic compounds in the book Sep 9 Functional Groups and Infrared I From the quiz octet rule is more important than all other rules in choosing resonance structures Van der Waals forces linear isomer has much higher boiling point than branched one because of increased surface area Geckos have really finely divided high surface area claws with which to climb glass We are doing classes of Organic Compounds and Functional Group and We are doing Infrared Spectroscopy in Chapter 12 Classes and functional groups 1 Hydrocarbons a Alkanes eg methane ethane propane butane called saturated Cycloalkanes rings with all single bonds b Alkenes there are trans and cis units and cycloalkenes as Well which are unsaturated c Alkynes there are the terminal and intemal ones the terminal having a C H group on the end which you can see in the spectroscopy d Aromatic ie benzene rings 2 Ocontaining compounds are alkaloid a Primary alcohol chain Secondary alcohol OH group branched off Tertiary alcohol three branches from OH group b ether See movie The Great Moment about anesthetics Important property of ether is that it only takes and does not give as far as H bonding is concerned It has two lone pairs of electrons and can form an H bond with alcohol alcohol can only be an acceptor as it has no lone pairs to donate c aldehydes CCOH carbonyl group d ketones CCOC carbonyl group e carboxylic acids CH3COOH or CH3CO2H can be H bond donors and acceptors at the same time derivatives include acid chloride and ester RCOOR primary secondary and tertiary amines RCONR In 0 Chem 1 means primary 3 N Containing Compounds a Amines b Amides have important resonance structures Amndes 0 neh3quota G I39zrrarr 7 C h39s39 5 ll1 quotx I 3 II E NH O gth39Inurife1r39 54 C gt r n V s 4 r t J KW 111 39 1 3 h1 quot NH t O r 39m1quot 39vu J39u1n C 9 339 38 E 39 quot2 39 HN NH G 0 u 2 a no I 0 I3939 1 4 rr 3 E c39 J I rM39 c N1tr1les Polar due to polarity of CEN bond toward N Example acetonitrile H3C3N which is a very important solvent in the lab very distinctive peak on infrared Infrared Part One Infrared means below the red as it s the EM wave you can almost see Longer wavelength than visible light cause molecular vibration Infrared use is limited can only give us indication of functional groups Measures the bond vibration frequencies in a molecule and is used to determine functional group Mass spectrometry gives us an idea of molecular weight Nuclear Magnetic Resonance NMR analyzes the environment of the hydrocarbons in a compound gives useful clues as to the alkyl and other functional groups present this is the most important tool in 0 Chem UV spectroscopy used for conjugation or aromatic rings uses electronic transitions to determine bonding patterns Don t need to remember all the math but that frequency and wavelength are inversely proportional Ie the shorter the frequency the longer the wavelength Infrared Spectroscopy generated from molecular stretching and compressing 9 vibration If a molecule is not at all polar we will not likely be able to detect its Infrared Spectroscopy Also remember that frequency decreases with increasing atomic mass And frequency increases with increasing bond energy Fingerprint Region of the Spectrum used to be important between 600 and 1400 cm 1 We need the 16003500 region now September 18 Nomenclature continued The class began with some practice of naming and drawing molecular compounds Try for yourself First draw what 23dimethyl 4propylnonane looks like Also name the compounds on the lecture slides Energy conformation for butane ll I5 kl Kb lxfllll polcnliul cncrg l1 LI ls lL39lll 0 00quot 120 lH0 340 300 360 2I kl 38 kl I5 kl l0C IS H 38 U 21 kl 5 kcul 09 kcull 3 kc1l 3a kcull 09 kcul 5 kcull totally l0l1lll39 culipnctl guuc1c eclipsed umi eclipsed gauche g g llp5L Ll totally eclipsed is the highest energy totally anti means the CH3 groups are exactly opposite and that is the lowest energy state Steric Strain in Butane You can explain most o chem in two terms steric and electronic If you can use these on an exam you are going the right direction The interference between two bulky groups is called steric strain Cycloalkanes These are cyclic alkanes and the nomenclature is simple Count the number of carbons and name it as you would a linear alkane but add cyclo prefix Junctions are CH2 s Physical properties of all alkanes Nonpolar Relatively inert Boiling and melting points depend on molecular weight They are low because there are no dipoledipole or H bonds If there is only one alkyl group present there is no need to number the carbon 1 ehtylcyclopentane and ethylcyclopentane mean the same thing The alphabet rule applies to the TYPE prefixes rather than the number prefixes so between ethyl and dimethyl ethyl comes first because E comes before M in the alphabet So we can have 3ethyl1ldimethylcyclohexane for example Geometric isomers Same side cis opposite side trans Play with a molecular model if you don t understand why a molecule is trans Stabilities Five and six membered rings most common The further the carboncarbon bond is from 1095 the less stable the cycloalkane so cyclopropane is very unstable cyclobutane a little more stable because the carboncarbon bond changes from 60 to 90 degrees Also in cycobutane the atoms can move more Sp3 hybridized cyclopropanes behave as if they were sp2 hybridized There is no ring strain on a sixmember ring it is the most stable See table of ring strain when Chapter 3 Part 2 slides are posted Hexane conformations there are two peaks on the surface of a hexane due to bond angles you can see them when you build a model Boat has the peaks in the same directions whereas chair is opposite and clearly the lowest energy state Naming bonds All the horizontal bonds in a drawing are called equatorial All the vertical bonds in a drawing are called axial Does not matter which direction they go but what angle they are in comparison to each other Trans means that one side group is on top and one is below Thus cis can be two equatorial CH3 s on the top side Cis and trans have nothing to do with whether the side group is equatorial or axial Cis defined exactly means that both side groups are above the plane of the ring compared to their corresponding C H groups Equatorial also is lower energy than axial Whenever you have a bulky substituent always draw it as equatorial Professor is looking into a good explanation for why that is lower enery September 20 Chair conformation of cyclohexane is ideal because of the lack of Baeger strain Also all of the C C bonds are staggered There are pairs of collinear C atoms in your drawing of cyclohexane try to make the lines parallel Make a molecular model of tertbutylcyclohexane If the tertbutyl group is axial at the C1 position there is interaction between the butyl and the hydrogens at the 1 and 3 carbons of the ring This is unfavored If the tertbutyl group is equatorial all the C C bonds are staggered and this is much more energetically favorable Boat conformation is much less favorable than chair for a cyclohexane Sometimes with the tertbutyl group the cyclohexane goes somewhere in between the boat and chair a twisted conformation The energy is the highest at halfchair a little bit lower at twisted then higher again at boat Analogy the equatorial metyl is a major contributor whereas the axial is minor Tbutyl is an extreme case of preferring to be equatorial because the energy difference is huge due to the crowdedness The equilibrium coefficient for the conversion reaction is always gt1 Cistrans isomerism in cyclohexanes Example with two methyl groups The cyclohexane has two faces If both methyl groups are on top regardless of whether they are equatorial or axial the molecule is cis If the methyl groups are one on the top face and one on the bottom regardless of whether they are equatorial or axial the molecule is trans To make a twisted chair cyclohexane really recommend trying this with a model make a chair conformation then take the hydrogens that are at the peaks and scew them away from each other The whole point is to avoid the interference between those two hydrogens When those two hydrogens are alkyl groups instead the twisting is more important and if they are tbutyl twisting is pretty much guaranteed A note about drawing chairs and boats as demonstrated in the book you no longer need wedges and dashes to represent the 3D structure The plane you have shown is good enough Between two trans isomers the one with equatorial alkyl groups is ALWAYS more stable same with two cis isomers Sep 23 Bicyclic alkanes and Stereochemistry Going over the quiz topics Naming 1 For the main group pick the one that has the most Cs 2 Start with the first substituent that should be the lowest C number of the substituents Recognizing isomers 1 You can draw cis butane with the two single C C bonds facing up or down and it is still cis butane 2 Remember single bonds can rotate note from the notetaker I look more at the order the bonds go in like a single and then two doubles Bicyclic Alkanes There are three kinds 1 Fused bicyclic alkanes the two cycloalkanes share a side 2 Bridged bicyclic alkanes 3 Spirocyclic The best way to describe this is Visually Here is a picture of the types 3lt1ltIgt LG Spiro Fused Bridged Intro to Chapter Five Stereochemistry Definitions ConstitutionalStructural isomers isomers with a different bonding sequence Stereoisomers isomers with the same bonding sequence and different atomic orientation in space Chirality handedness Re ect an object over a mirror on paper a y axis is the mirror If the image is the same as the object it s achiral If the image is different it s chiral A carbon with four different substituents would be asymmetrical and chiral An enantiomer is asymmetrical chiral and cannot be superimposed If you find an internal mirror within a molecule it is achiral Draw the molecule out if the substituents are different then it is chiral If you have one chiral atom the whole thing is chiral Sep 25 More Stereochemistry IUPAC Nomenclature Ways of naming chiral CahnIngold Prelog system adopted by IUPAC RS configuration means rightleft respectively RS two possible spatial arrangements Not to be confused with conformation When you talk about two conformations it s the same molecule with different arrangements configuration means two different molecules with different bonding sequence 0 For naming 1 Assign a relative priority a Larger atomic number 9 higher priority lgtBrgtClgt S 13C would be higher than no A proton is always the lowest b What if there s a tie We have a tiebreaker next of kin That means the next atom in the chain c Multiple vs single bond Eg what do we do with a carboxylic acid Equates situations where carbon is attached to oxygen three times In some cases double and triple bonds increase the priority But if a double bond is found it s treated as two substituents of that atom similar with a triple bond 2 Put 4th priority group away from you and draw arrow from the first priority group through the medium priority group to the third priority group If the arrow is clockwise R If it s counterclockwise S In class exercise 0 In molecule A the chiral center is the third carbon because it has two different substituents The CH3 group has higher priority than the H group of course 0 But both the other C5 are attached to 2 CS and one H 0 Next of kin rule the left C is attached to CS that are attached to Carbon three times so it has higher priority it is therefore 1 0 The arrow ends up being clockwise so it s R 0 Notetaker numbered above arrows for clarity 0 In molecule B the carbon at the very back is a chiral group The carbonyl group gets first priority because it is considered as if the carbon were bound to two oxygen The left carbon is bound to three carbons and one hydrogen so it comes next note I drew in a double bond Professor Li added later The top carbon is bound to two Cs and one H so it is 3rd priority Obviously the H is the 4th priority substituent This ends up being R Interesting example In the l960 s a German company put out the drug Thalidemyde to treat pregnant women s morning sickness It was a very powerful medicine so all the pregnant women who took it had no symptoms but the baby came out with ipper arms However Americans were spared because a lady in the FDA dragged her feet approving the drug then when she collected enough data the disaster was happening and she refused the drug Why was this occurring When you use the R enantiomer it s a potent teratogen but if you have only S there is no problem at all you can treat the disease with no side effects How do we separate them We found some simple ways to determine the chirality and how much the chirality is If we pass light through a polarizing filter it gets only one direction the chiral compound rotates the polarized light If the light is tumed to the right you call it D and if it s to the left it s called L If you mix two different enantiomers together the mixture is called a racemate racemic mixture and if light passes through that there is no rotation To calculate specific rotation there is a formula there must be a standard to measure the degree to which the compound is rotating light 1 the length of the compound cell usually 1 la Uobserved Sep 27 Optical Activity Optical Activity Conformationally mobile systems Anything with an internal plane of symmetry is achiral Model 12 trans dimethylcyclohexane You ll nd that it is chiral whereas cis is achiral At room temperature the conformation A and B of the cis dimethylcyclohexane are interconverted very rapidly so always as a rule of thumb use the most symmetric possible conformation The individual cis conformations are still chiral but they don t exist due to the constant and rapid inter conversion Nonmobile conformers If two covalently linked benzene rings have all proton substituents they can rotate freely The first position from the position you re concemed about is called ortho the second is meta and the one right across is para If there ortho positions of the benzene rings are each one bromine and one iodine the benzene rings cannot rotate because the eclipsed conformation is impossible The staggered conformations are chiral because they are not superimposable These are called atropic isomers by the generation before us Fischer Projection Can be considered a founding father of organic chemistry Lost his IPR to the Allied forces after WWII He invented a convention for how to write stereochemistry He made a convention where the chiral atom is at the crosspoint of a cross and the horizontal substituents are coming toward you out of the wall and the vertical substituents are going into the wall E s i A J F The above shows a 3D model left a Fischer projection bottom right and what you should think when you see a Fischer projection top right inside the bubble The exercise for this is In Classroom Exercise 3 In Classroom Exercise 3 Sept 26 2013 1 Draw a Fischer projection for each compound Remember that the cross represents an asymmetric carbon atom and the carbon chain should be along the Vertical with the IUPAC numbering from top to bottom a Spropane12diol b R2bromobutan 1ol c S 1 2 dibromobutane d Rbutan2ol II For each Fischer projection label each asymmetric carbon atom as R or S CHZCH3 CO2H CH2OH a H Br b H2NH 0 BrC CH3 CH3 CH3 If it is confusing build a model Then it becomes easier to see Answers to part II aR bS c S Sep 30 Diastereomers and Meso Compounds Note that there will be questions on intermolecular forces on the exam Systems with multiple chiral centers a Diastereomers By definition they are not their own mirror images There are two types i Geometric Cistrans in double bonds Cistrans in rings such as dimethylcyclohexane the cis and trans are diastereomers to each other ii Compounds with two or more chiral centers One or more of the diastereomers may have an enantiomer as well The enantiomers have the opposite RS configurations from each other One diastereomer will have enantiomers that are RS and SR whereas the other diastereomer may have SS and RR enantiomers These days when we refer to diastereomers we aren t generally referring to geometric isomers Enantiomers have the same physical properties bp mp d except speci c rotation Diastereomers however have different physical properties b Meso compounds Achiral molecules with chiral centers This may be caused by an intemal plane of symmetry that internal plane of symmetry can cut through the middle of the rings InClass Exercise 4 For each structure 1 star any asymmetric carbon atoms 2 label each asymmetric carbon as R or S 3 draw any internal mirror planes of symmetry 4 label the structure as chiral or achiral 5 label any meso structures Cninr 41413 H Blquot H Br H Br Br H CHEB1 CH3Br On the left There are two chiral centers and the top one is S and the bottom one is R On the right Again there are two chiral centers and the top one is S and the bottom one is S The left one is meso of U The one on the left is simply achiral The one on the right has a chiral center at the C Br and the configuration is S
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