Organic Chemistry I Structure and Reactivity
Organic Chemistry I Structure and Reactivity CHEM 30A
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Date Created: 09/04/15
Scale of Leaving Groups if RQ R gt O R Sulfonate ion 0 0 Excellent leaving groups RLl gt 1 Iodide ion R rkr gt Br Bromide ion R 1 gt c1 Chlon39de ion H R H20 Water Moderate leavmg groups H About equal CHz Rg gt CHEOH Alcohol H gt Fluoride ion Rarely leave Decreasing leaving group strength RhMCHgg gt NCl33 Amine R CH3 39OCH Alkoxide ion About equal Leave only under R H gt OH Hydromdelon special circumstances A V R NH2 gt NHZ Nitranion r1quot Leavesonlyin R gt H Hydndelon Chichibabinreaction EH Never leaves unless R 3 gt CH Carbamon highly stabilized Lecture Supplement Ionic Substitution SN2 Effect of Steric Hindrance on 8N2 Reaction Rate R Br 1 a R I Brquot Verify with models R CH3 Carbon bonded to leaving group 2 methyl i gt a R CHCH2 Carbon bonded to leaving group 2 primary 2 1quot a i R CHgCHQ Carbon bonded to leaving group secondary 20 R CgCHQ Carbon bonded to leaving group tertiary 3quot g i Reaction rate trend Methyl primary secondary tertiary Lecture Supplement Ionic Substitution S N2 5 Solvent Effect on Rate Solvent A substance in which other substances are dissolved Role of Solvent in Organic Reactions Dissolve reactants so they can mingle and achieve transition state orientation 39 Disperse heat 39 Precipitate a product driving equilibrium to the right 39 Other laboratory considerations Demonstration Alkarseltzer tablets mix dry I O HO COZH O W NaHC03 HO OH Cimc zfcid Sodium bicarbonate a solid 3 solid mix aqueous Solvent Effects on SN2 Reaction Rate Reaction rate 0t Ed 2 energy difference between reactants and transition state solvent A solvent C 39 39 solvent A Energy solventB Energy Reaction coordinate Reaction coordinate Solvent C provides little stabilization of Solvent B provides good stabilization of reactants but lots for transition state reactants but little for transition state Eu B E A E C E A rate B rate A rate C rate A Lecture Supplement Ionic Substitution SN2 How Does Solvent Stabilize Interact with Reactants and Transition State gt Noncovalent forces electrostatic effects Review Chem 14C notes gt Stronger interaction larger charges more stabilization Dipoleidipole Ionidipole Hydrogen bonding London forces weak insignificant stabilization ignored for simplicity L L I J LULC 9 or B g 9 F surrounded by CH3CHZOH solvent shell provides soivation for F Important Solvent Properties Dielectric constant a Ability to insulate unlike charges from each other Higher a greater number and magnitude of Yo39 on solvent stronger attraction to solute molecules Strength of interactions stabilization 7 gt Yo39 gt neutral Greater number andor magnitude of polar bonds higher a Example CH30H a 33 versus CH3CHZOH a 25 a gt 20 polar solvent a lt 20 nonpolar solvent Proticity Ability to donate hydrogen atom for hydrogen bond Review Chem 14C notes Protic solvent can donate hydrogen atom 07H in common solvents Aprotic solvent not protic Example CH3CHZOH protic versus CHZCI2 aprotic Lecture Supplement Ionic Substitution 5N2 7 Structures and Properties of Common Organic Solvents Polar solvents s gt 20 Name Water Dimethylsulfoxide N JV Dimethylformamide DMF Methanol MeOH Ethanol EtOH Acetone 2 propanone Nonpolar solvents s lt 20 Dichloromethane methylene chloride Tetrahydrofuran THF Acetic acid HOAc Ethyl ether Hexane What Facts Must I Know About Solvents 39 Know structures but no need to memorize Structure 396 HquotH 0 s CH3 CH3 0 i H iiCH32 CH3 H CH3CH2 H O k CH CH 25101251 9 20 L CH3 9H CHgCHZ CHchg CH3CH24CH3 Proticity protic aprotic aprotic protic protic aprotic aprotic aprotic protic aprotic aprotic 39 Know relationship between structure and polarity 39 Know if a given solvent is protic or aprotic Dielectric constant 80 49 37 33 25 21 91 76 62 43 19 39 Learn all this by doing lots of problems with reference to this table when needed Lecture Supplement Ionic Substitution SN2 Which Solvent is Best For SNZ Most common 8N2 reaction involves charged nucleophile neutral electrophile and transition state with partial charges R I Nuc39 R3C LG Nuc no LG 6 gt NuC CR3 LG Fastest 8N2 lowest Ea need solvent that stabilizes transition state dispersed charges more than reactants Nuc so nonpolar solvent is best Problem Nuc usually ionic such as NaCl usually not very soluble in nonpolar solvents such as hexane Therefore need polar solvent such as CH30H or DMF to dissolve Nuc Hydrogen bonding stabilization must be sacrificed to achieve transition state costs energy so use aprotic solvent 8N2 slower but usually not stopped in polar protic solvent such as CH3OH CH3OH Fquot hidden by CH30H CH3I In order for Fquot and CH3I to reach the 8N2 transition state some CH3OH molecules arrows of the Fquot solvent shell must move out of the way This desolvation reduces the stability of Fquot raises E and slows the reaction Lecture Supplement Ionic Substitution S N2 9 Is Hydrogen Bonding Equal for All Solvents and Nucleophiles Effect of solvent DMF versus CH3OH on the 8N2 rate k of halide ion nucleophiles 25 C Nuc H3C I gt Nuc CH I solvent DMF aprotic E 37 CHSOH protic E 33 Nucleophile k M 1 s391 log k 97 Completion k M 1 s 1 log k 97 Completion F gt 3 gt 048 lt 12 sec 50 X 10 3 73 22 years Cl 25 040 14 sec 30 X 106 55 13 days Er 13 011 87 sec 80 X 10 5 41 12 hours I39 04 040 87 sec 34 X 10393 25 17 minutes log k L Conclusion What solvent characteristics give the fastest rate for a typical 8N2 reaction Nuc Elec Polar Nonpolar Protic Aprotic 10 chture Supplement Ionic Substitution SN2 The SNZ Reaction Checklist D Nucleophile Must be moderate or better D Leaving group Must be moderate or better D Steric hindrance Carbon bearing leaving group cannot be 3 D Solvent Polar aprotic usually preferred Polar protic acceptable Nonpolar solvent rarely useful Consider charges of nucleophile electrophile and transition state SNZ Examples Example 1 S Adenosylmethionine SAM biological methylator 0 CH3 M S Methionine O a 39H3 39 SAM involved in wide range of biological methylations 39 Example during the stress response norepinephrine is converted into epinephrine adrenaline Lecture Supplement Ionic Substitution SN2 11 Nomenclature Guide Chem 30A Fall 2008 Primary Alcohol OH Secondary Alcohol OH Tertiary Alcohol XOH Primary Amine NH2 Secondary Amine H N Tertiary Amine N Alkane Alkene N Alkyne Ketone i Aldehyde 150 Molecular Geometry How Do We Determine Molecular Shape oEmpirically spectroscopy etc oPredict via theory Example H H H CH30H Lewis structure implies planarity H But other geometn39es can be conceived So what is the real structure of this molecule oBonds vibrating bending etc oBy geometry we usually mean most common geometry Most common geometry geometry with lowest energy What Controls Molecular Geometry Most common geometry geometry with lowest energy Lowest energy results from minimizing electron repulsion nonbonded interactions van derWaals repulsions resonance conjugation and aromaticity How does the molecule minimize electron repulsion oMoving electron clouds electron groups away from each other oElectron groups Atoms Groups of atoms Lone pairs Vaence Shell Electron Pair Repulsion theory VSEPR Molecular Geometry Survey Best geometry largest distance between electron groups Composition Geometries Example AB Linear A B H H ABZ Linear B A B c Bent KB quotJ4 B A B H B H A83 Tshape l l B H B H Tri onal anar g pl B AB H BH Molecular Geometry Survey Best geometry largest distance between electron groups Composition Geometries Example i i A84 Squareplanar B IIA B H IZ H B H Receding back B B H Tetrahedral ANNE A B B B Blt H H Projecting out c1 39 wci A85 Tn39gonalbipyramid cl llquotcl Cl Explore and verify molecular geometry examples with molecular models Bond Angles Best geometry largest distance between electron groups Are bond angles always egual oBond angles controlled by magnitude of electron group repulsions H I an E Q b2mH c m c H H 3 H H Electron groups equal size All HCH 1095O HNH I07O HcH lt 10950 l Electron group repulsions equal Electron group repulsions unequal Which electron groups have greatest repulsion Which electron groups are largest IApproximate size ranking H F lt lone pair Cl Br lt all groups of atoms I Bond Angles In general oTwo electron groups linear oThree electron groups trigonal planar or distorted trigonal planar oFour electron groups tetrahedron or distorted tetrahedron Warning Atomic geometry may also be in uenced by resonance covered in Chem 30A conjugation Not a consideration in Chem 30A aromaticity Four attachments Trigonal planardue to resonance Example O N H l Chem 30A F Ol Prof Garrell Stereochemistry of E2 Reactions I atoms involved nucleophilic atom H C C leaving group lying in the same plane with the HCCL dihedral angle 180 This is called an anti periplanar transition state It provides the proper overlap of orbitals for the developing TE bond Aside For certain rigid molecules that cannot assume the anti conformation the E2 reaction can proceed through a syn periplanar transition state HCCL dihedral angle 0 I The preferred transition state geometry for the E2 reaction has the ve B EV 0 3919 I This geometric requirement affects the outcomes of E2 reactions As shown through the example described on the next two pages E2 reactions can be stereoselective In a stereoselective reaction one stereoisomer is formed or destroyed in preference to all others Case studies Neomenthyl chloride and menthyl chloride Neomenthyl chloride and menthyl chloride are diastereomers In neomenthyl chloride the chlorine and isopropyl groups are cis and two elimination products are possible In menthyl chloride the chlorine and isopropyl groups are trans This species can only undergo elimination when in its less stable chair conformation so the reaction is slow Attaining the less stable conformation requires energy Furthermore only one elimination product can be formed Elimination is thus stereoselective with neomenthyl chloride reacting faster than menthyl chloride Elimination is also regioselective with a neomenthyl chloride showing a 3l preference to form the more stable alkene Zaitsev s ruleand b menthyl chloride forming exclusively the less stable alkene Chem 30A F Ol Prof Gmell mm m two 13 hydmgen atoms In sml posmons ant and coplmmw h the chlorine Thus two 12 pmducts L A Path 2 rule m 15 formed N3 1 mm 2 H3 1 CHCH32 3 2 E2 enthene 78 more stable alkene Cl b H3O mom neomenthyl chloride enthene 22 less stable alkene The chloride Notice N191 omonvl Inthns omumm 5 A L conf not possible Neomenthyl cloride more stable conformtion less stable conformtion CFQ amp PP Elimination Reactions Reading Brown and Foote 51 52 88 811 Suggested Text Exercises Brown and Foote Chapter 5 2 5 13 14 16 17 19 23 24 Chapter 8 6 8 36 45 47 Lecture Supplement Elimination Reactions page 24 this Thinkbook Optional Interactive Organic Chemistry CD and Workbook Supporting Concepts Elimination Reactions p 72 SNlSN2 versus E1E2 p 69 Mechanisms Alkyl Halide Dehydrohalogenation p 20 Dehydration of 2 Butanol p 24 Dehydration of an Unbranched Primary Alcohol p 24 E1 Reaction of 2 Bromo 2 methylpropane p 25 Concept Focus Questions 1 Define quotelimination reaction quot Give a specific example 2 Give an example of an E2 reaction including a curved arrow mechanism and the transition states for each step of the mechanism 3 Brie y explain the E2 transition state geometry requirement 4 What structural and reactivity factors are necessary for an E2 reaction to occur 5 Give an example of an E1 reaction including a curved arrow mechanism and the transition states for each step of the mechanism 6 What structural and reactivity factors are necessary for an E1 reaction to occur 7 What is the rate determining step of the E1 mechanism Brie y explain your choice 8 Why is acid necessary to eliminate water from an alcohol to form an alkene 9 Why can skeletal rearrangement occur in an E1 reaction but not an E2 reaction Give an example of an E1 reaction with mechanism in which rearrangement occurs 10 If multiple alkene products are possible in an elimination reaction which alkene will be the major product What factors control this CF Q amp PP Elimination Reactions 165 11 When deciding if an elimination reaction proceeds via the E2 or E1 mechanism why is the E2 mechanism considered before the E1 mechanism Concept Focus Questions Solutions 1 An elimination reaction is a reaction in which portions of a molecule are lost usually resulting in the formation of a new 7 bond W NaOCH3 1 gt CH3 OH NaC Example c1 CHSOH 1 e 6 CH3O H CH30 39H 2 Ar C15 CH3OH C 3 The energy of the E2 transition state is lower if the carbon leaving group and carbon hydrogen bonds are parallel incipient 1 note parallel pl orbitals This arrangement is termed antiperiplmlar and is preferred over a syn periplanar arrangement to minimize van der Waals repulsions As the bonds are breaking the carbons are changing from spj to spz hybridization and p orbitals are beginning to replace the bonds If the incipient p orbitals are parallel they can immediately begin to form the new 7 bond The energy from the incipient 7 bond helps replace the bonding energy lost from the carbon leaving group and carbon hydrogen bonds thus stabilizing the transition state 4 There are three fundamental requirements for the E2 reaction a moderate or better leaving group a strong base and a hydrogen atom 5 to the leaving group the H C C L arrangement if a CC bond is formed The first two factors are interdependent With a better leaving group a weaker base can be used With a stronger base a poorer leaving group can be eliminated The most commonly used strong bases for E2 reactions are hydroxide alkoxides such as CH3039 and amide ion HZN39 166 CF Q amp PP Elimination Reactions 0 gt1 gt0 rds osozcr3 51 quotOSOZCF3 6 9 gt Hf H OH2 V Some students include a strong base such as H039 or CH 3039 in an E1 mechanism I f these species are not part of the given reactants or they are not generated in any reasonable concentration as part of the reaction then they are not present in the reaction and cannot be part of the mechanism There are three fundamental requirements for the E1 reaction a moderate or better leaving group a stable carbocation and a polar solvent These three factors interact For example when a good leaving group and highly polar solvent are present a less stable carbocation can be formed There are two steps found in every E1 mechanism ionization of the carbon leaving group bond to form a carbocation and loss of a proton by the carbocation to form a 7 bond There may be other steps as well such as protonation of an OH group These extra steps are not considered in this question because they are not present in every E1 reaction Formation of a carbocation is energetically expensive because a bond is lost and no new bond is formed In the proton loss step one 0 and one 7 bond are gained while one 0 bond is lost for a net bonding energy increase Recall that activation energy controls rate and that to a reasonable approximation activation energy is controlled by bond energy changes Therefore the more energetically expensive step ionization will be slower Acid is necessary to transform a poor leaving group hydroxide ion into a better but still moderate leaving group water Hydroxide ion is a poor leaving group due to the small atomic radius of oxygen and the oxygen atoms gains a charge as it departs Water is a better leaving group because the positive charge on oxygen is neutralized when it departs Rearrangement requires a carbocation intermediate Because carbocations are not formed in the E2 mechanism there can be no rearrangement CF Q amp PP Elimination Reactions 167 HTBH 7 2 9 CH3 OH 3H2 H 39OH2 CH3 gt gt Q gt gt CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 10 Q When the base or leaving group is large the least substituted alkene is favored Hofmann orientation Otherwise the most highly substituted alkene is favored Zaitsev orientation Q The most stable alkene is favored 11 The E2 mechanism avoids the energetically expensive carbocation formation step of E1 Thus we consider E2 before E1 However there may be cases where E2 is not disallowed but E1 is faster because the reaction conditions and reactants favor E1 over E2 Practice Problems 1 Select the most stable alkene of each set wUUUo 2 Select the most and least stable 0 h gtlt gt1 3 Select the molecule that is less stable and briefly explain If or Dr 4 Consider this carbocation mechanism step Hc1 39D G ClAH H Will this mechanism step proceed as written and nothing else is more likely to occur or will this mechanism step not occur as written and something else is more likely to occur Brie y explain your answer 168 CF Q amp PP Elimination Reactions U 0 gt1 9 gt0 State Zaitsev39s Rule Write an E2 reaction including mechanism that clearly illustrates Zaitsev39s Rule Consider this elimination reaction Br lt CH3OH l Methylcyclopentene NaOCH3 CH3OH O Methylenecyclopentane a Write an E2 mechanism for the formation of methylenecyclopentane b Write an E1 mechanism for the formation of 1 methylcyclopentene Examination of the elimination reactions of menthyl and neomenthyl chlorides was important in establishing the mechanism of ionic substitution and elimination reactions Give the E2 products formed in the following reactions along with a brief explanation CH3 CH3 9Q CH33CO K and gt Ellmmation products 5 c1 5quot c1 Menthyl chloride Neomenthyl chloride Give the major alkene products that result when menthyl and neomenthyl chlorides are subjected to E1 elimination water heat For the reaction shown a Write the mechanism for this reaction b Brie y explain your choice of mechanism NaBr DMF CH3 CH3 For the reaction shown a Write a detailed curved arrow mechanism for this reaction b Brie y explain your choice of reaction mechanism Llt H20 gt DMF CF Q amp PP Elimination Reactions 169 11 Provide a complete and detailed curved arrow mechanism for this reaction H20 1 ethanol 12 Provide a complete and detailed curved arrow mechanism for the following reaction OH H2804 gt H20 13 Very brie y explain why the previous reaction does not occur in the absence of a strong acid such as HZSO4 or H3PO4 14 Provide a complete and detailed curved arrow mechanism for the following reaction H3PO4 1 OH H20 15 Provide a complete mechanism Label the rate determining step with quotrdsquot OH H2804 gt H20 16 Which reaction is faster Write a very similar reaction that occurs by the same mechanism but is obviously faster NaOCH3 NaOCH3 gt 01 gt CH3OH CH3OH OH c1 17 For the reaction shown a Write the transition state for the slowest step in its E2 mechanism b What is the most likely elimination mechanism for this reaction Brie y explain NaOCH3 gt CH3OH Br 170 CF Q amp PP Elimination Reactions 18 For the reaction shown a Write an E1 mechanism for this reaction b Write an E2 mechanism for this reaction c What is the most likely mechanism for this reaction Brie y explain 1 CH3CONa gt CH3OH a Select the major product b Write the curved arrow mechanism for the major product of this reaction c Very brie y explain your choice for the reaction mechanism gtlt NaOH Br CH3OH For the reaction shown For the reaction shown a Select the major product b Write the curved arrow mechanism for the major product of this reaction c Very brie y explain your choice for the reaction mechanism For the reaction shown a Write the major product of this reaction b Provide a mechanism including all transition states for the major product c Very brie y explain your choice of mechanism for this reaction CH3 KOH gt CH3OH c1 For the reaction shown a Write all the products of this reaction b Provide a mechanism that clearly shows how all of your products are formed c For each reaction mechanism listed give a single brief reason why it was not chosen for this reaction SNl E1 8N2 and E2 r gt CH3CHZOH CF Q amp PP Elimination Reactions 171 23 For the reaction shown a Select the most likely mechanism E1 E2 SNl 8N2 b Brie y explain your mechanism choice 1 6 K OCCH3gtj gt HOCCH3gtj 24 The following reactions could occur by the E2 8N2 E1 or SNl mechanisms Predict the major products in each case If no reaction occurs write NR Provide a brief explanation for your mechanism choices Kl a CHSBr 1 CH3OH gt CHSCHZOH acetone 99 CH3 sto4 CH33CO K e H 0 c1 CH33COH OH 2 c1 0 1 0 H20 CH3OH 25 Consider the dehydration of citrate by the enzyme aconitase OH 9 e 9 CO2 9 CO2 02c aconitase 02c H2 0 e 9 C02 C02 citrate cis aconitate a What is the most probable mechanism for this reaction b Brie y explain your mechanism choice Clearly state any assumptions you make about the structure of aconitase c Draw a curved arrow mechanism for the dehydration of citrate by aconitase 172 CF Q amp PP Elimination Reactions 26 Consider the 39 39 of J 39 a 391 building block of nucleic acids The final step in adenosine biosynthesis is an elimination reaction HOZCl HN COZH NH2 H02 N N enyzme N N quotpmHe lt J lt J N N N N COZH I deoxynbose deoxynbose adenosine fumaric acid a What is the mechanism of this reaction b Brie y explain your mechanism choice You may make any assumptions that you want as long as the are logical and are clearly stated c Provide the detailed and complete curved arrow mechanism for this elimination reaction You may use quotpurinequot as an abbreviation for the fused aromatic ring system with the four nitrogen atoms 27 Why are E1 and SNl reactions faster in the Arctic and Antarctic than elsewhere Hint This is a joke so think like Dr H Practice Problems Solutions 1 a Alkene stability increases with the number of carbons directly attached to the spz carbon atoms of the alkene The selected structure is the only one with three C C bonds b The left and middle alkenes are both internal and trisubstituted The alkene on the right is terminal and disubstituted and therefore the least stable of the three We saw in lecture that large alkene substituents can destabilize an alkene due to severe van der Waals eclipsing interactions recall tetra tert butylethylene In this problem the left alkene has a methyl group eclipsing with the alkene hydrogen whereas the middle alkene has a much larger tert butyl group eclipsing with the alkene hydrogen CF Q amp PP Elimination Reactions 173 N The most stable alkene is 23 dimethyl 2 butene tetrasubstituted H The least stable alkene is 2 propylcyclopropene disubstituted with ring strain E The ring of 1 methylcyclobutene the molecule on the right consists of two S173 carbons and two spz carbons The bond angles in a four membered ring are roughly 90 whereas the optimal bond angles are 10947 for an S173 carbon and 120 for an S172 carbon We can quantify although not rigorously the angle strain in 1 methylcyclo butene by summing the differences between the actual and optimal bond angles for this structure 2 X 120 90 2 X 10947 90 2 99 of angle strain Similar analysis for methylenecyclobutane the molecule on the left gives 1 X 120 90 3 X 10947 90 2 88 of angle strain Thus the angle strain is more for 1 methyl than for There are less H H eclipsing interactions torsional strain in 1 methylcyclobutene than in methylenecyclobutane However in the latter case puckering the ring can significantly reduce these Thus 1 methylcyclobutene has more ring strain less stability than methylenecyclobutane 4 This mechanism step will not occur as written This mechanism step is another one of the three carbocation fates lose a proton to form a 7 bond In this case proton loss leads to a disubstituted alkene when loss of a different proton could lead to a trisubstituted alkene A trisubstituted alkene is more stable than an isomeric disubstituted alkene isomer so this proton loss is unlikely to occur 5 In a S elimination reaction the most highly substituted alkene will be the major product The E2 example must have a strong base moderate or better leaving group and hydrogen The mechanism is a single step To clearly illustrate Zaitsev39s Rule there must be at least two possible alkene products with different levels of substitution and the most highly substituted must be the major product 9 CH3OquotH IN major product minor product trisubstituted disubstituted Br 9 6 a gt HOCH3 Br K e HquotOCH3 Br 6 b gt gt CH3OH2 MCH3 174 CF Q amp PP Elimination Reactions Carbocations are very strongly driven to lose a proton they have very low pKa that just about any species in the reaction with an electron pair is sufficient to deprotonate them In this case the carbocation can be deprotonated by CH3OH as shown or by Br CHSOH is the reaction solvent and therefore present in much higher concentration than Br Thus the carbocation is more likely to encounter and thus be deprotonated by CH3OH than by Br The mechanism is written to reflect this fact 7 Recall that an E2 reaction requires the C H bond and C Cl bond to be anti to each other Review to the E2 transition state discussion in the Elimination Reactions lecture handout Recall also that cyclohexane rings are not flat but rather prefer a chair conformation Thus we must see if there are any chair conformations that have an anti periplanar arrangement of the C H and C Cl bonds Make a model of cyclohexane to convince yourself that this can only occur when the C Cl and C H bonds are both axial CH3 CH3 CH3 Cl Menthyl chloride 39 same as H CCH3 3 Thus we see only one alkene product from this reaction because there is only one hydrogen atom that is anti periplanar to the leaving group The more stable trisubstituted alkene cannot form because the C H bond next to the isopropyl group cannot become anti periplanar to the C Cl bond CH3 hf Neomenthyl H CCH33 same as chloride WOW WCHs c1 ii 6 CH3 HPOCCH35 Also W W same as CH3 V CH3 c1 Thus we expect to see two alkene products from this E2 reaction But which will be the major one Because the base is large we will see Hofmann elimination which means the least substituted alkene will be the major one CF Q amp PP Elimination Reactions 175 9 E1 elimination proceeds via a carbocation Once the carbocation is formed think about the three carbocation fates Since we are considering only elimination and not substitution in this case we need only think about lose a proton to make a 7 bond and rearrange we can ignore capture a nucleophile CH3 CH3 Rearrange 2 to 3 in this case IIII I gt A We must now consider the three fates of this new tertiary carbocation A the last carbocation shown above Once again we will ignore capture a nucleophile A more stable carbocation 2 with resonance or 3 with resonance is not available via rearrangement This leaves only lose a proton to form a 7 bond Deprotonation to form the more stable alkene is favored CH3 CH3 CH3 9 versus a 9 H PIAOHZ I U 2 Pathway B CH3 Pathway C A tetrasubstituted alkene is more stable than an isomeric tritsubstitued alkene so deprotonation pathway B occurs more readily than pathway C 176 CF Q amp PP Elimination Reactions Cl 9 9 a gt gtO HBr CH3 CH3 CH3 HbBre b This is obviously an elimination reaction so the mechanism choices are E2 or E1 We consider E2 before E1 first because is avoids the energetically expensive step of carbocation formation The three E2 requirements are 39 Strong base Absent 39 Moderate or better leaving group Cl is a moderate leaving group and 39 H C C L Present The absence of a strong base prevents this elimination from occurring via the E2 mechanism With weak bases such as bromide ion the leaving group would have to be truly exceptional for the E2 mechanism to occur Because the E2 mechanism cannot occur the E1 mechanism is the only other possibility To verify that E1 is a reasonable choice consider the three E1 requirements 39 Moderate or better leaving group Cl is a moderate leaving group 39 Stable carbocation A secondary carbocation is formed and 39 Polar solvent DMF has a high dielectric constant Therefore E1 is reasonable under these reaction conditions although it may be sluggish because the leaving group is only moderate and the carbocation intermediate is only secondary Carbocations are generally so strongly driven to ll the octet of the carbon bearing the positive charge that even a weak base can deprotonate it forming a new 7 bond In this case either Cl39 or Br39 is adequate for the job 10 a V H30ea Any base that was present at the start of the reaction or generated during the reaction can be used for the deprotonation step Carbocations are very reactive so even weak bases iodide ion water or even DMF can perform the deprotonation that produces a full octet on carbon b When considering elimination reaction mechanisms we consider E2 before E1 The absence of strong base indicates this cannot be an E2 reaction The E1 requirements are 39 Moderate or better leaving group iodide ion is among the best 39 Stable carbocation the carbocation intermediate is 3 with resonance and CF Q amp PP Elimination Reactions 177 39 Polar solvent the solvent is a mixture of water a 80 and DMF s 37 so it will have an s somewhere between the two The exact value of 8 depends upon the waterDMF ratio The reaction therefore will occur by the E1 mechanism 11 9 I gt gt H3 De HAOHZ Note This is an E1 reaction because there is no strong base present 12 In water HZSO4 pKa 10 is completely ionized to form H3O pKa 178 Only in instances in which there is no water or in which the number of moles of water is less than the number of moles of HZSO4 will there be any non ionized HZSO4 in solution Thus in aqueous H2804 the material that protonates the alcohol is H302 not H2804 3 gHHQEHZ e OH2 e gt gt 6 H3 De Hquot0H2 HquotOH2 gt e gt H306 13 The reaction involves elimination of water from an alcohol to form an alkene In the absence of acid the leaving group would be hydroxide ion This is a very poor leaving group due to the small atomic radius of oxygen and the change from neutral to negative charge and in conjunction with the energetically expensive process of carbocation formation would not allow the reaction to proceed Protonation of the alcohol by acid converts the leaving group into water Water is a moderate leaving group because the charge of the oxygen atom changes from positive to neutral when it leaves That water is a better leaving group than hydroxide ion helps to overcome the difficulty in forming the carbocation so the reaction can proceed 178 CF Q amp PP Elimination Reactions 9 H OH2 14 g g H OH2 Hquot OH2 gt H3O gt gt H3 06 HhOHZ Hof HgH2 H20 0 gt63 15 gt gt gt m 0H2 HZOquotH 9 gt same as gt W I 6 Of 16 These are obviously elimination reactions When analyzing elimination mechanisms we consider E2 before E1 The E2 requirements are 39 Moderate or better leaving group H0 is very poor Cl is moderate 39 Good base CH3O is a strong base and 39 H C C L arrangement present Thus we predict that cyclohexyl chloride reaction occurs by the E2 mechanism and cyclohexanol reaction cannot The cyclohexanol reaction cannot occur by the E1 mechanism under these conditions because hydroxide ion is a poor leaving group Thus the cyclohexyl chloride reaction is faster because the cyclohexanol reaction cannot occur at all under the given reaction conditions regardless of mechanism We can accelerate this E2 reaction by using a better leaving group illustrated below NaOCH3 Faster reaction gt CH3OH I CF Q amp PP Elimination Reactions 179 CH3 a 57 9cm 1 III I 17 a I I I I I I Br 67 Br 5 b The E1 reaction mechanism involves the energetically expensive step of carbocation formation E2 avoids this step and so is energetically cheaper Thus if other reaction conditions allow an elimination reaction will usually proceed by the E2 mechanism in preference to the E1 mechanism Under conditions where both are allowed both mechanisms may operate simultaneously For an elimination to proceed via the E2 mechanism there are three requirements strong base methoxide is present moderate or better leaving group Br and the leaving group must be beta to the hydrogen being removed L C C H arrangement All of these requirements are met so the E2 mechanism will predominate in this case I G 18 a gt gt U CH30H2 H HOCH3 1 b gt HOZCCH3 19 H 902CCH3 c An E2 elimination requires a moderate or better leaving group iodide ion H C C L arrangement and a good base Acetate ion is a modest base but iodide is an excellent leaving group Therefore the reaction could occur by the E2 mechanism The E1 mechanism requires a moderate or better leaving group iodide ion polar solvent methanol and stable carbocation tertiary These requirements are met so the reaction could occur by the E1 mechanism Given a choice we predict the reaction will occur by the E2 mechanism because this avoids the energetically expensive step of carbocation formation However E2 is slowed in this case because acetate is a modest base Thus the reaction might occur by either mechanism 19 a The first alkene is more substituted and is therefore the major product IP39QOH b gt AbHZOBre 180 CF Q amp PP Elimination Reactions c This is an elimination reaction The common elimination mechanisms are E2 and E1 We consider E2 before E1 because E2 is less energetically expensive no carbocation formation E2 requires a strong base HO a moderate or better leaving group Br and L C C H arrangement These are all present so the reaction proceeds via the E2 mechanism 20 a The most substituted alkene the alkene on the right is the major product 9 e b 6 b CH3OH2 I CH3OH H Iodide ion could also be the base that deprotonates the carbocation but it is less prevalent than CH3OH the solvent The carbocation will most likely encounter a molecule of methanol before an iodide ion c The reaction mechanism cannot be SNl or 8N2 because this is an elimination not substitution reaction When deciding between E2 and E1 we examine E2 first because it is less energetically demanding no carbocation is formed E2 requires a strong base moderate or better leaving group and L C C H arrangement There is no strong base present so E2 is ruled out CH3OH is a weak base CH3O is a strong base but is not present E1 requires moderate or better leaving group stable carbocation and polar solvent all of which are present in this case CH3 CH3 21 a E 0 c1 CH3OH b For an E2 reaction the C H and C leaving group bonds being broken must be periplanar In a cyclohexane ring this can only be achieved when both bonds are axial The C H next to the methyl cannot be axial at the same time as the C Cl bond thus the trisubstituted alkene 1 methylcyclohexene cannot form in this case CH3 CH3 H H H 0 same as lekCHs H quot391 Cl H Cl Cl No CH antiperiplanar to Only one CH is CCl E2 not possible antiperiplanar to CCl CF Q amp PP Elimination Reactions 181 i Ho 6 I H CH3 CH3 I H CH3 I39 H I HH same as H I H I C1 67 c This is an elimination reaction so the mechanism choices are E2 and E1 We consider E2 before E1 because E2 avoids the energetically expensive step of carbocation formation E2 requires strong base H0 is a strong base H C C L present and moderate or better leaving group Cl is a moderate leaving group The E2 requirements are satisfied so we predict the E2 mechanism will dominate 22 a Mechanism analysis suggests this to be an E1 SNl reaction Br gt OCHZCH3 CH3CHZOH SNl product E1 product HTRDCHZCH3 gamed OCHZCH3 H 4 e D gt IDCHZCH3 gt OCHZCH3 H HOCHZCH3 c Q Ethanol is a poor base Bromide is not a good enough leaving group to overcome this poor basicity so E2 is ruled out EH2 Ethanol CH3CH20H is a poor nucleophile Bromide is not a good enough leaving group to overcome this poor nucleophilicity so 8N2 is ruled out l E1 not ruled out 23 a The most likely mechanism for this reaction is E2 b This is an elimination reaction so the mechanism choices are E1 and E2 Of these we consider E2 first because it avoids the energetically expensive process of carbocation formation The E2 requirements are strong base moderate or 182 CF Q amp PP Elimination Reactions better leaving group and the L C C H arrangement Tert butoxide CH33CO is a strong base similar to methoxide CH3O Chloride is a modest leaving group The L C C H arrangement that would lead to the product shown is present Therefore E2 is the most likely mechanism for this reaction 24 Recall that we consider the substitutionelimination mechanism possibilities in the order listed below In each case the requirements for that mechanism to occur are noted If one of the requirements is not met then the reaction cannot occur by that mechanism Strong base L C C H moderate or better leaving group SH Moderate to good nucleophile moderate or better leaving group carbon undergoing substitution not 3 E1Sl Stable carbocation moderate or better leaving group polar solvent a Q Lacks the L C C H arrangement SEQ Has good nucleophile cyanide has little steric hindrance CH3 and a moderate leaving group Br Therefore this will be an SN2 reaction 9 ax e CEN gtH3C CEN Br b E Strong base tert butoxide moderate or better leaving group Cl H C C L arrangement present Therefore this will be an E2 elimination Tert butoxide is a large base so the Hofmann elimination product less substituted alkene is favored H CCH3 3 0 c Q No strong base lt gt 9 HOCCH33 c1 S No good nucleophile ElSNl Carbocation easily formed 3 with resonance moderate or better leaving group I and polar solvent water Thus the reaction will proceed via the ElSNl mechanisms new CF Q amp PP Elimination Reactions 183 Remember the three carbocation fates This carbocation cannot rearrange to become more stable Iodide capture returns it to starting material So we need to consider water capture OH2 H OH gt gt Hquot0H2 Or losing a proton to form a 7 bond Hquot OH2 gt d E No strong base sag No moderate or better leaving group ElS l The H3C carbocation is too unstable to form None of these three mechanisms can operate so this is a case of NR e No strong base 2 No strong nucleophile m 1 Sui Protonation of the OH affords water a moderate leaving group loss of which leads to a secondary carbocation Water is a polar solvent So we conclude ElSNl to be reasonable in this case 39 H20 H CH3 CH3 CH3 CH3 gt gt V 9 HZ quotI OH OH J 2 major product 184 CF Q amp PP Elimination Reactions CH3 CH3 gt minor product 6 UHz CH3 H CH3 CH3 UHZ Hzmll CH2 CH2 9 G minor product f E2 Strong base HO H C C L present and moderate or better leaving group Cl is moderate The E2 requirements are met so we predict E2 will be the major mechanism C1 gt major trisubstituted e UH c1 M minor disubstituted e HOH 25 a The most probable mechanism is E2 The answer to this question depends upon your assumptions b Given a choice between E1 and E2 an elimination reaction will generally proceed via the E2 mechanism because this avoids the energetically expensive step of carbocation formation Thus we can make some reasonable assumptions that will allow the E2 mechanism to proceed E2 requires Moderate or better leaving group Hydroxide is a poor leaving group but protonation by aconitase would convert the hydroxide to water a much better leaving group Strong base No strong base is present but water is a moderate leaving group so we can get away with a weaker base LG C C H arrangement This is present in citrate CF Q amp PP Elimination Reactions 185 Resonance Worksheet Background What is the purpose of resonance Resonance structures are necessary because the line structures we draw are only useful for representing the connectivity of the atoms but don t accurately represent the electronics distributions of electrons in the molecule Understanding the electronic nature of the molecule is important because the driving forces of most reactions can be explained by electronics In short if you can draw resonance structures you can explain all of organic chemistry alright perhaps this is an overstatement but it isn t far off When drawing resonance structures keep in mind that you are not drawing different molecules but that the average of all the structures represents the real molecule Resonance structures highlight features such as conjugation neighboring atoms with overlapping p orbitals see example 1 and the pull of electronegative atoms example 2 that may not be apparent in the given structure GHQ Example 1 All of the carbon atoms in benzene are sp2 hybridized The electrons in the pi bonds ow freely around the ring We represent this through resonance structures The hybrid structure true structure of benzene is a ring where all carbon atoms have 15 bonds Drawing resonance structures iei Example 2 The electronegative nature of oxygen is represented through these resonance structures The carbon oxygen bond is a double bond in nature planer and sp2 hybridized but the oxygen has a partial negative charge and the carbon has a partial positive charge This will be very important in explaining the reactivity of carbonyls There are a few key concepts when drawing resonance structures 1 Don t break single bonds 2 Don t violate the octet rule E Electrons move from an electron rich area pi bond lone pairs to electron poor areas empty orbitals pos charges atoms that can loose a pi bond 9399 Remember to include formal charges Be careful if you have more than one electron deficient atom less than 8 electrons in your resonance structure it is not a reasonable structure Chem 30A Prof Garrell Lecture Handout Reaction Mechanisms 1 Neighboring group participation in SNZ 2 E1 3 E2 1 Neighboring group participation in SNZ illustrated by the reaction of a nitrogen mustard gas with water STEP 1 An internal 8N2 reaction SLOW I CINNWIACIVNV Cl 1 395 a c clic ammonium ion bis2chloroethylmethylamine a ems acid A second rapid 8N2 reaction NipmbH FAST C39N 7 clVNWH H STEP 3 PI OtOI39I transfer to water H30 Note that STEP 1 is the ratedetermininq step and does not depend on the concentration of H20 80 kinetically the reaction would appearto be firstorder depending only on the concentration of the amine Water is involved in a second nucleophilic process This step is fast however so the concentration of water doesn39t appear in the rate equation The ratedetermining step is clearly SN2Iike with concerted bondmaking and breaking Chem 30A Prof Garrell 2 E1 mechanism illustrated by the reaction of tert butyl chloride with 80 aqueous ethanol at 25 C CH3 CH3 5N1 H30 OH H30 CHZCH3 80 CZHSOH CH3 CH3 CH3 H 04 C 3 20 H20 CH3 25 OC tertbutyl alcohol tertbutyl ethyl ether V 83 0 H3 C H3 2methylpropene 7 H20 0 The E1 reaction CH33CC1 HZOgt CH2CCH32 H3O Cl39 Step 1 CI 1 C H3 quot S OW H3C Cq Jo H ZCgt H3O 091 0 CH3 CH3 H CH3 H H H C H3 Step 2 039 H nglt gt oo H C C H FII CH3 H 39 H CH3 Key features First step aided by the polar solvent the halide leaves forming a carbocation This is the slow rate determining step The carbocation and halide are solvated and stabilized by surrounding solvent water molecules The second step is proton abstraction by water the Bronsted base in this example This step is more rapid and results in formation of the alkene and in this case hydroniurn ion The ratedetermining step is first order depending only on the concentration of the alkyl halide Notice that for both 8N1 and E1 processes the first step is formation of the carbocation Whether substitution or elimination occurs depends on the next fast step In this example if a solvent molecule reacts as a nucleophile at the positive carbon atom of the tertbutyl cation the product is tertebutyl alcohol or tertebutyl ether and the reaction is SNl A 39 J if the solvent molecule reacts as a base t quot a 5 hydrogen atom as a proton 2 methylpropene is formed In this event the reaction is El In most unimolecular reactions 8N1 is favored especially at low temperatures Increasing temperature favors E1 at the expense of 8N1 E1 reactions are favored With substrates that form stable carbocations 3 halides by use of poor nucleophiles weak bases and by use of polar solvents Because it s hard to in uence the relative partition of SN and El products to direct a synthesis towards the elimination product it is more convenient to add a strong base and force an E2 reaction instead Chem 30A W 07 Prof Garrell Stereochemistry of E2 Reactions The preferred transition state geometry for the E2 reaction has the five atoms involved nucleophilic atom H C C leaving group L lving in the same plane with the H C C L dihedral angle 180 This is called an anti coplanar or anti periplanar transition state It provides the proper overlap of orbitals for the developing TE bond Beg B37 H lt H and L are H 3139 anti coplanar L3ltl This geometric requirement affects the outcomes of E2 reactions As shown through the example described on the next two pages E2 reactions can be stereoselective In a stereoselective reaction one stereoisomer is formed or destroyed in preference to all others Consider cyclohexyl halides such as bromocyclohexane The more stable conformation has bromine in an equatorial position As shown at left below in this conformation there are no 5 hydrogens anti to the bromine In this conformation an E2 elimination of HX to form cyclohexane is not possible H Br H H H H Br H H H H H In the less stable chair conformation illustrated above right the bromine is in an axial position There are hydrogen atoms anti to the bromine on both of the adjacent 5 carbons so E2 elimination of HX is possible In this particular example it doesn t matter to which 5 carbon we form the double bond we get the same product If one of the two 5 carbons had a substituent the preference will be to form the double bond to that carbon Zaitsev s rule Aside For certain rigid molecules that cannot assume the anti conformation the E2 reaction can proceed through a syn coplanar or syn periplanar transition state H7C7C7L dihedral angle 0 Chem 30A W390 7 Prof Gmell Summarv of 439 Nenmemhvl LI 4 L chlnn39de Neomemhyl chlon39de andmemhyl chlon39de are diaslereomers As you39ll see below in neomemhyl chlorideL Lquot 439 mlnare 39r A quot 39 In rhemhylellohdeL quot quot TL39 39 39 bod 39 39 39 quot 39 quot 39 eorhorhrrrloh 39 39 39 39 quot 39 39 LI 39Funhermnm l quot 39 39 39 39 39 39 quot quot ulmuucuulyl Medan L L preference to fonn the more stable alkene Zaitsev39s rule ahdb memhyl chlon39de forming exclusivelyL L39 quot L 39 quot product 39 ro Emplel Neomenthyl chloride Neomemhyl quot 39 I gmup equatorial A L L39 39 quot 39 39 her areewors 39 39 39 quot 39 anti aholeoymhr L39 Thuston2pr0ducIsare si ul 39 quotller quot 39 39 39 quot39 Palhagives the more highly substituted alkehe whichis preferredZailseV39s rule A mix of products is formed 31 mo 3 H30 4 1 CHCH32 2 E2 lrmenthene 78 more stable alkene 11 H30 IIIII39CHCH32 heorhemhyl chloride 2rmemhene 22 less stable alkene The uurr hlnn39dp 39 l l 39 eorrorhmlorr and quot 39 39 39 quot 39 39 eorhorrherrlo l 39 39 quot Iheisopropylgroupis L 1 the u equatorialrequaloxial so an E2 reactionis not possible Chem 30A Lecture 3 Molecular 10207 Structure amp Functional Groups Hammond s postulate review 0 the structure of the transition state for an exothermic step looks more like the reactants of that step TO than the products for an endothermic step looks more like the products of that step than the reactants o This postulate applies equally well to the transition state for a onestep reaction and to each transition state in a Representations of molecules mm39Step react39m Transition state 7 Transition state Products Wrap up introduction to reaction energetics lntroduction to functional groups He39ac39i39ariis 39 quot Reactants I Reaction coordinate a Highly exothermic reaction Reaction coordinate b Highly endothermic reaction Rules of thumb Energy diagram for multistep reaction 7 o A twostep reaction with one intermediate Reaction intermediate ener minimum between m iransition states 9 Refer to examples on Monday s reaction energetics handout Transition Rate determining st or ra e mu my 939 10 increase in reaction temperature approximately 5 quot 39quot 39quotquotquotquot quot quot quot doubles the reaction rate Activation barrier the slowest step Here The rst step has the highest barrier so is slowest Every 14 kcalmol 57 kJmol increase in activation energy results in a 10fold decrease in reaction rate The intermediate reacts fastest It might go forward to products or back to reactants depending on the relative barrier heights Suggestion briefly review covalent bond energies from gen chem How strong is a hydrogen bond How strong is a CO bond with the HIGHEST energy 4 l l Heat of H reaction c D Experiments can be designed to trap intermediates Transition states can39t be trapped Reaction coordinate Chem 30A Lecture 3 Molecular Structure amp Functional Groups 10207 Benzene Imlecuhr fornllla CsHs amuseleafed arowrnmcates resonance structures it is understood tnat oaroon formS 4 bonds Hydrogen atoms are often not snown Unlabeled vertexes are assumed to be carbon atoms gt60 3 lE l n Em em Aspirin acetylsalicylic acid nnlecular formula C 9H804 Lewis structure Aspirin acetylsalicylic acid I Framework lnlnieexample gray orblack C red O blueH y cylindrical bonds space lling C PKquot Chem 30A Lecture 3 Molecular 10207 Structure amp Functional Groups 139 Other Computer Models CH4 methane HiH I Electron density maps Electrostatic potential surfaces E g benzene d of methane How do we draw this and why c n w Hlmssn of methane V of urethane map for methane Red indicates regions of highest electron density Blue indicates regions of lowest electron density Macromolecules Tetrahedral arrangement predicted by VSEPR and Framework eg w or We amino acid side chains molecular orbital MO theory more on this next class Ribbon eg protein viewed through RasMoI Physical Models Ballandstick Without hydrogens With hydrogens Bondonly Simple Showing multiple bonds eg Dreiding Spacefilling CPK In opomiwrin 1 will I 3 morn Chem 30A Lecture 3 Molecular Structure amp Functional Groups 10207 Categorizing Carbon Atoms 1s PRIMARY 20 SECONDARY 39 TEUMRY A QUA39IERNARY Functional Groups I Functional group an atom or group of atoms within a molecule that shows a characteristic set of physical and chemical properties Functional groups are important for 3 reasons They are e units by wnicn We dlvld organic compounds into classes 2 the Sites of characteristic cnernicai reactions 3 the basis fornamlng organic compounds Functional Groups h drox l rou The compound containing it is an alcohol ETHANOL A primary 1 alcohol Named from ethane ol we ve wntteanH Note 3 ewnnour snowmg me OH bonds Smilch eas H307 to Show me cc connecziwry Be careful H Isa We The compmmd Conmlnlng it is an a 99110 OH is hydroxide ion M 1 Hamil1 m For example oxlde OH chloride Cl hydride H methoxide CH30 formed by deprotonatjng C 30H methanol CFQ amp PP Multistep Organic Synthesis Reading Brown and Foote Sections 1010 Appendix 7 Reagents and Their Uses Appendix 8 Summary of Methods for the Synthesis of Functional Groups Lecture Supplement Multistep Organic Synthesis age 9 of this Thinkbook Optional Web Site Reading The 1990 Nobel Prize in Chemistry httpwwwnobelsechemistrylaureatesl990 indexhtml Review Topics to review from Chem 30A as needed SNZSNl elimination radical halogenation addition to carboncarbon 7 bonds Suggested Text Exercises Brown and Foote Chapter 8 46 50 Chapter 9 41 43 46 Chapter 10 6 l93l Chapterll 3239 Miscellaneous Projects Create reaction ash cards Brown and Foote Appendix 8 very useful for this Concept Focus Questions 1 De ne quotretrosynthesisquot 2 Describe the retrosynthesis process used to design a multistep organic synthesis Use the example shown below Convert into Concept Focus Questions Solutions Retrosynthesis is the process of thinking backwards in synthesis design We consider how a given target molecule is made from some precursor molecule instead of starting with the given starting material N We start by examining the aldehyde target structure Can it be made in a single step from the given starting material No 74 CFQ amp PP Mum step Organic Synthesis Therefore we must decide what sequence of reactions is appropriate There is no chance in the basic carbon skeleton number or sequence of carbons so we need only be concerned with functional group changes What functional groups are in the product and how are they prepared 0 Benzene ring A benzene ring is present in the target and the starting material so we don t have to be concerned about adding it Aldehyde We know several ways to make an aldehyde Alkene ozonolysis Alkene ozonolysis involves changing the carbon skeleton Since the carbon skeleton is unchanged in this synthesis problem ozonolysis would require adding this extra carbon at some point The route employs extra steps and is therefore less efficient Alkyne hydroboration Alkyne hydroboration cannot be used here because the alkyne that would lead to this aldehyde would have to have a pentavalent carbon Primary alcohol oxidation The only viable choice left The more reactions you know the more flexibility you have and thus synthesis problem are easier This is an excellent time to bring your ash cards up to date and start using them every night We have identi ed a primary alcohol as a reasonable precursor to the aldehyde target This is written as H oxidation 0 OH The fat retrosynthesis arrow means quotthe aldehyde can be made from the alcohol by oxidationquot While it is not necessary to write the reaction above the retrosynthesis arrow it helps you to remember what you planned to do transformation oxidation retrosynthesis arrow Now we repeat the process with the primary alcohol as the new target molecule CFQ amp PP Mum step Organic Synthesis 75 Can the target molecule be made from the given starting material the alkene in a single step Yes by hydroboration hydroboration gt 0H Thus the complete twostep retrosynthetic scheme is H oxidation hydroboration gt O OH quot3 quot101951419 given starting material The process is not yet complete The last step is to write out the synthesis in the forward direction This is when specific reagents for each step are chosen It is also a good time to critically review each step to make sure it will proceed exactly as expected Forward direction k 1 BH3 PCC N H 2 H202 NaOH OH 0 We must use PCC for the alcohol oxidation to stop at the aldehyde and avoid over oxidation to the carboxylic acid Practice Problems Show how the following transformations may be carried out Include your retrosynthetic reasoning 1 X into V I M CEN Br CH3 2 OH 39 mto W OH 3 I lt into O 76 CFQ amp PP Mum step Organic Synthesis 4 OH i 39 AX into I OCH3 6 into Practice Problems Solutions There may be more than one solution to each synthesis problem If you have questions about the viability of your synthesis consult a TA or Dr H ardinger 1 Retrosynthesis 5 Forward direction NaOCH L is w CH3OH H202 Br DMSO CEN Br 2 Retrosynthesis CH3 hydroboration elimination OH gt I gt 399 OH Forward direction CH3 OH H2 so 4 1 BH3 gt gt 2 H202 NaOH l OH CFQ amp PP Multi step Organic Synthesis 77 3 Retrosynthesis oxidation hydroboration COZH CHZOH gt Forward direction 1 BH3 Cro3 gt CHZOH gt COZH 2 H202 NaOH aq st04 4 Retrosynthesis Williamson hydroboration OCH3 OH dehydration OH gtlt Forward direction VOH sto4 Ax 1 3H3 1 NaH gt gt gt 2 H202 2 CH3I NaOH 0H OCH3 5 Retrosynthesis Br Br E2 twice Brz E2 gt gt gt Br radical bromination gt 78 CFQ amp PP Mum step Organic Synthesis Chem 30A Fall 2007 ll Prof TOPICSZ39 GO Garre Chem 30A WEEKO 1 SYNOPSIS GOALS AND ASSIGNMENTS FOR SEPT28 OCT 5 2007 MOLECULAR STRUCTURES 8c REACTION ENERGETICS ALS MOLECULAR POLARITY RESONAN CE Lewis structure recap Formal charges Oxidation and reduction in organic chemistry Reaction energetics Gibbs free energy change activation energy reaction mechanisms Overview of reaction types Representations of molecules from molecular formulas to threedimensional visualization Functional groups Bond and molecular dipoles polarity Electrophilic nucleophilic Lewis structures and resonance Refresh your Lewis structuredrawing skills including assignment of formal charges Be able to recognize oxidation and reduction Read text pp 232235 this will not be covered explicitly in lecture but the concept will be used Understand the following terms Gibbs free energy exergonic endergonic exothermic endothermic potential energy diagram reaction coordinate transition state activation energyEa AGI reaction rate functional group carbocation reaction mechanism homolytic heterolytic radical bond dipole moment molecular dipole moment polar amp nonpolar bonds polar amp nonpolar molecules electrophile electrophilic nucleophile nucleophilic resonance structure resonance hybrid reaction intermediate ratedetermining step Become familiar with the various two and threedimensional representations of molecules and the structural information each reveals Be able to identify functional groups 0 Review topics and oldnew skills practice Lewis structures formal charges resonance Stereochemistry an introduction Chem 30A Fall 2002 Grazia Piizzi Steve Hardinger Stereochemistry of Tetrahedral Carbons We need Il one Carbon sp3hybridized at least II to represent molecules as 3D objects For example T 7 2D drawing H C C HC Not appropriate for Stereochem IBr Br H H Mg H quotCCI E H quotCI Appropriate for Stereochem Br Br Let s consider some molecules First pair H Br same molecular formula CH2BrCl same atom connectivity H H Br Cl CI H Superposable A B identical same compound Second pair H F same molecular formula CHFBrCl m w same atom connectivity 39339 Cl H c nonsuperposable D Br stereoisomers two different compounds C Thus we can define C Stereoisomers isomers that have same formula and connectivity M differ in the position of the atoms in space C Stereochemistry chemistry that studies the properties of stereoisomers Historical perspective Christiaan Huygens 1629 1695 Dutch astronomer mathematician and physicist He discovers plane polarized light Horizontally polarized light Horizontal filter Direction of light Historical perspective Carl Wilhelm Scheele 1742 1786 Oh how happy I am No care for eating or drinking or dwelling n0 care for my pharmaceutical business for this is mere play to me But to watch newphenomena this is all my care and how glad is the enquirer when discovery rewards his diligence then his heart rejoicesquot H COZH In 1769 he discovers Tartaric Acid from tartar the potassiuIn salt of tartaric acid deposited on barrels and corks during fermentation of grape juice Tartaric Acid H COZH Historical perspective Jean Baptiste Biot 1774 1862 In 1815 he notes that certain natural organic compounds liquids or solutions rotate plane polarized light Optical Activity IN m0 ecu 392 OUT geculgx tube containing plane polarized plane a llquld organlc polarized l ht compound or light lg solution Definitions Optically Active the ability of some compounds to rotate plane polarized light iDextrorotatory an optically active compound that rotates plane polarized light in a clockwise direction pLevorotatory an optically active compound that rotates plane polarized light in a counterclockwise direction H3O 2H Methamphetamine H IN Nicotine I CH3 CH N 3 8 Historical perspective In 1819 Racemic Acid was HO COZH discovered Later shown to have I the same formula as Tartaric Acid HO cozH In 1832 Biot notes that Tartaric Acid from grape juice fermentation rotates plane polarized light in a clockwise direction plane tube coytaining plane polarized light polarized 50mm of rotated clockwise light Tartarlc ACId TA TA is dextrorotatory 9 Historical perspective In 1819 Racemic Acid was HO COZH discovered Later shown to have I the same formula as Tartaric Acid HO cozH In 1838 Biot notes that Racemic Acid does not rotate plane polarized light IN OUT 39gt plane tube coytaining plane polarized light polarized R501ut19nA0fd unchanged acemlc 01 llght RA RA 1s not optlcally active 10 Historical perspective Louis Pasteur 18221895 In 1847 he repeats earlier work on Racemic Acid Crystallization of sodium ammonium salt gives mirror image crystals that he separated by hand Equimolar solutions of separated crystals have egual but opposite optical activity HO cogD Na 0LD 127 Tartaric Acid I dextrorotatory natural crystals HO 0amp2D 71 4 0LD 127 Tartaric Acid Racemic acid salt levorotatory unnatural 11 Historical perspective In 1853 Pasteur studies Mesotartaric Acid same formula as Racemic and H0 COzH Tartaric Acid but fails to separate I 1nto and crystals HO COZH In 1854 he notes that certain plant mold metabolizes tartaric acid but not tartaric acid 12 Historical perspective Jacobus H van t Hoff 1852 1930 Joseph A LeBel 1847 1930 In 1874 they propose Carbon with 4 attachments is Tetrahedral A molecule having a tetrahedral carbon with 4 different attachments may exist as a pair of isomers Therefore Stereoisomers isomers that differ only in the position of atoms in space and that cannot be interconverted by rotation around a single bond Stereocenter a carbon atom bearing 4 different atoms or group of atoms H F H CD are a palr of 3 Cl EN Cl Carbon is a r r c D 14 another example Stereoisomers of 2chlorobutane AB are stereoisomers j lt E if Carbons are stereocenters lI C E C quot H AB are nonsuperposable A E B mirror images Enantiomers Enantiomers stereoisomers that are nonsuperposable mirror images Chiral any molecule that is nonsuperposable with its mirror image ie A and B are chiral Achiral any molecule that is not chiral Racemic mixture a 11 equimolar mixture of two enantiomers 15 Unsolved Issues H COZH H COZH Mesotartaric Acid could not be separated into crystals and crystals Joseph A Jacobus H LeBel van t Hoff 1847 1930 1852 1930 Carbon with 4 attachments is Tetrahedral A molecule having a tetrahedral carbon with 4 different attachments may exist as a pair of isomers In 1877 Hermann Kolbe one of the best organic chemist of the time wrote Not long ago I expressed the view that the lack of general education and of through training in chemistry was one of the reasons of the causes of the deterioration of chemical research in GermanyWill anyone to whom my worries seem exaggerated please read if he can a recent memoir by a Herr van t Hoff on The Arrangement of Atoms in Space a document crammed to the hilt with the outpouring of childish fantasyThis Dr J H van t Hoff employed by the Veterinary College at Utrecht has so it seems no taste for accurate chemical research He finds it more convenient to mount his Pegasus evidently taken from the stables of the Veterinary College and to announce how on his bold ight to Mount Parnassus he saw the atoms arranged in in space In 1901 van t Hoff received the first Nobel Prize in Chemistry 17 Takehome problem Stereoisomers E a 9 of 2chlorobutane 5 H c C H A B Remember Enantiomers stereoisomers that are nonsuperposable mirror images Racemic mixture a 11 equimolar mixture of two enantiomers Explain why A and B cannot be physically separated a racemic mixture of A and B has no optical activity no rotation of plane polarized light 18 Summary Stereoisomers isomers that have same formula and connectivity M differ in the position of the atoms in space They possess one or more stereocenters Stereocenter a carbon atom bearing 4 different atoms or group of atoms Chiral any molecule that is nonsuperposable with its mirror image Enantiomers stereoisomers that are non superposable mirror images Racemic mixture a 11 equimolar mixture of two enantiomers Optically Active the ability of some compounds to rotate plane polarized light 19 Configuration of Stereocenters Enantiomers of 2chlorobutane The CahnIngoldPrelog CIP 3lt 5 9f rule assigns R or S If C E C quot H configuration to the two A i B enantiomers 1 Assign the priorities to the groups attached to the stereocenter Priority is based on the atomic number ie H has lower priority than Cl But methyl and ethyl both are attached to the stereocenter through carbon In these cases priority assignments proceed outward to the next atoms The Methyl carbon has 3 Hs attached While the Ethyl carbon has 2Hs and and a carbon the terminal methyl group Therefore the latter gets higher priority 20 Configuration of Stereocenters 2 3 2 Orient the molecule so that the group of priority four lowest priority H Cl points away from the observer A 2 3 3 Draw a circular arrow from 939 s O 4R the group of first priority to the 01 group of second priority 4 If this circular motion is clockwise the enantiomer is the R enantiomer If it is counterclockwise it is the S enantiomer Thus A is the R enantiomer of 2chlorobutane 21 Con guration of Stereocenters Ibuprofen an COZH antiinflammatol39y H3 CH3 agent Molecules with multiple stereocenters l eiLQELL QMEQN eevu eemwex 23 Tartaric Acid HO COZH 4possible 2 stereocenters steremsomers HO Mirror D I CO 39 i 2 Enantiomers D 1 i a S a t C02H a St s 139 e 0 C02H I39 O m e A 3 m 1 COZH e S I39 S 24 Remember Enantiomers stereoisomers that are non superposable mirror images Diastereomers stereoisomers that are not mirror images For example S STartaric acid S RTartaric acid DIASTEREOMERS 25 R S Tartaric acid 3 S RTartaric acid Enantiomers 9 26 Why not Enantiomers Same compound same molecular formula Enantiomers same connectivity mirror images Xnonsuperposable Superposable Achiral compound 27 Why not Enantiomers plane of symmetry Meso compound A compound with at least 2 stereocenters that is achiral due to the presence of a plane of symmetry 28 Properties of Stereoisomers Enantiomers have same chemical and physical properties in an achiral environment m they differ on the sign of rotation of plane polarized light For example Enantiomers of Epinephrine Adrenaline Same meltingboiling point same rate of reaction with achiral reagents same degree of rotation of plane polarized light thus difficult to separate 29 Properties of Stereoisomers Carvone exists as a pair of enantiomers CaI39V0ne carvone smells like caraway smells like spearmint OCD 625 OCD 625 Note No relationship exists between the SIR configuration and the sign or the magnitude of rotation of plane polarized light A 11 mixture of enantiomers racemic mixture has always no optical activity rotation equal to zero because the rotation of 50 of one enantiomer is cancelled out by the rotation equal but opposite of 50 of the other enantiomer 30 Properties of Stereoisomers Diastereomers have different chemical and physical properties in any type of environment SSTartaric Acid Mesotartaric Acid 0LD 127 0 achiral Melting p 00 171 174 146 148 Density gcm3 17598 1660 Solubility in H20 139 125 31 Biological Signi cance of Chirality Since most of the natural biological environment consists of enantiomeric molecules amino acids nucleosides carbohydrates and phospholipids are chiral molecules then enantiomers will display different properties Then in our body Drug ED Enzyme 3 3 Tight Binding Biological Significance of Chirality Enantiomers of Epinephrine H3C H H N Ia H H Anionic Flat area Anilnic I Flat area site Site Not Occu ied Occupied Epinephrine Epinephrine Poorer Fit Less Active Better Fit More Active 33 The case of Thalidomide Thalidomide was synthesized in West Germany in 1953 by Chemie GrL39inenthal It was marketed available to patients from October 1 1957 West Germany into the early 1960 s Sold in at least 46 countries US not included Thalidomide was hailed as a quotwonder drug that provided a quotsafe sound sleep It was a sedative that was found to be effective when given to pregnant women to combat many of the symptoms associated with morning sickness No clinical testing was available to show that Thalidomide molecules could cross the placental wall affecting the fetus until it was too late 34 The case of Thalidomide Thalidomide was a catastrophic drug With tragic side effects Not only did a percentage of the population experience the effects of peripheral neuritis a devastating and sometimes irreversible side effect but Thalidomide became notorious as the killer and disabler of thousands of babies When Thalidomide was taken during pregnancy particularly during a speci c Window of time in the first trimester it caused startling birth malformations and death to babies Any part of the fetus that was in development at the time of ingestion could be affected 35 The case of Thalidomide 1 stereocenter 2 stereoisomers O O Ii H S N R N O O H O O H S thalidomide Rthalidomide to calm nervousness to cause birth defects 36 Why did the two enantiomers display different biological activity Emwmtiomcm differ in the arrangement of atoms in space Therefore the ff zvl of Thalidomide can t the active site of a specific enzyme like a key for a specific lock producing the desired effect sedative On the other hand the 17c cannot interact with the same site due to a different arrangement of atoms 3D shape As consequence it fits a different enzyme active pocket triggering a different biological effect toxic How to solve this problem racemic mixture Separate enantiomers Resolution Resolution of Enantiomers Enantiomers are temporarily converted into a pair of diastereomers by adding a chiral reagent gt same proPertws Add a chiral C 1 or more reagent C stereocenters diastereomers I R439 SC different properties i 39 ie different boiling points Separate by distillation Cleave off the chiral reagent Searated enantiomers Carbocations Based on a Fall 2001 Chemistry 30BH Honors project by Patricia Young Tutorial Contents A Introduction B Carbocation Classi cation C Carbocation Stability D Carbocation Formation E Three Fates of a Carbocation F Rearrangement Causing a Change in Ring Size Exercise Solutions A Introduction A carbocation is molecule in which a carbon atom bears three bonds and a positive charge Carbocations are generally unstable because they do not have eight electrons to satisfy the octet rule 3 H II open octet on carbon B Carbocation Classi cation In order to understand carbocations we need to learn some basic carbocation nomenclature A primary carbocation is one in which there is one carbon group attached to the carbon bearing the positive charge A secondary carbocation is one in which there are two carbons attached to the carbon bearing the positive charge Likewise a tertiary carbocation is one in which there are three carbons attached to the carbon bearing the positive charge H Methyl carbocations I I No C7C bonds CH3O C H H H Primary 1 carbocations I l One crt bond CH3 6 I39 H H Organic Chemistry Tutorials C arbocations Page 1 H a o 39 I Secondary 2 carbocations CH3C 9 Two C7C borials CH3 ICH3 Tertiary 3 carbocations Three C7C bonds CH3 CH3 Exercise 1 Exercise solutions can be found at the end of the tutorial Label each carbocation as primary secondary or tertiary 69 3 a CH3 f CH2CH3 b c O H If the carbon bearing the positive charge is immediately adjacent to a carboncarbon double bond the carbocation is termed an allylic carbocation The simplest case all R H is called the allyl carbocation R R 43 R A69 HZC CH2 R R Generic allylic carbocation The allyl carbocation If the carbon bearing the positive charge is immediately adjacent to a benzene ring the carbocation is termed a benzylic carbocation The simplest case is called the benzyl carbocation R H lt gt ltG lt gt lt R H Generic benzylic carbocation The benzyl carbocation If the carbon bearing the positive charge is part of an alkene the carbocation is termed a vinylic carbocation The simplest case is called the vinyl carbocation Note that the carbon bearing the positive charge has two attachments and thus adopts sp hybridization and linear geometry Organic Chemistry Tutorials C arbocatioris Page 2 R H CC R c C H R H Generic vinylic carbocation The Vinyl carbocation If the carbon bearing the positive charge is part of a benzene ring the carbocation is termed an aryl carbocation The simplest case is called the phenyl carbocation R R Re a RR Generic aryl carbocation The phenyl carbocation C Carbocation Stability The stability of carbocations is dependent on a few factors The first factor to look at when deciding the stability of a carbocation is resonance Resonance is a stabilizing feature to a carbocation because it delocalizes the positive charge and creates additional bonding between adjacent atoms Decreasing the electron deficiency increases the stability Having trouble with resonance Consult the resonance tutorials on the course web site webchemuclaeduNhardingtutorialstutorialshtml Consider the following H H H 9 CH3 B CH3O 3 lt gtCH3O H H H No resonance Resonance The structure on the left does not have any resonance contributors in which electrons are donated to the carbon with the open octet Compare this with the carbocation that has resonance and a delocalized positive charge Charge delocalization imparts stability so the structure with resonance is lower in energy Organic Chemistry Tutorials C arbocations Page 3 In the example shown above an oxygen atom lone pair is involved in resonance that stabilizes a carbocation In general any adjacent lone pair or 7 bond can also be involved in resonance delocalization of a carbocation positive charge Allylic and benzylic carbocations enjoy resonance stabilization by delocalization of the positive charge to the adjacent 7 bonds Vinylic and aryl carbocations do not enjoy resonance stabilization because their 7 electron clouds are perpendicular to the vacant p orbital of the carbocation Recall that resonance requires the interacting orbitals to be parallel so they can overlap Without overlap there can be no resonance Note the in uence of inductive effect versus resonance on the energies of these molecules The oxygen atom that is bonded to the carbocation on the right is more electronegative than the corresponding hydrogen atom in the lefthand structure We would think that the inductive effect would pull electron density away from the carbocation making it higher in energy In actuality resonance usually but not always outweighs other factors In this case carbocation stabilization by resonance electron donation is a more signi cant factor than carbocation destabilization by inductive electron withdrawal Methyl and primary carbocations without resonance are very unstable and should never be invoked in a reaction mechanism unless no other pathway is possible More stable carbocations secondary or tertiary with resonance or any carbocation with resonance is sufficiently stable to be formed in a mechanism under reasonable reaction conditions Exercise 2 Draw all significant resonance contributors for the following carbocations ea 20 6 0 CH2 b 9 9CH3 d The second factor that should be considered when thinking about carbocation stability is the number of carbons attached to the carbon carrying the positive charge We look at the number of bonding electrons that are attached to the carbocation because those bonding electrons will help in alleviating the positive charge Bonding electrons from adjacent 0 bonds may overlap with the unoccupied p orbital of the carbocation Organic Chemistry Tutorials C arbocations Page 4 Hyperconjugative overlap Empty pz orbital of carbocation This phenomenon is termed hyperconjugation Since the overlap supplies electron density to the electronde cient carbocation carbon we predict that increasing the number of hyperconjugative interactions increases carbocation stability Extending this idea we predict that increasing the number of bonds adjacent to the carbocation by increasing the number of alkyl groups attached to the carbocation carbon results in an increase in carbocation stability For example a tertiary carbocation should be more stable than a secondary carbocation This prediction is accurate Our simple prediction suggests that any adjacent bonding electron pair will participate in carbocation hyperconjugation However only CiH and C bonds provide a signi cant level of increased stability When considering the importance of hyperconjugation versus resonance as the more important stabilizing feature resonance usually wins out For example a primary carbocation with resonance is more stable than a secondary carbocation without resonance A secondary carbocation with resonance is usually more stable than a tertiary carbocation without resonance The general rules for carbocation stability can be summarized as follows a Increasing substitution increases stability CH methyl least stable lt RCHf 1quot lt RZCH 2quot lt R3C 3quot most stable b Resonance is more important than substitution For example a secondary carbocation without resonance is generally less stable than a primary carbocation with resonance Exercise 3 Rank the relative stability of the three carbocations in each set 9 B G a HZC CH2 H3C c CH3 and CH33C H Organic Chemistry Tutorials C arbocatioris Page 5 b 6 x e 9 In vinylic carbocations the positive charge is assigned to a carbon with Sp hybridization How does this in uence the carbocation s stability An Sp orbital has more S character than an sz orbital Electrons in an S orbital are closer to the nucleus and therefore more tightly held than electrons in a p orbital This can be taken to mean that the electronegativity of carbon increases with increasing S character Thus Sp carbon most S character most electronegative gt sz gt Sp3 least S character least electronegative Electronegativity is a measure of electron attraction So the stability of a cation is in uenced by the electronegativity of the atom bearing the positive charge The more electronegative the atom the less stable the cation A vinylic carbocation carries the positive charge on an Sp carbon which is more electronegative than an Sp2 carbon of an alkyl carbocation Therefore a primary vinylic carbocation is less stable than a primary alkyl carbocation Similar reasoning explains why an aryl carbocation is less stable than a typical secondary alkyl carbocation such as cyclohexyl carbocation Because of their reduced stability vinyl and aryl carbocations are not often encountered D Carbocation Formation Even though carbocations can be found in many organic reaction mechanisms most carbocations are formed by one of only two basic mechanism steps ionization of a carbon leaving group bond or electrophilic addition to a 7 bond Ionization of a Carbon Leaving Group Bond When a bond between a carbon atom and a leaving group ionizes the leaving group accepts the pair of electrons that used to be shared in the covalent bond This may leave the carbon atom with an open octet resulting in a carbocation The ionization is indicated with a curved arrow starting at the bond and pointing to the leaving group atom that accepts the electron pair Better leaving groups or formation of a more stable carbocation result in lower activation energy and faster ionization Carbon leaving group bonding ionization is illustrated using an oxonium ion CH33cC9H2 gt CH33C quotH2 Organic C hemiStry TutorialS C arbocationS Page 6 Carbocation formation by ionization of a leaving group occurs in many organic reactions such as the SNl and El mechanisms Not every leaving group ionization affords a carbocation for example the SN2 reaction Electrophilic Addition to a it Bond When an electrophile attacks a 7 bond the 7 electron pair may form a new 0 bond to the electrondeficient atom of the electrophile Not all additions to 7 bonds involve electrophiles or carbocations The other 7 bond carbon no longer shares the 7 electron pair resulting in a carbocation This addition is indicated with a curved arrow starting at the 7 bond and ending at the electron deficient atom of the electrophile More powerful electrophiles or the formation of more stable carbocations result in lower activation energy and faster addition Electrophilic addition to a 7 bond is illustrated by the reaction of HBr an electrophile with styrene PhCHCHz Note that the more stable carbocation secondary with resonance is formed This is a key mechanistic feature of Markovnikov s Rule HIE a ltgt gt 13 Electrophilic addition to a 7 bond occurs in many reactions of alkenes alkynes and benzene rings Note every addition reaction forms a carbocation for example catalytic hydrogenation or ozonolysis Exercise 4 Suggest the products of each mechanism step 9 I HOAH a gt c gt G b O HLB T E Three Fates of a Carbocation Now we consider how carbocations behave in reaction mechanisms Generally speaking carbocations are unstable due to their open octets and positive charges Thus their reactions will be strongly in uenced by filling the octet of the carbon bearing the positive charge or at least making this positive charge more stable There are three common mechanism pathways or fates by which carbocations may achieve this stability These Organic Chemistry Tutorials C arbocations Page 7 fates are a capture a nucleophile b lose a proton to form a 7 bond and c rearrange Note in each case that the carbon bearing the open octet gains a pair of electrons thus completing its octet Capture a nucleophile The carbocation is electrophilic because it has a positive charge and in most cases a carbon atom with an open octet The positive charge is neutralized when an electron pair is accepted and a new covalent bond is formed By de nition a species that donates a pair of electrons to form a new covalent bond is a nucleophile Because carbocations are very reactive even weak nucleophiles such as water can be captured with ease ICH3 EH3 H3C C C 2 gt H3C C CH2 I 399 H2 I I H H 0H2 Lose a proton to form a II bond Accepting an electron pair from an adjacent bond to a hydrogen atom neutralizes the positive charge or lls the open octet and forms a new 7 bond The carbocation carbon now has four bonds and a full octet so its formal charge is zero The hydrogen atom must be removed by a base but because carbocations are generally very reactive species and very strongly driven to dispose of the positive charge even a weak base such as water or iodide ion can accomplish this deprotonation CH3 CH3 6 H3C Cj CH2 gtH3C CCH2 H30 I 6 H2 When carbocation deprotonation can lead to more than one product the more stable product is major Rearrangement The bonding electrons of a carbocation may shift between adjacent atoms to form a more stable carbocation For example rearrangement will occur if a secondary carbocation can be formed from a primary carbocation because a secondary carbocation is more stable than the primary carbocation There can be two types of rearrangements Shift of an alkyl group is called a l2alkyl shift CH3 CH3 12 alkyl shift 69 H3C H2 gtH3C C CH2 I H H Primary Secondary Shift of a hydrogen atom is called a l2hydride shift Hydride ion Hf Organic Chemistry Tutorials C arbocations Page 8 CH CH3 4 3e 12 hydride shift 4 H3C CH2 gtH3C CH2 L1 9 H H Primary Tertiary Of these two examples hydride shift leads to a tertiary carbocation whereas alkyl shift leads to a secondary carbocation Because a tertiary carbocation is more stable than a secondary carbocation the hydride shift is favored in preference to the alkyl shift Any C H or C C bond adjacent to a carbocation may shift including C C bonds that are part of a ring but only C C and C H bonds can migrate during carbocation rearrangement The most common carbocation rearrangements involve a carbocation rearranging into a more stable carbocation such as 20 gt 30 with resonance So use these rearrangements with impunity Rearrangements that transform a carbocation into another of apparently equal stability are less common but they do occur So before invoking this kind or rearrangement ask yourself if a better rearrangement or some other mechanism step is possible Rearrangement to a less stable carbocation is very unusual but also does occur This is the pathway of last very last resort All other reasonable options must be ruled out rst Exercise 5 For each carbocation draw the best rearrangement Include curved arrows H H cH3 a H3C C CH3 c H CH3 6 AA Vinylic carbocations generally do not rearrange even if they can become more stable For example the rearrangement shown below does not occur even though a secondary carbocation would rearrange to become a more stable allylic carbocation primary with resonance H H 9 H2C3CCH2 erc CCH2 Organic Chemistry Tutorials C arbocations Page 9 This resistance to rearrangement is probably due to orbital alignment restrictions during the rearrangement transition state E Rearrangement Causing a Change in Ring Size Rearrangement may lead to a change in ring size For example Q 12 alkyl shift V4 39 9 Primary Secondary The driving force for this rearrangement is formation of a more stable secondary carbocation from a less stable primary carbocation What often puzzles students is how to draw the structure of the product Here is a little trick that might help Begin by redrawing the starting structure aware Next number the ring to keep track of the atoms then make the bond changes suggested by the curved arrows but leave the atoms in place This may lead to a funky structure but this will be fixed in the next step In this case the Cl CS bond shifts taking a pair of electrons from C1 leaving Cl with an open octet and a positive formal charge The former carbocation carbon gained an electron pair so its formal charge becomes one unit more negative 1 to zero The fivemembered ring has expanded to a sixmembered ring 2 2 3 1 3 3 1 gt 4 4 5 5 Now redraw the rearranged product to make it look better using the numbering scheme to keep the substituent positions in order Organic Chemistry Tutorials C arbocations Page 10 3 redraw Exercise 6 Draw the rearranged carbocations based on the given curved arrows Indicate how the carbocation stability has increased by this rearrangement H3C e gt gt a JHZ b O H3C Exercise 7 Draw the products based on the curved arrows Name the carbocation fate illustrated a gt OCH3 Cle b 9 c W gt H Exercise 8 Provide a carbocation to complete each reaction Draw the curved arrows 9 H20 0H2 a gtX H O b W 2 rk H5 Organic Chemistry Tutorials C arbocations Page 11 12 alkyl shift c gt 6 Exercise 9 Illustrate the three carbocation fates using any molecules you want Use curved arrows and give the products Exercise Solutions Exercise 1 The carboncarbon bonds that determine the carbocation type are shown in bold a CH3 C CHZCH3 secondary b y tertiary Gr c O pnmary Exercise 2 a CH3 a QCH3 8cm Ho G c CH2 4 CH2 lt gt CH2 H H d No other signi cant resonance contributors can be drawn for this carbocation Exercise 3 Use the general carbocation stability rules i Methyl least stable lt 1 lt 2 lt 3 most stable and Organic Chemistry Tutorials C arbocations Page 12 ii Resonance is more signi cant than substitution pattern 9 Cgt a H3C f CH3 lt HZC CCH2 lt CH33C H H Secondary Primary with resonance Tertiary least stable most stable b A lt Cgt lt e 6 Tertiary Tertiary with resonance Tertiary with more resonance least stable most stable Exercise 4 Recall that if the curved arrow starts at a bond that bond is broken A curved arrow that points form an atom or bond to another atom or the space between two atoms indicates that those two atoms become bonded If the atoms are already bonded then the bond order increases For example a single bond becomes a double bond 1 e e a gt I Br Problems a and b show that the same carbocation cyclohexyl can be formed by two very different reaction pathways 9 HOAH c gt Y H20 C16 6 Recall that ionization of a carbon leaving group bond does not always form a carbocation In this case the electron pair lost to the leaVing group is replaced by the Organic Chemistry Tutorials C arbocations Page 13 electron pair gained from scission of the adjacent C H bond The carbon that starting with the leaving group never has an open octet Exercise 5 The best rearrangement will provide the greatest increase in stability For example a carbocation that can rearrange into a secondary or tertiary carbocation will preferentially rearrange into the tertiary carbocation because everything else being equal a tertiary carbocation is more stable than a secondary carbocation H H CH3 H H CH3 a H3C b b b CH3 gtH3C Ib CH3 H CH3 H CH3 H b M 6 3 H 0 GB 9 d This carbocation is tertiary with resonance and cannot gain greater stability by rearrangement Exercise 6 a The rearranged carbocation is tertiary whereas the starting carbocation is primary Rearrangement also relieves the strain associated with the cyclobutane ring H3C H3C 9 redraw djoZ gt 6 CH2 Igt b The rearranged carbocation is secondary with resonance whereas the starting carbocation is secondary 13m O O O H3C e H3C 6399 H3c H3C Organic Chemistry Tutorials C arbocations Page 14 Exercise 7 6 a Lose proton form 7 bond gt E HZOCH3 mcm 19 b Capture nucleophile gt ltgtltCl R G gt c earrange W W H Exercise 8 311 a X 0H2 gt 2 B H OH2 b gt H3643 69 12 alkyl shift 0 M gt M Gr Exercise 9 Any carbocation and any other reactants are acceptable as long as the carbocation fates are accurately illustrated Capture nucleophile H H H CH 3 I W I I I CH2 HOCH3 gt CH2 H H Organic Chemistry Tutorials C arbocations Page 15 DISCUSSION 7 CHEM 30A SPRING O8 JESUS C 1 Give the products that would be obtained from the reaction of cis2butene and trans2 butene with each of the following reagents If the products can exist as stereoisomers show which stereoisomers are obtained 31 I HA 1 r I a HCI MA I b BH3followed byNaOHH202 all I we l f c BrzinCHzCIfv quot 73 I M P quot 5 Br A 13 d BrZH20 N W Mg M3 61 we 2 I J 4 A H3 Na kg I I J e HCI CH OH We my I 4 I A 3 a 7M9 I39I W IA9V M fill 39 V 311Hgt A A 4 k r f H 4 1TU I3 4 II 4 K v g t qwj S zme mx lwe 2 Predict the major products of the following reacthns Include stereochemistry where appropriate 3 WA HCI Brz gt b DCM I mymc N 5xquot I39vf KMnO4 OH If C gt Mk 2 term I I Mt cold dil A 7 A G ug 1t W W 1 HI H20 39 L l 7 3 m I rcUA Llwa 1 H9OAC2 H2O 13H H rF j J I L J KMnO4 OH tquot 1H 0 lt3 cw warm conc H J vawm C Manashi Challerjee PhD Winter2008 Chem 30A Winter 2007 Week 7 and 8 Home Work Chapter 5 51 A and B Structure of alkene and calculating index of hydrogen de ciency Home Work problems and example 51 5 Home work End of Chapter problems 59 510 511 512 532 51 C Cis and trans isomerization Cis has more steric strain and less stable 52 Nomenclature E and Z system dienes trienes and polyenes E Z in dienes and trienes Home Work problems and example 53 54 56 58 Home work End of Chapter problems 513 a c e g i l m 514 515 516 517 518 519 520 523524 54 Isoprene units terpene Vitamin A has 4 isoprene units Home work End of Chapter problems 529 530 531 533 534 5 Biological connection Importance of Cis double bonds Fats Vs Oils Triglycerides and their hydrolysis HomeWork Chpater 6 62 Read Home work End of Chapter problems613 63A Addition of HX Mechanism and energy diagram included Regioselectivity and stability of carbocation inductive effect and h yperconju gation Home Work problems and example 61 62 63 Home work End of Chapter problems 615 28 631 635 633 Addition of Water Acid catalysed hydration Mechanism and energy diagram included Regioselectivity and stability of carbocation nductive effect and hyperconjugation Home Work problems and exam le 64 65 Home work End of Chapter problems 617 627 629 634a 63C Carbocation rearrangement Home Work problems and example 66 63D Addition of Br or Cl in presence of CH 2C1 Mechanism and energy diagram included via Bridged Halonium intermediate Anti steroselectivity Home Work problems and example 67 Home work End of Chapter problems 623 625 620b 626 634b 63EHalohydrin formation Addition of Br or Cl in presence of water Mechanism and energy diagram included Anti and regioselectivity 0H goes to the carbon that has less h ydrogens Home Work example and problem 68 Home work End of Chapter problems 630 632 633 63F OxymercurationRedu ction Mercuric acetate water and NaBH4 sodium borohydride Overall reaction is hydration addition of HOH Hg0Ac is the electrophile
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