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Chem E2b - Organic Chemistry II Read syllabus before asking any questions…please. 3 midterms (100 pts each) 6-7 quizzes or take-home test questions Final exam Lab grade There are NO make-up exams for the midterms!!! There are NO make-up quizzes!!! What is Organic Chemistry? The chemistry of molecules associated with living organisms The Chemistry of C, H, N, O, S, & PContaining molecules Natural Products Loads of FUN!! Unnatural Bioactive Molecules, eg Drugs Polymers, Materials 1 Consumer Products: From Food to to the Package that Contains it! What Molecules Are Important? All of them… H3C O CH3 O2N H3C Me NO2 O-Na+ MeO O NO2 O naproxen sodium (alleveTM) tri-nitrotoluene (TNT) testosterone NH2 Cl O O O OH HO Cl OH N N Cl OH OH NH HO N O sucralose (SplendaTM) O HO N NH2 HO tryptophan OH adenine short-hand drawings: CH3CH2CH3 propane H H H C C C H H H H H H3C CH3 E2a Review: Things You Should Already Know for this Class 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Hybridization Molecular Orbitals: orbital types and interactions, diagrams Conformational analysis: drawing chair conformations and Newman projections Functional group names and nomenclature Stability of carbocation and anion intermediates Resonance Curved arrow mechanism, identifying electrophiles and nucleophiles Acidity and Basicity Kinetics and thermodynamics: reaction coordinate diagrams Relative Reactivity Stereochemistry Reagents and Reactions of: alkenes, dienes, alkynes, alkyl halides and alcohols, substitution and elimination reactions 2 Review - Molecular Orbitals Bonding in ethene: Review - Acidity Brønsted–Lowry Acids and Bases: acid donates a proton, base accepts a proton • Strong reacts to give weak • Weaker base = stronger conjugate acid • Stable bases (anions) are weak bases Lewis Acids and Bases • Lewis acid: non-proton-donating acid; will accept two electrons • Lewis base: electron pair donors 3 Review - Acidity and Anion Stability Factors that influence anion stability (more stable conjugate anion is a better acid): 1) Size of atom - applies only to comparison within columns of periodic table 2) Electronegativity - applies to comparison in rows, as well as carbon hybridization 3) Resonance - more resonance is more stable 4) Electron-withdrawing groups stabilize (inductive effect), electron-donating groups destabilize 5) Aromaticity - an anion is more stable if it is aromatic CH3CH2OH most acidic pKa = 15.5 Strongest conjugate base Weakest conjugate base Review: Carbocation Stability • Carbocations are important intermediates in electrophilic additions to alkenes, and in SN1and E1 reaction mechanisms. 1) 2) 3) 4) hyperconjugation: more substituted carbocation is more stable Resonance/conjugation: carbocation with more resonance is more stable Electron-donating groups stabilize, electron-withdrawing groups destabilize Aromaticity: an aromatic carbocation is very stable H + H + CH3 H3C N > H3C H + CH3 O > H3C CH3 H + CH3 > CH3 H3C O O more electron-donating substituent less electron-donating substituent 4 Review - Resonance Localized vs. delocalized electrons: Examples of resonance contributors: Phenol is more acidic than cyclohexanol because the phenoxide ion is stabilized by resonance: (conjugate base of phenol) Review - Conformational Analysis 0° Conformations of n-Butane: (Newman projections) A staggered conformer is more stable than an eclipsed conformer H3C CH3 H H H CH3 H H H HCH3 H H CH3 H H H eclipsed CH3 gauche A B H CH3 H H CH3 H CH3 H H H3C H H D H H H H gauche anti C CH3 H3C E F Chair conformations: Steric strain of 1,3-diaxial interactions makes axial conformer less stable 5 Review: Thermodynamics and Kinetics Reaction coordinate diagrams: ∆G° = ∆H° – T∆S° ∆G° = Gibbs standard free energy change Enthalpy (∆H°) = the heat given off or absorbed during a reaction Entropy (∆S°) = a measure of freedom of motion (usually ∆S° is small compared to ∆H° and ∆G° ~ ∆H°) ∆G‡ = (∆G of transition state) – (∆G of reactants) ∆G‡ = energy of activation Two “Rules of 1.4”: 1) Increasing ∆G‡ by 1.4 kcal/mol decreases the rate by a factor of 10 2) changing ∆G˚ by 1.4 kcal/mol changes the product ratio by a factor of 10 The rate-limiting step controls the overall rate of the reaction The highest hill on the reaction coordination diagram is the rate-limiting step Review: Stereochemistry Cl Cl CH3 CH3 Enantiomers Cl Cl Diastereomers Diastereomers Cl Cl CH3 Enantiomers CH3 Cl Cl Cl CH3 H3C lowest priority group oriented behind 1 Cl Enantiomers Br Cl Diastereomers CH3 Diastereomers Cl MESO CH3 H3C Cl Cl 6 3 ' 1-2-3 counterclockwise = S Cl CH3 H H3CH2C Cl H3C 2 CH3 H3C Functional Group Review H CH3 R R alkane F R N R Cl R Br R alkyl chloride I R alkyl bromide O epoxide benzene diene alkyne alkene alkyl fluoride R R R SH R alkyl iodide CN R nitrile (cyano group) thiol Me R R O OH R H amine O R alcohol R R ether R S O O sulfonate ester (OTs, tosylate) R aromatic compounds (substituted benzenes) carbonyl compounds O R O N H amide O R R O ester O O O O O R R OH carboxylic acid R H aldehyde R R ketone R Cl acid chloride R O R anhydride Review: Polar Organic Reactions (anions and cations) • Reactions involving cations: electrophilic addition to an alkene, SN1 and E1, etc • Reactions involving anions: anything with a nucleophile (SN2, E2, etc) • Curved arrow mechanism: from an electron-rich center to an electron-poor center (Robinhood rule) Movement of a pair of electrons Movement of one electron “fish-hook arrow” 7 Reactions of Benzene Z + Y Z R Y R Z- R Y (a) H H H Y Z + Y H addition (path a) Z- (b) substitution (path b) H Z Y H benzene +Y-Z (nonaromatic) H Y + Hbenzene (- H+) + (Y+) (aromatic) Electrophilic Substitution is Thermodynamically Favored First step is rate determining 8 Sidenote: Synthesis of Aspirin from Crude Oil O OH OH O CO2/base [O] O OH Me Me O O O O OH Me acetyl salicylic acid (Aspirin) oil well Willow Trees: Relieves Fever, Upsets Stomach Summary of Electrophilic Additions FeX3/X2 Halogenation HNO3/H2SO4 Nitration H2SO4/∆ Sulfation X NO2 SO3 O Friedel-Crafts Acylation RCOCl/AlCl3 Friedel-Crafts Alkylation RCl/AlCl3 9 R R Halogenation of Benzene • • • In general, benzene is less reactive than most alkenes. Alkenes generally react with neutral electrophiles, e.g. Br2, Hg(OAc)2, etc. Benzene generally reacts with cationic electrophiles, which are more reactive due to the positive charge Br R Br Br R Br H Br Br X Br FeBr3 H Br + Br Br Fe Br + H Br Br Br Fe Br Br Br Nitration of Benzene O H O N+ O O O- H2SO4 HSO4- H O+ HNO3 N+ O- +N + H O H O H H O N + O + O N+ H O N+ - O 10 HSO4- O- + H2SO4 Friedel-Crafts Acylation O R Cl Cl Cl Al O Cl Cl Al Cl+ - Cl R + O O Cl + AlCl4- + R R an acyl chloride an acylium ion + O O H H O R R R AlCl4- AlCl3 Friedel-Crafts Alkylation R R Cl Cl Cl Al R Cl R H Cl Cl Al Cl Cl + R an alkyl chloride + R + AlCl4- a carbocation H R R H R + R H R R AlCl4- H+ AlCl4- 11 Carbocation Rearrangements Occur During Friedel-Crafts Alkylations R H Cl R Cl Cl Al R Cl + R + AlCl4- 1,2 alkyl shift 1,2 hydride shift ring expansion Me Cl Me Me Me Me O% H 100% + H Me Me Me Me Me Me AlCl3 Me + Me Me Me Friedel-Crafts Acylation/Reduction: Net Alkylation H H R Cl AlCl3 + H H R X H2; Pd/C O O R • • Cl AlCl3 + R Because of the problems associated with rearrangements, etc, primary alkyl groups are best introduced by a Friedel-Crafts Acylation, followed by reduction to replace the carbonyl group with two H’s The use of H2/Pd only works with carbonyl groups with benzene rings on one or both sides. 12 Disubstituted Benzenes Br Br Br Br Br 1,2-dibromobenzene ortho-dibromobenzene o-dibromobenzene 1,3-dibromobenzene meta-dibromobenzene m-dibromobenzene Br 1,4-dibromobenzene para-dibromobenzene p-dibromobenzene NO2 ortho Z meta NO2 ortho Br meta para Br 1-bromo-3-nitrobenzene meta-nitrobenzene Cl 2-bromo-4-chloro1-nitrobenzene Reactions of Benzene Substituents Br NaOH Me KOt-Bu SN2 (& SN1) OH Br E2 (& SN1) H2; Pd/C Me H H2; Pd/C OH O- H2; Pd/C Alkene Reduction: O Carbonyl Reduction: O N+ Nitro Reduction: 13 NH2 Effect of Substituents on Reactivity R R R H E + E+ E -H+ R H E + R R E • • Electrophilic Substitution reactions proceed via carbocation intermediates. Carbocation formation is the rate limiting step, so the stability of the carbocation that is formed determines the speed of the reaction Substituent Effects on Reactivity R = Electron Withdrawing (Slower) R=H R = Electron Donating (FASTER) R + H E R • • Electron donating groups help stabilize positive charge, so they make benzene rings more reactive Electron withdrawing substituents destabilize positive charge, so they make benzene rings less reactive 14 Inductive Effects on Benzene Reactivity electron donating group electron withdrawing group Z H Y > > LESS REACTIVE (Electrophilic Substitution) MORE REACTIVE (Electrophilic Substitution) • • • • • • • • As with acidity, inductive effects are generally WEAKER than resonance effects Z = NR2, OR: Strongly Activating (resonance) Z = NHCO2R, OCO2R: Moderately Activating (inductive) Z = R (Alkyl, vinyl): Weakly Activating (inductive) Y = F, Cl, Br, I: Weakly Deactivating (inductive withdrawal/resonance donation) Y = CO2R: Moderately Deactivating (resonance) Y = CN, SO2R, NO2: Strongly Deactivating (resonance) Y = NH3+: Strongly Deactivating (inductive, positively charged) Effects of Substituents on Orientation 1) All activators are o,p-directors Me Me Me Me Br FeBr3/Br2 Br Br p-bromotoluene o-bromotoluene m-bromotoluene NOT OBSERVED 2) Weak deactivators (halogens) are o,p-directors. Br Br Br Br Br FeBr3/Br2 Br o-dibromobenzene Br p-dibromobenzene m-dibromobenzene NOT OBSERVED 3) All moderate and strong de-activators are m-directors O Me O Me O O Me Me NO2 HNO3/H2SO4 NO2 m-nitroacetophenone 15 o-nitroacetophenone NOT OBSERVED NO2 p-nitroacetophenone NOT OBSERVED Substituent Effects: Ortho/Para Directors MeO + MeO MeO MeO H + H H MeO NOT OBSERVED H E E H MeO + MeO MeO -H+ + para: + + H H E MeO -H+ + H E -H+ H E + MeO MeO E para MeO H MeO -OR- H E + + H E H E + MeO H E + MeO H meta + H E meta: -OR- ortho MeO + H E + E+ stable resonance form ortho: MeO MeO H E H H E H H E H E E stable resonance form Substituent Effects: Meta Directors + + E+ NO2 -OR- H E + -OR- + H meta -H+ H E + E para NO2 NO2 H E + NO2 H NO2 H E + H E ortho unstable resonance form ortho: NO2 NO2 NO2 E NOT OBSERVED NO2 meta: NO2 H E + + H NO2 H E H NO2 NO2 -H+ + H E E H NO2 NO2 NO2 -H + para: + + H H E + H H E H H E E unstable resonance form 16 NOT OBSERVED Hill Diagram Summary: Activation and Direction R O H + N+ R H R +N H R O S R O OO N R = Deactivating/m-directing R o,p - m frontier R = weakly deactivating/o,p-directing activation frontier R=H F, Cl, Br, I R = Activating/o,p-directing R + R H N R R O R O H E R + H E O N H OR R Halogens are Ortho/Para Directors Cl Cl Cl E+ + Cl H E -H+ -OR- + H Cl E -OR- H E E MeO + + Cl H E NO2 H E H E + MeO + Cl + NO2 + H H E H H E Cl behaves like MeO for directing substitution (resonance) Cl behaves slightly like NO2 for activation (induction) As with acidity, resonance is more important than induction! 17 H H E Sample Problem: Planning A Synthesis with Aromatic Substitution SO3H Br Sample Problem: Planning A Synthesis with Aromatic Substitution O Me NO2 18 Arenediazonium Salts: Selective Monosubstitution CN NO2 CuCN X H2; Pd/C CuX N2+ NH2 NaNO2/HCl (X = Cl or Br) I Cl- ("HONO") KI (X = Cl or Br) OH HCl/H2O Synthesis with Arenediazonium Salts Me Me Me Cl FeCl3/Cl2 Cl o-chloroethylbenzene Me p-chloroethylbenzene Me Me NaNO2/HCl Cl- CuCl ("HONO") NH2 N2+ Cl p-chloroethylbenzene 19 Nucleophilic Aromatic Substitution (SNAr) R R R E E+ + H+ E X Nu Nu+ X- Cl OH NaOH/∆ pH 14 "Nu-" NO2 NO2 Cl O2N NO2 H2O pH 7 OH O2N NO2 NO2 NO2 Mechanism of Nucleophilic Aromatic Substitution HO- X OH NaOH/∆ pH 14 NO2 NO2 addition elimination HO X HO X - NO2 NO2 other resonance forms • X needs to be Small (F and Cl are best), and the benzene ring needs to be highly activated (at least one NO2). 20 Benzyne Cl NH2 NaNH2 SNAr? Cl H2N H - NH2 * H NaNH2 * AND H elimination NH3 - H2N * H2N NH2 - * NH3 -NH2 Benzyne Intermediate H2N * - Structure of Benzyne R R Alkyne Benzyne 21 * Heteroaromatics 5-membered heterocycles N S O H pyrrole furan thiophene 6-membered heterocycles N N N pyridine isoquinoline quinoline 6,5-fused heterocycles N indole O H S benzofuran benzothiophene All of these heterocycles are aromatic (recall rules of aromaticity & Hückels Rule) They all participate in Electrophilic Aromatic Substitution (EAS) Reactions EAS: 5-Membered Heterocycles O O H3C O CH3 3 4 EAS: 2 5 O CH3 N N1 + H3C OH O H 2-acetylpyrrole H 4 3 2 5 EAS occurs preferentially at the C2-position Br2 O1 Br O 2-bromofuran 4 3 HNO3 2 5 H2SO4 S 1 S NO2 2-nitrothiophene Consider Mechanism: EAS at C2 H 20-allylic carbocation E+ N N 20-allylic carbocation E E H H H EAS at C3 N N H H 20-alkyl carbocation 22 E N H H H + E E N H H Relative Reactivities of 5-Membered Heterocycles in EAS > > > O N H S Pyrrole, furan, and thiophene are all more reactive than benzene as heteroatom lp can better stabilize the carbocation intermediate Relative reactivities are reflected in L.A. needed to mediate Friedel-Crafts acylations O O + AlCl3: Strong L.A. AlCl3 H3C O SnCl4: Weaker L.A. + S H3C SnCl4 H2O Cl O + BF3: Weak L.A. O CH3 H2O Cl O BF3 H3C H2O Cl CH3 S CH3 O O O NO L.A. + N H O H3C O no cat. CH3 required CH3 N H less reactive acylating agent O Reactions of Six-Membered Heterocycles + SN2: H3C I N-methylpyridinium iodide N N I CH3 + N-Oxidation: + H2O HO OH N OH N N OH O Pyridine-N-Oxide + Br2 EAS: N Br FeBr3 Preferential substitution @C3, why ?? 300 °C N 30% High Temps required due to the fact that electron withdrawing N-atom destablizes carbocation intermediates. B Mechanism: Y + N Y+ slow H N 23 Y fast + N HB+ Electrophilic and Nucleophilic Substitutions on Pyridine What about Nucleophilic Aromatic Substitutions ? 4 3 5 2 6 N 1 OCH3 NaNH2 ∆ N NH2 SNAr takes place ONLY at C2 and C4 positions OMe Br Why ??? N NaOMe ∆ N Regiochemistry of SNAr Regiochemistry of Nu Addition: 2-Different LG’s: If LG’s are different, Nu will preferentially substitute at LG which is weaker base (better LG) better LG Br- is weaker base than CH3O- Br N NH2 OCH3 NaNH2 ∆ CH3 N N OCH3 CH3 Cl 24 NaOMe ∆ N OMe Other Reactions of Pyridine NBS Benzylic Bromination ∆, ROOR R N R N Br Reactions of Pyridine Diazonium Salts H2O NaNO2 / HCl N NH2 0 °C N N Cl N N OH enolic form NH2 N NaNO2 / HCl N N Cl- OH N O H keto form (more stable) α-pyridone O H2O 0 °C N N enolic form N H keto form (more stable) γ-pyridone Summary Slide • Aromatic compounds undergo electrophilic aromatic substitution (EAS), (mechanistically related to alkene additions) in which aromaticity is restored in the product • Electron donating substituents activate benzene toward EAS, and direct the electrophiles to the ortho and para positions • Electron withdrawing substituents deactivate benzene toward EAS, and direct electrophiles to the meta positions • Halogens are deactivating (inductive effect) and ortho/para directing (resonance) • Nucleophiles can add to arenediazonium salts, strongly activated aryl halides (SNAr), or benzyne intermediates • 5-Membered heteroaromatics participate in EAS at C2position. • 6-Membered heteroaromatics participate in EAS (C3-position) and SNAr reactions (C2 and C4 positions) 25 Neutral Organic Reactions - Radicals heterolytic cleavage H H H C C H H H H H H + C C formation of polar, charged species H H H methyl anion methyl cation homolytic cleavage H H H C H C H H H H H + C C formation of non-polar, neutral species H H H methyl radical methyl radical Structure of radical is somewhere between a cation and anion, half-filled p-orbital is a SOMO (Singly Occupied Molecular Orbital) both are electron-deficient Thermodynamics • Bond breaking is endothermic (∆H˚ is positive) • Bond formation is exothermic (∆H˚ is negative) H H H H H C H C H + C H C H H ∆H˚ = 90 kcal/mol H H Energy of activation (∆G‡) = ∆H˚ when bonds are broken homolytically, but no bonds are formed, IF no solvation is involved (in the gas phase) formation of bond: cleavage of bond: σ∗ sp3 H H C H H H C H H C HH sp3 H C H sp3 H sp3 energy H H H C C H H H σ stabilization energy = 90 kcal/mol reaction progress ∆H˚ is negative 26 90 kcal/mol required to break bond ∆H˚ is positive Bond Dissociation Energy (BDE) X + Y ∆H˚ = ∆G≠ = BDE energy Bond Dissociation Energy (BDE) = energy required to break a bond homolytically X Y reaction progress Bond Cl Br I RO Cl Br I I Br Cl Bond Dissociation Energy (kcal/mol) Cl Br I OR H H H CH3 CH3 CH3 Bond Dissociation Energy (kcal/mol) Bond 58 46 36 38 103 87 71 84 70 57 H H3C H3C H2C H3C H3C H2N CH3O HO 104 105 88 67 (π bond only) 92 85 107 104 119 H H CH3 CH2 OH NH2 H H H See Bruice Table 3.1, page 129 Two Factors Can Decrease the BDE (make the bond easier to break) 1) Make bond less stable (raise the energy of the reagent) H2 C H2C H3C CH2 H2 C CH3 ∆H˚ = 65 kcal/mol ∆H˚ = 85 kcal/mol CH2 CH2 H2C energy H3C CH2 + CH3 ∆H˚ = 65 kcal/mol CH2 H2 C CH2 ∆H˚ = 85 kcal/mol CH3CH2CH3 cyclopropane has strain energy that raises the energy of the C–C bond, therefore making it easier to break reaction progress 2) Make radical more stable (lower the energy of the transition state) BDE (kcal/mol) -from cleavage of a C–H bond, with R = CH3 85 105 85 96 27 99 101 104 Stability of Radicals: Just Like Carbocation Trends 1) Hyperconjugation: more substituted radicals are more stable 2) Conjugation/Resonance: more resonance, more stable radical H C most resonance CH2 > > CH2 > H3C no resonance CH2 3) Hybridization: more s character of the SOMO, less stable radical sp3-hybridized, least s character H3C CH2 > H H > > R C H C sp-hybridized, most s character Hyperconjugation in Carbocations and Radicals Carbocation stabilization: antibonding MO A σ C–H bond donates into the empty p-orbital to stabilize the carbocation bonding MO σ C–H = donor (HOMO) empty p-orbital = acceptor (LUMO) Cationic, 2-electron system Radical stabilization: A σ C–H bond donates into the SOMO (singly occupied molecular orbital) to stabilize the radical half-filled p orbital H H H C C antibonding MO H H σ C–H = donor (HOMO) SOMO = acceptor bonding MO Radical, neutral, 3-electron system Hyperconjugation is less stabilizing for a radical because the one electron in the antibonding MO has a destabilizing effect. (recall He2+ with a similar 3-electron system) 28 Radical Reactions with Alkenes reaction proceeds through a carbocation intermediate, rearrangements are possible reaction proceeds through a radical intermediate, NO rearrangements are possible hv or ∆ Formation of bromine radical: ∆H˚ = 38 kcal/mol Initiation steps to create radicals This is NOT a general reaction. The peroxide effect does NOT work with any other HX. Mechanism: More Stable Radical Forms Faster carbocation mechanism: + H3C CH3 H3C H Br δ+ δ- + Br- Br H+ adds to primary carbon to form a secondary carbocation, which is more stable than a primary carbocation radical mechanism: H3C + .Br CH3 H3C secondary alkyl bromide H-abstraction and formation of new radical H3C Br Br adds to primary carbon to form a secondary radical, which is more stable than a primary radical 29 H Br H3C Br primary alkyl bromide + .Br Synthesis Using Primary Alkyl Bromides A primary alkyl bromide is a good substrate for an SN2 reaction: O HBr, ROOR H3C hv NaO Br H3C O CH3 O H3C CH3 nucleophilic substitution Recall the previous general method that we used to convert an alkene to a primary SN2 substrate with a good leaving group: H3C 1. BH3 2. NaOH, H2O2 H3C OH 1. TsCl/pyridine 2. Nucleophile hydroboration and oxidation H3C Nu formation of good leaving group followed by nucleophilic substitution Examples of nucleophiles = NaCN, NaOH, NaOMe, NaSMe, NaOAc, NaCl, NaBr Radical Mechanism Sample Problem HBr Provide the mechanism for this reaction. ROOR, hv ∆H˚ = 38 kcal/mol Problem-solving strategy: 1. Draw a step that breaks the weak bond in the initiator 2. Draw a reaction of the initiator with one of the starting material (SM) 3. Draw a reaction of SM radical with another SM (repeat if needed) 4. Draw a termination step 30 Br Radical Reactions with Alkanes Not practical or useful with F2 or I2 Major product because secondary radical is more stable Remember, rearrangements are not possible with radicals!! Mechanism of Radical Reactions with Alkanes H-abstraction goes back to starting material very minor by-product desired product!! The rate-determining step of the overall reaction is hydrogen abstraction (Same mechanism for monobromination - practice at home…) 31 Again, More Stable Radical Forms Faster H-abstraction at benzylic and allylic positions is easier because benzyl and allyl radicals are stabilized by resonance NBS as a Milder Brominating Reagent for Allylic Bromination Advantage: the low concentration of Br2 and HBr present cannot be added to the double bond NBS provides a good source of HBr and Br2 in low concentration O H3C Br hv or ∆ + N Br O RO- OR (peroxide) + H3C Br H3C Unsymmetrical alkenes will form two different products because the allyl radical intermediate has two resonance contributors that are not equivalent H H3C + .Br H3C 32 H3C CH2 + HBr Phenolic Compounds as Anti-Oxidants • Radical reactions occur in the body, usually initiated by metal ions in enzymes. Unwanted radicals cause damage to cells, leading to disease. • Phenolic compounds, such as BHT, BHA and Vitamins A & E are “anti-oxidants” that act as radical scavengers (inhibitors). They are typically used as food preservatives. OH CH 3 CH 3 CH 3 CH 3 OH H 3C H 3C CH 3 CH 3 CH 3 CH3 HO CH3 H3C OCH 3 O CH3 CH 3 CH3 CH3 CH3 CH3 vitamin E (α-tocopherol) BHA BHT (butalyted hydroxyanisole) (butalyted hydroxytoluene) Radical scavenger mechanism: H R unstable free radical Reactive!! O CH3 CH3 CH3 + O OCH3 radical scavenger CH3 CH3 CH3 + R H OCH3 very stable radical (resonance) Unreactive!!! Phenolic compounds donate an H to the free radical to form two stable unreactive species Natural “anti-oxidants” that serve as radical scavengers in vitro can be found in chocolate, green tea, wine, grape juice, fruits and vegetables, etc. Recent studies have shown that dark chocolate has 2x as many phenolic compounds as milk chocolate, 2x as many as green tea or wine, and 20x as many as in tomatoes. Stereochemistry of Radical reactions CH3 H3C Br Br2, hv Racemic mixture of products (a pair of enantiomers) CH3 H3C • Both enantiomers are formed because the radical intermediate is planar, like a carbocation • If there is already an asymmetric center present in the molecule, then diastereomers will be formed. Cl H3C CH3 Cl Br2, hv Cl CH3 H3C Br 33 + CH3 H3C Br a pair of diastereomers Summary of Radical Reactions Know practical information of which radical reactions work (see below) Know thermodynamics and effects of bond and radical stability on BDEs Know radical stability to identify which product will form and be able to explain why a radical is stable - be able to draw resonance structures!! Know stereochemistry of radical reactions • • • • Br Br2 Br2, ∆ CH3 CH3 Br hv or ∆ Br NBS, ∆ HBr H3C CH3 H3C secondary alkyl bromide HBr ROOR H3C 1. 2. 3. 4. ROOR Br Br allyl bromide primary alkyl bromide Addition of H-Br to alkenes with peroxides Halogenation (Cl2 or Br2) of alkanes Substitution of benzylic hydrogens with Cl2 or Br2 Substitution of allylic hydrogens with NBS Radicals: Sample Problems 1) Based on the BDEs given, explain why an acyl peroxide forms radicals more easily than an alkyl peroxide? (include structures in your answer) RO OR alkyl peroxide CH3 CH3 O O ∆H˚ = 38 kcal/mol O O ∆H˚ = 29 kcal/mol acyl peroxide 2) How many products are possible for the reaction of 3-methyl-1-cyclohexene with NBS? Draw the structures of these products. (your answer does not need to include enantiomers) CH3 O + N Br O hv or ∆ RO- OR (peroxide) 3) In order to demonstrate why BHA is such a good radical scavenger, draw the 5 major resonance structures for the radical formed upon the reaction of BHA with a free radical. OH OCH3 CH3 CH3 CH3 BHA 34 Summary of Reactive Intermediates Name Structure Stability Properties R carbocation 3˚ > 2˚ > 1˚ > Me C R R R radical 3˚ > 2˚ > 1˚ > Me C R R C R Nucleophilic, electron-rich, strong base, lone pair (HOMO) is a good donor all very reactive Neutral divalent carbon, empty porbital is a good acceptor and lone pair is a good donor, so it is both electophilic AND nucleophilic R R carbene electron-deficient, singly occupied MO (SOMO) is reactive, can be either electrophilic or nucleophilic Me > 1˚ > 2˚ > 3˚ R carbanion Electrophilic, electron-deficient, strong acid, empty p-orbital (LUMO) is a good acceptor C R Carbene Formation A carbene is a neutral species containing a divalent carbon Br Br CH3 H3C CH3 C Br empty 2pz orbital (carbocation) KO Br Br C Br (or KOH) H 103˚ C Br dibromocarbene filled sp2 hybrid orbital (lone pair/carbanion) treatment with a strong base Mechanism of formation: Br Br Deprotonation with strong base Br Br Br C α-elimination Br C C Br H Br dibromocarbene OH 35 Formation of Carbenes from a Diazo-Compound empty 2pz orbital (carbocation) CH2N2 = diazomethane (a yellow gas) H C N N H hv or ∆ H C H + 103˚ N2 H C H carbene and N2 gas filled sp2 hybrid orbital (lone pair/carbanion) resonance structures of diazomethane: H H C N N H C N N H C N N H H carbon has anion character carbon has cation character Examples of other diazo-compounds: (brightly colored, but dangerous!!) O H3C N2 N2 H3C 4,4-dimethyldiazocyclohexa2,5-diene (a purple liquid) H3 C diazocyclopentadiene (an iridescent orange liquid) O N2 methyl diazoacetate (a yellow liquid) Carbenes react with Alkenes to form Cyclopropanes H CH2N2 CH2 hv or ∆ H3C H H CH3 N2 hv or ∆ H H3C H Stereochemistry of the alkene is retained in the cyclopropane product H CH3 H CHBr3, KOH Br C H Br One-step stereospecific syn addition mechanism 36 Carbenes: Molecular Orbital Interactions with Alkenes Consider what orbitals are available to interact for bonding: H alkene π∗ = LUMO H (acceptor) H H H H alkene π = HOMO H empty 2pz orbital = LUMO (acceptor) H C H filled sp2 hybrid orbital = HOMO (donor) (donor) H Two stabilizing (bond forming) interactions are possible: Stabilizing Interaction #1: Stabilizing Interaction #2: empty p orbital of carbene (acceptor) H H C H H H H H H empty p orbital alkene π C lone pair of carbene (donor) H H H H alkene π * sp2 carbene alkene π* (acceptor) alkene π (donor) There are two destabilizing interactions that DO NOT occur: 1) interaction between alkene π and filled sp2 of carbene (because both are donors) 2) interaction between alkene π* and empty p orbital of carbene (because both are acceptors) Carbenes: Sample Problems 1) Draw at least 5 of the major resonance structures for 4,4-dimethyldiazocyclohexa-2,5-diene (a purple liquid): H3C CH3 N2 2) Fill in the products or reagents for these reactions: H3C H3C H3C H3C O H3CO N2 hv 37 H Cl Cl H Determining the Structures of Organic Molecules How do you identify the products you synthesize? OH H3C 1) BH3 2) NaOH, H2O2 H2O, H2SO4 H3C CH3 H3C OH How can you tell which reagents give which products? How can you tell if only one major product is formed? CH3 CH3 CH3 NO2 HNO3 Does the nitro group add once or twice? -OR- H2SO4 Is addition selective for the ortho and para positions? NO2 NO2 Chromatography - Good for separations, but not for identification…. Chromatogram: What is the structure of this compound? 2.05 min 4.37 min All we can tell here is that the compound is more polar 3.22 min Time Gas chromatography (GC) and Liquid chromatography (LC) only give information about polarity (based on retention time), not the structure (Recall, last semester in lab, you performed silica gel column chromatography to purify a solid compound and you ran gas chromatography to check your SN2 reaction) 38 4 Techniques to Identify Structures of Organic Compounds (Analytical Chemistry) 1. UV/vis spectroscopy - information about conjugated π-systems 2. Mass spectroscopy - identify molecular mass of the compound and some structural features (functional groups) 3. IR spectroscopy - tells you important functional groups present 4. NMR spectroscopy - tells you functional groups, connectivity of atoms (framework), some stereochemistry, etc. http://www.spectroscopynow.com Spectroscopy and the Electromagnetic Spectrum • Spectroscopy is the study of the interaction between matter and electromagnetic radiation high frequency = short wavelength = high energy • A visible spectrum is obtained if visible light is absorbed • An ultraviolet (UV) spectrum is obtained if UV light is absorbed • An infrared (IR) spectrum is obtained if infrared light is absorbed • An nuclear magnetic resonance (NMR) spectrum is obtained if radiowaves are absorbed 39 UV/vis Spectroscopy: Background Info LUMO (lowest unoccupied molecular orbital) HOMO (highest occupied molecular orbital) LUMO Electronic Transitions hv HOMO • When a molecule absorbs light, an electron is promoted to a higher energy MO (from the HOMO to the LUMO), and the molecule is in an “excited state” • Although several electronic transitions exist between the MOs, only two transitions are low enough in energy to occur with UV and Visible light. electronic transition with the lowest energy Only organic compounds with π-electrons can produce UV/Vis spectra!! 40 Effects of Conjugation LUMO π∗ LUMO LUMO HOMO HOMO HOMO π orbitals of ethene CH2 H2C π orbitals of 1,3-butadiene 165 217 H2C CH2 λmax (nm) π CH2 H2C π orbitals of 1,3,5-hexatriene 256 See table 8.3 Bruice p. 325 Conjugation raises the energy of the HOMO and lowers the energy of the LUMO As conjugation increases, the HOMO-LUMO gap decreases More conjugation = less energy required for electronic transition = longer wavelength UV/Vis with Conjugated Carbonyl Compounds Two peaks are observed in the spectrum because two electronic transitions can occur. The HOMO-LUMO gap decreases with conjugation 41 Functional group effects Two structural features will show an increase in the wavelength of the chromophore: 1) Increased conjugation 2) A substituent with a lone pair attached to the chromophore (an auxochrome) An auxochrome is a substituent in a chromphore that alters the λmax and the intensity of the absorption CH3 H3C N N N amine-substituted azobenzene - a yellow dye (approx. 400nm) Sidenote #1: Polyenes and Vision or “do carrots help you see better?” • Vitamin A is a source of 11-cis-retinal • Opsin, a vision protein, binds 11-cis-retinal to form the Rhodopsin complex (rods) • When Rhodopson absorbs light, 11-cis-retinal isomerizes to 11-trans-retinal, causing it to be released from opsin. Upon release, a nerve impulse is generated and perceived by our brain as light in black or white vision • Same mechanism exists with iodopsin, another vision protein, to give us color vision (cones) • 1 carrot = 2000 mg of retinal equiv. (sweet potato and mango have 1200 and 800 mg each) cis-alkene H2N opsin lysine side-chain H Rhodopsin (500 nm) O H N 11-cis-retinal H opsin hv 11-trans-retinal + H release N H opsin trans-alkene nerve impulse = vision 42 opsin UV/Vis Spectroscopy: Sample Problems 1) Match the following UV absorption maxima with the corresponding compounds: 353 nm, 313 nm, 256 nm, 227nm, 180nm CH3 H3C CH3 CH3 H3C H3C CH3 2) How can you use UV/Vis spectroscopy to identify if this reaction has consumed the starting reagents and produced the desired cyclohexene structure shown? O O CH3 + ∆ CH3 Mass Spectrometry 70 eV R X R X electron beam dislodges electron fragmentation + R cation fragment observed in MS molecular ion radical cation observed in MS X radical fragment NOT observed in MS A molecular ion (radical cation) is recorded in a mass spectrum. High-Resolution Mass Spectrometry will give you the exact molecular formula - useful data for identifying an unknown structure. 43 Fragmentation 70 eV CH3CH2CH2CH2CH3 (electron beam pentane MW = 72 Only cations are recorded in the mass spectrum m/z = mass to charge ratio of the fragment The base peak at m/z = 43 is the most abundant cation, which is not usually the same as the molecular ion Fragmentations Give the Most Stable Cations Fragmentation of the molecular ion occurs one of two different ways to give the most stable cations: a) A C–X bond is cleaved heterolytically, where all electrons go to the more electronegative atom (usually X) b) A C–X bond is cleaved homolytically, at the α-position to give a stabilized cation across the C–X bond Examples of homolytic cleavage to give stable cations: 70 eV H2C CH CH2 R H2C CH CH2 + H2C R R m/z = 41 70 eV R Z CH2 CH3 R Z CH2 CH3 + R Z CH2 Z = N, O, S R can also be H 70 eV R R C O R C O R R C O acylium ion 44 + R CH3 Fragmentations at Functional Groups • The weakest bond is the C–Cl bond heterolytic cleavage • Both heterolytic and homolytic cleavage of the C–Cl bond occur. Cl isotope has 1/3 the abundance α-cleavage homolytic cleavage positive charge shared by C and Cl atoms Fragmentations of an Ether Group 45 Fragmentations of an Alcohol Group γ-abstraction Formation of new radical cation by formation of small neutral molecule (such as H2O, ROH, NH3, H2, ethene, etc) Fragmentations Occur to Give Stable Cations acylium ions stabilized by resonance γ-abstraction Formation of new radical cation and a small neutral molecule 46 Summary • • • • 1) 2) 3) 4) Fragmentations occur to give cations recorded in the mass spectrum; only positively charged fragments are recorded. The base peak is the peak with the greatest intensity, due to its having the greatest abundance Weak bonds break in preference to strong bonds Bonds that break to form more stable fragments break in preference to those that form less stable fragments Alkanes, alkenes and aromatics: cleave to give the most stable carbocations Alcohol: loss of water or α-cleavage to give stabilized cation Ethers: loss of an alkyl group or α-cleavage to give stabilized cation Ketones and aldehydes: loss of alkyl group to give stabilized acylium ion or McLafferty rearrangement (γ-abstraction) Be able to propose or identify favorable fragmentations for 2 or 3 of the largest peaks in the spectrum. You DO NOT need to account for all of the peaks in a spectrum. Mass Spectroscopy Sample Problems CH3 1) Predict the cation structure and base peaks for toluene 2) Account for the peaks at m/z 87, 111 and 126 in the mass spectrum of 2,6-dimethyl-4-heptanol. CH3 OH CH3 H3C CH3 47 Introduction to Infrared Spectroscopy The covalent bonds in molecules are constantly vibrating Stretching vibrations Bending vibrations It takes more energy to stretch a bond than to bend a bond Each stretching and bending vibration of a bond occurs with a characteristic frequency that gives a specific absorption band (peak) in the IR spectrum An Infrared Spectrum High frequency = short wavelengths (inversely proportional) Wavelength (µm) Wavenumber (cm-1) Functional group region (4000-1400 cm-1) Fingerprint region (1400-600 cm-1) Peaks are similar for each functional group Pattern is unique for each compound High frequency = large wavenumbers (directly proportional) - high energy 48 What Determines the Intensity of an IR peak? Greater change in dipole moment = more intense absorption A symmetrical bond will have no dipole moment and will therefore be infrared inactive (have no absorption band) Cl H H H H Cl H3C C C CH3 unsymmetrical unsymmetrical symmetrical symmetrical symmetrical Remember, you are looking only at each bond, not the entire molecule What Determines the Position of an IR peak? 1) Smaller atomic mass = larger wavenumbers (higher frequency) C H 3000 cm-1 C D 2200 cm-1 C O 1100 cm-1 C Cl 700 cm-1 increasing wavenumber 2) Stronger bonds = larger wavenumbers (higher frequency) a) Higher bond order = stronger C N 2200 cm-1 double bond C N -1 triple bond single bond C N b) Hybridization: more s character = stronger sp 1600 cm 2 sp -1 3 1100 cm sp C C H 3300 cm-1 C C H 3050 cm-1 C C H 2900 cm-1 c) Other factors: electron delocalization, the electronic effect of neighboring substituents, and hydrogen bonding Any effect that makes a bond stiffer and harder to stretch will increase the wavenumber 49 Carbonyl Compounds: Resonance and Inductive effects stronger, less flexible C=O bond O O O > > 1788 cm-1 C=O stretch 1718 cm-1 1691 cm-1 (resonance) (ring strain) stronger C=O, more double bond character C=O stretch (approx.) R O O O O R weaker, more flexible C=O bond O > > > R H R R R N R weaker C=O, less double bond character H ester aldehyde ketone amide 1740 cm-1 (inductive withdrawl) 1730 cm-1 1720 cm-1 1660cm-1 (resonance) Stronger C=O because No resonance to weaken C=O bond Resonance weakens C=O bond Alcohol and Acid Characteristic Peaks After a carbonyl peak, the broad O-H peak is the most characteristic peak to look for O-H of alcohol O-H of acid An O-H bond of an alcohol or acid are both very broad and intense. The O-H of an acid is even more broad, usually covering the C-H peaks Hydrogen-bonding effect: 3500-3200 cm-1 3300-2500 cm-1 An O–H bond is weaker and easier to stretch when it is hydrogen-bonded Reality check: water contamination in your sample can make it look like there is an OH present when there is not. 50 Alcohol group Broad OH peak Alcohol group + carbonyl group Broad OH peak Carboxylic acid group Super broad OH peak In addition to position, identify type of carbonyl by looking at secondary peaks… Ketone: no secondary peaks Aldehyde: C-H stretch at 2720 and 2820 cm-1 Amide: N-H stretch in 3500-3300 cm-1 range, and/or C-N stretch in 1200-1000 cm-1 range (similar for an ester with a C-O stretch and an acid with an O-H stretch) 51 Do not confuse an alkene or alkyne with a carbonyl - the intensities are much weaker than a carbonyl peak Note the differences in the sp, sp2 and sp3 C-H stretches (both the intensities and the positions). sp2 C-H stretch at 3080 cm-1 Also note the intensity and position difference of the CC double and triple bond. C=C stretch at 1650 cm-1 sp3 C-H stretch at 2850-2950 cm-1 Even when the intensity is diminished, the shift of the sp3 C-H is distinct from the C-H of an aldehyde because it is higher frequency and no carbonyl peak is present. CC triple bond stretch at 1650 cm-1 O-H stretch 3 C-H sp3sp C-H stretch stretch -1 -1 at 2800-2950 at 2850-2950 cmcm sp C-H stretch at 3300 cm-1 Be able to identify compounds with benzene rings (sp2 vs. sp3 carbons) 52 How to develop a “6th sense” for analysis of IR spectra: Problem Solving Strategy 1. 2. 3. 4. 5. 6. 7. 8. Look in 1800-1600 cm-1 range for strong sharp peak indicating a carbonyl, consider its relative position. Look for secondary peaks to distinguish between carbonyl compounds. For example, a broad OH peak (3300-2500 cm-1) to indicate a carboxylic acid. Look in 3650-3200 cm-1 range for strong broad peak indicating alcohol or amine. Look for C-O and C-N peaks in 1250-1000 cm-1 range, indicating an ether or tertiary amine. Look in 2800-3100 cm-1 range for sp2 vs. sp3, indicating alkene or benzene ring, look for C=C bond (1680-1600 cm-1) or triple bond (2100-2160 cm-1) Do NOT confuse C=C bond or triple bond with a carbonyl - the carbonyl has a strong intensity and the others have medium or small intensity Consider symmetry that would account for “missing” absorption bands Look in 1800-2800 cm-1 region - usually desolate, but has very characteristic peaks for CN and CC triple bonds Look in Appendix VI in your Bruice textbook to help with practice problems IR Spectroscopy Sample Problems - Practice!! Common Question Types: 1. 2. 3. 4. 5. 6. Given an IR spectra for an unknown molecule, identify three functional groups present in the molecule Rank and/or explain which bond has a lower frequency (lower wavenumber) for a series of bonds given For several carbonyl compounds given, be able to identify which carbonyl group exhibits the highest wavenumber. Identify which compound has a vibration that is IR inactive Know how to distinguish between a pair of compounds using IR data. Use IR data, along with mass spectrometry and NMR data, to propose a structure for an unknown organic molecule (in some cases, you will be given a molecular formula) Rank the relative frequencies of the following C=O bonds, where 1 = larger wavenumber and 3 = smaller wavenumber O H3C O CH3 Cl 53 H3C O CH3 F H3C CH3 CH3 Introduction to NMR Spectroscopy NMR = Nuclear Magnetic Resonance MRI = Magnetic Resonance Imaging Nobel Prize (Physics) in 1954 Nobel Prize (Chemistry) in 1991 Nobel Prize (Chemistry) in 2002 Nobel Prize (Physiology and Medicine) in 2003 for MRI NMR spectroscopy provides 4 pieces of data to identify the about the carbon-hydrogen framework of an organic molecule: 1) 2) 3) Number of peaks: tells the number of different types of protons (or carbons) Relative area of the peaks: tells the relative types of different protons Position of the peaks: tells the chemical environment of the proton, ie. the neighboring functional groups Peak splitting pattern: tells the number of protons on adjacent atoms 4) NMR Theory A nucleus must have a nuclear spin of +1/2 or -1/2 to be NMR active Examples include: 1H, 13C, 19F, 29Si, 15N, and 31P Any spinning charged particle generates a magnetic field; therefore, think of a nucleus as a mini-magnet that can be affected by an applied magnetic field (Bo) - - energy + aligned against the field: higher energy β nuclei - - + - + + + - + In the absence of an applied magnetic field Bo - + + + - + - - - + + + + - - - aligned with the field: lower energy α nuclei (slightly favored) In the presence of an applied magnetic field Bo + pole Bo - pole 54 NMR Theory - - + + + + - + + - + + + - - + Energy supplied + - Energy released + - - - - + + + + + + + Lower energy α nuclei - + Higher energy β nuclei + pole Absorption of energy can “flip” the nuclear spin - converts low energy to high energy Bo - pole Spin “relaxes” to equilibrium state through the magnetic field and energy is released, which can be detected as a signal Signals in 1H NMR spectra - How many? • Each set of chemically equivalent protons in a compound gives rise to a different signal/peak in the 1H NMR spectrum - Look for symmetry! propyl group isopropyl group ethyl group + methyl group methyl group tert-butyl group + methyl group “vinyl” protons “aryl or aromatic” protons 55 + energy Radiowaves supply enough energy to “flip the spin” of a nucleus NMR Time-scale IR spectroscopy is like a fast camera - you get an instantaneous picture of all the vibrations of a molecule (10-13 s) NMR spectroscopy is like a slow camera - you get a blurry picture that is timeaveraged for the molecule (10-3 s) Cyclohexane example: chair-chair interconversion H H (ring flip) (12.1 kcal/mol energy barrier) H H 1 peak @ 25 ˚C (fast ring flip) 2 peaks @ –90 ˚C (slow ring flip) the rate of chair–chair conversion is temperature dependent What Determines the Position of an NMR signal? Less electron density = less shielded more electron density = more shielded 56 NMR - Chemical Shifts • The chemical shift is a measure of how far the peak/signal is from the reference signal (TMS) • The common scale for chemical shifts = δ Position of TMS = Internal reference signal (0 ppm) CH3 H3C Si CH3 CH3 TMS = tetramethylsilane distance downfield from TMS (Hz) δ= operating frequency of the spectrometer (MHz) NMR - Chemical Shifts 1) Presence of an electronegative atom: more electronegative = less shielded Electron withdrawing effects cause a proton to be deshielded and the NMR signals appear at higher frequency (further downfield, at larger δ values) 2) Presence of an adjacent π-bond: C=C and C=O are also electron-withdrawing CH3 CH3 2.6 ppm O CH3 CH3 2.4 ppm 2.2 ppm CH3 2.1 ppm H3C CH3 1.2 ppm 3) Hydrogens directly attached to a π-bond: sp2 carbon of alkene has high s character that is electron-withdrawing, deshielding the hydrogen 7.5 ppm H R H R R 57 5.0 ppm but isn’t this a rather big effect for just an sp2 carbon center? Is there another effect? Diamagnetic Anisotropy - Effects of π-electrons The π electrons are less tightly held by the nuclei than are σ electrons; they are more free to move in response to a magnetic field. This creates an induced magnetic field Benzene (7-8 ppm) Alkene (5-6 ppm) Alkyne (2.5 ppm) Bo - Binduced Bo + Binduced Protons are shielded (at lower frequency) Protons are deshielded (at higher frequency) The induced magnetic field (Bo + Binduced) creates a unique environment for hydrogens that are bonded directly to carbons that form π bonds NMR - Chemical Shifts Protons in more electron poor environment R C C H 2.4 ppm Protons in more electron dense environment What about protons attached directly to a triple-bonded carbon? Proton on an alkyne experiences [Bo - Binduced], therefore, proton is shielded 58 Chemical Shift Sample Problem 1) 2) How many signals would you expect to see in the 1H NMR spectrum for each of the following compounds? For each molecule, indicate which protons are least shielded, ie. will give the NMR signal with the highest chemical shift value (furthest downfield). F C H2 H2 C C H2 Br H3C O C H2 H2 C C H2 H N O CH3 Hb H3C CH3 Ha 3) The hydrogens attached directly to the carbon of an aldehyde are very distinct because they occur at especially high chemical shift values, usually greater than 9 ppm. Explain why this hydrogen is so deshielded? NMR - Integration • The area under each peak is proportional to the number of protons that give rise to that signal • The height of each integration step is proportional to the area under a specific signal • The integration tells us the relative number of protons that give rise to each signal, not the absolute number 59 NMR - Splitting of the Signals • The splitting of signals, caused by spin–spin coupling, occurs when different kinds of protons are close to one another • An 1H NMR signal is split into N + 1 peaks, where N is the number of equivalent protons bonded to adjacent carbons • The number of peaks in a signal is called the multiplicity of the signal Splitting is observed if the protons are separated by three σ-bonds OR three σ-bonds and a π-bond (alkene or alkyne) What is the Theory Behind Splitting Patterns? Splitting for a doublet (N = 1): 1:1 ratio Splitting for a quartet (N = 3): 1:3:3:1 ratio The ways in which the magnetic fields of three protons can be aligned 60 NMR - Splitting of the Signals An 1H NMR signal is split into N + 1 peaks, where N is the number of equivalent protons bonded to adjacent carbons Singlet, N = 0 Doublet, N = 1 Triplet, N = 2 Quartet, N = 3 Quintet, N = 4 Sextet, N = 5, etc Quartet (N = 3) Triplet (N = 2) Doublet (N = 1) Quintet (N = 4) NMR - Coupling Constants (J) The coupling constant (J) is the distance between two adjacent peaks of a split NMR signal in hertz Coupled protons have the same coupling constant quartet doublet The magnitude of the coupling constants is determined by the angle between the two C-H bonds with the coupled protons – known as the Karplus relationship. Why are coupling constants important? 1) They give you information about which protons are coupled to each other 2) They can help determine some stereochemical elements 61 NMR - Coupling Constants with Alkenes Coupling constants for alkenes can be used to identify the cis and trans geometry of an alkene - trans coupling constant is always larger!! In general…. You can always tell alkene stereochemistry by using NMR You can almost always tell regioisomers apart by using NMR You can usually tell diastereomers apart by using NMR You can NEVER tell enantiomers apart by using NMR NMR Protons attached to benzene or an alkene are at relatively high frequency because of diamagnetic anisotropy The signals for the Hc, Hd, and He protons overlap Notice benzene ring does not have “ideal” splitting pattern - all signals are similar and overlap - multiplets ethyl splitting pattern 62 NMR - Samples The signals for the Ha, Hb, and Hc protons do not overlap Notice the difference in frequency for each proton signal - Look at resonance structures of nitrobenzene to assign each hydrogen. NMR - Protons Bonded to Oxygen and Nitrogen Hydrogen-bonding and proton exchange cause O-H and N-H signals to be broad without acid No proton exchange, splitting is observed (ethyl pattern with additional splitting) with acid proton exchange occurs, splitting is NOT observed (isolated ethyl pattern with NO additional splitting) What happens if you add a drop of D2O to your NMR sample? The D exchanges for H and the peak disappears (this is a great experiment to test for an N-H or O-H peak) 63 Carbon (13C) NMR Spectroscopy • The number of signals reflects the number of different kinds of carbons in a compound • Integration and signal splitting are typically not used • The overall intensity of a 13C (an isotope of 12C) signal is about 6400 times less than the intensity of an 1H signal • The chemical shift ranges over 220 ppm • The reference compound is TMS Same resonance and electron withdrawing effects determine chemical shift : O O Electron deficient carbon of a carbonyl group is shifted downfield to high frequency (165 – 220 ppm) H 110 - 170 ppm H 100 - 150 ppm H3 C Z 40 - 80 ppm Diamagnetic anisotropy accounts for the chemical shifts of aryl and vinyl carbons at higher frequencies (downfield) Electron-withdrawing effects of an electronegative atom will cause smaller downfield shifts (Z = OH, OR, NH2, NHR, Cl, Br, etc.) Carbon NMR Spectroscopy Proton-coupled 13C spectra – splitting is observed (uncommon) Proton-decoupled 13C spectra – NO splitting is observed (very common) Example of a proton-decoupled 13C spectra: Each chemically equivalent carbon atom is a singlet TMS 64 1H 1H and 13C NMR Shift Comparison NMR shifts: δ (ppm) 13C NMR shifts: δ (ppm) From Organic Chemistry Textbook by Wade Types of NMR Questions 1) 2) 3) 4) 5) How many signals will be in an 1H NMR spectra for a given molecule (how many protons are chemically equivalent)? How many signals will be in an 13C NMR spectra for a given molecule (how many carbons are chemically equivalent)? Relative chemical shift values, for example, identify which proton will give a signal with the highest chemical shift value (farthest downfield) for a given molecule. Assign a spectra - which proton accounts for which peak, and what is the splitting pattern Identify an unknown compound using NMR and IR data (see examples in the practice problems handed out in class) Do lots of practice problems!!! 65 NMR Sample Problems 1) Label the splitting pattern that will be observed in the 1H NMR spectrum for each of the indicated hydrogens. H2 C C H2 CH3 3) An unknown compound C4H8Br2, gave the following 1H NMR data. What is the compound? Singlet at 1.97 ppm (6H integration) Singlet at 3.89 ppm (2H integration) NMR Sample Problems How would you use NMR data to determine which is the major product in each of the following reactions? CH3 X H3C X X2, hv 1) -OR- X = Br or Cl H3C CH2 CH3 Cl KOH, EtOH 2) -OR- CH3 H3C CH3 Cl HCl -OR- 3) 66 Cl “One C=O bond to Rule them All” O Carbonyl structure and reactivity: nucleophilic O C electrophilic Substitutent (Z) on carbonyl determines reactivity: O O O R Cl acid chloride > R H > R aldehyde R R > R O anhydride ketone Most electrophilic carbon Most inductive electron-withdrawing O ester or acid R O O O O O > R > R N R > H amide R O carboxylate ion Most nucleophilic oxygen increasing reactivity with nucleophiles Most resonance Introduction to Carbonyl Compounds H3C CH3 H3C 1. O3 H3C 2. H2O2 H CH3 + O H3C H H2O, H2SO4 H3C C C CH3 1. disiamylborane Ketone or aldehyde synthesis using ozonolysis O Ketone synthesis from an internal alkyne O H3C CH3 2. H2O2, NaOH O H2O, H2SO4 HgSO4 H3C C C H 1. disiamylborane 2. H2O2, NaOH H3C Ketone or aldehyde synthesis from a terminal alkyne O H3C H O O H3C CH3 Cl + AlCl3 CH3 67 Synthesis of an aromatic ketone using Friedel-Crafts acylation with an acid chloride Introduction to Carbonyl Compounds O R O H R aldehyde R R O R OH carboxylic acid ketone O O O R N R R H amide ester O O O R R Cl acid or acyl chloride R O R C nitrile anhydride carbonyl oxygen O O R O R' O O Cyclic ester = lactone Cyclic amide = lactam NH lactone lactam carboxyl oxygen Systematic nomenclature is rarely used for carbonyl compounds, but common names are somewhat consistent for a series of carbonyl compounds: O O H3C H H3C acetaldehyde O OH H3C acetic acid Cl H3C acetyl chloride O O O CH3 O H3C acetic anhydride NH2 acetamide H3C C N acetonitrile Carbonyl Structure and Bonding C=O bond (1.23 Å) is shorter and stronger than a C=C bond (1.33 Å) C O δ+ δ- Large dipole moment O O O C C high energy structure, do not ever draw this!!! minor contributor major contributor Carbon is an electrophile (has Lewis acid character) Oxygen is a nucleophile (has Lewis base character) R Cl acid chloride O O O > R H aldehyde Most electrophilic carbon > R R ketone R O anhydride R > O ester or acid R increasing reactivity with nucleophiles 68 O O O O O > R > R N H amide R > N R O carboxylate ion Most nucleophilic oxygen Molecular Orbitals of C=O Double Bond Compare the molecular orbitals of the C=C and C=O double bond: alkene π* (LUMO) C=O π* (LUMO) The C=O bond is unsymmetrical!! alkene π (HOMO) C=O π (HOMO) energy π* C O C 2p O 2p π C O The 2p orbital of oxygen is lower energy (because oxygen is more electronegative) and contributes more to the π molecular orbital. The higher energy 2p orbital of carbon contributes more to the π∗ molecular orbital. Resonance vs. Inductive Effects O R Cl Most inductive electron withdrawing effects Resonance structure shows that carbonyl oxygen is more Lewis basic than the carboxyl oxygen (OCH3 or OH) Most resonance donation Most Lewis basic oxygen High double bond character in amide (as seen in resonance structures) causes trigonal planar geometry of nitrogen, indicating sp2-hybridization Recall relative IR stretch frequencies based on resonance and inductive effects… 69 General Comments About Acid-Catalyzed Reactions of Carbonyl Compounds O R H+ O X R Protonation of the carbonyl oxygen makes any carbonyl a better electrophile (more susceptible to nucleophilic attack) because the C=O LUMO is lowered H X aldehydes ketones esters amides Complexation of a carbonyl oxygen (a lewis base) with a Lewis acid will also make the carbonyl a better electrophile by lowering the LUMO BF3 O R X O R BF3 Lewis acids = BF3, AlCl3, ZnCl2, FeBr3 X The proton and Lewis acid both have an electron-withdrawing effect. Any electronwithdrawing effect will lower the LUMO and increase the electrophilicity of the C=O. Physical Properties: High Boiling Points intermolecular hydrogen-bonds and dipole-dipole interactions Hydrogen-bonding accounts for the super broad OH peak in the IR spectra of an acid Amides have the highest boiling points N-H of an amide is also broad in the IR spectra, but not as much 70 Reactivity of Carbonyl Compounds Recall order of reactivity of carbonyls with nucleophiles: R O O O Cl acid chloride > R H > aldehyde R R ketone R R O O O O O > > anhydride O ester or acid R R > O N R H amide R > R O carboxylate ion Increasing reactivity with a nucleophile addition Nucleophilic Acyl Substitution: Y = good leaving groups elimination carbonyl product Z = oxygen, nitrogen, hydride and carbon nucleophiles addition Nucleophilic Addition: Z = oxygen, nitrogen, hydride and carbon nucleophiles protonation a tetrahedral intermediate tetrahedral product Nucleophilic Acyl Substitution Reactions General mechanism: addition elimination Y:- must be a good leaving group RDS is elimination RDS is addition (a) the nucleophile (Z:–) is a weaker base (k–1 >> k2) (b) the nucleophile (Z:–) is a stronger base (k2 >> k–1) (c) the nucleophile (Z:–) and the leaving group have similar basicities 71 Nucleophilic Acyl Substitution For nucleophilic acyl substitution, a carbonyl compound must have a good leaving group (a weak base) that can be replaced by a nucleophile in nucleophilic acyl substitution reactions: O Cl < O R < OR ~ OH < weakest base best leaving group A weaker base is easier to eliminate NH2 strongest base poor leaving group See Bruice Table 17.1 on page 672 O R O H R R' (R’ = alkyl or aryl group) Ketones and aldehydes DO NOT undergo nucleophilic acyl substitution because there is not a good leaving group that can eliminate. (However, remember that they are great electrophiles and we will discuss other addition reactions with them soon.) MO diagram of Nuclelophilic Acyl Substitution acceptor Donor (on nucleophile) Examples of nucleophiles (Lewis bases) for acyl substitution reaction: H 2O CH3OH OH NH2CH3 H3C 72 H N O CH3 O CH3 Reactions of Acid Chlorides esters acids O H3C O O Cl + O H3C CH3 O O CH3 anhydrides amides Reactions of Acid Chlorides with Amines Not a good nucleophile • Requires 2 equivalents of amine because HCl is formed and amine will be protonated and no longer be a good nucleophile for the reaction • Only primary and secondary amines can be used to synthesize an amide O H3C + Cl H3C N CH3 No reaction to give amide products CH3 (2 equivalents) But a tertiary amine still has a lone pair to serve as a donor, why doesn’t it react ? 73 Tertiary Amines Still Promote the Reaction O + H3C Cl H3C + NH2 H3C (1 equiv) H O N N H CH3 N + Cl- (1 equiv) A tertiary amine will still add to a carbonyl, but since NO deprotonation can occur, it will always be the best leaving group O H3C Cl Cl- + H3C Cl H3C NH2 (1 equiv) O O N + H3C N H CH3 + O (1 equiv) H N H CH3 H Cl- N + + H3C N CH3 Cl- H3C O N H3C NH2 O+ N H OH N CH3 Cl- H3C N H2 CH3 Sample Problem: Mechanism Provide a mechanism for the following conversion of an acid chloride to an ester: O + H3C Problem solving strategy: O CH3OH Cl H3C When nucleophile is neutral: Step 1 = addition Step 2 = deprotonation Step 3 = elimination OCH3 When nucleophile is an anion: Step 1 = addition Step 3 = elimination Must draw three steps and show tetrahedral intermediate!! 74 Reactions of Anhydrides ester + acid acid (2 equivalents) amide Anhydrides are still reactive enough to synthesize esters, amides and acids; however, they are actually more stable than an acid chloride making them more practical to use. Reactions of Esters 1) Amidolysis 2) Hydrolysis Reaction is reversible and both acid and ester will be present at equilibrium 3)Trans-esterification Reaction is reversible and both esters will be present at equilibrium 75 Acid-catalyzed Hydrolysis of an Ester No negatively charged species Several of proton-transfer steps Need to drive equilibrium by adding excess water or removing the alcohol product Acid catalyzed Hydrolysis of an Ester Esters with tertiary alkyl groups undergo faster hydroylsis by SN1-related mechanism CH3 O H3C H3C O H R H H3C +O H3C O R H3C H2O Provides both an acid and an alcohol product CH3 H3C H3C OH CH3 SN1 OH + H3C + CH3 O tertiary carbocation CH3 H3C H3C B: O H H Once the carbonyl oxygen is protonated, the carboxylic acid is an excellent leaving group for an SN1 reaction 76 R Hydroxide Ion-Promoted Hydrolysis of an Ester: Saponification Hydroxide ion increases the rate of formation of the tetrahedral intermediate This reaction is NOT reversible Under basic conditions, the carboxylic acid will be deprotonated. The carboxylate ion is NOT electrophilic and the reverse addition will not occur Hydrolysis of an Amide Requires Harsh Conditions Amides are very stable and hydrolysis conditions require acid or base, with heat • Requires both heat and acid to hydrolyze an amide • Nitrogen must be protonated to become a good leaving group H3C O H NHCH2CH3 O + H3C HO HCl, H2O N O H HNCH2CH3 H+ H2NCH2CH3 OH O H2N O H+ OH H3N OH β-lactam A cyclic amide (a lactam) will be easier to hydrolyze if there is ring strain 77 Sidenote: Importance of the Amide Bond in Biological Systems amide bond R1 O N H H N N H R2 O R3 O H H N O R4 N H R5 O N H R4 H H N O N R3 O R5 N O H R2 O N amino acids linked by amide bonds: Linear Structure N H O Amide bonds are difficult to hydrolyze, making them perfect for proteins. H N O R1 O alpha helix with hydrogen-bonds Of course, there are enzymes that can hydrolyze the amide bond quite readily!! protein structure Reactions of Carboxylic Acids 1) Acid-base reactions (recall pKa’s) O H3C OH + H N O CH3 CH3 H3C + O H2N CH3 CH3 No amides will form because the amine will be deprotonated 2) Activated by converting to an acid chloride Converts carboxylic acid to a very electrophilic (activated) carbonyl to use in nucleophilic acyl substitition reactions Mechanism: 78 Reactions of Carboxylic Acids cont’d 3) Lactonization with Br2 or I2 OH O I I2 O + CH2Cl2 H I O I I OH O I+ O I I- O+ H I- The carbonyl oxygen is more nucleophilic than the carboxyl OH and will add to to the halonium intermediate (electrophile) Sidenote: Activated Carboxylic Acid Derivatives for Biosynthesis Similar to anhydrides, great leaving groups for synthesis: ATP is a great energy storage device: thermodynamically reactive and kinetically unreactive Good sterics, good leaving group, not as much energy OK sterics, good leaving group, gives more energy AMP Reactions are determined by sterics, leaving group ability and “energy requirements” of the biochemical pathway. Enzymes play a big role! 79 Reactions of Nitriles H3C Br NaCN H3C S N2 O HCl/H2O CN H3C ∆ butanenitrile (propyl cyanide) OH Butanoic acid (butyric acid) Acid-catalyzed hydrolysis to a carboxylic acid is even more difficult than an amide hydrolysis Reduction of a nitrile to give an amine: H2, Pd/C CN H3C H3C NH2 (2 equiv of H2 is required for reaction, but since H2 is added in the gas form, excess is always added) Sample Problems 1) Rank the reactivity of the following esters towards hydrolysis, where 1 = most reactive and 3 = least reactive. Hint: Since the carbonyl is the electrophile, the rate is increased by anything electronwithdrawing and decreased by anything electron-donating O CH3 Cl O O 2) Rank the relative energy level of the LUMO for the following C=C and C=O bonds, where 1 = lowest LUMO (most electrophilic) and 3 = highest LUMO CH3 CH3 O O R O OCH3 R H OCH3 3) Show three ways to prepare an amide from N,N-dimethylamine 80 OCH3 O O CH2 H3C CH3 Lidocaine Synthesis: Nucleophilic Acyl Substitution vs. SN2 Nucleophilic acyl subsitution O O R NH2 R Cl Cl (excess amine) H N R N H SN2 If only one equivalent of amine is used, nucleophilic acyl substitution will be faster than an SN2 because acyl halides are better electrophiles (more reactive) than alkyl halides CH3 Step 1: O + NH2 CH3 NaOAc Cl Cl Cl N H CH3 1 equivalent amine Step 2: CH3 O CH3 O Cl CH3 NaI, Et2NH CH3 O N N H CH3 N H CH3 Lidocaine CH3 Synthesis of Aspirin An experiment from Chem E2a lab: Acyl transfer with acetic anhydride OH H3C OH O O O O CH3 CH3 O O + H3PO4 H3C OH O O Salicyclic acid Aspirin™ (acetylsalicylic acid) • Active ingredient in wart removal medicine (40 %) • Active ingredient in some acne treatment medicine (1.5%) How to get rid of excess acetic anhydride in a reaction? hydrolysis gives acetic acid: O H3C H2O O O O H3C CH3 Acetic anhydride O H + O H Acetic acid 81 O CH3 O H Sidenote: Aspirin Mechanism of Action The reverse acyl transfer from Chem E2a lab: Cyclooxygenase O CH3 Ser O OH Cyclooxygenase OH OH OH enzyme-mediated O O salicylic acid Aspirin™ + Ser H3C O O Aspirin is a nonsteroidal anti-inflamatory drug (NSAID) that inhibits the biosynthesis of prostaglandins (which cause inflammation) by PGH2 synthase. Specifically, Aspirin inactivates the cyclooxygenase (Cox) activity of PGH2 synthase by an irreversible covalent modification mechanism: acylation of the Ser530 amino acid, a strategic location which blocks entry of the enzyme substrate into the binding site Ibuprofen is an NSAID that also inactivates cyclooxygenase activity of PGH2 synthase, but inactivation with Ibuprofen occurs by competitive substrate binding (a non-covalent, reversible mechanism) β-Lactam Antibiotics: Medicine and Molecular Orbitals Penicillin mechanism of action: Inhibits Transpeptidase to prevent cross-linking of small peptide chains (petidoglycan strands) in the cell wall of gram+ bacteria. But how is penicillin reactive if amides are so unreactive to hydrolysis? N O O N H O C N H H β-lactam Planar structure allows good overlap between lone pair on N (in p-orbital) and the C=O π bond to give good resonance stabilization 3-D picture = planar H N O CH 3 H S N O penicillin G CH 3 CH 3 O H S O C N CH 3 CO 2 H OH 3-D picture = bent 82 Ring constraints of fused β-lactam structure force lone pair to be almost perpendicular to the C=O π bond and resonance stabilization is lost Sample Problems 1) 2) 3) 4) General question types: Provide reagents or products in a reaction scheme (or short syntheses) Know mechanisms and relative reactivity trends for Nucleophilic Acyl Substitution Remember IR and NMR data relating to the C=O bond Be able to draw resonance structures and understand how they influence reactivity 1) The β-lactam of penicillan G is fairly labile in your stomach (acidic conditions). Provide a curved-arrow mechanism showing how it is rendered inactive in your stomach. H R S N O CH3 CH3 CO2H penicillin G β-Lactam Antibiotics: Medicine and Sterics Bacteria become resistant to penicillan G by producing an enzyme (penicillanase) that catalyzes the hydrolysis of the amide bond. However, simple structural modifications of penicillan G can make a drug that is effective against penicillan G-resistant bacteria. H N O CH3 H N H S N CH3 O CH3 O O penicillin G CH3 H S N CH3 O OH O More bulky aryl group CH3 OH methicillin Methicillan is effective in treating patients infected with penicillan G-resistant bacteria because the bulky aryl group prevents it from binding in the active site of the penicillinase. It does not, however, prevent it from inhibiting the transpeptidase target. 83 Sample Problem 2) Provide a synthesis of methyl butyrate from bromopropane: O Br H3C H3C starting material OCH3 desired product Reactivity of Aldehydes and Ketones Recall order of reactivity of carbonyls with nucleophiles: R O O O Cl acid chloride > R H aldehyde > R R > R R O > anhydride ketone O O O O O ester or acid R R > R O N H amide R > R Increasing reactivity with a nucleophile O O > H H formaldehyde O > R H an aldehyde R R a ketone An aldehyde is more electrophilic because an H atom is electron-withdrawing compared to an alkyl group: more electron-withdrawing effect = greater positive charge on carbon (less stable) 84 O carboxylate ion Two Factors: Electronic and Sterics Very reactive H O O > H H3C H > H3C O O O > CH3 H3C CH3 CH3 Reactivity based on stability and sterics: stability more important than sterics > H3C CH3 Less reactive CH3 CH3 Reactivity determined by sterics: more sterics = less reactive Electronic effect (stability): A hydrogen is more electron-withdrawing than an alkyl group. EWG = Greater positive charge = more electrophilic = Lower LUMO (Remember, Electron-withdrawing groups always lower the LUMO) Sterics has two effects: 1) The carbonyl carbon of an aldehyde is more accessible to the nucleophile 2) Ketones have greater steric crowding in their transition states, so they have less stable transition states Aldehydes and Ketones - Nomenclature Aldehydes - remove “e” from hydrocarbon name and add “al” Ketones - remove “e” from hydrocarbon name and add “one” 85 Nucleophilic Addition Reactions Recall Nucleophilic Acyl Substitution: addition elimination carbonyl product Nucleophilic Addition: addition protonation a tetrahedral intermediate tetrahedral product (Both general reactions can also be acid-catalyzed) Addition of Carbon Nucleophiles: Grignard Reagents Grignard reagents react with aldehydes, ketones, and carboxylic acid derivatives to form new C–C bonds Recall: Synthesis of a Grignard reagent from an alkyl bromide: Very polar C–Mg bond! Grignard reactions with aldehydes give secondary alcohols: Grignard reactions with ketones give tertiary alcohols: 86 Grignard Reactions with an Ester - Double Addition MgBr O Methyl benzoate OH 1) 2.5 equivalents OCH3 Triphenylmethanol 2) H2O PhMgBr (1st equiv) Nucleophilic acyl substitution H2O PhMgBr (2nd equiv) O OMgBr Nucleophilic addition Addition of the first equivalent of Grignard reagent will form a ketone Addition of the second equivalent of Grignard reagent will form an alcohol Grignard Addition to CO2: Synthesis of a Carboxylic Acid O Mg, Et2O Br MgBr 1. CO2 OH 2. HCl, H2O bromobenzene benzoic acid Useful synthetic methods to recall: 1. BH3 H3C H3C H3C 2. H2O2, NaOH OH Br Br OH H3C PBr3 NaCN Radical bromination CH3 O HCl, H2O CN H3C (or NBS) Alkyl bromide formation Br H3C Br2, hv Hydroboration/oxidation Two methods to synthesize a carboxylic acid H3C CH3 Br2 FeBr3 OH Br Electrophilic aromatic substitution: bromination 87 SN2 and Nitrile hydrolysis Synthesis Sample Problem 1) How would you synthesize the following compound from any carbonyl compound and any alkyl bromide? You may use any inorganic reagents. OH H3C CH3 2) Provide a synthesis of this carboxylic acid from benzene: H3C CO2H CH3 Other Carbon Nucleophiles: Acetylide anion Recall: Previous reactions of acetylide anion with electrophiles: H H3C 2) H H3C 1) NaNH2 (or nBuLi) H3C CH3 H3C Br 1) NaNH2 (or nBuLi) 2) OH H3C O 88 SN2 with alkyl halides Epoxide opening Hydride Ion as a Nucleophile General mechanism: reduction of ketone to secondary alcohol by addition of a hydride ion, followed by protonation Tetrahedral products CH3 What are sources of hydride ion “H:–”? NaBH4 = sodium borohydride - four “H:–” equivalents Dibal-H = diisobutylaluminum hydride - one “H:–” equivalent LiAlH4 = lithium aluminum hydride (LAH) - four “H:–” equivalents CH3 CH3 Al H3C H Dibal-H Relative ease of reduction: Most reactive O O O > R H O > R R R OR R O > > NR2 R Least reactive O NaBH4 LiAlH4 - most reactive NaBH4 Reductions H Na H B H H • Only works with aldehydes and ketones • Each molecule of NaBH4 provides 4 hydride equivalents (therefore, you only need one molecule of NaBH4 for every four molecules of carbonyl compound) Aldehydes give primary alcohols Ketones give secondary alcohols Because NaBH4 is less reactive, it can be very selective. You can selectively reduce a ketone and the ester will remain untouched. Only the ketone will be reduced!! 89 H LiAlH4 reductions Li H Al H H 1) LiAlH4 is a stronger reducing agent than NaBH4 2) LiAlH4 is used to reduce compounds that are nonreactive toward NaBH4 (basically, LiAlH4 will reduce any carbonyl compound) 3) Double hydride addition will occur to give the completely reduced product Amine substitution depends on what amide substitution started as An ester is reduced to a primary alcohol (double hydride addition) A carboxylic acid is reduced to a primary alcohol (double hydride addition) An amide is reduced to an amine (double hydride addition) LiAlH4 Double Addition Mechanism with an Ester elimination addition addition protonation The first hydride addition is a nucleophilic acyl substitution and a carbonyl product is obtained. The second hydride addition, a nucleophilic addition reaction, follows immediately because the aldehyde is even more reactive. 90 Dibal-H Reductions Dibal-H allows the addition of only one equivalent of hydride to an ester because it is less reactive than LiAlH4 LiAlH4 reduction gives alcohol (double hydride addition): Dibal-H reduction gives aldehyde (single hydride addition): CH3 Dibal-H H3C CH3 Why is Dibal-H less reactive than LiAlH4? 1) As a neutral species, it is less nucleophilic than the anionic aluminum species 2) It is more sterically hindered CH3 Al H Reduction Sample Problems Provide products for each of the following reactions: 1. NaBH4 O 2. HCl, H2O H 1. NaBH4 2. H2O 2. HCl, H2O 1. LiAlH4 O H 1. Dibal-H 1. LiAlH4 CH3 O 2. HCl, H2O O OCH3 O 1. LiAlH4 O H3CO 2. HCl, H2O N CH3 2. HCl, H2O CH3 O O 91 1. LiAlH4 2. HCl, H2O Alcohols Can be Oxidized to Carbonyl Compounds [O] Oxidation Reduction: Oxygen content decreases and/or hydrogen content increases R OH Reduction [H] Oxidation: Oxygen content increases and/or hydrogen content decreases O R H Oxidation with chromic acid reagents: Secondary alcohols can be oxidized to ketones Aldehydes and ketones can be oxidized to carboxylic acids Oxidation Mechanism with Chromic Acid What are the different Chromium reagents? All three are variations of the same reagent, (chromic acid): H2CrO4 = Chromic acid O CrO3/H2SO4 = H2CrO4 HO Cr OH Na2Cr2O7/H2SO4 = H2CrO4 Mechanism: Proceeds by substitution and an E2 elimination 92 O PCC Oxidation of Alcohol to an Aldehyde Chromic acid will oxidize a primary alcohol all the way to a carboxylic acid: H2CrO4 OH H3C H+ O H3C H2O H2O H Primary alcohol OH H3C H2CrO4 OH O H3C H aldehyde OH hydrate Carboxylic acid The oxidation of a primary alcohol can be stopped at the aldehyde if pyridinium chlorochromate (PCC) is used as the oxidizing agent CrO3 + HCl H N N + CrO3ClŠ PCC The oxidation with PCC stops at the aldehyde because PCC can be used in a non-aqueous solvent to avoid formation of the hydrate. The hydrate is what oxidizes further to the carboxylic acid Swern Oxidation Swern is also a useful method to oxidize a primary alcohol to an aldehyde because over-oxidation can be avoided….. And less toxic than chromium reagents Mechanism: Me O O Cl Cl Me O oxalyl chloride S R OH Me Cl Cl Me S Me SN2 Cl Cl O O Cl O O DMSO Cl O S Me Cl O S Me Me Me Me O S R Me O Me S R H NEt3 triethylamine Me dimethylchlorosulfonium ion R H S O Me S H H O R CH2 E2 + Proceeds by substitution and an E2 elimination Note: also converts secondary alcohols to Ketones 93 H aldehyde Me S Me dimethylsulfide Dimethylsulfide is: 1) A great leaving group 2) Super Stinky!!! Sidenote: Analysis of Blood Alcohol Content H3C OH H2SO4 + Na2Cr2O7 H3C O + H2Cr2O3 OH The ethanol from your lungs (in equilibrium with your blood) will be oxidized by chromic acid. The chromate ion produced can either be detected by the a simple color change that indicates if any alcohol is present, or a quantitative Breathalyzer™ can be used for a more accurate reading of the alcohol content. However, a direct blood test is the most accurate measurement. H2SO4 Na2Cr2O7 reduced chromate ion ROH Red-orange color green color Synthesis Sample Problems 1) Provide a synthesis of Ibuprofen from the starting matieral provided: H3C H3C CH3 CH3 O OH CH3 Starting material Ibuprofen (AdvilTM) 2) Provide a synthesis of the ketone below, starting with any alcohol that is 4 carbons or less, and using any inorganic reagents. CH3 O H3C CH3 CH3 Desired product 94 Reactions of Aldehydes and Ketones with Nitrogen Nucleophiles: Imines from 1° amines H H R O H2N aldehyde R H2O + R N R "Imines" aldimine 1° amine -ORR R R O H2N ketone R H2O + R N "Schiff Bases" R ketimine 1° amine Bonding in an imine: CH3 CH3 C N CH3 sp2-hybridized carbon sp2-hybridized nitrogen Reactions of Aldehydes and Ketones with Nitrogen Nucleophiles: Enamines from 2° Amines R R HN O R R H2 O R aldehyde or ketone + R N R 2° amine enamine recall... R R R R R O "keto" tautomer OH "enol" tautomer 95 Mechanism of Imine Formation addition protonation deprotonation protonation deprotonation elimination Mechanism of Enamine Formation 96 Carbonyl Group Removal via Imine (hydrazone) Formation: The Wolf-Kischner Reduction Me Works with any ketone H2N O NH2 Me ketone Me NaOH hydrazine N NH2 H ∆ hydrazone H alkane Mechanism to form hydrazone is EXACTLY as with imines Reduction mechanism: This ketone reduction is more general than the H2 with Pd/C reduction conditions that ONLY work with benzylic ketones Reactions of Aldehydes and Ketones with Oxygen Nucleophiles: Gem-Diols (Hydrates) from Water O R H2O R ketone or aldehyde Water HO OH R R gem-diol or "hydrate" Recall... R R OH OsO4/H2O2 R R OH vic-diol gem, geminal => gemini, "twins"...attached to same carbon vic, vicinal => vicinity, "neighbors"...attached to neighboring carbons 97 Carbonyl Reactivity Predicts the Amount of Hydrate Present most stable, least reactive least stable, most reactive most stable Acetals and Ketals O R H aldehyde R H+ OH H2O alcohol R ketone R RO OR R H acetal O R + H+ OH H2O alcohol + RO OR R R ketal Examples of “H+”acids = HCl, H2SO4, p-TsOH SO3H p-TsOH = H3C 98 Mechanism is Similar to Hydrate Formation, conditions are Similar to Imine Formation (dehydrating) This reaction does not happen unless you get rid of the water! The Dean-Stark Apparatus H O 5g condenser cold water HO MeO Dean-Stark Trap O pTsOH water level O OH benzene (reflux) O reaction flask 6.4g MeO 99 + H2O (0.54 mL) O Applications of Ketals in Synthesis OH O Me LiAlH4 OH O Me O ?? OMe BOTH carbonyl groups reduced Me LESS REACTIVE carbonyl group reduced? NaBH4 OH O Me OMe MORE REACTIVE carbonyl group reduced Ketals are Useful as “Protecting Groups” O Me HO O O OMe OH O Me ?? pTsOH D/S trap benzene Me OH H2O, HCl LiAlH4 O O O Me OMe more reactive carbonyl group protected more reactive carbonyl group de-protected O OH LESS reactive carbonyl group reduced 100 OH Applications of Protecting Groups with Alcohols Re-call from E2a: TMS ethers are useful protecting groups for alcohols OMe Me Me OH OMe MeI/NaH (excess) OMe Me Me BOTH Alcohols converted to methyl ethers OH MeI/NaH ?? Me Me OH LESS REACTIVE alcohol converted to methyl ether OH Me Me OMe MORE REACTIVE alcohol converted to methyl ether Protecting Groups Are Important for Organic Synthesis Protecting groups are often necessary, but remember, each protecting group adds 2 steps to your synthesis, so they should be minimized or avoided. 101 Sidenote: Acetals in Aesthetic Chemistry: The Nanoputians… Addition of Sulfur Nucleophiles to Aldehydes and Ketones: Formation of Thioketals + H2O Thioketals Offer a Milder Alternative to the Wolf-Kischner Reduction 102 The Wittig Reaction: Conversion of Aldehydes and Ketones to Alkenes O X Y +X Yylide resonance contributor Ph Ph P Ph Me S OMe Me S+ Me DMSO H CH3 Ylide Formation and the Mechanism of the Wittig Reaction 103 Wittig Reaction and Alkene Stereochemistry stabilized ylide O + Ph3P - UNstabilized ylide + Ph3P OMe O - H OMe OMe R O R Me - O O + Ph3P + Ph3P OMe E-alkene predominates H Me H H + Ph3P Z-alkene predominates Me - R H Sample Problem Propose a synthesis of the following product from the indicated starting material Ph OH Ph 104 OH Nucleophilic Addition to α,β-Unsaturated Aldehydes and Ketones Mechanism of Conjugate Addition 105 Nucleophiles That are Weak Bases Will Generally Form Conjugate Addition Products O Nu OH Carbonyl C=O bond broken Nu H O O R R SH H Br Br S O Carbonyl C=O left intact R NH R O R H CN O N N R C Grignard Reagents Prefer 1,2-addition (direct), Cuprates Prefer 1,4-addition (conjugate) O R O HO 1) R2CuLi 1) RMgBr 2) H3O+ 2) H3O+ 106 R Summary • Amine nucleophiles add to aldehydes and ketones to form imines and enamines • Alcohols and thiols (diols and dithiols) add to aldehydes and ketones to form acetals and thioacetals • Acetals are useful as protecting groups so that chemistry can take place on less reactive carbonyl groups • Thioacetals are useful for two-step reduction of carbonyls to alkanes as a mild alternative to the Wolf-Kischner reduction • Unsaturated carbonyl compounds can undergo direct (1,2) or conjugate (1,4) addition • 1,4 addition is preferred for weak nucleophiles, 1,2 addition is preferred by very strong (eg Grignard) nucleophiles. Cuprates allow 1,4 addition of carbon nucleophiles to predominate Carbonyl Compounds Part III: Reactions at the αCarbon So far, carbonyl groups have only served as electrophiles at the carbon bound to oxygen, or the terminal carbon of an unsaturated carbonyl. We will now see that the α-carbon of a carbonyl compound can act as a nucleophile. alcohol: pKa = 16 carboxylic acid: pKa = 5 O O H3C O H H3C O- + H+ H3C O H H3C O H3C H H + H+ ketone: pKa = 20!! alkane: pKa = 50 H O- H3C - H + H+ H O H H3C H 107 H H3C H H + H+ Electron-Withdrawing Groups Stabilize α-Anions For a more extensive table see: http://daecr1.harvard.edu/pdf/evans_pKa_table.pdf The Effect of Stabilization is Additive: More Groups, More Stability 108 Enols and Enolates are Nucleophiles Enols are like very reactive alkenes... OH + H3C O Br Br H Br Br H3C Enolates are carbon nucleophiles O + H3C O R X O H3C H3C O + H R O R H3C OH R Keto-Enol Tautomerization 109 + X Enol-Keto Interconversion is Catalyzed by Base or Acid Halogenation of Ketones: Acid Catalysis Only 1 α-hydrogen is replaced by a halogen 110 Halogenation of Ketones: Base Promotion Stoichiometric Formation of an Enolate With Lithium Diisopropylamide (LDA) Preparation of LDA from diisopropylamine and n-butyllithium Me Me Me N H Me Me + H3C Li DIA Me Me N Li LDA 111 Me + H3C CH3 Alkylation of Ketones, Esters, and Nitriles Note Similarity to… R n-BuLi R H Li H3C Br R CH3 Sample Problem: Ketone Alkylation Propose a synthesis of the following branched ketone from a linear ketone starting material: O H3C CH3 CH3 112 Enolates of Unsymetrical Ketones: Regiochemistry Problem with α-Branched Ketones Thermodynamic and Kinetic Enolates 'kinetic' enolate CH3 CH3 H3C N O CH3 H H 'thermodynamic' enolate OLi LDA CH3 OLi CH3 H Li LDA 1.1 equiv LDA: major (Low Temperature) minor 0.95 equiv LDA: minor (Higher Temperature) Mechanism of Equilibration: CH3 + major CH3I O H H H OLi CH3 + O CH3 H3C CH3I O CH3 CH3 CH3 OLi O CH3 + CH3 The kinetic enolate is formed more quickly because of reduced steric hindrance, and when there is no excess ketone around, it cannot equilibrate. When there is excess ketone, the kinetic enolate can deprotonate the ketone and form more of the thermodynamic enolate. 113 Hydrazones Give Alkylation Products That Would Result from the Kinetic Enolate +Li Bu- Ketone Alkylation Can Be Problematic: Polyalkylation, Regiochemistry, and O-Alkylation are Possible Enamines are monoalkylated, and are a good alternative when over-alkylation is a problem, especially in the case of methylation. 114 Carbonyl Alkylation: Stereochemistry and the Evans Asymmetric Alkylation New Stereocenter Formed (product is racemic) products cannot easily be separated! O O MeO O N O LDA CH3I O Li H CH3 MeO H Ph O Ph O O O CH3 N O H H Ph Ph i-Pr i-Pr major minor (diastereomeric proiducts can be easily separated) Br O CH3 H CH3 i-Pr H3C single enantiomer: chiral auxilliary derived from valine + Ph O N O CH3 MeO PhCH2Br O CH3 O LDA CH3 CH3 N Sm(OTf)3/MeOH O O Me Cl N O Et3N O H CH3 + MeO H i-Pr Ph single enantiomer Synthesis of Evans Auxilliaries O CH3 O H3C CH3 OH LAH H3C NH2 OH NH2 • • OR: Cl O N O O Cl Cl H i-Pr Cl O O Cl "triphosgene" (0.33 equiv) O Me Ph OH NH2 norephedrine • Cl Cl occurs in nature as one enantiomer. Synthesis of the other enantiomer of chiral ester would require the enantiomeric chiral auxilliary... Cl phosgene (1 equiv) + Et3N O Ph N Me Chiral auxilliaries are useful for making enantiomerically pure products by a variety of reactions, not just alkylation. Since the two products are diastereomers, purification can be used to increase the purity of the product, and thus the final enantiomeric purity once the auxilliary is cleaved. The auxilliary can be recovered by purification and used in subsequent reactions. 115 H The Aldol Addition • • It was first observed as a dimerization of aldehydes to produce a molecule that was named “aldol” because it was a βhydroxy aldehyde These two examples represent homoaldol reactions, or reactions where both the electrophile and nucleophile are derived from the same molecule. The Cross Aldol Addition: The Nucleophile and Electrophile Emanate from Two Different Carbonyl Compounds A-A A-B B-A B-B 116 The Cross Aldol Reaction • • • • While the dimerization of aldehydes and ketones through the aldol reaction is not terribly useful, the cross aldol reaction is very useful. The cross aldol reaction involves an electrophile and nucleophile derived from different carbonyl compounds. In the cross aldol reaction, the enolate nucleophile usually comes from an ester or a ketone, and the electrophile is usually an aldehyde. Because the carbonyl that supplies the nucleophile and the electrophile are both acidic, the nucleophile must be generated in the absence of the electrophile. Intramolecular Aldol Addition: 5- and 6-Membered Rings are Favored. 117 Stereochemistry of the Aldol Reaction: The Acetate and Propionate Aldol Addition Reactions O CH3O OLi LDA CH3 O RCHO CH3O OH CH3O R one new stereocenter created ("acetate" aldol product) OLi O CH3O O CH3O MeO CH3 E(O)-enolate LDA R CH3 anti-aldol product RCHO OR CH3 OH OLi O OH CH3 CH3O MeO Z(O)-enolate R CH3 syn-aldol product two new stereocenters created ("propionate" aldol product) Polyketide Natural Products are Synthesized by Successive Aldol Additions O H3C H3C CH3 OH H3C OH Me O O O CH3 OH OH O OH OH CH3 HO OH H3C H3C H3C OH CH3 CH3 CH3 OH CH3 7 "propionate" units make up the skeleton of erythronolide Erythronolide B O O RO H Me Me O O OH CH3 RO CH3 CH3 H CH3 118 O OP OH OP CH3 RO CH3 CH3 Update: The Aldol Reaction is Nothing More than a Special Case of Carbonyl Addition… "Carbonyl Addition" O R Nu Nu = H R Li Nu R The Aldol Reaction OLi Nu = O R OH R H R O R Sample Problem 119 H- RMgBr OH The Aldol Condensation: Elimination of Aldol Products to α,β-Unsaturated Carbonyl Compounds Condensation reactions involve the loss of water or an alcohol so that the molecular weight of the product is less than the sum of the two starting materials. Aldol Addition vs Condensation O R O LDA OH R RCHO Me aldol addition ∆ • • • HEAT HCl/H2O O R NaOH/H2O ∆ Me aldol condensation NaOH/RHCO ∆ Use of LDA to form an enolate, followed by addition of an aldehyde will produce the aldol addition product If the aldol addition product is then treated with either aqueous acid OR base, the condensation product will be formed If the aldol components are combined with base, it is likely that the condensation product will predominate 120 The Aldol Condensation Produces “Wittig-like” Products…when is it useful? O O O O Ph Ph Ph Ph O Ph Ph Ph Ph3P PPh3 Ph O O Ph Ph Ph Wittig Ph Aldol The aldol condensation is one step, not two (like the Wittig), so in cases where control of alkene geometry is not important, aldol condensation is OK In many cases, such as cyclizations, only one product is possible from aldol condensation, so it is the way to go… O Ph Ph KOH O Ph O K Ph + O Ph Ph Ph H Ph O O O Ph Ph O Ph Ph K Ph O Ph Ph O Ph O H Ph Ph H EtOH H Ph O O K Ph EtOK O O Ph The Michael Reaction: Conjugate Addition of an Enolate When an enolate nucleophile reacts with a conjugated carbonyl compound, it is called the Michael Reaction. This reaction works well with very stable enolates, ie those derived from βcarbonyl compounds 121 Ph The Michael Reaction: Conjugate Addition of an Enolate If one reactant in the Michael reaction (or other reaction involving an alkoxide base) is an ester, it is important to match your alkoxide base with the ester component. If the enolate component in a Michael Reaction is an unstabilized ketone, it works better if you use the corresponding enamine, which is called the Stork Modification after Gilbert Stork. The Claisen Condensation: Reaction of Enolates with Esters When an enolate nucleophile reacts with an ester to give the product of nucleophilic acyl substitution, it’s called the Claisen Condensation. 2 O H3C O NaOCH3 OCH3 O H3C OCH3 CH3 122 + CH3OH The Driving Force for the Claisen Condensation is the Formation of a Stable Anion A full equivalent of base is employed, and the product is the dicarbonyl anion and alcohol from which the alkoxide is derived. Mixed (Cross) Claisen Condensation and the Dieckman Condensation O H3C O CH3 O H H H3C O O NaOEt O H3C O Me α-proton removed NO α-protons! to form enolate H H O O O H3C CH3 O H O NaOEt H α-protons of ketone are more acidic... the enolate of the ketone will be formed preferentially The intramolecular Claisen condensation is called the Dieckman condensation 123 CH3 Update: The Claisen Condensation is Nothing More than a Special Case of Nucleophilic Acyl Substitution… "Nucleophile Acyl Substitution" O R O Nu Nu = X R Nu ONa Nu = R O R X R O R Sample Problem 124 NH R The Claisen condensation O R R OH H2O Michael + Aldol = Robinson Annulation Decarboxylation of Dicarbonyl Compounds β-carbonyl acids will lose CO2, or “decarboxylate,” when heated… O O R O OH H OH O R O Recall... R C R O CO2 O O R O O HCl/H2O OMe R OH Hydrolysis of a β-keto ester results in decarboxylation… O R CO2 O O HCl/H2O OMe R 125 O O OH R CH3 CH3 Malonic Ester Synthesis: Alkylation of a Dicarbonyl followed by Decarboxylation " X " O + H3C R OH O X + H3C OH R O Base H3C O R Malonic Ester Synthesis: Can be Used Iteratively to Incorporate Two Electrophiles 126 Acetoacetic Ester Synthesis is Analogous to the Malonic Ester Synthesis O 2 H3C O NaOCH3 OCH3 O H3C OCH3 CH3 Sample Problem 127 Sample Problem Biosynthesis of Polyketide Natural Products by Claisen Condensations and Decarboxylations O O O O O AT H3C SCoA -O S Acyl Transferase S H3C ACP ACP Ketoreductase acyl carrier protein O NADPH OH S Dehydrogenase ACP Enoylreductase O H3C H3C S ACP ER O NAD+ KR DH H3C NADPH S ACP 128 NAD+ Carbonyl Chemistry on One Slide O R O O OMe R [H-] O R R H [H-] [O] R R N R1 N R R [O] OH OH OH R1 R R R R R OM O O Nu Nu R X R O R1 O Claisen R1 R OM OH R R O Nu Nu R OH R1 R R R OM Nu R O O Nu R R R1 O Aldol R1 O 1 R Michael O R R Sample Problem 129 R R R1 R1O OR1 R R1 R Sample Problem The Diels-Alder Reaction • • • • Conjugated dienes react with alkenes in a process known as the Diels-Alder Reaction. The product is a substituted cyclohexene The mechanism is concerted and pericyclic (concerted = bond formation and bond breakage occurs simultaneously in the same step) The hydroboration reaction is the other concerted, pericyclic reaction that we have seen previously. diene 2 3 1 CH2 CH2 4 dienophile CH2 6 1 2 ∆ CH2 5 6 3 5 4 3 π-bonds 4 σ-bonds 1 π-bond 6 σ-bonds 2 π-bonds were "traded" for 2 σ-bonds ∆= heat: in this case temperatures of 30-250 °C are generally necessary 130 Molecular Orbitals of the DA Transition State The Diels-Alder reaction is a [4π + 2π] cycloaddition reaction. Diene= 4π component Dienophile= 2π component The Diels Alder Reaction is Accelerated by Lowering the LUMO of the Dienophile LUMO Energy LUMO is lowered by the electron-withdrawing group (CO2Me) lower ∆G(HOMO/LUMO) = lower ∆G‡(reaction) HOMO ethene H 1,3butadiene methyl acrylate O H OMe H Dienophiles with EWGs (electron withdrawing groups) are good dienophiles. EWGs = cyano, ester, aldehyde, ketone, etc 131 The Diels-Alder Reaction is a Stereospecific Syn Addition: Dienophile Stereochemistry O H H H ∆ OMe Me H O Me Me H CO2Me Me cis Diels-Alder products (enantiomers) cis-dienophile (Z-alkene) H H CO2Me H ∆ OMe H H CO2Me H Me CO2Me H Me trans Diels-Alder products (enantiomers) trans-dienophile (E-alkene) Diene Stereochemistry/Syn Addition Transition States: Top Face or Bottom face Addition gives Enantiomers H H MeO2C Me H H H H H MeO2C H H H Me H CO2Me H H Me H H CO2Me H H Me H H H MeO2C Me MeO2C H H H H H H H H Me H H 132 CO2Me Me The Diels-Alder Reaction is a Syn Addition: Diene Stereochemistry Me H Me CO2Me Me CO2Me H CO2Me H H Me CO2Me CO2Me CO2Me Me E-E-diene Me cis Diels-Alder products (enantiomers) Me Me H CO2Me Me CO2Me CO2Me H CO2Me Me H CO2Me CO2Me Me H Me trans Diels-Alder products (enantiomers) E-Z-diene Note: Alkynes can serve as dienophiles, but a double bond is produced, so no new stereocenters are formed from the dienophile The Diels Alder Reaction is Accelerated by Raising the HOMO of the Diene Energy LUMO HOMO is RAISED by adding an Electron Donating Group (OMe) HOMO lower ∆G(HOMO/LUMO) = lower ∆G‡(reaction) ethene 1,3butadiene 1-methoxy-1,3butadiene H H MeO H H 133 H Regiochemistry of the Diels-Alder Reaction is Predicted by the Substituents of the Diene and Dienophile 1. Draw out resonance contributors + OMe H OMe H H H H H OMe H H H + H H H - H O H OMe H - H O H OMe + O OMe O O OMe H OMe - CO2Me H H + OMe H H OMe H + H OMe H H - H + H H 2. Match up charges (opposites attract) H - OMe H + H OMe X + - X - OMe CO2Me O Diene Conformation: “s-cis” Required for the Diels-Alder Reaction H H H H H H H free rotation H s-trans conformation H Diels Alder Reaction H H H s-cis conformation NO Diels-Alder REACTION diene is locked in s-trans conformation (no free rotation around σ-bond) FAST DIELS-ALDER REACTION diene is locked in s-cis conformation More s-cis conformer of diene results in a faster Diels-Alder reaction 134 The “Endo” Rule: Endo and Exo Transition States Produce the Endo and Exo Diastereomers of Product H H H H H H CO2Me "exo" (substituent of dienophile is oriented away from double bond) CO2Me Me H H CO2Me CO2Me H Me H O H H H H H H OMe H H H MeO2C H MeO2C Me CO2Me H H H H Me H H H Me "endo" (substituent of dienophile is oriented toward double bond) endo transition state, and thus the endo product, is favored Sample Problem For the following three reaction, draw the DA product that results, including RELATIVE stereochemistry. Circle the reaction that proceeds fastest. CO2Me + CO2Me + O CO2Me + 135 H H H CO2Me Introduction to Pericyclic Reactions 1) Polar/Ionic Reactions: Heterolytic bond cleavage, formation of anionic or cationic intermediates O Aldol reaction O H3C H O CH3 OH H3C CH3 2) Radical Reactions: Homolytic bond cleavage, formation of radical intermediates Cl H 2Cl Cl H H HCl + H Cl Cl H H H 3) Pericyclic Reactions: Cyclic transition structures where all bond-breaking and bond forming occurs at the same time (in concert), reorganization of electrons in the reactants, without formation of an intermediate O H O O H O H O H3C H BH2 H3C O BH2 Hydroboration Diels- Alder reaction transition state Pericyclic Reactions A concerted, single barrier, one-transition state reaction energy starting material(s) A B (intramolecular) A + C B (intermolecular) product progress of reaction • Cyclic transition structures (therefore, SN2 is not a pericyclic reaction) • Can be thermal or photochemical reactions - A photochemical reaction takes place when a reactant absorbs light - A thermal reaction takes place without the absorption of light - In a thermal reaction the reactant is in its ground state; in a photochemical reaction, the reactant is in its excited state • Orbitals MUST have the same symmetry (be in-phase) to overlap and form products (“conservation of orbital symmetry” theory) • Can have suprafacial or antarafacial orbital interactions (bond formation or rearrangements); always suprafacial for 6 atoms or less in transition state 136 3 Types of Pericyclic Reactions 1) Cycloaddition Reaction Usually intermolecular diene dienophile (Note: look for conjugated alkenes) 2) Electrocyclic Reaction Always intramolecular (Note: look for conjugated alkenes) 3) Sigmatropic Rearrangement Always intramolecular Molecular Orbitals: Thermal vs. Photochemical (LUMO) (HOMO) Thermal HOMO and photochemical HOMO are different! (HOMO) (ground state) thermal energy levels (excited) photochemical • The normal electronic state of a molecule is known as its ground state • The ground state electron can be promoted from its HOMO to its LUMO by absorption of light (excited state) • In a thermal reaction the reactant is in its ground state; in a photochemical reaction, the reactant is in its excited state 137 Molecular Orbitals of 1,3-Butadiene: Thermal vs. Photochemical (LUMO) (LUMO) (HOMO) (HOMO) The ground state HOMO and the excited HOMO have opposite symmetry Pericyclic Type 1: Cycloaddition Reactions Two π-bonds are converted into two σ-bonds and a new ring is formed (usually intermolecular) thermal reaction ∆ forms new cyclohexene ring photochemical reaction (doesn’t work with thermal conditions) forms new cyclobutane ring [# + #] refers to the number of π-electrons in the cycloaddition reaction: [4 + 2] = [4π + 2π] = 4π electrons + 2π electrons [2 + 2] = [2π + 2π] = 2π electrons + 2π electrons 138 Review of Diels Alder Reaction (Bruice chapter 8: pages 313-321) •Conjugated dienes react with alkenes (dienophiles) in a process known as the Diels Alder Reaction. The product is a substituted cyclohexene • The mechanism is concerted and pericyclic - proceeds through a cyclic transition state • The reaction is stereospecific: stereochemistry of the starting materials is conserved in the products ∆ CH3 + dieneophile CH3 CH3 diene cyclohexene product transition state O H3CO CH3 trans-dieneophile O ∆ + H3CO H3C diene trans-cyclohexene product A dienophile with an electron-withdrawing group (carbonyl or CN), will make the Diels-Alder reaction faster because the LUMO of the dienophile is lowered to have better overlap with HOMO of the diene Molecular Orbitals for [4 + 2] Cycloaddition (Diels-Alder) Consider the HOMO and LUMO (frontier molecular orbitals) of both reactants: (This orbital interaction is most common) Suprafacial bond formation = formation of both bonds on the same face (side) of the π-system You can consider EITHER combination of HOMO and LUMO interactions, as long as the overlapping orbitals are in-phase 139 Antarafacial vs. Suprafacial • Only suprafacial orbital interactions will occur if the transition state has six or fewer atoms in the ring Migrating group = C, H or other atoms [2 +2] Cycloaddition Reactions H3C CH3 H3C CH3 hv + H3C H3C CH3 CH3 Two π-bonds are converted to two σ-bonds Must be a photochemical reaction for in-phase orbital overlap [2+2] cycloaddition/dimerization reactions of your DNA: DNA photolyase • Ultraviolet light (hv) can cause skin cancer by promoting thymine dimerization, a structural modification of DNA that can lead to mutations • DNA photolyase is an enzyme that repairs damaged DNA by reversing the [2+2] cycloaddition reaction to regenerate the original thymine residues Note: you will not be responsible for stereo- and regiochemistry for [2+2] cycloadditions 140 MO Analysis of the [2 + 2] Cycloaddition Reaction Orbitals are NOT in-phase = NO reaction X Orbitals ARE in-phase = cycloaddition Pericyclic Type 2: Electrocyclic Reactions • Electrocyclic reactions are reversible: ring closing or opening • Look for conjugation in reagents and/or products Ring closing: An intramolecular reaction in which a new σ bond is formed between the ends of a conjugated π system to give a cyclic ring product Ring opening: An intramolecular reaction in which a σ bond of a ring breaks and a conjugated π system is formed in an acyclic product 141 Electrocyclic Reactions: Conrotatory vs Disrotatory Ring Closure The symmetry of the HOMO of the compounds undergoing ring closure determines the stereochemical outcome of the electrocyclization. HOMO (ground state) H H H3C Me Me CH3 H ∆ HOMO (excited state) H H H3C Me Me H CH3H The ground state HOMO is symmetric, so disrotatory closure occurs The excited state HOMO is asymmetric, so conrotatory closure occurs Woodward-Hoffman Rules Relate the Number of π-Bonds to the Mode of Cyclization For systems with an EVEN # of π-bonds, the ground state undergoes conrotatory closure Me Me ∆ Fortunately, a set of selection rules prevents the need to draw out the HOMO for each specific case: These are known as the Woodward-Hoffman Rules 142 Pericyclic Type 3: [m,n] Sigmatropic Rearrangements a [1,5] sigmatropic rearrangement (a 1,5-hydrogen shift) 4 4 5 3 3 5 2 1 2 1 2 1 2 3 1 (note: do not worry about the 1,3 shifts in the book or previous lecture note copy, the book has an error and the 1,3-shifts are less useful so we will not discuss them.) 3 2 1 3 1 3 2 Suprafacial rearrangement = migrating group remains on the same face (side) of the π-system • Sigmatropic rearrangements have cyclic transition states • Rearrangement must be suprafacial if the transition state has six or fewer atoms in the ring [3,3] sigmatropic reactions Count Carbons: 1) Count Carbons (1,2,3-3,2,1) in the starting material to see where alkenes will shift, and what the product will look like 2) Count carbons in the product (1,2,3,4,5) to help identify IF a [3,3] sigmatropic rearrangment can be used to synthesize the product What is the driving force for these rearrangements? More substituted double bonds in Cope rearrangement (usually) Stronger bonds formed (C=O vs. C=C) in Claisen rearrangement 143 Ireland-Claisen Rearrangement General Claisen reaction forms an alkene and an aldehyde: Ireland-Claisen rearrangement forms an alkene and a carboxylic acid: O O H3C HO O O acetylation of alcohol with acetic anhydride forms ester 1. LDA CH3 CH3 OH 2. H3O+ workup O O Ireland-Claisen Rearrangement Mechanism: O O O CH3 LDA O CH2 O Claisen Rearrangement H3O+ workup O OH O enolate formation (H starts at C-1 and ends up on C-n) [1, n] Hydrogen Shifts 4 1,5-hydrogen shift 4 3 3 2 2 1 1,7-hydrogen shift 5 5 3 One C–H σ bond broken, one C–H σ bond formed 1 2 3 1 4 4 5 5 2 1 6 6 7 144 7 Vitamin D Synthesis Sunlight (hv) converts a cholesterol steroid to vitamin D using two pericyclic reactions: (Precursor to vitamin D) Summary: How to Identify Different Pericyclic Reactions Are two molecules coming together to form a new ring? That’s a cycloaddition Is a ring opening or closing in an intramolecular fashion? That’s an electrocyclic process Is something shifting from one point in the molecule to another? Sometimes products do not look like starting materials? That’s a sigmatropic rearrangement (Hardest to identify - count carbons) - Be able to identify different pericylic reaction types, (especially specific [1,5] and [3,3] sigmatropic rearrangments) - Know thermal vs. photochemical HOMO - Be able to provide products from given starting materials - you do NOT need to know how to draw the cyclic transition state, but you should be familiar with the mechanistic details of the process Vocabulary: Concerted, Pericyclic, Electrocyclic, Suprafacial , Antarafacial Conservation of orbital symmetry 145 Sample Problems 1. a) What kind of pericyclic reaction is this? b) What is the driving force of this reaction? 120°C 2. a) What is the product of this reaction? b) Describe, as specifically as possible, what type pf pericyclic reaction this is. 170°C O H3C CH3 3. Provide products and indicate what type of pericyclic reaction occurs. O 170°C CH2 CH2 + hv O O Introduction to Carbohydrates: Polyfunctional Compounds general molecular formula = Cn(H2O)n Carbohydrates (saccharides) are named based on their molecular formula, which make them appear to be “hydrates of carbon” Monosaccharide: the simplest carbohydrate unit or “single sugar”, classified by its number of carbons (trioses, tetroses, pentoses and hexoses) α- and β-notations: relative stereochemical designation at the anomeric center D- or L-notations: the stereochemical designations for the configuration of a sugar, based on glyceraldehyde (instead of R- and S-designations) Reducing Sugars: have an aldehyde or hemiacetal group that can be oxidized 146 Classification by Carbonyl Type and # of Carbons Aldoses: polyhydroxy aldehydes (an aldehyde sugar) Ketoses: polyhydroxy ketones (a ketone sugar) # of Carbons: triose, tetrose, pentose, and hexose 1 CHO 2 CHOH 3 CH2OH an aldotriose O Nomenclature and Structure: Glyceraldehyde H OH OH Glyceraldehyde (an aldotriose) has a single stereogenic center and can exist in two enantiomeric forms: H CHO O H H HO H H HO CH2OH CH2OH (R)-glyceraldehyde CHO HO (S)-glyceraldehyde D-notation: hydroxy is on the right side D-glyceraldehyde (Fischer projection) O CH2OH OH Fischer projections: Horizontal lines are taken as coming forward and vertical lines are going backward H CH2OH L-notation: hydroxy is on the left side L-glyceraldehyde (Fischer projection) Note: “CHO” is a common shorthand for drawing an aldehyde group 147 Nomenclature and Structure: Enantiomers and Epimers • Same D- and L-notation for aldoses with more carbons - look at lowest OH group • Aldehyde group is always written at the top of the fischer projection and the primary alcohol is at the bottom. enantiomers “L-series” of sugars “D-series” of sugars Epimers = diastereomers that differ in configuration at only one stereogenic carbon center; epimers are NOT enantiomers (C-1 epimers are called anomers) Hydroxy-Aldehydes Form Cyclic Hemiacetals Intramolecular 5- and 6-membered ring (cyclic) hemiacetals are easily formed, and are usually more stable than the open (acyclic) form. a “pyran” ring H 2O H O H a new stereocenter is formed in the hemiacetal product OH O HO hydroxy-aldehyde (open form) cyclic hemiacetal (favored) Cyclic product is the predominant form H 2O H O H H 2O H O O H O OH OH derived from the alcohol Equilibrium exists in aqueous solution, or with catalytic acid 148 derived from the aldehyde The Hemiacetal (cyclic) Form Predominates with Aldoses β-glucose is favored because the anomeric OH is in an equatorial position H HO OH OH H H OH H OH O HO HO CHO O OH CH2OH (0.02%) H D-glucose open/acyclic form OH β-D-glucopyranose OH H β−D-glucose (64%) OH HO HO beta = hydroxy up, equatorial (more favored) OH anomeric center at C1 OH HO HO O H OH α-D-glucopyranose OH alpha = hydroxy down, axial α-D-glucose (more steric strain) (36%) The new chiral stereocenter of a hemiacetal is called the anomeric center (carbon-1). Therefore, the two resulting α- and β-stereoisomers are called anomers (anomers are specifically a C-1 epimers at an hemiacetal or acetal center) Equilibrium (open-close-open) between the two anomers is called mutarotation - this is NOT simply a chair flip - acetals can NOT mutarotate, only hemiacetals Ketoses also exist predominantly in cyclic forms Same Structures, Different Representations: How to Draw Cyclic Monosaccharides • hemiacetal (cyclic) form is predominant • β-anomer is more favored (α-D-glucopyranose) (β-D-glucopyranose) • Six-membered rings are called pyranoses • Five-membered rings are called furanoses 149 Anomeric Effect HO HO HO O HO HO HO mutarotation H OH OH O OH OH H α-anomer, OH axial (34%) β-anomer, OH equatorial (64%) If the α-anomer (axial) is disfavored because of steric interactions, why do we even see any of it?? We see even more α-anomer (axial) for more electron withdrawing groups: HO HO HO O OH X HO HO HO H O 34 67 86 94 OH OMe OAc Cl X OH H X %axial increasing e- withdrawing ability Reason? Although an equitorial position is favored for steric reasons, the axial position is favored for molecular orbital reasons Anomeric Effect: Molecular Orbitals axial lone pair equatorial lone pair O O X H good overlap between axial lone pair of O and axial σ* C-X bond Look at σ* C–X bond in Newman projections: must have antiperiplanar orientation between lone pair and σ* C–X bond X σ* C-X H H no overlap with equatorial C-X bond! X X σ* C–X σ* C–X H No overlap!! good overlap!! • The lone pair of oxygen stabilizes the axial anomer through donation of electron density into the empty anti bonding orbital (σ* C-X) • This effect is strongest for the most electronegative substituents (X) because there is better overlap between the orbitals (This alignment is similar to the conformation that is necessary for E2 elimination) 150 Reactions of Carbohydrates: Reductions Reduction of a cyclic hemiacetal proceeds through a small amount of open (acyclic) form. As long as an equilibrium exists between the cyclic and acylic form, the reduction will take place and more open form will be regenerated until the reaction is complete Reduction of an aldose: OH O HO HO CHO OH H OH HO HO OH OH OH H β−D-glucose HO O (0.02%) H H CH2OH OH H H 1. NaBH4 OH 2. H2O H OH HO H H OH OH H CH2OH OH CH2OH D-glucitol Reduction of a ketose: CH2OH CH2OH HO CH2OH O HO H HO HO H OH H α-D-fructofuranose H O NaBH4 H HO OH H OH H HO H HO + OH H OH H H H OH OH CH2OH CH2OH D-glucitol D-mannitol CH2OH open chain ketone CH2OH OH A mixture of sugar alcohols (alditols) More Reductions of Carbohydrates (H2/Ni can also be used) CHO H HO OH H H OH H OH CH3 H H2NNH2 HO NaOH, heat H H CH2OH OH H OH OH CH2OH Wolff-Kishner reduction (via hydrazone) 151 Reduces the aldehyde group all the way to a methyl group Oxidations of Carbohydrates Oxidation to gluconic acids: Only the aldehyde will be oxidized This method will NOT oxidize an alcohol or a ketose, therefore, it is a useful test to distinguish between aldoses and ketoses Oxidation to glucaric acids: BOTH the aldehyde AND the primary alcohol will be oxidized with HNO3 Oxidations of Carbohydrates: Reducing Sugars A sugar with an aldehyde, a hemiacetal, a ketone or a hemiketal group is considered a reducing sugar and can be oxidized. Hemiacetals and hemiketals MUST be in equilibrium with the open form for oxidation. Benedict’s and Tollen’s reagent will oxidize BOTH aldoses and ketoses, therefore, they are good tests for reducing sugars Oxidation to gluconic acids: Benedict’s reagents = Na2CO3, CuSO4, and sodium citrate Cu(OH)2 + (Benedict’s solution) 1. Ag2O, NH3 2. H3O+ workup + (Tollen’s test) Cu2O A red solid Ago Silver mirror Look for color change or mirror to show proof of an aldehyde (a positive test). these will NOT oxidize the primary alcohol Tollen’s reagent = Ag(NH3)2+ -OH 152 Base-Catalyzed Enediol Rearrangments How is a ketose a reducing sugar? How can a ketone be oxidized? In a basic solution, ketoses are converted into aldoses (an enediol rearrangement). This rearrangement explains why a ketose (and a hemiketal) can be a reducing sugar - the ketose can still provide an aldehyde for oxidation. The conditions for the Benedict’s and Tollen’s test are basic enough to promote the endiol rearrangement. Enediol Rearrangement Mechanism A “must know” mechanism 153 Base-Catalyzed Epimerization of an Aldose This is known as the “Lobry de Bruijin-Alberda van Ekenstein reaction” resonance stabilized enolate Deprotonation of the alpha-proton will give an achiral enolate. Now, reprotonation can occur from either face to give a mixture of epimers. (must know mechanism) Reactions of Carbohydrates: Alcohols The OH groups of monosaccharides show the chemistry of typical alcohols. For example, the OH groups can be acylated or alkylated: Ester formation: acylation with acetic anhydride Ether formation: alkylation with an alkyl iodide and Ag2O Ag2O is a milder method to alkylate instead of using a base (NaH) to deprotonate the OH. Ag2O associates with the iodide to make the CH3I a better electrophile. 154 Kiliani-Fischer Synthesis The carbon chain of an aldose can be increased by one carbon in a Kiliani–Fischer synthesis Addition of the cyano group is not selective and it will make two epimers. This means that a mixture of aldose products will result in this synthesis O- and N-Glycoside Formation A glycoside is an acetal or ketal, which do not equilibrate with the open form. Therefore, glycosides cannot be oxidized and they are NON-reducing sugars Acetal (O-glycoside) formation with glucose and ethanol: Reducing sugar NON-Reducing sugars N-glycoside formation with ribofuranose and phenylamine: A mixture of α- and βglycosides will form 155 Mechanism of Glycoside Formation Step 1: Protonation of anomeric OH at C-1 Step 2: Elimination of water to form oxygen-stabilized cation (electrophile) Step 3: Nucleophile (O, N or other) can attack from either the top or the bottom face. Step 4: Deprotonation A mixture of anomers will be produced - acetals DO NOT mutarotate A “must know” mechanism O-Glycoside Formation to make a Disaccharide A disaccharide is composed of two monosaccharide subunits hooked together by an acetal linkage OH OH 4 4 6 H+ O 5 HO HO 1 3 2 OH α-D-glucose HO α-1,4'-glycosidic linkage 6 O 5 HO HO 1 2 3 H OH H 4' OH 6' O 5' O HO 2' 3' (two equivalents) 1' H OH HO β-1,4’-glycosidic linkage HO OH OH O HO 4 + HO HO OH OH H β-D-galactose (a C-4 epimer of glucose) 6 H+ O 5 1 3 2 OH OH H β-D-glucose HO OH HO OH 6 4 O 5 3 2 4 1 6 O OH H lactose O 5 1 HO 3 2 OH OH H 1,2’- and 1,6’-glycosidic linkages are also possible If a disaccharide glycoside still has an end with a hemiacetal group, then it is still a reducing sugar - both maltose and lactose are reducing sugars. 156 Polysaccharides: Biopolymers Amylose (a linear structure) is a component of starch Amylopectin is another polysaccharide component of starch that has a branched structure Carbohydrate Summary • Sugars (carbohydrates) are polyfunctional compounds that have similar reactivity to their alcohol, aldehyde and ketone components (such as reductions, oxidations, acetal reactions, ether and ester formation) • Reducing Sugars (aldoses and ketoses) are hemiacetals (since they exist predominantly in their cyclic form) and can mutarotate • Non-reducing sugars (glycosides) are acetals (which do not equilibrate with an open chain form) and they do not mutarotate • Tollens’s and Bendedict’s test will distinguish between a reducing and non-reducing sugar. The Br2/H2O oxidation will distinguish between aldoses and ketoses. 1) 2) 3) 4) 5) Be able to draw the cyclic hemiacetal form from a Fischer projection Be able to identify and draw anomers and epimers; identify D- and L-sugars Be able to fill-in products or reagents for a given reaction Mechanisms you need to know: hemiacetal formation, mutarotation, enediol rearrangement, epimerization and glycoside formation Know molecular orbital interactions for anomeric effect Vocabulary: Epimers, Anomers, Anomeric center, Ketose, Aldose, Mutarotation, D- and L-notations, alpha- and beta-notations, glycosides, glycosidic linkage 157 Summary of Reducing and Non-reducing Sugars (Reducing gives positive test with Tollen’s or Benedict’s reagent) Reducing sugars: Look for an aldehyde, hemi-acetal, ketone or hemiketal. Hemiketal MUST be able to access the open form for enediol rearrangement of the ketone to an aldehyde HO O HO H HO OH OH CH2OH OH HO HO O HO HO OH O H OH OH O O OH HO β−D-glucose OH maltose H If a disaccharide glycoside still has a hemiacetal, then it is still a reducing sugar OH H α-D-fructofuranose Non-Reducing sugars: Alcohols and acetals are non-reducing and ketals cannot access the open form for enediol rearrangement of the ketone to an aldehyde. CH2OH OH HO HO HO O HO O OCH3 OH H HO H methyl β−D-glucopyranoside CH2OH H HO OCH2CH3 H H ethyl α-D-fructofuranoside All monosaccharide glycosides are NON-reducing sugars Sucrose R OH H2N R O α-amino acid (prominant in nature) H2N O H N R R N H OH O oligo/poly-peptide: amino acids linked by amide bonds O R β α OH β-amino acid (uncommon in nature, studied by synthetic chemists) protein (large peptide) 158 OH D-glucitol α-amino acid residue (amino acid -H2O) O H OH CH2OH Amino Acids, Peptides, and Proteins H2N α OH Amino Acids are Chiral and Charged O H2N Most amino acids have one stereocenter OH All “natural” amino acids are L-configuration, most have the absolute configuration S R Amino acids have at least one basic and one acidic functional group, so as a result they are always charged...positively at low pH (in acid) and negatively at high pH (in base). At neutral pH, amino acids are net neutral, with one negative and one positive charge (zwitterionic) net charge = +1 net charge = +0 net charge = -1 Amino Acids 1: Hydrocarbon, Alcohol, Thiol, Acid and Amide side chains O H2N O H2N OH O H2N OH CH3 glycine (Gly/G) H3C alanine (Ala/A) OH H2N CH3 CH3 OH O HS H3C cysteine (Cys/C) OH S methionine (Met/M) O OH HO H2N CH3 isoleucine (Ile/I) H2N OH O H2N OH H3C O threonine (Thr/T) O OH CH3 leucine (Leu/L) H2N OH HO serine (Ser/S) O H2N H3C O HO H2N OH valine (Val/V) O H2N O H2N O OH H2N OH H2N O aspartic acid (Asp/D) HO O O glutamic acid (Glu/E) 159 asparagine (Asn/N)) H2N O Glutamine (Gln/Q) Amino Acids 2: Basic, Aromatic, and Heterocyclic Sidechains O H2N O H2N OH O H2N OH O H2N OH HN H2N H2N lysine (Lys/K) HO NH phenylalanine (Phe/F) arginine (Arg/R) tyrosine (Tyr/Y) O H O N H2N O H2N OH OH OH N HN NH proline (Pro/P) histidine (His/H) tryptophan (trp/W) The Acid-Base Properties of Amino Acids are Important in Many Circumstances 160 OH The Isoelectric Point (pI) • When the AA has no ionizable sidechain (no acidic or basic groups) the pI is calculated by averaging the pKas of the amino and acid groups. • In the case of alanine, when the pH is below 2.34, both groups are mostly protonated. When the pH is above 9.69, both groups are largely deprotonated. The Isoelectric Point (pI) 161 Isoelectric Point (pI) for Amino Acids with Ionizable Sidechains For groups with ionizable sidechains, the pI is the point at which the two groups with similar pKas share one charge, which is then balanced with the charge of the third group. For lysine (basic side chain), at pH 9.87, the two amino groups are 50% protonated [2 x (+0.5)], and the carboxylic acid is 100% deprotonated (-1), so the net charge is 0. For glutamic acid (acidic side chain), at pH 3.22, the two carboxylic acid groups are 50% deprotonated [2 x (-0.5)], and the amino group is 100% protonated (+1), so the net charge is 0. Electrophoresis Separates Amino Acids and Peptides Based on their pI Values Amino acids will migrate toward electrodes when applied to a stationary surface that is held at constant pH with a buffer. Amino acids with pIs higher than the buffer pH migrate toward the cathode (- electrode), and amino acids with pIs lower than the buffer pH move toward the anode (+ electrode). The bigger the difference between pI and pH, the faster the amino acids will migrate. A few drops of a mixture of amino acids was placed here Arginine, alanine, and aspartic acid separated by electrophoresis at pH 5. 162 Amino Acids are visualized in Electrophoresis with Ninhydrin Ninhydrin is also a useful stain for distinguishing primary (purple), secondary (yellow), and tertiary (no color) amines in TLC experiments Analysis of Amino Acids and Peptides By Ion Exchange Chromatography •Charged resins are used to separate amino acids based on charge, in analogy to silica gel chromatography separating organic molecules based on polarity •Cation exhange resin (eg Dowex 50, shown below) has a negatively charged group bound to the resin and an “exchangeable” cation (Na+). •Anionic amino acids and peptides (low pI) flow through quickly, while cationic amino acids and peptides (high pI) flow through slowly. 163 Automated Ion Exchange Chromatography: Determination of Relative Amounts of Individual Amino Acids Analysis of Amino Acid Content is Useful for Structure Determination (based on the known structures of these amino acids) A typical chromatogram obtained from the separation of amino acids using an automated amino acid analyzer. Determination of the Structure of Peptide Natural Products N OH HO O H2N OH alanine (Ala/A) O (6N HCl/reflux) H2N HO O H2N OH HN 2 tyrosine (Tyr/Y) H2N O H2N OR O aspartic acid (Asp/D) (a cyclic peptide natural product) OH arginine NH (Arg/R) O H2N OH HO Cyclonellin D OH CH3 2 proline (Pro/P) CH3 threonine (Thr/T) Hydrolysis O H O H2N OH H2N O asparagine (Asn/N)) Readout from Amino Acid Analyzer 164 Synthesis of Amines: Reduction of Amides H H O R R N H primary amines (R1=R2=H) H OH SOCl2 H O 1 R R N R Cl H H O R2 LiAlH4 R2 N Et3N R N R2 secondary amines (R1=H) H R1 H H R N R2 tertiary amines R1 Ammonia Cannot be used to Synthesize Amines from Alkyl Halides H3C NH3 Br H3C H3C NH2 Br H3C (n-BuBr) H3C Br+ H3C N H3C CH3 H3C CH3 N CH3 H3C O O phthalimide; an "NH3" synthon, or synthetic equivalent R NH2 amide O NH O 1) KOH 2) n-BuBr 1) H+/H2O 2) NaOH 165 H3C NH2 R N H imide O R NH The Gabriel Amine Synthesis of Primary Amines Synthesis of Primary Amines with Azide and Cyanide Azide is also an NH3 Synthon (replaces Br with NH2)… H3C Br NaN3 H3C N3 H2; Pd/C H3C NH2 Cyanide is a CH2NH2 Synthon (replaces Br with CH2NH2)… H3C Br NaCN H3C 166 H2; Pd/C N H3C NH2 Reductive Amination: Conversion of Aldehydes and Ketones to Amines H O H + H2N NaHB(OAc)3 CH3 N CH3 "H-" N CH3 H imine intermediate O + H3C N CH3 NaHB(OAc)3 CH3 N H CH3 H3C N CH3 "H-" enamine intermediate Synthesis of Amino Acids: From Acids and α-Keto Acids O O OH 1) Br2/PBr3 O OH 2) H2O NH3 ONH3+ Br (+/-)-phenylalanine O O OH NH3 O OH O H2; Pd/C ONH3+ NH (+/-)-phenylalanine Note: these syntheses will give racemic products 167 Synthesis of Amino Acids 2: From Aldehydes…The Strecker Reaction O H O- 2) H+/H2O O Imine formation 1) NH3/HCN NH3+ (+/-)-phenylalanine NH3 H+/H2O H H H N N H-CN H H N+ H N H -CN H Nucleophilic addition Synthesis of Amino Acids 3: Amino Malonic Ester Alkylation O O O EtO OEt N O O 1) KOEt/BnBr O- 2)H+/H2O NH3+ CO2 (+/-)-phenylalanine O O HO KOEt Br O OK EtO O OEt N O OEt Ph H2N O EtO O EtOH O N OEt Ph O O H+/H2O H2N phthalic acid 168 EtO O OEt Ph Preparation of Enantiomerically Pure Amino Acids: Kinetic Resolution •All of the amino acid syntheses discussed are racemic, but generally, enantiomerically pure amino acids are desired. •While they can sometimes be resolved (Bruice 5.14), or prepared using a chiral auxilliary , a kinetic resolution can also be effective. •A kinetic resolution is a reaction in which a catalyst (synthetic or enzyme) reacts more quickly with one enantiomer than the other. Unlike a normal resolution, where a full equivalent of a chiral reagent is required, a kinetic resolution uses only a small amount of catalyst or enzyme. Peptide Bond Synthesis: Regiochemical Problem Normally, making amides from acids and amines is easy… O 1 R O SOCl2 OH 1 R H2N Cl R2 Et3N O 1 R N R2 H Since amino acids are bifunctional, there is a problem with regiochemistry… which amine group reacts with which acid? 169 Peptide Synthesis Step 1: N-Protection One standard protecting group for peptide synthesis is the t-butoxy carbonyl group, also called a t-Boc or Boc group: Peptide Synthesis Step 2: C-Activation The imidate intermediate is an activated ester “equivalent,” similar to an acid chloride, but produced under milder, more selective conditions 170 Peptide Synthesis Step 3: Amide bond Formation Peptide Coupling Agents Accomplish a Net Dehydration O BocNH H OH CH3 N C H O N O OH CH3 BocNH CH3 CH3 OH N H + H2O O O N + H2O N N H H DCC is one of many different coupling reagents, but in all cases they work by basically the same mechanism…activation of the carboxylic acid so that a nucleophilic acyl substitution can take place. In all cases, the peptide coupling agents produce a product that results from the addition of one equivalent of water. 171 Peptide Coupling Reactions Can Be Repeated to Make Long Chains Peptide Couplings Are NOT Well-Suited to Large Peptide Synthesis Problems with peptide synthesis: • Low yields from incomplete reactions •Difficult separation of product, unreacted starting materials, and DCU •Difficulty of purification. Chromatography is difficult with very polar molecules and crystallization is limited to large scale Bruce Merrifield received the Nobel Prize for solving all of these problems with a new synthetic technique known as “solid phase synthesis” which is still widely used today. The first step involves attaching the first amino acid to the resin: 172 Solid Phase Peptide Synthesis In the next step, the N-protecting group is removed, and the amine is coupled to a protected amino acid. reaction Solution resin beads (90 µM) fritted glass filter stopcock The incoming acid and DCC can be used in gross excess to ensure complete reaction; The excess acid, and DCU are washed away! Primary Structure of Peptides: Sequence Determination by Edman Degradation Edman degradation removes one amino acid residue from the N-terminus O R R H H N N R' H O H+ S + H N N H S PTH-amino acid thiazolinone derivative + H O N H3N+ O R'' peptide without the original N-terminal amino acid The Edman degradation can be performed a maximum of 50 times, so it is suitable only for short peptides 173 Longer Peptides are Sequenced by Partial Hydrolysis Dilute acid is used to perform a random partial hydrolysis, but various reagents and enzymes can used to perform site-specific cleavage reactions: Partial Hydrolyses Lead to Fragments Which are Lined up to Determine the Complete Structure Ala-Lys-Phe-Gly-Asp-Trp-Ser-Arg-Met-Val-Arg-Tyr-Leu-His Trypsin: Ala-Lys Phe-Gly-Asp-Trp-Ser-Arg Chymotrypsin: Ala-Lys-Phe Elastase: Ala Cyanogen Bromide: Ala-Lys-Phe-Gly-Asp-Trp-Ser-Arg-Met Gly-Asp-Trp Lys-Phe-Gly Met-Val-Arg Tyr-Leu-His Ser-Arg-Met-Val-Arg-Tyr Leu-His Asp-Trp-Ser-Arg-Met-Val-Arg-Tyr-Leu-His Val-Arg-Tyr-Leu-His Modern peptide sequencing is largely done by Mass Spectroscopy. New advances in MS have fueled many discoveries in Proteomics, or the study of the protein products of gene expression. 174 Secondary Structure of Proteins: Amide Bond Rotation The peptide chains that make up proteins might look like they are very flexible, but in reality many factors contribute to their rigid structure Cysteine Residues Form Disulfide Bonds (Br2) 175 Disulfide Bonds Contribute to Secondary Structure Hydrogen Bonding is an Important Secondary Structural Element: The α-Helix 176 Hydrogen Bonding is an Important Secondary Structural Element : The β-Pleated Sheet Coil/Loop Conformation: >50% of Most Globular Peptide Secondary Structure 177 Tertiary Structure: The 3-D Structure That results From the Composition of various Secondary Structural Elements Summary of Protein Structure 178 Enzyme Active Site Nucleic Acids Make up DNA and RNA A nucleic acid is a polymer of ribofuranoside rings, phosphate ester groups and nucleobases NH2 7 N 5 6 8 a ribosephosphate backbone 9 5' HO N O 4' 3' OH 4 N 3 1' 2' A nucleoside 179 N1 2 Ribofuranoside Rings from Ribose CHO H OH H OH H OH HO H HO CH2OH CHO H H OH H OH CH2OH D-2-deoxyribose H OH nitrogen base R R N H HO O H HO OH H β-D-2-deoxyribofuranose R HO glycosidic bond formation O R a ribonucleotide nitrogen base R R N H glycosidic bond formation R HO O H HO N R H a deoxyribonucleotide Phosphoester and Nucleobase Structures 180 N H OH H HO β-D-ribofuranose D-ribose H O OH Aromatic Amines and Basicity: 6-membered Rings piperidine + H+ pyridine N+ H H + H+ + N H pKa sp morpholine O O + H+ 5.2 + N pKa=9.3 N H H N 3 H N sp2 pKa=11.2 N H The lone pair of pyridine is NOT part of the π-system of the aromatic ring. The nitrogen is sp2 hybridized, and therefore less basic. Basicity = piperidine > morpholine > pyridine nb: do not confuse acidity, basicity, low pKa, and stability of the conjugate base…they are all ways of asking the same question! Whenever ranking acidity or basicity, consider 4 effects: aromaticity, resonance, hybridization, and induction (EWGs). Nitrogen Atoms are Electron-Withdrawing in an Aromatic Ring pyridine pyrimidine N + N+ N H pKa 5.2 2 sp H sp2 1.0 Pyrimidine is significantly less basic than pyridine due to the electron withdrawing effect of the added nitrogen 181 Aromatic Amines and Basicity: 5-Membered Rings Recall: EAS of 5-membered heterocycles (E+ = H+) H N+ H H N H H+ H+ + N 2 sp H sp2 H The lone pair of pyrrole IS part of the π-system of the aromatic ring. Protonation destroys aromaticity, and is therefore highly disfavored. N H pyrrolidine pyrrole N N sp2 17 pKa H+ + N 11.3 pKa -3.8 H N+ H H sp3 H DEprotonation of pyrrole to form an anion is facile. The resultant anion is very stable. 36 Imidazole is Aromatic, and Behaves like a hybrid of Pyridine AND Pyrrole pyridinium protonated pyrrole protonated imidazole + N H pKa = 5.16 N+ H H N+ H H H pKa = 6.8 imidazole N N N H H pKa = 17 Aromaticity Retained N pKa = -3.8 pyrrole H N+ N Imidazole can serve as a weak acid or as a moderate base/nucleophile. It is responsible for many important biological functions. pKa = 14.4 O H2N OH N NH 182 histidine (His/H) The Nucleosides of RNA and DNA Nucleoside = ribofuranoside + nucleobase NucleoTIDE = Nucleobase + Ribofuranoside + Phosphate 183 Monomeric Nucleotides are Important Biological Molecules The Phosphate Linkage is Reactive…Kind of Like an Anhydride O OH OH O O O HO HO O -O P O -O HO P O-O O HO HO OH OH ∆G = + 3.3 kcal/mol + H2O OH Not a favorable reaction O P P P -O O O O -O -O -O O A O + H2O O P P O O O -O -O A O ∆G = - 7.3 kcal/mol + HO P O-O OH OH ATP O ADP Reaction becomes favorable: HO HO O OH -O P O -O O + ATP OH HO HO OH OH 184 + ADP O OH ∆G = - 4.0 kcal/mol ATP is Thermodynamically Unstable (reactive), but KINETICALLY Quite Stable (UNreactive) Enzymes are necessary for the phosphoryl transfer reactions of ATP Without ATP, No Phosphorylation Would Take Place This phenomenon is reminiscent of alcohol activation by conversion to a tosylate…you must have a good leaving group! Nu: R OH x R R Nu Unstable leaving group TsCl/ Pyr Nu: OH- TsO- OTs R Nu Stable leaving group 185 ATP’s Function in Biosynthesis The following reaction is impossible in the absence of ATP… Activation with ATP allows thioester formation to take place Kinases are an Important Class of Enzymes that Mediate Phosphoryl Transfer Reactions Kinase + ATP P Active Protein Inactive Protein Amino acids with alcohol groups in their side-chains are subject to phosphorylation by kinases… O H2N HO serine (Ser/S) O OH H2N HO O H2N OH CH3 threonine (Thr/T) HO tyrosine (Tyr/Y) 186 OH Inhibition of Kinases Could lead to Cures for Many Diseases, including some kinds of Cancer Molecules that “look” like ATP can occupy the binding site and prevent phosphorylation The problem is specificity: all kinases (and many other enzymes) have ATP binding sites, so most kinase inhibitors are non-specific GleevecTM (Novartis) is the one of the first SPECIFIC Kinase Inhibitors Approved for the Treatment of Cancer 187 Several Other Nucleotides Play Important Biological Roles Oligomerization of Nucleotide Triphosphates gives DNA 188 Hydrogen Bonding of Nucleobases N H O N keto H-Bond donor, d enol H N X H N H N B H-Bond acceptor, a OH NH N H H a N a R H N a N R NH2 amine imine d O N a N Hd O a N N a OH d R cytosine Keto-enol and imine-amine tautomerization of the nucleobase determines the type of hydrogen-bonding that can occur Base Pairing in DNA Occurs through Hydrogen Bonding 189 Complimentary Strands are Anti-parallel; π-Stacking Interactions are Important The attractive force between two proximal aromatic rings is referred to as "π-stacking" The Double Helical Structure of DNA 190 DNA Structural Unit: Chromatin Histones combine with DNA to form nucleosomes, the fundamental structural units of chromatin. All of this points to the conclusion that chromatin in intact nuclei is highly dynamic, with different folding conformations that reflect its activity. (Figure from Biology: Concepts and Connections by Campell, Mitchell, and Reece. Text from Introduction to Cell and Molecular Biology by Stephen L. Wolfe) How Can Drugs React with DNA if it is Packed in a Nucleosome? 1) Outer surface of DNA is still accessible to small molecules 2) Nucleosomes are in dynamic equilibrium with uncoiled DNA so that the drug can bind after uncoiling, which also interferes with the binding of the DNA to the histone 3 Types of DNA binders: 1) Reversible DNA binders External electrostatic binding Groove binding Intercalation 2) Irreversible covalent reactions with DNA binders by Alkylation 3) Irreversible DNA strand breakers by radical reactions 191 Reversible DNA Binders 1) 2) 3) External electrostatic binding: cationic complexes, usually metals, bind the negatively charged phosphate backbone and lead to disruption of the DNA structure Groove binding: interactions with minor groove by electrostatic and hydrogen bonding Intercalation: flat, aromatic molecules insert in between base pairs, stabilized by π-stacking and charge transfer interactions (anti-tumor agent) Intercalation: Ethidium Bromide Ethidium bromide is a common fluorescent stain used with double-stranded DNA. Ethidium intercalates between DNA bases. In the intercalated state, ethidium exposed to short-wave UV light (302nm) will fluoresce bright orange (595nm). The intercalation of ethidium into double-stranded DNA causes the helix to extend and unwind. 192 DNA as a Nucleophile: Alkylation most nucleophilic site in DNA double helix O NH N O N NH2 N O H O N H3C H O H3C N H I O N NH2 O H alkylation O P O- NH O H H O P O- O O Most reactive nucleophilic sites of DNA: N-7 of guanine > N-3 of adenine > N-7 of adenine > N-3 of guanine > N-1 of adenine > N-1 of cytosine. The amine on C-2, the O-6 of guanine and phosphate groups can also be alkylated. This reactivity is strongly controlled by a combination of steric, electronic and hydrogen-bonding effects. For example, nucleophilicity is diminished by hydrogenbonding and nucleophilic sites on the interior of the DNA double helix are sterically less accessible. Recall: DNA Damage with other electrophiles epoxide hydrolase p450 HO O benzo[a]pyrene OH O N O NH N O O P O p450 NH2 N HO O O- OH diol epoxide O O N N O O P O NH HO N Cytochrome p450 N H O O OO covalently modified, damaged DNA 193 benzene benzene oxide Recall: Nitrogen Mustards as bis-Alkylating Agents that Crosslink DNA CH3 S Cl Nu: Cl N Cl Nu intermolecular N Nu Nu: intramolecular substitution (SN1) N CH3 SN2 CH3 intermolecular H3C Cl Cl mechlorethamine treatment of advanced Hodgkin's disease Nu: SN2 CH3 N N Cl sulfur mustard toxic nerve gas used in World War II Two methods: Cl Cl CH3 H3C Cl N Cl Nu N Cl aziridinium ion is a great electrophile!! Nu Nu: CH3 Nu N Nu bis-alkylation product Recall: Bis-Alkylation of DNA = Crosslinking DNA H3C N Cl Nu: intramolecular substitution (SN1) Cl CH3 H3C N Cl Cl N aziridinium ion intramolecular substitution (SN1) DNA O Nu: -O P O O H H H O CH3 HN N H2N N O O P O N O OH O H H N N O O N H CH3 N -O O P O O N N NH2 Crosslinked G-G residues of DNA double-helix O P OO 194 DNA DNA DNA is Transcribed into RNA, Which is Used in the Template Synthesis of Peptides The DNA code: Each amino Acid is Described by a 3-letter Codon 195 DNA is Very Stable (encyclopedia), RNA is hydrolyzed very readily (post-itTM note) Cell Biology Experiments Involve the Ability to Control Gene Expression DNA Genetics (mutation/deletion) RNA Block Transcription (eg triple helix) Block Translation (Antisense/siRNA) Peptide Chemical Genetics (small molecules/drugs) Function/Phenotype 196 Antisense Oligomers and siRNA bind to RNA and Prevent Translation into a Protein DNA-like strand still around 5'-T-A-C-G-T-A-T-T-5' A 3'-A-U-G-C-A-U-A-A-5' Antisense Oligonucleotide (DNA-like) U C G T A A A RNAse (recognizes RNADNA duplex, destroys the RNA strand) 5'-T-A-C-G-T-A-T-T-5' 3'-A-U-G-C-A-T-A-A-5' X Ribosome, tRNA, etc Ribosome Protein Chemically Modified RNA Analogs Are not Hydrolyzed, and are More Effective at Blocking RNA Translation O O O O base O P OO O O O base O base O P SO O S O base O base O H C H O base N O base O P O Me N Me O base N O Phosphodiesterlinked DNA O O Phosphothioatelinked DNA 197 Thioformacetallinked DNA Morpholinolinked “DNA” RNA Was Probably the First Biological Molecule: Ribozymes The Ribozymes are oligomers of RNA that can catalyze reactions…Tom Czech received the Nobel Prize for this discovery. X-ray crystal structure of the Tetrahymena ribozyme. The secondary structure is highly pre-organized for substrate binding, much like an enzyme. (T. Czech, et al, Sciene 282, p282, 1998) RNA Was Probably the First Biological Molecule: The Ribosome The Ribosome is made up of mostly RNA, unlike most enzymes which are made of amino acids. Gray=RNA, Yellow=Peptide 198 How, then, Did Life Begin, Chemically Speaking? The Miller experiment demonstrated that simple organic molecules would spontaneously form amino acids with the help of lightning O H O O H2N OH OH OH O O H2N HO O O H3C OH CH3 OH H3C OH O H2N OH O OH H3 C O OH H2N O HO O N H H OH H3C O HO CH3 OH O OH H3C N H OH O N O NH2 H2N NH2 CH3 O H2N O OH H2N OH CO2H CO2H The Oro Experiment Demonstrated That Prebiotic Nucleobase Synthesis was Possible In the presence of UV light, HCN forms adenine… NH2 HCN N N H N N Adenine; C5H5N5 or 5 x HCN! The conditions can be varied to produce other heterocycles, including other nucleobases 199 The Formose Reaction Provides Ribose O O H HO H Base H OHO H Glycoaldehyde Formaldehyde (the simplest carbohydrate) O H H OOH O HO HO H HO O H HO H OH Ribozymes HO Ribose Threose NH2 NH2 N N HO O OH N H OH N N N N HO N O OH OH OH The yield of ribose is quite low, and many other sugars are formed at the same time. The addition of borate to the reactions greatly increases the selectivity for ribose formation (Science, 303, 196, 2004). How Did we End Up as Single Enantiomers? This is a raging origin of life debate, but the answer is probably not through the influence of magnetic fields… O OH H RMgBr R New Stereocenter Created... racemic (no magnetic field) up to 90% ee (magnetic field) Angew. Chem. Int. Ed. Engl. 1994, 33, 454; (original paper); Angew. Chem. Int. Ed. Engl. 1994, 33, 1458; (results not reproducible); Angew. Chem. Int. Ed. Engl. 1994, 33, 1459; (results not reproducible) ; Angew. Chem. Int. Ed. Engl. 1994, 33, 1376; (paper withdrawn); Angew. Chem. Int. Ed. Engl. 1994, 33, 1457; (paper withdrawn); Angew. Chem. Int. Ed. Engl. 2004, 43, 2194. (Ph.D. revoked) “On February 7, 1996, as recommended by the commission, the committee of the Faculty of Mathematics and Natural Sciences decided to strip Guido Zadel of his doctorate because of grave deception during (and after) his doctoral studies.” “The doctoral certificate must be withdrawn and Guido Zadel is no longer allowed to use the title of Doctor.” 200 Biosynthesis, Natural Products and Drug Action Basic organic chemistry concepts can be applied to biological systems in two main topics of discussion: 1) Biosynthesis of natural products: How does nature synthesize all of her molecules? Recall ATP…but there are also other methods. Paul M. Dewick (2002) Medicinal Natural Products: A Biosynthetic Approach John Wiley and Sons, New York, NY. 2) Drug design and drug action: How do small molecules have such a huge impact on biological systems? Recall aspirin… Richard B. Silverman (2004) The Organic Chemistry of Drug Design and Drug Action Elsevier Academic Press, Burlington, MA. Sample Problem: Aflatoxin Aflatoxin is formed in peanuts as a result of a mold that can grow inside the shell. One mode of toxicity involves covalent modification of DNA by a metabolite of aflatoxin, namely aflatoxin epoxide. O O O O O O cytochrome p450 H O O O O O O H H aflatoxin aflatoxin a) Enzyme-mediated oxidation generally take place on electron-rich alkenes. In the case of aflatoxin, why is only one of the 3 possible alkenes epoxidized? O H O O H O H 201 Sample Problem, continued b) Aflatoxin epoxide damages DNA by covalent modification of N7 of adenine. Draw the product. NH2 N N HO N N O O O P OO- c) N7 of adenine is not the most nucleophilic nitrogen. What binding property might aflatoxin have that would cause it to interact more selectively with DNA? Chlorismate Mutase: Biosynthesis of Phenylalanine and Tyrosine Using a Claisen Rearrangement decarboxylative aromatization and transamination CO2H NH2 L-Phe O OH chorismate chlorismate mutase mutase OH O OH HO HO O Claisen rearrangement chlorismic acid O O O OH OH O O CO2H NH2 OH OH acid prephenic Count carbons to see which alkenes are involved in the [3,3]-sigmatropic rearrangement L-Tyr decarboxylative aromatization, oxidation, then transamination OH [3,3] sigmatropic rearrangements such as the Claisen rearrangment normally occur under thermal reaction conditions. The rate of the Claisen rearrangment is increased 106-fold in the presence of the enzyme due to stabilization of the transition state. 202 Synthesis of Vitamin C from D-glucose OH HO2C O HO HO OH HO HO NAD+ O OH OH OH NADH D-glucuronic acid D-glucose HO2C NAD+ = oxidant; NADH = reductant OH OH HO OH HO O EKT O HO HO OH OH OH L-gluconic acid HO HO O O [O] HO O OH O CH2OH OH HO O 2-oxogluconolactone Ascorbic Acid (Vitamin C) H L-gluconolactone HO Plants and most animals can convert glucose into ascorbic acid (vitamin C) by the pathway shown. Humans and primates are deficient in the enzyme oxidizing gluconolactone to the ketolactone, and are thus dependent on a dietary source of Vitamin C. H H OH H OH OH CO2H Sample Problem: Ascorbic Acid’s Acidity OH HO O HO O Ascorbic Acid (Vitamin C) is not a carboxylic acid, yet it is acidic enough to get “acid” in it’s name. Which proton is most acidic and why? OH Ascorbic Acid (Vitamin C) 203