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, & P- Containing molecules

Natural Loads of FUN!! Products Unnatural Polymers, Consumer Bioactive Materials Products: Molecules, From Food to eg Drugs to the Package that Contains it!

1 What Molecules Are Important? All of them…

H C O CH 3 3 Me O2N NO2 - + H3C MeO O Na

O NO2 tri-nitrotoluene naproxen sodium testosterone TM (TNT) (alleve ) NH2 Cl OH N O N O O N O Cl N OH OH OH O HO HO Cl NH NH2 HO sucralose HO OH TM tryptophan (Splenda ) adenine

short-hand drawings: H H C H CH3CH2CH3 H C C H3C CH3 propane H H H H

E2a Review: Things You Should Already Know for this Class

1. Hybridization 2. Molecular Orbitals: orbital types and interactions, diagrams 3. Conformational analysis: drawing chair conformations and Newman projections 4. Functional group names and nomenclature 5. Stability of carbocation and anion intermediates 6. Resonance 7. Curved arrow mechanism, identifying electrophiles and nucleophiles 8. Acidity and Basicity 9. Kinetics and thermodynamics: reaction coordinate diagrams 10. Relative Reactivity 11. Stereochemistry 12. 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 Weakest conjugate base conjugate base

Review: Carbocation Stability

• Carbocations are important intermediates in electrophilic additions to alkenes, and in

SN1and E1 reaction mechanisms. 1) hyperconjugation: more substituted carbocation is more stable 2) Resonance/conjugation: carbocation with more resonance is more stable 3) Electron-donating groups stabilize, electron-withdrawing groups destabilize 4) Aromaticity: an aromatic carbocation is very stable

H H H H + + + + CH3 >>>CH3 CH3 CH3

H3C H3C O N O H3C H3C

CH3 O more electron-donating less electron-donating substituent 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°

CH HCH3 CH3 H CH CH H C CH3 3 H 3 3 3 H CH3 H H3C H Conformations of n-Butane: H H H CH H H H H H H H 3 H H3C H H (Newman projections) H H H CH3 H H A staggered conformer eclipsed gauche anti gauche is more stable than an A BCDEF eclipsed conformer

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 Enantiomers CH3

Cl Cl

Diastereomers Diastereomers

Cl Cl Enantiomers CH3 CH3

Cl Cl lowest priority group oriented behind 1 Cl Cl Enantiomers Br CH3 CH3 2 H H3C H3C H3CH2C Cl Cl CH3 3 ' Diastereomers Diastereomers 1-2-3 Cl Cl counterclockwise = S MESO CH3 CH3 H3C H3C Cl Cl

6 Functional Group Review

H CH 3 R R R R R O alkane alkene alkyne diene benzene epoxide

R F R Cl R Br R I R SH R CN alkyl fluoride alkyl chloride alkyl bromide alkyl iodide thiol nitrile (cyano group)

Me R R R R N O H R OH R O R S R O R amine alcohol ether O sulfonate ester aromatic compounds (OTs, tosylate) (substituted benzenes)

carbonyl compounds O O O O O O O O R R N R R O R OH R H R R R Cl R O R H amideester carboxylic aldehyde ketone acid anhydride acid chloride

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 Movement of one electron pair of electrons “fish-hook arrow”

7 Reactions of Benzene

Z + Y Y R Y Z R Z- R

(a) H H addition Z H H Y (path a) Y Y Z + H Z- H

(b) benzene +Y-Z (nonaromatic) substitution (path b)

H Y + H-

benzene (- H+) + (Y+) (aromatic)

Electrophilic Substitution is Thermodynamically Favored

First step is rate determining

8 Sidenote: Synthesis of Aspirin from Crude Oil

OH OH O O O Me O O

[O] CO2/base OH Me O Me OH

acetyl salicylic acid (Aspirin) oil well Willow Trees:

Relieves Fever, Upsets Stomach

Summary of Electrophilic Additions

X Halogenation FeX3/X2

NO HNO /H SO 2 Nitration 3 2 4

SO3 H2SO4/∆ Sulfation

Friedel-Crafts RCOCl/AlCl3 R Acylation

R Friedel-Crafts RCl/AlCl3 Alkylation

9 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 Br R Br Br R

H Br Br Br X

FeBr3

H Br H Br + - - Br Br Fe Br + Br Br Fe Br Br Br

Nitration of Benzene

O O O O + - + + H N - H SO HSO4 H N - +N H H O O 2 4 O+ O O HNO3 H

H O O O + + + N - N - N + O O H SO H HSO - + 2 4 O 4

10 Friedel-Crafts Acylation

+ O Cl O Cl Cl O O - R Cl Al Al + + AlCl4 Cl Cl R Cl+ - Cl R R an acyl chloride an acylium ion

+ H O O O R R H

R - AlCl4

AlCl3

Friedel-Crafts Alkylation

R Cl H R Cl Cl + - R Cl Al Al + AlCl4 Cl Cl R Cl+ - Cl R R an alkyl chloride a carbocation

H R R H + R R R R H - AlCl4

+ - H AlCl4

11 Carbocation Rearrangements Occur During Friedel-Crafts Alkylations

R H Cl + + AlCl - Al R R 4 R Cl Cl Cl

1,2 alkyl shift 1,2 hydride shift ring expansion

Me Me Me Me AlCl3 Me Cl Me Me Me Me

O% 100% H Me + H Me Me + Me Me Me

Friedel-Crafts Acylation/Reduction: Net Alkylation

H H H H AlCl3 + R R Cl X

H2; Pd/C

O O + AlCl3 R R Cl

• 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 Br 1,2-dibromobenzene 1,3-dibromobenzene 1,4-dibromobenzene ortho-dibromobenzene meta-dibromobenzene para-dibromobenzene o-dibromobenzene m-dibromobenzene p-dibromobenzene

NO2 ortho Z ortho NO2 Br

Br Cl meta meta 1-bromo-3-nitrobenzene 2-bromo-4-chloro- meta-nitrobenzene 1-nitrobenzene para

Reactions of Benzene Substituents

Br NaOH OH SN2 (& SN1)

Me KOt-Bu E2 (& SN1)

H ; Pd/C Me Alkene Reduction: 2

Carbonyl Reduction: H H2; Pd/C OH

O + NH2 N - Nitro Reduction: O H2; Pd/C

13 Effect of Substituents on Reactivity

R R R H -H+ E E+ + E

R H + E

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 electron withdrawing group group

Z H Y >>

MORE REACTIVE LESS REACTIVE (Electrophilic Substitution) (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 o-bromotoluene p-bromotoluene m-bromotoluene NOT OBSERVED 2) Weak deactivators (halogens) are o,p-directors. Br Br Br Br Br FeBr3/Br2 Br Br o-dibromobenzene p-dibromobenzene m-dibromobenzene NOT OBSERVED 3) All moderate and strong de-activators are m-directors O Me O Me O Me O Me

NO2 HNO3/H2SO4

NO2 NO2 m-nitroacetophenone o-nitroacetophenone p-nitroacetophenone NOT OBSERVED NOT OBSERVED

15 Substituent Effects: Ortho/Para Directors MeO MeO MeO MeO H + + E+ E H -OR-+ -OR- + E H H H E stable resonance form ortho meta para + MeO MeO MeO MeO MeO -H+ E ortho: H H H H + E E E E + +

MeO MeO MeO MeO -H+ NOT meta: + + OBSERVED + H H H E E E E H H H

MeO MeO MeO + MeO MeO

+ + -H para: + + H H H H H E H E H E H E E

stable resonance form

Substituent Effects: Meta Directors

NO2 NO2 NO2 NO2 H + E + E+ -OR- + H -OR- + E H H E unstable resonance form ortho meta para

NO NO2 NO2 NO2 2 -H+ E ortho: H H H + E E E + + NOT OBSERVED

NO2 NO2 NO2 NO2 -H+ meta: + + + H H H E E E E H H H

NO NO2 NO2 NO2 2 + + -H para: + + H H H H E H E H E E NOT OBSERVED unstable resonance form

16 Hill Diagram Summary: Activation and Direction

O O R H + H + N+ S R N O- R R O H O N R = Deactivating/m-directing R

o,p - m frontier

R = weakly deactivating/o,p-directing F, Cl, Br, I R = H activation frontier R = Activating/o,p-directing H N R O R R R H O O R + E + R H N OR E H R

Halogens are Ortho/Para Directors

Cl Cl Cl Cl Cl E+ H -H+ E + -OR- -OR- E + H H E E + MeO +Cl NO2 H H H + E E E

MeO + + Cl NO2

H H H H E H E H E Cl behaves like MeO for directing substitution (resonance) Cl behaves slightly like NO2 for activation (induction) As with acidity, resonance ismoreimportant than induction!

17 Sample Problem: Planning A Synthesis with Aromatic Substitution

SO3H

Sample Problem: Planning A Synthesis with Aromatic Substitution

O Me

NO2

18 Arenediazonium Salts: Selective Monosubstitution

CN NO2 CuCN

X H2; Pd/C CuX + (X = Cl or Br) NH2 N2 NaNO /HCl 2 Cl- I ("HONO") KI (X = Cl or Br)

HCl/H2O

Synthesis with Arenediazonium Salts

Me Me Me

Cl FeCl3/Cl2

Cl o-chloroethylbenzene p-chloroethylbenzene

Me Me Me

NaNO /HCl CuCl 2 Cl- ("HONO") + NH2 N2 Cl p-chloroethylbenzene

19 Nucleophilic Aromatic Substitution (SNAr)

RRR E+ E + H+

E XNu Nu- + X-

Cl OH NaOH/∆ pH 14 "Nu-"

NO2 NO2

Cl OH H2O O2NNO2 O2NNO2 pH 7

NO2 NO2

Mechanism of Nucleophilic Aromatic Substitution

X OH HO- 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 NH2 H N- H * H * 2 NaNH2

AND H elimination H N - NH 2 * H2N 3 NH2 * - *

NH3

- -NH2 H2N * Benzyne Intermediate

Structure of Benzyne

R R Alkyne Benzyne

21 Heteroaromatics

5-membered heterocycles N O S H pyrrole furan thiophene

6-membered heterocycles N N N pyridine quinoline isoquinoline

6,5-fused heterocycles N O S H indole 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 4 3 O H3C O CH3 EAS: 5 2 CH + 3 H C OH N 1 N 3 H H O 2-acetylpyrrole EAS occurs 4 3 preferentially Br2 5 2 at the C2-position O1 O Br 2-bromofuran

4 3 HNO3 5 2 NO S H2SO4 S 2 1 2-nitrothiophene

Consider Mechanism: EAS at C2 E E E N H N H N H H H H 20-allylic carbocation 20-allylic carbocation + E+ N E E H H H EAS at C3 N N H H 20-alkylcarbocation

22 Relative Reactivities of 5-Membered Heterocycles in EAS

> > > N O S H

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 + CH3 AlCl3: Strong L.A. H3C Cl H2O O SnCl4 + SnCl4: Weaker L.A. H3C Cl H O CH3 S 2 S O O BF3 BF : Weak L.A. + 3 H3C Cl H O CH3 O 2 O O O O no cat. NO L.A. + CH H3C O CH3 required 3 N N H H less reactive O acylating agent

Reactions of Six-Membered Heterocycles

+ SN2: H3C I N-methylpyridinium iodide N N I

CH3

+ HO OH + H2O N-Oxidation: N N OH N OH O Pyridine-N-Oxide

Br FeBr3 Preferential substitution @C3, why ?? EAS: + Br2 300 °C N N High Temps required due to the fact that 30% electron withdrawing N-atom destablizes carbocation intermediates. B Mechanism: Y Y slow fast + Y+ H + HB+ N N N

23 Electrophilic and Nucleophilic Substitutions on Pyridine

What about Nucleophilic Aromatic Substitutions ?

4 5 3 NaNH2 2 ∆ 6 N OCH N NH2 1 3 SNAr takes place ONLY at C2 and C4 positions Br OMe Why ??? NaOMe ∆ N 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 Br- is weaker NH2 - base than CH3O NaNH2 ∆ N OCH3 N OCH3

CH3 CH3

NaOMe ∆ N Cl N OMe

24 Other Reactions of Pyridine

NBS Benzylic Bromination R ∆, ROOR R N N Br Reactions of Pyridine Diazonium Salts

NaNO2 / HCl H2O Cl N NH2 0 °C N N N OH N O N H enolic form keto form (more stable) α-pyridone

NH2 N OH O N Cl- NaNO2 / HCl H2O

N 0 °C N N N H enolic form 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 C2- position. • 6-Membered heteroaromatics participate in EAS (C3-position) and SNAr reactions (C2 and C4 positions)

25 heterolytic cleavage Neutral Organic Reactions - Radicals

H H H H H CC H + C H C H formation of polar, H H H H charged species methyl anion methyl cation homolytic cleavage

H H H H H H + CCH C C H formation of non-polar, H H H H neutral species methyl radical methyl radical

both are Structure of radical is somewhere between a cation and anion, electron-deficient half-filled p-orbital is a SOMO (Singly Occupied Molecular Orbital)

Thermodynamics

• Bond breaking is endothermic (∆H˚ is positive) • Bond formation is exothermic (∆H˚ is negative)

H H H H H H CC C + C H H ∆H˚ = 90 kcal/mol H H 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: H H H ∗ 3 σ sp3 C C sp H H H H H H C C energy 90 kcal/mol required H 3 H to break bond sp3 sp H H H H CC σ H H H ∆H˚ is positive stabilization energy = 90 kcal/mol reaction progress ∆H˚ is negative

26 Bond Dissociation Energy (BDE)

X + Y

∆H˚ = ∆G≠ = BDE energy

Bond Dissociation Energy XY (BDE) = energy required to break a bond homolytically reaction progress

Bond Dissociation Bond Dissociation Bond Energy (kcal/mol) Bond Energy (kcal/mol) Cl Cl 58 HH 104 Br Br 46 H3CH 105 II 36 H3CCH3 88 RO OR 38 H2CCH2 67 (π bond only) Cl H 103 H3COH 92 Br H 87 H3C NH2 85 IH 71 H2NH 107 ICH3 84 CH3OH 104 Br CH3 70 HO H 119 Cl CH3 57 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 ∆H˚ = C CH 65 kcal/mol 2 H C CH H2C CH2 2 2 CH ∆H˚ = energy 2 H C CH 65 kcal/mol H2 ∆H˚ = 2 2 C H3C 85 kcal/mol CH + CH H3C CH3 2 3 ∆H˚ = CH3CH2CH3 85 kcal/mol cyclopropane has strain energy that raises the energy reaction progress of the C–C bond, therefore making it easier to break

2) Make radical more stable (lower the energy of the transition state)

BDE (kcal/mol) 85 85 96 99 101 104 -from cleavage 105 of a C–H bond,

with R = CH3

27 Stability of Radicals: Just Like Carbocation Trends 1) Hyperconjugation: more substituted radicals are more stable

2) Conjugation/Resonance: more resonance, more stable radical

H most C CH2 no > > CH2 > CH2 resonance resonance H3C

3) Hybridization: more s character of the SOMO, less stable radical

3 H sp -hybridized, CH2 > H > > R CC sp-hybridized, least s character H3C most s character H

Hyperconjugation in Carbocations and Radicals

Carbocation antibonding MO stabilization: A σ C–H bond donates into the empty p-orbital to stabilize the carbocation

σ C–H = donor (HOMO) bonding MO empty p-orbital = acceptor (LUMO) Cationic, 2-electron system

half-filled Radical H p orbital stabilization: antibonding MO A σ C–H bond donates H H into the SOMO (singly CC occupied molecular H H orbital) to stabilize the radical σ C–H = donor (HOMO) bonding MO SOMO = acceptor 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 hv or ∆ possible

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:

CH3 - CH + HBr H3C +Br 3 H3C H3C δ+ δ- Br H+ adds to primary carbon to form a secondary carbocation, secondary whichismorestablethanaprimarycarbocation alkyl bromide radical mechanism: H-abstraction and formation of new radical

+.Br H3C Br H3C Br +.Br H3C HBr primary alkyl bromide Br adds to primary carbon to form a secondary radical, which is more stable than a primary radical

29 Synthesis Using Primary Alkyl Bromides

A primary alkyl bromide is a good substrate for an SN2 reaction:

O O HBr, ROOR NaO CH3 H C Br O CH 3 hv H3C H3C 3 nucleophilic substitution

Recall the previous general method that we used to convert an alkene to a primary SN2 substrate with a good leaving group:

1. BH3 2. NaOH, H2O2 1. TsCl/pyridine H3C H3COH H3CNu 2. Nucleophile hydroboration formation of good leaving group and oxidation followed by nucleophilic substitution

Examples of nucleophiles = NaCN, NaOH, NaOMe, NaSMe, NaOAc, NaCl, NaBr

Radical Mechanism Sample Problem

HBr Br 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 Radical Reactions with Alkanes

Major product because Not practical or useful with F or I secondary radical is 2 2 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 Br hv or ∆ + N Br + H C H C H C Br 3 RO- OR 3 3 O (peroxide)

Unsymmetrical alkenes will form two different products because the allyl radical intermediate has two resonance contributors that are not equivalent

H CH +.Br 2 +HBr H3C H3C H3C

32 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 CH3 CH3 OH CH3 CH3

CH3 H3C CH3 HO

CH3 H3C CH3 CH3 H C O 3 CH 3 CH3 CH3 CH3 OCH3 CH3 CH3 BHA BHT vitamin E (butalyted hydroxyanisole) (butalyted hydroxytoluene) (α-tocopherol) Radical scavenger mechanism: H O CH 3 O CH3 CH3 CH Phenolic compounds R 3 + CH 3 CH3 + RH donate an H to the unstable free radical to form free radical OCH two stable unreactive Reactive!! 3 OCH3 radical very stable radical (resonance) species scavenger Unreactive!!!

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

Br Br2, hv CH3 CH3 Racemic mixture of products H3C H3C (a pair of enantiomers)

• 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 Br , hv Cl Cl 2 a pair of CH3 CH3 + CH3 H3C H3C H3C diastereomers Br Br

33 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 Br Br Br2, ∆ 2 CH3 CH3 hv or ∆

Br HBr NBS, ∆ H C CH Br 3 3 H3C ROOR secondary allyl bromide alkyl bromide HBr ROOR

Br primary alkyl bromide H3C

1. Addition of H-Br to alkenes with peroxides 2. Halogenation (Cl2 or Br2) of alkanes 3. Substitution of benzylic hydrogens with Cl2 or Br2 4. 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 ∆H˚ = 38 kcal/mol

CH3 CH3 ∆H˚ = 29 kcal/mol O OO O 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 hv or ∆ + N Br RO- OR O (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 CH3

CH3

CH3 BHA OCH3

34 Summary of Reactive Intermediates

Name Structure Stability Properties

R Electrophilic, electron-deficient, carbocation 3˚ > 2˚ > 1˚ > Me strong acid, empty p-orbital C (LUMO) is a good acceptor R R

R electron-deficient, singly occupied MO (SOMO) is reactive, can be radical C 3˚ > 2˚ > 1˚ > Me R R either electrophilic or nucleophilic

R Nucleophilic, electron-rich, strong base, lone pair (HOMO) carbanion C Me > 1˚ > 2˚ > 3˚ R R is a good donor

R Neutral divalent carbon, empty p- carbene all very reactive orbital is a good acceptor and C lone pair is a good donor, so it is R both electophilic AND nucleophilic

Carbene Formation A carbene is a neutral species containing a divalent carbon

empty 2pz orbital (carbocation) KO CH3 Br H C CH Br Br Br 3 3 C 103˚ C C (or KOH) Br Br H Br dibromocarbene filled sp2 hybrid orbital treatment with (lone pair/carbanion) a strong base

Mechanism of formation:

Deprotonation with strong base α-elimination Br Br Br Br Br C C C Br H Br Br dibromocarbene OH

35 Formation of Carbenes from a Diazo-Compound

CH2N2 =diazomethane(ayellowgas) empty 2pz orbital (carbocation) H hv or ∆ H H +N 103˚ CNN C 2 C H H H carbene and N2 gas filled sp2 hybrid orbital (lone pair/carbanion) resonance structures of diazomethane:

H H H CNN CNN CNN H H H carbon has carbon has cation character anion character

Examples of other diazo-compounds: (brightly colored, but dangerous!!)

H3C O N2 N2 H3C N2 H3C O

4,4-dimethyldiazocyclohexa- diazocyclopentadiene methyl diazoacetate 2,5-diene (a purple liquid) (an iridescent orange liquid) (a yellow liquid)

Carbenes react with Alkenes to form Cyclopropanes

H CH2N2 CH2 hv or ∆ H

N2 H Stereochemistry of the H3C H H3C alkene is retained in the cyclopropane product H CH3 hv or ∆ H CH3

H CHBr3,KOH Br C Br H

One-step stereospecific syn addition mechanism

36 Carbenes: Molecular Orbital Interactions with Alkenes Consider what orbitals are available to interact for bonding:

H H alkene π∗ = LUMO empty 2pz orbital = LUMO (acceptor) (acceptor) H H H C H H alkene π = HOMO H filled sp2 hybrid (donor) H H orbital = HOMO (donor)

Two stabilizing (bond forming) interactions are possible: Stabilizing Interaction #1: Stabilizing Interaction #2: empty p orbital of H carbene (acceptor) H lone pair of carbene (donor) C H C H

empty p alkene π* H H orbital H H alkene π 2 H H sp H H carbene alkene π (donor) alkene π* (acceptor)

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:

H H3C H3C Cl

Cl H3C H3C H

N2 H3CO

37 Determining the Structures of Organic Molecules

How do you identify the products you synthesize?

1) BH3 OH H O, H SO 2) NaOH, H2O2 2 2 4 OH H C 3 H3C H3C CH3

How can you tell which reagents give which products? How can you tell if only one major product is formed?

CH CH CH3 3 3 Does the nitro group add once NO2 or twice? HNO3 -OR- H2SO4 Is addition selective for the

NO2 NO2 ortho and para positions?

Chromatography - Good for separations, but not for identification….

2.05 What is the structure Chromatogram: min of this compound?

All we can tell here is 4.37 that the compound is min 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

CH2 CH2 H2CCH2 H2C H2C π orbitals π orbitals of π orbitals of of ethene 1,3-butadiene 1,3,5-hexatriene See table 8.3 Bruice p. 325 λmax (nm) 165 217 256

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 N H3C 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

Rhodopsin H H N H O (500 nm) 11-cis-retinal opsin hv

H 11-trans-retinal release + N opsin opsin H

nerve impulse trans-alkene = vision

42 UV/Vis Spectroscopy: Sample Problems

1) Match the following UV absorption maxima with the corresponding compounds: 353 nm, 313 nm, 256 nm, 227nm, 180nm

CH 3 CH3 CH3 H C 3 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 fragmentation R X R X R + X electron beam dislodges electron molecular ion cation fragment radical fragment radical cation observed in MS NOT observed in MS 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 CH CH CH CH CH 3 2 2 2 3 (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 H C + R H2C CH CH2 R H2CCHCH2 R 2 m/z = 41

70 eV RZCH2 CH3 RZCH2 CH3 RZCH2 + CH3 Z = N, O, S R can also be H

R 70 eV R C O C ORCO+ R R R acylium ion

44 Fragmentations at Functional Groups

• The weakest bond is the C–Cl bond

• Both heterolytic and homolytic cleavage heterolytic cleavage of the C–Cl bond occur.

Cl isotope has 1/3 α-cleavage homolytic cleavage the abundance

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

• 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

1) Alkanes, alkenes and aromatics: cleave to give the most stable carbocations 2) Alcohol: loss of water or α-cleavage to give stabilized cation 3) Ethers: loss of an alkyl group or α-cleavage to give stabilized cation 4) 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

H3CCH3

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 Fingerprint region (4000-1400 cm-1) (1400-600 cm-1) Peaks are similar for Pattern is unique for each functional group 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)

H H

Cl Cl H3C CCCH3 H H 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)

CH 3000 cm-1 CD 2200 cm-1 increasing wavenumber CO 1100 cm-1 CCl 700 cm-1

2) Stronger bonds = larger wavenumbers (higher frequency)

a) Higher bond order = stronger b) Hybridization: more s character = stronger -1 triple bond CN 2200 cm sp CCH 3300 cm-1 -1 2 double bond CN 1600 cm sp CCH 3050 cm-1 -1 3 single bond CN 1100 cm sp CCH 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

OO stronger, less O weaker, more flexible C=O flexible C=O bond > > bond

C=O stretch 1788 cm-1 1718 cm-1 1691 cm-1 (ring strain) (resonance)

stronger C=O, O O O O weaker C=O, > more double R > > R less double bond RH RR RN bond character RO character H ester aldehyde ketone amide -1 -1 C=O stretch 1740 cm-1 1730 cm-1 1720 cm 1660cm (approx.) (inductive (resonance) withdrawl)

Stronger C=O because Resonance No resonance to weakens weaken C=O bond 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 Also note the intensity -1 C=C stretch at 3080 cm and position difference at 1650 cm-1 of the CC double and sp3 C-H stretch triple bond. at 2850-2950 cm-1

CC triple bond Even when the intensity stretch at 1650 cm-1 is diminished, the shift of the sp3 C-H is distinct from the C-H of an aldehyde because it is O-H sp3spC-H3 C-H stretch stretch higher frequency and no stretch at 2800-2950at 2850-2950 cm cm-1 -1 carbonyl peak is present.

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. Look in 1800-1600 cm-1 range for strong sharp peak indicating a carbonyl, consider its relative position. 2. Look for secondary peaks to distinguish between carbonyl compounds. For example, a broad OH peak (3300-2500 cm-1) to indicate a carboxylic acid. 3. Look in 3650-3200 cm-1 range for strong broad peak indicating alcohol or amine. 4. Look for C-O and C-N peaks in 1250-1000 cm-1 range, indicating an ether or tertiary amine. 5. 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) 6. 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 7. Consider symmetry that would account for “missing” absorption bands 8. 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. Given an IR spectra for an unknown molecule, identify three functional groups present in the molecule 2. Rank and/or explain which bond has a lower frequency (lower wavenumber) for a series of bonds given 3. For several carbonyl compounds given, be able to identify which carbonyl group exhibits the highest wavenumber. 4. Identify which compound has a vibration that is IR inactive 5. Know how to distinguish between a pair of compounds using IR data. 6. 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 O O

H3C H3C H3C CH3 CH3 CH3 Cl FCH3

53 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) Number of peaks: tells the number of different types of protons (or carbons) 2) Relative area of the peaks: tells the relative types of different protons 3) Position of the peaks: tells the chemical environment of the proton, ie. the neighboring functional groups 4) Peak splitting pattern: tells the number of protons on adjacent atoms

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)

- - - + - - - aligned against the field: +

+ + + + higher energy β nuclei

+ + + + - +

+ energy aligned with the field:

+ -

- - - + - - lower energy α nuclei + (slightly favored) In the absence of an In the presence of an + pole applied magnetic field B o applied magnetic field Bo Bo - pole

54 NMR Theory

Radiowaves supply enough energy to “flip the spin” of a nucleus

Higher energy - - - Energy - - - - Energy - - - β nuclei

+ + + supplied + + + + released + + + + + + + + + + + + + + + +

energy

------

- - Lower energy - α nuclei

+ pole Absorption of energy Spin “relaxes” to equilibrium Bo can “flip” the nuclear state through the magnetic - pole spin - converts low field and energy is released, energy to high energy 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 ethyl group methyl group + methyl group isopropyl group tert-butyl group + methyl group

“vinyl” protons “aryl or aromatic” protons

55 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 time- averaged for the molecule (10-3 s)

Cyclohexane example:

H chair-chair interconversion H H (ring flip) (12.1 kcal/mol energy barrier) 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 more electron density = less density = more shielded 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 CH CH CH3 CH3 3 CH3 3 O H3C

2.6 ppm 2.4 ppm 2.2 ppm 2.1 ppm 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 HHR 5.0 ppm but isn’t this a rather big effect for just an sp2 carbon center? R R Is there another effect?

57 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 Alkene (5-6 ppm) Alkyne (7-8 ppm) (2.5 ppm)

Bo -Binduced B + B o induced Protons are shielded Protons are deshielded (at lower frequency) (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 Protons in more electron poor electron dense environment environment R CCH 2.4 ppm 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) How many signals would you expect to see in the 1H NMR spectrum for each of the following compounds? 2) For each molecule, indicate which protons are least shielded, ie. will give the NMR signal with the highest chemical shift value (furthest downfield).

O Hb H2 H2 H F C Br O C N C C H3C C C CH3 H3C CH3 H2 H2 H2 H2 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 Quartet Triplet, N = 2 (N = 3) Doublet Quartet, N = 3 (N = 1) Quintet, N = 4 Sextet, N = 5, etc

Triplet (N = 2) 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

doublet Coupled protons have the quartet same coupling constant

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 Diamagnetic anisotropy accounts for the chemical shifts of aryl and vinyl carbons at H higher frequencies (downfield) 100 - 150 ppm

Electron-withdrawing effects of an electronegative Z 40 - 80 ppm H3C 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 and 13C NMR Shift Comparison

1H NMR shifts:

δ (ppm)

13C NMR shifts:

δ (ppm)

From Organic Chemistry Textbook by Wade

Types of NMR Questions

1) How many signals will be in an 1H NMR spectra for a given molecule (how many protons are chemically equivalent)? 2) How many signals will be in an 13C NMR spectra for a given molecule (how many carbons are chemically equivalent)? 3) 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. 4) Assign a spectra - which proton accounts for which peak, and what is the splitting pattern 5) 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 CH C 3 H2

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?

X CH3 H3C X X2, hv 1) -OR- X = Br or Cl

CH H3C Cl CH3 2 KOH, EtOH 2) -OR-

CH 3 CH H3C Cl 3 HCl Cl 3) -OR-

66 “One C=O bond to Rule them All”

nucleophilic Carbonyl structure O O and reactivity: C electrophilic

Substitutent (Z) on carbonyl determines reactivity:

O O O O O O O O R R RCl> RH>>RR ROR >>R O R N > RO acid ester H carboxylate chloride aldehyde ketone anhydride or acid amide ion Most Most electrophilic increasing reactivity nucleophilic carbon with nucleophiles oxygen Most inductive electron-withdrawing Most resonance

Introduction to Carbonyl Compounds

H3CCH3 1. O3 H3C CH3 Ketone or aldehyde synthesis O + O using ozonolysis 2. H O H3C H 2 2 H3C H

H2O, H2SO4 O Ketone synthesis from an internal alkyne H3C CCCH3 H3C 1. disiamylborane CH3

2. H2O2, NaOH O

H2O, H2SO4 H3CCH3 HgSO4 Ketone or aldehyde synthesis H3C CCH O from a terminal alkyne 1. disiamylborane H3C 2. H2O2, NaOH H

O O Synthesis of an aromatic ketone AlCl3 CH using Friedel-Crafts acylation H C Cl + 3 3 with an acid chloride

67 Introduction to Carbonyl Compounds

O O O O O O O O N R R C RH RR ROHR O R N RCl ROR R carboxylic H acid or acyl aldehyde ketone acid ester amide chloride anhydride nitrile

carbonyl oxygen O O O Cyclic ester = lactone R' RO O NH Cyclic amide = lactam 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 O O O O N C H3CHH3COHH3CClH3COCH3 H3C NH2 H3C acetaldehyde acetic acid acetyl chloride acetic anhydride acetamide 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

major minor high energy structure, contributor contributor do not ever draw this!!!

Carbon is an electrophile (has Lewis acid character) Oxygen is a nucleophile (has Lewis base character)

68 Molecular Orbitals of C=O Double Bond

Compare the molecular orbitals of alkene π* C=O π* the C=C and C=O double bond: (LUMO) (LUMO)

The C=O bond is unsymmetrical!! alkene π C=O π (HOMO) (HOMO)

π* CO

The 2p orbital of oxygen is lower C 2p energy (because oxygen is more electronegative) and contributes energy O 2p more to the π molecular orbital. The higher energy 2p orbital of carbon contributes more to the π∗ π CO molecular orbital.

Resonance vs. Inductive Effects

O Most inductive electron withdrawing effects RCl

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

H Protonation of the carbonyl oxygen makes any carbonyl O + O H a better electrophile (more susceptible to nucleophilic RX RX attack) because the C=O LUMO is lowered 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 BF3 O Lewis acids = BF3, AlCl3, ZnCl2, FeBr3 RX R X

The proton and Lewis acid both have an electron-withdrawing effect. Any electron- withdrawing 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:

O O O O O O O O R R RCl> RH>>RR ROR >>R O R N > RO acid ester H carboxylate chloride aldehyde ketone anhydride or acid amide ion Increasing reactivity with a nucleophile

Nucleophilic Acyl addition elimination Substitution: Y = good leaving groups carbonyl product Z = oxygen, nitrogen, hydride and carbon nucleophiles

Nucleophilic Addition: addition Z = oxygen, nitrogen, hydride and carbon protonation nucleophiles a tetrahedral tetrahedral intermediate product

Nucleophilic Acyl Substitution Reactions

General mechanism: Y:- must be a good leaving addition elimination 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 A weaker base is Cl <

See Bruice Table 17.1 on page 672

O O Ketones and aldehydes DO NOT undergo nucleophilic acyl substitution because there is not a good leaving group that R H R R' can eliminate. (However, remember that they are great electrophiles and we will (R’ = alkyl or aryl group) 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:

O OH H H2O CH3OH NH2CH3 N H3C CH3 OCH3

72 Reactions of Acid Chlorides

esters

acids

O O O O + anhydrides H3CClOCH3 H3COCH3

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 CH3 + N No reaction to give amide products H CCl 3 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

H O N O + N Cl- + + H CCl H3CNH2 H3CNCH3 3 H (1 equiv) (1 equiv)

A tertiary amine will still add to a carbonyl, but since NO O deprotonation can occur, it will always be the best leaving group H3CCl

Cl- O

O N O + + + H CNH NCH3 H3CCl 3 2 H3CNCH3 H Cl- (1 equiv) (1 equiv)

H3CNH2

H H O O+ N Cl- N OH

+ + N H3CNCH3 H3CNCH3 N CH H H 3 H2 Cl- H3C

Sample Problem: Mechanism Provide a mechanism for the following conversion of an acid chloride to an ester:

O O + CH3OH H3CCl H3COCH3

Problem solving When nucleophile is neutral: When nucleophile is an anion: strategy: Step 1 = addition Step 1 = addition Step 2 = deprotonation Step 3 = elimination 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

H CH OH CH O H C +O S 1 3 3 H 3 N + H3C H3C H CCH+ O R O R O R 3 3 H3C H3C tertiary carbocation H2O Provides both an acid and an alcohol product CH B: CH 3 3 H C H C 3 H 3 H C O H C OH 3 3 H

Once the carbonyl oxygen is protonated, the carboxylic acid is an

excellent leaving group for an SN1 reaction

76 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

O H O H+ H3C NHCH2CH3 + HNCH2CH3 H2NCH2CH3 HO H3COH

HCl, H O O O 2 H+ N H NOH H NOH O H 2 3 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 H bond N R5 H O R1 O R3 O R5 H H H H R4 O N N N N N N N N H O H 2 H 4 H 2 O R O R O R3 O R N H amino acids linked by amide bonds: Linear Structure N H O O N R1 Amide bonds are difficult to hydrolyze, making O them perfect for proteins. alpha helix Of course, there are with hydrogen-bonds enzymes that can hydrolyze the amide bond quite readily!! protein structure

Reactions of Carboxylic Acids

1) Acid-base reactions (recall pKa’s)

No amides will O O CH H CH3 3 + N + H2N form because H COH H CO the amine will 3 CH 3 CH3 3 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 I 2 O + H I O CH2Cl2 O I I I

OH O+ O O H I+ I I- 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

O Br NaCN CN HCl/H2O H3C H3C H3C OH SN2 ∆ butanenitrile Butanoic acid (propyl cyanide) (butyric acid)

Acid-catalyzed hydrolysis to a carboxylic acid is even more difficult than an amide hydrolysis

H2, Pd/C Reduction of a nitrile to give an amine: CN H CNH H3C 3 2

(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. OCH O Cl O 3 Hint: Since the carbonyl is the O electrophile, the rate is CH3 O CH O CH3 O increased by anything electron- 3 withdrawing and decreased by anything electron-donating

H O O CH 2) Rank the relative energy level of the 2

LUMO for the following C=C and C=O ROCH3 ROCH3 H3CCH3 bonds, where 1 = lowest LUMO (most electrophilic) and 3 = highest LUMO

3) Show three ways to prepare an amide from N,N-dimethylamine

80 Lidocaine Synthesis: Nucleophilic Acyl Substitution vs. SN2

Nucleophilic O O acyl subsitution RNH2 H Cl R N Cl N R (excess H amine)

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

CH 3 CH3 O Step 1: O NaOAc Cl NH2 + Cl N Cl H CH 3 CH3 1 equivalent amine

Step 2: CH CH3 3 O CH3 O NaI, Et NH Cl 2 NCH3 N N H H Lidocaine CH 3 CH3

Synthesis of Aspirin

An experiment from Chem E2a lab: Acyl transfer with acetic anhydride

O O O CH3 O OH H3COCH3 O + H OH H3CO H3PO4 OH O O Salicyclic acid Aspirin™ • Active ingredient in wart (acetylsalicylic acid) 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 O H2O O O + H H H3COCH3 H3CO O CH3 Acetic anhydride Acetic acid

81 Sidenote: Aspirin Mechanism of Action

The reverse acyl transfer from Chem E2a lab:

Cyclooxygenase O CH3 Ser Cyclooxygenase O OH OH + Ser OH enzyme-mediated OH O O O H3C O Aspirin™ salicylic acid

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?

Planar structure allows good C N overlap between lone pair on N (in N N O H O H OH p-orbital) and the C=O π bond to give good resonance stabilization β-lactam 3-D picture = planar

CH3 H N H H S S Ring constraints of fused β-lactam CH3 CH3 structure force lone pair to be almost O C N CO2H N CH O O 3 perpendicular to the C=O π bond OH and resonance stabilization is lost O penicillin G 3-D picture = bent

82 Sample Problems

General 1) Provide reagents or products in a reaction scheme (or short syntheses) question 2) Know mechanisms and relative reactivity trends for Nucleophilic Acyl Substitution types: 3) Remember IR and NMR data relating to the C=O bond 4) 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.

R H S CH3

N CH O 3 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.

CH3 H H N H H S CH3 N S CH3 O N CH CH3 O O 3 N CH O 3 OH O OH O penicillin G methicillin More bulky aryl group

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 H3COCH3 starting material desired product

Reactivity of Aldehydes and Ketones

Recall order of reactivity of carbonyls with nucleophiles:

O O O O O O O O R R RCl> RH>>RR ROR >>R O R N > RO acid ester H carboxylate chloride aldehyde ketone anhydride or acid amide ion

Increasing reactivity with a nucleophile

O O O > > HH RH RR formaldehyde an aldehyde 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 Two Factors: Electronic and Sterics

O O O O O Very > > > > Less CH3 H3C CH3 reactive HHH3CHH3CCH3 H3C reactive CH3 CH3 CH3

Reactivity based on stability and sterics: stability more Reactivity determined by sterics: important than 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 tetrahedral intermediate 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 Very polar an alkyl bromide: 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 1) OH Methyl 2.5 equivalents OCH3 Triphenylmethanol benzoate 2) H2O

Nucleophilic PhMgBr H O acyl substitution (1st equiv) 2

O PhMgBr OMgBr (2nd equiv)

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 Br Mg, Et O MgBr 1. CO2 2 OH 2. HCl, H2O

bromobenzene benzoic acid

Useful synthetic methods to recall: Two methods

1. BH3 to synthesize H3C OH Hydroboration/oxidation H3C a carboxylic 2. H2O2, NaOH acid PBr OH 3 Br Alkyl bromide formation H3C H3C O SN2 and NaCN HCl, H O Br CN 2 Nitrile H3C H3C H3COHhydrolysis

CH Br CH3 Br Br2, hv 3 2 FeBr (or NBS) 3 Br

Radical Electrophilic aromatic bromination substitution: bromination

87 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.

H3CCH3

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 1) NaNH2 (or nBuLi) S 2 with CH3 N alkyl halides H3C 2) H3C H3CBr

H 1) NaNH2 (or nBuLi) OH Epoxide H3C 2) O H3C opening

88 Hydride Ion as a Nucleophile

General mechanism: reduction of ketone to secondary alcohol by addition of a hydride ion, followed by protonation

Tetrahedral products

What are sources of hydride ion “H:–”? CH3 – NaBH4 = sodium borohydride - four “H: ” equivalents CH3 CH3 Dibal-H = diisobutylaluminum hydride - one “H:–” equivalent Al H C – 3 H LiAlH4 = lithium aluminum hydride (LAH) - four “H: ” equivalents Dibal-H Relative ease of reduction:

Most O O O O O Least reactive >>> > reactive RH RR RORRNR2 RO

NaBH4

LiAlH4 -most reactive

H NaBH4 Reductions Na HHB 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 Li LiAlH4 reductions HHAl 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

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 Why is Dibal-H less reactive than LiAlH ? Dibal-H 4 CH3 CH3 1) As a neutral species, it is less nucleophilic than Al the anionic aluminum species H3C H 2) It is more sterically hindered

Reduction Sample Problems

Provide products for each of the following reactions:

1. NaBH4 OO 1. LiAlH4

2. HCl, H2O HCH3 2. HCl, H2O

1. NaBH4 OO 1. LiAlH4

2. HCl, H2O 2. HCl, H O HOCH3 2

1. Dibal-H OO 1. LiAlH4 2. H O CH3 2 H3CO N 2. HCl, H2O

CH3

O 1. LiAlH4 O 2. HCl, H2O

91 Alcohols Can be Oxidized to Carbonyl Compounds

[O] Reduction: Oxygen Oxidation: Oxygen content decreases Oxidation O content increases OH R and/or hydrogen R and/or hydrogen content increases Reduction H content decreases [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 HO Cr OH CrO3/H2SO4 = H2CrO4 O Na2Cr2O7/H2SO4 = H2CrO4

Mechanism:

Proceeds by substitution and an E2 elimination

92 PCC Oxidation of Alcohol to an Aldehyde

Chromic acid will oxidize a primary alcohol all the way to a carboxylic acid:

H+ OH H2CrO4 O H2CrO4 OH H C O 3 H3C OH H3C H3C H O H2O H 2 H OH Primary aldehyde hydrate Carboxylic acid alcohol

The oxidation of a primary alcohol can be stopped at the aldehyde if pyridinium chlorochromate (PCC) is used as the oxidizing agent

H N N Š CrO3 + HCl + CrO3Cl

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 S O Cl Cl Cl O O Me Cl S Cl O Me S Me Me Cl O Cl S Me Me O O oxalyl DMSO Me dimethylchlorosulfonium chloride ion R Cl H Cl H O O R O R O H S NEt3 R H R OH Me Me S S H S Me Me Me triethyl- Me CH2 aldehyde SN2E2amine

S + Me Me dimethyl- Proceeds by substitution sulfide and an E2 elimination Dimethylsulfide is: Note: also converts secondary alcohols to Ketones 1) A great leaving group 2) Super Stinky!!!

93 Sidenote: Analysis of Blood Alcohol Content

H SO 2 4 + H Cr O H3COH+ Na2Cr2O7 H3CO 2 2 3

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 green color color

Synthesis Sample Problems

1) Provide a synthesis of Ibuprofen from the starting matieral provided:

H3C CH3 H3C CH3 O

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 CH3 H3C CH3 Desired product

94 Reactions of Aldehydes and Ketones with Nitrogen Nucleophiles: Imines from 1° amines

H H R H O + R R O H2N 2 R N "Imines" aldehyde 1° amine aldimine -OR- R R "Schiff R R R O H2N H2O + R N Bases"

ketone 1° amine ketimine

Bonding in an imine:

CH3 CH3 CN CH3

sp2-hybridized carbon sp2-hybridized nitrogen

Reactions of Aldehydes and Ketones with Nitrogen Nucleophiles: Enamines from 2° Amines

R R R HN R R R O R H2O + N aldehyde 2° amine R or ketone enamine

recall...

R R R R O OH "keto" tautomer "enol" tautomer

95 Mechanism of Imine Formation

deprotonation addition protonation

protonation

deprotonation elimination

Mechanism of Enamine Formation

96 Carbonyl Group Removal via Imine (hydrazone) Formation: The Wolf-Kischner Reduction

NH Me 2 H2N Me Me hydrazine NaOH Works with O H N NH2 any ketone ∆ H

ketone hydrazone 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 HO OH H O R R 2 R R

ketone or Water gem-diol or aldehyde "hydrate"

Recall... OH OsO /H O R 4 2 2 R 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 + RO OR OH H H O + R H R 2 R H

aldehyde alcohol acetal

O + RO OR OH H R R R H2O + R R

ketone alcohol 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 OH condenser HO

cold water MeO O

Dean-Stark pTsOH benzene Trap (reflux)

water level O O reaction flask 6.4g + H2O (0.54 mL)

MeO O

99 Applications of Ketals in Synthesis

OH O O O LiAlH ?? Me OH 4 Me OMe Me OH BOTH carbonyl groups LESS REACTIVE reduced NaBH carbonyl group 4 reduced?

OH O

Me OMe

MORE REACTIVE carbonyl group reduced

Ketals are Useful as “Protecting Groups”

O O O ?? Me OMe Me OH

pTsOH H O, more reactive OH D/S trap 2 HO HCl carbonyl group benzene de-protected

O LiAlH4 O O O O Me OMe Me OH

more reactive LESS reactive carbonyl group carbonyl group protected reduced

100 Applications of Protecting Groups with Alcohols

Re-call from E2a: TMS ethers are useful protecting groups for alcohols

OMe OH OMe MeI/NaH ?? Me OMe Me OH Me OH Me (excess) Me Me BOTH Alcohols converted MeI/NaH LESS REACTIVE to methyl ethers alcohol converted to methyl ether OH

Me OMe Me

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 O- XY +X Y- S S+ ylide resonance Me Me Me Me contributor DMSO

Ph H Ph P -

Ph CH3

Ylide Formation and the Mechanism of the Wittig Reaction

103 Wittig Reaction and Alkene Stereochemistry

stabilized ylide UNstabilized ylide - O O + + + Ph P Ph P Ph P Me 3 OMe 3 OMe 3 - -

O O + H Ph P OMe 3 OMe E-alkene predominates - O R H

R H Me

H + Z-alkene predominates Ph3P Me - R H

Sample Problem Propose a synthesis of the following product from the indicated starting material

Ph OH Ph OH

104 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 Br HBr RSH S O

R Carbonyl C=O NH HCN left intact R O O

R N C N R

Grignard Reagents Prefer 1,2-addition (direct), Cuprates Prefer 1,4-addition (conjugate)

O O HO R

1) R2CuLi 1) RMgBr

+ + R 2) H3O 2) H3O

106 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 α- Carbon 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 H H H3C O H3C O- + H+ H3C O H3C O- + H+

alkane: pKa = 50 ketone: pKa = 20!! O O H H H H H3C H3C - H3C H3C - H + H+ H + H+ H H H H

107 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 O + Br Br HBr H3C Br H3C

Enolates are carbon nucleophiles

O O + H3C R X + X H3C R

O O O OH + H3C H R H3C R

Keto-Enol Tautomerization

109 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 Me + + CH Me N Me H C Li Me N Me H C 3 H 3 Li 3 DIA LDA

111 Alkylation of Ketones, Esters, and Nitriles

Note Similarity to…

R n-BuLi R R H3C Br H Li

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' 'thermodynamic' enolate enolate CH CH 3 3 O OLi OLi H H H3C N CH3 CH3 LDA CH3 CH3 Li H + LDA 1.1 equiv LDA: major minor (Low Temperature) 0.95 equiv LDA: minor major Mechanism of Equilibration: (Higher Temperature)

CH3I CH3I O OLi H H O O CH3 CH 3 H C CH CH H + 3 3 3 CH3 OLi O CH 3 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 O

CH LDA CH3 CH3 MeO 3 MeO + MeO PhCH Br 2 H H Ph Ph

Ph O O O O Li O O O O CH CH3 CH3 3 LDA Br O N O N O N CH3 CH I O N H H 3 Ph Ph CH3 H i-Pr i-Pr i-Pr major minor H3C (diastereomeric proiducts can be easily separated) single enantiomer: chiral auxilliary Sm(OTf)3/MeOH derived from valine O O O Me Cl H CH3 O N + MeO Et3N H i-Pr Ph single enantiomer

Synthesis of Evans Auxilliaries

O phosgene (1 equiv) O CH3 O CH3 Cl Cl + Et3N H LAH O N H3C OH H3C OH OR: Cl O Cl NH NH Cl Cl 2 2 i-Pr occurs in nature as one Cl O O Cl enantiomer. "triphosgene" (0.33 equiv)

O Synthesis of the other enantiomer Me H of chiral ester would require the Ph O N enantiomeric chiral auxilliary... OH NH2 norephedrine Ph 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 The Aldol Addition

• It was first observed as a dimerization of aldehydes to produce a molecule that was named “aldol” because it was a β- hydroxy 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 OLi O OH LDA RCHO

CH3O CH3 CH3O CH3O R one new stereocenter created ("acetate" aldol product)

OLi O OH CH3O

CH3 MeO R O E(O)-enolate CH LDA RCHO 3 anti-aldol product CH3O CH3 OR

OLi O OH CH3 CH3O MeO R

Z(O)-enolate CH3 syn-aldol product two new stereocenters created ("propionate" aldol product)

Polyketide Natural Products are Synthesized by Successive Aldol Additions

H3C CH3

H C CH3 O OH OH O OH OH 3 OH OH CH3 H C Me HO 3 O OH H3C H3C H3C OH CH3 CH3 CH3 O OH CH3 7 "propionate" units make up the skeleton of erythronolide Erythronolide B

O Me H O OH O OP O OH OP O CH3 CH3 CH3 Me RO H RO RO CH3 CH3 CH3 CH3

118 Update: The Aldol Reaction is Nothing More than a Special Case of Carbonyl Addition…

"Carbonyl Addition" - RMgBr H O Nu OH Nu = Li R H R Nu R

The Aldol Reaction OLi O Nu = R OH O

R H R R

Sample Problem

119 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 O OH O LDA R HCl/H2O R R RCHO NaOH/H2O Me ∆ Me aldol addition aldol condensation

NaOH/RHCO ∆ HEAT ∆

• 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 O O Ph Ph Ph Ph Ph3P PPh3 Ph Ph Ph Ph Wittig 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…

K O O O O KOH + Ph H Ph Ph Ph Ph Ph Ph Ph O O Ph K O EtOH O O O H Ph Ph Ph K Ph Ph Ph Ph EtOK O O O Ph Ph Ph O O H Ph Ph H 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 β- carbonyl compounds

121 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.

O O O 2 NaOCH3 H3C H3C + CH3OH OCH3 OCH3

CH3

122 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 O O O CH3 NaOEt H3C O H3C O H H H3C O Me

α-proton removed NO α-protons! to form enolate

O O O H O CH CH3 NaOEt 3 H H3C O H 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 Update: The Claisen Condensation is Nothing More than a Special Case of Nucleophilic Acyl Substitution…

"Nucleophile Acyl Substitution"

O Nu O R NH R Nu = OH H2O R X R Nu R

The Claisen condensation

ONa O Nu = R O O

R X R R

Sample Problem

124 Michael + Aldol = Robinson Annulation

Decarboxylation of Dicarbonyl Compounds

β-carbonyl acids will lose CO2, or “decarboxylate,” when heated…

H CO O O O O OH O 2 O C R OH R O R R CH O 3

Recall... O O HCl/H2O R OMe R OH

Hydrolysis of a β-keto ester results in decarboxylation…

CO2 O O O O O HCl/H2O

R OMe R OH R CH3

125 Malonic Ester Synthesis: Alkylation of a Dicarbonyl followed by Decarboxylation

""X O + R H3C OH

O X O Base + H3C OH R 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 O O 2 NaOCH3 H3C H3C OCH3 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 H3C S NADPH Acyl Transferase ACP ACP

acyl carrier protein Keto- reductase KR

O OH O DH NADPH NAD+ H3C S H3C S Dehydrogenase ACP ACP

Enoyl- ER reductase O

H C S NAD+ 3 ACP

128 Carbonyl Chemistry on One Slide

O O O R1 R1 R1 R OMe R H R R N N R - R R R [H-] [O] [H ] [O] R1 O OH OH R1O OR1

R OH R R R R R R R

O Nu O R1 O O Claisen R Nu R X R R1 OM

OH O 1 OH O Nu R Aldol R Nu R R R R1 R R OM O Nu O O Nu R1 R1 O Michael R R R R R R

Sample Problem

129 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 dienophile

1 1 2 CH2 6 2 6 CH2 ∆

3 CH2 CH2 3 5 5 4 4

3 π-bonds 1 π-bond 4 σ-bonds 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

LUMO is lowered by the electron-withdrawing group

y (CO2Me)

Energ lower ∆G(HOMO/LUMO) = lower ∆G‡(reaction) HOMO

ethene H O methyl H OMe acrylate 1,3- H butadiene

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 CO Me CO Me OMe ∆ 2 2

H Me Me Me H H cis-dienophile cis Diels-Alder products (Z-alkene) (enantiomers)

O H H H CO Me CO Me OMe ∆ 2 2

Me H H H Me Me trans-dienophile trans Diels-Alder (E-alkene) products (enantiomers)

Diene Stereochemistry/Syn Addition Transition States: Top Face or Bottom face Addition gives Enantiomers

H H H H MeO2C H H CO2Me MeO2C H H H Me H Me H Me H H H H CO2Me H H Me H

H MeO2C H MeO2C H CO2Me Me H H H H Me H H Me H H H

132 The Diels-Alder Reaction is a Syn Addition: Diene Stereochemistry

Me Me Me CO2Me H CO Me CO Me H 2 2 H H CO2Me CO2Me CO2Me Me Me Me E-E-diene cis Diels-Alder products (enantiomers)

Me Me Me CO2Me CO2Me CO2Me H H Me CO2Me CO2Me H CO2Me Me Me H trans Diels-Alder products E-Z-diene (enantiomers) 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

LUMO

y HOMO is RAISED by adding an Electron Donating Group (OMe) Energ

lower ∆G(HOMO/LUMO) = HOMO lower ∆G‡(reaction)

ethene H H 1-methoxy-1,3- H butadiene 1,3- MeO H butadiene H

133 Regiochemistry of the Diels-Alder Reaction is Predicted by the Substituents of the Diene and Dienophile

1. Draw out resonance contributors + OMe OMe OMe O - O O H H H H H H + H OMe H H - OMe OMe H H H H H H H H - - H + H H + H H H H

2. Match up charges (opposites attract) OMe O OMe + - OMe CO2Me

OMe O - + H H H OMe H OMe H H H OMe + + H X X OMe - - CO2Me O

Diene Conformation: “s-cis” Required for the Diels-Alder Reaction

H H H H H H Diels Alder Reaction H free rotation H H H H H s-trans conformation 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 CO2Me "exo" (substituent of dienophile is oriented away from double bond) H H CO2Me H CO Me H 2 Me H Me CO2Me H H H O H H H H OMe

H H H CO2Me H H H H MeO2C H Me H CO2Me H H MeO2C H "endo" (substituent of dienophile Me H Me H is oriented toward double bond) H 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 +

135 Introduction to Pericyclic Reactions

1) Polar/Ionic Reactions: Heterolytic bond cleavage, formation of anionic or cationic intermediates Aldol O O reaction OOH H C 3 HCH3 H3CCH3

2) Radical Reactions: Homolytic bond cleavage, formation of radical intermediates

H H H H Cl Cl 2Cl HCl + 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 H BH H O O H C 2 3 H3C BH2

O H O Hydroboration Diels- Alder reaction

transition state Pericyclic Reactions

A concerted, single barrier, one-transition state reaction

starting AB(intramolecular) material(s) A + C B (intermolecular) energy

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

•OrbitalsMUST 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)

energy levels (ground state) (excited) thermal 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

CH CH3 ∆ CH3 3 +

dieneophile diene cyclohexene transition state product O O

H3CO ∆ H CO + 3

CH3 H3C trans-dieneophile 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 Suprafacial bond formation = most common) 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 Two π-bonds are H3C CH3 H C CH 3 3 hv converted to two + H3C CH3 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 CH3 H Me Me H3C H ∆

HOMO (excited state)

H H

H3C CH3H H Me Me

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

4 4 (note: do not worry a [1,5] sigmatropic rearrangement 3 5 3 about the 1,3 shifts in the book or previous 5 (a 1,5-hydrogen shift) lecture note copy, the 2 1 2 book has an error and the 1,3-shifts are less 1 useful so we will not discuss them.)

2 2 1 3 1 3

1 3 1 3 2 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

H COCH CH3 1. LDA 3 3 OH HO acetylation of O O 2. H3O+ alcohol with workup O acetic anhydride Ireland-Claisen forms ester Rearrangement Mechanism: O O Claisen O OH Rearrangement H3O+ OCH LDA 3 OCH2 O workup O enolate formation

[1, n] Hydrogen Shifts (H starts at C-1 and ends up on C-n)

4 4 1,5-hydrogen shift 5 5 3 3

2 2 1 1 One C–H σ bond broken, one C–H σ bond formed

1,7-hydrogen shift 3 2 32 1 4 4 1

5 5

6 7 6 7

144 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? 120°C b) What is the driving force of this reaction?

2. a) What is the product of this reaction? b) Describe, as specifically as possible, what type pf pericyclic reaction this is.

O 170°C

H3C CH3

3. Provide products and indicate what type of pericyclic reaction occurs.

170°C CH2 hv + O CH2 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

Nomenclature and Structure: Glyceraldehyde HOH OH Glyceraldehyde (an aldotriose) has a single stereogenic center and can exist in two enantiomeric forms:

Fischer projections: Horizontal lines are taken as HO CHO D-notation: coming forward and vertical HOHhydroxy is on lines are going backward H CH OH CH OH HO 2 2 the right side (R)-glyceraldehyde D-glyceraldehyde (Fischer projection)

HO CHO HO H L-notation: H hydroxy is on CH2OH CH2OH HO the left side (S)-glyceraldehyde 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

“D-series” “L-series” of sugars 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 a new stereocenter H is formed in the H H2O O hemiacetal product OH HO O Cyclic product is the hydroxy-aldehyde cyclic hemiacetal (open form) (favored) predominant form

H2O OOH H H H H derived from derived from O H2O O the alcohol the aldehyde O OH

Equilibrium exists in aqueous solution, or with catalytic acid

148 The Hemiacetal (cyclic) Form Predominates with Aldoses

β-glucose is favored because the anomeric OH is in an equatorial position beta = hydroxy up, equatorial OH (more favored) O CHO HO β-D-glucopyranose HO OH H OH OH OH H HO H OH HO β−D-glucose H OH HO O (64%) H OH OH anomeric center at C1 H OH CH2OH (0.02%) D-glucose O HO H open/acyclic form HO OH α-D-glucopyranose OH α-D-glucose alpha = hydroxy down, axial (36%) (more steric strain)

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 mutarotation HO If the α-anomer (axial) is O O HO H HO OH disfavored because of HO HO OH OH steric interactions, why do OH H we even see any of it?? α-anomer, β-anomer, OH axial (34%) OH equatorial (64%)

We see even more α-anomer (axial) for more electron withdrawing groups: increasing e- withdrawing ability HO HO X %axial O O HO H HO HO HO X OH 34 OH OH OMe 67 X H OAc 86 Cl 94

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 H X good overlap between X σ* C-X H no overlap axial lone pair of O and with equatorial axial σ* C-X bond C-X bond!

Look at σ* C–X bond in H X σ* C–X Newman projections: must have antiperiplanar X H orientation between lone σ* C–X No overlap!! pair and σ* C–X bond 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: CHO CH OH OH OH 2 H OH H OH O OH 1. NaBH HO HO HO H 4 HO H HO OH HO O H OH H OH OH OH 2. H2O H H H OH H OH β−D-glucose (0.02%) CH2OH CH2OH D-glucitol Reduction of a ketose: CH2OH CH2OH CH2OH O H OH HO H HO CH2OH O HO H NaBH4 HO H HO H HO + H OH H OH H OH H OH HO H OH H OH H OH CH OH CH OH α-D-fructofuranose CH2OH 2 2 D-glucitol open chain D-mannitol ketone A mixture of sugar alcohols (alditols)

More Reductions of Carbohydrates

(H2/Ni can also be used)

CHO CH3 H OH H OH H2NNH2 Reduces the aldehyde HO H HO H group all the way to a H OH NaOH, heat H OH methyl group H OH H OH CH OH 2 CH2OH

Wolff-Kishner reduction (via hydrazone)

151 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 + Cu O (Benedict’s 2 A red solid solution) Look for color change or mirror to show proof of an aldehyde (a positive test). these will NOT oxidize the 1. Ag O, NH 2 3 primary alcohol + o 2. H3O workup + Ag Silver mirror (Tollen’s test)

+ - 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 β- glycosides 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 α-1,4'-glycosidic linkage 4 6 6 5 O 4 H+ 5 O HO OH HO HO 1 H 1 2 ' 3 ' 6 HO 2 H OH 4 3 OH 5' O O HO 1' α-D-glucose HO ' H ' 2 (two equivalents) 3 OH HO β-1,4’-glycosidic linkage OH OH OH HO HO 4 6 + 6 OH 5 O H O + 45O 4 6 HO 1 5 O OH HO OH 1 O HO 2 HO 1 3 OH 2 OH 3 OH HO 2 OH H H H 3 OH β-D-galactose β-D-glucose H (a C-4 epimer of glucose) lactose

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) Be able to draw the cyclic hemiacetal form from a Fischer projection 2) Be able to identify and draw anomers and epimers; identify D- and L-sugars 3) Be able to fill-in products or reagents for a given reaction 4) Mechanisms you need to know: hemiacetal formation, mutarotation, enediol rearrangement, epimerization and glycoside formation 5) 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 OH OH HO CH OH O 2 O O HO HO HO HO H OH HO OH H OH OH OH O O HO H HO OH α-D-fructofuranose β−D-glucose OH maltose H If a disaccharide glycoside still has a hemiacetal, then it is still a reducing sugar 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 CH OH O 2 H OH O HO HO HO H HO OCH 3 H OCH2CH3 H OH OH HO H H OH methyl β−D-glucopyranoside ethyl α-D-fructofuranoside CH2OH All monosaccharide glycosides are NON-reducing sugars Sucrose D-glucitol

Amino Acids, Peptides, and Proteins

α-amino acid residue (amino acid -H2O)

O R O R H α H2N N OH OH H2N N H R O R O α-amino acid oligo/poly-peptide: (prominant in nature) amino acids linked by amide bonds

H2N O

β R α OH β-amino acid (uncommon in nature, studied by synthetic chemists)

protein (large peptide)

158 Amino Acids are Chiral and Charged

O Most amino acids have one stereocenter H2N OH All “natural” amino acids are L-configuration, R most have the absolute configuration S

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 O O O O H N H N H N H N H N 2 OH 2 OH 2 OH 2 OH 2 OH

CH3 H3C H3C H3C CH3 CH3

glycine alanine valine CH3 leucine isoleucine (Gly/G) (Ala/A) (Val/V) (Leu/L) (Ile/I)

O O O O H N H N H N H N 2 OH 2 OH 2 OH 2 OH S HO HO CH3 HS H3C serine threonine cysteine methionine (Ser/S) (Thr/T) (Cys/C) (Met/M)

O O O O H N H N H N H N 2 OH 2 OH 2 OH 2 OH

HO H2N

O O HO O H2N O aspartic acid glutamic acid asparagine Glutamine (Asp/D) (Glu/E) (Asn/N)) (Gln/Q)

159 Amino Acids 2: Basic, Aromatic, and Heterocyclic Sidechains

O O O O H N H N H N H N 2 OH 2 OH 2 OH 2 OH

HN HO

H2N H2N NH phenylalanine tyrosine lysine arginine (Phe/F) (Tyr/Y) (Lys/K) (Arg/R)

O O H O H N H N 2 OH 2 OH N OH N HN NH

proline histidine tryptophan (Pro/P) (His/H) (trp/W)

The Acid-Base Properties of Amino Acids are Important in Many Circumstances

160 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

O H O O H N N H N 2 OH OH 2 OH

CH3 HO CH3 2 proline threonine (Pro/P) alanine (Thr/T) (Ala/A) Hydrolysis O O

(6N HCl/reflux) H2N H N OH 2 OH

HO HN 2 tyrosine arginine (Tyr/Y) H2N NH (Arg/R)

O O H2N OH H2N OH HO OR H2N O Cyclonellin D O aspartic acid asparagine (a cyclic peptide (Asp/D) (Asn/N)) natural product) Readout from Amino Acid Analyzer

164 Synthesis of Amines: Reduction of Amides

H H H primary amines O R N (R1=R2=H) H R OH

SOCl2 H O N O H H 1 2 2 R R R2 LiAlH4 R secondary amines R Cl R N R N (R1=H) Et N 3 R1 H

H H R2 tertiary amines R N R1

Ammonia Cannot be used to Synthesize Amines from Alkyl Halides

NH3 H3C Br H3C Br H3C NH2 H3C NH (n-BuBr) H C Br- 3 + H3C N CH3 H3C N CH3

H3C CH3 H3C

O O O phthalimide; an "NH " synthon, 3 R NH R N R or synthetic equivalent 2 H amide imide

O 1) KOH 2) n-BuBr NH H3C NH2 1) H+/H2O 2) NaOH O

165 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)…

NaN3 H2;Pd/C H3C Br H3C N3 H3C NH2

Cyanide is a CH2NH2 Synthon (replaces Br with CH2NH2)…

NaCN H2;Pd/C NH H C Br H C H C 2 3 3 N 3

166 Reductive Amination: Conversion of Aldehydes and Ketones to Amines

O H

NaHB(OAc) N CH3 H 3 + H2N CH3

"H-" N CH3

imine intermediate

O CH H C N CH NaHB(OAc)3 3 + 3 3 N H CH3

H3C N CH3 "H-"

enamine intermediate

Synthesis of Amino Acids: From Acids and α-Keto Acids

O O O 1) Br /PBr NH OH 2 3 OH 3 O- + 2) H2O Br NH3 (+/-)-phenylalanine

O O O NH H ; Pd/C OH 3 OH 2 O- + O NH NH3 (+/-)-phenylalanine

Note: these syntheses will give racemic products

167 Synthesis of Amino Acids 2: From Aldehydes…The Strecker Reaction

O H 1) NH /HCN 3 O- O 2) H+/H O + 2 NH3 Imine (+/-)-phenylalanine NH formation 3 + H H /H2O N H N H H-CN N H H H -CN N+ H H Nucleophilic addition

Synthesis of Amino Acids 3: Amino Malonic Ester Alkylation

O O O

1) KOEt/BnBr - EtO OEt O 2)H+/H O NH + O N 2 3 CO2 O (+/-)-phenylalanine

O O

HO OEt KOEt H2N Ph Br O OK O O EtOH EtO OEt O O EtO OEt N O O H+/H2O O N Ph EtO OEt O H2N Ph

phthalic acid

168 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 O R2 O SOCl H2N 2 R2 R1 OH R1 Cl R1 N Et3N 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 H O O CH3 BocNH N BocNH OH OH H OH N + H2O H CH3 CH3 CH3 O

N O C 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

R O R' H O R O H + S + N + N H H+ H3N N H N N O R'' H H S peptide without the original N-terminal amino acid PTH-amino acid thiazolinone derivative 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 Met-Val-Arg Tyr-Leu-His

Chymotrypsin: Ala-Lys-Phe Gly-Asp-Trp Ser-Arg-Met-Val-Arg-Tyr Leu-His

Elastase: Ala Lys-Phe-Gly Asp-Trp-Ser-Arg-Met-Val-Arg-Tyr-Leu-His Cyanogen Bromide: Ala-Lys-Phe-Gly-Asp-Trp-Ser-Arg-Met 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 5 6 N N 1 8 2 N 4 N 9 a ribose- 5' 3 HO O phosphate 4' 1' 3' backbone OH 2' A nucleoside

179 Ribofuranoside Rings from Ribose

nitrogen base CHO R R R R HO OH HO H OH O N N H O H OH glycosidic H H H H H OH bond formation HO OH HO OH CH2OH β-D-ribofuranose a ribonucleotide D-ribose

nitrogen base CHO R R R R HO OH HO H H O N N H O H OH glycosidic H H H H H OH bond formation HO HO CH2OH β-D-2-deoxyribofuranose D-2-deoxyribose a deoxyribonucleotide

Phosphoester and Nucleobase Structures

180 Aromatic Amines and Basicity: 6-membered Rings

piperidine

+ H+ pyridine N+ N pKa=11.2 H H H sp3 + H+ + N N morpholine H sp2 O O pKa 5.2 + H+ + N N pKa=9.3 H H H The lone pair of pyridine is NOT part of the π-system of 2 N the aromatic ring. The nitrogen is sp 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 sp2 H sp2 pKa 5.2 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 + + H + H+ N N+ N + + H N N H H H 2 H H H sp H sp2 pKa -3.8 11.3

The lone pair of pyrrole IS part of the π-system of N H the aromatic ring. Protonation destroys aromaticity, and is therefore highly disfavored.

pyrrole pyrrolidine

N N sp3 DEprotonation of pyrrole to H sp2 H form an anion is facile. The resultant anion is very stable. pKa 17 36

Imidazole is Aromatic, and Behaves like a hybrid of Pyridine AND Pyrrole

pyridinium protonated protonated pyrrole imidazole H N N+ Aromaticity + + N N N+ N Retained H H H H H H pKa = 5.16 pKa = -3.8 pKa = 6.8

pyrrole imidazole Imidazole can serve as a weak N acid or as a moderate base/nucleophile. It is N N responsible for many important biological functions. H H pKa = 17 pKa = 14.4 O H N 2 OH histidine N (His/H)

182 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 -O P O O O -O HO P O ∆G = + 3.3 HO - HO + H2O HO O kcal/mol OH -O HO OH OH OH Not a favorable reaction

O O O P P P A O O -O O O O ∆G = - 7.3 -O -O -O O P P A O O O O + H2O O kcal/mol -O -O P OH + HO O- -O ATP OH ADP

Reaction becomes O favorable: OH P O -O O HO -O ∆G = - 4.0 HO + ATP O + ADP kcal/mol OH HO OH HO OH OH

184 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 OH- R Nu

TsCl/ Unstable leaving group Pyr

TsO- R Nu Nu: R OTs

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 + P ATP

Inactive Protein ActiveProtein

Amino acids with alcohol groups in their side-chains are subject to phosphorylation by kinases…

O O O H2N H2N H N OH OH 2 OH

HO HO CH3 serine threonine (Ser/S) (Thr/T) HO

tyrosine (Tyr/Y)

186 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 O N OH N NH N NH H H 2 keto enol imine amine

d H-Bond H H H a H a donor, d N N N X a H d a N N N H aad N O N O N OH B R R R H-Bond cytosine acceptor, a

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) External electrostatic binding: cationic complexes, usually metals, bind the negatively charged phosphate backbone and lead to disruption of the DNA structure 2) Groove binding: interactions with minor groove by electrostatic and hydrogen bonding 3) 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 O H3C N NH N NH N N NH2 N N NH2 O O H CI 3 O O H H O H H H alkylation O H O P O- 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 hydrogen- bonding and nucleophilic sites on the interior of the DNA double helix are sterically less accessible.

Recall: DNA Damage with other electrophiles

epoxide p450 hydrolase

HO O benzo[a]pyrene OH O p450

N NH O

N O N NH2 HO O P O O O- OH O diol epoxide O

N NH HO Cytochrome O p450 O N N N H O P O O O- benzene benzene covalently modified, oxide O damaged DNA

193 Recall: Nitrogen Mustards as bis-Alkylating Agents that Crosslink DNA CH3 S N Cl Cl Cl Cl sulfur mustard mechlorethamine toxic nerve gas used treatment of advanced Two methods: in World War II Hodgkin's disease

Nu: Nu: CH CH CH3 3 SN2 3 SN2 N N N Cl Cl Cl Nu Nu Nu intermolecular intermolecular

Nu: intramolecular CH H3C H3C 3 substitution (SN1) N N N Cl Cl Cl Cl Nu

aziridinium ion is Nu: a great electrophile!!

CH3 N Nu Nu bis-alkylation product

Recall: Bis-Alkylation of DNA = Crosslinking DNA Nu: intramolecular CH H3C H3C 3 substitution (SN1) N N N Cl Cl Cl Cl DNA

aziridinium ion intramolecular substitution (SN1) O DNA Nu: -O P O CH3 H O N H H DNA -O O O P O CH3 N O HN N O N N N H2N N NH2 O N N O H O P O O O- H H Crosslinked G-G residues of DNA double-helix O H OO-P O

194 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)

Block Transcription (eg triple RNA 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' U C T A A G A A RNAse (recognizes RNA- DNA duplex, destroys the Antisense RNA strand) Oligonucleotide 5'-T-A-C-G-T-A-T-T-5' (DNA-like) 3'-A-U-G-C-A-U-A-A-5' 3'-A-U-G-C-A-T-A-A-5' Ribosome, tRNA, etc X Ribosome

Protein

Chemically Modified RNA Analogs Are not Hydrolyzed, and are More Effective at Blocking RNA Translation

O O O base base base O O O O O base

H O O O O S N P O- P S- C H O P O O O O Me N base base base O base O O O Me

N O O O

Phosphodiester- Phosphothioate- Thioformacetal- Morpholino- linked DNA linked DNA linked DNA linked “DNA”

197 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 O O H N HO H OH 2 OH OH O O O

H2N H C O OH 3 OH

CH3 OH H3C OH O O

H3C O H2N OH OH H N 2 OH O O H O N CH HO N OH H C OH 3 H 3 O O O HO OH H3C N NH2 H2N NH2 H CH3 O O H N H N 2 OH 2 OH

CO2H

The Oro Experiment Demonstrated That Prebiotic Nucleobase Synthesis was Possible

In the presence of UV light, HCN forms adenine…

NH2

N N HCN N N H 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 O- Base HO HO H H H H

Formaldehyde Glycoaldehyde O (the simplest carbohydrate) H H O- OH O HO O H HO H HO H HO OH HO Ribozymes Ribose Threose NH2 NH2 N N N N N N N N HO OH H HO O O

OH 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 New Stereocenter Created... RMgBr H R 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 H cytochrome H p450 O

O O O O H H O aflatoxin aflatoxin O a) Enzyme-mediated oxidation generally take place on

O electron-rich alkenes. In the case of aflatoxin, why is only H one of the 3 possible alkenes epoxidized?

O 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 O

O O P O- O- 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

CO2H decarboxylative aromatization and NH2 transamination

L-Phe OOH O O O OH chlorismatechorismate HO OH CO H mutasemutase HO O 2 OH O NH O Claisen 2 OH O rearrangement OH OH chlorismic acid prephenic acid L-Tyr decarboxylative OH Count carbons to see which alkenes are aromatization, involved in the [3,3]-sigmatropic rearrangement oxidation, then transamination

[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 O HO NAD+ HO OH HO OH HO OH OH NADH D-glucuronic acid D-glucose

HO2C OH NAD+ = oxidant; NADH = reductant HO HO OH OH OH OH OH L-gluconic acid HO HO HO OH O HO O HO O O EKT O [O] O CH2OH HO OH O L-gluconolactone H OH Ascorbic Acid 2-oxogluconolactone HO H (Vitamin C) H OH Plants and most animals can convert glucose into ascorbic acid H OH (vitamin C) by the pathway shown. Humans and primates are deficient CO H in the enzyme oxidizing gluconolactone to the ketolactone, and are thus 2 dependent on a dietary source of Vitamin C.

Sample Problem: Ascorbic Acid’s Acidity

OH Ascorbic Acid (Vitamin C) is not a carboxylic acid, yet HO it is acidic enough to get “acid” in it’s name. Which O O proton is most acidic and why?

HO OH Ascorbic Acid (Vitamin C)

203