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Organometallic Reagents: Grignard Reagents Chapter 13-1 Grignard Reagents Grignard reagents are organometallic reagents derived from an alkyl halide and magnesium R—X + Mg diethyl ether Rδ−—Mgδ+X Grignard reagent Since the carbon carries a partial negative charge, the carbon is a strong base and a good nucleophile. CH3CH2—Br + Mg diethyl ether CH3CH2δ−—Mgδ+Br HOH CH3CH2—H Reaction with epoxides O H C H BrMgO CH3MgBr H C H H H C C H H CH3 HO H+ H C H H C H CH3 CH3–MgBr+ Because carbonyl pi bonds are polarized, they can undergo a reaction called nucleophilic addition: the addition of a nucleophile to an electron deficient pi bond. O– Oδ− O– C+ Cδ+ C R1 R1 R1 R1 R1 Nu Nucleophilic Addition R1 Nu: R1 O– Oδ− O– MgX+ C+ Cδ+ C R1 R1 R2 Mg-X R1 R1 R2 R1 Nucleophilic Addition Chem 66H Preparation of Alcohols from Grignard Reagents Chapter 13-2 Chem 66H A Grignard reaction with 1. formaldehyde produces a primary alcohol O 1. CH3 C H Mg-X + 2. H2O, H H Oδ− OH H C Cδ+ H H CH3 2. an aldehyde produces a secondary alcohol C CH3CH2 H 2. H2O, H+ H C Oδ− CH2CH3 H δ− C CH3CH2 1.(CH3)CHMgBr + CH3 2. H2O, H HC H 3C OH CH3 C H H H C H H CH2CH3 C H CH(CH3)2 C CH(CH3)2 CH2CH3 δ+ Mg-X CH3 CH(CH3)2 O CH3MgBr C OMgBr C – CH3O–MgBr+ OCH3 CH3 O OH 1. 2 CH3MgBr OCH3 2. H2O, H+ CH3 C CH3 O CH3MgBr C CH3 5. ethylene oxide produces a primary alcohol OMgBr C CH3 CH3 1. C6H5MgBr 2. H2O, H+ C H H HO H+ CH2CH3 OCH3 C H CH2CH3 4. an ester produces a tertiary alcohol (addition of two molecules of Grignard reagent) O C H BrMgO Cδ+ CH(CH3)2 3. a ketone produces a tertiary alcohol O H C HO H+ Mg-X OH 1. (CH3)CHMgBr H H δ+ δ− CH3 O BrMgO OH OH CH3 C CH3 H+ Organolithium Reagents Chapter 13-3 Organolithium Reagents Organolithium reagents are organometallic reagents derived from an alkyl halide and lithium metal R—X + 2 Li diethyl ether Rδ−—Liδ+ = LiX organolithium reagent Since the carbon carries a partial negative charge, the carbon is a strong base and a good nucleophile. CH3CH2—Br + 2 Lidiethyl ether CH3CH2δ−—Liδ+ HOH CH3CH2—H n-Butyl Lithium is a common commercially available organolithium reagent which is used primarily as a strong organic base. It also acts as a nucleophile to add to carbonyl compunds, much like a Grignard reagent. O C CH3CH2 1. CH3CH2CH2CH2Li + CH3 2. H2O, H OH CH3 C CH2CH3 CH2CH2CH2CH3 n-Butyl Lithium can also be used to generate aryl and vinyl lthium reagents by lithium-halogen exchange CH3CH2CH2CH2Li Br Li + CH3CH2CH2CH2Br (n-BuLi) CH3CH2CH2CH2Li Br Li + BuBr OCH3 Br 2 CH3CH2CH2CH2Li OCH3 Li + BuBr + Butane OH OLi Chem 66H Alkynes: Acetylides Chapter 13-5 H C + base C C H pKa = 45 C C – .. C + base—H H + base H C . C. – + base—H pKa = 25 The relatively high acidity of the alkyne —C≡C—H bond is associated with the large degree of s character in the sp C—H bond (50% compared with 33% in sp2 bonds). The carbon atom is more electronegative in the sp state; thus the C—H bond is more acidic. The acetylide ion may be formed by such strong bases as —:NH2 (pKa 33), RMgX or RLi (pKa 45-50). R CH CH2 NaNH2 No reaction NH3 R C C H NaNH2 NH3 R C C H R C C:– Na+ + NH3 acetylide ion RMgX R C C:– MgX+ R C C:– Li+ + RH n-BuLi R C C H + RH Chem 66H Alkynes: Acetylides Chapter 13-6 SN2 reaction with acetylide ion NH3 R'—C ≡C:– Na+ + R' R—CH2—X O – C: MgBr+ + C R—CH2—C≡C—R' R' C – + C—CH 2—CH2—O MgBr H+ R' C C—CH 2—CH2—OH Nucleophilic addition reaction with acetylide ion. O – Li+ O R' – + C: Li C R C H R' C C C R H n-BuLi R' C H+ OH C—H R' C C C H O R' C – C: Li + + O – Li+ C C R' H+ OH C C R' R Chem 66H Chem 66H Organocuprates Chapter 13-7 Preparation: Organocopper reagents can be prepared from organo lithium reagents and Grignard reagents Cuprates: Me2 CuLi, Bu2 CuLi for common readily available organolithium reagents. R—X + 2Li R—Li + LiX CuI R—Cu R—Li CO2Et O LiCu HO H CO2Et 90% H R2 CuLi O Higher Order Cuprates: somewhat more stable than dialkylcuprates 2 R—Li + CuCN Epoxides PhCH2O OH Me Me2CuLi OH PhCH2O R2 CuCNLi2 OH 6 Substitution 6:1 OH Organocopper reagents react with alkyl halides, epoxides, allylic halides, propargylic halides, vinyl halides to give substitution products R1 —X + 1 Me R—R1 R2 CuLi O X = I, Br, Cl, OTs Me Me2CuCNLi2 oxidative addition HO 90% R R Cu X R1 O O I t-Bu MeO2C reductive R—R1 + RCuX elimination OTs OTs CO2Me (C6 H5)2 CuLi O (4 equ) O (t-Bu) 2 CuLi t-Bu Ph Ph 47% CO2Me 76% NHCOC6H5 (4 equ) NHCOC6H5 X Bu2 CuLi n-Bu X = OTf 100% X = OP(O)(OPh) 60% 2 OH PhCH2O t-Bu OCH2Ph MeO2C OCH2Ph Organometallic Reactions: An Overview Chapter 13-8 Organometallic complexes contain a metal and coordinated ligands. The type and number of ligands will depend on the metal and its oxidation state. Typical ligands of organometallic complexes Ligand Charge H– Cl–, Br–, I– R–, Ar– No. of electrons donated -1 2 -1 2 -1 2 0 2 2 0 R3P: 2 0 :O≡C: 6 -1 Oxidation state of metal is the difference between the overall charge on the complex and the sum of the charges for each ligand. Cl CH3 Pd Ph3P PPh3 H Ph3P Rh Cl Cl Cl: -1 CH3: -1 Ph3P: 0 Pd: +2 CO PPh3 Ph3P: 0 H: -1 CO: 0 Cl: -1 (X2) Rh: +3 Chem 66H Organometallic Reactions: An Overview Chapter 13-9 Organometallic complexes undergo three basic reactions Oxidative addition Migratory insertion reductive elimination Oxidative addition The oxidation state and the coordnation number of the metal ion both increase (usually by two) Br R L L L M + R–Br M L L L L L CN = 4 CN = 6 L = generic ligand reductive elimination The oxidation state and the coordnation number of the metal ion both decrease (usually by two) H L M L R L L L M L + R–H L CN = 4 L CN = 6 Migratory insertion L L L H 2C L L L M H CH2 L Rh L solvent L L CH2–CH3 L Migratory insertion no change in the metal ion oxidation state L Rh S CH2–CH3 Chem 66H Hydrogenation with Wilkinison's Catalyst Chapter 13-10 Wilkinson's Catalyst, (Ph3P)3RhCl functions as a catalyst in the presence of hydrogen to convert alkenes into alkanes, ie. hydrogenation. Ph3P Cl Rh Ph3P CH3CH2OH ligand exchange Cl H Ph3P H CH2 Cl Rh H Ph3P S Rh = +3 Rh = +3 H H Cl Ph3P H Ph3P Rh Ph3P ethanol Cl Rh C–C H Migratory insertion H Ph3P oxidative addition Rh = +1 – ethanol Rh Ph3P S CH2 H Ph3P H2 Cl Rh Ph3P PPh3 Rh = +1 Ph3P Ph3P Ph3P Cl Rh Ph3P S C–C H reductive elimination Ph3P Cl Rh S + H C–C H Chem 66H Chapter 13-11 Palladium Catalyzed Carbon-Carbon Bond Formation Suzuki Reaction OR Pd(Ph3P)4 R–X + R–R' + R' B NaOH Ph3P Pd Ph3P Pd Ph3P Ph3P R–X PPh3 oxidative addition Ph3P Pd PPh3 Ph3P PPh3 Ph3P OR – PPh3 PPh3 + NaX HO B R'–B(OR)2 R Pd Ph3P X "transmetallation" step R Pd R–R' + Ph P 3 R' Ph3P Pd PPh3 reductve elimination Example Pd(PPh3)4 Br + (RO) 2B Pd(PPh3)4 + N B(OH) 2 Br N CH3 N N 70% CH3 Chem 66H Spectroscopy Chapter 14-1 Molecular Spectroscopy electromagnetic radiation: energy that is transmitted through space in the form of waves wavelength: (λ): the distance from the crest of one wave to the crest of the next wave frequency: (ν): the number of complete cycles per second ν= c λ where c = speed of light Electromagnetic radiation is transmitted in particle-like packets called photons or quanta. The energy is inversely proportional to the wavelength and directly proportional to frequency. Ε= hc λ where c = speed of light; h = Planck's constant Ε = hν ultraviolet visible h = Planck's constant infrared radio decreasing energy Absorbtion of ultraviolet light results in the promotion of an electron to a higher energy orbital. Absorbtion of infrared results in increased amplitudes of vibration of bonded atoms. Intensity of radiation is proportional to the number of photons. Chem 66H Mass Spectrometry Chapter 14-2 MassSpectrometry useful for determining molecular weight, presence of specific atoms and also certain molecular fragments an organic molecule can be ionized by a number of methods such as bombardment by electron s or other high energy species. usually the ionization results from loss of a single electron and the production of a cation radical. The princple of mass spectrometry is based on the fact that depending on the mass to charge ratio of a particular cation radical, it will travel along a different curved path when exposed to a magentic field. m = z H2r2 2V m = mass of cation radcal z = charge (usually +1) H + strength of the magnetic field r = radius of the path V = accelerating potential placing a detector at some point along the flight path of the ion allows its mass to charge ratio to be calculated. Since almost all the ions will have a charge of +1, the mass to charge ratio is also the mass. A mass spectrum produces a series of peaks which correspond to different mass of different molecular frgaments and their relative abundance 100 M+ M+1 0 0 10 20 30 40 50 60 70 80 M+2 90 100 Chem 66H Chapter 14-3 Mass Spectrometry The molecular ion is the result of loss of one electron from the parent molecule. Sometimes the molecular ion is too unstable to be detected, but it usually is present in the mass spectrum. Chem 66H M+ M+ M+ M-1+ The molecular ion also fragments into various other fragments by bond breaking processes in the gas phase. M+ Each of the fragments which reach the detector will produce a peak in the spectrum corresponding to its mass. M+1 + A peak in the region of highest m/z in a mass spectrum often corresponds to the moecular ion. In addition to the peak for the molecular ion, there will also be peaks of M+1 and M+2 mass which correspond to similar ions which contain other isotopes of specific elements, For example, the mass spectrum of 2-butanone contains a peak at 72 for the molecular ion 12C41H816O and a peak at 71 for other ions such as 13C12C31H816O or 12C42H1H716O or 12C41H817O The base peak is the largest peak in the spectrum corresponding to the ion which is present in the greatest abundance. The base peak is often the molecular ion, but not always. The base peak can be the result of a fragmentation of the molecular ion into two other species. To determine the molecular weight of a compound from the mass spectrum, first look at the region of hghest m/z ratio. It is usually a reasonable assumption that one of these peaks will be the molecular ion. If the molecular ion is present and no Cl, BR or S are present in the molecule, one of four patterns are most common. no M+2+ present M+1+ M+2+ M-1+ M+1+ M+2+ M-2+ M+1+ M+2+ Mass Spectrometry Chapter 14-4 To determine the molecular weight of a compound from the mass spectrum, first look at the region of hghest m/z ratio. Using the known relative abundance of isoptopes of different elements, the molecular formula can be deduced. It is usually a reasonable assumption that one of these peaks will be the molecular ion. For example: if the molecular ion is 68, there are three reasonable possibilities If the molecular ion is present and no Cl, BR or S are present in the molecule, one of four patterns are most common. M + M + M-1 M+ formula + M M-1 M+1+ M+2+ M+1+ + 1 2 H H % 99.98 0.01 Isotope 31 P 32 12 M+1+ M+2+ 13 14 C C 98.89 1.11 N N 99.63 0.37 15 35 75.53 24.47 Cl Cl 17 O O 18 O 99.76 0.04 0.20 79 19 100.0 127 F 100.0 95.00 0.76 4.22 0.01 37 16 % S S 34 S 36 S 33 Br 81 Br I M+1 M+2 C 3H 4N 2 4.07 0.06 C 4H 4O 4.43 0.28 C5H8 5.53 0.12 + Assumes M+ is 100% otherwise it would be the specified pecentage of the M+ intensity. no M+2+ present Isotope Chem 66H 50.54 49.46 100.0 M-2+ M+1+ M+2+ Mass Spectrometry Chapter 14-5 Presence of Nitrogen Determining the presence of nitrogen is very simple if there are an odd number of nitrogens present, because the molecular ion will be an odd mass. For even numebrs of nitrogens, the M+, M+1, and M+2 intensties can be examined as illustrated before. Presence of Sulfur Determining the presence of sulfur can usually be determined the presence of a slightly large M+2 peak since 34S is 4.22% abundant. Presence of Bromin and Chlorine Determining the presence of bromine and chlorine can also be determined from the M+2 peak since 37Cl is 24.47% abundant and 81Br is 49.46% abundant. Thus the M+ and M+2 peaks in a compound containing chlorine will be about a 3:1 ratio and for one containing bromine M+ to M+2 will be about 1:1. Fragmentation Patterns cleavage at branches R +. + + . R because of cation stability, cleavage to produce stable cations is common Chem 66H Mass Spectrometry Chapter 14-6 Fragmentation Patterns α,β-cleavage +. O+ O + . cleavage of a bond alpha and beta to a heteroatom such as oxygen is common in carbonyl compounds Loss of a neutral molecule Loss of H2O, CO, HCN, HCl, NO, etc is common due to the stability of the neutral species CO + + + CO McLafferty Rearrangement O H H +. R O H +. H R + R R McClafferty rearrangment is very common in carbonyl compounds with a gamma hydrogen atom. Chem 66H Infrared Spectroscopy Chapter 14-7 Infrared Spectroscopy Infrared is recorded as %T versus wavelength or frequency When a sample absorbs at a particular wavelength or frequency, %T is reduced and a peak or band is displayed in the spectrum. Infrared is recorded as %T versus wavelength or frequency When a sample absorbs at a particular wavelength or frequency, %T is reduced and a peak or band is displayed in the spectrum. 100 %T 0 frequency Nuclei of bonded atoms undergo vibrations similar to two balls connected by a spring. Depending on the particular atoms bonded to each other (and their masses) the frequency of this vibration will vary. Infrared energy is absorbed by molecules resulting in an excited vibrational state. Vibrations occur in quantized energy levels and thus a particular type of bond will absorb only at certain frequencies. Both stretching and bending vibrations can be observed by infrared. O CH3 CH3 stretching O CH3 CH3 bending Chem 66H Infrared Spectroscopy Chapter 14-8 Frequency of vibration will be inversly proportional to the masses of the atoms. C S C 1350 cm-1 O C 1700 cm-1 H C 3000 cm-1 D 2200 cm-1 Frequency of vibration will be directly proportional to the strength of the bonds C C C 2150 cm-1 C C C 1200 cm-1 1650 cm-1 Some vibrations are coupled when atoms of similar masses are involved such as two or mopre C–H bonds such as in a methyl group where there are symmetric and antisymmetric stretches H H C C H H H C–Hsym = 2872 cm-1 C–Hasym = 2962 cm-1 O– H C N H H + O H N: H coupled vibrations are common as in functional groups above which each have a symmetric and antisymmetric vibration. These can help dentify certain functional groups Chem 66H Infrared Spectroscopy Chapter 14-9 4000 cm-1 to 1300 cm-1 is known as the functional group region. 400 cm-1 to 1300 cm-1 is known as the fingerprint region since it is unique for every compound. Interpretation of Infrared Spectra Correlation tables Infrared spectra of thousands of compounds have been tabulated and general trends are known. Some common functional groups are shown below. C=O str OH and NH str CH str C≡N str C—O str C=N str C=C str NH bend Chem 66H C—C and C—H Bonds sp3 C—C sp2 C=C sp2 C—C (aryl) sp C≡C sp3 C—H sp2 C—H sp C—H C(CH3)2 2500 2000 1500 1640-1820 cm–1 Aldehydes C=O; C—H(O) 1640-1820 cm–1 Carboxylic acids C=O; C(O)—OH 1640-1820 cm–1 Esters C=O ; C(O)—OR 1640-1820 cm–1 1100-1300 cm–1 2800-3000 cm–1 3000-3300 cm–1 3300 cm–1 1360-1385 cm–1 (two peaks) Alcohols and Amines 3000-3700 cm–1 900-1300 cm–1 1000 800 –1 2820-2900 and 2700-2780 cm (weak but characteristic) 3330-2900 cm–1 2100-2250 cm–1 CH bend Carbonyls One of the most useful absorbtions in infrared 1640-1820 cm-1 Ketones (saturated) C=O 1450-1600 cm–1 C—C str O—H or N—H C—O or C—N 3000 1600-1700 cm–1 C—N str OH bend 3500 weak, not useful Ethers C—O 1050-1260 strong Nuclear Magnetic Resonance Spectroscopy Chapter 15-1 Nuclear Magnetic Resonance (NMR) Spectroscopy Some atomic nuclei (1H, have a nuclear spin. 13 C, others) behave as if they are spinning...they Spinning of a charged particle creates a magnetic moment. If an external magnetic field is applied, these small magnetic moments (of the nuclei) either align with the field α () or against the field β (), about 50% with and 50% against the field at any one time. β Ho ∆E hν α β Ho ∆E α Ho = the external magnetic field Resonance: the flip of the magnetic moment from parallel to antiparallel to the external magnetic field. Irradiation at the frequency equal to the energy difference,∆E, causes resonance. ∆E depends on the external magnetic field. Protons (or other nuclei) in different magnetic environments resonate at different field strengths. A proton which resonates at a higher field is in a stronger magnetic environment or shielded. A proton which resonates at a lower magnetic field is said to be deshielded. Different magnetic environments are created by different electron densities in the vicinity of a proton. Chem 66H Nuclear Magnetic Resonance Spectroscopy Chapter 15-2 Chem 66H Adjacent electron withdrawing groups, highly electronegative atoms, or the hybridization of the carbon to which the proton is bonded can alter the magnetic environment. The pi system of benzene creates a magnetic field or ring current which deshields the protons attached to the ring. The local electrons create a small electric and magnetic field around a proton and shield it. Similarly, pi electrons in a C=O bond create a field which deshields the proton bonded to the C=O of an aldehyde. This is also affected by the inductive effect of the C=O. The more electron density present around the proton, the greater the field and the greater the shielding. Resonances are reported in chemical shifts (δ) downfield from tetramethylsilane (TMS) (CH 3) 4Si. δ= distance from TMS in Hz MHz of spectrum ppm In methyl halides, the more electronegative the halogen, the more deshielded the protons on the methyl. This is because F is inductively more electron withdrawing, causing the carbon to be more positive and thus pulling more electrons away from the hydrogen and causing it to be less shielded. In methyl halides, the more electronegative the halogen, the more deshielded the pr H3C—F δ H3C—Cl 3.0 4.3 H3C—Br H3C—I 2.7 2.1 Pi electron effects Magnetic fields created by pi electrons are directional and said to have an anisotropic effect. R Ho C H O H H deshielded H deshielded Nuclear Magnetic Resonance Spectroscopy Chapter 15-3 Equivalent and Nonequivalent Protons Protons that are in the same magnetic environment are equivalent and have the same chemical shifts. Protons in different magnetic fields are nonequivalent and have different chemical shifts. Magnetic equivalence is usually the same as chemical equivalence. Equivalence can be established by symmetry operations such as rotation, mirror planes and centers of symmetry Chemically equivalent protons have the same chemical shifts. To determine if protons are chemically equivalent, replace one by a different group, e.g. D or Br. Then replace a different one by the same group and compare the two compounds. If they are identical, the protons are equivalent. H H H C C H H OH H H H H C C C H Cl H H equivalent, but not to CH3 protons all six are equivalent equivalent Equivalent protons can be on different carbons. Protons which are homotopic or enantiotopic resonate at the same chemical shift in the NMR. If protons are interconverted by rotation about a single bond, they will average out on the NMR time scale and a single resonance will be observed. ClH2CCH2Cl anti and gauche forms rapidly interconvert and a single resonance is observed. Axial and equatorial hydrogens in cyclohexane average to a single peak because of rapid ring inversion. Diastereotopic hydrogens are chemically nonequivalent and thus give different chemical shifts in the NMR Chem 66H Nuclear Magnetic Resonance Spectroscopy Chapter 15-4 Intergration The spectrometer can integrate and determine the relative number of hydrogens associated with each resonance in the NMR spectrum by determining the area under the peaks. Spin-Spin Coupling for example... 3 3 CH3CH2OCH3 2 TMS If a proton (H a) is bonded to a carbon which is bonded to a carbon that has one proton (Hb), Ha will appear as a doublet Since in half the molecules, Hb will be in the α state and in half will be in the β state, Ha will experience two different magnetic fields and two peaks (a doublet) will appear for Ha. Ha without an adjacent hydrogen For one adjacent hydrogen α or β Ha with Hb adjacent in theβ state Ha with Hb adjacent in theα state Chem 66H Chapter 15-5 Nuclear Magnetic Resonance Spectroscopy For two adjacent hydrogens: Hb, Hc At any one time Hb or Hc could be in the α or β state (50:50) thus 4 combinations for Hb, Hc exist: αbαc αbβc βbαc βbβc gives 1:2:1 triplet When both Hb and Hc are α, a different field is observed than if both are β or one is α and one is β. When one is α and one is β, the field is the same. That is, βbαc and αbβc produce the same field and a single signal for Ha is observed with twice the intensity. Thus three signals are observed in a 1:2:1 ratio: a so-called triplet For three adjacent protons: ααα ααβ αββ βββ 1:3:3:1 quartet αβα βαβ βαα ββα Thus the splitting pattern of a particular proton or equivalent protons will be a pattern with n+1 lines where n is the number of adjacent equivalent protons. singlet 0 neighboring protons doublet 1 neighboring protons triplet 2 neighboring protons quartet 3 neighboring protons quintet 4 neighboring protons sextet 5 neighboring protons septet 6 neighboring protons The separation of the peaks in a splitting pattern is called the coupling constant, J. Chem 66H Nuclear Magnetic Resonance Spectroscopy Chapter 15-6 Splitting Diagrams Splitting patterns for protons can be constructed in diagram form by starting with one line to represent the unsplit proton resonance. If an adjacent proton Hb affects Ha it is split into a doublet; if another equivalent proton to Hb is present, each line of the double will be split into a doublet, since the coupling constant J is the same, the two center lines overlap and a only three lines are observed with the center line twice the height. This can be repeated for additional adjacent protons. Ha without an adjacent hydrogen 1 1 1 splitting diagram Ha split by one adjacent hydrogen 1 2 Ha split by a second adjacent hydrogen Ha split by a third adjacent hydrogen 1 3 3 1 Chemical Exchange and Hydrogen Bonding CH3OH, methanol would be expected to give an NMR spectrum of a doublet for the CH3 and a quartet for the OH. For a dilute sample at -40° in CCl4 this is the case. If the NMR spectrum is run at 25° as a more concentrated sample only two singlets are observed. This is because the intermolecular hydrogen bonding in methanol allows the rapid exchange of the OH proton from one CH3OH molecule to another, effectively averaging the spin states of the OH proton and resulting in no change in the magnetic field due to the OH. Amines and other compounds which can undergo hydrogen bonding can also show this effect. Thus the NMR spectra of alcohols, amines and carboxylic acids are temperature, concentration and solvent dependent. Chem 66H Nuclear Magnetic Resonance Spectroscopy Chapter 15-7 CHEMICAL SHIFTS Functional Group Shift,δ Primary alkyl, RCH3 Secondary alkyl, RCH2R Tertiary alkyl, R3CH 0.8-1.0 1.2-1.4 1.4-1.7 Allylic, R2C=C—CH2R 1.6-1.9 Benzylic, ArCH2R Iodoalkane, RCH2I Bromoalkane, RCH2Br Chloroalkane, RCH2Cl Ether, RCH2OR Alcohol, RCH2OH Ketone, RCH2C(=O)R 2.2-2.5 3.1-3.3 3.4-3.6 3.6-3.8 3.3-3.9 3.3-4.0 2.1-2.6 Aldehyde, RCH(O) Terminal alkene, R2C=CH2 Internal alkene, R2C=CHR Aromatic, Ar—H Alkyne, RC≡C—H Alcoholic hydroxy, ROH Amine, RNH2 9.5-9.6 4.6-5.0 5.2-5.7 6.0-9.5 1.7-3.1 0.5-5.0 (variable) 0.5-5.0 (variable) Chem 66H The Carbonyl group Chapter 17-1 Carbonyl Group sp3 hybridized, trigonal planar carbonyl carbon; partial positive C and partial negative (Lewis basic) oxygen. pi bond : C O lone pairs δ+ δ− : C O .. + C – .. O .. : : C–O pi* is low-lying and therefore interacts well with high-lying filled-nonbonding orbitals: thus nucleophilic, not electrophilic addition reactions are charactersistic of carbonyl compounds. Because of the polar C=O bond, boiling points are higher than nonpolar compounds of similar molecular weights. Aldehydes and ketones are capable of hydrogen bonding to water, alcohols and acids Spectral Properties Infrared: Ketones(C=O) 1660-1750 cm-1 shifted by 25 cm-1 if aromatic or unsaturated : PhCHO, CH2=CHCOCH3 Adlehydes (C=O) 1700-1740 1 (C—H) 2850 H NMR R—CHO 9-10 ppm RCH2COR 2.0 - 2.6 ppm due to inductive deshielding Chem 66H The Nucleophilic Addition Reaction Chapter 17-2 Nucleophilic addition reactions Addition Reactions of Aldehydes and Ketones Carbonyl group can be attacked by nucleophiles R sp2 C R : .. O. . R sp3 R C .. _ O .. : Nu Nu: or undergo addition of reagents to the pi bond by electrophiles adding first R C R .. O. . H+ R C R R +H O.. R R C R aldehyde O R C H .. O .. H Nu Nu: ketone O C formaldehyde O H C H increasing reactivity due to steric and electronic effects Ketones are more sterically hindered since they have two alkyl groups; aldehydes have one H and one alkyl group. Alkyl groups are electron releasing and make the C=O carbon less positive. Reaction with H2O: Formation of Hydrates hydrates are normally transient, unstable species which are in equilibrium with the carbonyl compound Chem 66H The Nucleophilic Addition Reaction Chapter 17-3 Acid catalyzed mechanism H+ .. .. O .. + H O H+ H 2O CH3 C CH3 H2O: CH3 C CH3 . H ..O. O - H+ H CH3 C CH3 hydrate O H CH3 C CH3 O+ H H the equilibrium constant for the formation of hydrates is dependent on the carbonyl substituents: the more sterically hindered and the more electron rich the carbonyl, the less of the hydrate that will be present, conversely, the less sterically hindered and the more electron deficient the carbonyl carbon, the more hydrate that will be present. O O F3C CF3 H 22,000 O H H 3C O H (H3C)3C 1.8 X 10 -2 41 O H H 3C 4.1 X 10 -3 CH3 2.5 X 10 -5 Khydration Base catalyzed mechanism O CH3 C CH3 HO – H 2O . ..O. – CH3 C CH3 O H H–O–H O H CH3 C CH3 O H hydrate + HO – Chem 66H Formation of Acetals Chapter 17-4 with alcohols: formation of acetals .. .. O O—H H+ CH3 C H O—CH3 CH3 C H + CH3OH CH3OH, H+ a hemiacetal O—CH3 CH3 C H O—CH3 an acetal mechanism .. .. O H+ CH3 C H CH3 C H O+ CH3 H CH3 C H H+ .. H O H . H . +O CH3 C H . O. CH3 . . CH3 C H O CH3 CH3 .. H O+ CH3 C H O CH3 . H ..O. .. CH3OH .. .. + H O H+ .. CH3OH .. - H2 O CH3 C H O+ CH3 O—CH3 CH3 C H O—CH3 if H2O is present in a large amount, the carbonyl compound will be favored. if H2O is not present, but ROH is present, the acetal will be favored O HO OH p-toluenesulfonic acid toluene, heat H2O, HCl or H2SO4, acetone O O Chem 66H Acetals and Cyanohydrins Chapter 17-5 Acetals as protecting groups many times multifunctional compounds must be treated to convert one functional group selectively. In the example below, direct treatment of the keto-ester with LiAlH4 would result in reduction of both carbonyls, so the ketone must be protected prior to reduction of the ester O O CH2OH CO2CH3 HO OH p-toluenesulfonic acid toluene, heat O O H2O, H2SO4 O LiAlH4 CO2CH3 O CH2OH Formation of cyanohydrins R R C R .. O.. .. .. O Ph C H – R C O–H NC .. .. O Ph C H HCN - O—H CN Ph C H CN + HCN .. ... – O. Ph C H CN H—CN a cyanohydrin . ..O H Ph C H CN CN forms a new C–C bond and introduces a functional group which can be converted to a carboxylic acid or an amine Chem 66H Preparation of Alcohols Chapter 17-6 Reduction of Carbonyl Compounds a) Hydrogenation C=O bond can be hydrogentated much like a C=C bond, C=O usually requires harsher conditions O H OH Pt, H2 O O CH2=CHCH2CH2 C H Ni, H2, 25°C CH3CH2CH2CH2 C H OH Ni, H2 heat, pressure CH3CH2CH2CH2 C H H b) Metal Hydrides LiAlH4, lithium aluminum hydride NaBH4, sodium borohydride OH O CH3CH2CH2 C CH3 1. LiAlH4 CH3CH2CH2 C CH3 H 2. H2O, H+ OH 1. NaBH4 CH3CH2CH2 + 2. H2O, H C CH3 H LiAlH4 much more reactive also reduces esters, carboxylic acids, nitriles, amides NaBH4 sodium borohydride reduces only aldehydes and ketones: more selective Chem 66H Preparation of Alcohols Chapter 17-7 For example O H CH2CH2CO2Et OH CH2CH2CO2Et 1. NaBH4 2. H2O, H+ O H OH CH2CH2CH2OH CH2CH2CO2Et 1. LiAlH4 2. H2O, H+ LiAlH4 and NaBH4 do not reduce isolated double bonds Mechanism for NaBH4: R R δ+ C O δ− H R H B– H R C O—B – H3 H H R R R C O)4B H 2O – R C OH + H3BO3 H H Mechanism for LiAlH4: Li + H – O H Al H H H O H + Al–H2(OR)Li H2 O H OH H H Chem 66H Preparation of Alcohols fromo Grignard Reagents Chapter 17-8 A Grignard reaction with 1. formaldehyde produces a primary alcohol O 1. CH3 C H 2. H2O, H+ H OH Mg-X H C H CH3 2. an aldehyde produces a secondary alcohol O OH 1. (CH3)CHMgBr C CH3CH2 H 2. H2O, H+ H C CH2CH3 CH(CH3)2 3. a ketone produces a tertiary alcohol O C CH3CH2 1.(CH3)CHMgBr + CH3 2. H2O, H OH CH3 C CH2CH3 CH(CH3)2 4. an ester produces a tertiary alcohol (addition of two molecules of Grignard reagent) O OH 1. 2 CH3MgBr C OCH3 2. H2O, H+ CH3 C CH3 5. ethylene oxide produces a primary alcohol O 1. C6H5MgBr 2. H2O, H+ OH Chem 66H Carbohydrates Chapter 17-9 Carbohydrates are naturally occurring compounds with C, H, O; often with the emperical formula CH2O. D and L sugars In the 19th century (+)-glyceraldehyde was arbitrarily assigned the configuration below and designated D. Monosaccharides...simple sugars which cannot be broken down by hydrolysis; e.g. glucose, fructose, ribose, galactose, deoxyribose, etc. CHO Disaccharides...dimers of monosaccharides units; e.g.sucrose is made up of glucose and fructose H Oligosaccharides...two to eight monosaccharides units Projection is designated D. If the OH is projected to the left, the sugar is an L sugar. for simplicity carbohydrates are often represented by Fischer Projections. in a Fischer projection the horizontal bonds are always out and the vertical bonds are always in. HO C H CH3 HO H CH3 HO CHO OH H CHO H Fischer Projections may be rotated 180° but not 90°. A 90° rotation creates the enantiomer. CHO H D-glyceraldehyde Now all carbohydrates with a hydroxyl to the right on the last carbon in the Fischer Fischer Projections CH2OH OH CH2OH Polysaccharides...more than eight monosaccharide units; e.g. cellulose is polyglucose CH2OH Chem 66H or H C OH HO C H H OH H C OH H OH H C OH CH2OH Classification of Carbohydrates The ending-ose indicates a carbohydrate An aldose contains and aldehyde; a ketose contains a ketone A triose has three carbons, a tetrose has four carbons, a pentose has five carbons and a hexose has six carbons. A ketohexose is a six carbon sugar containing a ketone. CH2OH HO OH H CHO HO H H OH H OH H OH H OH H OH HO H H OH HO H CH2OH D-glucose CHO CH2OH CH2OH L-glucose D-ribose Aldoses Chapter 17-10 Chem 66H Aldohtetroses Aldohexoses The aldotetroses have 2 asymmetric carbons and thus 22 or 4 stereoisomers. The D- seres is shown below. Each has a corresponding L-isomer The aldohexoses have 4 asymmetric carbons and thus 24 or 16 stereoisomers. The D- seres is shown below. Each has a corresponding L-isomer CHO CHO CHO H OH HO H OH H H OH CH2OH CH2OH D-erythrose D-threose CHO CHO CHO H OH HO H OH H OH HO H OH H OH H D-ribose H CH2OH D-arabinose H HO H OH H OH HO H OH H OH H OH H OH H OH H OH H OH H OH H HO H H HO H H HO H H CH2OH D-xylose OH CHO OH H OH CH2OH D-lyxose H OH CH2OH D-galactose H CH2OH D-altrose CHO HO OH CHO OH CH2OH D-allose The aldotetroses have 3 asymmetric carbons and thus 23 or 8 stereoisomers. The D- seres is shown below. Each has a corresponding L-isomer CHO H Aldopentoses CH2OH CHO H HO H HO H H OH CH2OH D-talose HO H H HO H CH2OH D-glucose CH2OH D-mannose CHO CHO HO OH CHO H OH HO H OH H HO H H OH CH2OH D-gulose HO H H OH H OH CH2OH D-idose Cyclic forms of Carbohydrates Chapter 17-11 The open chain representations of the sugars shown above is for simplicity. Sugars normally exist as cyclic hemiacetals. R C R H C Pyranose Forms O OH ROH, H+ O OR + ROH, H H R OR C C H H OR acetal hemiacetal H CH2OH CH2OH O H H O OH H H or OH H OH H HO OH HO H H OH H OH H OH HO Furanose Forms Chem 66H H H OH H OH diastereomers CH2OH D-glucose CHO HO OH O H H H H H OH H OH H OH OH HO HO H O H H H OH β-D-ribofuranose α-D-ribofuranose D-ribose CH2OH O OH HO CH2OH Haworth projection HO OH hydrogen if not labeled OH OH α-D-glucopyranose six membered ring glucose O N N HO O H O H H O H P O NH N NH NH2 N HO O H O- Oa DNA nucleotide HO HO H O H H O H OH P O- Oan RNA nucleotide CH2OH O OH OH O CH2OH up at C-5 OH at C-1 down Chapter 17-12 Anomers and Mutarotation Anomers: monosaccharides which differ only in their configuration at C-1 anomeric carbon HO HO CH2OH O H OH OH α−D-glucopyranose HO HO CH2OH OH C O OH H HO HO OH H β−D-glucopyranose CH2OH O HO OH OH OH CH2OH O OH CH2OH O OH HO OH OH Mutarotation α-D-glucose has a melting point of 146° and a specific rotation of +112°. β-D-glucose has a melting point of 150° and a specific rotation of 18.7°. The specific rotation of a solution of either α or β-D-glucose slowly changes until it reaches an equilibrium value of +52.6°. This is mutarotation and is due to the conversion of α-D-glucose to β-D-glucose or the reverse. The two forms are in equilibrium in solution: 36%α and 64% β. CH3OH, CH3OH, OH CH + 2 CH2OH O O H+ CH2OH O H HO HO HO HO H HO H HO OCH3 OH OH OH OH OCH3 H methyl-β-D-glucopyranoside α−D-glucopyranose methyl-α-D-glucopyranoside an acetal Glycosides are stable under neutral and basic conditions but are interconverted in acid. Chem 66H Chapter 17-13 Oxidation of Monosaccharides Oxidation Aldoses are readily oxidized because the hemiacetal is in equilibrium with an aldehyde, a readily oxidized functional group. The product of the oxidation of the aldehyde of an aldose to a carboxylic acid is called an aldonic acid. HO HO CH2OH O HO HO H OH OH Ag(NH3)2+ CH2OH OH H OHC O HO – or Br2, H2O HO HO CH2OH OH O– C OH O Even ketoses can be oxidized since they are in equilibrium with the aldose through an enediol tautomer CH2OH O HO H HO CHOH CHO C CHOH OH H HO CO2H CHOH Ag(NH3)2+ H H OH H OH H OH H OH H OH H OH CH2OH CH2OH CH2OH enediol ketose aldose HO – or Br2, H2O HO H H OH H OH CH2OH aldonic acid The product of oxidation of the aldehyde and the terminal primay alcohol to form a diacid is called an aldaric acid. CHO H CO2H OH H H dil. HNO3 H OH heat H OH HO CH2OH D-glucose HO OH H H OH H OH CO2H D-glucaric acid A uronic acid results when the terminal primary hydroxyl has been oxidized to a carboxylic acid. This generally only occurs enzymatically. Chem 66H Chapter 17-14 Reactions of the Hydroxyl Groups Reactions of the Hydroxyl Groups The alcohols in sugars react much like any hydroxyl functional group. Ester formation from acetic anhydride O HO HO CH2OH O CH3 H OH C O O C CH3 NaOAc, cold OH O CH2OAc O C CH3 O CH3 O H C O O CH3 O CH3 C C O O penta-O-acetyl-α-D-glucopyranose Ether Formation The hydroxyls of carbohydrates are more acidic than normal alcohols because of the inductive effects of the adjacent oxygens. These hydroxyls can be deprotonated to form alkoxides with NaOH. The alkoxides can be alkylated (SN2 displacement) by dimethyl sulfate...sulfate is an excellent leaving group because of the resonance stabilization of the anion. HO HO CH2OH O NaOH, H OH OH CH3OSO3CH3 CH3O CH3O CH2OCH3 O H CH3O CH3O all hydroxyls are converted to methyl ethers Acetal Formation HO HO CH2OH O C6H5CHO, H+ H OH OH O C6 H5 O HO O H OH OH Chem 66H Disaccharides Chapter 17-15 Disaccharides Maltose α−linkage H2O, H+ CH2OH O HO HO H CH2OH O O HO α-maltose: glycoside between C-4 OH of glucose and glucose anomeric carbon α-1,4-glucan Starch 2 glucose OH or enzymes OH H OH α-glucosidase maltose enzymes glucose beer maltohydrolase Cellobiose principal disaccharide of cellulose HO HO CH2OH O OH H H2O, H+ CH2OH O O HO OH H OH β−linkage 2 glucose or β-glycosidase Lactose only in mammals, 5% in human milk; one unit of glucose, one of galactose HO H HO CH2OH O OH H H2O, H+ CH2OH O O HO β−linkage OH H OH or enzyme glucose + galactose Chem 66H Biological Oxidation of Alcohols Chapter 17-16 O alcohol dehydrogenase CH3 CH2OH CH3 ethanol acetaldehyde – O OO O– P P O O OO HO HO N N O OH HO N + NAD+ + N O NH2 H 2N N O alcohol dehydrogenase CH2OH + NAD CH3 CH3 ethanol O N H C H H R N H 3C + NAD+ C lactic acid CH3 + H+ NADH O OH H O H 2N NAD+ C + H CH3 H 2N O H O H + CH + NADH + H acetaldehyde + R CH OH lactic acid dehydrogenase O H 3C C O OH pyruvic acid + NADH + H+ Chem 66H Oxidation of Aldehydes and Ketones Chapter 17-17 Oxidation of Aldehydes Aldehydes can be oxidized to carboxylic acids by KMnO4 or H2CrO4. Ketones cannot ordinarily be oxidized further. CH3CH2CH2CH2CO2H CH3 O H2CrO4 CHO CO2H Aldehydes are oxidized to carboxylic acids throught their hydrates O O H 2O CH3 CH O Cr OH CH3 CH2OH CH3 CH H O H2O: CH3 H O CH3 C O Cr OH OH O OH CH OH CH3 O C—OH + HCrO3- Baeyer-Villiger Oxidation oxidation of a ketone to an ester (cyclic ketone to a lactone) Ar O O ArCO3H H O O O O O + ArCO2H most substituted carbon (best able to stabilize a positive charge; i.e. one with highest electron density) will migrate. O O CH3 m-CPBA O CH3 retention of configuration CH3 CH3 Strained systems can be oxidized with H2O2, HO – or t-BuOOH, HO – and do not require RCO3H t-BuOOH, NaOH CH3CH2CH2CH2CHO + KMnO4, H+, H2O CH3 Chem 66H O O Chapter 18-6 Sulfur Ylides Unstabilized sulfur ylides reaction with unsaturated ketones to give epoxides: O O Me2S+—-CH2 O O- CH3 S+ CH3 O Explanation: k1 O + Me2S=CH2 CH2—S+(CH3)2 k2 O O- k-1 unstabilized sulfur ylides: k-1 < k2 in stabilized sulfur ylides k-1 > k2; k'2 > k'-1 O O O + k1 Me2S+—-CH2 k-1 k' 1 CH2—S+(CH3)2 k2 O- O k' -1 O- k' 2 O (CH3)2S+—CH2 Cyclopropylidine sulfur ylides O Ph + Ph2S Ph O O + Ph2S Ph Ph O Trost J. Am. Chem. Soc.1973, 95, 5298, 5307, 5311, 5317. Chem 66H Chapter 18-5 Sulfur Ylides Sulfur Ylides Sulfur ylides react with aldehydes and ketones to give epoxides rather than alkenes CH3—I CH3—S—CH3 ICH3 S+ CH3 CH3 sulfide — NaH CH3 S+ CH2 DMSO CH3 sulfonium salt O CH3—S—CH3 CH3—I I- O CH3 S+ CH3 CH3 slower sulfide CH3 S CH2 CH3 sulfur ylide O NaH O — CH3 S+ CH2 DMSO CH3 S CH2 CH3 CH3 sulfoxonium salt oxygen anion in intermediate displaces the sulfide to form an epoxide — O O– CH3 S+ CH2 + CH3 O CH3 S+ CH2 CH3 + CH3—S—CH3 Examples: O CH3 S CH2 CH3 O CHO H OH BF3-OEt2 O+B-F3 H H O+B-F3 H OB-F3 Chem 66H Chapter 18-4 Wittig Reaction Stereocontrol in the Wadsworth-Emmons Me Ph3P CHO CO2Et CO2Et O O Me Me Me (MeO) 2(O)P Me 70% 95:5 E:Z Kishi JACS1979, 101, 259. CO 2Me CHO Me KO t-Bu, THF Me Me CO2Me 5:95 E:Z Me CHO (MeO) 2(O)P CO 2CHMe2 CO2CHMe2 KO t-Bu, THF Me Me Me 95:5 E:Z Kishi Tetrahedron Lett.1981, 37, 3873. O CF3CH2O P CF3CH2O CO2Et R THF 18-crown-6 RCHO Me O CF3CH2O P CF3CH2O KN(SiMe3)2 CO2Et Me 30-50:1 Z:E CO2Me KN(SiMe3)2 R CO2Me THF 18-crown-6 RCHO 4-50:1 Z:E Still Tetrahedron Lett.1983, 24, 4405. O CH3CH2O P CH3CH2O CO2Et MgBr2 THF Et3N RCHO R CO2Me >97:3 E:Z Rathke J. Org. Chem 1986, 50, 2624. Chem 66H Chapter 18-3 Wittig Reaction Special Ylides one carbon homologation: H2O; H+ Ph3P=CHOCH3 + O OHC OCH3 Corey-Fuchs CBr4 + Ph3P Zn° RCHO Ph3P=CBr2 E+ R C C Li RCH=CBr2 2 BuLi E+ = H, ClCO2CH3, Cl—SiMe3 R C C E Wadsworth-Emmons Reaction Uses phosphonate anions instead of ylides Nucleophile is an anion, not an ylide and it is thus more reactive Phosphonate is formed by the reaciton of a trialkyl phosphite and and alkyl halide: the Arbuzov Reaction: BrCH3CH2 O (EtO) 3P: + Br—CH2CO2Et O base (EtO) 2P CH2CO2Et (EtO) CH2CO2Et 2P + O (EtO) 2P CH—CO 2Et _ phosphonate anion O CO2Et CO2Et P(OEt) 2 O– O + O (EtO) 2P O water soluble Chem 66H Wittig Reaction Chapter 18-2 Chem 66H The Wittig Reaction The position of the alkene is unambiguous in the product. R1 – + R1 O R1 R2 R2 C PR3 R2 + R1 carbonyl O P R R R2 R phosphorous ylide R1 R2 R1 R2 + Ph3P=O H 3C C O H 3C H 3C CH2CH3 C CH H 3C Driving force is the formation of the very strong P—O bond Phosphonium salts are readily formed from triphenyl phosphine and primary or secondary alkyl halides. Tertiary alkyl halides are not useful since the reaction is an SN2 reaction. R1 X C R2 H + R2 :PPh3 R2 Ph3P C H R2 BASE + pKa = 23 t-BuO- K+ BuLi NaH EtO- Na+ CH3SOCH2- R2 Ph3P C R2 - R2 Ph3P C R2 dπ−pπ Phosphorane (yilde) Acidity of carbon adjacent to+PPh3 is due to a combination of inductive and resonance effects. C lone pair P-antibonding overlap. Mechanism: oxaphosphetanes R1 O H + – + R2 C PR3 H R1 R2 O P R R R Otrans R2 C C H H R1 R1 O P R R R + R2 Ocis + CH2CH3 HC O better since ylide (phosphorane is derived from a primary halide) ylide derived from a secondary halide: substitution will be more difficult. R1 Ph C P+ Ph H Ph X– X = I, Br, OTs + H 3C C PPh3 H 3C CH2CH3 + HC PPh3 + H R1 C C H R2 Addition Elimination Reactions Chapter 18-1 Addition-Elimination Reactions secondary amines produce enamines primary amines produce Imines O H C N H 2N CHO Chem 66H CH2CH3 + N H CH2CH3 CH2CH3 N– CH2CH3 H+ enamine an imine H+ mechanism of enamine formation if the reaction is too acidic the amine is completely protonated and will not add if the reaction is not acidic, OH2 is not eliminated ..O.. mechanism of imine formation .. .. O O - H2 O R R R C R R N .. H H+ R Hydrazones and Oximes O + H2N—NH2 H+ R - H+ + H2N—OH R R C N+ H R N N—NH2 hydrazone O H+ addition H O +—H C elimination .. HNR2 N—OH oxime oximes, phenyl hydrazones, etc. are often solid and can be used to characterize carbonyl compounds by melting points N .. R H H+ O + N R H R - H2 O + R N H H addition .. H2N—R O +—H – R C R R C R H – - H+ H elimination N R + R R N R enamine Nitriles Chapter 19-15 Nitriles Chem 66H Base CH3CO2H acetic acid CH3CH2CH2CH2CO2H pentanoic acid —C≡N: pKb = 24 CH3CN acetonitrile CH3CH2CH2CH2CN pentanonitrile HO – NH3 pKb = 4.5 electrons more tightly held in sp orbital, 50% s character N more electronegative Preparation H HO– R C N—H O H 2O H—O H R C N—H O– R C N—H O—H - H+ transfer – R C O—H O NH2 R C O – + NH3 O SN2 with cyanide ion Ph—CH2—Br – R C N H—O R C N: Na+ - CN Ph—CH2—CN Reduction From benzenediazonium salts NH2 N2+ X – NaNO2 HCl CH2—CN CN CuCN or H2, Ni KCN Dehydration of amides O CH3CH2CH C NH2 CH3 SOCl2 CH3CH2CH C N: CH3 Reactions: Hydrolysis Acid R C N: R C N+—H R C N—H O+H2 H2O: H + R C N—H O—H H2O: H R C N—H +O—H - H+ H—O H R C N—H H—O R C N + H3 + O—H O—H 1. LiAlH4 2. H2O - H+ transfer – H+ H2O+ H R C N—H O—H H transfer R C OH O + NH4+ CH2—CH2—NH2 Amides Chapter 19-14 proteins O N H N N H O O N H O H N O N H Polyamides x HO2C(CH2)4CO2H + X H2N(CH2)6NH2 O O C (CH2)4 C NH (CH2)6—NH— nylon 66 Compounds related to amides CH3NH O C OCH3 carbamate H 2N O C NH2 urea X NH C H 2N NH2 guanidine CH3 O C O C NH CH3 imide Spectra α-hydrogens in 1H NMR O CH3 C N CH3 C OCH3 2.00 2.03 O CH3 C NH2 2.08 O O CH3 C OH CH3 C Cl 2.10 2.67 O C=O stretch about 1800 cm-1 R C Cl O -1 R C OCH3 C=O stretch about 1740 cm ; C—O 1200 O R C OCOR C=O stretch doublet; C—O 1100 O R C NH2 O R C NHR O R C NR2 —C≡N C=O stretch about 1700 cm-1; NH bend 1515 - 1670; NH doublet 3500 C=O stretch about 1700 cm-1; NH bend 1515 - 1670; NH 3300 C=O stretch about 1700 cm-1; no NH bend; no NH stretch 2200 cm-1 Chem 66H Hoffmann Rearrangement Chapter 19-13 Hoffmann Rearrangement O R H2O, HO- NaOBr R—NH2 R N C O NH2 isocyanate H R'OH R N OR' carbamate O O R O Br+ R NH2 .. Br N O HO – R H .. Br N .. HOH R N C O HO R N C O– – OH HO – R NH C O OH R NH C O O– CO2 R—NH2 or via the nitrene O R O N Br R N: .. R N C O H CONH2 BrO – NH2 Chem 66H Amides Chapter 19-12 Reactions Acid Hydrolysis LiAlH4 CH3 Not reversible since amine forms ammonium salt O C +O H C NHCH3 H+ NHCH3 :OH2 OH NHCH3 C OH NHCH3 C OH – H+ OH2 + OH + NH2CH3 C – H+ OH O C H+ OH + CH NH 3 2 H+ CH3NH3+ Base Hydrolysis CH3CH2CH2 O C O– NH2 HO – CH3CH2CH2 O C – O–H NH2 CH3CH2CH2 C NH2 O H O C + NH3 CH3CH2CH2 O– Reduction CH3CH2CH2 O C Chem 66H O– NHCH3 H3A–—H CH3CH2CH2 C NHCH3 H H3A –—H O—AlH3 CH3CH2CH2 C NHCH3 H CH3CH2CH2 C NHCH3 + H H CH3CH2CH2 C NHCH3 H N O CH3 N H H Amides Chapter 19-11 Amides O C NH2 CH3 acetamide O C O C NHCH3 CH3 N(CH3)2 CH3 N-methylacetamide N,N-dimethylacetamide Amides are not as basic as amines due to the overlap of the lone pair on nitrogen with the carbonyl pi bond. Amide pKb's: 15 - 16; CH 3NH2: pKb: 3.34. The result is a partial double bond between the nitrogen and the carbonyl carbon. The barrier to rotation is about 18 kcal/mol This is evident from the difference in chemical shifts of the two methyl groups in dimethyl formamide in the 1H NMR. O C H O– C + CH3 N H CH3 CH3 N CH3 different chemical shifts due to restricted rotation Preparation R R R O C O C O C R2NH Cl R2NH OCH3 O O C R O C N R R R2NH R Reactions Acid Hydrolysis Not reversible since amine forms ammonium salt Chem 66H Esters Chapter 19-10 Lactones Reaction with Ammonia CH3CH2 O C OCH3 + H3N CH3CH2 O C NH2 + HOCH3 COOH O C 1. LiAlH4 2. H2O OCH3 or Na, CH3CH2OH H CH3(CH2)5 C OH + HOCH3 H reduction always produces one primary alcohol plus the alcohol from the ether linkage of the ester Reaction with Grignard reagents Preparation of tertiary alcohols with at least two identical R groups CH3CH2 CH3CH2 O C O C 1. 2 CH3CH2CH2MgBr OCH3 2. H2O, H+ OCH3 CH3CH2CH2– + MgBr CH3CH2 Lactones are cyclic esters and react like esters. They are formed from acyclic hydroxy acids or hydroxy esters O O Reduction CH3(CH2)5 Chem 66H O C CH2CH2CH3 CH3CH2 CH2CH2CH3 CH3CH2 C OH + HOCH3 CH2CH2CH3 O– C OCH3 CH2CH2CH3 CH3CH2 O – MgBr+ C CH2CH2CH3 CH2CH2CH3 H+ H 2O CH3CH2CH2– + MgBr O—H CH3CH2 C CH2CH2CH3 CH2CH2CH3 Addition to the ketone is faster than addition to the ester and therefore two additions occur A secondary alcohol results from addition to a formate HCO2R since hydrogen was already bonded to the carbonyl carbon. OH + CO2CH3 H+, heat H , heat O OH O Esters Chapter 19-9 Acid hydrolysis: the reverse of esterification CH3 O C + H O C OCH3 CH3 OCH3 : O—H CH3 H2O18: H O + CH3 C OCH3 HO18 H +O H C 18 O H CH3 C OCH3 + O18H2 + CH3OH CH3 O C O18H labelled water results in labelled carboxylic acid; no label in the alcohol Base Hydrolysis O C HO – O C CH2CH3 O C H CH3 O—H + – O– CH2CH3 C O C H OH CH3 O C CH2CH3 O C H CH3 O– + CH2CH3 HO C H CH3 The product of the base hydrolysis is the carboxylate salt and the reaction is irreversible. If the alcohol is chiral; retention of configuration is observed. Thus the C—O bond of the ester is broken, not the C—O bond of the alcohol. Transesterification one alcohol is used in excess to drive the equilibrium in the desired direction O C H+, heat OCH3 + HOCH2CH3 O C OCH2CH3 + HOCH3 Chem 66H Esters Chapter 19-8 Chem 66H Esters Acid hydrolysis: the reverse of esterification Preparation (CH3)2CH (CH3)2CH (CH3)2CH (CH3)2CH O C O C O C O C H+, heat + HOCH2CH3 OH (CH3)2CH Cl + HOCH2CH3 O O C (CH3)2CH HOCH2CH3 (CH3)2CH CH(CH3)2 O – Na+ O C BrCH2CH3 (CH3)2CH O C O C O C OCH2CH3 + H2O OCH2CH3 + HCl OCH2CH3 + RCO2H (CH3)2CH OCH2CH3 (CH3)2CH O C OCH2CH3 + NaBr H OCH2CH3 Nu: In alkaline solution strong nucleophiles can effect addition-elimination (CH3)2CH Nu : O C OCH2CH3 O– (CH3)2CH C OCH2CH3 Nu OCH3 O—H + CH3 C OCH3 HO 18H CH3 O C H CH3 OCH3 O—H C OCH3 +O 18 H2 (CH3)2CH + – O C Nu OEt +O CH3 C H + CH3OH O 18 H CH3 O C O 18 H labelled water results in labelled carboxylic acid; no label in the alcohol Base Hydrolysis Acid can protonate the carbonyl oxygen and make the carbonyl carbon more susceptible to attack by nucleophiles + + H2O18: Reactions of Esters O C CH3 O C Acid Anhydrides Chapter 19-7 Anhydrides Preparation O C CH3CH2 O C CH3CH2 + Cl OH Na+ – O + (CH 3 O C O C CH2CH3 CH3CH2 heat CH3CH2 )2O O C O C O C O O O C CH2CH3 CH2CH3 + CH3CO2H distilled off to drive the equilibrium Reactions Same reactions as acid chlorides but with somewhat slower rates due to the poorer leaving group ability of RCO2 – R O C O O C O– O R C C O R Nu R Nu: R O C + Nu – Nucleophiles: H2O, ROH, ArOH, NH3, RNH2, R2NH CH3CH2CH2 O C O O C O C O O C H 2O CH2CH2CH3 CH3OH CH3CH2CH2 O C OCH3 Pyridine O C O O C NH3 O C O C NH2 OH O O C R Chem 66H Acid Halides Chapter 19-6 Fredel Crafts Reactions + CH3CH2COCl O C AlCl3 CH2CH3 Reaction with Grignard Reagents O C CH3CH2 H 2O Cl Rδ−Mgδ+X CH3CH2 O– Rδ−Mgδ+X O CH3CH2 C Cl C CH3CH2 R R tetrahedral intermediate O MgX H+, H2O C R R CH3CH2 O—H C R R Reaction with Lithium Dialkyl Cuprates 4 Li 2 R—X CH3CH2 (CH3)2CH O C O C 2 R—Li + 2 LiX R2CuLi Cl Cl CH3CH2 O C CuI R2CuLi + LiI R O (CH3CH2)2CuLi C (CH3)2CH CH2CH3 Reduction O C Cl OC(CH3)3 Li+H Al- OC(CH3)3 OC(CH3)3 or H2, Pd/BaSO4 O C H LiAl[OC(CH 3) 3] 3 is less reactive than LiAlH4 due to steric hindrance and the electron withdrawing effects of the oxygens Chem 66H Acid Halides Chapter 19-5 Acid Halides Chem 66H Amide formation Preparation O CH3CH2CH2 C OH O C OH SOCl2, heat O CH3CH2CH2 C Cl R R O C Reactions of Acid Halides most reactive of the carboxylic acid derivatives since the halide ion is a good leaving group. R O C O– O elimination + Cl R C Cl C R Nu Nu overall net substitution tetrahedral intermediate of Nu for Cl Nu: R H 2O Cl H2O: O– O R C Cl C +H R O +OH2 H tetrahedral intermediate - H+ + Cl – R Ester formation Cl CH3CH2CH2OH N O C OCH2CH2CH3 O C OH + HCl Rate decreases with increasing size of R since water solubility decreases O C CH3 O– R C Cl +NH3 + O – -H + + Cl C R NH3 tetrahedral intermediate NH2 O C Cl + HCl NH3 NH4Cl + CH3NH2 Cl 2 (CH3)2NH CH3 O C O C NHCH3 + CH3NH3 N CH3 + (CH3)2NH2Cl CH3 Use of an added tertiary amine avoids the loss of a second equivalent of nucleophilic amine Hydrolysis O C Cl O C addition Cl H 2O H3N: O C Cl PCl3, heat O C + N+ Cl H pyridine reacts with HCl to remove it from the reaction Carboxylic Acid Derivatives Chapter 19-4 Chem 66H Derivatives of Carboxylic Acids Reacticvity is also related to the resonance donating ablity of the acyl Any compound which yields a carboxylic acid on hydrolysis (acid or base) with water substituent CH3 O O O C OCH3 CH3 C NH2 CH3 C Cl ester amide CH3 O O C O C CH3 CH3 C N acid chloride anhydride R CH3 R leaving groups X = OR, Cl, NH2, OCOR X = H, R, Ar not leaving groups Aldehydes and ketones undergo nucleophilic addition Carboxylic acid derivatives undergo nucleophilic substitution due to the presence of a leaving group on the carbonyl carbon Nucleophilic acyl substitution CH3 Nu: X – < O C X O – R C O – < OR < – NH2 < – Cl R O C O C O O O CH3 R O– O C + R O R O– C + CH3 O R O R R O C CH3 N: CH3 O– + C O R O– C + CH3 N R CH3 best resonance donor since nitrogen is least electronegative and better Lewis base: O– O CH3 C X CH3 C X Nu tetrahedral (sp3) intermediate O– C + Cl R nitrile Reactivity O C X O C + X– CH3 increasing basicity of acyl substituent (leaving group) decreasing reactivity (leaving group ability) Reactivity of carboxylc acid derivatives decreases with increasing basicity of leaving group Acid chlorides and anhydrides react readily with water while esters and amides are fairly stable toward water and require acid or base to effect hydrolysis Amides have significant C=N double bond character and hindered rotation about the N–Cacrbonyl bond. Esterification of Carboxylic acids Chapter 19-3 Chem 66H Reduction of Carboxylic Acids Decarboxylation carboxylic acids can be reduced to primary alcohhols with LiAlH4 b-keto acids lose CO2 (decarboxylate) on heating CH3CH2CH2 O C OH CO2H H CH3CH2CH2 C OH H 2. H2O, H+ 1. LiAlH4 CH2OH 1. LiAlH4 2. H2O, H+ O H O O R R O H H H O O R H O enol H R = CH3, alkyl, OH, R = CH3, alkyl, OH, OR Polyfunctional Carboxylic Acids Dicarboxylic acids are called dibasic or diprotic acids O O CO2H heat the acidity of the first COOH to lose a proton is increased by the electron withdrawing ability of the other COOH, but the acidity of the second is lower (pKa increased) because of the adjacent negative charge created by the first COOacid structure oxalic acid malonic acid succinic acid glutaric acid adipic acid H pKa 1 1.2 2.8 4.2 4.3 4.4 HO2C—CO2H HO2C—CH2—CO2H HO2C—(CH2)2—CO2H HO2C—(CH2)3—CO2H HO2C—(CH2)4—CO2H 2 4.2 5.7 5.6 5.4 5.4 difference between pKa1 and pKa 2 decreses as the length of the chain increases since induction is directly dependent on distance Anhydride formation if a 5 or 6 membered ring can form, dicarboxylic acids form cyclic anhydrides with loss of water upon heating CO2H CO2H O heat or O O CH3 C O C CH3 O O + H2O O CH3CH2O O heat CO2H CH3 ketone CH3CH2O CH3 Esterification of Carboxylic acids Chapter 19-2 O R1 + C OH acid HO R2 CH3CH2 C R1 alcohol CH3CH2 CO2H + HOCH2C6H5 O O H+ + HO CH OH 3 Rate of esterification: O–R C—O bond of the acid is broken, not the C—O bond of the alcohol that is, the alcohol oxygen is incorporated into the ester not the oxygen from the acid —OH. + H2 O 2 ester O C O H+ C + H2O C O—CH3 CO2CH2C6H5 + H2 O HO2C R C OH . . O H .. + H O R 1 C R1 C CH3O OH H + R1 C CH3O + O—H H +O R 1 C O OH O O O 25 macrolactone linkage 21 OH R1 C O Me O OH OH OH .. .. O H OCH3 H+ OH Me R1 C OH CH3O + H .. .. O H OH O18CH3 Me CH3OH .. .. O H C Many naturally occurring macrolactones are known and many have important biological activities such as the antibiotic cytovarycin Me Mechanism of the esterification reaction: 1 HO18CH3 intramolecular ester are called lactones formation of five and six membered lactones is very fast: yielding stable esters O steric hindrance controls the rate of the reaction Estierification proceeds through a series of reversible steps involving protonation and deprotonation . . H OH H+ CH3OH > 1° > 2° > 3° + O H+ R3CCO2H < R2CHCO2H < RCH2CO2H < CH3CO2H < HCO2H .. .. O Chem 66H Me OCH3 OHO H O MeO Me Me OH Me OH Carboxylic Acids Chapter 19-1 Carboxylic acids contain both a carbonyl and a hydroxyl function .. .. O C .. R O—H .. H + .. .. O C .. – R O .. : + .. . – ..O . C .. R R O .. O C O O: H O C R carboxylic acid dimer: R C O H :O: results in higher melting and boiling points Spectral Properties —OH stretch in the infrared is intense due to dimers...3300 - 3000 C=O stretch 1700 - 1725 shifted to 1680 - 1700 if conjugated —COOH in 1H NMR at about 10 - 13 ppm as a broad singlet Preparation 1. Hydrolysis of carboxylic acid derivatives 2. oxidation of alcohols, aldehydes, or alkenes 3. Grignard reactions Hydrolysis of Carboxylic acid derivatives CH3 CH3 O C O C H2O, O—CH2CH3 Cl H+ or HO- CH3 O C O C O CH3 CH3 O C + HOCH2CH3 O—H CH3C≡N also yield acetic acid on hydrolysis Oxidation alcohols CH3CH2CH2CH2OH CH2OH H2CrO4 H2CrO4 CH3CH2CH2COOH COOH Chem 66H