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1 Organic Chemistry Mario Lintz 303-946-5838 1.

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1 1 Organic Chemistry Mario Lintz

2 2 Organic Chemistry I Functional Groups Molecular Structure Hydrocarbons Substitution and Elimination Oxygen Containing Compounds Amines 2

3 3 Functional Groups List #1- Critical for the MCAT Alkane Alkene Alkyne Alcohol Ether Amine Aldehyde Ketone Carboxylic Acid Ester Amide 3

4 4 Functional Groups List #2- Memorize as well Alkyl Halogen Gem-dihalide Vic dihalide Hydroxyl Alkoxy Hemiacetal Hemiaketal Mesyl group Tosyl group Carbonyl Acetal Acyl Anhydride Aryl Benzyl Hydrazine Hydrazone Vinyl Vinylic Allyl Nitrile Epoxide Enamine Imine Nitro Nitroso 4

5 55

6 6 Bonds Types: Ionic: complete transfer of electrons Covalent: shared electrons Coordinate covalent bonds- One atom provides both electrons in a shared pair. Polar covalent: unequal sharing of electrons Hydrogen Bonds: bonds between polar molecules containing H and O, N, or F 6

7 7 Bonds In the pi bond of an alkene, the electron pair have: A. 33% p character and are at a lower energy level than the electron pair in the o bond. B. 33% p character and are at a higher energy level than the electron pair in the o bond. C. 100% p character and are at a lower energy level than the electron pair in the o bond. D. 100% p character and are at a higher energy level than the electron pair in the o bond. 7

8 8 Covalent Bonds Sigma Pi 8

9 9 Covalent Bonds Sigma Between s orbitals Small, strong, lots of rotation Pi 9

10 10 Covalent Bonds Sigma Between s orbitals Small, strong, lots of rotation Pi Between p orbitals Discreet structure, weaker than sigma, no rotation 10

11 11 Covalent Bonds Sigma Between s orbitals Small, strong, lots of rotation Pi Between p orbitals Discreet structure, weaker than sigma, no rotation Always add to sigma bonds creating a stronger bond 11

12 12 When albuterol I dissolved in water, which of the following hydrogen-bonded structures does NOT contribute to its water solubility? 12

13 13 Dipole Moments (Solely responsible for Intermolecular Attractions) Charge distribution of bond is unequal Molecule with dipole moment = polar Molecule without dipole moment = nonpolar Possible to have nonpolar molecules with polar bonds Induced Dipoles Spontaneous formation of dipole moment in nonpolar molecule Occurs via: polar molecule, ion, or electric field Instantaneous Dipole Due to random e- movement Hydrogen Bonds Strongest dipole-dipole interaction Responsible for high BP of water London Dispersion Forces Between 2 instantaneous dipoles Responsible for phase change of nonpolar molecules 13

14 14 Lewis Dot Structures Rules for writing Find total # valence e- 1 e- pair = 1 bond Arrange remaining e- to satisfy duet and octet rules Exceptions Atoms containing more than an octet must come from the 3 rd period, (vacant d orbital required for hybridization) Not very popular on the MCAT Formal Charge # valence e- (isolated atom) - # valence e- (lewis structure) Sum of formal charge for each atom is the total charge on the molecule (actual charge distribution depends on electronegativity) 14

15 15 Structural Formulas Dash Formula Condensed Formula Bond-line Formula Fischer projection Newman projection Dash-line-wedge Ball and stick All Images courtesy of Exam Krackers 15

16 16 Hybridization 16

17 17 Hybrid Bonds SuffixC bondsHybridiz ation Percent S:P Bond Angle Bond Length Bond Strength -ane -ene -yne -yl 17

18 18 Hybrid Bonds SuffixC bondsHybridiz ation Percent S:P Bond Angle o Bond Length (pm) Bond Strength (kJ/mol) -aneC-Csp325: eneC=Csp233: yneC=CC=Csp50: yl Side chain 18

19 19 Hybrid Bonding in Oxygen and Nitrogen Nitrogen- Lone pair occupies more space than N-H Causes compression of the bond angle. Bond angles are as opposed to Oxygen- 2 sets of lone pair electrons Causes greater compression than in Nitrogen. H2O bond angles are vs

20 20 A) sp2, sp2 B) sp2, sp3 C) sp3, sp3 D) sp3, sp2 For the molecule 1,4 pentadiene, what type of hybridization is present in carbons # 1 and # 3 respectively? 20

21 21 VSEPR valance shell electron pair repulsion Prediction of shape Minimize electron repulsion 1. Draw the Lewis dot structure for the molecule or ion 2. Place electron pairs as far apart as possible, then large atoms, then small atoms 3. Name the molecular structure based on the position of the atoms (ignore electron pairs) 21

22 22 moleculeLewis structureShapemoleculeLewis structureShape BeCl 2 Linear, sp SF 4 SeesawSO 3 Trigonal planar, sp 2 ICl 3 T shapedNO 2 - Bent CH 4 Tetrahedral, sp 3 NH 3 Trigonal Pyramid al PCl 5 Trigonal bipyramidal, dsp 3 SF 6 Octahed ral, d 2 sp 3 IF 5 Square Pyramidal ICl 4 - Square Planar VSEPR 1. Draw the Lewis dot structure for the molecule or ion 2. Place electron pairs as far apart as possible, then large atoms, then small atoms 3. Name the molecular structure based on the position of the atoms (ignore electron pairs) 22

23 23 Delocalized e - and Resonance passage 25 Resonance forms differ only in the placement of pi bond and nonbonding e- Does not suggest that the bonds alternate between positions Neither represent the actual molecule, rather the real e assignment is the intermediate of the resonant structures. The real structure is called a resonance hybrid (cannot be seen on paper) 23

24 24 Organic Acids and Bases Organic Acids- Presence of positively charged H+ Two kinds present on a OH such as methyl alcohol present on a C next to a C=O such as acetone Organic Bases- Presence of lone pair e to bond to H Nitrogen containing molecules are most common Oxygen containing molecules act as bases in presence of strong acids 24

25 25 Stereochemistry Isomers: same elements, same proportions. Different spatial arrangements => different properties. Structural (constitutional): Different connectivity. Isobutane vs n-butane Both C 4 H 10 Conformational (rotational): Different spatial arrangement of same molecule Chair vs. boat Gauche vs Eclispsed vs Antistaggered vs Fully Eclipsed 25

26 26 Stereochemistry-isomers Stereoisomers: different 3D arrangement Enantiomers: mirror images, non- superimposable. Same physical properties (MP, BP, density, solubility, etc.) except rotation of light and reactions with other chiral compounds May function differently; e.g. thalidomide, sugars, AA Have chiral centers 26

27 27 Stereochemistry-isomers Stereoisomers: different 3D arrangement Diastereomers: not mirror images (cis/trans) Different physical properties (usually), Can be separated Chiral diastereomers have opposite configurations at one or more chiral centers, but have the same configuration at others. 27

28 28 Stereochemistry-isomers What kind of isomers are the two compounds below? A. Configurational diastereomers B. Enantiomers C. Constitutional isomers D. Cis -trans diastereomers 28

29 29 Stereochemistry-polarization of light Excess of one enantiomer causes rotation of plane-polarized light. Right, clockwise, dextrarotary (d), or + Left, counterclockwise, levarotary (l), or – Racemic: 50:50 mixture of 2 enantiomers, no net rotation of light RELATIVE Configuration: configuration of one molecule relative to another. Two molecules have the same relative configuration about a carbon if they differ by only one substituent and the other substituents are oriented identically about the carbon. Specific rotation [ : normalization for path length (l) and sample density (d). o cm 3 /g [ ] = / (l*d) 29

30 30 Stereochemistry-Chiral molecules passage 27 Achiral=plane or center of symmetry ABSOLUTE Configuration: physical orientation of atoms around a chiral center R and S: 1. Assign priority, 1 highest, 4 lowest H < C < O < F higher atomic #, higher priority If attachments are the same, look at the atoms (ethyl beats methyl) 2. Orient 4 away from the observer 3. Draw a circular arrow from 1 to 2 to 3 R = clockwise S = counterclockwise This has nothing to do with the rotation of light! E and Z: Different than cis and trans Z= same side of high priority groups E=opposite side of high priority groups 30

31 31 IUPAC Naming Conventions IUPAC Rules for Alkane Nomenclature 1. Find and name the longest continuous carbon chain. 2. Identify and name groups attached to this chain. 3. Number the chain consecutively, starting at the end nearest a substituent group. 4. Designate the location of each substituent group by an appropriate number and name. 5. Assemble the name, listing groups in alphabetical order. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing. 31

32 32 Hydrocarbons # of CRoot Name# of CRoot Name 1meth6hex 2eth7hept 3prop8oct 4but9non 5pent10dec 32

33 33 Hydrocarbons Saturated: C n H (2n+2)\ Unsaturated: C n H [2(n-u+1)] ; u is the # of sites of unsaturation Primary, secondary, tertiary, and quaternary carbons Know and be able to recognize the following structures n-butylsec-butyl iso-butyltert-butyl n-propyl Iso-propyl 33

34 34 Alkanes Physical Properties: Straight chains: MP and BP increase with length (increased van Der Waals interactions) C1-4: gas C5-17: liquid C18+: solid Branched chains: BP decreases (less surface area, fewer vDW) When compared to the straight chain analog, the straight chain will have a higher MP than the branched molecule. BUT, amongst branched molecules, the greater the branching, the higher the MP. 34

35 35 Alkanes-Important Reactions Very Unreactive Combustion: Alkane + Oxygen + High energy input (fire) Products: H 2 O, CO 2, Heat Halogenation Initiation with UV light Homolytic cleavage of diatomic halogen Yields a free radical Propagation (chain reaction mechanisms) Halogen radical removes H from alkyl Yields an alkyl radical Termination Radical bonds to wall of container or another radical Reactivity of halogens: F > Cl > Br >>> I Selectivity of halogens (How selective is the halogen in choosing a position on an alkane): I > Br > Cl > F more electronegative (Cl) means less selective (Br) Stability of free radicals: more highly substituted = more stable aryl>>>alkene> 3 o > 2 o > 1 o >methyl 35

36 36 Halogenation In the halogenation of an alkane, which of the following halogens will give the greatest percent yield of a tertiary alkyl halide when reacted with 2- methylpentane in the presence of UV light. A. F2 B. Cl2 C. Br2 D. 2-methylpentane will not yield a tertiary product 36

37 37 Cycloalkanes General formula: (CH2)n or CnH2n As MW increases BP increases though MP fluctuates irregularly because different shapes of cycloalkanes effects the efficiency in which molecules pack together in crystals. Ring strain in cyclic compounds: Bicyclic Molecules: 37

38 38 Cycloalkanes Naming Find parent Count Cs in ring vs longest chain. If # in ring is equal to or greater than chain, then name as a cycloalkane. Number the substituents and write the name Start at point of attachment and number so that subsequent substituents have the lowest # assignment If two or more different alkyl groups are present, number them by alphabetic priority If halogens are present, treat them like alkyl groups Cis vs Trans Think of a ring as having a top and bottom If two substituents both on top: cis It two substituents and 1 top, 1 bottom: trans 38

39 39 Cycloalkanes Ring Strain Zero for cyclohexane (All C-C-C bond angles: 111.5°) Increases as rings become smaller or larger (up to cyclononane) Cyclohexane Exist as chair and boat conformations Chair conformation preferred because it is at the lowest energy. Hydrogens occupy axial and equatorial positions. Axia (6)l- perpendicular to the ring Equatorial (6)- roughly in the plane of the ring Neither energetically favored When the ring reverses its conformation, substituents reverse their conformation Substituents favor equatorial positions because crowding occurs most often in the axial position. 39

40 40 Cyclohexanes In a sample of cis-1,2-dimethylcyclohexane at room temperature, the methyl groups will: A. Both be equatorial whenever the molecule is in the chair conformation. B. Both be axial whenever the molecule is in the chair conformation. C. Alternate between both equatorial and both axial whenever the molecule is in the chair conformation D. Both alternate between equatorial and axial but will never exist both axial or both equatorial at the same time 40

41 41 Substitutions Substitution: one functional group replaces another Electrophile: wants electrons, has partial + charge Nucleophile: donates electrons, has partial – charge 41

42 42 Substitution SN1: substitution, nucleophilic, unimolecular Rate depends only on the substrate (i.e. leaving group) R=k[reactant] Occurs when Nu has bulky side groups, stable carbocation (3 o ), weak Nu (good leaving group) Carbocation rearrangement Two step reaction 1. spontaneous formation of carbocation (SLOW) 2. Nucleophile attacks carbocation (chiral reactants yield racemic product mixtures) 42

43 43 Substitution SN2: substitution, nucleophilic, bimolecular Rate depends on the substrate and the nucleophile R=k[Nu][E] Inversion of configuration Occurs with poor leaving groups (1 o or 2 o ) One step reaction 1. Nu attacks the C with a partial + charge 43

44 44 Which of the following carbocations is the most stable? A.A.CH 3 CH 2 CH 2 CH 2 B.B.CH 3 CH 2 CH 2 CHCH 3 C.C.(CH 3 ) 3 C D.D.CH 3 44

45 45 Benzene Undergoes substitution not addition Flat molecule Stabilized by resonance Electron donating groups activate the ring and are ortho-para directors Electron withdrawing groups deactivate the ring and are meta directors Halogens are electron withdrawing, however, are ortho-para directors 45

46 46 Benzene Substituent Effects 46

47 47 Oxygen Containing Compounds Alcohols Aldehydes and Ketones Carboxylic Acids Acid Derivatives Acid Chlorides Anhydrides Amides Keto Acids and Esters 47

48 48 Alcohols One of the most common reactions of alcohols is nucleophilic substitution. Which of the following are TRUE in regards to S N 2 reactions: I. Inversion of configuration occurs II. Racemic mixture of products results III. Reaction rate = k [S][nucleophile] A. I only B. II only C. I and III only D. I, II, and III 48

49 49 Alcohols Physical Properties: Polar High MP and BP (H bonding) More substituted = more basic (CH3)3COH: pKa = CH3CH2OH: pKa = CH3OH: pKa = Electron withdrawing substituents stabilize alkoxide ion and lower pKa. Tert-butyl alcohol:pKa = Nonafluoro-tert-butyl alcohol:pKa = 5.4 IR absorption of OH at ~3400 cm - General principles H bonding Acidity: weak relative to other O containing compounds (CH groups are e - donating = destabilize deprotonated species) Branching: lowers BP and MP 49

50 50 Alcohols Naming Select longest C chain containing the hydroxyl group and derive the parent name by replacing –e ending of the corresponding alkane with –ol. Number the chain beginning at the end nearest the –OH group. Number the substituents according to their position on the chain, and write the name listing the substituents in alphabetical order. 50

51 51 Alcohols-Oxidation & Reduction Oxidation Reduction 51

52 52 Alcohols-Oxidation & Reduction Common oxidizing and reducing agents Generally for the MCAT Oxidizing agents have lots of oxygens Reducing agents have lots of hydrogens Oxidizing AgentsReducing Agents K 2 Cr 2 O 7 LiAlH 4 KMnO 4 NaBH 4 H 2 CrO 4 H 2 + Pressure O2O2 Br 2 52

53 53 Reduction Synthesis of Alcohols Reduction of aldehydes, ketones, esters, and acetates to alcohols. Accomplished using strong reducing agents such as NaBH 4 and LiAlH 4 Electron donating groups increase the negative charge on the carbon and make it less susceptible to nucleophilic attack. Reactivity: Aldehydes>Ketones>Esters>Acetates Only LiAlH 4 is strong enough to reduce esters and acetates 53

54 54 Alcohols: Pinacol rearrangement Starting with Vicinal Diol Generate ketones and aldehydes Formation of most stable carbocation Can get ring expansion or contraction 54

55 55 Alcohols-Protection Alcohol behaves as the nucleophile. (As is often the case) OH easily transfer H to a basic reagent, a problem in some reactions. Conversion of the OH to a removable functional group without an acidic proton protects the alcohol 55

56 56 56

57 57 Alcohols to Alkylhalides via a strong acid catalyst R-OH + HCl RCl + H 2 0 Alcohol is protonated by strong acid, (it takes a strong acid to protonate an alcohol). -OH is converted to the much better leaving group, H 2 O Occurs readily with tertiary alcohols via treatment with HCl or HBr. Primary and secondary alcohols are more resistant to acid and are best converted via treatment with SOCl 2 or PBr 3 57

58 58 Alcohols to Alkylhalides reactions with SOCl 2 and PBr 3 Halogenation of alcohols via SN1 or SN2 OH is the Nu, attacking the halogenating agent It is not OH that leaves, but a much better leaving group -OSOCl or –OPBr2, which is readily expelled by backside nucleophilic substitution. Does not require strong acids (HCl, HBr) 58

59 59 Alcohols-preparation of mesylates and tosylates OH is a poor leaving group, unless protonated, but most Nu are strong bases and remove such a proton Conversion to mesylates or tosylates allow for reactions with strong Nu Preparation SN1: no change of stereogenic center. Reaction SN2: inversion of configuration 1st/380,36,Tosylates allow control of stereochemistry and 59

60 60 Alcohols: Esterification Fischer Esterification Reaction: Alcohol + Carboxylic Acid Ester + Water Acid Catalyzed- protonates –OH to H 2 O (excellent leaving group) Alcohol performs nucleophilic attack on carbonyl carbon These bonds are broken 60

61 61 Alcohols: Inorganic Esters passage 30 Esters with another atom in place of the carbon 1. Sulfate esters: alcohol + sulfuric acid 2. Nitrate esters: alcohol + HNO3 (e.g. nitroglycerine) 3. Phosphate esters: DNA 61

62 62 Upon heating 2,3-Dimethyl-2,3-butanediol with aqueous acid, which of the following products would be obtained in the greatest amount? a) 3,3-Dimethyl-2-butanone b) 2,2-Dimethyl-3-butanone c) 2,3-Dimethyl-3-butanone d) 2,3-Dimethyl-2-butanone 62

63 63 In the reaction above, what is the purpose of using the 1,2-ethanediol in the first step? a) Heterogeneous catalyst b) Homogeneous catalyst c) Alcohol protection d) Oxidizing agent 63

64 64 In the reaction above, if the reagents in the first step were replaced with LiAlH4, what product would result? a)c) b)d) 64 OH O HO

65 65 Carbonyls Carbon double bonded to Oxygen Planar stereochemistry Partial positive charge on Carbon (susceptibility to nucleophilic attack) Aldehydes & Ketones (nucleophilic addition) Carboxylic Acids (nucleophilic substitution) Amides 65

66 66 Aldehydes and Ketones Physical properties: Carbonyl group is polar Higher BP and MP than alkanes because of dipole-dipole interactions More water soluble than alkanes Trigonal planar geometry, chemistry yields racemic mixtures IR absorption of C=O at ~1600 General principles: Effects of substituents on reactivity of C=O: e - withdrawing increase the carbocation nature and make the C=O more reactive Steric hindrance: ketones are less reactive than aldehydes Acidity of alpha hydrogen: carbanions, unsaturated carbonyls-resonance structures 66

67 67 Aldehydes and Ketones Naming Naming Aldehydes Replace terminal –e of corresponding alkane with –al. Parent chain must contain the –CHO group The –CHO carbon is C1 When –CHO is attached to a ring, the suffix carbaldehyde is used. Naming Ketones Replace terminal –e of corresponding alkane with –one. Parent chain is longest chain containing ketone Numbering begins at the end nearest the carbonyl C. 67

68 68 Aldehydes and Ketones- Acetal and Ketal Formation nucleophilic addition at C=O bond 68

69 69 Aldehydes and Ketones- Imine Formation nucleophilic addition at C=O bond Imine R2C=NR Primary amines (RNH2) + aldehyde or ketone R2C=NR Acid Catalyzed protonation of –OH H2O 69

70 70 Aldehydes and Ketones- Enamine Formation nucleophilic addition at C=O bond Enamine (ene + amine) R2N-CR=CR2 Secondary amine (R2N) + aldehyde or ketone R2N- CR=CR2 Acid catalyzed protonation of –OH H2O 70

71 71 Aldehydes and Ketones- reactions at adjacent positions Haloform: trihalomethane Halogens add to ketones at the alpha position in the presence of a base or acid. Used in qualitative analysis to indicate the presence of a methyl ketone. The product, iodoform, is yellow and has a characteristic odor. 71

72 72 Aldehydes and Ketones- reactions at adjacent positions Aldol (aldehyde + alcohol) condensation: Occurs at the alpha carbon Base catalyzed condensation Alkoxide ion formation (stronger than –OH, extracts H from H 2 O to complete aldol formation) Can use mixtures of different aldehydes and ketones 72

73 73 Aldehydes are easy to oxidize because of the adjacent hydrogen. In other words, they are good reducing agents. Potassium dichromate (VI): orange to green Tollens reagent (silver mirror test): grey ppt. Prevents reactions at C=C and other acid sensitive funtional groups in acidic conditions. Fehlings or benedicts solution (copper solution): blue to red Ketones, lacking such an oxygen, are resistant to oxidation. Aldehydes and Ketones-Oxidation (Aldehydes Carboxylic acids) 73

74 74 Aldehydes and Ketones Keto-enol Tautomerism: Keto tautomer is preferred (alcohols are more acidic than aldehydes and ketones). 74

75 75 Aldehydes and Ketones Internal H bonding: 1,3-dicarbonlys Enol tautomer is preferred (stabilized by resonance and internal H-bonding) 75

76 76 Guanine, the base portion of guanosine, exists as an equilibrium mixture of the keto and enol forms. Which of the following structures represents the enol form of guanine? 76

77 77 Aldehydes and Ketones Organometallic reagents: Nucleophilic addition of a carbanion to an aldehyde or ketone to yield an alcohol 77

78 78 Acetoacetic Ester Synthesis Alkyl Halide + Acetoacetic Ester Methyl Ketone Acetoacetic ester synthesis: Use acetoacetic ester (ethyl acetoacetate) to generate substituted methyl ketones Base catalyzed extraction of α H 78

79 79 Aldehydes and Ketones Wolff-Kishner reduction: Nucleophilic addition of hydrazine (H 2 N-NH 2 ) Replace =O with 2 H atoms + H 2 O 79

80 80 In which of the following reactions would the formation of an imine occur? a) Methylamine + propanol b) Methylamine + propanal c) Dimethylamine+ propanal d) Trimethylamine + propanal 80

81 81 In which of the following reactions would the formation of an enamine occur? a) Methylamine + propanol b) Methylamine + propanal c) Dimethylamine+ propanal d) Trimethylamine + propanal 81

82 82 In an organic chemistry class a group of students are trying to determine the identity of an unknown compound. In the haloform reaction the reaction mixture turned yellow indicating a positive result. Which of the following is true of the unknown compound? a) It contains an aldehyde b) It contains an alcohol c) It contains a methyl ketone d) It contains a carboxylic acid 82

83 83 Oxygen Containing Compounds-Carboxylic Acids Physical Properties: Acidic Trigonal planar geometry Higher BP and MP than alcohols Polarity, dimer formation in hydrogen bonding increases size and VDW interactions Solubility: small (n<5) CA are soluble, larger are less soluble because long hydrocarbon tails break up H bonding IR absorption of C=O at ~1600, OH at ~3400 General Principles: Acidity Increases with EWG (stabilize carboxylate) Acidity decreases with EDG (destabilize carboxylate) Relative reactivity Steric effects Electronic effects Strain (e.g. -lactams: 3C, 1N ring; inhibits bacterial cell wall formation) 83

84 84 Carboxylic Acids Naming Carboxylic acids derived from open chain alkanes are systematically named by replacing the terminal –e of the corresponding alkane name with –oic acid. Compounds that have a –CO2H group bonded to a ring are named using the suffix –carboxylic acid. The –CO2H group is attached to C #1 and is not itself numbered in the system. 84

85 85 Carboxylic Acids-important reactions Carboxyl group reactions: Nucleophilic attack: Carboxyl groups and their derivatives undergo nucleophilic substitution. Aldehydes and Ketones undergo addition because they lack a good leaving group. Must contain a good leaving group or a substituent that can be converted to a good leaving group. 85

86 86 Carboxylic Acids-important reactions Reduction: Form a primary alcohol LiAlH 4 is the reducing agent Unlike oxidation, cannot isolate the aldehyde CH 3 (CH 2 ) 6 COOH CH 3 (CH 2 ) 6 CH 2 OH LiAlH4 86

87 87 Carboxylic Acids-important reactions Carboxyl group reactions: Decarboxylation: 87

88 88 Carboxylic Acids-important reactions Fischer Esterification Reaction: Alcohol + Carboxylic Acid Ester + Water Acid Catalyzed- protonates –OH to H 2 O (excellent leaving group) Alcohol performs nucleophilic attack on carbonyl carbon These bonds are broken H+H+ 88

89 89 Carboxylic Acids- reactions at two positions Substitution reactions: keto reactions shown, consider enol reactions To make - > SOCl 2 or PCl 3 Heat, -H 2 O R'OH, heat, H+ - R 2 NH heat HO - 89

90 90 Carboxylic Acids- reactions at two positions passage 26 Halogenation: enol tautomer undergoes halogenation 90

91 91 Acid Derivatives- Acid Chlorides, Anhydrides, Amides, Esters Physical Properties: Acid chlorides: acyl chlorides React violently with water Polar Dipole attractions (no H bonds) Higher BP and MP than alkanes, lower than alcohols Anhydrides Large, polar molecules Dipole attractions (no H bonds) Higher BP than alkanes, lower than alcohols 91

92 92 Acid Derivatives Physical Properties: Amides: Highest BP and MP Soluble in water (H bonds) Esters: Poor to fair H bond acceptors Sparingly soluble in water Weakly basic H on alpha C weakly acidic 92

93 93 Acid Derivatives Naming Acid Halides (RCOX) Identify the acyl group and then the halide Replace –ic acd with –yl, or –carboxylic acid with –carbonyl Acid Anhydrides (RCO2COR) Symmetrical anhydrides or unsubstituted monocarboxylic acids and cyclic anhydrides of dicarboxylic acids are named by replacing the word acid with anhydride. 2 acetic acid acetic anhydride Anhydrides derived from substituted monocarboxylic acids are named by adding the prefix –bis to the acid name. 2 chloroacetic acid bis(chloroacetic) anhydride Unsymmetrical anhydrides- those produced from two different carboxylic acids- are named by citing the two acids alphabetically. Acetic acid + benzoic acid acetic benzoic anhydride Amides (RCONH2) Amides with an unsubstituted –NH2 group are named by replacing the –oic acid or ic acid ending with amide, or by replacing the –carboxylic acid ending with carboxamide. Acetic acid acetamide If the nitrogen atom is further substituted, the compound is named by first identifying the substituent groups and then the parent amide. The substituents are preceded by the letter N to identify them as being directly attached to nitrogen. Propanoic acid + methyl amine N-Methylpropanamide Esters (RCO2R) Identify the alkyl group attached to oxygen and then the carboxylic acid. Replace the –ic acid ending with -ate 93

94 94 Acid Derivatives- Relative Reactivity and Reactions of Derivatives A more reactive acid derivative can be converted to a less reactive one, but not vice versa Only esters and amides commonly found in nature. Acid halides and anhydrides react rapidly with water and do not exist in living organisms Hydrolysis- +water carboxylic acid Alcoholysis- +alcohol ester Aminolysis- +ammonia or amine amide Reduction- + H- aldehyde or alcohol Grignard- + Organometallic ketone or alcohol 94

95 95 Acid Derivatives-important reactions Preparation: replace OH Nucleophilic Substitution: 95

96 96 Acid Derivatives Hoffman Degredation Hoffman degradation (rearrangement) of amides; migration of an aryl group 1° Amides + Strong basic Br or Cl soln 1° Amines + CO2 /resources/name_reactions/hofmann.s html 96

97 97 Acid Derivatives Transesterification Transesterification: exchange alkoxyl group with ester of another alcohol Alcohol + Ester Different Alcohol + Different Ester 97

98 98 Acid Derivatives- Saponification Saponification- ester hydrolysis in basic solutions 98

99 99 Acid Derivatives- Hydrolysis of Amides passage 33 Hydrolysis of amides: Acid or base catalyzed 99

100 100 Acid Derivatives Strain (e.g., β-lactams) Lactams- cyclic amides Although amides are most stable acid derivative, β-lactams are highly reactive due to ring strain. Subject to nuclephilic attack. Found in several types of antibiotics Inhibits bacterial cell wall formation. 100

101 101 Keto-Acids and Esters Keto acids contain a ketone and a carboxyl group (alpha and beta) Amino acids degraded to alpha keto acids and then go into the TCA Esters have distinctive odors and are used as artificial flavors and fragrances Beta-keto esters have an acidic alpha hydrogen Consider keto-enol tautomerism Naming Esters Esters are named by first determining the alkyl group attached to the oxygen and then the carboxylic acid from which the ester is derived. EX: Methyl Propanoate is derived from propanoic acid and a methyl group 101

102 102 Keto Acids and Esters- important reactions Decarboxylation Acetoacetic ester synthesis: see aldehydes and ketones 102

103 103 Amines Important functions in amino acids, nucleotides, neurotransmitters 1 o, 2 o, 3 o, 4 o based on how many carbons bonded to Can be chiral, rarely have 4 side groups 103

104 104 Amines Important functions in amino acids, nucleotides, neurotransmitters 1 o, 2 o, 3 o, 4 o based on how many carbons bonded to Can be chiral, rarely have 4 side groups Physical properties: Polar Similar reactivity to alcohols Can H bond, but weaker H bond than alcohols MP and BP higher than alkanes, lower than alcohols IR absorption:

105 105 Amines Important functions in amino acids, nucleotides, neurotransmitters 1 o, 2 o, 3 o, 4 o based on how many carbons bonded to Can be chiral, rarely have 4 side groups Physical properties: Polar Similar reactivity to alcohols Can H bond, but weaker H bond than alcohols MP and BP higher than alkanes, lower than alcohols IR absorption: General principles: Lewis bases when they have a lone electron pair NR 3 > NR 2 > NR > NH 3 (least basic) Stabilize adjacent carbocations and carbanions Effect of substituents on basicity of aromatic amines: Electron withdrawing are less basic Electron donating are more basic 105

106 106 Amines-major reactions Amines are basic and fairly nucleophilic Amide formation: proteins 106

107 107 Amines-major reactions Reactions with nitrous acid (HONO): Distinguishes primary, secondary, and tertiary Primary: burst of colorless, odorless N 2 gas Secondary: yellow oil, nitrosamine-powerful carcinogen Tertiary: colorless solution, amine forms an ion, e.g. (CH 3 ) 3 NH + 107

108 108 Amines- Alkylation Alkylation: SN2 with amine as the nucleophile and alkyl halides as the electrophile Reaction with 1° alkyl halide Alkylation of 1° and 2° are difficult to control and often lead to mixtures of products Alkylation of 3° amines yield quaternary ammonium salts 108

109 109 Amines- Hoffman Elimination Elimination of amine as a quaternary ammonium salt to yield an alkene. Does not follow Zaitsevs rule. Less highly substituted alkene predominates 109

110 110 For Next Time… Last Slide (Hooray!!!!) Functional Group Quiz Spectra Separations and Purifications Biological Molecules Carbohydrates Amino acids and proteins Lipids Phophorous containing compounds 110

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