Organic Chemistry IMario Lintz

1st Year MD/PhD CandidateMario.Lintz@UCDenver.edu

303-946-5838Topics Covered

Functional Groups Molecular Structure Hydrocarbons Substitution and Elimination Oxygen Containing Compounds Amines

Functional Groups- List #1- Critical for the MCAT

Alkane Alcohol Aldehyde EsterAlkene Ether Ketone AmideAlkyne Amine Carboxylic Acid

Functional Groups- List #2- Memorize as wellAlkyl Halogen Gem-dihalide Vic-dihalideHydroxyl Alkoxy Hemiacetal HemiketalMesyl group Tosyl group Carbonyl AcetalAcyl Anhydride Aryl BenzylHydrazine Hydrazone Vinyl VinylicAllyl Nitrile Epoxide EnamineImine Nitro Nitroso

Bonds Types:

o Ionic: complete transfer of electronso Covalent: shared electrons

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

Problem 1In 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.

Covalent Bondso Sigma σ

Between s orbitals Small, strong, lots of rotation

o Pi π Between p orbitals Discreet structure, weaker than sigma, no

rotation Always add to sigma bonds creating a

stronger bond

Problem 2When albuterol I dissolved in water, which of the following hydrogen-bonded structures does NOT contribute to its water solubility?

Dipole Moments- Solely responsible for Intermolecular Attractions Charge distribution of bond is unequal

o Molecule with dipole moment = polaro Molecule without dipole moment = nonpolar o Possible to have nonpolar molecules with polar bonds

Induced Dipoleso Spontaneous formation of dipole moment in nonpolar moleculeo Occurs via: polar molecule, ion, or electric field

Instantaneous Dipole o Due to random e- movement

Hydrogen Bondso Strongest dipole-dipole interactiono Responsible for high BP of water

London Dispersion Forceso Between 2 instantaneous dipoleso Responsible for phase change of nonpolar molecules

Lewis Dot Structures Rules for writing

o Find total # valence e-o 1 e- pair = 1 bondo Arrange remaining e- to satisfy duet and octet rules

Exceptionso Atoms containing more than an octet must come from the 3rd period, (vacant d orbital required for

hybridization)o Not very popular on the MCAT

Formal Chargeo # valence e- (isolated atom) - # valence e- (lewis structure)o Sum of formal charge for each atom is the total charge on the moleculeo Actual charge distribution depends on electronegativity

Structural Formulas

Dash Formula Condensed Formula Bond-line formula

Fischer projection Newman Projection Dash-line-wedge formula

Ball and stick

Hybridization

Hybrid Bonds

Suffix C bonds Hybridization Bond angle

Notes Bond length (pm)**

Bond Energy(kJ/mol)**

-ane C-C sp3 109.5o Has 4 things around it, only molecules that can be chiral

154 346

-ene C=C sp2 120 o Has 3 things around it 134 612-yne C C Sp 180 o Has 2 things around it 120 835-yl Side

chainAlkyl or enyl

Hybrid Bonding in Oxygen and Nitrogen Nitrogen-

o Lone pair occupies more space than N-H o Causes compression of the bond angle. Bond angles are 107.3 as opposed to 109.5

Oxygen-o 2 sets of lone pair electronso Causes greater compression than in Nitrogen. H2O bond angles are 104.5 vs 109.5.

Problem 3For the molecule 1,4 pentadiene, what type of hybridization is present in carbons # 1 and # 3 respectively?A) sp2, sp2B) sp2, sp3C) sp3, sp3D) sp3, sp2

VSEPR Valance Shell Electron Pair Repulsion Prediction of shape Minimize electron repulsion

1. Draw the Lewis dot structure for the molecule or ion2. Place electron pairs as far apart as possible, then large atoms, then small atoms3. Name the molecular structure based on the position of the atoms (ignore electron pairs)

Molecule Lewis structure Shape molecule Lewis structure Shape

BeCl2 Linear, sp

SF4 Seesaw SO3Trigonal

planar, sp2

ICl3 T shaped NO2- Bent

CH4Tetrahedral,

sp3 NH3Trigonal

Pyramidal

PCl5

Trigonal bipyramidal,

dsp3

SF6Octrahedral,

d2sp3

IF5Square

PyramidalICl4

- Square Planar

Delocalized e- and Resonance 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)

Passage 25Organic Acids and Bases

Organic Acids- Presence of positively charged H+o 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 Ho Nitrogen containing molecules are most commono Oxygen containing molecules act as bases in presence of strong acids

Stereochemistry- Isomers Isomers: same elements, same proportions. Different spatial arrangements => different properties.

o Structural (constitutional): Different connectivity. Isobutane vs n-butane Both C4H10

o Conformational (rotational): Different spatial arrangement of same molecule Chair vs. boat Gauche vs Eclispsed vs Antistaggered vs Fully Eclipsed

Stereoisomers: different 3D arrangemento 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

Diastereomers: not mirror images (cis/trans)o Different physical properties (usually), o Can be separatedo Chiral diastereomers have opposite configurations at one or more chiral centers, but have the same

configuration at others.

Problem 4What kind of isomers are the two compounds below?

A)Configurational diastereomers B) Enantiomers C) Constitutional isomersD) Cis -trans diastereomers

Stereochemistry- Plane Polarization of light Excess of one enantiomer causes rotation of plane-polarized light

o Right, clockwise, dextrarotary (d), or +o 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 [α]: normalization for path length (l) and sample density (d). ocm3/g[α] = α / (l*d)

Stereochemistry-Chiral molecules 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 b atoms (ethyl beats methyl)

2. Orient 4 away from the observer3. 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

o Z= same side of high priority groupso E=opposite side of high priority groups

Passage 27

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.

Hydrocarbons# of C Root Name # of C Root Name

1 Meth- 6 Hex-2 Eth- 7 Hept-3 Prop- 8 Oct-4 But- 9 Non-5 Pent- 10 Dec-

Saturated: CnH(2n+2)\

Unsaturated: CnH[2(n-u+1)] ; u is the # of sites of unsaturation o Add the number of halogens to the number of hydrogenso Ignore the number of oxygenso Subtract the number of nitrogens from the number of halogens

Primary, secondary, tertiary, and quaternary carbonsKnow and be able to recognize the following structures

Alkanes- Physical Properties Straight chains: MP and BP increase with length (increased van Der Waals interactions)

o C1-4: gaso C5-17: liquido C18+: solid

Branched chains: o BP decreases (less surface area, fewer vDW)o 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.

Alkanes-Important Reactions Very Unreactive

o Combustion: Alkane + Oxygen + High energy input (fire) Products: H2O, CO2, Heat

o 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

o Reactivity of halogens: F > Cl > Br >>> Io 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)

o Stability of free radicals: more highly substituted = more stable o aryl>>>alkene> 3o > 2o > 1o >methyl

Problem 5In 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) F2B) Cl2C) Br2D) 2-methylpentane will not yield a tertiary product

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:

Cycloalkanes- Naming1) Find parent2) Count C’s in ring vs longest chain. If # in ring is equal to or greater than chain, then name as a cycloalkane. 3) Number the substituents and write the name

4) Start at point of attachment and number so that subsequent substituents have the lowest # assignment5) If two or more different alkyl groups are present, number them by alphabetic priority6) If halogens are present, treat them like alkyl groups7) Cis vs Trans8) Think of a ring as having a top and bottom9) If two substituents both on top: cis 10) It two substituents and 1 top, 1 bottom: trans

Cycloalkanes Ring Strain

o Zero for cyclohexane (All C-C-C bond angles: 111.5°)o Increases as rings become smaller or larger (up to cyclononane)

Cyclohexane o Exist as chair and boat conformationso Chair conformation preferred because it is at the lowest energy.o Hydrogens occupy axial and equatorial positions.o Axia (6)l- perpendicular to the ringo Equatorial (6)- roughly in the plane of the ringo Neither energetically favoredo When the ring reverses its conformation, substituents reverse their conformationo Substituents favor equatorial positions because crowding occurs most often in the axial position.

Problem 6In 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 conformationD) Both alternate between equatorial and axial but will never exist both axial or both equatorial at the same time

Substitutions and Eliminations Substitution: one functional group replaces another

o Electrophile: wants electrons, has partial + chargeo Nucleophile: donates electrons, has partial – charge

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 (3o), weak Nu (good leaving group) Carbocation rearrangement Two step reaction

o 1)spontaneous formation of carbocation (SLOW) 2) Nucleophile attacks carbocation (chiral reactants yield racemic product mixtures)

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

o 1) Nu attacks the C with a partial + charge

Problem 7Which of the following carbocations is the most stable?

A) CH3CH2CH2CH2*B) CH3CH2CH2CH*CH3

C) (CH3)3C*D) CH3*

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

Benzene- Substituent Effects

Oxygen Containing Compounds Alcohols Aldehydes and Ketones Carboxylic Acids Acid Derivatives

o Acid Chlorideso Anhydrideso Amides

Keto Acids and Esters

Problem 8One of the most common reactions of alcohols is nucleophilic substitution. Which of the following are TRUE in regards to SN2 reactions:

I. Inversion of configuration occursII. Racemic mixture of products results

III. Reaction rate = k [S][nucleophile]A) I onlyB) II onlyC) I and III onlyD) I, II, and III

Alcohols Physical Properties:

o Polaro High MP and BP (H bonding)o More substituted = more basic

(CH3)3COH: pKa = 18.00 CH3CH2OH: pKa = 16.00 CH3OH: pKa = 15.54

o Electron withdrawing substituents stabilize alkoxide ion and lower pKa. Tert-butyl alcohol: pKa = 18.00 Nonafluoro-tert-butyl alcohol: pKa = 5.4

o IR absorption of OH at ~3400 cm-

o 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

Alcohols- Naming1. Select longest C chain containing the hydroxyl group and derive the parent name by replacing –e ending of the

corresponding alkane with –ol.2. Number the chain beginning at the end nearest the –OH group.3. Number the substituents according to their position on the chain, and write the name listing the substituents in

alphabetical order.

Alcohols-Oxidation & Reduction

Common oxidizing and reducing agentso Generally for the MCAT

Oxidizing agents have lots of oxygens Reducing agents have lots of hydrogens

Oxidizing Agents Reducing AgentsK2Cr2O7 LiAlH4

KMnO4 NaBH4

H2CrO4 H2 + PressureO2

Br2

Reduction Synthesis of Alcohols Reduction of aldehydes, ketones, esters, and acetates to alcohols. Accomplished using strong reducing agents such as NaBH4 and LiAlH4

Electron donating groups increase the negative charge on the carbon and make it less susceptible to nucleophilic attack.

o Reactivity: Aldehydes>Ketones>Esters>Acetates Only LiAlH4 is strong enough to reduce esters and acetates

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

Protection Involves 3 steps:

o 1) introduce protecting group to block interfering functiono 2) carry out desired reactiono 3) remove the protecting group

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 One common method of alcohol protection is the reaction of chlorotrimethylsilane to yield a trimethysilyl (TMS)

ether. The reaction is carried out in the presence of a base (often triethyl amine) to facilitate formation of the alkoxide anion from the alcohol and to remove the HCl by-product from the reaction

Alcohols to Alkylhalides via a strong acid catalyst R-OH + HCl à RCl + H20 Alcohol is protonated by strong acid, (it takes a strong acid to protonate an alcohol). -OH is converted to the much better leaving group, H2O 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 SOCl2 or

PBr3

Alcohols to Alkylhalides: Reactions with SOCl2 and PBr3

Halogenation of alcohols via SN1 or SN2 Best for primary and secondary alcohols 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 of the displaced halide ion. Does not require strong acids (HCl, HBr)

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

Esterification Fischer Esterification Reaction: Alcohol + Carboxylic Acid à Ester + Water Acid Catalyzed- protonates –OH to H2O (excellent leaving group) Alcohol performs nucleophilic attack on carbonyl carbon

Inorganic Esters Esters with another atom in place of the carbon

1. Sulfate esters: alcohol + sulfuric acid2. Nitrate esters: alcohol + HNO3 (e.g. nitroglycerine)3. Phosphate esters: DNA

Passage 30

Problem 8Upon 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-butanoneB) 2,2-Dimethyl-3-butanoneC) 2,3-Dimethyl-3-butanoneD) 2,3-Dimethyl-2-butanone

Problem 9

In the reaction above, what is the purpose of using the 1,2-ethanediol in the first step?A) Heterogeneous catalystB) Homogeneous catalystC) Alcohol protectionD) Oxidizing agent

Problem 10In the same reaction above, if the reagents in the first step were replaced with LiAlH4, what product would result?

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

Aldehydes and Ketones Physical properties:

o Carbonyl group is polar

o Higher BP and MP than alkanes because of dipole-dipole interactionso More water soluble than alkanes o Trigonal planar geometry, chemistry yields racemic mixtures

IR absorption of C=O at ~1600 General principles:

o Effects of substituents on reactivity of C=O: e- withdrawing increase the carbocation nature and make the C=O more reactive

o Steric hindrance: ketones are less reactive than aldehydes o Acidity of alpha hydrogen: carbanions o α, β unsaturated carbonyls-resonance structures

Naming Naming Aldehydes

1) Replace terminal –e of corresponding alkane with –al.2) Parent chain must contain the –CHO group3) The –CHO carbon is C14) When –CHO is attached to a ring, the suffix carbaldehyde is used.

Naming Ketones 1) Replace terminal –e of corresponding alkane with –one. 2) Parent chain is longest chain containing ketone 3) Numbering begins at the end nearest the carbonyl C.

Acetal and Ketal Formation Nucleophilic addition at C=O bond

Imine Formation Nucleophilic addition at C=O bond Imine R2C=NR Primary amines (RNH2) + aldehyde or ketone à R2C=NR Acid Catalyzed protonation of –OH à H2O

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

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.

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 H2O to complete aldol formation)

Can use mixtures of different aldehydes and ketones

Oxidation (Aldehydes à Carboxylic acids) 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.

Keto-enol Tautomerism

Keto tautomer is preferred (alcohols are more acidic than aldehydes and ketones).

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

Problem 11Guanine, 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?

Organometallic reagents

Nucleophilic addition of a carbanion to an aldehyde or ketone to yield an alcohol

Acetoacetic Ester Synthesis Alkyl Halide + Acetoacetic Ester à Methyl Ketone

Use acetoacetic ester (ethyl acetoacetate) to generate substituted methyl ketones Base catalyzed extraction of α H

Wolff-Kishner reduction Nucleophilic addition of hydrazine (H2N-NH2)

Replace =O with 2 H atoms

Problem 12In which of the following reactions would the formation of an imine occur?

A) Methylamine + propanol B) Methylamine + propanal C) Dimethylamine+ propanal D) Trimethylamine + propanal

Problem 13In which of the following reactions would the formation of an enamine occur?

A) Methylamine + propanol B) Methylamine + propanal C) Dimethylamine+ propanal D) Trimethylamine + propanal

Problem 14In 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 alcoholC) It contains a methyl ketone D) It contains a carboxylic acid

Carboxylic Acids Physical Properties:

o Acidico Trigonal planar geometryo Higher BP and MP than alcohols

Polarity, dimer formation in hydrogen bonding increases size and VDW interactionso Solubility: small (n<5) CA are soluble, larger are less soluble because long hydrocarbon tails break up H

bondingo IR absorption of C=O at ~1600, OH at ~3400o General Principles:o Acidity Increases with EWG (stabilize carboxylate)o Acidity decreases with EDG (destabilize carboxylate)o Relative reactivityo Steric effectso Electronic effectso Strain (e.g. b-lactams: 3C, 1N ring; inhibits bacterial cell wall formation)

Naming1) Carboxylic acids derived from open chain alkanes are systematically named by replacing the terminal –e of the

corresponding alkane name with –oic acid.2) Compounds that have a –CO2H group bonded to a ring are named using the suffix –carboxylic acid.3) The –CO2H group is attached to C #1 and is not itself numbered in the system.

Nucleophilic attack at Carboxyl group Carboxyl groups and their derivatives undergo nucleophilic substitution.

o 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.

Reduction Form a primary alcohol LiAlH4 is the reducing agent

o Unlike

oxidation, cannot isolate the aldehyde

Decarboxylation Removal of COO-

Fischer Esterification Reaction Alcohol + Carboxylic Acid à Ester + Water

o Acid Catalyzed- protonates –OH to H2O (excellent leaving group)o Alcohol performs nucleophilic attack on carbonyl carbon

Reactions at Two Positions Substitution reactions The substitution reactions shown involve the keto forms, consider the reactions of the enol forms

Passage 26Reactions at Two Positions

Halogenation: enol tautomer undergoes halogenation

Acid Derivatives- Acid Chlorides, Anhydrides, Amides, Esters Physical Properties:

o Acid chlorides: acyl chlorides React violently with water Polar Dipole attractions (no H bonds) Higher BP and MP than alkanes, lower than alcohols

o Anhydrides Large, polar molecules Dipole attractions (no H bonds) Higher BP than alkanes, lower than alcohols

o Amides: Highest BP and MP Soluble in water (H bonds)

o Esters: Poor to fair H bond acceptors Sparingly soluble in water Weakly basic H on alpha C weakly acidic

Naming Acid Halides (RCOX)

1) Identify the acyl group and then the halide2) Replace –ic acd with –yl, or –carboxylic acid with –carbonyl

Acid Anhydrides (RCO2COR’)1) Symmetrical anhydrides or unsubstituted monocarboxylic acids and cyclic anhydrides of dicarboxylic acids

are named by replacing the word acid with anhydride. 2 acetic acid à acetic anhydride

2) Anhydrides derived from substituted monocarboxylic acids are named by adding the prefix –bis to the acid name.

2 chloroacetic acid à bis(chloroacetic) anhydride3) Unsymmetrical anhydrides- those produced from two different carboxylic acids- are named by citing the two

acids alphabetically. Acetic acid + benzoic acid à acetic benzoic anhydride

Amides (RCONH2)1) 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 à acetamide

2) 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 à N-Methylpropanamide Esters (RCO2R)

1) Identify the alkyl group attached to oxygen and then the carboxylic acid. 2) Replace the –ic acid ending with -ate

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 à carboxylic acid Alcoholysis- +alcohol à ester Aminolysis- +ammonia or amine à amide Reduction- + H- à aldehyde or alcohol Grignard- + Organometallic à ketone or alcohol

Preparation of Acid Derivatives Replace OH

Nucleophilic Substitutiono Starting with an acid chloride

o Starting with an acid anhydride

Hoffman Degredation Hoffman degradation (rearrangement) of amides; migration of an aryl group 1° Amides + Strong basic Br or Cl soln à 1° Amines + CO2

Transesterification Transesterification: exchange alkoxyl group with ester of another alcohol Alcohol + Ester à Different Alcohol + Different Ester

Saponification Saponification- ester hydrolysis in basic solutions

Hydrolysis of Amides Acid or base catalyzed

Passage 33

Strain (e.g., β-lactams) Lactams- cyclic amides Although amides are most stable acid derivative, β-lactams are highly reactive due to ring strain.

o Subject to nuclephilic attack. Found in several types of antibiotics

o Inhibits bacterial cell wall formation.

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 Esters1) 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

Decarboxylation

Acetoacetic ester synthesis: see aldehydes and ketones

Amines

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

o Polaro Similar reactivity to alcoholso Can H bond, but weaker H bond than alcoholso MP and BP higher than alkanes, lower than alcohols

IR absorption: 2800-3000 General principles:

o Lewis bases when they have a lone electron pair NR3 > NR2 > NR > NH3 (least basic)

o Stabilize adjacent carbocations and carbanions o Effect of substituents on basicity of aromatic amines:

Electron withdrawing are less basic Electron donating are more basic

Major reactions Amines are basic and fairly nucleophilic Amide formation: proteins

Reactions with nitrous acid (HONO) Distinguishes primary, secondary, and tertiary

Primary: burst of colorless, odorless N2 gas

Secondary: yellow oil, nitrosamine-powerful carcinogen

Tertiary: colorless solution, amine forms an ion, e.g. (CH3)3NH+

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

Hoffman Elimination Elimination of amine as a quaternary ammonium salt to yield

an alkene.

Does not follow Zaitsev’s rule. Less highly substituted alkene predominates