Organic Chemistry I
Mario Lintz
1 Year MD/PhD Candidate
Mario.Lintz@UCDenver.edu
303-946-5838
st
Topics Covered
 Functional Groups
 Molecular Structure
 Hydrocarbons
 Substitution and Elimination
 Oxygen Containing Compounds
 Amines
Functional Groups- List #1- Critical for the MCAT
Alkane
Alkene
Alkyne
Alcohol
Ether
Amine
Aldehyde
Ketone
Carboxylic Acid
Ester
Amide
Functional Groups- List #2- Memorize as well
Alkyl
Halogen
Hydroxyl
Alkoxy
Mesyl group
Tosyl group
Acyl
Anhydride
Hydrazine
Hydrazone
Allyl
Nitrile
Imine
Nitro
Gem-dihalide
Hemiacetal
Carbonyl
Aryl
Vinyl
Epoxide
Nitroso
Vic-dihalide
Hemiketal
Acetal
Benzyl
Vinylic
Enamine
Bonds
 Types:
o Ionic: complete transfer of electrons
o Covalent: shared electrons
 Coordinate covalent bonds- One atom provides both electrons in a shared pair.
o Polar covalent: unequal sharing of electrons
o Hydrogen Bonds: bonds between polar molecules containing H and O, N, or F
Problem 1
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.

Covalent Bonds
o 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 2
When 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 = polar
o Molecule without dipole moment = nonpolar
o Possible to have nonpolar molecules with polar bonds
 Induced Dipoles
o Spontaneous formation of dipole moment in nonpolar molecule
o Occurs via: polar molecule, ion, or electric field
 Instantaneous Dipole
o Due to random e- movement
 Hydrogen Bonds
o Strongest dipole-dipole interaction
o Responsible for high BP of water
 London Dispersion Forces
o Between 2 instantaneous dipoles
o Responsible for phase change of nonpolar
molecules
Lewis Dot Structures
 Rules for writing
o Find total # valence eo 1 e- pair = 1 bond
o Arrange remaining e- to satisfy duet and octet rules
 Exceptions
o 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 Charge
o # valence e- (isolated atom) - # valence e- (lewis structure)
o Sum of formal charge for each atom is the total charge on the molecule
o Actual charge distribution depends on electronegativity
Structural Formulas
Dash Formula
Condensed Formula
Bond-line formula
Fischer projection
Ball and stick
Hybridization
Newman Projection
Dash-line-wedge formula
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
o
Has 2 things around it
120
835
-yne
C C
-yl
Side
chain
Sp
180
Alkyl or enyl
Hybrid Bonding in Oxygen and Nitrogen
 Nitrogeno 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
 Oxygeno 2 sets of lone pair electrons
o Causes greater compression than in Nitrogen. H2O bond angles are 104.5 vs 109.5.
Problem 3
For the molecule 1,4 pentadiene, what type of hybridization is present in carbons # 1 and # 3 respectively?
A) sp2, sp2
B) sp2, sp3
C) sp3, sp3
D) sp3, sp2
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)
Molecule
Lewis structure
Shape
BeCl2
Linear, sp
SF4
Seesaw
molecule
SO3
Lewis structure
Shape
Trigonal
planar, sp2
ICl3
T shaped
NO2-
Bent
CH4
Tetrahedral,
sp3
NH3
Trigonal
Pyramidal
PCl5
Trigonal
bipyramidal,
dsp3
SF6
Octrahedral,
d2sp3
IF5
Square
Pyramidal
ICl4-
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 25
Organic 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 H
o Nitrogen containing molecules are most common
o 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 arrangement
o 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 separated
o Chiral diastereomers have opposite configurations at one or more chiral centers, but have the same
configuration at others.
Problem 4
What kind of isomers are the two compounds below?
A)Configurational diastereomers
B) Enantiomers
C) Constitutional isomers
D) 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 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
o Z= same side of high priority groups
o 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
several groups of the same kind, are not considered when alphabetizing.
to designate
Hydrocarbons
# of C
1
2
3
4
5
Root Name
MethEthPropButPent-
# of C
6
7
8
9
10
Root Name
HexHeptOctNonDec-


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 hydrogens
o Ignore the number of oxygens
o Subtract the number of nitrogens from the number of halogens
 Primary, secondary, tertiary, and quaternary carbons
Know 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: gas
o C5-17: liquid
o 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 >>> I
o 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 5
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
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- Naming
1) Find parent
2) 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 # assignment
5) If two or more different alkyl groups are present, number them by alphabetic priority
6) If halogens are present, treat them like alkyl groups
7) Cis vs Trans
8) Think of a ring as having a top and bottom
9) 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 conformations
o Chair conformation preferred because it is at the lowest energy.
o Hydrogens occupy axial and equatorial positions.
o Axia (6)l- perpendicular to the ring
o Equatorial (6)- roughly in the plane of the ring
o Neither energetically favored
o When the ring reverses its conformation, substituents reverse their conformation
o Substituents favor equatorial positions because crowding occurs most often in the axial position.
Problem 6
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
Substitutions and Eliminations

Substitution: one functional group replaces another
o Electrophile: wants electrons, has partial + charge
o 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 7
Which 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 Chlorides
o Anhydrides
o Amides
Keto Acids and Esters
Problem 8
One 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 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
Alcohols

Physical Properties:
o Polar
o 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 cmo 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- Naming
1. 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 agents
o Generally for the MCAT
 Oxidizing agents have lots of oxygens
 Reducing agents have lots of hydrogens
Oxidizing Agents
K2Cr2O7
KMnO4
H2CrO4
O2
Br2
Reducing Agents
LiAlH4
NaBH4
H2 + Pressure
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 function
o 2) carry out desired reaction
o 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  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  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 acid
2. Nitrate esters: alcohol + HNO3 (e.g. nitroglycerine)
3. Phosphate esters: DNA
Passage 30
Problem 8
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
Problem 9
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
Problem 10
In 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 interactions
o 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 group
3) The –CHO carbon is C1
4) 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  R2C=NR
 Acid Catalyzed protonation of –OH  H2O
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
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  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
11
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?
Organometallic reagents
 Nucleophilic addition of a carbanion to an aldehyde or ketone to yield an alcohol



Acetoacetic Ester Synthesis
Alkyl Halide + Acetoacetic Ester  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 12
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
Problem 13
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
Problem 14
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
Carboxylic Acids

Physical Properties:
o Acidic
o Trigonal planar geometry
o Higher BP and MP than alcohols
 Polarity, dimer formation in hydrogen bonding increases size and VDW interactions
o Solubility: small (n<5) CA are soluble, larger are less soluble because long hydrocarbon tails break up H
bonding
o IR absorption of C=O at ~1600, OH at ~3400
o General Principles:
o Acidity Increases with EWG (stabilize carboxylate)
o Acidity decreases with EDG (destabilize carboxylate)
o Relative reactivity
o Steric effects
o Electronic effects
o Strain (e.g. b-lactams: 3C, 1N ring; inhibits bacterial cell wall formation)
Naming
1) 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  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 26
Reactions 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 halide
2) 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  acetic anhydride
2) Anhydrides derived from substituted monocarboxylic acids are named by adding the prefix –bis to the acid
name.
 2 chloroacetic acid  bis(chloroacetic) anhydride
3) 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)
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  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  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  carboxylic acid
Alcoholysis- +alcohol  ester
Aminolysis- +ammonia or amine  amide
Reduction- + H-  aldehyde or alcohol
Grignard- + Organometallic  ketone or alcohol
Preparation of Acid Derivatives
 Replace OH

Nucleophilic Substitution
o 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  1° Amines + CO2
Transesterification
 Transesterification: exchange alkoxyl group with ester of another alcohol
 Alcohol + Ester  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 Esters
1) 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 Polar
o Similar reactivity to alcohols
o Can H bond, but weaker H bond than alcohols
o 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