Organic Synthesis
• Series of reactions by which a set of organic starting materials is converted to a more complicated structure
• A successful synthesis:• Provides the desired product in maximum yield with a maximum control of
stereochemistry at all stages• Should not produce or release byproducts harmful to the environment (“green”
synthesis)
• Best strategy is to work backward from the target product
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Retrosynthetic Analysis: Analysis of Target Molecule1. Count the carbon atoms of the carbon skeleton of the target molecule
• Challenge - Determining how to build the carbon skeleton from available starting materials
• Only method to date for forming a new C—C bond is the alkylation of acetylide anions with methyl or 1° halides
2. Analyze the functional groups• What are they, how can they be changed to facilitate formation of carbon skeleton,
and how can they then be used in forming the final set of functional groups in the target molecule?
3. Work backward• Use retrosynthesis
• Retrosynthesis: A process of reasoning backward from a target molecule to a set of suitable starting materials• An open arrow to symbolize a step in a retrosynthesis
targetmolecule
startingmaterials 3
Reactions of Functional Groups in Complex Molecules• Synthesis of fexofenadine illustrates two
important points about how functional group analysis is used to develop systematic syntheses of complex molecules
• Alkyne was the only functional group to react under the conditions used, and it reacted as expected
• Other functional groups often have an influence over the regiochemistry or stereochemistry of a reaction
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Sample Synthesis Problem (from old exam) • You should be able to find at least three completely different synthetic routes for the following problem!
Show all reagents and intermediates for the multi-step synthesis below. You may use any organic or inorganic reagents in your synthesis, but you must begin with the indicated starting material. You do not need to show any mechanisms. The correct answer will require fewer than 5 steps. • Starting material
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Organic Synthesis Using Alkynes • Provide a synthetic route to the following product from starting materials that contain no
more than 2 carbons:
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Example
• An acid-catalyzed domino reaction that ultimately leads to the synthesis of progesterone is depicted in the image
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• We focus on the acid-catalyzed step that leads to cyclization of the four-ring system, which is characteristic of steroids
• Assume that this reaction is initiated by protonation of the 3° alcohol followed by loss of water to give a carbocation
• The series of reactions initiated by this carbocation gives compound B
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Tandem reactions, cascade sequencesJohnson’s elegant cation-π cyclization
Solvents in Organic Chemistry • The vast majority of organic reactions are carried out in a solvent, a liquid
that is used to dissolve the reactants, reagents, and products. Each solvent can be characterized based on three properties: whether it is polar, whether it is protic, and whether it is a donor:
• polar or nonpolar:polar - dielectric constant greater than 15 nonpolar - dielectric constant less than 15
• protic or aprotic:protic - solvent acts as hydrogen bond donor aprotic - solvent cannot act as hydrogen bond donor
• donor or nondonor:donor - solvent can share (donate) lone pair nondonor - solvent cannot share (donate) lone pair
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Haloalkane, Haloalkene, and Haloarene• Haloalkane (alkyl halide): Contains a halogen atom covalently bonded to
an sp3 hybridized carbon• General symbol - R—X
• Haloalkene (vinylic halide): Contains a halogen atom bonded to one of the carbons of a carbon-carbon double bond
• Haloarene (aryl halide): Contains a halogen atom bonded to a benzene ring
• General symbol - Ar—X
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Nomenclature - common name and IUPAC System • Haloalkanes
• Treated as an alkyl group; fluoro-, chloro-, bromo-, and iodo- and list them in alphabetical order with other substituents
• Haloalkenes• Number the parent chain to give carbon atoms of the double bond and substituents
the lower set of numbers
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IUPAC nameCommon name
Common Names of Haloalkanes and Haloalkenes
• Haloform: Compound of the type CHX3 where X is a halogen
• Perhaloalkanes or perhaloalkenes - Hydrocarbons in which all hydrogens are replaced by halogens
CH2Cl2 CHCl3 CH3CCl3 CCl2═CHClDicholormethane Trichloromethane 1,1,1-Trichloroethane Trichloroethylene
(Methylene chloride) (Chloroform) (Methyl chloroform) (Trochlor)
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Test Yourself• Write the IUPAC name and, where possible, the common name of each
compound• Show stereochemistry where relevant
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Test Yourself
• What is the IUPAC name of the following molecule?
1. 2,4-Dimethylheptyl-6-chloride2. 2,4,6-Trimethyl-6-chlorohexane3. 2-Chloro-4,6-dimethylheptane4. 1-Chloro-1,3,5-trimethylhexane5. 2-Chloro-4-methyl-isoheptane
Cl
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Polarity of Haloalkanes• Magnitude of a dipole moment depends on:
• Size of the partial charges• Distance between the partial charges • Polarizability of the unshared electrons pairs on each halogen
C Xδ+ δ-
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Haloalkanes - Boiling Point• Haloalkanes are associated in the liquid state by van der Waals forces
• van der Waals forces: Group of intermolecular attractive forces • As molecules are brought closer and closer, van der Waals attractive forces are
overcome by repulsive forces between electron clouds of adjacent atoms• Energy minimum - Point where attractive and repulsive forces are balanced
• Nonbonded interatomic and intermolecular distances at minima can be measured by x-ray crystallography, and each atom and group of atoms can be assigned a van der Waals radius
• van der Waals radius: Minimum distance of approach to an atom that does not cause nonbonded interaction strain
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Haloalkanes - Trends in Boiling Points• Constitutional isomers with branched chains have lower boiling points
than their unbranched-chain isomers• Have a more spherical shape and decreased surface area, leading to
smaller van der Waals forces between their molecules
• For an alkane and a haloalkane of comparable size and shape, the haloalkane has a higher boiling point
• Polarizability: Measure of the ease of distortion of the distribution of electron density about an atom in response to interaction with other molecules and ions
• Fluorine has a very low polarizability, whereas iodine has a very high polarizability• Larger the halogen, the greater its polarizability
2-Bromo-2-methylpropanebp 72°C
Br1-Bromobutane
bp 100°C
BrCH3CH3 CH3Brbp -89°C bp 4°C
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• Boiling points of fluoroalkanes are comparable to those of hydrocarbons of similar molecular weight and shape
• Low boiling points of fluoroalkanes are the result of the small size of fluorine, the tightness with which its electrons are held, and their particularly low polarizability
CH3CHCH3
CH3CH3CHCH3
F
2-FluoropropaneMW 62.1, bp -11°C
2-MethylpropaneMW 58.1, bp -1°C
Hexane(MW 86.2, bp 69°C)
F1-Fluoropentane
(MW 90.1, bp 63°C)
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Haloalkanes - Density• Liquid haloalkanes are more dense than hydrocarbons of comparable
molecular weight• A halogen has a greater mass-to-volume ratio than a methyl or methylene group
• All liquid bromoalkanes and iodoalkanes are more dense than water• Liquid chloroalkanes are less dense than water
• Substitution of chlorine for hydrogen increases the density of di- and polychloroalkanes in comparison to water
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Haloalkanes - Bond Lengths and Strengths• Bond dissociation enthalpy - Measure of bond strength
• Amount of energy required to break a bond homolytically into two radicals in the gas phase at 25°C
• Radical (free radical): Any chemical species that contains one or more unpaired electrons
• Homolytic bond cleavage: Cleavage of a bond so that each fragment retains one electron, producing radicals
• Heterolytic bond cleavage: Cleavage of a bond so that one fragment retains both electrons and the other has none
• Fishhook arrows are used to indicate:• Change in position of single electrons • Homolytic mechanism
• As the size of the halogen atom increases, the C—X bond length increases and its strength decreases
• C—F bonds are stronger than C—H bonds
• C—Cl, C—Br, and C—I bonds areweaker than C—H bonds
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Test Yourself
• Dipole moments of CH3—X are 1.85, 1.87, 1.81, and 1.62 D for F, Cl, Br, and I, respectively
• What is the best explanation of the small differences in dipole moments?1. The electronegativity of halogens in alkyl halides is nearly constant2. The inductive effects are counterbalanced by hyperconjugation3. The electronegativity effects are counterbalanced by bond strengths4. Lone-electron pairs are more delocalized on smaller halogens5. The charge-separation and bond-lengths trends have opposite effects
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Halogenation of Alkanes• Haloalkanes can be prepared by:
• Addition of HX and X2 to alkenes• Replacement of the —OH group of
alcohols by halogen
• Simpler low-molecular-weight haloalkanes are prepared by the halogenation of alkanes by Cl2 or Br2
• If a mixture of methane and chlorine is kept in the dark at room temperature, no change occurs
• If the mixture is heated or exposed to visible or ultraviolet light, reaction begins at once with the evolution of heat
• This is a substitution reaction• Substitution reaction: An atom or group
of atoms is replaced by another atom or group of atoms 36
Regioselectivity• High for bromination, but not as high for chlorination• Takes the order of 3° > 2° > 1°• Approximately 1600:80:1 per hydrogen for bromination• Approximately 5:4:1 for chlorination
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Exercise - Free-Radical Halogenation
• Name and draw structural formulas for all monobromination products formed by treating 2-methylpropane with Br2
• Predict the major product based on the regioselectivity of the reaction of Br2 with alkanes
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Energetics• Heat of any reaction (enthalpy change) can be calculated by comparing
the bond strengths in the reactants and products
• Bond Dissociation Enthalpies for Selected C—H Bonds
0 = BDEs(broken) BDEs(formed)HD S - S0
4 2 3 CH Cl CH Cl + HCl = 96 kJ/mol ( kcal/mol)BDE, kJ/mol +439 +247 351 431
H+ ® D - - 23 - -
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Formation of Radicals - Examples
• Following reactions result in homolytic cleavage to give radicals
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Energetics of Chain Propagation Steps• After the radical chain is initiated, the heat of reaction is derived entirely
from the heat of reaction of the individual chain propagation steps• Sum of the heats of reaction for each propagation step is equal to the
observed heat of reaction
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Example - Enthalpy of Reactions• Using the bond dissociation enthalpies, calculate ΔH0 for each propagation
step in the radical bromination of propane to give 2-bromopropane and HBr
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• Here are the two chain propagation steps along with bond dissociation enthalpies for the bonds broken and the bonds formed
• The first chain propagation step is endothermic, the second is exothermic, and the overall reaction is exothermic by 71 kJ (17 kcal)/mol
Test Yourself
• Which of the following radicals is expected to be the least selective?
1. HO•2. F•3. Cl•4. Br•5. I•
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Regioselectivity of Bromination vs Chlorination
• Regioselectivity in halogenation of alkanes caused by the relative stabilities of radicals (3° > 2° > 1° > methyl)
• More stable radical products are formed with a lower activation energy
• To account for the greater regioselectivity of bromination (1600:80:1) compared with chlorination (5:4:1), Hammond’s postulate should be considered.
• Structure of the transition state is: • More similar to the structure of reactants or products depends on the energy
• Provides a reasonable way of deducing something about the structure of a transition state by examining the structure of reactants and products and heats of reaction
• Applies equally well to the transition state for one-step reactions and to each transition state in a multistep reaction
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Hammond’s Postulate - Relative Regioselectivities of Chlorination vs Bromination of Alkanes
• Abstraction of hydrogen by chlorine is exothermic• Transition state is reached early in the course of the reaction, and its
structure resembles the alkane and the chlorine atom• There is little radical character on carbon in the transition state
• Regioselectivity is only slightly influenced by the relative stabilities of radical intermediates
• Products are determined by whether a chlorine atom happens to collide with a 1°, 2°, or 3° H
• Abstraction of hydrogen by bromine is endothermic• There is significant radical character on carbon in the transition state
• Reasons for larger regioselectivity• Later transition state and correspondingly larger difference in activation energies
(ΔΔG‡) that causes a large difference in reaction rates 51
Transition States and Energetics for Hydrogen Abstraction in the Radical Chlorination and Bromination of 2-Methylpropane (Isobutane)
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Stereochemistry of Radical Halogenation
• When radical halogenation produces a chiral center or takes place at a hydrogen on a chiral center, the product is an racemic mixture of R and Senantiomers
• In the case of sec-butyl radical, the carbon bearing the unpaired electron is sp2
hybridized and the unpaired electron occupies the unhybridized 2p orbital
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Test Yourself
• How many constitutional monochloride isomers will form in the radical-chain chlorination of 2,3-dimethylpentane?
1. 3 2. 4 3. 5 4. 6 5. 7
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Test Yourself
• The selectivity of chlorine radical is 1.0:3.5:5.0 for 1°, 2°, and 3°hydrogens, respectively
• If only monochlorides form in the radical chlorination of 1,3-dimethylcyclobutane, what is the expected yield of the tertiary chloride? 1. 1/62. 1/53. 1/44. 1/35. 1/2
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Test Yourself
• Radical chain chlorination of (2R)-fluorobutane yields 2-chloro-3-fluorobutane as one of the products
• Which is the best accounting for the stereochemistry of this product?1. Two enantiomers in equal amounts2. Two diastereomers in different amounts3. Two diastereomers in equal amounts4. Four stereoisomers in different amount each5. Four stereoisomers with enantiomeric pairs in equal amounts but
diastereomeric pairs in unequal amounts
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Test Yourself
• Which mechanistic step decides the regiochemistry of the radical-chain chlorination of 2-methylpropane?
1. Protonation step2. Hydrogen abstraction step3. Chlorine abstraction step4. Chloride addition step5. Initiation step
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Test Yourself• The reaction shown below produces 5 different bromides
• Which of the compounds listed is one of the expected products?
NBS
CCl4
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Allylic Halogenation
• Allylic carbon: Carbon adjacent to a C-C double bond• Allylic substitution: substitution at an allylic carbon
• When propene and a halogen are allowed to react at a high temperature, substitution of a halogen occurs at the allylic C
• An allylic C—H bond is weaker than a vinylic C—H bond
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Allylic Halogenation - Radical Reaction
• It is possible to change the product(s) by changing the mechanism through a change in reaction conditions
• Under high temperatures, concentration of bromine radicals becomes much higher than at room temperature
• Accelerates the substitution reaction, which occurs by the radical halogenation mechanism
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Allylic Bromination
• Can be executed in the laboratory at or slightly above room temperature by using the reagent N-bromosuccinimide (NBS) in dichloromethane
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Allylic Bromination of Propene Using NBS
• A radical chain mechanism • Chain initiation
• Chain propagation
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• Chain termination
• The Br2 necessary for chain propagation is formed by reaction of NBS and HBr
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Allyl Radical
• Hybrid of two equivalent contributing structures
• Position of the radical electron in the two contributing structures predicts that radical reactivity will occur at carbons 1 and 3 but not at carbon 2
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Molecular Orbital Model of Covalent Bonding in the Allyl Radical• Combination of three 2p orbitals gives three p molecular orbitals
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Example - Allylic Bromination• Account for the fact that allylic bromination of 1-octene by NBS gives these
products
• Explain the results:
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Test Yourself
• Which of the following is not the product of a radical chain-reaction of 2-pentene with NBS?
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Test Yourself
• What is the best reagent to carry out the transformation below?
1. PBr3
2. HBr3. NBS/hv/CCl44. Br2/hv5. NBS/H2O/DMSO
Br
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Test Yourself
• Which of the following molecules would not yield the desired organolithium reagent when treated with Li in pentane?
1. Iododecane2. 4-Bromo-1-pentanol3. Phenyl bromide4. 4-Bromobutyl methyl ether5. Cyclohexylbromide
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Radical Autoxidation
• Autoxidation: Oxidation requiring oxygen, O2, and no other oxidizing agent
• Occurs by a radical chain mechanism
• Autoxidation of the hydrocarbon chains of polyunsaturated fatty acid esters• Polyunsaturated fatty acid esters have chains of 16 or 18 carbons containing
1,4-diene functional groups• Radical abstraction of a doubly allylic hydrogen of a 1,4-diene forms a
particularly stable radical• Doubly allylic C—H is weaker than an allylic C—H bond, which is in turn
weaker than a corresponding alkane C—H bond
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Radical Autoxidation• Begins when a radical initiator, X•, abstracts a doubly allylic hydrogen to
form a radical
• Radical is delocalized through resonance with both double bonds
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Radical Autoxidation
• The doubly allylic radical reacts with O2, a diradical, to form a peroxy radical• The peroxy radical reacts with the CH2 of another 1,4-diene fatty acid
ester to give a new radical, R•, and a hydroperoxide
• The new radical reacts again with oxygen, causing a radical chain reaction in which hundreds of molecules of fatty acid ester are oxidized for each initiator radical
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Radical Autoxidation
• Some autoxidation products degrade to short-chain aldehydes and carboxylic acids with unpleasant “rancid” smells
• Some are toxic and/or carcinogenic• Oils lacking the 1,4-diene structure are less easily oxidized• Many natural and unnatural compounds are referred to as radical inhibitors
• Radical inhibitor: Compound that selectively reacts with radicals to remove them from a chain reaction and terminate the chain
• Example - Phenol
How to deal with the unwanted autoxidation?
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Naturally Occurring Antioxidant - Vitamin E
• Reacts preferentially with the initial peroxy radical to give a resonance stabilized phenoxy radical, which is less reactive and survives to scavenge another peroxy radical
• Resulting peroxide is relatively stable
• Known as α-tocopherol (a phenol)
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Example - Radical Autoxidation
• What products would you expect from the following reaction?• Indicate the major one and specify stereochemistry if relevant
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Radical Addition of HBr to Alkenes
• Addition of HBr to alkenes gives either Markovnikov addition or non-Markovnikov addition depending on reaction conditions
• Markovnikov addition occurs when radicals are absent• Anti-Markovnikov addition occurs when peroxides or other sources of radicals are
present
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Similarity in Polar and Radical Mechanisms• Regiochemistry of each reaction is dominated by the reactions that
proceed through the most stable reactive intermediates
• Reactive intermediates in both cases are tertiary
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Example - Free-Radical Addition of HX to an Alkene• Predict the product of the following reaction:
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Test Yourself
• What is the product of the following sequence of reactions?
I2Li
pentane
CuI
ether
I
ether
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