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• Addition is the opposite of elimination.
• A pi bond is converted to a sigma bond.
9.1 Addition Reactions
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -1
• A pi bond will often act as a Lewis base (as a nucleophile or as a Brønsted-Lowry base).
• Why are pi bonds more reactive in this sense
than sigma bonds?
9.1 Addition Reactions
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -2
• Because an addition is the reverse of an elimination, often the processes are at equilibrium.
• An equilibrium is a thermodynamic expression. • We assess ΔG (the free energy) to determine
which side the equilibrium will favor.
9.2 Addition/Elimination Equilibria
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -3
• To determine which side the equilibrium will favor, we must consider both enthalpy and entropy.
9.2 Addition/Elimination Equilibria
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -4
• Typical addition reactions have a –ΔH. WHY?
9.2 Addition/Elimination Equilibria
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -5
• Typical addition reactions have a –ΔH. • Will heat be absorbed by or released into the
surroundings? • What will the sign (+/-) be for ΔSsurr? • Will the enthalpy term favor the reactants or products? • The heat change (ΔH) will remain roughly constant,
regardless of temperature.
9.2 Addition/Elimination Equilibria
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -6
• Having a –ΔH (or a +ΔSsurr) favors the addition reaction rather than the elimination reaction.
• To get ΔG (or ΔStot) and make a complete assessment, we must also consider the entropy of the system (ΔSsys).
• What will the sign (+/-) be for ΔSsys? WHY? • What will the sign (+/-) be for -TΔSsys? • Will the enthalpy term favor the reactants or products?
9.2 Addition/Elimination Equilibria
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -7
• Plugging into the formula gives:
• To favor addition, a –ΔG (or a +ΔStot) is needed. – How can the temperature be adjusted to favor addition?
• To favor elimination (the reverse reaction in this example), a +ΔG (or a –ΔStot) is needed. – How can the temperature be adjusted to favor
elimination?
9.2 Addition/Elimination Equilibria
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -8
• Note the temperature used in this addition reaction.
• Does it matter whether the Br adds to the right side of the C=C double bond or whether it adds to the left?
9.3 Hydrohalogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -9
• Regiochemistry becomes important for asymmetrical alkenes.
– In 1869, Markovnikov showed that in general, H atoms tend to add to the carbon already
bearing more H atoms.
9.3 Hydrohalogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -10
• Markovnikov’s rule could also be stated by saying that, in general, halogen atoms tend to add to the carbon that is more substituted with other carbon groups.
• This is a regioselective reaction, because one constitutional isomer is formed in greater quantity than another.
• Draw the structure of the minor product.
9.3 Hydrohalogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -11
• Anti-Markovnikov products are observed when reactions are performed in the presence of peroxides such as H2O2.
• Why would some reactions yield Markovnikov products, while other reactions give anti-Markovnikov products? – The answer must be found in the mechanism.
• Practice with CONCEPTUAL CHECKPOINT 9.1.
9.3 Hydrohalogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -12
• The mechanism is a two-step process. – Which step do you think is rate determining? – Write a rate law for the reaction.
9.3 Hydrohalogenation – Mechanism
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -13
• Explain the FREE energy changes in each step.
9.3 Hydrohalogenation – Mechanism
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -14
• Recall that there are two possible products, Markovnikov and anti-Markovnikov products.
• Which process looks more favorable? WHY?
9.3 Hydrohalogenation – Mechanism
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -15
• Practice with SKILLBUILDER 9.1.
9.3 Hydrohalogenation – Mechanism
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -16
• In many addition reactions, chirality centers are formed.
• There are two possible Markovnikov products:
• Which step in the mechanism determines the stereochemistry of the product?
9.3 Hydrohalogenation – Stereochemical Aspects
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -17
• Recall the geometry of the carbocation.
• Practice with CONCEPTUAL CHECKPOINT 9.6.
9.3 Hydrohalogenation – Stereochemical Aspects
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -18
• Rearrangements (hydride or methyl shifts) occur for a carbocation if a shift makes it more stable.
9.3 Hydrohalogenation – Rearrangements
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -19
• A mixture of products limits synthetic utility. • With an INTRAMOLECULAR rearrangement, WHY isn’t
the rearrangement product an even greater percentage? • How might [Cl-] be used to alter the ratio of products? • Practice with SKILLBUILDER 9.2.
9.3 Hydrohalogenation – Rearrangements
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -20
• Predict the major product(s) for the reaction below.
9.3 Hydrohalogenation – Example
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -21
HCl
• The components of water (–H and –OH) are added across a C=C double bond.
• The acid catalyst is often shown over the arrow because it is regenerated rather than being a reactant.
9.4 Hydration
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -22
• Given the data below, do you think the acid catalyzed hydration goes through a mechanism that involves a carbocation?
9.4 Hydration
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -23
• Why does the hydrogen add to this carbon of the alkene?
9.4 Hydration – Mechanism
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -24
• Could a stronger base help promote the last step? • Practice with CONCEPTUAL CHECKPOINT 9.10.
9.4 Hydration – Mechanism
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -25
• Similar to hydrohalogenation, hydration reactions are also at equilibrium.
• Explain HOW and WHY temperature could be used to shift the equilibrium to the right or left.
9.4 Hydration – Thermodynamics
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -26
• How could Le Châtelier’s principle be used to shift the equilibrium to the right or left?
• Practice with CONCEPTUAL CHECKPOINT 9.11.
9.4 Hydration – Thermodynamics
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -27
• Similar to hydrohalogenation, the stereochemistry of hydration reactions is controlled by the geometry of the carbocation.
• Draw the complete mechanism for the reaction above to show WHY a racemic mixture is formed.
• Practice with SKILLBUILDER 9.3.
9.4 Hydration – Thermodynamics
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -28
• Ethanol is mostly produced from fermentation of sugar using yeast, but industrial synthesis is also used to produce ethanol through a hydration reaction.
• Predict the major product(s) for the reaction below.
9.4 Hydration – Examples
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -29
H3O+
H2O
• Because rearrangements often produce a mixture of products, the synthetic utility of Markovnikov hydration reactions is somewhat limited.
• OXYMERCURATION-DEMERCURATION is an alternative process that can yield Markovnikov products more cleanly.
9.5 Oxymercuration-Demercuration
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -30
• OXYMERCURATION begins with mercuric acetate.
• How would you classify the mercuric cation? – As a nucleophile or an electrophile? – As a Lewis acid or Lewis base?
• How might an alkene react with the mercuric cation?
9.5 Oxymercuration-Demercuration
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -31
• Similar to how we saw the alkene attack a proton previously, it can also attack the mercuric cation.
• Resonance stabilizes the mercurinium ion. Can you draw a reasonable resonance hybrid?
9.5 Oxymercuration-Demercuration
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -32
• The mercurinium ion is also a good electrophile, and it can easily be attacked by a nucleophile, even a weak nucleophile such as water.
• NaBH4 is generally used to replace the –HgOAc group with a –H group via a free radical mechanism.
9.5 Oxymercuration-Demercuration
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -33
• To achieve anti-Markovnikov hydration, hydroboration-oxidation is often used.
• Note that the process occurs in two steps.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -34
• Hydroboration-oxidation reactions achieve SYN addition.
• ANTI addition is NOT observed.
• To answer WHY, we must investigate the mechanism.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -35
• Let’s examine how this new set of reagents might react. • The BH3 molecule is similar to a carbocation but not as
reactive, because it does not carry a formal charge.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -36
• Because of their broken octet, BH3 molecules undergo intermolecular resonance to help fulfill their octets
• The hybrid that results from the resonance (diborane) involves a new type of bonding called BANANA BONDS.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -37
• In the hydroboration reaction, BH3•THF is used. BH3•THF is formed when borane is stabilized with THF (tetrahydrofuran).
• What general role do you think BH3•THF is likely to play in a reaction?
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -38
• Let’s examine the first step of the hydroboration.
9.6 Hydroboration-Oxidation – The Hydroboration Step
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -39
• What evidence is there for a concerted addition of the B-H bond across the C=C double bond?
• Use sterics and electronics to explain the regioselectivity of the reaction.
• Practice with CONCEPTUAL CHECKPOINT 9.17.
9.6 Hydroboration-Oxidation – The Hydroboration Step
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -40
9.6 Hydroboration-Oxidation – The Oxidation Step
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -41
9.6 Hydroboration-Oxidation – The Oxidation Step
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -42
• When ONE chirality center is formed, a racemic mixture results. – WHY? What is the geometry of the alkene as the borane
attacks?
• The squiggle bond above shows two products, a 50/50 mixture of the R and the S enantiomer.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -43
• When TWO chirality centers are formed, a racemic mixture results.
– WHY aren’t the other stereoisomers formed?
• Practice with SKILLBUILDER 9.4.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -44
• Predict the major product(s) for the reactions below.
9.6 Hydroboration-Oxidation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -45
1) BH3
THF
2) H2O2
/ NaOH
1) BH3
THF
2) H2O2
/ NaOH
• The addition of H2 across a C=C double bond:
• If a chirality center is formed, SYN addition is observed. – Draw the other stereoisomers
that are NOT produced.
9.7 Hydrogenation – Catalytic
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -46
• Analyze the energy diagram. – Why is a catalyst
necessary? – Does the catalyst affect
the spontaneity of the process?
• Typical catalysts include Pt, Pd, and Ni.
9.7 Hydrogenation – Catalytic
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -47
• The metal catalyst is believed to both adsorb the H atoms and coordinate the alkene.
• The H atoms add to the same side of the alkene pi system.
9.7 Hydrogenation – Catalytic
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -48
• Draw product(s) for the reaction below. Pay close attention to stereochemistry.
• How many chirality centers are there in the alkene reactant above?
• How does the term MESOCOMPOUND describe the product(s) of the reaction?
• Practice with SKILLBUILDER 9.5.
9.7 Hydrogenation – Catalytic
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -49
• If catalysis takes place on the surface of a solid surrounded by solution, the catalyst is HETEROGENEOUS. WHY?
• HOMOGENEOUS catalysts also exist.
• What advantage might a homogeneous catalyst have?
9.7 Hydrogenation – Catalytic
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -50
• In 1968, Knowles modified Wilkinson’s catalyst by using a chiral phosphine ligand.
• A chiral catalyst can produce one desired enantiomer over another. HOW?
• Why would someone want to synthesize one enantiomer rather than a racemic mixture?
9.7 Hydrogenation – Asymmetric
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -51
• A chiral catalyst allows one enantiomer to be formed more frequently in the reaction mixture.
• Some chiral catalysts give better enantioselectivity than others. WHY?
9.7 Hydrogenation – Asymmetric
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -52
• BINAP is a chiral ligand that gives very pronounced enantioselectivity.
• For any reaction, stereoselectivity can only be achieved if at least one reagent (reactant or catalyst) is chiral.
9.7 Hydrogenation – Asymmetric
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -53
• Predict the major product(s) for the reactions below.
9.7 Hydrogenation – Asymmetric
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -54
H2
Pt
H2
Pt
• Halogenation involves adding two halogen atoms across a C=C double bond.
• Halogenation is a key step in the production of PVC.
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -55
• Halogenation with Cl2 and Br2 is generally effective, but halogenation with I2 is too slow, and halogenation with F2 is too violent.
• Halogenation occurs with ANTI addition
• Given the stereospecificity, is it likely to be a concerted or a multi-step process?
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -56
• Let’s look at the reactivity of Br2. Cl2’s reactivity is similar.
• It is nonpolar, but it is polarizable. WHY?
• What type of attraction exists between the Nuc:1- and Br2?
• Does the Br2 molecule have a good leaving group attached to it?
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -57
• We know alkenes can act as nucleophiles. • Imagine an alkene attacking Br2. You might imagine the
formation of a carbocation.
• However, this mechanism DOES NOT match the stereospecificity of the reaction. HOW? WHY?
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -58
• Only ANTI addition is observed. WHY? • Prove to yourself that the products are enantiomers
rather than identical.
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -60
• Only ANTI addition is observed.
• Can you design a synthesis for ?
• Practice with CONCEPTUAL CHECKPOINT 9.26.
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -61
• Predict the major product(s) for the reactions below.
9.8 Halogenation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -62
Br2
CCl4
Cl2CCl4
• Halohydrins are formed when halogens (Cl2 or Br2) are added to an alkene with WATER as the solvent.
• The bromonium ion forms from Br2 + alkene, and then it is attacked by water.
• Why is the bromonium attacked by water rather than a Br1- ion? Is water a better nucleophile?
9.8 Halogenation – Halohydrin Formation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -63
• A proton transfer completes the mechanism producing a neutral halohydrin product.
• The net reaction is the addition of –X and –OH across a C=C double bond.
9.8 Halogenation – Halohydrin Formation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -64
• The –OH group adds to the more substituted carbon.
• The key step that determines regioselectivity is the attack of water on the bromonium ion.
9.8 Halogenation – Halohydrin Regioselectivity
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -65
• When water attacks the bromonium ion, it will attack the side that goes through the lower energy transition state.
• Water is a small molecule that can easily access the
more sterically hindered site. • Practice with SKILLBUILDER 9.6.
9.8 Halogenation – Halohydrin Regioselectivity
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -66
• Predict the major product(s) for the reactions below.
9.8 Halogenation – Halohydrin Regioselectivity
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -67
Br2
H2O
Cl2H2O
• Dihydroxylation occurs when two –OH groups are added across a C=C double bond.
• ANTI dihydroxylation is achieved through a multi-step process.
9.9 Anti Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -68
• First, an epoxide is formed.
• Replacing the relatively unstable O–O single bond is the
thermodynamic driving force for this process. • Is there anything unstable about an epoxide? • Is an epoxide likely to react as a nucleophile (Lewis
base) or as an electrophile (Lewis acid)?
9.9 Anti Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -69
• Water is a poor nucleophile, so the epoxide is activated. with an acid
9.9 Anti Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -70
• Note the similarities between three key intermediates.
• Ring strain and a +1 formal charge makes these structures GREAT electrophiles.
• They also each yield ANTI products because the nucleophile must attack in an SN2 fashion.
• Practice with SKILLBUILDER 9.7.
9.9 Anti Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -71
• Like other syn additions, SYN dihydroxylation adds across the C=C double bond in ONE step.
9.10 Syn Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -72
• Because OsO4 is expensive and toxic, conditions have been developed where the OsO4 is regenerated after reacting, so only catalytic amounts are needed.
9.10 Syn Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -73
• MnO41- is similar to OsO4 but more reactive.
• SYN dihydroxylation can be achieved with KMnO4 but only under mild conditions (cold temperatures).
• Diols are often further oxidized by MnO41-, and MnO4
1-
is reactive toward many other functional groups as well. • The synthetic utility of MnO4
1- is limited. • Practice with CONCEPTUAL CHECKPOINT 9.33.
9.10 Syn Dihydroxylation
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -74
• C=C double bonds are also reactive toward oxidative cleavage.
• Ozonolysis is one such process.
• Ozone exists as a resonance hybrid of two contributors.
9.11 Oxidative Cleavage with O3
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -75
• Common reducing agents include dimethyl sulfide and Zn/H2O.
• Practice with SKILLBUILDER 9.8.
9.11 Oxidative Cleavage with O3
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -76
• Predict the major product(s) for the reaction below.
• Predict the reactant used to form the product below.
9.11 Oxidative Cleavage with O3
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -77
1) O3
SS2)
O
H O
O
H
O
H
1) O3
SS2)
1. Analyze the reagents used to determine what groups will be added across the C=C double bond.
2. Determine the regioselectivity (Markovnikov or anti-Markovnikov).
3. Determine the stereospecificity (syn or anti addition) – Each step can be achieved with minor reagent memorization
and a firm grasp of the mechanistic rational. – The more familiar you are with the mechanisms, the easier
predicting products will be.
• Practice with SKILLBUILDER 9.9.
9.12 Predicting Addition Products
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -78
• Predict the major product(s) for the reaction below.
9.12 Predicting Addition Products
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -79
• To set up a synthesis, assess the reactants and products to see what changes need to be made.
• Label each of the processes below.
9.13 Application – One-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -80
• To set up a synthesis, assess the reactants and products to see what changes need to be made.
• Give reagents and conditions for the following.
• Practice with SKILLBUILDER 9.10.
9.13 Application – One-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -81
+ En
BrOH
• Multistep syntheses are more challenging, but the same strategy applies.
• This is not a simple substitution, addition or elimination, so two processes must be combined.
9.13 Application – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -82
• For the strategy to work, the regioselectivity must be correct.
• A smaller base should be used to produce the more stable Zaitsev product.
9.13 Application – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -83
• For the strategy to work, the regioselectivity must be correct.
• Will the regioselectivity for the HBr reaction give the desired product?
9.13 Syntheses – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -84
• Multistep syntheses are more challenging, but the same strategy applies.
• This is not a simple substitution, addition or elimination, so two processes must be combined.
9.13 Syntheses – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -85
• How can the alcohol be eliminated to give the less stable Hoffmann product?
• H3O+ will give the Zaitsev product. • OH- is too poor of a leaving group to use the bulky
base, t-BuOK. • The OH must first be converted to a better leaving
group, and then t-BuOK can be used.
9.13 Syntheses – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -86
• In the last step, –H and –OH must be added across the C=C double bond.
• Is the desired addition Markovnikov or anti-Markovnikov?
9.13 Application – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -87
• Use reagents that give anti-Markovnikov products.
• Is stereochemistry an issue in this specific reaction?
• Practice with SKILLBUILDER 9.11.
9.13 Application – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -88
• Solve the multistep syntheses below.
– This is not a simple substitution, addition or elimination, so
two processes must be combined.
– What reagents should be used?
• Practice with SKILLBUILDER 9.12.
9.13 Application – Multi-step Syntheses
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 9 -89