Control In Organic
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1
C5: Control in Organic Chemistry
Part II Chemistry 2003
Control in Organic Chemistry Lectures 7-12 Dr David Spring [email protected] Tel. 01223 336364 Lab 180
O
O O Me
BrMg CuBr•SMe2
O
O Me O
Course Outline 1. Introduction to Stereoselectivity 2. Stereochemical Control in Four- and Five-Membered Rings 3. Stereochemical Control in Six-Membered Rings 4. Stereochemical Control with Bicyclic Compounds 5. Open Chain Stereochemical Control
Essential Reading
Clayden, Greeves, Warren, Wothers “Organic Chemistry” Oxford, 2001.
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C5: Control in Organic Chemistry
Introduction to Stereoselectivity How are rxns controlled? Kinetically and thermodynamically controlled rxns: If treatment of a starting material can give more than one product then the amount of each product is determined by its rate of formation (kinetic control). However, if the rxn is reversible then the product distribution is determined by the relative thermodynamic stability of each product. O O S OH H2SO4 80 ˚C
O H2SO4
O S
OH
160 ˚C
∆G
O
O S
H2SO4
O ∆G˚ O S OH
OH
H2O
H2O
3
C5: Control in Organic Chemistry
The majority of this course will be concerned with kinetically controlled rxns that involve the formation of tetrahedral (sp3) carbon atoms within a molecular framework.
What type of selectivity are we talking about? Chemoselectivity – functional group discrimination Regioselectivity – product structural isomer discrimination Stereoselectivity – product stereoisomer discrimination O
O O Me
O
BrMg
Me
CuBr•SMe2
O
O
Stereochemistry involves diastereomers and enantiomers. Therefore stereoselectivity encompasses diastereoselectivity (product diastereomer discrimination) and enantioselectivity (product enantiomer discrimination). This course will focus on diastereoselectivity. Enantioselective rxns will be covered in detail next year. So from now on, when we talk about stereoselectivity, more specifically we mean diastereoselectivity.
E
Enantiomeric transition states must be the same energy
OLi Ph Bu H
O H
E
Diastereomeric transition states are not necessarily the same energy
OLi BuLi Ph
Rxn Coordinate
OLi H Bu Ph
Bu
O
Ph Me
H
BuLi Ph * Me
OLi Bu
Ph Me
Rxn Coordinate
As you know enantiomers are equal in energy, therefore enantiomeric transition states are also equal in energy. So it is impossible to achieve any selectivity, we get a racemic mixture. However, if we have a chiral centre in the substrate then we form diastereomers, then the transition state energies need not be equal and we should observe some selectivity. This simple statement in fact forms the basis for all diastereoselectivity (and enantioselectivity! – except that the chirality is not in the substrate).
4
C5: Control in Organic Chemistry
Selective or Specific? If a rxn proceeds via a mechanism with a strict stereochemical requirement, then the rxn is stereospecific. So in a stereospecific rxn, the starting material goes to only one possible product stereoisomer, there is NO selectivity. For example, substitution by the SN2 mechanism must involve an inversion of configuration of the carbon atom that is reacting:
Me Si Me
Me O
O (±)
N3-
Me
O
Me Si Me
Me O
O S O Me
Me
O N N+ N-
(±)
What does the other enantiomer look like?
Me Si Me
Me O
NaN3
O
Me
O O
Me Si Me
Me
Me
O N N+ N-
O S O Me
Stereospecific rxns are enantiospecific for chiral starting materials where a stereospecific rxn occurs at all stereocentres. For example:
Br Me
HO-
OH Me
Me
Me 95% e.e.
95% e.e.
Stereospecific rxns are also possible with achiral starting materials:
5
C5: Control in Organic Chemistry
In contrast, for a stereoselective rxn, the mechanism does not prevent the formation of two (or more) stereoisomeric products. Even if the rxn is completely selective giving only one stereoisomer, it still NOT ‘stereospecific’. As we will only be discussing diastereoselectivity, the selectivity will be between diastereomeric products and we shall talk about diastereomeric ratios (d.r.) and diastereomeric excesses (d.e.). [Enantioselectivity concerns selectivity between product enantiomers]
Two common cases of diastereoselective rxns are:
1. Rxns that involve the formation of one (or more) new stereocentres in a chiral substrate:
O
OH
NaBH4
OH
+ Me
Me
Me
Me
Me
90%
10%
si face addition
re face addition
Me
Chiral Starting Material
Diastereomeric Products d.r. = 9:1 d.e. = 80%
2. Rxns in which two prochiral substrates react to give a product where two new stereocentres are formed, where one new stereocentre originated from each substrate. OH O Me Me
Me Me
O H
+
Me
Li
+
Me
Me O
OH O Me
Me Me
Me Me
44%
Me 6%
Me
Me OH O +
Me Me
OH O Me
Me
Me 44%
+
Me Me
Me Me
Me 6%
6
C5: Control in Organic Chemistry
Stereoselective rxns of cyclic compounds Revision of Conformational Analysis
DRAW IN 3-D!! H
H H
H
H
H H H H
H H
H
H H
H H H
H
H H
H
H H
H
Ring Flip
H H
H H H
H
H
H
H
H
H
HH
HH
H
H H
Chair
H
H H
H
H H
H H H Boat
H H
H
H
H H
H
Twist-Boat
Learn how to draw a cyclohexane on p. 459-460 of Clayden et al.
C5: Control in Organic Chemistry
OH
Me
Me Me
H
OH
H
H
H
OH
7
8
C5: Control in Organic Chemistry
Stereochemical Control in Four- and Five-Membered Rings Four-membered rings Just one sp2 centre O
OH NaBH4
R
R
More than one sp2 centres
O O
Me Me
O
(i) LDA, -78 ˚C, THF (ii) BnBr O
Me Me
(±) d.r. >98 : 2
9
C5: Control in Organic Chemistry
Five-membered rings Just one sp2 centre O
OH Me
LiAlH4
OH Me
Me
+
THF (±)
O Me
77%
23%
OH
OH
LiBH(s-Bu)3
Me
Me
+
THF (±)
1.5%
98.5%
10
C5: Control in Organic Chemistry
More than one sp2 centre M
O
O
O R2 X
(R1)2CuLi TBSO
R1
TBSO
R2 TBSO
R1
O CO2H Me
HO OH PGE2
O Li
+
Me H
CuI, Bu3P, THF then Ph3SnCl, HMPA
OTBS
TBSO
O I
O
O
OMe
OMe HMPA, Et3N -78 to 23 ˚C
Me
TBSO OTBS
Suzuki, Noyori et al. J. Am. Chem. Soc. 1985, 107, 3348-3349.
11
C5: Control in Organic Chemistry
Chiral Memory: O
O + HO OH
Me
H+ H
Me Me Me
Me Me
O
(i) LDA (ii) PrBr
O
O
How do we make both epoxides stereoselectively? Me Me Me Me Me Me Me
O
Me Me Me Me Me
O
Me Me Me
O O
O Me
12
C5: Control in Organic Chemistry
Stereochemical Control in Six-Membered Rings Just one sp2 centre-addition to cyclohexanones
NaBH4 Me Me
Me
Me OH
O
OH
MeMgBr Me Me
Me
Me Me
Me
13
C5: Control in Organic Chemistry
Me Me
OH
O
OH
Al(Oi-Pr)3
LiBH(s-Bu)3
Me Me
Me
O
Me
Me
equitorial diastereomer H Me OH
[H]
Me Me
Me
axial diastereomer OH Me H
+
Me Me
0
Me Me
i-PrOH
Me Me
% axial diastereomer 10
Li/NH3 (99:1) LiAlH4 (93:7) LiAlH(Ot-Bu)3 (92:8)
20
30
40
50
70
60
80
90
100%
Me DIBAL (72:28) NaBH4 (79:21)
Me M+
H B Me
Me Me
Me
K-Selectride (3:97)
L-Selectride (8:92)
14
C5: Control in Organic Chemistry
More than one sp2 centre
O Me Me
(i) n-PrI (ii) H+, H2O
N
N H
Me Me
Me
O Me Me
Me
Me
MgBr Me Me Me
O
O
Me
O
Me
CuBr•SMe2
Me Me Me
O
O
O Me Me
Me > 80%
Me
Me
15
C5: Control in Organic Chemistry
Trans-diaxial ring opening (Fürst-Plattner Rule)
O
Me Me
OH
NuMe Me
Me
Nu Me
95%
But:
O
Me Me Me
Nu
NuMe Me
OH Me
95%
16
C5: Control in Organic Chemistry
Br Br2 Me Me
Me Me
Br Me
Me
MeOH Me
Br
OMe
Br2 Me Me
100%
+
Me Me
Br Me
Pasto, J. Am. Chem. Soc. 1970, 92, 7480.
47%
Me Me
OMe Me
53%
17
C5: Control in Organic Chemistry
Chelation can reverse the stereoselectivity. O O O
O
O
H
Me
O O
Me
Cl O
O O OH
O
H OH
Cl O
18
C5: Control in Organic Chemistry
OH
OH
t-BuOOH VO(acac)2
O
Dihydroxylation: OH
OH OH Me Me
OH Me
OsO4,NMO Acetone,H2O
OH
OsO4, TMEDA CH2Cl2, -78 ˚C
Me Me Me
OH Me Me
OH Me
19
C5: Control in Organic Chemistry
Prévost and Woodward Methods: O Me
O OH
Me Me
I2, C6H6
OH Me
then NaOH
O
Me
Ag
Me
O
AcOH, I2, H2O
Me Me Me
Me
Ag
then NaOH
OH Me Me
OH Me
20
C5: Control in Organic Chemistry
Iodolactonisation in Synthesis O
O
O
I2, H2O OH Me NaHCO3
HO
O
Me mCPBA
O
O
Me
Me HBr
O
O
Me
Br
O
O
O
Me MnO 2
O
HO
I
O O
Me DBN
O
O
O
O R2CuLi O
Me R
(i) Zn (ii) TsOH, MeOH
OMe Me O
R
21
C5: Control in Organic Chemistry
O
O OMe Me
O
(i) O3 then Me2S (ii) [O] O
O
OMe Me
O
Me O
MeONa MeOH OMe
O
OMe O
O O
OMe Me
O
O
(i) NaOH (ii) HCl
Me O
O O
H
OMe
O
H
22
C5: Control in Organic Chemistry
Stereochemical Control with Bicyclic Compounds
Cyclopentane
O
Fused Bicyclic
Spiro-cyclic
LiAlH4
H
THF
OH
Bridged Bicyclic
OH
+
H
norbornane
Me
Me Me O
LiAlH4 THF
camphor
Me
Me Me
Me H
OH
+
Me Me
OH
H
23
C5: Control in Organic Chemistry
Fused rings
cis-fused only
trans possible, but difficult
trans more stable than cis
24
C5: Control in Organic Chemistry
H
O R
R
R=H
R
Li NH3
O Me
EtI LiO
LiO Me
H Me
H Et Me Me
R = Me O
H Et Me
25
C5: Control in Organic Chemistry
H O O
(i) NaH (ii) (MeO)2C=O
O
O
H
H O
H
H
NaBH4
O
H OH
O H MeO
mCPBA H
H
O
H
H
H O
O
H MeO
O
26
C5: Control in Organic Chemistry
Open Chain Stereochemical Control Nucleophilic Attack on Carbonyls with α–Chiral Centres Me
O
Me Me
Me
NaBH4
OH
Me Me
Me
Me Me
Me
Molecules with a high degree of flexibility tend to react unselectively.
But carbonyls with adjacent chiral centres show some selectivity. Why?
O
OH
LiAlH4
Et
Et Me
Me
O
OH
+
Et
anti : syn 3:1
OH
EtMgBr
Me
OH
+
Et
H Me
Me
Et
anti : syn 1:3
Me
Well, as with the cyclic molecules we have been discussing previously, the key to understanding this selectivity is conformation: so you still must draw in 3-D to explain things.
Me
O
O Ph
H
Me
H
Ph
Ph
O
H
H
O Me
H
H
H
Ph
Me
Me
O
Ph
O H
H
H
Ph
Me H
H
The trajectory for attack on a carbonyl group was determined by two crystallographers Bürgi and Dunitz in the 1970s: the ‘Bürgi-Dunitz angle’ of attack is about 109˚, not 90˚ as previously thought.
Nu109˚ O
27
C5: Control in Organic Chemistry
The major product results from the most reactive conformer.
Me
O
O Ph
H
Nu-
-Nu
H
H
Ph H
Nu-
Me
-
Nu
Not all ‘flight paths’ for the nucleophile are equally favourable.
This is called the ‘Felkin-Anh’ model and correctly predicts major products in such rxns; however we need to be aware of two other possibilities: electronegative groups and chelating groups.
Orbital Control in the Felkin-Anh model Consider Reetz’s synthesis of an unusual amino acid present in the anticancer compound Dolastatin.
Me Et
O
O H
NBn2
Li Me OMe
OH
Me
O
OH O
+ Et
OMe
Et
OMe NBn2
NBn2
anti : syn 24 : 1 92% d.e.
28
C5: Control in Organic Chemistry
H
O
Bn
Bn
N
Et H Me
Me O Et
N Bn
Bn
H
H
Me Bn
Et
N H
Bn
Why is the stereocontrol so good when NBn2 and CH(Me)Et are fairly similar in size to each other? Why does NBn2 want to be perpendicular to C=O for the model to work? The reason is that conformations where electronegative groups are perpendicular to C=O are more reactive to nucleophilic attack. Why? new LUMO O Me
X
π* of the C=O bond
X is an electronegative group but not a leaving group (OR, NR2, SR, etc.)
X σ* of the C-X bond
X combine O
O
-Nu
σ*C-X
Energy
O X
π*C=O π* + σ*
Orbital overlap is best with this conformation
New LUMO Lower energy More reactive
Note that the most heavily populated conformations will be those where the bulky groups are perpendicular to the C=O bond, but the more reactive conformations will be when the most electronegative atom is perpendicular to the C=O bond (cf. Curtin-Hammett principle).
29
C5: Control in Organic Chemistry
Chelation Control Sometimes the result depends on the reagent: O
Nu OH
HO Nu OMe
OMe
OMe
+
Me
Me (±)
Me
NaBH4
73%
27%
Me2Mg
1%
99%
When chelation is possible the direction of stereocontrol inverts and the level of stereocontrol is higher (usually). Note this fits nicely with the ideas presented earlier in the course: stereoselectivity is likely to be high if a cyclic transition state is involved. Chelation involves just such a transition state, so it should be no surprise that it lets us achieve much higher levels of control than the acyclic Felkin-Anh model does. Transition states with different metal counterions: With Mg2+
With Na+
Mg2+ Me
O
O OMe
H
H-
OMe
Me
Ph
Ph
H
-Me
Two things are required for chelation control: 1. A heteroatom with lone pairs available for coordination to a metal ion. 2. A metal ion that favour coordination to both C=O and the heteroatom. For example: Mg2+, Zn2+, Al3+, Ce3+ and Ti4+ are excellent Li+ is sometimes okay Na+ and K+ are bad Note that the best ones are usually more highly charged ions. Summary: O Z Y
R X
Is there a heteroatom at the chiral centre? NO Use Felkin-Anh model: consider rxns on conformations with the largest group perpendicular to C=O
YES
Is there a metal ion YES capable of chelation with the heteroatom? NO Use Felkin-Anh model: consider rxns on conformations with the most electronegative atom perpendicular to C=O
Use chelation model: consider rxns on conformation with C=O and heteroatom held close together in space
30
C5: Control in Organic Chemistry
Diastereoselective Rxns of Acyclic Alkenes As we discovered before molecules with a high degree of flexibility tend to react unselectively; but carbonyls with adjacent chiral centres show selectivity. Me Me
O
mCPBA
SiMe2Ph
Me
Me
O
+
Me
SiMe2Ph
Me SiMe2Ph
>95%
As with rxns at C=O, conformation is the key. The Houk Model Calculations by Ken Houk found that low energy conformations of alkenes with allylic substituents have one substituent eclipsing the double bond. Lets look at the alkene below. The lowest energy conformation is the one that has the hydrogen atom (the smallest group) eclipsing the double bond. Another low energy conformation (only 3.1 kJ mol-1 higher) has one of the methyl groups eclipsing the double bond, so that when we start to look at rxns of this type of alkene (including trans alkenes) we should consider both conformations.
E H H Me Me Ψ Me
Me H H Me E=3.1
Me H
H E=0
E > 8 kJmol-1 This is NOT a minimum!
0
Ψ
0
60
180
120
The allylic 1,3-strain between H and a methyl group is not large so a significant proportion of the slightly higher energy conformation is present at equilibrium. So what happens when we try to epoxidise the alkene shown below?
Me
Me SiMe2Ph
mCPBA
Me
O
Me
+
Me
SiMe2Ph 61 : 39 22% d.e.
O
Me SiMe2Ph
31
C5: Control in Organic Chemistry
Me H Me
Me Me Si H
Me major conformer: H eclipsing C=C
Me SiMe2Ph Me H
H Me
Si Me minor conformer: Me eclipsing C=C Me
What about the conformations of cis alkenes? Me H Me Me E = 16.8 kJmol-1
E
H Me Me Me E = 14.4 kJmol-1
Ψ Me
Me Me H E=0 0
Ψ
0
60
180
120
So for cis alkenes only one conformer is heavily populated and this leads to the high diastereoselectivity we observed earlier. What if we have a directing group? Me
Me Me (±)
OH
mCPBA
Me
O Me
Me + Me
O Me
OH 95 : 5
Me OH
32
C5: Control in Organic Chemistry
Summary: To explain the stereoselectivity of rxns of chiral alkenes: • Draw the conformer with H eclipsing the double bond. • Allow the reagent to attack the less hindered face (or syn to a coordinating group). • Draw product in same conformation as starting material. • Redraw the product as a ‘2-D’ structure with the longest chain in the plane of the paper. Stereoselective enolate alkylation on carbonyls with β-chiral centres The same model can be used to explain the diastereoselectivity of enolates with an adjacent chiral centre.
PhMe2Si
OEt Me
O
LDA
PhMe2Si
OEt Me
O
Li
Me MeI
PhMe2Si
Me OEt
Me
+
O
PhMe2Si
OEt Me
95 : 5
O
33
C5: Control in Organic Chemistry
Diastereocontrol in Aldol Rxns We have now come the second common case of a diastereoselective rxn: i.e. rxns in which two prochiral substrates react to give a product where two new stereocentres are formed, where one new stereocentre originated from each substrate. So with substituted enolates we need to consider which diastereomer is the major product, and why!
O Ph
Me
LDA, THF -78 ˚C
Li
O
O
O Me
Ph
Me
H
cis-enolate
Ph
O
OH
Et Me syn aldol major
+
Ph
OH
Et Me anti aldol minor
Note that the rxn is diastereoselective not because of attack onto one of two diastereotopic faces, but because of the way in which two prochiral reagents, each with enantiotopic faces, come together. Also note that aldol rxns can be under thermodynamic or kinetic control depending on the conditions. We will consider cases under kinetic control. With substituted enolates we have the possibility of two geometrical isomers: c i s– or trans–enolates, and this is an important factor controlling the diastereoselectivity. A very approximate generalisation (with many exceptions) is that: cis–enolates give mainly syn aldol products while trans–enolates give mainly anti aldol products
[This is an oversimplification since enolates of some metals (Sn, Zr, Ti) give syn aldols regardless of enolate geometry; however, you will not meet these in this course.] How do we know this is the case? O O
Li LDA, THF -78 ˚C
O
H
only the transenolate can form
O
OH
anti aldol
34
C5: Control in Organic Chemistry
The Zimmerman–Traxler Transition State The reason why enolate geometry is so important is due to the aldol rxn having a cyclic transition state. The six-membered ring transition state was proposed by Zimmerman and Traxler, hence the name of the transition state model. cis-enolates give syn aldol products: R O
Li Me
R
H
R H
O Ph
Phenyl group is pseudoaxial: disfavoured
O
H
or
O Ph Me
H
Li O
H Me
Phenyl group is pseudoequitorial: favoured
A
R
Ph O
Li
B
R H O
H
H
Li H
O
Li
O
O
redraw
Ph
R
O
Ph Me
OH
Ph Me
syn aldol
Me
A
trans-enolates give anti aldol products:
R O
Li H
R
Ph
Me
R H
O Me
or
O Ph C
Ph O
Li
O
H
Phenyl group is pseudoaxial: disfavoured
Phenyl group is pseudoequitorial: favoured
Me
Li O
H
H D
R H
Li
O
Me
O H
Ph C
Li
O O
O
redraw R
Ph
anti aldol
35
C5: Control in Organic Chemistry
Stereoselective Enolization: Boron Enolates The cyclic transition state explains how the enolate geometry controls the stereochemical outcome of the aldol rxn (cis gives syn; trans gives anti). But what controls the geometry of the enolate? For lithium enolates of ketones the most important factor is the size of the group that is not enolized (the R group in diagrams above). Large groups force the enolate to adopt the cis geometry; small groups allow the trans-enolate to form. But we cannot separate lithium enolates so really we need a better way to make each enolate geometry. O
OLi
LDA Me
R
OLi Me
R
+
R Me
R = t-Bu
98%
2%
R = Et
30%
70%
The answer to our problem is boron enolates. By choosing which boron triflate we use:
B
O
OH
O Me
PhCHO
Ph Me
B
OTf
syn aldol product > 94% d.e.
cis-enolate
(i-Pr)2NEt O Me
B
OTf
O
O
B
OH
PhCHO
(i-Pr)2NEt
Ph Me
Me
trans-enolate
anti aldol product 72% d.e.
Aldol rxns of boron enolates are frequently more diastereoselective than lithium enolates. One reason for this is because of the relatively short B–O bond length (B–O = 1.36-1.47 Å, Li–O = 1.922.00 Å) which exacerbates unfavourable 1,3-diaxial interactions in the transition state.
36
C5: Control in Organic Chemistry
Evans’ oxazolidinone chiral auxiliary An extremely important example of diastereoselectivity is the use of chiral auxiliaries. This topic will be covered in courses next year. But you should not feel intimidated by asymmetric synthesis, it follows all the principles we have learnt in this course: if we want to make chiral centres selectively we need to have diastereomeric transition states (under kinetic control). One widely used chiral auxiliary is the oxazolidinone shown below: Li O O
O
O Me
N
LDA O
Me
Me
N
O
BnBr O
Me
Me
N Me
Me
Me from L-Valine
O
O
Me d.r. 120:1
Note high diastereoselectivity results from: • • •
Exclusive formation of the cis-enolate Chelation results in a single rigid conformation π–Facial selectivity results from steric hindrance with i–propyl group
Added advantages are: • • • •
Auxiliary is easily made from L-Valine Removal of the oxizolidinone is easy (compared to an amide) without racemisation Products are often crystalline: recrystallisation can give diastereomerically pure material Opposite enantiomer can be made with an auxiliary derived from norephedrine
C5: Control in Organic Chemistry
37
Definitions: Stereoisomer
Compounds made up of the same atoms bonded by the same sequence of bonds but having different three-dimensional structures (configurations) that are not interchangeable.
Enantiomer
Molecule that is non–superimposable on its mirror image.
Diastereomer
Stereoisomers that are not enantiomers.
Epimer
Two diastereomers that have a different configuration at only one chiral centre.
Racemic
Mixture of equal amounts of enantiomers.
Conformations
Different three-dimensional arrangements in space of atoms in a molecule that are interconvertible by free rotation about bonds.
38
C5: Control in Organic Chemistry
Supervision Work: Essential Reading: Clayden, Greeves, Warren and Wothers “Organic Chemistry” Oxford, 2001. Chapters 16, 18, 32, 33 and 34. READ THEM!
Additional Reading: Carey and Sundberg “Advanced Organic Chemistry” Kluwer Academic, 2000. March “Advanced Organic Chemistry” Wiley, 2001. Procter “Stereoselectivity in Organic Synthesis” Oxford Primer, 1998. Eliel, Wilen and Mander “Stereochemistry of Organic Compounds” Wiley, 1994.
1. Write brief notes, which will be valuable for your revision, and include an example on the following keywords (how about looking them up in several books?): a. Chemoselectivity. b. Regioselectivity. c. Diastereoselectivity. d. Stereospecific reactions. e. Thermodynamic and kinetic control. f. Axial attack (with six membered rings). g. Fürst-Plattner rule. h. Felkin-Anh model. i. Houk model. j. Zimmerman-Traxler transition state.
2. (i) Explain the stereochemical control in the following sequence. Perhaps you would like to represent these 2-D structural representations in 3-D. O
OH R1
HO
NaBH4
OMs R1
MsCl
NH3
R1
R1
HN
Et3N R2
HO
R2
R2 MsO
R2
(ii) What does Ms stand for in MsCl? What is the structure? What is the mechanism of mesylation (you should probably look it up even if you think you know)?
3. Explain how the stereo- and regiochemistry of these compounds are controlled. To do this properly you must work out the lowest energy conformation of the starting materials or intermediates (you might like to use molecular models) and draw it clearly in 3-D. CO2Me mCPBA
CO2Me O
RNH2
HO RHN
CO2Me ∆
HO O N R
Why is the epoxidation only moderately stereoselective? Why does the amine attack where it does?
39
C5: Control in Organic Chemistry
4. In the previous half of this course you met the following rxn. You should now be able to explain the diastereoselectivity. Me
O
Me
NaBH4 O
OH
O
5. How would you make each diol from the same alkene? Give the mechanisms of all rxns and explain the stereochemical control. HO Me C8H17 H Me
HO
Me
Me
HO HO
H
H
???? H
HO
Me
H HO
H
Start by working out the lowest energy conformation of the steroid (you might like to use molecular models) and draw it clearly in 3-D. Then think of different ways to achieve dihydroxylation (some methods involve more than one step). Now put them together and bear in mind stereochemical control.
6. Consider the following sequence: OH
O
AcO AcOH, H+ OBn
(i) (ii) (iii) (iv)
O (i) MsCl, Et3N (ii) K2CO3, MeOH
OBn
(i) PhSeSePh, NaBH4 (ii) H2O2
HO
O
HO Reagent?
OBn
OBn
A
B
OBn C
Explain the formation of the stereochemistry of A. Reduction of PhSeSePh with NaBH4 gives the nucleophile PhSe-. Give mechanisms for the steps involved in forming B. Suggest a reagent to make C and explain why you think it will give the product with the correct stereochemistry. Explain the regioselectivity of the two epoxide opening rxns.
40
C5: Control in Organic Chemistry
7. Explain the stereoselectivities of the following rxns (you will need to draw Newman projections). O Me
Me
Et OH
EtMgBr
Me
Me
Cl SPh Ph
Cl SPh L-Selectride
Ph
Ph
Ph OH
O SPh Ph
Zn(BH4)2
SPh Ph
Ph
Ph OH
O
8. Explain the stereochemical outcomes of the following rxns. OH Me
OH Me CH I , Zn/Cu 22
OH Me +
Me
Me
Me 99 : 1
OH Me
OH
CH2I2, Zn/Cu Me
Me
OH Me
+ Me
Me
55 : 45
The rxn involved is the Simmons-Smith rxn. Draw a sensible mechanism (p.1067 of CGW&W should help).
9. Predict the structures of the products in the following rxns, showing their stereochemistry clearly. Explain your answer. H O
(i) LDA (ii) MeI
O (i) Me2CuLi (ii) Allyl Bromide
H OH
mCPBA
H
mCPBA H
Me O H
H2, Pd/C
41
C5: Control in Organic Chemistry
10. Explain the formation of essentially one stereoisomer in the following rxn. O
O
(i) LDA
Me Me
Me (ii) Me
Me
OH
Me Me Me
Me
Me
H O (+)-S
11. Provide the stereostructure of the major product and rationalise the stereochemical outcome of the following rxn. Me Ph
H
+
BF3•OEt2 CH2Cl2, -78 ˚C
? d.r. 16 : 1
OTMS
O
12. Account for the differences in the aldol addition products between the two substrates. Me Me Me
O O
(i) LDA (ii) MeCHO
O Me
Me Me Me
O
O
O
Me OH
80% yield d.r. 4 : 1
Me
Me Me Me
O
(i) LDA (ii) MeCHO
O Me
O
Me Me Me
O
O
i-Pr OH
O
Me
52% yield d.r. 6 : 1
Me
13. Draw a rxn mechanism and explain the diastereoselectivity in the following rxn. O Ph Me
O
(i-Bu)3Al
HO Ph Me
O
(i-Bu)3Al is a Lewis acid, but also serves as the reducing agent (with loss of isobutylene). This is not an easy problem. Try to see what is going on first mechanistically then look at the key steps more closely and consider stereochemical issues. The rxn was published in J. Am. Chem. Soc. 1995, 117, 6395.
42
C5: Control in Organic Chemistry
14. Explain the following selectivities: Me Me2CuLi O
anti : syn 96 : 4 O
Me
Me Me
O
Me2CuLi
O
anti : syn 4 : 96
Me Me
Again this is not an easy problem. What you have to do is firstly identify the rxn we are looking at, and don’t be confused by the way I have drawn the second product. Secondly, find out the preferred conformation of the unsaturated ten-membered rings that minimises strain and transannular interactions (molecular models will definitely help). Draw the conformation clearly in 3-D. Once you have done this the explanation should follow. This work was published in Tetrahedron 1981, 37, 3981.
C5: Control in Organic Chemistry
Part II Chemistry 2003
Control in Organic Chemistry Lectures 7-12 Dr David Spring [email protected] Tel. 01223 336364 Lab 180
O
O O Me
BrMg CuBr•SMe2
O
O Me O
Course Outline 1. Introduction to Stereoselectivity 2. Stereochemical Control in Four- and Five-Membered Rings 3. Stereochemical Control in Six-Membered Rings 4. Stereochemical Control with Bicyclic Compounds 5. Open Chain Stereochemical Control
Essential Reading
Clayden, Greeves, Warren, Wothers “Organic Chemistry” Oxford, 2001.
2
C5: Control in Organic Chemistry
Introduction to Stereoselectivity How are rxns controlled? Kinetically and thermodynamically controlled rxns: If treatment of a starting material can give more than one product then the amount of each product is determined by its rate of formation (kinetic control). However, if the rxn is reversible then the product distribution is determined by the relative thermodynamic stability of each product. O O S OH H2SO4 80 ˚C
O H2SO4
O S
OH
160 ˚C
∆G
O
O S
H2SO4
O ∆G˚ O S OH
OH
H2O
H2O
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C5: Control in Organic Chemistry
The majority of this course will be concerned with kinetically controlled rxns that involve the formation of tetrahedral (sp3) carbon atoms within a molecular framework.
What type of selectivity are we talking about? Chemoselectivity – functional group discrimination Regioselectivity – product structural isomer discrimination Stereoselectivity – product stereoisomer discrimination O
O O Me
O
BrMg
Me
CuBr•SMe2
O
O
Stereochemistry involves diastereomers and enantiomers. Therefore stereoselectivity encompasses diastereoselectivity (product diastereomer discrimination) and enantioselectivity (product enantiomer discrimination). This course will focus on diastereoselectivity. Enantioselective rxns will be covered in detail next year. So from now on, when we talk about stereoselectivity, more specifically we mean diastereoselectivity.
E
Enantiomeric transition states must be the same energy
OLi Ph Bu H
O H
E
Diastereomeric transition states are not necessarily the same energy
OLi BuLi Ph
Rxn Coordinate
OLi H Bu Ph
Bu
O
Ph Me
H
BuLi Ph * Me
OLi Bu
Ph Me
Rxn Coordinate
As you know enantiomers are equal in energy, therefore enantiomeric transition states are also equal in energy. So it is impossible to achieve any selectivity, we get a racemic mixture. However, if we have a chiral centre in the substrate then we form diastereomers, then the transition state energies need not be equal and we should observe some selectivity. This simple statement in fact forms the basis for all diastereoselectivity (and enantioselectivity! – except that the chirality is not in the substrate).
4
C5: Control in Organic Chemistry
Selective or Specific? If a rxn proceeds via a mechanism with a strict stereochemical requirement, then the rxn is stereospecific. So in a stereospecific rxn, the starting material goes to only one possible product stereoisomer, there is NO selectivity. For example, substitution by the SN2 mechanism must involve an inversion of configuration of the carbon atom that is reacting:
Me Si Me
Me O
O (±)
N3-
Me
O
Me Si Me
Me O
O S O Me
Me
O N N+ N-
(±)
What does the other enantiomer look like?
Me Si Me
Me O
NaN3
O
Me
O O
Me Si Me
Me
Me
O N N+ N-
O S O Me
Stereospecific rxns are enantiospecific for chiral starting materials where a stereospecific rxn occurs at all stereocentres. For example:
Br Me
HO-
OH Me
Me
Me 95% e.e.
95% e.e.
Stereospecific rxns are also possible with achiral starting materials:
5
C5: Control in Organic Chemistry
In contrast, for a stereoselective rxn, the mechanism does not prevent the formation of two (or more) stereoisomeric products. Even if the rxn is completely selective giving only one stereoisomer, it still NOT ‘stereospecific’. As we will only be discussing diastereoselectivity, the selectivity will be between diastereomeric products and we shall talk about diastereomeric ratios (d.r.) and diastereomeric excesses (d.e.). [Enantioselectivity concerns selectivity between product enantiomers]
Two common cases of diastereoselective rxns are:
1. Rxns that involve the formation of one (or more) new stereocentres in a chiral substrate:
O
OH
NaBH4
OH
+ Me
Me
Me
Me
Me
90%
10%
si face addition
re face addition
Me
Chiral Starting Material
Diastereomeric Products d.r. = 9:1 d.e. = 80%
2. Rxns in which two prochiral substrates react to give a product where two new stereocentres are formed, where one new stereocentre originated from each substrate. OH O Me Me
Me Me
O H
+
Me
Li
+
Me
Me O
OH O Me
Me Me
Me Me
44%
Me 6%
Me
Me OH O +
Me Me
OH O Me
Me
Me 44%
+
Me Me
Me Me
Me 6%
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C5: Control in Organic Chemistry
Stereoselective rxns of cyclic compounds Revision of Conformational Analysis
DRAW IN 3-D!! H
H H
H
H
H H H H
H H
H
H H
H H H
H
H H
H
H H
H
Ring Flip
H H
H H H
H
H
H
H
H
H
HH
HH
H
H H
Chair
H
H H
H
H H
H H H Boat
H H
H
H
H H
H
Twist-Boat
Learn how to draw a cyclohexane on p. 459-460 of Clayden et al.
C5: Control in Organic Chemistry
OH
Me
Me Me
H
OH
H
H
H
OH
7
8
C5: Control in Organic Chemistry
Stereochemical Control in Four- and Five-Membered Rings Four-membered rings Just one sp2 centre O
OH NaBH4
R
R
More than one sp2 centres
O O
Me Me
O
(i) LDA, -78 ˚C, THF (ii) BnBr O
Me Me
(±) d.r. >98 : 2
9
C5: Control in Organic Chemistry
Five-membered rings Just one sp2 centre O
OH Me
LiAlH4
OH Me
Me
+
THF (±)
O Me
77%
23%
OH
OH
LiBH(s-Bu)3
Me
Me
+
THF (±)
1.5%
98.5%
10
C5: Control in Organic Chemistry
More than one sp2 centre M
O
O
O R2 X
(R1)2CuLi TBSO
R1
TBSO
R2 TBSO
R1
O CO2H Me
HO OH PGE2
O Li
+
Me H
CuI, Bu3P, THF then Ph3SnCl, HMPA
OTBS
TBSO
O I
O
O
OMe
OMe HMPA, Et3N -78 to 23 ˚C
Me
TBSO OTBS
Suzuki, Noyori et al. J. Am. Chem. Soc. 1985, 107, 3348-3349.
11
C5: Control in Organic Chemistry
Chiral Memory: O
O + HO OH
Me
H+ H
Me Me Me
Me Me
O
(i) LDA (ii) PrBr
O
O
How do we make both epoxides stereoselectively? Me Me Me Me Me Me Me
O
Me Me Me Me Me
O
Me Me Me
O O
O Me
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C5: Control in Organic Chemistry
Stereochemical Control in Six-Membered Rings Just one sp2 centre-addition to cyclohexanones
NaBH4 Me Me
Me
Me OH
O
OH
MeMgBr Me Me
Me
Me Me
Me
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C5: Control in Organic Chemistry
Me Me
OH
O
OH
Al(Oi-Pr)3
LiBH(s-Bu)3
Me Me
Me
O
Me
Me
equitorial diastereomer H Me OH
[H]
Me Me
Me
axial diastereomer OH Me H
+
Me Me
0
Me Me
i-PrOH
Me Me
% axial diastereomer 10
Li/NH3 (99:1) LiAlH4 (93:7) LiAlH(Ot-Bu)3 (92:8)
20
30
40
50
70
60
80
90
100%
Me DIBAL (72:28) NaBH4 (79:21)
Me M+
H B Me
Me Me
Me
K-Selectride (3:97)
L-Selectride (8:92)
14
C5: Control in Organic Chemistry
More than one sp2 centre
O Me Me
(i) n-PrI (ii) H+, H2O
N
N H
Me Me
Me
O Me Me
Me
Me
MgBr Me Me Me
O
O
Me
O
Me
CuBr•SMe2
Me Me Me
O
O
O Me Me
Me > 80%
Me
Me
15
C5: Control in Organic Chemistry
Trans-diaxial ring opening (Fürst-Plattner Rule)
O
Me Me
OH
NuMe Me
Me
Nu Me
95%
But:
O
Me Me Me
Nu
NuMe Me
OH Me
95%
16
C5: Control in Organic Chemistry
Br Br2 Me Me
Me Me
Br Me
Me
MeOH Me
Br
OMe
Br2 Me Me
100%
+
Me Me
Br Me
Pasto, J. Am. Chem. Soc. 1970, 92, 7480.
47%
Me Me
OMe Me
53%
17
C5: Control in Organic Chemistry
Chelation can reverse the stereoselectivity. O O O
O
O
H
Me
O O
Me
Cl O
O O OH
O
H OH
Cl O
18
C5: Control in Organic Chemistry
OH
OH
t-BuOOH VO(acac)2
O
Dihydroxylation: OH
OH OH Me Me
OH Me
OsO4,NMO Acetone,H2O
OH
OsO4, TMEDA CH2Cl2, -78 ˚C
Me Me Me
OH Me Me
OH Me
19
C5: Control in Organic Chemistry
Prévost and Woodward Methods: O Me
O OH
Me Me
I2, C6H6
OH Me
then NaOH
O
Me
Ag
Me
O
AcOH, I2, H2O
Me Me Me
Me
Ag
then NaOH
OH Me Me
OH Me
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C5: Control in Organic Chemistry
Iodolactonisation in Synthesis O
O
O
I2, H2O OH Me NaHCO3
HO
O
Me mCPBA
O
O
Me
Me HBr
O
O
Me
Br
O
O
O
Me MnO 2
O
HO
I
O O
Me DBN
O
O
O
O R2CuLi O
Me R
(i) Zn (ii) TsOH, MeOH
OMe Me O
R
21
C5: Control in Organic Chemistry
O
O OMe Me
O
(i) O3 then Me2S (ii) [O] O
O
OMe Me
O
Me O
MeONa MeOH OMe
O
OMe O
O O
OMe Me
O
O
(i) NaOH (ii) HCl
Me O
O O
H
OMe
O
H
22
C5: Control in Organic Chemistry
Stereochemical Control with Bicyclic Compounds
Cyclopentane
O
Fused Bicyclic
Spiro-cyclic
LiAlH4
H
THF
OH
Bridged Bicyclic
OH
+
H
norbornane
Me
Me Me O
LiAlH4 THF
camphor
Me
Me Me
Me H
OH
+
Me Me
OH
H
23
C5: Control in Organic Chemistry
Fused rings
cis-fused only
trans possible, but difficult
trans more stable than cis
24
C5: Control in Organic Chemistry
H
O R
R
R=H
R
Li NH3
O Me
EtI LiO
LiO Me
H Me
H Et Me Me
R = Me O
H Et Me
25
C5: Control in Organic Chemistry
H O O
(i) NaH (ii) (MeO)2C=O
O
O
H
H O
H
H
NaBH4
O
H OH
O H MeO
mCPBA H
H
O
H
H
H O
O
H MeO
O
26
C5: Control in Organic Chemistry
Open Chain Stereochemical Control Nucleophilic Attack on Carbonyls with α–Chiral Centres Me
O
Me Me
Me
NaBH4
OH
Me Me
Me
Me Me
Me
Molecules with a high degree of flexibility tend to react unselectively.
But carbonyls with adjacent chiral centres show some selectivity. Why?
O
OH
LiAlH4
Et
Et Me
Me
O
OH
+
Et
anti : syn 3:1
OH
EtMgBr
Me
OH
+
Et
H Me
Me
Et
anti : syn 1:3
Me
Well, as with the cyclic molecules we have been discussing previously, the key to understanding this selectivity is conformation: so you still must draw in 3-D to explain things.
Me
O
O Ph
H
Me
H
Ph
Ph
O
H
H
O Me
H
H
H
Ph
Me
Me
O
Ph
O H
H
H
Ph
Me H
H
The trajectory for attack on a carbonyl group was determined by two crystallographers Bürgi and Dunitz in the 1970s: the ‘Bürgi-Dunitz angle’ of attack is about 109˚, not 90˚ as previously thought.
Nu109˚ O
27
C5: Control in Organic Chemistry
The major product results from the most reactive conformer.
Me
O
O Ph
H
Nu-
-Nu
H
H
Ph H
Nu-
Me
-
Nu
Not all ‘flight paths’ for the nucleophile are equally favourable.
This is called the ‘Felkin-Anh’ model and correctly predicts major products in such rxns; however we need to be aware of two other possibilities: electronegative groups and chelating groups.
Orbital Control in the Felkin-Anh model Consider Reetz’s synthesis of an unusual amino acid present in the anticancer compound Dolastatin.
Me Et
O
O H
NBn2
Li Me OMe
OH
Me
O
OH O
+ Et
OMe
Et
OMe NBn2
NBn2
anti : syn 24 : 1 92% d.e.
28
C5: Control in Organic Chemistry
H
O
Bn
Bn
N
Et H Me
Me O Et
N Bn
Bn
H
H
Me Bn
Et
N H
Bn
Why is the stereocontrol so good when NBn2 and CH(Me)Et are fairly similar in size to each other? Why does NBn2 want to be perpendicular to C=O for the model to work? The reason is that conformations where electronegative groups are perpendicular to C=O are more reactive to nucleophilic attack. Why? new LUMO O Me
X
π* of the C=O bond
X is an electronegative group but not a leaving group (OR, NR2, SR, etc.)
X σ* of the C-X bond
X combine O
O
-Nu
σ*C-X
Energy
O X
π*C=O π* + σ*
Orbital overlap is best with this conformation
New LUMO Lower energy More reactive
Note that the most heavily populated conformations will be those where the bulky groups are perpendicular to the C=O bond, but the more reactive conformations will be when the most electronegative atom is perpendicular to the C=O bond (cf. Curtin-Hammett principle).
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C5: Control in Organic Chemistry
Chelation Control Sometimes the result depends on the reagent: O
Nu OH
HO Nu OMe
OMe
OMe
+
Me
Me (±)
Me
NaBH4
73%
27%
Me2Mg
1%
99%
When chelation is possible the direction of stereocontrol inverts and the level of stereocontrol is higher (usually). Note this fits nicely with the ideas presented earlier in the course: stereoselectivity is likely to be high if a cyclic transition state is involved. Chelation involves just such a transition state, so it should be no surprise that it lets us achieve much higher levels of control than the acyclic Felkin-Anh model does. Transition states with different metal counterions: With Mg2+
With Na+
Mg2+ Me
O
O OMe
H
H-
OMe
Me
Ph
Ph
H
-Me
Two things are required for chelation control: 1. A heteroatom with lone pairs available for coordination to a metal ion. 2. A metal ion that favour coordination to both C=O and the heteroatom. For example: Mg2+, Zn2+, Al3+, Ce3+ and Ti4+ are excellent Li+ is sometimes okay Na+ and K+ are bad Note that the best ones are usually more highly charged ions. Summary: O Z Y
R X
Is there a heteroatom at the chiral centre? NO Use Felkin-Anh model: consider rxns on conformations with the largest group perpendicular to C=O
YES
Is there a metal ion YES capable of chelation with the heteroatom? NO Use Felkin-Anh model: consider rxns on conformations with the most electronegative atom perpendicular to C=O
Use chelation model: consider rxns on conformation with C=O and heteroatom held close together in space
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C5: Control in Organic Chemistry
Diastereoselective Rxns of Acyclic Alkenes As we discovered before molecules with a high degree of flexibility tend to react unselectively; but carbonyls with adjacent chiral centres show selectivity. Me Me
O
mCPBA
SiMe2Ph
Me
Me
O
+
Me
SiMe2Ph
Me SiMe2Ph
>95%
As with rxns at C=O, conformation is the key. The Houk Model Calculations by Ken Houk found that low energy conformations of alkenes with allylic substituents have one substituent eclipsing the double bond. Lets look at the alkene below. The lowest energy conformation is the one that has the hydrogen atom (the smallest group) eclipsing the double bond. Another low energy conformation (only 3.1 kJ mol-1 higher) has one of the methyl groups eclipsing the double bond, so that when we start to look at rxns of this type of alkene (including trans alkenes) we should consider both conformations.
E H H Me Me Ψ Me
Me H H Me E=3.1
Me H
H E=0
E > 8 kJmol-1 This is NOT a minimum!
0
Ψ
0
60
180
120
The allylic 1,3-strain between H and a methyl group is not large so a significant proportion of the slightly higher energy conformation is present at equilibrium. So what happens when we try to epoxidise the alkene shown below?
Me
Me SiMe2Ph
mCPBA
Me
O
Me
+
Me
SiMe2Ph 61 : 39 22% d.e.
O
Me SiMe2Ph
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C5: Control in Organic Chemistry
Me H Me
Me Me Si H
Me major conformer: H eclipsing C=C
Me SiMe2Ph Me H
H Me
Si Me minor conformer: Me eclipsing C=C Me
What about the conformations of cis alkenes? Me H Me Me E = 16.8 kJmol-1
E
H Me Me Me E = 14.4 kJmol-1
Ψ Me
Me Me H E=0 0
Ψ
0
60
180
120
So for cis alkenes only one conformer is heavily populated and this leads to the high diastereoselectivity we observed earlier. What if we have a directing group? Me
Me Me (±)
OH
mCPBA
Me
O Me
Me + Me
O Me
OH 95 : 5
Me OH
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C5: Control in Organic Chemistry
Summary: To explain the stereoselectivity of rxns of chiral alkenes: • Draw the conformer with H eclipsing the double bond. • Allow the reagent to attack the less hindered face (or syn to a coordinating group). • Draw product in same conformation as starting material. • Redraw the product as a ‘2-D’ structure with the longest chain in the plane of the paper. Stereoselective enolate alkylation on carbonyls with β-chiral centres The same model can be used to explain the diastereoselectivity of enolates with an adjacent chiral centre.
PhMe2Si
OEt Me
O
LDA
PhMe2Si
OEt Me
O
Li
Me MeI
PhMe2Si
Me OEt
Me
+
O
PhMe2Si
OEt Me
95 : 5
O
33
C5: Control in Organic Chemistry
Diastereocontrol in Aldol Rxns We have now come the second common case of a diastereoselective rxn: i.e. rxns in which two prochiral substrates react to give a product where two new stereocentres are formed, where one new stereocentre originated from each substrate. So with substituted enolates we need to consider which diastereomer is the major product, and why!
O Ph
Me
LDA, THF -78 ˚C
Li
O
O
O Me
Ph
Me
H
cis-enolate
Ph
O
OH
Et Me syn aldol major
+
Ph
OH
Et Me anti aldol minor
Note that the rxn is diastereoselective not because of attack onto one of two diastereotopic faces, but because of the way in which two prochiral reagents, each with enantiotopic faces, come together. Also note that aldol rxns can be under thermodynamic or kinetic control depending on the conditions. We will consider cases under kinetic control. With substituted enolates we have the possibility of two geometrical isomers: c i s– or trans–enolates, and this is an important factor controlling the diastereoselectivity. A very approximate generalisation (with many exceptions) is that: cis–enolates give mainly syn aldol products while trans–enolates give mainly anti aldol products
[This is an oversimplification since enolates of some metals (Sn, Zr, Ti) give syn aldols regardless of enolate geometry; however, you will not meet these in this course.] How do we know this is the case? O O
Li LDA, THF -78 ˚C
O
H
only the transenolate can form
O
OH
anti aldol
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C5: Control in Organic Chemistry
The Zimmerman–Traxler Transition State The reason why enolate geometry is so important is due to the aldol rxn having a cyclic transition state. The six-membered ring transition state was proposed by Zimmerman and Traxler, hence the name of the transition state model. cis-enolates give syn aldol products: R O
Li Me
R
H
R H
O Ph
Phenyl group is pseudoaxial: disfavoured
O
H
or
O Ph Me
H
Li O
H Me
Phenyl group is pseudoequitorial: favoured
A
R
Ph O
Li
B
R H O
H
H
Li H
O
Li
O
O
redraw
Ph
R
O
Ph Me
OH
Ph Me
syn aldol
Me
A
trans-enolates give anti aldol products:
R O
Li H
R
Ph
Me
R H
O Me
or
O Ph C
Ph O
Li
O
H
Phenyl group is pseudoaxial: disfavoured
Phenyl group is pseudoequitorial: favoured
Me
Li O
H
H D
R H
Li
O
Me
O H
Ph C
Li
O O
O
redraw R
Ph
anti aldol
35
C5: Control in Organic Chemistry
Stereoselective Enolization: Boron Enolates The cyclic transition state explains how the enolate geometry controls the stereochemical outcome of the aldol rxn (cis gives syn; trans gives anti). But what controls the geometry of the enolate? For lithium enolates of ketones the most important factor is the size of the group that is not enolized (the R group in diagrams above). Large groups force the enolate to adopt the cis geometry; small groups allow the trans-enolate to form. But we cannot separate lithium enolates so really we need a better way to make each enolate geometry. O
OLi
LDA Me
R
OLi Me
R
+
R Me
R = t-Bu
98%
2%
R = Et
30%
70%
The answer to our problem is boron enolates. By choosing which boron triflate we use:
B
O
OH
O Me
PhCHO
Ph Me
B
OTf
syn aldol product > 94% d.e.
cis-enolate
(i-Pr)2NEt O Me
B
OTf
O
O
B
OH
PhCHO
(i-Pr)2NEt
Ph Me
Me
trans-enolate
anti aldol product 72% d.e.
Aldol rxns of boron enolates are frequently more diastereoselective than lithium enolates. One reason for this is because of the relatively short B–O bond length (B–O = 1.36-1.47 Å, Li–O = 1.922.00 Å) which exacerbates unfavourable 1,3-diaxial interactions in the transition state.
36
C5: Control in Organic Chemistry
Evans’ oxazolidinone chiral auxiliary An extremely important example of diastereoselectivity is the use of chiral auxiliaries. This topic will be covered in courses next year. But you should not feel intimidated by asymmetric synthesis, it follows all the principles we have learnt in this course: if we want to make chiral centres selectively we need to have diastereomeric transition states (under kinetic control). One widely used chiral auxiliary is the oxazolidinone shown below: Li O O
O
O Me
N
LDA O
Me
Me
N
O
BnBr O
Me
Me
N Me
Me
Me from L-Valine
O
O
Me d.r. 120:1
Note high diastereoselectivity results from: • • •
Exclusive formation of the cis-enolate Chelation results in a single rigid conformation π–Facial selectivity results from steric hindrance with i–propyl group
Added advantages are: • • • •
Auxiliary is easily made from L-Valine Removal of the oxizolidinone is easy (compared to an amide) without racemisation Products are often crystalline: recrystallisation can give diastereomerically pure material Opposite enantiomer can be made with an auxiliary derived from norephedrine
C5: Control in Organic Chemistry
37
Definitions: Stereoisomer
Compounds made up of the same atoms bonded by the same sequence of bonds but having different three-dimensional structures (configurations) that are not interchangeable.
Enantiomer
Molecule that is non–superimposable on its mirror image.
Diastereomer
Stereoisomers that are not enantiomers.
Epimer
Two diastereomers that have a different configuration at only one chiral centre.
Racemic
Mixture of equal amounts of enantiomers.
Conformations
Different three-dimensional arrangements in space of atoms in a molecule that are interconvertible by free rotation about bonds.
38
C5: Control in Organic Chemistry
Supervision Work: Essential Reading: Clayden, Greeves, Warren and Wothers “Organic Chemistry” Oxford, 2001. Chapters 16, 18, 32, 33 and 34. READ THEM!
Additional Reading: Carey and Sundberg “Advanced Organic Chemistry” Kluwer Academic, 2000. March “Advanced Organic Chemistry” Wiley, 2001. Procter “Stereoselectivity in Organic Synthesis” Oxford Primer, 1998. Eliel, Wilen and Mander “Stereochemistry of Organic Compounds” Wiley, 1994.
1. Write brief notes, which will be valuable for your revision, and include an example on the following keywords (how about looking them up in several books?): a. Chemoselectivity. b. Regioselectivity. c. Diastereoselectivity. d. Stereospecific reactions. e. Thermodynamic and kinetic control. f. Axial attack (with six membered rings). g. Fürst-Plattner rule. h. Felkin-Anh model. i. Houk model. j. Zimmerman-Traxler transition state.
2. (i) Explain the stereochemical control in the following sequence. Perhaps you would like to represent these 2-D structural representations in 3-D. O
OH R1
HO
NaBH4
OMs R1
MsCl
NH3
R1
R1
HN
Et3N R2
HO
R2
R2 MsO
R2
(ii) What does Ms stand for in MsCl? What is the structure? What is the mechanism of mesylation (you should probably look it up even if you think you know)?
3. Explain how the stereo- and regiochemistry of these compounds are controlled. To do this properly you must work out the lowest energy conformation of the starting materials or intermediates (you might like to use molecular models) and draw it clearly in 3-D. CO2Me mCPBA
CO2Me O
RNH2
HO RHN
CO2Me ∆
HO O N R
Why is the epoxidation only moderately stereoselective? Why does the amine attack where it does?
39
C5: Control in Organic Chemistry
4. In the previous half of this course you met the following rxn. You should now be able to explain the diastereoselectivity. Me
O
Me
NaBH4 O
OH
O
5. How would you make each diol from the same alkene? Give the mechanisms of all rxns and explain the stereochemical control. HO Me C8H17 H Me
HO
Me
Me
HO HO
H
H
???? H
HO
Me
H HO
H
Start by working out the lowest energy conformation of the steroid (you might like to use molecular models) and draw it clearly in 3-D. Then think of different ways to achieve dihydroxylation (some methods involve more than one step). Now put them together and bear in mind stereochemical control.
6. Consider the following sequence: OH
O
AcO AcOH, H+ OBn
(i) (ii) (iii) (iv)
O (i) MsCl, Et3N (ii) K2CO3, MeOH
OBn
(i) PhSeSePh, NaBH4 (ii) H2O2
HO
O
HO Reagent?
OBn
OBn
A
B
OBn C
Explain the formation of the stereochemistry of A. Reduction of PhSeSePh with NaBH4 gives the nucleophile PhSe-. Give mechanisms for the steps involved in forming B. Suggest a reagent to make C and explain why you think it will give the product with the correct stereochemistry. Explain the regioselectivity of the two epoxide opening rxns.
40
C5: Control in Organic Chemistry
7. Explain the stereoselectivities of the following rxns (you will need to draw Newman projections). O Me
Me
Et OH
EtMgBr
Me
Me
Cl SPh Ph
Cl SPh L-Selectride
Ph
Ph
Ph OH
O SPh Ph
Zn(BH4)2
SPh Ph
Ph
Ph OH
O
8. Explain the stereochemical outcomes of the following rxns. OH Me
OH Me CH I , Zn/Cu 22
OH Me +
Me
Me
Me 99 : 1
OH Me
OH
CH2I2, Zn/Cu Me
Me
OH Me
+ Me
Me
55 : 45
The rxn involved is the Simmons-Smith rxn. Draw a sensible mechanism (p.1067 of CGW&W should help).
9. Predict the structures of the products in the following rxns, showing their stereochemistry clearly. Explain your answer. H O
(i) LDA (ii) MeI
O (i) Me2CuLi (ii) Allyl Bromide
H OH
mCPBA
H
mCPBA H
Me O H
H2, Pd/C
41
C5: Control in Organic Chemistry
10. Explain the formation of essentially one stereoisomer in the following rxn. O
O
(i) LDA
Me Me
Me (ii) Me
Me
OH
Me Me Me
Me
Me
H O (+)-S
11. Provide the stereostructure of the major product and rationalise the stereochemical outcome of the following rxn. Me Ph
H
+
BF3•OEt2 CH2Cl2, -78 ˚C
? d.r. 16 : 1
OTMS
O
12. Account for the differences in the aldol addition products between the two substrates. Me Me Me
O O
(i) LDA (ii) MeCHO
O Me
Me Me Me
O
O
O
Me OH
80% yield d.r. 4 : 1
Me
Me Me Me
O
(i) LDA (ii) MeCHO
O Me
O
Me Me Me
O
O
i-Pr OH
O
Me
52% yield d.r. 6 : 1
Me
13. Draw a rxn mechanism and explain the diastereoselectivity in the following rxn. O Ph Me
O
(i-Bu)3Al
HO Ph Me
O
(i-Bu)3Al is a Lewis acid, but also serves as the reducing agent (with loss of isobutylene). This is not an easy problem. Try to see what is going on first mechanistically then look at the key steps more closely and consider stereochemical issues. The rxn was published in J. Am. Chem. Soc. 1995, 117, 6395.
42
C5: Control in Organic Chemistry
14. Explain the following selectivities: Me Me2CuLi O
anti : syn 96 : 4 O
Me
Me Me
O
Me2CuLi
O
anti : syn 4 : 96
Me Me
Again this is not an easy problem. What you have to do is firstly identify the rxn we are looking at, and don’t be confused by the way I have drawn the second product. Secondly, find out the preferred conformation of the unsaturated ten-membered rings that minimises strain and transannular interactions (molecular models will definitely help). Draw the conformation clearly in 3-D. Once you have done this the explanation should follow. This work was published in Tetrahedron 1981, 37, 3981.
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