Stereochemistry Question Consider cis-1,2-dimethylcyclohexane, shown below in its chair form.
CH3 cis-1,2-dimethylcyclohexane
H CH3 H i.) Is this conformation chiral? ii.) Is this molecule chiral? Why or or why not?
Answer Yes, this conformation is chiral because its mirror image is non-superimposable. The molecule is not considered chiral because these mirror image conformations can be converted into each other simply by ring inversion. You can also orient this molecule in the following conformation
H3C
CH3 H
H
mirror plane
. In this conformation it is obvious that this is a meso compound. A good rule to remember is that if a molecule has any has any conformation in which it is achiral, then it is an achiral molecule. Question You dissolve 65.3 mg of camphor in 100. mL of methanol and measure (in a 1.00 dm sample tube) the optical rotation to be -2.88°. What is the specific rotation of camphor? Watch your significant figures, folks. Answer o
[! ]D
=
!
o
[! ]D
lc
=
"2.88° 0.0653g .0653g / 100mL• 1. 00dm 00dm
=
"44.1°
Question For each of the following pairs of molecules, identify the stereochemical relationship. Are they enantiomers, diastereomers, constitutional isomers (isomers with differing connectivity), or the same molecule. Building models of these will be a real help. ALSO IDENTIFY ANY MESO COMPOUNDS AS SUCH. a.)
CHO
H3C
OH
CH3 H
CHO
H
OH
As you can tell, these are Fischer projections.
c.)
H
HO
Cl
HO
Cl
H
d.)
Br Br Br Br H
b.)
H
H Cl
1
H
H
Cl Cl
Cl H
e.) H F
CH2OH HOCH2
H F
HO
H
OH
H
H
H
g.) H3C H3C Answer a.)
Cl H3C
H
H3C
Cl
H CHO
H3C
f.)
H F
CH3
H
HO
CH3
HO
H
h.)
F CH3 H
H
F
F
CH3
H
CH3
OH
H
CHO
These are enantiomers. enantiomers. You may have noticed that one diagram can be converted into the other simply by rotating rotating 90°. You can't do that with with H OH Fischer projections. project ions. Remeber what how these As you can tell, these are Fischer projections. projections are defined.
CHO H3C
CH3
OH
H OH
b.)
c.)
Rotate 90°
CHO
H
HO
CHO
H H H Cl Cl
Cl e.) H F H
trans-1,3-dichlorocyclobutane. Cl One is cis and the other trans-1,3-dichlorocyclobutane. They are stereoisomers in that the connectivities are the same. They are obviously obviously not enantiomers, enantiomers, so H they must be diastereomers. this question and use it every year. As H I love this F drawn these clearly are mirror images of each
CH2OH HOCH2
H
OH
H
These are enantiomers. Note that the other chair conformers of these molecules have both bromines axial, but these conformers are also going to be enantiomers.
Br Br Br Br d.) H
OH
!
H
H
H
OH H3C
These are constitutional constitutional isomers. isomers. In the first first molecule, the double bond is closer to the HO group; in the second it is closer to the the Cl group. The connectivities connectivities Cl are therefore different.
HO
Cl H
H3C
CHO
HO
H
other. But if you look carefully, there is no stereocenter in this molecule. This is obvious H if you draw it another way:
H F
CH2OH
These represent the same molecule. This CH2OH is a wonderful example of how drawings can mislead you. If you got this wrong, and most of you will, will, take comfort: I blew it too.
2
f.)
H F
CH3
HO
H
F The top stereocenter is different, the bottom one CH3 is the same. These are diastereomers.
H
CH3 HO
H
CH3
g.)
H3C
H These are planar molecules and have no
Cl H3C
H3C
stereocenters. E/Z isomerism is not possible since the left end of the double bond has two H3C Cl identical substituents. If you flip one of these over, it will be obvious that these are the same molecule. These are different drawings of the same molecule. F H This molecule lacks stereocenters and has an internal mirror plane of symmetry that runs through the CF bond and chops the ring in half.
H
h.)
H
F
Question Give the configurational symbols ( E/Z, R/S ) for all of the stereocenters in each of the following molecules:
a.)
CHO
H3C
b.) H C 3
Cl
OH H
H
H
c.)
CH2CH3
Br Br
d.)
HO
O H
H
Answer 2
a.)
The Parity rule works well for Fischer projections. 1 Recall that the horozontal bonds are coming towards you. This will OH give the sequence 1432 as one possibility and 3421 as the other. This molecule is S. H4
CHO
H3C 3
b.)
1
1
H3C
Cl
H
After prioritizing each end of the double bond, we find both top priority groups on the same side. This molecule is Z.
CH2CH 3 2
H R
c.)
The best way here is to build a model and figure this out. Note that for this molecule, both symbols had to be the same. If they were opposite, then this would be a meso-compound.
Br Br R
H
The problem here is distinguishing between the 2nd and 3rd O priority groups. I guess the easiest way to explain this is to point out 4 that in one case you follow the sequence H R 2 C-O-C (as you go around the ring) HO 3 and the other way you go C-O-H. Since C has priority over H, the former path has higher priority.
d.)
1
3
Question Assign configurational symbols to the following. H3 C CH 3 N H H3 C
CH2 C C
H2C
H
H
O
OH
H
OCH 3
C
H3 C
H O
O
H
CH3 Isomenthol: a terpene, whatever that is.
CH3
H3 C
Darvon, a painkiller
OH
O ClCH2 CH2 C
C
COOH H
C
CH3 CH2
N H
H
H2 N
COOH
H
OH H
CH2 OH
-Proline
L
(-)-Serine, one of those nifty amino acids
Cl
HO H
CH 2 CH 3
OH H
H3 C
O
H
H
Br
CH3
H H
OH H
HO H
O
O
C
CH 3
H NHCOCHCl2
F F
CH2 OH Chloramphenicol, an antibiotic used in the treatment of typhoid.
4
CH=CH-CH=CH 2
Answer H3 C
N
H
CH3
CH2
H3 C
CR C S
H2 C
H
H
R
O
OH H
Darvon, a painkiller
H
R S O
S
OCH 3
C
H3 C
R
O
H
CH 3 Isomenthol: a terpene, whatever that is.
CH3
H3 C
O ClCH2 CH2
COOH
Z
C
C
CH3 CH2
N H
Z
H
H2 N
COOH
H
-Proline
L
S
OH H
CH2 OH
R
(-)-Serine, one of those nifty amino acids
Cl
H H3 C
C
S
OH
HO
CH 2 CH3
H
OH O
R
H
H
R
R
S
H
H
Br
R
Z CH3 H
OH H R
HO H
R
O
O
C R
H
R R
NHCOCHCl2
F F
CH2 OH
Chloramphenicol, an antibiotic used in the treatment of typhoid.
5
CH3
R
CH=CH-CH=CH 2
Question What is the stereochemical relationship between the following pairs of molecules? i) ii) Br Cl
Br
H Cl
H
H
CH 3CH 2
iii)
CH 2CH 3
H
CH 3
CH 3
iv) Cl
CH 3 H
Cl
OH
H
HO
H
H
CH 3
COOH
COOH
HO
H
v)
H
OH
vi) H
H
H
H
H
H 3C
CH 3
H 3C
F
F
H
CH 3
H
H
vii)
F F
H
F
viii)
O
CH 3
O O
Cl
CH 3 H
Cl
Cl
H
H
O
H
H
H 3C
CH 3
H
H Cl
ix)
H
x) Cl
Cl
Cl H
Cl
CH 3
CH 3
CH 3 H
H
H
CH 3
H
H 3C H
H
Cl
H H
CH 2Cl
xi) H 3CH 2C
H 3C
CH 3
CH 2CH 3
H 3C
CH 2CH 3
H 3CH 2C
CH 3
Answer i) identical molecules ii) enantiomers iii) enantiomers v) same molecule (a meso compound) vi) diastereomers viii) rotational isomers of the same molecule ix) enantiomers
6
H
iv) enantiomers vii) enantiomers
F
x) constitutional isomers xi) identical molecules Question (7 marks) Examine the following pairs of molecules. a.) What is the stereochemical relationship between the following pairs of molecules? Are they constitutional isomers, the same molecule (note: changing the conformation of a molecule does not make it a different molecule), enantiomers or diastereomers. b.) Clearly identify all meso compounds.
a.) H3CH2C H c.)
CH3
OH
CH3
H H3CH2C
CH3
Br
b.) OH
H
CH3
H3C
H
Br H
Br
H
Br Cl
Cl
H
CH3
H3C
H CH3
d.) H Cl
CH3
CH3 Br
e.)
H
Cl
H H
Br
Answer a) same molecule b) enantiomers c) diastereomers, the left hand molecule is a meso compound d) same molecule, a meso compound e) same molecule Question The isolation of Cordycepic acid from a fungus was reported in the Journal of the American Pharmaceutical Association, 46, 114, (1957). It was found to be optically active with [ !]26 = +40.3°. It was assigned the structure shown below in which all four hydroxyl groups are equatorial. HO
HO HO
O OH OH
Should you believe everything you read? Elaborate making reference to the above example.
7
Answer This compound has several (two) chiral centers but also has an internal mirror plane of symmetry. It is achiral and cannot possibly be optically active.
HO
HOCO2H
O OH
HO
OH
HO
HO
OH OH
Question 1.) Convert the following structure of Sorbitol, a sugar derivative and artificial sweetener, into its corresponding Fischer diagram. CH2OH OH H
OH H
HO
CH2OH CH2 H
H OH
OH CH2OH
Is Sorbitol chiral, yes or no? (three marks) Answer
CH2OH OH H HO
OH H
CH2 H
OH
H
H
CH2OH
OH
H
HO
H
CH2OH HO H
OH
H
H
OH
HO
OH
H
HO
CH2OH
or
OH H CH2OH
These are the same: take one and rotate by 180°. Yes, it is chiral.
8
Question a) Complete the Fischer projection diagrams for the two compounds shown below.
H HOCH2
OH
O H
H
OH
H
H
Reduction
HOCH2
OH
CH2OH H
D-(+)-Xylose
OH
OH
H
OH
"Xylitol"
CHO
CH2OH
CH2OH
CH2OH
b) Reduction of D-(+)-Xylose (a sugar, [!]24 = -9.3°) gives Xylitol (an artificial sweetener) which has no measurable optical activity. Why not? The product belongs to a special stereochemical class. What might that be? Answer Xylitol has an internal mirror plane of symmetry that xylose does not. Xylitol is a meso compound and therefore lacks any optical activity.
D-(+)-Xylose
"Xylitol"
CHO H
OH
HO
H
H
OH CH2OH
CH2OH H HO H
OH H OH CH2OH
Question Five (3 marks) a/) When looking at the structure of a molecule, how does one determine the maximum number of possible stereoisomers? b.) When is the actual number of stereoisomers less than that predicted by the procedure outlined in (a)? Answer a) One counts all of the chiral centers and double bonds that are E or Z. If we let n = the sum of these two numbers, then the maximum number of stereoisomers is 2 n. b) If some of the stereoisomers are meso compounds, then this will reduce the total number of possible stereoisomers. Question Six (4 marks) a.) Examine the molecule below (drawn in 2-D). Is this molecule really flat? If not, draw it in three dimensions using dash/wedge bonds to present a more accurate descriptiton. Building a model is highly recommended. i.) a primary (1°) H atom b.) Label (single examples will suffice; you ii.) an sp2 hybridized C atom need not label all examples): iii.) a vinyl H atom
9
iv.) an allylic C atom c.) Is this molecule chiral?
H
CH3
H3C
H
Answer
vinyl H
1° H atoms
sp2 C
H
CH3
H3C
H allylic C
c) This molecule is not superimposable on its mirror image, therefore it is chiral (even though it has no chiral centers - remember chirality is a property of objects). Question Assign configurational symbols (R/S or E/Z) to the following molecules but do not name them. H H
OH H3C
Br
H
H
C
O H
C
H
C
H3CH 2C
H
CH2CH 3
H CH3
Answer H 1
OH
3
H3C R
1
Br
H
H
4
4
2
H H3CH 2C
PR: 3124 = 2 = R
1
C
S
3
2
H3C
H O
PR: 4123 = 3 = S
C 1 C
Z
H
H 2
CH2CH 3
H 2 CH3
1
2 R
H
Br
H 4 3
PR 4321 = 6 = R
PR means using the parity rule. You may have gotten a different sequence BUT the first number should have been the same.
10
Question Three (7 marks) Examine the following pairs of molecules. a.) What is the stereochemical relationship between the following pairs of molecules? Are they constitutional isomers, the same molecule (note: changing the conformation of a molecule does not make it a different molecule), enantiomers or diastereomers. b.) Clearly identify all meso compounds.
H
OH
a.)
b.)
H
H H H
H3C CH3 H3CH 2C
H3CH 2C H c.)
CH3
Br Br
OH
CH3
H3C
Br Br
Cl H
HCl
CH3
d.) CH3
H Cl
CH3 Br
e.)
CH3 H3C
H Cl
H H Br
Answer
a.) H3CH 2C H
H
OH CH3
b.)
H3C H3CH 2C
CH3
H H H
Br Br Br Br These are mirror images of the same molecule, a meso compound.
OH
Enantiomers c.)
H
CH3
H3C
Cl H
Cl H
CH3
d.) CH3
H Cl CH3 H3C Cl H Diastereomers. The molecule on the right is a meso compound.
CH3
Enantiomers e.)
Br
H H
Br
These are conformational isomers of the same achiral molecule. This molecule has an internal mirror plane of symmetry, but that does not make it a meso compound.
11
Question Six (2 marks) Optically pure R-Glycidol has a specific rotation, [ !] of +12° (neat, i.e. without solvent). What would be the measured rotation of a sample of R-glycidol that is contaminated by its enantiomer such that 25% of the sample is S -glycidol? Answer
CH2OH R-Glycidol
H
O
You could reason this out. When pure the rotation is 12°. When 75% pure, the rotation would drop to 75% of 12 which is 9°. But since the impurity is its enantiomer , the impurity will rotate the plane polarized light 25% of -12° or -3°. 9°-3°= 6°, the correct answer. Another way to reason this out: 50% of each is racemic (0°). 100% is +12. Half way between the two must be 6°.
Question Ten (3 marks) The enzyme aconitase catalyzes the hydration of aconitic acid to two products, citric acid (which is achiral) and isocitric acid (which is chiral). O
O C
OH
C
HO
Aconitic acid O
C
H
OH
The respective products result from the Markovnikov and anti-Markovnikov addition of water to the carbon-carbon double bond. Identify the structures of citric and isocitric acid. Answer O
O C
C
HO
Aconitic acid O
C
O C HO HO O
O
O C
OH
C
HO HO
C OH
O C
OH
H2O
O
O
C
H
OH
H2O
O
OH
HO
H
H C
Citric Acid
OH
H
O
OH
C * C
*
OH H
OH Isocitric Acid
There are two possible regioisomeric structures that result from the addition of water to the double bond in aconitic acid. They can be distinguished by the fact that the one on the left (citric acid) has no assymetric atoms and is therefore achiral. The other (isocitric acid) has two assymetric centers and is chiral. You cannot predict the absolute configuration of the two assymetric centers.
12
Question Three (seven marks) For each of the following: i) assign group priorities and configurational symbols (R or S, E or Z) to all stereocenters where possible and ii) indicate whether the molecule is chiral or achiral. Part (c) is a Fischer projection of C-13 labeled glycerol.
a)
Br
Br
H
b) Br
CH3
c) O
d)
OH
CH3
C
H H3CH 2C
H
T D
C
T = 3H H D = 2H H = 1H H3C
Answer
a)
Br R
S Br
H
b) Br
Z
CH3
c) O
d)
OH
CH3
C
H H3CH2C
Achiral (meso)
C T D R
H
Achiral
T = 3H 2 H D= H H = 1H
Chiral H3C Chiral
Question Five (eight marks) Examine each of the following pairs of molecules and indicate whether they are constitutional isomers, enantiomers, diastereomers or identical.
i)
CH2OH
CH2OH
H
OH
HO
H
H
OH
HO
H
CH2OH
Fischer ProjectionS
iii)
F
F
ii)
CH2OH
F
F
HO
CH3
H3C
OH
iv) H3C
H C
C
Br Answer i) enantiomers ii) enantiomers iii) diastereomers iv) constitutional isomers
13
H3C
F C
F
H
C Br
Question Six (two marks) Optically pure S-Glycidol has a specific rotation, [ !] of -12° (neat, i.e. without solvent). What would be the measured rotation of a sample of S -glycidol that is contaminated by 25% of R-Pinene ([ !] = +50.7° (neat))? Explain your answer.
H
CH3
CH2OH
O
(S)-(- )- Glycidol
(R)-(+)-Pinene Answer The net rotation is: (0.75 x -12°) + (0.25x50.7°) = 3.7° Question Nine (eight marks) i) Using the following molecules, explain how classical resolution (the separation of enantiomers) works.
F 3C CH
CH3
OCH3 C
OH C
NH2
O R-Mosher's Acid (Optically pure)
Racemic
ii) Why is the preparation of optically pure compounds important? Answer i) The racemic mixture of the amine consists of molecules that are R or S. When mixed with the optically pure R-Mosher's acid one obtains a 1:1 mixture of salts: F3C OCH3 R S CH3 CH3 C OH C C C H2N
H2N
H
H
O
Racemic
C H3N
CH 3 H
R-Mosher's Acid (Optically pure)
F3C
OCH3 C
C
F3C
O
O
CH3
C H3N
H SS salt
RR salt
14
OCH3 C
C O
O
The RR and SR salts are diastereomeric and can therefore be separated (in principal) by crystallization etc. ii) Phthalidomide is an excellent example of the importance of separating enantiomeric compounds. Often, one enantiomer has a desireable biological activity and the other has an undesireable or unknown effect. Question Three (four marks) For reactions that are stereoselective (i.e. they form an asymmetric center but not a racemic mixture), the degree of selectivity is usually reported in terms of enantiomeric excess. D
enantiomeric excess =
measured [! ]o of mixture D
[! ]o for pure enantiomer
!
D
[ ! ]
100%
x
o
=
l "c
For the following reaction:
O C
H3C
OH
H CH2CH3
Chiral reducing agent
For the enantiomerically pure R isomer,
C
H3C !
HO +
CH2CH3
H3C
H C CH2CH3
° = -13° D
Given that the measured specific rotation of the mixture is -6.5°: a) What is the enantiomeric excess? b) What is the % of the mixture that is R and what is the % of the mixture that is S? c) What would be the measured specific rotation of the reaction mixture if an achiral reducing agent had been used? Explain. Answer a) 50%.
b) 75% is R and 25% is S. Recall that the S isomer rotates plane polarized light in the opposite direction (+3.25° in this case). The 75% R rotates plane polarized light -9.75°. The sum of the rotations is -6.5°. c) An achiral reducing agent would give a racemic mixture: equal proportions of the R and S products. It would not have any net rotation of plane polarized light. Question Two (3 marks) a) Label, where appropriate, all of the stereocenters/double bonds in the following molecules with the correct configurational symbol. Write down your priority assignments if you want part marks. b) How many possible stereoisomers are there for this molecule? H 1
Cl
4H
Cl 1
1
H3C
H3C 3 2
2
H
H
2
The stereocenter is R.
The double bond is Z.
15
To get the number of possible stereoisomers you must find the number of double bonds capable of being E or Z and the total number of asymmetric atoms capable of being R or S. No. stereoisomers = 2n: four in this case.
16
Question Four (five marks) Examine the following pairs of molecules. a.) What is the stereochemical relationship between the following pairs of molecules? constitutional isomers, the same molecule, enantiomers or diastereomers. b.) Clearly identify all meso compounds, if any.
Are they
Answer a: In these two molecules, one asymmetric center is the same and one is different. These are diastereomers. b: These molecules lack asymmetric carbon atoms, but then, so do golf clubs. These two molecules are non-superimposable mirror images of one another so they must be enantiomers.
a.) CHO OH H H H
H HO
b.)
CHO
H H H Cl C C C C C C Cl H Cl Cl
OH OH CH2OH
CH2OH
(Fischer Projections) Question a.) Give the structures of all possible stereoisomeric products of the following reaction. b.) What is the stereochemical relationship between all of the products? c.) Are the products formed in equal proportions? O
1) NaBH 4, CH 3OH
O
C
+
2) H , H 2O
C
H3C
CH3
Answer All products will have the same connectivity, but there there are different stereochemical outcomes.
H OH HO H3C
H
HO H CH3 H3C
H
OH
H OH
CH3 H3C
H
OH CH3
A meso-diastereomer Enantiomers, formed in exactly equal amounts. Three different stereoisomers are formed: a racemic mixture of enantiomeric compounds and a meso diastereomer. The amounts of the two enantiomers will be equal to each other, but the meso-compound can, and will likely be formed in some differing amount. Question What is/are the product(s) produced upon the catalytic hydrogenation of Z-3,4-dimethyl-3-hexene? Compare them to the product(s) formed when E-3,4-dimethyl-3-hexene is reduced under the same conditions. H3C
H3C
CH3
CH3
17
Answer Hydrogenation is a syn-addition and H 2 is equally likely to add to the top or bottom face of the double bond. The resulting products are enantiomers of each other and are formed in equal amounts - in other words the reaction gives a racemic mixture.
H
H3C
H
H3C
H
Top face
CH3
H3C H3C
CH3 H3C
CH3 Bottom face
H3C
H
H
H CH3 CH3
H3C H3C
CH3
H
CH3
CH3
H
If the E isomer is hydrogenated, a single product is produced - a meso compound.
H3C
H
H
H CH3
H3C
Top face
H3C H3C
CH3 CH3
Bottom face H3C H3C
CH3 CH3
CH3
H3C
CH3
H3C
CH3 H
H
H
H
H
H
H
H3C H3C
CH3 CH3
Internal mirror plane of symmetry. Question Ethylene and many other alkenes react with bromine in an ionic reaction to form vicinal dibromides shown in the following mechanism. The central intermediate is called a bromonium ion. Due to symmetry, attack on either end of this ion is equally likely though only attack on the right end is shown the example below. Br
Br
Bromonium ion Br
+
Br H
H
H
H H
H H
H
H
H
H
H
Br
-
Br
This mechanism also applies to substituted olefins. Keeping this in mind, show the import stereochemical differences in the nature of the products obtained when one reacts bromine with Z butene and (separately) with E -2-butene to get 2,3-dibromobutane.
18
Answer This is quite complicated due to the number of possibilities. Bromine can add to either side of the molecule to form a bromonium ion. The two possible bromonium ions are enantiomers of each other. Each bromonium ion can undergo attack by bromide to either carbon (the two possibilities are represented by solid and dashed arrows) with equal likelihood. So, there are four possible paths in this case. Interestingly, they all give exactly the same product, a meso-compound. Br Bromonium ion
Br
CH3
Br
H +
Br CH3
H H3C
H
H
CH3
H3C
H H3C
Br
H -
Br
H
Br
H3C CH3
Br
H Br
H CH3
H3C
-
Br
CH3
H H3C
H
H
Br
H
H3C H
CH3 +
Br
H3C
Br
H H
Br Br
CH3
Br
Cis-2-butene gives an identical bromonium ion no matter which side the bromine attacks. Subsequent attack by bromide can, again, take two possible paths. In this case, however, one obtains a racemic mixture of dibromides (diastereomeric to the dibromide formed from the trans-2-butene.
19
Br Bromonium ion
Br
Br Br H
H H3C
CH3
+
H
CH3
H
H H3C
H
H3C
R,R
Br
CH3
BrHas an internal plane of symmetry.
H
Br
H3C Br
S,S
H CH3
Question A 1.00 M solution of 2-chloropentane in chloroform in a 10 cm cell gives an observed a of +3.64°. Calculate [ !]D the specific rotation. The molar mass of chloropentane is 106 g/mol. Answer
[! ]
!
D o
=
c =
l"c
1.0
mol / L x 106 g / mol 1000 mL /
=
L
0.106 g /
mL
The specific rotation is 34.3°. Question A mixture of diastereomeric butenes, 60% E and 40% Z, is treated with cold, dilute, basic KMnO 4. Knowing what you do about the stereochemistry of the formation of 1,2-diols from alkenes, a.) What are the products produced and in what proportions? b.) What is the stereoisomeric relationship between all of the products? (6 marks) H3C
H3C
H
H
CH3
H
CH3 H
Z-2-Butene
E-2-Butene
Answer This reaction results in the syn-addition of two hydroxyl groups to the double bond of the alkene. Addition to either face of Z-2-butene gives a meso-diol. Since the meso-diol is derived from the Zisomer, it constitutes 40% of the final mixture. O
O Mn
O
H C
O Mn
H H3C
O
O
C CH3
O
O
H
H C
C
H3C
CH3
HO H C H3C
OH H C CH3
meso-diol
In contrast, addition to the top face of E-2-butene gives one enantiomeric product, while addition to the bottom face gives the opposite enantiomer. The two enantiomers will be formed in equal amounts and each will constitute 30% of the entire mixture of three stereoisomeric products. 20
O
O Mn
O
O
O
O Mn
H C
CH3
C
O
O
H C
C
CH3
H3C
H
H3C
H
H3C
H
H3C
H
C
C
C
H
CH3
C
H O
O Mn
O
O Mn
O
O
O
O
21
CH3
HO H C R H3C H3C
OH CH3 C R H
S
S
C
C
H HO
H
CH3 OH
