Stereochemistry
Question
Consider cis-1,2-dimethylcyclohexane, shown below in its chair form.
CH3
H cis-1,2-dimethylcyclohexane
CH3
H
i.) Is this conformation chiral?
ii.) Is this molecule chiral? Why 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 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
[! ]oD =
!
lc
[! ]oD =
"2. 88°
= "44.1°
0.0653g / 100mL •1. 00dm
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
H
CH3
H
CHO
OH
As you can tell, these are Fischer projections.
c.)
H
H
Br Br
Br Br
H
b.)
HO
d.)
H
1
Cl
HO
Cl
H
H H
Cl
Cl
Cl Cl
H
e.) H
F
CH2OH HOCH2
H
g.)
H3C
H
HO
OH
H3C
H
H3C
Cl
Answer
a.)
CHO
H3C
OH
H
H
H
Cl H3C
H
f.)
H
F
H
F
HO
H
h.)
CH3
H
CH3
HO
H
F
CH3
H
F
F
CH3
H
CH3
H
CHO
OH
These are enantiomers. You may have noticed
that one diagram can be converted into the other
simply by rotating 90°. You can't do that with
As you can tell, these are Fischer projections. Fischer projections. Remeber what how these
projections are defined.
CHO
H3C
CH3
OH
H
CHO
OH
H
b.)
HO
c.)
CHO
Rotate 90°
H
H
H H
Cl Cl
Cl
e.) H
H
OH
H
Cl One is cis and the other trans-1,3-dichlorocyclobutane.
They are stereoisomers in that the connectivities are
the same. They are obviously not enantiomers, so
H they must be diastereomers.
H I love this question and use it every year. As
F drawn these clearly are mirror images of each
CH2OH HOCH2
F
H
H
H
OH
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! H3C
These are constitutional isomers. In the first molecule,
the double bond is closer to the HO group; in the
second it is closer to the Cl group. The connectivities
Cl are therefore different.
HO
Cl
H
H3C
CHO
HO
H
other. But if you look carefully, there is no
in this molecule. This is obvious
H stereocenter
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, take comfort: I blew it too.
2
f.)
H
F
CH3
HO
H
g.)
CH3 HO
H3C
F
top stereocenter is different, the bottom one
CH3 The
is the same. These are diastereomers.
H
H
CH3
are planar molecules and have no
H These
stereocenters. E/Z isomerism is not possible
Cl H3C
H3C
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.)
CH2CH 3
Br
Br
H
d.)
HO
O
H
Answer
2
a.)
CHO
1
H3C
OH
3
H4
b.)
1
1
H3C
Cl
H
2
Br
Br
R
After prioritizing each end of the double bond, we find both
top priority groups on the same side. This molecule is Z.
CH2CH 3
H R
c.)
The Parity rule works well for Fischer projections.
Recall that the horozontal bonds are coming towards you. This will
give the sequence 1432 as one possibility and 3421 as the other.
This molecule is S.
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.
H
The problem here is distinguishing between the 2nd and 3rd
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
O
3
Question
Assign configurational symbols to the following.
H3 C
CH3
N
H
H3 C
CH2
C
C
H2 C
H
H
O
OH
H
OCH3
C
H3 C
Darvon, a painkiller
OH
H
O
O
H3 C
CH3
H
CH3
Isomenthol: a terpene, whatever
that is.
O
ClCH2 CH2
C
C
COOH
H
C
CH3 CH2
N
H
H2 N
COOH
H
OH
H
CH2 OH
HO
(-)-Serine, one of
H
those nifty amino acids
L-Proline
H Cl
OH
H3 C
H Br
O
H
H
CH2 CH3
CH3
H
H
OH
H
HO
H
O
O
C
H
NHCOCHCl2
F
CH2 OH
Chloramphenicol, an antibiotic
used in the treatment of typhoid.
4
CH3
F
CH=CH-CH=CH2
Answer
H3 C
H
N
CH3
CH2
H3 C
CR
C
S
H2 C
H
O
OH
H
O
Darvon, a painkiller
R S
H
S
R
CH3
Isomenthol: a terpene, whatever
that is.
O
H
CH3
H3 C
O
ClCH2 CH2
COOH
Z
C
C
CH3 CH2
S
N
H
H
L-Proline
H Cl
H3 C
Z
C
H2 N S
COOH
H
O
R
R
H
Z
H
R
CH2 CH3
H Br
R
CH3
H
O
O
H
R
H
S
H
R
OH
CH2 OH
(-)-Serine, one of
HO
those nifty amino acids
H
OH
R
OH
HO
H
OH
OCH3
C
H3 C
H
R
C
R
H
NHCOCHCl2
R
R
F
CH2 OH
Chloramphenicol, an antibiotic
used in the treatment of typhoid.
5
CH3
F
R
CH=CH-CH=CH2
Question
What is the stereochemical relationship between the following pairs of molecules?
i)
ii)
Cl
Br
H
iii)
H
Cl
COOH
OH
H
H
COOH
CH 3
HO
HO
H
v)
H 3C
CH 3
H 3C
H
CH 3
vii)
viii)
O
O
Cl
Cl
H
H
Cl
F
F
H
H
Cl
H
H
Cl
CH 3
Cl
H
H
x)
H
H 3C
CH 3
CH 3
H 3C
H
H
CH 3
CH 3
CH 3
H
CH 3
xi)
H 3CH 2C
H
CH 3
Cl
H
H
Cl
O
H
H 3C
ix)
O
Cl
H
H
CH 2Cl
CH 2CH 3
H 3CH 2C
CH 3
H 3C
CH 2CH 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
H
F
F
H
OH
H
H
H
H
H
vi)
H
CH 2CH 3
CH 3
H
iv)
Cl
CH 3
H
CH 3
CH 3CH 2
Cl
H
Br
iv) enantiomers
vii) enantiomers
H
F
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.)
H3CH 2C
H
c.)
CH3
OH
CH3
H
H3CH 2C
CH3
Br
b.)
OH
H
CH3
H3C
H
Br
H
Br
H
Br
Cl
Cl
H
CH3
H3C
H
CH3
d.)
CH3
H
Cl
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 [α]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
HO
OH
H
CH2OH
CH2
H
OH
H
OH
CH2OH
Is Sorbitol chiral, yes or no? (three marks)
Answer
CH2OH
OH
H
HO
OH
H
CH2
H
OH
H
H
CH2OH HO
H
OH
H
CH2OH
OH
H
or
HO
H
HO
H
OH
H
OH
HO
CH2OH
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
Reduction
H
HOCH2
OH
CH2OH
H
D-(+)-Xylose
OH
OH H
"Xylitol"
CHO
OH
CH2OH
CH2OH
CH2OH
b) Reduction of D-(+)-Xylose (a sugar, [α] = -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.
24
D-(+)-Xylose
"Xylitol"
CHO
H
OH
CH2OH
HO
H
H
OH
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 2n.
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
OH
H3C
Br
H
H
O
H
H3CH 2C
H
H
C
C
H
C
H
CH2CH 3
CH3
Answer
H
1
3
H3C R
H
4
1
OH
Br
H
H
H
S
H3CH 2C
2
3
2
PR: 3124 = 2 = R
H3C
4
O
2
R Br
PR: 4123 = 3 = S
1
H
H
C
H
1C
Z
C
H
2
CH2CH 3
2 CH3
1
H4
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.)
H3C
CH3 H3CH 2C
H3CH 2C
H
c.)
CH3
b.)
OH
H3C
d.)
HCl
CH3
Br
e.)
H
H H
CH3
CH3
H
Br
Br
ClH
Br
Br
HCl
CH3 H3C
CH3
H Cl
H
H Br
Answer
a.)
H3CH 2C
H
H
OH
CH3
b.)
CH3
H
H H
H3C
H3CH 2C
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, [α] 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
You could reason this out. When pure the rotation is 12°.
CH2OH
When 75% pure, the rotation would drop to 75% of 12 which
9°. But since the impurity is its enantiomer, the impurity
R-Glycidol is
will rotate the plane polarized light 25% of -12° or -3°.
H
O
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
C OH
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 OH
HO
Aconitic acid
O C
OH
H2O
O
O
C
HO HO
O C
OH
O
O
C
C
HO HO
OH
O C
Citric Acid
OH
H
H2O
O
C OH
H
H
O
C
HO
H
O C
*
C OH
*
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
CH3 c) O
b) Br
C
H
H3CH 2C
H
T
D
C
OH
d)
CH3
T = 3H
H D = 2H
H = 1H
H3C
Answer
a)
Br R
S Br
H
b) Br
Z
CH3 c) O
C
H
C
H3CH 2C
H
Achiral
Achiral
(meso)
T
D R
Chiral
OH
d)
CH3
T = 3H
H D = 2H
H = 1H
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, [α] 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 ([α] =
+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.
F3C
CH
CH3
OCH 3
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 CH
S CH
3
3
C
OH
C
C
C
H2N
H2N
H
H
O
Racemic
C
H3N
CH3
H
R-Mosher's Acid
(Optically pure)
F3C
OCH3
C
C
O
F3C
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.
measured [! ]Do of mixture
enantiomeric excess =
x100%
[! ]Do for pure enantiomer
[ ! ]Do =
!
l "c
For the following reaction:
O
C
H3C
OH
H
CH2CH3
Chiral reducing agent
C
H3C
HO
+
CH2CH3
H3C
H
C
CH2CH3
For the enantiomerically pure R isomer, ! °D = -13°
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?
4H Cl 1
H Cl
1
1 H C
3
H3C
2
2
H
3
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? Are they
constitutional isomers, the same molecule, enantiomers or diastereomers.
b.) Clearly identify all meso compounds, if any.
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
CHO
OH
H
OH H
OH
HO
H H
CH2OH
CH2OH
(Fischer Projections)
b.)
H
H
H
C C C
C C C Cl
Cl
H
Cl
Cl
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?
H3C
O
O
C
C
1) Na BH 4, CH 3OH
2) H +, H 2O
CH3
Answer
All products will have the same connectivity, but there there are different stereochemical outcomes.
H OH HO H
H3C
HO H
CH3 H3C
H
OH
H OH H
OH
CH3 H3C
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 H2 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 H
H3C
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
Top face
CH3
H3C
H3C
CH3
CH3
Bottom face H3C
H3C
CH3
CH3
CH3
H3C
CH3
H3C
H3C
H
CH3
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 as
shown in the following mechanism. The central intermediate is called a bromonium ion. Due to its
symmetry, attack on either end of this ion is equally likely though only attack on the right end is shown in
the example below.
Br
H
H
Br
H
H
Bromonium ion
Br+
H
H
H
Br-
H
Br
H
H
H
H
Br
This mechanism also applies to substituted olefins. Keeping this in mind, show the important
stereochemical differences in the nature of the products obtained when one reacts bromine with Z-2butene 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
Br
Br+
CH3
H
H3C
H
H
CH3
H3C
CH3
H
H
H3C
Br
H
Br-
H
Br
H3C
Br
H
Br
CH3
H
CH3
H3C
-
Br
CH3
H
H3C
H
H
Br
H
H3C
H
CH3
+
Br
Br
H3C
H
Br
Br
H
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
CH3
+
Br
H
H
H3C
H
CH3
H
H3C
H
H3C
H
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 [α]D the specific rotation. The molar mass of chloropentane is 106 g/mol.
Answer
[! ]Do =
!
l"c
c =
1.0 mol / L x 106 g / mol
= 0.106g / mL
1000mL / L
The specific rotation is 34.3°.
Question
A mixture of diastereomeric butenes, 60% E and 40% Z, is treated with cold, dilute, basic KMnO4.
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
H3C
C
O
O
C
O
Mn
H
CH3
H
H3C
O
O
C
C
H
CH3
HO
OH
H
H
C C
H3C
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
H
O
O
O
C
Mn
CH3
C
H3C
H
H3C
H
H
C
C
O
Mn
O
CH3
O
O
H
H3C
H3C
H
O
O
C
C
C
C
O
O
Mn
O
O
O
21
CH3
H
H
CH3
HO
H
C
H3C R
OH
CH3
C
R H
S
S
C
H
HO
C
H3C
H
CH3
OH