4
Lesson 4
Lesson 4
Molecular Geometry and Isomers II
CH
HO
3
H
H
OH
CH
3
Organic Edge
A. Structural Isomers (Constitutional Isomers)
1. Structural isomers are molecules that share the same molecular formula but differ in the connectivity of
their atoms.
2. Structural isomers have different physical properties (i.e. boiling point, melting point, etc).
Draw three different structural isomers of C3H6O.
B. Stereoisomers
1. Stereoisomers are molecules that share the same molecular formula and connectivity of atoms, but differ
in the orientation of their atoms in space.
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Kaplan
Lesson 4
2. There are two classes of stereoisomers.
a. Conformational stereoisomers
i.
Conformational stereoisomers are generated by rotation about sigma bonds.
ii. Interconversion of conformational stereoisomers usually occurs spontaneously.
What are some examples of conformational stereoisomers?
b. Configurational stereoisomers
i.
Configurational stereoisomers differ in the three dimensional arrangement of their atoms.
ii. Interconversion of configurational stereoisomers requires the breaking of bonds.
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Organic Edge
iii. There are two classes of configurational isomers–enantiomers and diastereomers.
ISOMERS
Structural Isomers
Stereoisomers
Conformational Isomers
Configurational Isomers
Diastereomers
Enantiomers
Figure 4.1
Illustration of different isomer families
C. Enantiomers
1. Enantiomers are molecules that are nonsuperimposable mirror images of each other.
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Lesson 4
a. Notice that molecule B is the mirror image of molecule A.
A
B
mirror
Figure 4.2
Ball and stick models of molecules that are mirror images
b. Molecule B cannot be superimposed on molecule A, regardless of how the molecule is rotated.
Molecules A and B are enantiomers.
B
A
Figure 4.3
Ball and stick models of nonsuperimposable molecules
2. An object that cannot be superimposed on its mirror image is chiral.
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Organic Edge
a. A simple test to rule out chirality is to look for an internal plane of symmetry.
i.
Achiral objects possess an internal plane of symmetry. Objects with an internal plane of symmetry are superimposable on their mirror images.
Figure 4.4
Coffee mug demonstrating achirality by having an internal plane of symmetry
ii. The human hand is chiral. There is no internal plane of symmetry. The right hand is a nonsuperimposable mirror image of the left.
Figure 4.5
Human hands as an example of chiral objects
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Lesson 4
3. When a carbon atom is bonded to four different substituents, it is known as a chiral center (or stereocenter). The methane derivative below is chiral.
Draw the enantiomer
F
F
I
H
I
H
C
Cl
Cl
mirror
Circle the chiral center in the following molecules (if one exists).
1.
2.
OH
H
CH3CH2CCH2CH3
CH3CCO2H
H2N
H
H
4.
O
3.
CH
C
OH
CH3
5.
O
HO
O
OH
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Organic Edge
4. Enantiomers can be generated by interchanging any two substituents of a stereocenter.
A
B
B
A
Figure 4.6
Generating enantiomers by interchanging two substituents
5. The arrangement of substituents about a stereocenter can be represented two dimensionally by Fischer
projections.
Ball and stick model
Dash/wedge diagram
Fischer projection
F
F
F
H
I
I
H
C
Cl
Cl
H
I
Cl
The Fischer "bow tie"
horizontal line
out of page
vertical line
into page
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Lesson 4
D. Absolute Configuration: The R–S system
1. Chlorofluoroiodomethane is an ambiguous name for the molecule below.
Why?
F
H
I
Cl
2. The absolute configuration indicates the spatial arrangement of substituents bonded to a stereocenter.
3. Each enantiomer is given an R or S designation.
Assigning R and S designations:
a. Each of the chiral carbon’s substituents is assigned a priority. Priority is based on atomic number–
the higher the atomic number, the higher the priority.
Assign priorities (1=highest, 4=lowest) to the substituents in the molecule below.
F
H
I
Cl
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117
Organic Edge
b. When the atoms directly attached to a chiral carbon are identical, priority is assigned at the first
point of difference along the substituent groups.
Assign priorities to the substituents of the chiral carbon in 1,2-butanediol.
O
C
C
H
C
O
c. The molecule is rotated so that the substituent of lowest priority (4) is facing away.
3
4
F
4
H
H
I
1
Cl
1
I
F
3
Cl
2
2
Figure 4.7
Orienting stereocenters so the substituent of lowest priority is facing away
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Lesson 4
d. Trace a path from 1 (highest priority) to 3 (third highest priority).
i.
If the direction is clockwise, the enantiomer is designated R.
R
ii. If the direction is counter–clockwise, the enantiomer is designated S.
S
4
H
1
I
F
3
Cl
2
(S)–chlorofluroiodomethane
Figure 4.8
Assigning R and S designations
Determine the absolute configuration for the 1,2-butanediol enantiomer below.
O
C
C
H
C
O
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Organic Edge
4. Absolute configurations can be assigned using Fischer projections.
F
H
F
I
H
I
Cl
Cl
a. Assign priority as usual.
3
F
4
H
I
1
I
1
Cl
2
b. Trace a path from 1–3.
3
F
4
H
Cl
2
c. If the substituent with the lowest priority is in the vertical position of the Fischer projection, then it is
facing away. The R–S designation is assigned as usual.
R
120
S
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Lesson 4
If the substituent with the lowest priority is in the horizontal position of the Fischer projection, then it
is facing us. The R–S designation must be reversed.
3
F
4
H
I
1
Cl
2
R
S
Using a Fischer projection, assign an RS designation for the 1,2-butanediol enantiomer
below.
O
C
C
H
C
O
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Organic Edge
E. Physical Properties of Enantiomers
1. Enantiomers are nonsuperimposable mirror images and share many physical properties (e.g. boiling point,
melting point, density). As a consequence, a mixture of enantiomers cannot be separated (resolved) by
physical means alone.
R
S
S
R
mirror
Figure 4.9
Ball and stick models of an enantiomeric pair.
2. Enantiomers differ, however, in the way they rotate plane-polarized light.
a. Light is an electromagnetic wave.
z
x
y
Electric field
Magnetic Field
c
Figure 4.10
Electromagnetic wave depicting oscillating electric and magnetic fields
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Lesson 4
b. Chiral molecules cause a net rotation of plane-polarized light.
z
x
y
Light waves of
random orientation
Polarizer
(parallel to z axis)
Plane polarized
light
path length (l)
observed rotation
(α)
Sample containing
R or S enantiomer
Figure 4.11
Schematic of a polarimeter (for simplicity, light waves were drawn to indicate only the electric portion of the EM wave)
c. The amount of rotation is measured using a polarimeter. The observed rotation is standardized to a
quantity called specific rotation. The formula for specific rotation is:
αobserved
αspecific = c•l
where αspecific
αobserved
c
l
=
=
=
=
specific rotation
observed rotation
concentration of solution
path length
When plane polarized light is rotated clockwise, the specific rotation is positive (+ α).
When plane polarized light is rotated counter-clockwise, the specific rotation is negative (– α).
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Organic Edge
d. Solutions that rotate plane polarized light (non-zero α) are optically active.
e. Enantiomers rotate plane polarized light with equal magnitude, but in opposite directions.
αR enantiomer = –αS enantiomer
f.
There is NO correlation between a molecule’s absolute configuration and the direction of rotation
(+α or –α) of plane polarized light.
g. A solution that contains equal concentrations of both enantiomers is a racemic mixture. A racemic
mixture is optically inactive (no net optical activity).
αracemate = 0
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Lesson 4
Complete the following table.
Name
Fischer Projection
Boiling Point (°C)
(R)-2-butanol
99.5
Specific Rotation
–13.9°
(S)-2-butanol
F. Compounds with Multiple Stereocenters
1. Molecules with x number of stereocenters have < 2x distinct stereoisomers. Consider 1,2-dichloro-1propanol.
H
H
HO
CH3
Cl
Cl
How many stereoisomers are possible for 1,2-dichloro-1-propanol?
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125
Organic Edge
Draw the enantiomers for the given stereoisomers of 1,2-dichloro-1-propanol.
OH
Cl
H
Cl
H
CH3
2
1
mirror
OH
Cl
H
H
Cl
CH3
4
3
mirror
2. Molecules that are stereoisomers but are not mirror images are diastereomers.
a. Diastereomers have different physical properties, including different specific rotations. Diastereomers
can be seperated by physical means, e.g. precipitation.
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Lesson 4
Characterize the relationship between the following pairs of molecules.
i.
1 and 2 are
_____________________
ii.
1 and 3 are
_____________________
iII. 1 and 4 are
_____________________
iv. 2 and 3 are
_____________________
v.
3 and 4 are
_____________________
vi. 2 and 4 are
_____________________
b. Fischer projections help determine the stereo-specific relationship between two molecules.
i.
Enantiomers have opposite configurations on each chiral carbon.
R
S
R
S
mirror
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127
Organic Edge
ii. Diastereomers can have any other combination of configurations.
R
S
R
R
S
R
Determine the relationship between each of the following pairs of molecules.
1.
CHO
Br
CHO
CH2OH
H
HOH2C
OH
H
CH2OH
Br
H
CH2OH
CH2OH
OH
CH2OH
128
OH
CH2OH
CHO
2.
Br
H
OH
Br
CH2OH
CHO
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Lesson 4
3. Compounds that contain stereocenters, yet have an internal plane of symmetry, are meso compounds.
a. Consider the stereoisomers of 2,3-butanediol.
CH3
H
OH
HO
H
CH3
2
1
mirror
CH3
H
OH
H
OH
CH3
4
3
mirror
Is stereoisomer 1 superimposable on 2?
What type of stereoisomers are 1 & 2?
Is stereoisomer 3 superimposable on 4?
What type of stereoisomers are 3 & 4?
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Organic Edge
b. (2S,3R)-2,3-butanediol is superimposable on its mirror image.
mirror
Figure 4.12
Meso-(2S,3R)-2,3-butanediol is superimposable on its mirror image
c. (2S,3R)-2,3-butanediol has an internal plane of symmetry. One half of the molecule is a mirror
image of the other half.
mirror
Figure 4.13
Meso-(2S,3R)-2,3-butanediol demonstrating an internal plane of symmetry
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Lesson 4
d. Meso compounds are optically inactive.
G. Geometric Isomers
1. Geometric isomers are diastereomers that differ in the arrangement of substituents about a double bond.
2. The terms cis and trans are used to designate the stereochemistry between disubstituted alkene
diastereomers.
H3CHC
H3C
H
H3C
CH3
C
H
CHCH3
C
C
H
H
C
CH3
trans-2-butene
cis-2-butene
3. The (E)–(Z) system is used to designate the stereochemistry of tri– and tetrasubstituted alkene
diastereomers.
R1
R3
C
R2
Kaplan
C
R4
131
Organic Edge
4. Assigning (E) and (Z) designations.
a. Determine the higher priority substituent on each carbon of the double bond.
H3C
OH
C
C
Cl
CH2CH3
b. If the higher priority substituents are on the same side of the double bond, the compound is the (Z)
isomer.
R3
R1
C
C
R2
R4
c. If the higher priority substituents are on opposite sides of the double bond, the compound is the (E)
isomer.
R3
R1
C
R2
132
C
R4
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Lesson 4
Assign (E)–(Z) designations for each of the following compounds.
1.
H
OH
C
F
I
NH3
C
Cl
CH2CH3
H
4.
C
OH
C
Br
C
HO
H
C
H3C
C
H3C
3.
2.
H3CH2C
C
CH2OH
End of Lesson 4
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Organic Edge
Review Problems
1.
(Choose the best answer) Ethanol and dimethyl ether are:
A.
structural isomers
B.
stereoisomers
C.
enantiomers
D.
meso compounds
E.
diastereomers
Solution
Ethanol (CH3CH2OH) and dimethyl ether (CH3OCH3) are molecules that share the same molecular formula
(C2H6O), but differ in the connectivity of their atoms. Hence, they are structural (or constitutional) isomers, and
choice A is the correct answer. Choice B (stereoisomers) is incorrect because although the molecules share
the same molecular formula, they don’t have the same connectivity of atoms. Choice C (enantiomers) is incorrect because the molecules are not nonsuperimposable mirror images of each other. Since the molecules do
not have stereocenters, choice D (meso compounds) is incorrect. Finally, since the molecules are not
stereoisomers, they cannot be diastereomers (choice E).
2.
(Choose the best answer) Cis and trans alkenes are considered to be:
A.
structural isomers
B.
stereoisomers
C.
enantiomers
D.
meso compounds
E.
conformational isomers
Solution
Cis and trans alkenes are molecules that share the same molecular formula and connectivity of their
atoms, but differ in the orientation of their atoms in space. Thus, they are stereoisomers, and choice B is the
correct answer. They are not structural isomers (choice A) because the connectivity of their atoms is identical.
They are not enantiomers (choice C) because they are not nonsuperimposable mirror images of each other.
They are not meso compounds (choice D) because they do not have stereocenters. Finally, they are not conformational isomers (choice E) because they cannot be interconverted by rotation about a sigma bond. (Recall
that the double bond of an alkene is composed of a sigma AND a pi bond.)
What type of stereoisomers are cis and trans alkenes? Since they are not enantiomers, they must be
diastereomers–stereoisomers that are NOT mirror images of each other and are NOT superimposable. More
specifically, cis and trans alkenes are sometimes referred to as geometric isomers, a special class of diastereomers.
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Lesson 4
3.
Of the following pairs of molecules, which are considered to be stereoisomers?
(a)
two constitutional isomers
(b)
two enantiomers
(c)
two diastereomers
(d)
two geometric isomers
Solution
Recall that stereoisomers are molecules that share the same molecular formula and connectivity of atoms,
but differ in the orientation of their atoms in space. Of the pairs of molecules given, enantiomers (b), diastereomers (c) and geometric isomers (d) are stereoisomers. The atoms of constitutional (or structural) isomers (a)
are connected differently; therefore, constitutional isomers are not stereoisomers.
4.
(Choose the best answer) Which of the following is NOT true?
A.
Carbon atoms that are stereocenters are bonded to four different substituents.
B.
2-Chlorobutane contains a stereocenter.
C.
2-Bromopropane contains a stereocenter.
D.
A molecule can contain chiral carbons, yet not rotate plane polarized light.
Solution
Choice C is not true; 2-bromopropane does not contain a stereocenter. Choice B is true because carbon
2 of 2-chlorobutane is a chiral center. Choice D is true because while meso compounds contain chiral centers,
they do NOT rotate plane polarized light because of an internal plane of symmetry.
H
CH3
C
H
CH2CH3
Cl
2-chlorobutane
5.
CH3
C
CH3
Br
2-bromopropane
Draw the structures of the following compounds and determine whether each contains at least one chiral
carbon.
(a)
2-butene
(b)
2-chloropentane
(c)
benzene
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135
Organic Edge
(d)
2-bromocyclopentanol
(e)
3-pentyn-2-ol
Solution
A chiral carbon is a carbon atom bonded to four different substituents.
(a)
CH3CH
CHCH3
2-Butene contains no carbons bonded to four different substituents. It is an achiral molecule.
(b)
H
CH3
Cl
C
CH2CH2CH3
Cl
2-Chloropentane contains one chiral carbon; carbon #2 is attached to four different substituents.
(c)
Benzene contains no chiral carbons.
(d)
OH
Br
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Lesson 4
2-Bromocyclopentanol contains two chiral carbons. Both carbon #1 and carbon #2 are chiral centers.
(Remember IUPAC nomenclature: the –OH functional group has a higher priority than the –Br functional
group. Hence, the carbon with the –OH group is carbon #1.
(e)
OH
CH3
C
C
CH
CH3
3-pentyn-2-ol contains one chiral carbon. Carbon # 2 is attached to four different substituents: a hydroxy
group, a methyl group, a hydrogen, and a propynyl group.
6.
Provide the IUPAC name for the following compound.
CH2CH3
H
CH3
OH
Solution
Since the above compound contains a chiral carbon, it is necessary to use the R/S designation to identify
it correctly as an enantiomer. When determining R/S designation, it is first necessary to rank the substituents
by atomic number. The higher the atomic number, the higher the priority. When the atoms directly attached to
the chiral carbon are identical, priority is assigned at the first point of difference along the substituent group. In
this case, the hydroxyl group has the highest priority (1). Between the methyl and ethyl groups, the ethyl group
has the higher priority (at the point of difference along the two chains, ethyl has a carbon atom and methyl
only has a hydrogen atom). Hence, the ethyl and methyl groups are assigned priories (2) and (3) respectively.
The hydrogen atom has the lowest priority.
Next, rotate the molecule so that the substituent with the lowest priority is facing away, and trace a path
from the highest priority (1) to the third highest priority (3). For this molecule, the direction of the path is clockwise, so the enantiomer is designated R. The correct IUPAC name for the above molecule is (R)-2-butanol.
7.
Determine whether each of the following would influence the rotation of plane-polarized light as the light
passes through an enantiomeric solution.
(a)
the concentration of the enantiomer
(b)
the length of the polarimeter tube
(c)
the ratio of R and S enantiomers present in the solution
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Organic Edge
Solution
observed rotation = (specific rotation) x (concentration) x (length of polarimeter tube)
According to the equation, all of the above would influence the rotation of plane-polarized light passing
through an enantiomer solution. The rotation of plane-polarized light depends on both the concentration of the
substance (a) and the relative ratio of the enantiomers present (c) (recall that enantiomers rotate plane-polarized light in OPPOSITE directions, so one enantiomer can cancel out the other’s optical rotation). Also, the
longer the sample tube (b), the greater the rotation of plane-polarized light.
8.
(Choose the best answer) Which of the following does not show optical activity?
(a)
(R)-2-butanol
(b)
(S)-2-butanol
(c)
A solution containing 1 M (R)-2-butanol and 2 M (S)-2-butanol
(d)
A solution containing 2 M (R)-2-butanol and 2 M (S)-2-butanol
Solution
A racemic mixture of 2-butanol consists of equimolar amounts of (R)-2-butanol and (S)-2-butanol. The (R)2-butanol molecule rotates the plane of polarized light in one direction, and the (S)-2-butanol molecule rotates
it by the same angle but in the opposite direction. In choice D, the concentration of (R)-2-butanol molecules
equals the concentration of (S)-2 butanol molecules. Consequently, exact cancellation of rotation occurs, and
no net rotation of polarized light is observed. Hence, the correct answer is choice D.
Choice A is wrong because all the molecules of the (R)-2-butanol solution rotate the plane of light in the
same direction, so rotations do not cancel and optical activity is observed. In the same way, the (S)-2-butanol
solution also shows optical activity. Thus, choices A and B are incorrect. In choice C, the concentration of (S)2-butanol molecules is higher than (R)-2-butanol molecules. All the rotation produced by the (R)-2-butanol molecules is canceled by half of the (S)-2-butanol molecules; the rotation produced by the other half of (S)-2-molecules contributes to the optical activity observed in this solution. Thus, choice C is incorrect.
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Lesson 4
9.
Cholesterol, shown below, contains how many stereocenters?
CH3
CH2CH2CH2CH
CH3
CH
CH3
CH3
CH3
HO
Solution
Recall that to be a stereocenter, the carbon atom must have four different substituents. There are eight
stereocenters in the molecule, marked below.
CH3
CH3
CH3
CH2CH2CH2CH
CH
CH3
C
C
CH3
C
C
C
C
C
HO
10.
How many chiral carbons does morphine possess?
OH
O
N
H
CH3
H
OH
H
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139
Organic Edge
Solution
A chiral carbon is bonded to four different substituents. Morphine has five such carbons as indicated
below.
OH
O
C
C
H
C
C
N
CH3
H
C
OH
H
Note that the nitrogen atom is also a chiral center. If nitrogen’s lone pair of electrons is considered a fourth
substituent bonded to nitrogen, it can be seen that nitrogen can also act as a chiral center.
11.
(Choose the best answer) Which of the following pairs of compounds constitutes a racemic mixture?
III.
I.
CHO
CHO
HO
H
H
OH
HO
H
H
OH
CHO
II.
CHO
IV.
CHO
H
HO
OH
H
CHO
140
A.
I and III
B.
I and IV
C.
II and III
D.
II and IV
E.
I and II
CHO
HO
H
H
OH
CHO
Kaplan
Lesson 4
Solution
A racemic mixture contains equal quantities of two enantiomers. Compounds I and III are mirror images,
but if you rotate one of them by 180 degrees, you will see that you can superimpose one on the other. This is
because each compound has a plane of symmetry. So even though these molecules contain chiral carbons,
they are optically inactive, and would be called meso compounds, not enantiomers. In fact, compounds I and
III are the same compound, so choice A is wrong.
It is somewhat easier to eliminate choice B, because compounds I and IV are clearly not mirror images at
all. In fact, they are diastereomers. The same is true of compounds II and III in choice C, and compounds I and
II in choice E. The correct answer is choice D. Although compounds II and IV are mirror images, they can’t be
superimposed even if you rotate one of them by 180 degrees.
12.
(Choose the best answer) The following structures are
A.
enantiomers.
B.
diastereomers.
C.
meso compounds.
D.
constitutional isomers.
CH3
CH3
CH3
CH3
Solution
The correct answer is choice A, enantiomers. If you look at the two structures you can see that they are
mirror images of each other. Choice B is incorrect because diastereomers are stereoisomers which are not
mirror images of each other. Choice C is incorrect because in order for a compound to be designated as a
meso compound, it must have a plane of symmetry, which neither of these structures contains. Choice D is
wrong because constitutional or structural isomers are compounds with the same molecular formula but different atomic connections. These compounds do have the same atomic connections. The only difference is that
they differ in their spatial arrangement of atoms.
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13.
(Choose the best answer) Which of the following statements is true about the compounds pictured below?
CH3
CH3
H
Br
Br
H
H
Br
Br
H
CH3
A.
They are enantiomers and are optically active.
B.
They are diastereomers and are optically active.
C.
They are diastereomers and are optically inactive.
D.
They are meso structures and are optically active.
E.
They are meso structures and are optically inactive.
CH3
Solution
The two compounds are identical. Each compound is a meso compound since each has two chiral centers
but nonetheless possesses a plane of symmetry. Meso compounds are optically inactive, and therefore choice
D cannot be true. Choice E is the correct answer.
14.
(Choose the best answer) Which of the following statements is true about the compounds pictured below?
CH3
CH3
H
Cl
Cl
H
H
Cl
Cl
H
CH2CH3
142
CH2CH3
A.
They are diastereomers with different melting points.
B.
They are diastereomers with identical boiling points.
C.
They are enantiomers with different melting points.
D.
They are enantiomers which rotate plane-polarized light in different directions.
E.
They are identical meso structures.
Kaplan
Lesson 4
Solution
The two compounds are non-superimposable mirror images; hence, they are enantiomers. (Note that rotating one compound by 180° does not make it superimposable on the other because the two ends of the molecule are different (i.e. a methyl versus an ethyl group). Compare this to the question above.) Choices A and B
are incorrect because diastereomers are stereoisomers that are NOT mirror images. Choice C is incorrect
because enantiomers have identical physical properties. Choice D is the correct answer. Choice E is incorrect
because meso structures possess a plane of symmetry.
15.
(Choose the best answer) Which of the following compounds is optically inactive?
A.
B.
CH3
CH3
H
Cl
Cl
H
Cl
H
H
Cl
CH3
C.
CH3
D.
CH3
CH2Cl
H
Cl
H
Cl
H
Cl
H
H
CH3
CH3
Solution
The correct answer is choice C.The compound shown in choice C is an example of a meso compound: a
compound that contains chiral centers but is superimposable on its mirror image. It has an internal plane of
symmetry. As a result of this internal plane of symmetry, the molecule is achiral and hence optically inactive.
Choices A and B are enantiomers of each other and will show optical activity. Choice D, since it contains a chiral carbon, is optically active as well.
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143
Organic Edge
16.
Categorize the following pairs as enantiomers, diastereomers, structural isomers, molecules of the same
compound, or different compounds.
(a)
CH3
HO
CH3
Br
and
Br
CH3
OH
CH3
(b)
CH3
H
CH3
Br
and
H
Cl
Cl
Br
CH2CH3
Cl
(c)
HO
Br
HO
Cl
and
HO
CH2CH3
HO
Br
CH2CH3
CH2CH3
Solution
(a)
The two structures are molecules of the same compound. Notice the carbon atoms in both structures
are achiral. If one structure is rotated, it can be superimposed on the other. In order for a compound to
be chiral, at least one of its carbon atoms must be bonded to four different substituents. In both of these
structures, the carbon atoms are attached to two identical groups.
(b)
The two molecules are enantiomers. Both compounds look alike, except the Cl and the Br have been
interchanged. Remember, when two of the four substituents on a chiral carbon are “switched,” the enantiomer is produced. Thus, these two molecules are enantiomers.
(c)
The two molecules are diastereomers. Rotate the second compound in the plane of the paper by 180°.
The two compounds now are:
CH2CH3
CH2CH3
HO
Br
Br
OH
HO
Cl
CH3CH2
OH
CH2CH3
Cl
Assign (R) and (S) designations to the compounds.
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CH2CH3
HO
HO
S
CH2CH3
Br
R
Br
Cl
CH3CH2
CH2CH3
R
OH
R
OH
Cl
One compound is an S, R stereoisomer and the other is an R, R stereoisomer; hence, they are diastereomers.
17.
Assign absolute configurations to the following compounds.
(b)
(a)
O
Cl
H
Cl
CH3
H
Br
C
H
H
OH
CH2OH
Solution
a.
The compound is (R)-1,1-dichloro-2-bromopropane. The top carbon atom (C-1) is not chiral because it
has two Cl atoms bound to it. Only C-2 is chiral, so the molecule must be designated as (R) or (S) with
respect to C-2. Assign priority numbers to the atoms connected directly to C-2. In this compound, Br has
the highest atomic number; it is assigned priority 1. Next come two carbon atoms connected to the chiral
carbon. Since these have the same atomic number, we consider their substituents. C-1 has two Cl atoms
and one hydrogen, whereas C-3 (the carbon atom of the methyl group) has three hydrogens. Since chlorine has a higher atomic number than hydrogen, we assign a higher priority to C-1. Thus, C-1 is
assigned 2 and C-3 is assigned 3. The hydrogen is assigned the lowest priority.
The hydrogen is connected to the chiral carbon via a horizontal bond. According to the conventions of
Fischer projections, it is pointing towards us. If we draw a curved arrow from 1 to 2 to 3, the direction is
counterclockwise. This usually suggests an (S) configuration. However, since the hydrogen needs to be
pointing away from us, the actual absolute configuration of the molecule is the opposite: (R).
b.
The compound has an (R) configuration. Double- or triple-bonded atoms are considered to have the
appropriate number of single bonds to the same atom to which they are multiply-bonded. For instance,
in –C=O, C would be considered to have two single bonds with O, and O would be considered to have
two single bonds with carbon. In this question, –OH is assigned the highest priority 1; –C=O (here considered to be two [–C–O)’s] is assigned the next priority 2; –CH2OH is assigned priority 3; and hydrogen,
with the lowest atomic number, is assigned the lowest priority 4. Again, keeping in mind that as drawn,
the hydrogen atom is pointing towards the front, the counterclockwise direction on going from 1 to 2 to 3
does not indicate the actual configuration of the compound. The actual configuration is the reverse: (R).
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18.
How many different stereoisomers does the following compound possess?
OH OH OH OH
H
C
C
C
C
H
H
H
H
COOH
Solution
The maximum number of configurational stereoisomers possessed by a molecule with no double bond is
2n, where n is the number of stereocenters the molecule has. This formula arises because each stereocenter
can take on one of two configurations: R or S. Two stereocenters thus lead to four possible different combinations: RR, RS, SR, and SS. Three stereocenters lead to eight possible combinations: RRR, RRS, RSR, RSS,
SRR, SRS, SSR, SSS. The number obtained by this formula, however, is a MAXIMUM. The actual number of
different configurational stereoisomers may be reduced from this maximum when meso compounds exist.
Meso compounds possess an internal plane of symmetry and are therefore superimposable on their mirror
images, even though they possess stereocenters. In the case of a molecule with two stereocenters, for example, the RS isomer may be a meso compound, in which case it is identical to (superimposable on) the SR isomer. The number of distinct stereoisomers is then only three: RR, RS (= SR), and SS.
In this case, however, there is no possibility of a meso compound, since no matter how we arrange the
groups about each individual stereocenter, there will not be an internal plane of symmetry. The molecule has
three stereocenters, labeled with asterisks below, and therefore has eight stereoisomers:
OH OH OH OH
H
C
*
C
C
H
H
H
*
C * COOH
H
Notice how each of the three labeled carbon atoms is attached to four different substituents. The leftmost
carbon atom is NOT a stereocenter because it is bonded to two hydrogen atoms. The rightmost carbon atom is
not a stereocenter because it is an sp2 hybridized carbon bonded to only three groups: an oxygen atom (via a
double bond), a hydroxy group, and the carbon chain.
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19.
Identify the following pairs of structures as enantiomers, diastereomers, structural/constitutional isomers,
or conformational isomers.
(a)
H
CH3
C
CH3
CH3CH2CH2OH
and
OH
(b)
CHO
H
CHO
OH
OH
H
and
H
OH
H
OH
CH2OH
CH2OH
(c)
C
and
C
C
C
(d)
CH3
H
C2H5
OH
C2H5
and
H
OH
CH3
Solution
(a)
The two compounds have the same molecular formula: C3H8O. However, the connectivity among the
atoms is different. For example, the OH group in the first compound is attached to the central carbon
atom, whereas the OH group is attached to a terminal carbon atom in the molecule on the right. The two
compounds are constitutional isomers or structural isomers. Note that in this case, both molecules are
alcohols. In general, however, the difference between two constitutional isomers may be more drastic:
They may have different functional groups and therefore belong to different families of organic compounds. CH3CH2OCH3, for example, also has the molecular formula C3H8O, but it is an ether and does
not have a hydroxy group. It is another structural isomer to the two given.
(b)
The two compounds have the same molecular formula. Furthermore, they differ only in the arrangement
of atoms in space. They are therefore configurational stereoisomers. We need to determine if they are
enantiomers or diastereomers. Both compounds have the same two stereocenters. However, the two
Fischer structures shown are not related to each other by a mirror image: They differ only in the configuration about one, not both, of the stereocenters. They are therefore diastereomers.
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Organic Edge
(c)
The two compounds differ only by the rotation about the carbon-carbon single bond. They are therefore
conformational isomers (or conformers). In particular, the structure on the left is in an eclipsed conformation, while the structure on the right is in a staggered conformation.
(d)
The two Fischer projections shown can be interconverted by exchanging the methyl and the ethyl substituents. When two substituents are interchanged in a Fischer projection, we have reversed the configuration about the stereocenter. Hence, the two molecules have opposite configurations about the stereocenter and are enantiomers. The compound on the left is the S isomer while the compound on the right
is the R isomer.
20.
MSG ([S]-monosodium glutamate) is a compound widely used as a flavor enhancer. MSG has the following structure.
H
+Na–OOCH CH
2
2
O
COH
NH2
This enantiomer of MSG has a specific rotation of +24°. In a racemic mixture of MSG, what would be the
specific rotation? What is the specific rotation of (R)-monosodium glutamate?
Solution
The specific rotation of a racemic mixture of MSG, or any other racemic mixture, is zero. A racemic mixture, by definition, is a mixture that contains equal amounts of the (+) and (–) enantiomers, which cancel each
other’s optical rotation. The specific rotation of (R)–monosodium glutamate is –24°. An enantiomer of a compound rotates plane-polarized light by the same amount but in the opposite direction.
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