1
Lecture 16
Problem 1 - Draw a 3D structure and its mirror image for each of the following molecules. Are they
different (enantiomers) or identical (superimposable)? Build models of each and see if your pencil and
paper analysis is correct. See if you can use your hands to help your analysis.
a. 1-bromopentane
b. 2-bromopentane
c. 3-bromopentane
d. 1,1-dibromocyclopentane
e. cis-1,2-dibromocyclopentane
f. trans-1,2-dibromocyclopentane
Problem 2 - Which molecules below have stereogenic centers? How many? Are they all chiral centers?
a.
d.
c.
b.
e.
OH
Cl
Br
Br
f.
Br
Cl
g.
Br
h.
Br
j.
i.
Br
Br
Br
RandSNomemclature
Problem 3 - Classify the absolute configuration of all chiral centers as R or S in the molecules below.
Use hands (or model atoms) to help you see these configurations whenever the low priority group is
facing towards you (the wrong way). Find the chiral centers, assign the priorities and make your
assignments.
a.
c.
b.
Cl
H
H3C
H
H
d.
CH3
H
H3C
H
Br
Cl
C
HO
C
I
H
e.
H
f.
O
H
h.
Br
H
Br
OH
H
H
C
CH3
C
H
CH3
H
CH3
H
H3 C
H
H
H3C
Cl
i.
j.
l.
k.
Cl
CH2CH3
Br
H3CH2C
H
C
H3C
g.
H3C
H3C
CH3
H
H
C
Br
CH3
H
CH3
H
Cl
S
D
CH3
H
H
CH3
CH3
O
2
Lecture 16
Pi Bond Priority
Problem 4 - Evaluate the order of priority in each part.
a.
H
* = path to chiral center
H
C
*
C
C
H
*
C
C
*
*
CH2
C
H
ethynyl
H
C
C
CH3
CH3
C
C
CH3
CH3
H
phenyl
t-butyl
2-propenyl
b.
CH3
O
*
H2
C
C
H2
C
*
H2
C
F
H2
C
*
C
N
*
H
H2
C
C
I
OH
c.
H
H
O
*
C
H
H
O
*
C
CH3
O
O
*
H
C
O
O
*
CH3
H
C
H
O
CH3
d.
*
*
*
*
Br
Cl
How to draw R and S absolute configurations from a name
Occasionally, you will have to draw absolute configurations from a name. The following strategy should
prove helpful.
1. Write out a two dimensional structure from the name.
2. Locate all chiral centers (4 different groups at sp3 atoms) and assign the priorities of the groups at each
chiral atom.
3. Draw a generic tetrahedral center and place the low priority group away.
C
4
4. Fill in any convenient (obvious) group, …say 1.
1
C
4
3
Lecture 16
5. Add in the other two groups in specified order (R would have 2 to the right and S would have 2 to the
left.)
1
1
R
C
4
C
4
2
S
2
6. If a second stereogenic center (3rd, 4th...etc.) exists, add in groups in a similar way. With one group
fixed, it may be difficult to follow the above procedure. In such cases, you can randomly put in groups 2
and 3 and then do a quick check of absolute configuration. This should be easy because you placed “4”
away from you in the first step (or you can use your arms and hands if “4” is towards you). If the
assignment is correct, you are finished. If it is incorrect, then interchange any two convenient groups and
it will be correct.
Problem 6 - Write a three dimensional structure for each of the following names.
a. (4R,6S)-4,6-dimethylnonane
b. (1R,3R)-3-methyl-1-cyclohexanol
c. (4R)-4-phenyl-5-methyl-1-hexyne
d. (2R)-2,3-dihydroxypropanal
e. (3S)-4-methyl-3-bromo-1-pentene
f. methyl (2R)-2-aminopropanoate
g. (1R,3S)-1,3-cyclohexanediol
h. (1R)-1-fluoro-1-chloropropane
i. (2S)-2-amino-2-phenylethanoic acid
j. (2R)-2-isopropylcyclohexanone
k. (2R,3R,4S,5R)-2,3,4,5-tetrahydroxyhexane
l. (1R,2S)-1-chloro-2-ethenylcyclopentane
Molecules with more than one chiral center and Fischer Projections
Basic Rules of Fischer Projections
1. Place the longest chain in vertical direction
2. Put the highest priority (nomenclature priority) in the top half of your representation.
3. Horizontal groups project toward the front (in front of the page/surface)
4. Vertical groups project away from the viewer (in back of the page/surface)
5. A carbon atom is indicated at each intersection of vertical and horizontal lines.
Problem 7 - Place glyceraldehyde (2,3-dihydroxypropanal) in the proper orientation to generate a Fischer
projection. Can you find any stereogenic centers? Is the molecule as a whole chiral? If so, draw the
enantiomer and classify all stereogenic centers as R or S.
4
Lecture 16
Problem 8 - Draw a 3D Newman projection and a sawhorse representation for each the following Fischer
projections. Redraw each structure in a sawhorse projection of a stable conformation. Identify
stereogenic atoms as R or S.
a.
b.
CH3
c.
NH2
HO
H
H3C
H
H
H
H
CH3
OH
Cl
Br
H3C
H
H
H
H3C
H
D
CH3
d.
F
CH3
OH
Problem 9 - Determine how many switches it takes to make the second stereogenic center appear identical
to the first and identify whether the configurations at the two stereogenic centers are identical or opposite.
Specify the absolute configurations as R or S. (There is another way to compare absolute configurations.
(See the next paragraph.)
a
H
c
b
D
OH
CO2H
CH3
H
CH3
CO2H
HO
CH3CH2O
CH3CH 2O
H
C
OCH3
CH3
H
OCH3
CH 2CH3
O
H
C
CH3
NH2
D
O
CH3
H
H
d
CH2CH 3
NH2
CH3
H
Problem 10 - Decide which are identical and which are enantiomers using your hands and arms. Specify
absolute configuration as R or S. First assign the correct priorities.
b.
a.
H 3C
H
CH3
H 2C
CH
CH3
C
CH
CH
C
CH
CH
CH3
H 3C
C
CH2CH 3
H 3C
H 3C
CH
CH2
CH
CH
C
C
C
C
CH3
OH
H 3C
OH
CH2CH 3
CH3
C
H
d.
c.
CH3
H
C
H
H
O
C
H
HO
H
O
C
H
C
C
H3C
N
OH
H
CH3
C
N
H3C
H
5
Lecture 16
Problem 11 – For a single chiral center, Fischer projections seem almost more trouble than they are
worth. There are 12 different representations of enantiomeric pairs having a single chiral center. Use the
first Fischer projection drawn as a reference structure and compare the 24 different representations drawn
to determine if they are identical or the enantiomer.
1
2
3
2
4
1
3
3
1
2
3
3
4
4
1
1
2
2
4
3
3
4
2
1
4
4
3
2
4
3
Reference
Structure
2
1
4
2
4
3
3
1
3
1
1
4
4
4
4
4
3
2
3
3
4
1
3
2
1
1
4
1
1
3
2
2
1
1
2
2
2
2
1
1
2
2
3
3
4
4
4
2
2
4
4
3
3
3
1
1
4
4
2
2
1
3
4
1
1
2
2
3
1
3
Problem 12 - Rearrange the Fischer projections below to their most acceptable form. You may have to
rotate the top and/or bottom atom(s). You also may have to twist a molecule around 180o in the plane of
the paper. Assign the absolute configurations of all stereogenic centers. Using your arm and fingers is
helpful here. Write the name of the first structure.
a.
b.
OH
H
CH3
H3C
c.
Br
CH3CH2
H
H3C
H
Br
HO
H3C
H
H
OH
H
Br
d.
H
HO
OH
HO
CHO
H
HO
H
OH
HO
H
CHO
H
H
H
OH
H
OH
H
Three Stereogenic Atoms
*H
*H
*H
Br
Cl
Cl
H3C
C
C
C
CH3
Three stereogenic centers with
23 = 8 possible stereoisomers.
Problem 13 - For each of the following structures, draw only the stereoisomer Fischer projection(s) with
all chlorines on one side. Is there a potentially stereogenic atom at the center of 2,3,4,5tetrachlorohexane? What about 2,3,4,5,6-pentachloroheptane? If so draw the two stereoisomers that
result from a switch only at that atom. Are the stereoisomers chiral or achiral? Are they meso,
enantiomers or diastereomers?
6
Lecture 16
Problem 14 - For the following set of Fischer projections answer each of the questions below by writing
the appropriate letter(s) or letter combination(s). Hint: Redraw the Fischer projections with the longest
carbon chain in the vertical direction.
CH3
OH
H
HO
H
CH3
CH3
H
OH
H
OH
H
H
H
OH
H
OH
HO
H
H
H3C
OH
H
OH
HO
H
HO
OH
a.
b.
c.
d.
e.
f.
g.
CH3
CH3
CH3
CH3
CH3
A
B
C
OH
HO
H
OH
CH3
CH3
H
E
D
Which are optically active?
Which are meso?
Which pairs are enantiomers?
Which pairs are identical?
Which pairs are diastereomers?
Which, when mixed as a 50/50 mixture, will not rotate plane polarize light (optically inactive)?
Draw any stereoisomers of 2,3,4-pentanetriol as Fischer projections, which are not shown above.
How does Nature do it?
Three carbon aldose carbohydrates
O
H
O
H
C
C
H
OH
HO
H
CH2OH
D-glyceraldehyde
CH2OH
L-glyceraldehyde
Four carbon aldose carbohydrates
O
H
O
H
C
O
C
OH
HO
H
HO
H
OH
HO
H
H
CH2OH
C
H
OH
CH2OH
H
OH
HO
CH2OH
L-erythrose
O
H
C
H
D-erythrose
H
H
CH2OH
D-threose
L-threose
Five carbon aldose carbohydrates
O
H
O
H
C
O
C
H
O
H
C
O
C
H
OH
HO
H
HO
H
OH
HO
H
H
OH
HO
H
HO
H
OH
HO
H
H
OH
HO
H
H
CH2OH
CH2OH
D-ribose
L-ribose
H
CH2OH
D-arabinose
H
H
OH
CH2OH
L-arabinose
H
O
C
OH
H
OH
CH2OH
D-xylose
O
H
C
HO
H
HO
H
H
O
H
C
C
HO
H
H
OH
OH HO
H
H
OH
H
OH
H
HO
H
CH2OH
CH2OH
CH2OH
L-xylose
D-lyxose
L-lyxose
7
Lecture 16
Six carbon aldose carbohydrates
O
H
O
H
C
O
C
H
O
H
C
O
C
H
O
H
C
O
C
OH
HO
H
HO
H
OH
HO
H
H
OH
HO
H
HO
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
HO
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
H
CH2OH
CH2OH
D-allose
O
H
O
C
H
H
O
H
C
OH
H
HO
H
H
OH
HO
H
H
OH
H
H
H
OH
HO
HO
H
HO
H
OH
HO
H
H
CH2OH
D-mannose
CH2OH
L-mannose
OH
HO
HO
CH2OH
H
H
H
OH HO
H
OH
H
OH HO
H
CH2OH
L-glucose
H
HO
OH
H
H
CH2OH
OH
H
HO
H
HO
CH2OH
L-idose
OH
OH HO
O
H
CH2OH
L-gluose
H
O
H
C
C
OH HO
H
H
OH
H
HO
H
H
OH
OH HO
H
H
OH
H
OH HO
CH2OH
D-idose
H
CH2OH
C
H
H
D-gluose
O
H
C
H
CH2OH
L-galactose
D-galactose
H
C
H
OH
HO
D-glucose
C
H
OH
H
O
HO
H
OH
CH2OH
L-altrose
D-altrose
O
H
OH
CH2OH
L-allose
C
H
O
H
C
H
H
H
H
CH2OH
D-talose
H
CH2OH
L-talose
Problem 16 – What would happen to the number of stereoisomers in each case above if the top aldehyde
functionality were reduced to an alcohol functionality (a whole other set of carbohydrates!)? A generic
structure is provided below to show the transformation.
O
H
C
CH2OH
Reduce
aldehyde
to alcohol
CH2OH
CH2OH
Cyclic Systems An extension of this approach to cyclic systems (rings) is straight forward, if you have understood
the material presented thus far. Some of our more commonly encountered ring systems include the small
(n = 3 and 4) and normal (n = 5, 6 and 7) sized rings. Remember, except for cyclopropane, rings have the
ability to make partial rotations (conformational changes) about their single bonds and have a certain
floppiness to them. These movements occur on the order of tens of thousands of times per second and no
single representation of our ring system will accurately describe all possible conformations. Our
approach in drawing rings depends on what we are emphasizing. If we want to try and show relationships
of the ring and its substituents, then we need to draw approximate 3D structures that focus on a single
possible conformation. Other conformations may be drawn as well for a more complete analysis.
However, if we just want to show that the structure is a ring with a top and a bottom, a flat 2D structure
may be sufficient. We often choose to represent the rings as time-averaged flat structures even though
none of them actually exists in a flat shape, except for cyclopropane.
8
Lecture 16
cyclopropane
cyclobutane
H2
C
H2C
CH2
Cyclopropane is a flat, three point planar
structure. No other shape is possible.
Cyclobutane has a puckered conformation
that flip-flops back and forth.
Average cyclobutane is flat.
2
cyclopropane
Average cyclopentane is flat.
cyclohexane
Average cyclohexane is flat.
1
3
5
4
boat 1
3
2
4
5
chair 1
chair 2
Mostly chair
conformations.
Cyclopentane has many envelope conformations
via pseudorotations. Tip of flap can be up or down.
boat 2
One can quickly evaluate whether substituents are on the same side (cis) or opposite sides (trans)
using this approach. The two additional substituents at each carbon atom of the ring are often drawn in
with simple straight lines perpendicular to the ring to indicate top and bottom (this is called a Haworth
structure in biochemistry). Of course, real 3D structures are much more complex than this simplistic
representation (i.e., cyclohexane chair conformations have axial and equatorial positions on both top and
bottom that can interchange, and boat, twist boat and half chair conformations are also possible).
Modified Fischer projections are workable with the ring systems. We can tilt the rings on their
sides and rotate the ring about an imaginary axis through the middle of the ring. The horizontal groups
will continually face you as they rotate by and the vertical groups (the ring connections) will always be
away. There is not really any longest, vertical carbon chain, rather each rotation is like a frame of a
moving picture.
rotate about
imaginary
center axis
9
Lecture 16
Meso Rings
If two bromine atoms were at vicinal positions (vicinal = adjacent, which comes from vicus, Latin
= village, neighbors) and on the same side of the ring (cis), we would have a very similar picture to the
meso 2,3-dibromobutane example presented earlier.
R
CH3
CH2
CH2
Br
H
Br
CH2
CH2
H
CH2
CH2
Br
S
H
Br
H
CH3
CH2
CH2
CH2
bisecting
mirror
plane
CH2
CH2
CH2
meso
2R,3S-dibromobutane
all of these cis-1,2-dibromo ring systems are meso
All of the cis-1,2-dibromo ring structures have two chiral atoms with mirror plane symmetry and
so are meso (achiral and superimposable on their mirror image). Confirm the top and bottom stereogenic
centers as R and S in the ring structures.
When two cis substituent groups are identical in any simple positional disubstituted cycloalkane
examples, the stereoisomers will either be meso (achiral with chiral atoms present) or achiral (no chiral
atoms present). All of these molecules have a bisecting mirror plane reflecting one half of the molecule
into the other half.
When there is an even number of carbon atoms in a ring and two cis substiturent groups are
exactly opposite one another, the substituted carbon atoms are stereogenic, but not chiral.
Cis possibilities - All are achiral, some are meso and some have no chiral centers
Br
Br
Br
Br
Br
Br
Br
Br
H
H
H
H
H
H
H
H
meso
(achiral)
meso
(achiral)
no chiral atoms
(achiral)
meso
(achiral)
Br
Br
Br
Br
H
meso
(achiral)
meso H
(achiral)
Br
Br
H
H
H
Br
H
H
Br
H
meso
(achiral)
no chiral atoms
(achiral)
10
Lecture 16
Enantiomeric Rings
The other 2,3-dibromobutane arrangement produced a (dl) enantiomeric pair. This turns out to be
true in the trans-1,2-dibromo ring systems as well.
R
CH3
CH2
CH2
CH2
Br
H
Br
H
H
Br
H
Br
CH2
CH2
CH2
CH3
CH2
CH2
CH2
CH2
CH2
R
CH2
mirror plane
enantiomers
(d/l or +/- pairs)
S
CH3
H
CH2
CH2
Br
H
CH2
CH2
Br
Br
H
H
Br
CH3
CH2
CH2
CH2
CH2
CH2
CH2
CH2
S
CH2
All of the trans-1,2-dibromo ring structures have two chiral atoms, with no mirror plane
symmetry. They are chiral, and come in enantiomeric pairs. Confirm the top and bottom chiral centers as
both R or both S in the ring structures.
When two trans substituent groups are identical in any simple positional disubstituted cycloalkane
examples, the stereoisomers will either be enantiomeric pairs (with chiral atoms present) or achiral (no
chiral atoms present). None of the enantiomeric pairs have a bisecting mirror plane reflecting.
Trans Possibilities - Pairs of Enantiomers (these mirror image pairs are different)
H
H
Br
H
Br
H
Br
H
Br
H
Br
H
Br
H
Br
Br
H
Br
H
Br
H
H
Br
H
Br
Br
H
Br
H
Br
H
Br Br
Br
H
H
Br
Br
H H
H
Br
Br
The left bromine is on the top in all examples.
H
H
Br
mirror
plane
H
Br
11
Lecture 16
As with the cis disubstituted isomers, when there is an even number of carbon atoms in a ring and
two trans substituent groups are exactly opposite one another, the substituted carbon atoms are
stereogenic, but not chiral.
Both cis and trans-1,3-disubstituted cyclobutanes and 1,4-disubstituted cyclohexanes illustrate
this feature. They have stereogenic centers in the cis/trans sense, but not in the R/S sense, since the two
paths traced about the ring are identical (two groups are the same). They are diastereomers that are also
classified as geometric isomers and cis/trans isomers.
4
4
Br
Br
3
H
H
2
6
5
Br
Br
H
1
4
3
4
1
1
Br
H
H
2
A
cis-1,3-disubstituted
cyclobutanes
6
Br
Br
1
5
H
B
trans-1,3-disubstituted
cyclobutanes
H
3
H
2
Br
3
2
D
C
cis-1,4-disubstituted
cyclohexanes
trans-1,4-disubstituted
cyclohexanes
These are achiral diastereomers. They
are also called cis/trans isomers and
geometric isomers
These are achiral diastereomers. They
are also called cis/trans isomers and
geometric isomers
Problem 17 - Draw the mirror image for each of the following structures. Is the mirror image identical or
different? (i.e. Is the molecule chiral?) Classify all stereogenic atoms as R or S absolute configuration.
Are any of these structures meso?
a.
b.
c.
d.
e.
H
f.
H
Problem 18 - a. Draw all possible isomers of dimethylcyclobutane (structural isomers and stereoisomers,
there should be 9 of them). Label your structures A, B, C, D…and indicate which, if any, are
enantiomers, diastereomers and/or meso structures. If stereogenic centers are present indicate the
absolute configuration as R or S or an achiral stereogenic center. Decide which conformation is most
likely preferred for each isomer.
b. How would the problem change if one of the methyl substituents had been changed to a Br substituent?
12
Lecture 16
Problem 19 - For the following set of stereoisomers answer each question by indicating the appropriate
letter(s) or letter combination(s). It may help to rotate and/or flip the rings to alternate perspectives.
Models would make this an easier problem, too. Another possibility is to assign absolute configurations.
OH
OH
H
OH
H
HO OH
H
HO
H
H H
H
HO H
H H
HO
A
OH
H
HO
OH
HO
B
H
C
OH
H
H
H
OH
OH
D
E
a. Which stereoisomers can rotate plane polarize light (are optically active)?
b. Which are meso?
c. Which pairs are enantiomers?
d. Which pairs are identical?
e. Which pairs are diastereomers?
f. Draw any stereoisomers of 2,3,4-cyclopentanetriol not shown above.
Problem 20 - Draw the mirror image structure for each of the following molecules. Is the mirror image
identical of different? Is the molecule chiral? Classify all stereogenic centers as R or S absolute
configuration. Are there any meso structures?
a.
c.
b.
CH3
d.
OH
H
CH3CH2
Br
H3C
CO2H
H
HO
OH
H
C
H
HO
NH2
e.
C
g.
f.
Br
Br
CH3
H3C
H
h.
Br
C
Br
C
H
Br
CH3
Br
E and Z Isomers
plane of
pi bond
plane of
pi bond
1
1
1
C
C
2
2
C
2
C
2
1
Z = zussamen
E = entgegen
The high priority groups are
on the same side of a plane
cutting through the pi bond.
The high priority groups are
on opposite sides of a plane
cutting through the pi bond.
"1" and "2" are the priorities
of the groups bonded to the
atoms of the pi bond.
13
Lecture 16
Problem 21 - Using the formulas provided, draw an example illustrating each of the listed types of
isomerism. To illustrate isomerism you have to draw at least two structures. (Hint - First calculate the
degree of unsaturation.)
Formula #1 = C6H10Cl2
Formula #2 = C8H16O2
a.
c.
e.
g.
chain or skeletal isomers
functional group isomers
diastereomers (not geometric)
conformational isomers
b.
d.
f.
h.
positional isomers
enantiomers
diastereomers (geometric)
a meso compound (only one structure needed)
Isomer Overview
Isomers - compounds that
have the same formula
Stereoisomers have their atoms
attached with the same connectivity,
but differ in their arrangements in
space.
Constitutional or sturctural
isomers have their atoms
joined together in different
arrangements.
chain or skeletal
isomers
positional
isomers
functional
group isomers
O
Cl
OH
H
Cl
O
O
Diastereomers are
stereoisomers that
are not mirror images
of one another.
Enantiomers are mirror
image reflections that
are nonsuperimposable
(different).
Conformational isomers
differ by rotation about a
single bond and are usually
easily interconverted.
mirror
plane
a.
OH
C
CH3CH2
C
H
CH2CH3
H
H3C
CH3
H
C
OH
H
C
OH
vs.
H
H
H
CH3
CH3
CH3
H
(dl) or (±) enantiomer pairs
H
C
OH HO
C
H
HO
C
H
H
C
OH
CH3
CH3
CH3
meso
H 3C
CH3
CH3
CH3
OH
(dl) enantiomers
b. geometric (cis/trans) isomers
Z or cis
E or trans
CH3
CH3
H
H
H
CH3
cis
CH3
H
trans
14
Lecture 16
Problem 22 - What is the relationship between the molecules in each of the following pairs? i. structural
isomers ii. functional group isomers iii. positional isomers iv. enantiomers v. diastereomers (not
geometric) vi. diastereomers (geometric, cis/trans) vii. conformational isomers viii. not isomers at all
a
c
b
Br
HO
D
HO
H
Cl
Cl
H
H
CH3
H
OH
H
H 3C
H
F
H 3C
Cl
H Cl
Cl
H
F
D
H
HO H
OH
H
Br
HO H
H
Br
H
i
h
H
CH3
Cl
D
f
CH3
g
H
Br
e
Cl H
H
Cl
H
D
CH3
d
H
H
H
H
F
Cl
OH
Cl
F
H
H
H3C
H
CH3
OH
CH2CH3
F
F
k
j
H
H
Br
Br
Br
H
H
H 3C
H 3C
Br
H 3C
CH3
l
H
CH 3
O
CH3
O
H CH3
CH3
CH3
O
O
CH3
Problem 23 - There are two possible stereoisomers at an E/Z double bond, just as there are two possible
absolute configurations at an R/S stereogenic center (atom). Using this knowledge, predict how many
stereoisomers are possible for each of the following structures. Draw a 3D structure of each stereoisomer.
Provide an acceptable name for your stereoisomers in part a.
a
H
H3C
CH
CH
CH
CH
C
OH
b
H3C
CH3
CH
CH
Cl
C
H
CH2CH3
O
c
d
H2C
CH
CH
CH
CH2
CH3
CH
CH
CH
CH
C
H