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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 d. 1,1-dibromocyclopentane
b. 2-bromopentane e. cis -1,2-dibromocyclopentane
c. 3-bromopentane f. trans -1,2-dibromocyclopentane
Problem 2 - Which molecules below have stereogenic centers? How many? Are they all chiral centers?
a. b. c. d. e.
Cl
Br Cl Br
OH
Br
Br
f. g. h. i. j.
Br Br Br^ Br
R and S Nomemclature
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.
Cl H H
a. b.
HO
H H
H 3 C H
H
c.
H 3 C
I
H
H
Cl CH 3
d.
C
Br
C
C
H
CH 3
CH 3
H
H 3 C H
H 3 C
e. O f. g. h.
H 3 C
H
Cl
H
Br
CH 3
H
C
C
H
OH
H
CH 3
H 3 C
Br
CH 3
H 3 C H
H
H
i. j. k. l.
H C Br
CH 2 CH 3
CH 3
Cl
HCl
H
H 3 CH 2 C
D
H
Br
CH 3
CH 3
S
O CH^3
Lecture 16
Pi Bond Priority
Problem 4 - Evaluate the order of priority in each part.
a.
C
CH 3
* CH 2
* = path to chiral center
*^ C^ C^ H
C C
C
C C
C
H H
H
H H
* C
CH 3
* CH 3
CH 3
ethynyl phenyl 2-propenyl t-butyl
b.
* HC^2 C N
HC 2 C
O
H
HC 2 HC 2 F C
CH 3
HC 2
OH
* * * I
c.
C
H
C * O
O
H
O
H H
C
O
H
O
H CH^3
C
O
CH 3
O
H CH^3
d.
* *^ *^ *
Cl
Br
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 sp 3 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. Fill in any convenient (obvious) group, …say 1.
C
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. CH 3 HO H H H CH 3
NH 2 H 3 C H H CH 3 D
F Cl Br H H CH 3
OH H 3 C H H 3 C H OH
a. (^) b. c. (^) d.
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.)
H OH
D
CH 3
HO H
CH 3
D
CO 2 H
NH 2
H
CO 2 H
NH 2
H
CH 3 CH 2 O OCH (^3)
CH 2 CH (^3)
CH 3
CH 3 CH 2 O CH 2 CH 3
OCH (^3)
CH 3
CH 3 H
C
O (^) H
H CH 3
C
O H
a (^) b c d
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. a. b.
C
H
C
C
H 3 C
CH
CH 2 CH 3
CH
CH 3 CH
H 3 C
CH
C
C
CH 3
CH 2 CH 3
C
H
H 2 C
CH
C
CH
CH 3
H 3 C OH
CH 3
CH
C
H 3 C
CH
OH
CH 3
H 3 C CH^2
c.
C
C
H
CH 3
C
d.
O
H
N
H
C
H 3 C
C
C
N
O
H
H
OH
CH 3
H
HO H
H
H 3 C
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 4 1
1
1 4 1
1
3 4 4
2 4 4
2 3
2
3 4
3
4 2
(^4 ) 2
(^24) 3
3 2
2 1
3 1
4 1 2
2 2
2
2 3 1
1
3 4 1
4 3
3
4 1 1 3
4
3
3 3
3
3 (^1 )
1 2
4
4 1
2 2 1
4
4 1 2
4
4 4
2
(^3 )
3
(^1 ) 3
2
(^1 )
3
4 1 4
2 3 2 1
4
Reference Structure
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. (^) OH b. c. d.
H CH 3
H 3 C
H
OH
Br
CH 3 CH 2 H
H Br
H
H 3 C OH
HO H
H 3 C H
Br
H OH
HO
H
CHO
H
HO CHO
HO H
HO H
H OH
H
H
OH
Three Stereogenic Atoms H^ HC 3 C Br
HC
Cl
HC
Cl
- *^ *^ CH (^3) Three stereogenic centers with 2 3 = 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 2,3,4,5- tetrachlorohexane? 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?
Lecture 16
C H OH
D-allose
O H
D-altrose
H OH
L-allose L-altrose D-glucose L-glucose D-gluose L-gluose
Six carbon aldose carbohydrates
H OH
H OH
CH 2 OH
C
HO H
H O
HO H
HO H
HO H
CH 2 OH
C
HO H
O H
H OH
H OH
H OH
CH 2 OH
C
H OH
H O
HO H
HO H
HO H
CH 2 OH
C
H OH
O H
HO H
H OH
H OH
CH 2 OH
C
HO H
H O
H OH
HO H
HO H
CH 2 OH
C
H OH
O H
H OH
HO H
H OH
CH 2 OH
C
HO H
H O
HO H
H OH
HO H
CH 2 OH
C
HO H
D-mannose
O H
D-galactose
HO H
L-mannose L-galactose D-idose L-idose D-talose L-talose
H OH
H OH
CH 2 OH
C
H OH
H O
H OH
HO H
HO H
CH 2 OH
C
H OH
O H
HO H
HO H
H OH
CH 2 OH
C
HO H
H O
H OH
H OH
HO H
CH 2 OH
C
HO H
O H
H OH
HO H
H OH
CH 2 OH
C
H OH
H O
HO H
H OH
HO H
CH 2 OH
C
HO H
O H
HO H
HO H
H OH
CH 2 OH
C
H OH
H O
H OH
H OH
HO H
CH 2 OH
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.
C
O H
CH 2 OH
Reduce aldehydeto alcohol
CH 2 OH
CH 2 OH
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.
Lecture 16
H 2 C
CH 2
HC 2
cyclopropane
Cyclopropane is a flat, three point planar structure. No other shape is possible.
cyclobutane
Cyclobutane has a puckered conformation that flip-flops back and forth. Average cyclobutane is flat. cyclopropane
Cyclopentane has many envelope conformationsvia pseudorotations. Tip of flap can be up or down.
1
3 2
4
5
2
3
4 5
Average cyclopentane is flat.
chair 1 chair 2
boat 1
boat 2
cyclohexane Average cyclohexane is flat.
Mostly chairconformations.
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
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.
CH (^3) Br H H Br CH (^3)
CH (^2) Br H H Br CH (^2)
CH (^2) CH (^2)
CH (^2) CH (^2) CH (^2)
CH (^2) CH (^2) CH (^2) CH (^2)
CH (^2)
R
R
CH (^3) H Br Br H CH (^3)
CH (^2) H Br H CH (^2)
Br
CH (^2) CH (^2)
CH (^2) CH (^2) CH (^2)
CH (^2) CH (^2) CH (^2) CH (^2)
CH (^2)
S
S
(d/l or +/- pairs)enantiomers^ mirror plane
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)
Br H
H Br
Br H
H Br
Br H
H Br
H BrBr H
Br H
H Br
Br
H H
Br
Br
H H
Br Br
H H
Br
Br H
H Br
Br
H H
Br H
Br
Br
H Br
H H
Br
The left bromine is on the top in all examples.
mirror plane
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.
Br
H
Br
H
Br
H
H
Br
Br
H
Br
H
Br
H
H
Br
cis-1,3-disubstituted cyclobutanes trans-1,3-disubstitutedcyclobutanes cis-1,4-disubstitutedcyclohexanes 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
1
2
3
(^4 5 ) 1
2
3
4
1
3 2
4
5 6
4
3 2
1
A B C D
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.
H
b. (^) c. d. (^) e. 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?
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 = C 6 H 10 Cl 2 Formula #2 = C 8 H 16 O 2
a. chain or skeletal isomers b. positional isomers c. functional group isomers d. enantiomers e. diastereomers (not geometric) f. diastereomers (geometric) g. conformational isomers h. a meso compound (only one structure needed)
Isomer Overview Isomers - compounds thathave the same formula
Constitutional or sturcturalisomers have their atoms joined together in differentarrangements.
Stereoisomers have their atomsattached with the same connectivity, but differ in their arrangements inspace.
chain or skeletalisomers positionalisomers functional group isomers
Conformational isomers differ by rotation about asingle bond and are usually easily interconverted.
Enantiomers are mirrorimage reflections that are nonsuperimposable(different).
Diastereomers arestereoisomers that are not mirror imagesof one another.
Cl
Cl O
O H O
OH
mirrorplane
(dl) or (±) enantiomer pairs
C
OH
CH 3 CHH 23 C H
C
OH
H CHCH 3 2 CH^3
H 3 C H
H CH 3
H CH 3
CH 3 H
H CH 3
CH 3 H
CH 3 CH 3
H H
C
CH 3
H OH
H C
CH 3
OH
C
CH 3
H OH
HO C
CH 3
H
C
CH 3
HO H
H C
CH 3
OH
(dl) enantiomers
vs.
meso
a.
b. geometric (cis/trans) isomers
Z or cis (^) E or trans
cis trans
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
H Cl
Cl H
H OH
H Cl
Br H
HO H (^) Br H
HO H
HO D
HO D
Br H
H Cl
H Br
H Cl
D H 3 C
H F
H 3 C H
F D
a b c
d CH^3 e^ f
Cl H
H Cl
CH 3
CH 3
F CH 2 CH 3
H CH 3
F
H Cl
H (^) OH
g h^ i
F
H
F
H
H Cl
H Cl H
OH H 3 C H
H CH (^3) OH
j k^ l H Br Br
H (^) Br
Br
H
H H^3 C^
H
CH 3 CH 3
H 3 C
H 3 C H
CH 3 CH 3
CH 3 CH 3
CH 3
O (^) O (^) O
O
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.
H 3 C CH CH CH CH C
H CH (^3) Cl
H 3 C CH CH C H
OH
CH 2 CH 3
H 2 C CH CH CH CH 2 CH 3
CH CH CH CH C
O H
a (^) b
c (^) d
