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 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

ClBr Br

OH

Br

Br

f. g. h. i. j.

Br Br Br Br

RandSNomemclatureProblem 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

HH

HH3C

H

c.

H3C

IH

HCl

CH3

d.

C

Br

C

C

HCH3

CH3H

H3C H

H3C

e. f. g. h.O

H3C

H

Cl

H

Br

CH3

H

C

C

HOH

H

CH3

H3C

Br

CH3

HH3C

H

H

i. j. k. l.

CH Br

CH2CH3

CH3

Cl

HCl

H

H3CH2C

DH

BrCH3

CH3

S

OCH3

2 Lecture 16 Pi Bond Priority Problem 4 - Evaluate the order of priority in each part. a.

C

CH3

CH2*

* = path to chiral center

C C H*

C C

C

CC

C

H H

H

HH

* C

CH3

CH3*

CH3

ethynyl phenyl 2-propenyl t-butyl b.

CH2C N*

H2C C

O

H

H2C

H2C F C

CH3

H2C

OH

* I**

c.

C

H

O*C

OH

O

HH

C

OH

O

CH3H

C

OCH3

O

CH3H

* * *

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 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.

4C

4. Fill in any convenient (obvious) group, …say 1.

4

1

C

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.)

4

1

C4

1

C2

2

R S

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.

CH3

HO H

HH

CH3

NH2

H3C H

H CH3

D

F

Cl Br

H H

CH3

OH

H3C H

H3C 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.)

OHHD

CH3

HHOCH3

D

CO2HNH2

H

CO2H

NH2

H

OCH3CH3CH2OCH2CH3

CH3

CH2CH3CH3CH2OOCH3

CH3

HCH3

CO H

H

CH3

CO

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.

b.a.

C

H

C

C

H3C

CH

CH2CH3

CHCH3 CH

H3C

CH

C

C

CH3

CH2CH3

C

H

H2C

CH

C

CH

CH3

OHH3C

CH3

CH

C

H3C

CH

OH

CH3

H3CCH2

c.

C

C

H

CH3

C

d.O

H

N

H

C

H3C

C

C

N

O

H

H

OH

CH3

H

HHOH

H3C

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.

12 3

4

1

1

1

4

1

1

34 4

24 4

2 3

243

324

4 32

423

3 2

21

3

1

4

1

2

2 2

2

2

13

143

134

314 1

34

3

3 3

3

31 2

12

4

41

2 21

4

4 12

4

4 4

23 1

321 12

3

21 3

3

4

1

4

2

32

14

ReferenceStructure

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. c. d.OH

H CH3

H3C

H

OH

Br

CH3CH2 H

H Br

H

H3C OH

HO H

H3C H

Br

H OH

HO

H

CHO

H

HO CHO

HO H

HO H

H OH

H

H

OH

Three Stereogenic Atoms

HCH3C

Br

HC

Cl

HC

Cl

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 2,3,4,5-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?

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.

OH

H CH3

HO H

H OH

CH3

CH3

H OH

H OH

H3C OH

CH3

CH3

H OH

H OH

H OH

CH3

CH3

H OH

HO H

HO H

CH3

CH3

HO H

H OH

HO CH3

H

A B C D E

a. Which are optically active? b. Which are meso? c. Which pairs are enantiomers? d. Which pairs are identical? e. Which pairs are diastereomers? f. Which, when mixed as a 50/50 mixture, will not rotate plane polarize light (optically inactive)? g. Draw any stereoisomers of 2,3,4-pentanetriol as Fischer projections, which are not shown above. How does Nature do it?

C

H OH

CH2OH

D-glyceraldehyde

O HC

HHO

CH2OH

OH

L-glyceraldehyde

Three carbon aldose carbohydrates

C

H OH

D-erythrose

O H

D-threose

H

CH2OH

OH

C

HHO

OH

H

CH2OH

HO

C

HO H

O H

H

CH2OH

OH

L-erythrose

C

OHH

OH

H

CH2OH

HO

L-threose

Four carbon aldose carbohydrates

C

H OH

D-ribose

O H

D-arabinose

H OH

L-ribose L-arabinose D-xylose D-lyxoseL-xylose L-lyxose

Five carbon aldose carbohydrates

H OH

CH2OH

C

HHO

OH

HHO

HHO

CH2OH

C

HO H

O H

H OH

H OH

CH2OH

C

OHH

OH

HHO

HHO

CH2OH

C

H OH

O H

HO H

H OH

CH2OH

C

HHO

OH

OHH

HHO

CH2OH

C

HO H

O H

HO H

H OH

CH2OH

C

OHH

OH

OHH

HHO

CH2OH

7 Lecture 16

C

H OH

D-allose

O H

D-altrose

H OH

L-allose L-altrose D-glucose D-gluoseL-glucose L-gluose

Six carbon aldose carbohydrates

H OH

OHH

CH2OH

C

HHO

OH

HHO

HHO

HO H

CH2OH

C

HO H

O H

H OH

H OH

OHH

CH2OH

C

OHH

OH

HHO

HHO

HO H

CH2OH

C

H OH

O H

HO H

H OH

OHH

CH2OH

C

HHO

OH

OHH

HHO

HO H

CH2OH

C

H OH

O H

H OH

HO H

OHH

CH2OH

C

HHO

OH

HHO

OHH

HO H

CH2OH

C

HO H

D-mannose

O H

D-galactose

HO H

L-mannose L-galactose D-idose D-taloseL-idose L-talose

H OH

OHH

CH2OH

C

OHH

OH

OHH

HHO

HO H

CH2OH

C

H OH

O H

HO H

HO H

OHH

CH2OH

C

HHO

OH

OHH

OHH

HO H

CH2OH

C

HO H

O H

H OH

HO H

OHH

CH2OH

C

OHH

OH

HHO

OHH

HO H

CH2OH

C

HO H

O H

HO H

HO H

OHH

CH2OH

C

OHH

OH

OHH

OHH

HO H

CH2OH

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

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

H2CCH2

H2C

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 conformations via pseudorotations. Tip of flap can be up or down.

1

23

4

5

2

3

45

Average cyclopentane is flat.

chair 1 chair 2

boat 1

boat 2

Average cyclohexane is flat.cyclohexane

Mostly chair conformations.

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 aboutimaginarycenter 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.

CH3

HBr

Br H

CH3

CH2

HBr

Br H

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

bisectingmirrorplane

all of these cis-1,2-dibromo ring systems are meso

mesoS

R

2R,3S-dibromobutane

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

H

Br

H

Br

H

Br

H

Br

H

Br

H

Br

H

Br

H

Br

H

Br

H

Br

H

meso(achiral)

meso(achiral)

meso(achiral)

no chiral atoms (achiral)

Br

H meso(achiral)

Br

H

Br

H

no chiral atoms (achiral)

meso(achiral)

meso(achiral)

Br

H

Br

H

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.

CH3

HBr

H Br

CH3

CH2

HBr

H Br

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

R

R

CH3

BrH

Br H

CH3

CH2

BrH

H

CH2

Br

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

S

S

enantiomers mirror plane(d/l or +/- pairs)

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

HH

Br

H

Br

Br

H

Br

HH

Br

The left bromine is on the top in all examples.

mirrorplane

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.

Br

H

Br

H

Br

H

H

Br

Br

H

Br

H

Br

H

H

Br

cis-1,3-disubstituted cyclobutanes

trans-1,3-disubstituted cyclobutanes

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

1

2

3

4 5 6

1

2

3

4

1

23

4

65

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

Hb. c. d. e. f.

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.

H

OH

H

OH

H

HO H

OH

OH

HH

OH

OH

H

H

HO

H

HO

OH

H

OH

H

H

HO

H

OHH

HO H

HO

A B C 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?

CH3

NH2

CH3CH2

CO2H

H

HOH

OHC

H3C

HBr

C

BrH

CH3

C

H3C

HBr

C

HBr

CH3

Br

Br

HO

HO

a. b. c. d.

e. f. g. h.Br

E and Z Isomers

C C

plane of pi bond

1 1

2 2

C C

plane of pi bond

1 2

2 1

Z = zussamen The high priority groups are on the same side of a plane cutting through the pi bond.

E = entgegen 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. 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 that have the same formula

Constitutional or sturctural isomers have their atoms joined together in different arrangements.

Stereoisomers have their atoms attached with the same connectivity, but differ in their arrangements in space.

chain or skeletal isomers

positional isomers

functional group isomers

Conformational isomers differ by rotation about a single bond and are usually easily interconverted.

Enantiomers are mirror image reflections that are nonsuperimposable (different).

Diastereomers are stereoisomers that are not mirror images of one another.

Cl

ClO

HO

O

OH

mirrorplane

(dl) or (±) enantiomer pairs

C

OH

HH3C

CH3CH2C

OH

HCH3

CH2CH3

H3C H

H CH3

CH3H

HCH3

CH3H

HCH3

CH3CH3

HH

C

CH3

H OH

CH

CH3

OH

C

CH3

H OH

CHO

CH3

H

C

CH3

HO H

CH

CH3

OH

(dl) enantiomers

vs.

meso

a.

b. geometric (cis/trans) isomers

Z or cisE or trans

transcis

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

H Cl

Cl H

H OH

H Cl

Br H

HO H HBr

HHO

HO D

HO

D

HBr

ClH

BrH

ClH

D

H3CHF

H3C

HFD

a b c

de fCH3

HCl

ClH

CH3

CH3

CH2CH3F

CH3H

F

H Cl

H OH

g h i

F

H

F

HH

Cl

HCl

H

OH

HH3C

HCH3 OH

j k lH Br

Br

H Br

Br

H

H H3C H

CH3

CH3

H3CH3C H

CH3CH3

CH3CH3

CH3

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.

CHH3C CH CH CH C

H

CH3

Cl

CHH3C CH C H

OH

CH2CH3

CHH2C CH CH CH2 CH3

CH CH CH CH C

O

H

a b

c d