STEREOCHEMISTRY NOTES Table of Contents Stereochemistry Section I · · 2015-10-14STEREOCHEMISTRY NOTES Section I (1) The following post deals with assigning the stereochemistry

STEREOCHEMISTRY NOTES

Table of Contents Stereochemistry

Section I

Rule 1 Describes the procedure for determining the configuration of a chiral center using a Fischer

projection, priority assignments and manipulation of assigned substituents.

Rule 2 Example of chiral center with substituents which are all of different atomic number.

Rule 3 Example of chiral center with two substituents, where the first atom out from the chiral center

has the same atomic number.

Priority assignment for alkyl substituents. Tetrahedral structure, molecular shape AX4.

Cyclic compounds with a chiral center.

Rule 4 Atoms other than carbon that can form chiral compounds, i.e., Sulfur, Phosphorus & Nitrogen.

Sulfonium ions and phosphine compounds. Molecular shape AX3E.

Quaternary ammonium ion. Benzyl group.

Rule 5 Compounds with two chiral centers.

Rule 6 Compounds with a double bond.

Carbon to carbon double bond.

Oxygen containing compounds, i.e., carbonyl group and alcohol priority rating.

methoxycarbonyl, carboxy, acetyl, formyl

Glyceraldehyde - R & S configurations. D & L designations.

Substituent ratings.

hydroxy, formyl, hydroxymethyl

Priority ratings for carbon, nitrogen and oxygen containing substituents:

alkoxy, hydroxy, amino, carboxy, formyl, hydroxymethyl, methyl

Lactic acid, R configuration.

Detailed examination of assigning configuration by manipulating four groups

around a Fischer projection.

Six steps for assigning a configuration to a chiral center unambiguously, a caveat.

Rule 7 Two nitrogen compounds: 2-butanamine; 2-amino-1-propanol

Rule 8 Basic Fischer projection of a generic amino acid.

Alanine - R & S configurations and D & L designations - R/D & S/L

Alternate 2-dimensional views of Alanine - R & S configurations.

Cysteine configurations: R/L & S/D (Cys is different from Ala).

Priority ratings including SH (sulfhydryl substituents).

Threonine and Isoleucine - amino acids with two chiral centers.

Rule 9 Sulfoxides.

Section II

(1) Other Stereochemistry related issues.

(i) Impurities in synthetic peptide drug products via solid phase peptide synthesis (SPPS)

(ii) Racemization of amino acids during acid hydrolysis of a peptide

(iii) Asparagine deamidation and aspartate isomerization

(2) Other chiral molecules: (1R, 2S)-ephedrine, (S)-Carboxymethylcsyteine, (S)-methylacetylcholine.


Section I

(1) The following post deals with assigning the stereochemistry to a chiral atom in a molecule.

(2) The configuration of a chiral atom (chiral center) can be assigned either R or S and these configurations are

non-superimposible mirror images. The following procedure is based on the method described in N.L.Allinger,

Organic chemistry, 2nd Edition, page 106-107.

(3) The are nine rules which will allow for many but not all assignments; thus, I will provide the basics.

Rule 1 describes a basic assignment and Rules 2-9 give specific examples.

RULE 1

Draw a Fischer projection for the chiral center. Place the four substituents around the chiral center. Then

assign priority numbers to each of the groups. The priority numbers are based on the Atomic Number of the

atom that is directly attached to the chiral atom, thus substituents will be labeled 1 (highest priority), 2, 3 & 4

(lowest priority).

The goal is to get substituent #4 at the bottom of the Fischer projection through a series of interchanges of the

substituents. One interchange will result in a configuration that is the enantiomer of the original configuration

and a second interchange will result in returning to the original configuration. The following is a generic

example (chemistry models can be used to help with visualizing the 3-D structure and to easily distinguish the

four different substituents).

(i) original configuration

A

- #4 & #1 are in the plane of

of the page

(ii) first change, interchange 3 & 4

- this is the enantiomer of B

the original configuration, A


of the page

(iii) second change, interchange

1 & 2

- this configuration is the C

same as the original

configuration, A


of the page, this is the R configuration.

1 2

4

3

1 2

3

4

2 1

4

3

3

4

2 1

4

3

2 1

3

4

1 2


(iv) The diagrams to the right of the Fischer projections are drawn to represent the tetrahedral structure, thus

in the original configuration (i), 4 and 1 are in the plane of the page and 3 is behind the page and 2 is in

front of the page. The next two tetrahedral structures follow the same orientation.

If you were to build models of (i), (ii) and (iii), you would find that (i) & (ii) are non-superimposible

mirror images (enantiomers) and (i) & (iii) have the SAME configuration, i.e., they are superimposible.

Thus, any two interchanges will always take you back to the original configuration.

(iv) In configuration (iii), 4 is at the bottom of the Fischer projection and if you read the priority numbers 1, 2

and 3, they read in a clockwise direction, thus is assigned the R configuration.

(iv) Another way of looking at (iii), is to rotate the model to your right, so that substituent 4 is directly behind

the chiral atom as in structure (a) below on the left, where the circle at the back represents substituent #4.

The structure on the right, i.e., (b) is an equivalent structure, with the substituents rotated clockwise.

(a) (b)

_________________________________________________________________________________________

RULE 2 - example with substituents of increasing atomic number (1 is highest and 4 is lowest priority):

Since the substituents 1, 2 and 3 read in a counter clockwise direction, this is assigned the S configuration.

________________________________________________________________________________________

3

1 2

2

3 1

3, CH3

2, Cl 1, Br

4 3

1

2

2 interchanges

1, Br 2, Cl

4, H

3, CH3

H, 4 CH3, 3

Br, 1

Cl, 2

3, CH3

4, H

2, Cl

1, Br


RULE 3a - example #1 with a chiral center that has two substituents with the same atom:

(i) If a chiral atom has two substituents that are similar, then the substituent with atoms of higher atomic

number will be assigned a higher priority number. Two substituents have a carbon atom attached to the

chiral center. Again, (1 is highest and 4 is lowest priority).

The Br atom is the highest priority and

the hydrogen atom is the lowest priority. The only question

arises with the methyl and ethyl group as to the priority

assignment. Since the ethyl group has 2 hydrogen atoms and

one carbon atom, it will be higher priority than the methyl

group, because the methyl group has only 3 hydrogen atoms

attached to it.

Rotating the tetrahedral structure to the right, which puts the

hydrogen atom behind the chiral center, one will see the following

Since the priority numbers read 1, 2 and 3 in a clockwise direction,

the assignment is R.

R configuration S configuration

Br, 1 H, 4

CH2CH3, 2

CH3, 3

Two interchanges

Br, 1

CH2CH3, 2 CH3, 3

Br, 1

H, 4

CH2CH3, 2 CH3, 3

Br, 1

H, 4

CH2CH3, 2 CH3, 3

Br, 1

H, 4

CH2CH3, 2 CH3, 3

Br, 1

H, 4

CH3CH2, 2

CH3, 3


(ii) If a chiral atom has two substituents with the same atom, example #2:

If the first carbon attached to the chiral center for both substituents is the same, then one has to

examine the second carbon atom of each substituent.

The Cl atom is highest (1) and the hydrogen atom is lowest (4) priority. With respect to the ethyl and

propyl group, the propyl group is assigned the higher priority, i.e., #2, because the second atom

"out" has 1 carbon atom and 2 hydrogen atoms versus 3 hydrogen atoms for the ethyl group.

Since the priority numbers read 1, 2 and 3 in a clockwise direction, the assignment is R configuration.

The following priority assignment, from highest to lowest, can be made for the following alkyl chains:

(highest) ----------------------------------------------------------> (lowest)

CH3CH— > CH3CH2CH2— > CH3CH2— > CH3— > H

iso-propyl > propyl > ethyl > methyl > hydrogen atom

reference: Allinger, 2nd Edition, Organic Chemistry, page 106-107.

It should be noted at this point that once a configuration has been assigned, one should always provide a

tetrahedral structure as in examples A, B & C in rule 1, so that there is an unambiguous description of the

orientation of the atoms. I have chosen to use the diagrammatic representations as in A, B and C because it is

the one that I can most easily "mentally" visualize, i.e., 2 atoms in the plane of the page, 1 atom behind the

page (lowest priority) and 1 atom in front of the page.

____________________________________________________

CH3

CH2CH3, 3

CH2CH2CH3, 2 Cl, 1

Cl, 1

H, 4

CH2CH3, 3

CH2CH2CH3, 2

H, 4

CH2CH3, 3

CH3CH2CH2, 2 Cl, 1

H, 4

Cl, 1

CH2CH3, 3

CH3CH2CH2, 2

2 interchanges


RULE 3b - example with a chiral center that has two substituents that are the same atom:

(iii) If a chiral atom has two substituents that are alkyl chains, example #3:

- substituents in clockwise direction: OH (hydroxyl), hexyl (C6H13), hydrogen atom, ter-butyl (CH3)3C-

The OH and hydrogen atom are the highest (1) and lowest (4) priority respectively.

There are two other groups, a hexyl and ter-butyl group. The carbon atom of the

of the ter-butyl group has 3 carbon atoms attached to it, whereas, the carbon atom of

the hexyl group has 2 hydrogen atoms and a carbon atom. Thus, the substituent

with the 3 carbon atoms, ter-butyl, will be higher priority.

Since the priority numbers 1, 2 and 3 read counter clockwise, the assignment is S. Furthermore, in

addition to the priority rules given in the previous example, a ter-butyl group will have a higher

priority than a iso-propyl group.

________________________________________________

RULE 3c - Three different ways to represent the tetrahedral structure: All of these representations are equal.

Using standard colors for the atoms and (atomic #): white = H (1), blue = N (7), red = O (8),

green = Cl (17) and assigning priority numbers, a configuration can be assigned. Since the priority

numbers read 1 → 3 in a clockwise direction, the configuration is R. Figure A is a Fischer

projection and the horizontal line indicates that the substituents are in front of the page and the

vertical line indicates that the substituents are behind the page. Figure B is a projection formula

and the grey triangles indicate that the substituents are in front of the page and the striped triangles

indicate that the substituents are behind the page. Figure C is a tetrahedral structure: green & blue

are in the plane of the page, white is behind the page and red is in front of the page.

A B C

H, 4

OH, 1

CH2CH2CH2CH2CH2CH3, 3 C, 2 CH3

CH3

CH3

1

4

3 2

OH, 1

(CH2)5 CH3, 3 (CH3 )3C , 2

OH, 1

(CH3 )3C , 2

H, 4

(CH2)5 CH3, 3

4

1

2

3

≡ ≡ 4

1

2 3

C

4

1

2

3


RULE 3 - example #4: The examples examine ring structures and the priority number assignment.

(iv) Cyclic compound with a chiral center, example #4, (a) is the chiral center:

First look at the chiral center (a) and then place the substituents around the chiral atom, i.e., hydrogen

atom, OH, 3 methylene groups in an alkyl chain and "iso-butyl" group (so termed because it does not

contain the methine group at the first carbon out).

The hydroxyl group and hydrogen atom will be the highest (1) and lowest (4) priority respectively.

The "iso-butyl" group at the second carbon atom out has 2 methyl groups, whereas, the 3 methylene

groups in an alkyl chain at the second atom out has one carbon atom and two hydrogen atoms, thus, the

"iso-butyl" group will be the higher priority (#2).

The dark triangle means the OH is in front of the page, the striped triangle means the H atom is behind

the page, drawn in perspective (Allinger, p.98). The other two substituents are in the plane of the page.

,

, 2

Since the priority numbers read 1, 2 & 3 in a counter clockwise direction, this is the S configuration.

(a )

HO, 1

H, 4

CH2CH2CH2 , 3

CH2 C

CH3

CH3

3

4

2 1

3

4

2 1

two interchanges

3, CH2CH2CH2

2, C4H8

1, OH

CH2CH2CH2

OH

H

CH2 C

CH3

CH3

1 2

4

3

H

CH3 CH3

OH


RULE 3 - example of another cyclic compound, example #5:

(iv) Cyclic compound with a chiral center, example #5, (a) is the chiral center:

A

Since the priority numbers read 1, 2 and 3 in a clockwise

direction, the assignment is R.

B

1,

C

Rotate structure B until the CH3 and CH2CH3

are in the same orientation as in structure A, i.e.,

behind the page and in front of the page, which is

structure C.

___________________________________________________________________________________________

RULE 4 - All molecules (or ions) with four electronic groups around a central atom adopt a tetrahedral

molecular shape (Silberberg, Chemistry, 5th Edition, page 391). Carbon is such an atom and all of the

previous examples showed how a carbon atom can be chiral.

Atoms other than carbon can adopt a similar molecular shape designated as AX3E and are termed trigonal

pyrimidal, where a lone pair of electrons forms one part of the tetrahedral structure (Silberberg, Chemistry, 5th

Edition, page 391), e.g., Sulfur, Phosphorus and Nitrogen and can form chiral compounds.

Sulfur atoms, in the form of sulfonium salts and Phosphorus atoms, in the form of either phosphine

(organophosphorus compounds) or phosphonium salts, are stable enough to form chiral enantiomers (Allinger,

2, CH2CH3

C

CH3 CH3 3, CH3

4, CH2

(a)

4 2

1

3

two interchanges

1 3

4

2

CH2CH3

CH3

CH3

CH3 CH3

CH2CH3 CH2

C

CH3 CH3

2, CH2CH3

3, CH3

4, CH2

C, 1

CH3

CH3


Organic Chemistry, 2 nd Edition, page 119). Note that the Nitrogen atom also forms the AX3E structure, but

only the ammonium ions form stable chiral compounds (Allinger, Organic Chemistry, 2 nd Edition, page 119).

RULE 4a - An example of a sulfonium ion (+ charge):

- example #1: in this example, the two dark dots represent a lone pair of electrons (and are assigned an atomic

number of zero). As well, the dark triangles indicate that the structures are in front of the page and the striped

triangles are behind the page (recall that these are in perspective, i.e., closer objects are large and distant

objects are small).

A B

Place all of the groups around the central atom. Assign priority numbers.

C

Since the priority numbers read 1, 2 & 3 in a counter-clockwise direction, the designation is S.

In structure C, substituent #2 and #3 are in the plane of the page and #4 is behind the page and #1 is in front of

the page. If a model of structure C is rotated to the right and you "pull" down on the lone pair of electrons, you

will end up with structure A or B, i.e., the CH3 and CH2CH2CH3 groups will be in front of the page and the

CH2CH3 and lone pair of electrons will be behind the page.

_______________________________________

RULE 4b - A second example of a sulfonium ion (+ charge):

In this case, the CH3 and the CH2CH3 are in the plane of the page and the CH(CH3)2 is in front of the page. It

should be noted that the lone pair of electrons are behind the page. Both structures are equivalent.

≡

CH2CH3 , 2

CH2CH2CH3, 1

●● , 4

CH3 , 3

CH(CH3)2

CH2CH3

S+

CH3 ●●

CH2CH3 CH3

●●

CH(CH3)2

S+

CH2CH3

CH2CH2CH3

●●

CH3

4

3

2

1

CH2CH3

CH2CH2CH3 S+

●●

CH3

CH2CH3

CH2CH2CH3 S+

CH3

●●


Place all of the groups around the central atom. Assign priority numbers. Perform 2 interchanges.

Since the priority numbers read 1, 2, and 3 in a

counter clockwise direction, the designation is S.

RULE 4c - A third example, a phosphine compound. Assign priority numbers, no interchanges needed.

Since the priority numbers read 1, 2, and 3 in a

clockwise direction, the designation is R.

________________________________________________________________________

RULE 4d - A quaternary ammonium ion.

In this example, the molecule has a benzyl group, i.e., (PhCH2—), thus, one needs to know how a benyl

group is analysed. A phenyl ring can be considered to be composed of 3 alternating double bonds.

At the point to which the phenyl group is attached to another group, the double bond for that carbon

2 interchanges

CH3CH2 H P

●●

CH3

CH3CH2 H P

CH3

●●

●●, 4

H, 3 1, CH3CH2

2, CH3

2, CH3

1, CH3CH2

●●, 4

H, 3

CH2CH3

CH(CH3)2

CH3

●●

1

2 3

4

3, CH3

1 ,CH(CH3)2

4, ●●

2, CH3CH2

1

2 3

4


and the corresponding carbon atom that it is double bonded to, is considered to be duplicated, see

below.

RULE 4d - A quaternary ammonium ion, continued,

The "Y" represents a group that is attached to the phenyl group. Consider the double bond to be a single bond

and add a carbon atom to each end of the double bond (Organic Nomenclature: A Programmed Study Guide

for Allinger, Organic Chemistry, 2nd Edition).

; below is an example of a quaternary

ammonium ion, which has attached to it a

benzyl group (PhCH2—),

----------------------------------------------

S

≡

Since the priority numbers 1 → 3 read in a counter clockwise direction, the configuration is S.

Thus, an addition can be made to the previous priority assignment, from highest to lowest:

(highest) ---------------------------------------------------------------------------> (lowest)

CH3CH— > PhCH2— > CH3CH2CH2— > CH3CH2— > CH3— > H

iso-propyl > benzyl > propyl > ethyl > methyl > hydrogen atom

____________________________________________________________________________________________

RULE 5 - Compounds that contain more than one chiral center, require assignment at each chiral center.

For example, the carbon atom that has a OH and the carbon atom that has a

Cl atom are both chiral.

OH

Cl

CH3CHCHCH3

C H3, 4

CH2CH2C H3, 2

CH2C H3, 3

1, PhCH2

CH3

C H3

CH2CH2C H3

CH2C H3

H

N+

CH2

C

C

Y Y

C

C H


RULE 5 - Compounds that contain more than one chiral center, require assignment at each chiral center.

a b

A B

Two interchanges for each center.

A1 A2 B1 B2

Since the priority numbers 1→ 3 read counter clockwise, both centers are the S configuration. Also, see appendix for

further discussion on changing one of the chiral centers to the R configuration and the resulting stereochemistry.

__________________________________________________________________________________________

RULE 6 - Compounds that contain double bonded atoms.

If a substituent has an atom that is in a double bond with another atom, each atom in the double bond are

replicated, i.e., for each atom in the double bond, consider the double bond to be a single bond and attach to

each atom, an atom corresponding to the atom to which it is double bonded to.

For example #1, if an ethenyl (vinyl) group is a substituent that is attached to another carbon atom (X), which

is a chiral center.

---------------------- continued on next page ---------------------------------------

H

X

C = C

H

H

C

H

C

H H

C C

X

a CH3

Cl H

OH H

CH3

b

1, Cl

2, Cb

3, CH3

4, H

4, H

1, Cl

2, Cb 3, CH3

4, H

1, Cl

3, CH3

2, Cb

1, OH

2, Ca

3, CH3

4, H

1, OH

4, H

2, Ca 3, CH3 4, H

1, OH

3, CH3

2, Ca


RULE 6a - Consider the following example. The diagram to the right will allow for the priority assignment.

The OH and hydrogen atom are highest (1) and lowest (4) priority respectively. A decision has to be made

with respect to the ethyl and ethenyl group as to assignment of priority numbers.

The first carbon of the ethyl group has one carbon atom and two hydrogen atoms, whereas, the first carbon of

the ethenyl group has two carbon atoms and one hydrogen atom, thus will be assigned a higher priority

number, i.e., #2 and the ethyl group will be assigned #3.

Since the priority numbers read 1, 2 & 3 in a counter clockwise direction, the designation is the S

configuration.

RULE 6b - A carbonyl group connected to a chiral center. Y is the chiral center and X is another substituent.

H

OH

CH2CH3

CH = CH2

C C

H

OH

CH2CH3

CH - CH2

H, 4

OH, 1

3, CH2CH3

2, CH = CH2 4 2

3

1

1

2

3

4

OH, 1

H, 4

2, CH = CH2

3, CH3CH2

C = O

X

Y

O C

X

Y

O C

2 interchanges


RULE 6b - Example #1, a carbonyl group connected to a chiral center.

The hydrogen atom are assigned #4. The carbon atom of the ethyl

group has 1 carbon and 2 hydrogen atoms, whereas, the methyl group

has 3 hydrogen atoms; thus, the methyl group is assigned priority #3.

The question arises with respect to the ethyl and carbonyl substituents.

Since the first carbon out in the ethyl group has 1 carbon atom

and 2 hydrogen atoms and the first carbon out for the carbonyl

substituent has two oxygen atoms and one carbon atom, the

carbonyl substituent will be assigned the higher priority number.

Thus, the chiral carbon atom is the R configuration.

________________________________________________________

RULE 6b - The following section examines several carbonyl containing substituents and assigns relative

priorities. The symbol Y represents a chiral center.

A B C D

CH3

C = O

CH3

C

H

CH3CH2

CH3

C - O

CH3

C

H

CH3CH2

O C

2, CH3CH2

4, H

3, CH3

C = O, 1

CH3

3, CH3

2, CH2CH3

4, H

C = O, 1

CH3

C = O

H

Y

C = O

CH3

Y

C = O

OH

Y

C = O

O - CH3

Y


A1 A2 If one compares A2 to B2, the carbon atom connected

to atom Y in A2 is connected to 3 oxygen atoms and

in B2, the carbon atom is connected to 2 oxygen

atoms and 1 carbon atom. Thus, A2 will have a

higher priority than B2.

If one compares A2 to C2, the carbon atom connected

to atom Y in A2 is connected to 3 oxygen atoms and

B1 B2 in C2, the carbon atom is connected to 2 oxygen

atoms and 1 hydrogen atom, thus, A2 will have a

higher priority than C2.

If one compares B2 to C2, the carbon atom connected

to atom Y in B2 is connected to 2 oxygen atoms and

1 carbon atom and in C2, the carbon atom is

connected to 2 oxygen atoms and 1 hydrogen atom,

C1 C2 thus, B2 will have a higher priority than C2.

If one compares D2 to A2, the carbon atom connected

to atom Y in D2 is connected to 3 oxygen atoms and

D1 D2 in A2, it is as well connected to 3 oxygen atoms.

Thus, one must examine the next atoms out. In D2,

the two oxygen atoms are connected to 2 carbon

atoms and in A2, the two oxygen atoms are connected

to a hydrogen atom and 1 carbon atom; thus, D2 will

have a higher priority than A2.

Thus, the relative order of priority of the groups will be:

D1 A1 B1 C1

> > >

(highest) ------------------------------------------------------------------------------> (lowest)

Methoxycarbonyl (D1) > carboxy (A1) > acetyl (B1) > formyl (C1)

(Carbomethoxy) ____________________________________________________________

C = O

OH

Y

C O

OH

Y

O C

C = O

CH3

Y

C = O

H

Y

C = O

O - CH3

Y

C O

H

Y

O C

O

CH3

Y

O C

C

C O

O - CH3

Y

O C

C = O

O - CH3

Y

C = O

OH

Y

C = O

CH3

Y

C = O

H

Y


RULE 6b - Example #2 - Glyceraldehyde. Glyceraldehyde has 3 carbon atoms and carbon #2 is chiral.

The hydrogen atom will be assigned the lowest priority, i.e., #4.

E The other 3 groups can be compared by their structure in order to determine their

priority number. In this case, Y is considered to be the chiral atom. This is also

designated D-glyceraldehyde (because the OH is drawn to the right, ref Mathews,

van Holde, Biochemistry, page 262).

E1

E2

E3

Since the first atom out for the hydroxyl group is an oxygen atom, it will be assigned the highest priority, #1.

For the formyl group (CHO), the first carbon atom out has a hydrogen and two oxygen atoms attached to it,

whereas, the hydroxymethyl group (—CH2OH) has two hydrogen atoms and one oxygen atom. Thus, the

formyl group will be assigned a higher priority number, i.e. #2 and the group (—CH2OH) is #3.

Thus: OH > CHO > CH2OH

E

D - glyceraldehyde {D = dextrarotatory, (+)} L - glyceraldehyde{L = levarotatory, (—)}

R - glyceraldehyde S - glyceraldehyde _________________________________

Y

2 interchanges

4, H 1, OH

3, CH2OH

2, CHO

3, CH2OH

4, H

1, OH

2, CHO

1, OH

3, HOCH2

2, CHO

4, H

1, OH

2, CHO

3, CH2OH 4, H

H OH

CHO

CH2OH

OH O H Y

CH2 CH2OH OH

Y Y

C = O

H

Y

C O

H

Y

O C


RULE 6b - Relative priority. Atomic numbers in decreasing order : oxygen (8), nitrogen (7) and carbon (6),

highest --------------------------------------------------------------------------------------------------------- lowest

—OR > —OH > —NH2 > — CO2H > — CHO > —CH2OH > —CH3 > H

alkoxy > hydroxy > amino > carboxy > formyl > hydroxymethyl > methyl > H

(ref: Mathews and van Holde, Biochemistry, page 263).

RULE 6c - Example #3 - Lactic acid, an example of another carbonyl containing compound. Structure F is

given in Allinger (2nd Edition, Organic Chemistry, page 118) and is designated the R configuration.

Place all four groups around the chiral center and assign priority number. The hydrogen atom will be assigned

the lowest priority number, i.e., #4.

The chiral atom is attached to an oxygen in the hydroxy group (OH), thus, it will be assigned the highest

priority. The methyl and carboxy group are attached to the chiral center through a carbon atom.

The carboxy group (COOH) will be higher priority than the methyl group, because the carboxy group is

connected to at least 1 oxygen atom; whereas, the methyl group is attached to only hydrogen atoms.

In structure F, the OH and COOH are in the plane of the page, hydrogen is behind and methyl is infront.

F F1

F2 F3 R

Since the priority numbers read 1, 2 & 3 in a clockwise direction, the designation is the R configuration. A

molecular model of structure F3 can be shown to be identical to structure F, upon rotation of F3 to obtain OH

and COOH in the plane of the page as in structure F. Alternatively, structure F already has the hydrogen atom

behind the page, thus, one can assigne priority numbers and directly determine the configuration.

The next page demonstrates the caveat of using the mathematical method of manipulating 4 different groups

around a Fischer projection in order to assign R and S configurations. The method allows one to randomly put

the 4 different groups around the Fischer projection, but after one has determined a designation, a tetrahedral

structure must be drawn (and preferably build a molecular model) in order to show that it is identical to the

chiral center in the molecule in question.

2 interchanges

H, 4

CH3, 3

OH, 1 COOH, 2 H, 4

OH, 1

COOH, 2

CH3, 3

OH

COOH

CH3 H

OH, 1

COOH, 2

CH3, 3 H, 4

C

COOH

OH

CH3

H


The following demonstrates that caveat.

B1 B2 B3 R

C1 C2 C3 S

D1 D2 D3 R

E1 E2 E3 S

The B1 to E1 series show 4 examples where four substituents are randomly placed around the Fischer

projection (cross hairs). The B2 to E2 series are manipulated to put the lowest priority number, #4, at the

bottom). The B3 to E3 series are just the 2dimensional depictions of the B2 to E2 series.

As one can see from the above diagrams, if one were to randomly place four substituents around a Fischer

projection, using the mathematical method of manipulating the groups, one has a 50/50 chance of getting an R

or S configuration. Thus, once you have obtained the tetrahedral structure (or molecular model), compare it

back to the chiral center in question, i.e., compound F.

You will find that both B3 and D3 are consistent with F (as well as F3); whereas, C3 and E3 are not consistent

with F (and neither with F3). Thus, molecule F in rule 6c has a R configuration, which is consistent with that

stated in Allinger.

OH, 1

COOH, 2

CH3, 3

H, 4

1

2 3

4

OH, 1

CH3, 3

H, 4

COOH, 2

2 interchanges

H, 4

CH3, 3

OH, 1

COOH, 2

CH3, 3

H, 4

OH, 1

COOH, 2

1

2 3

4

0 interchanges

CH3, 3 OH, 1

COOH, 2

H, 4

1

2

3

4

2 interchanges

H, 4

CH3, 3

OH, 1

COOH, 2

H, 4

CH3, 3

OH, 1

COOH, 2

OH, 1

COOH, 2

CH3, 3

H, 4

1

2 3

4

2 interchanges


Thus, a simple set of rules for using the mathematical method follows below:

STEPS: (1) Place four substituents around the Fischer projection.

(2) Assign priority numbers from 1 (highest) to 4 (lowest).

(3) Manipulate the numbers 1 through 4 so the number 4 is at the bottom of the

Fischer projection.

(4) Assign a configuration as either R (1 → 3 reads in a clockwise direction) or S (1 → 3

reads in a counter clockwise direction).

(5) Draw the 2 dimensional depiction of the tetrahedral (or build a model) and compare

back to the chiral center in the compound in question, e.g.,

- if one obtained the R configuration in steps 1-4 and the tetrahedral structure is

consistent with the chiral center in the compound in question, then the chiral center in

the compound in question is R.

- if one obtained the R configuration in steps 1-4 and the tetrahedral structure is not

consistent with the chiral center in the compound in question, then the chiral center in

the compound in question is S.

________________________________________________________

RULE 6d - Another carbonyl containing compound.

The following compound contains a Deuterium atom, D (higher mass isotope of hydrogen). In the

assignment of R or S, the higher mass isotope of an element will be assigned a higher priority number.

The hydrogen atom will be assigned the lowest priority number, #4

and the deuterium will be assigned #3.

The only question that arises is the priority of the other two

substituents. The —COOH substituent is a carboxy and the

—COOCH3 is a methoxycarbonyl, as described earlier in

rule 6b, thus, assigned #2 and #1 respectively.

Since the priority reads 1 → 3 clockwise, the configuration is R.

R

The following provides the priority numbering for a series of oxygen containing substituents and it is shown

that the following order is obtained:

highest ----------------------------------------------------------------------------- lowest

acetoxy > ethoxy > methoxy > hydroxy > methoxycarbonyl > carboxy

---------------------------- continued on next page --------------------------------

H COOCH3

D

COOH

2 interchanges

1

2

3

4

1

2 3

4

1, COOCH3

4, H

3, D

2, COOH


RULE 6e - Relative priority numbering for some oxygen containing substituents (groups):

(i) acetoxy

(ii) ethoxy

(ii) methoxy

(iii) hydroxy

(iv) methoxycarbonyl

(vi) carboxy

____________________________________________________________________________________

RULE 7a - Nitrogen containing groups, example #1: 2-butanamine (page 118, Allinger).

The lowest priority is the hydrogen atom, i.e., #4.

A Based on increasing atomic number in order C(6), N(7) & O(8), nitrogen will

be assigned the highest priority; thus, NH2 substituent is assigned #1.

From the previous discussion, ethyl will be assigned a higher priority than

methyl.

—OCH2CH3

O

—O —CH2CH3

O

O—CH3

—C = O

O—CH3

C O

O C

CH3

—O—C = O

CH3

C O

O C

— O

—O —CH3

O

—OCH3

O —OH —O—H

C = O

OH

C O

OH

O C

H, 4

NH2 , 1

CH2CH3 , 2

CH3 , 3 H, 4

NH2 , 1

CH2CH3 , 2

CH3 , 3

C

NH2

CH3

H

CH2CH3


RULE 7a - Nitrogen containing groups, example #1: 2-butanamine, continued,

S S

B C

which is equivalent to →

Since the priority numbers 1 → 3 read in a counter clockwise direction, the configuration is S.

If the tetrahedral structure B is rotated so that the ethyl and amino are in the plane, one obtains structure C.

Structure C is consistent with the compound depicted by structure A. Thus, the molecule is S-butanamine.

RULE 7b - Nitrogen containing substituents, example #2: 2-amino-1-propanol:

S

R

(R)-2-amino-1-propanol (S)-2-amino-1-propanol

______________________________________________________________________________________________

RULE 8a - Amino acids

Amino acids have the following composition of substituents:

- amino group (NH2)

- carboxyl group (COOH)

- α hydrogen atom

- a side chain (R)

1, NH2

3, CH3

4, H

2, CH2OH

CH2OH

H

CH3 OH, 1

COOH

CH3

H, 4

NH2

2, CH2OH

4, H

3, CH3

1, NH2 2, CH2OH

4, H

3, CH3

1, NH2

NH2 , 1

CH3 , 3

H, 4

CH2CH3 , 2 CH2CH3 , 2

H, 4

CH3 , 3

NH2 , 1

HOCH2 , 2

H, 4 CH3 , 3

NH2 , 1

R

H

COOH H2N

2 interchanges


RULE 8a - Amino acids, example #1, Alanine (Ala, A), where R = methyl

A

Since the priority numbers 1 → 3 read in a clockwise direction, the Fischer projection

in A is the R configuration.

A B

D - Alanine L - Alanine

R - Alanine S - Alanine

It should be noted that when R = H (hydrogen atom), the amino acid is Glycine (Gly, G) and since it has only 3

different substituents, it is not chiral (or is termed achiral).

The natural amino acids have the same configuration as natural S-glyceraldehyde, which is designated L. Of

the common 20 amino acids, another 18 have the L designation and the S configuration. The one exception in

the 20 common amino acids is Cysteine (Cys, C), (ref: Benoiton, Chemistry of Peptide Synthesis, 2006, page

1), which has the L designation, but the R configuration, as shown on the next page.

RULE 8a - Ala (A), a second way of visualizing the tetrahedral structure of Alanine.

D - Alanine L - Alanine

R - Alanine S - Alanine

H

COOH H2N

CH3 OH, 1

COOH

CH3

H, 4

H, 4

2, COOH 1, H2N

3, CH3

2 interchanges

H, 4

3, CH3

2, HOOC 1, NH2

3, CH3

2, HOOC

H, 4

1, NH2

3, CH3

H, 4

1, H2N

2, COOH

COOH

NH2 C

CH3

H

COOH

NH2 C

CH3

H


RULE 8b - Cysteine (Cys, C)

{carboxy} {side chain containing a sulfhydryl group}

;

The first atom out for each of the above substituents is a carbon atom. The

carboxy group is attached to 3 oxygen atoms; whereas, the carbon atom of the

side chain is attached to 2 hydrogens and a sulfur atom.

Based on the atomic numbers: C(6), N(7), O(8) & S(16), the sulfur atom will

be assigned the highest priority number. Thus, in Cys, the side chain will be

assigned priority #2 and the carboxyl group will be assigned #3.

L - Cysteine D - Cysteine

R - Cysteine S - Cysteine

Thus, the L & D designations do not necessarily correspond to the S & R configurations. This apparent

anomaly is also seen for fructose as well. Thus, the L & D system is a designation that is based on a

physicochemical measurement and is relative to a reference compound, whereas, the S & R system is absolute

(Mathews & van Holde, Biochemistry, page 263).

---------------------------- continued on next page -----------------------------------

C O

OH

O C

CH2 SH

2, CH2SH

H, 4

3, COOH

1, H2N

CH2SH

H, 4

COOH 1, H2N

2 interchanges

2, CH2SH

H, 4

3, COOH

1, H2N 2, CH2SH

H, 4

3, HOOC

1, NH2

1, NH2

3, HOOC

H, 4

2, CH2SH 2, HSCH2

H, 4

1, NH2

3, COOH


As well, an addition to the priority ranking for substituents is:

highest --------------------------------------------------------------------------- lowest

—SH > —OH > —NH2 > — CO2H > — CHO > —CH2OH

RULE 8c - Amino acids with two chiral centers: Threonine (T) and Isoleucine (I)

D-Threonine The amino acids are named as the carboxylic acids and the IUPAC/IUB name for threonine is:

2-amino-3-hydroxybutanoic acid.

(Nomenclature & Symbolism for Amino Acids & Peptides, 1984, Pure & Applied Chemistry,

56, p. 603).

The superscripts in red font designate the numbering of the carbon atoms for

the butanoic acid. Carbon #2 (C2) and carbon #3 (C

3) are chiral.

The Fischer projections for each chiral center are presented below.

For the chiral center at C2, the amino group and hydrogen atom are assigned

priority numbers #1 and #4 respectively.

Chiral center at carbon #2 A B

C2

Structure A is the carboxy group and structure B is carbon #3.

The carboxy substituent, C1, has 3 oxygens attached to it;

whereas, C3 has only 1 oxygen atom attached to

it. Thus, the carboxy group will be higher priority than C3.

Thus, C1OOH is assigned #2 and C

3 is assigned #3.

Chiral center at carbon #3

C3 For the chiral center at C3, the hydroxy group and hydrogen

atom are assigned priority numbers #1 and #4 respectively.

Since C4 has only hydrogen atoms attached to it,

it is assigned a lower priority than C2. See next page.

Thus, C2 is asigned #2 and C

4 is assigned #3.

1, OH

C4H3

C3 —

C2 — H2N —

H

H — OH

C1OOH

C 4H3

4, H

1, H2N C1OOH

C3

C3 — H — OH

CH3 OH, 1

COOH

CH3

H, 4

C1 O

OH

O C

4, H C2


RULE 8c: D-Threonine (T)

C2

R

C2 Since the priority numbers 1 → 3 read in a clockwise direction, the

configuration at C2 is the R configuration.

C3 C3 S

Since the priority numbers read in a counter clockwise direction, the configuration at C3 is the S configuration.

If one were to take C2 as it is drawn and rotate C3 so that C2 and C

4H3 are inverted, and then place C3 on

top of C2, one will get the figure below at the left. Combined, they result in the figure to the right.

C3: S D1

D-Threonine

C2: R Thus, will be named: (2R, 3S)-2-amino-3-hydroxybutanoic acid.

As reported in the IUPAC/IUB document, p. 603, is also named

D-Threonine.

2, C2

2 interchanges

4, H

1, NH2

3, C3

2, C1OOH

4, H

1, H2N 2, C1OOH

3, C3

4, H

1, H2N C1OOH

C3

2, C1OOH

3, C3

4, H

1, NH2

4, H

1, NH2

3, C3

2, C1OOH

1, OH

3, C4H3

4, H 2, C2

4, H

3, C4H3

2, C2

1, OH 4, H

1, OH

2, C2

3, C4H3

2 interchanges

3, C4H3

4, H

1, OH 1, OH

4, H 3, C4H3

4, H

1, NH2

2, C1OOH

R, C2

S, C3


RULE 8c - Amino acids with two chiral centers, D-Threonine continued:

If structure D1 above is rotated around the C2 to C

3 single bond so that the amino group and the methyl

group are in the plane of the page (amino trans to methyl), one can draw two alternate

representations, D1a (amino and methyl in the plane of the page) & D1b (as viewed from C3 to C

2).

(2R, 3S)-2-amino-3-hydroxybutanoic acid

D-Threonine

D1a D1b

≡

D-Threonine

------------------------------------------------------------------------------------------------------------------------

L-Threonine: Furthermore, it can be shown that L-Threonine (IUPAC/IUB, p. 603) would have the

following structures and would be named as: (2S, 3R)-2-amino-3-hydroxybutanoic acid.

L-Threonine E1

E1a E1b

L-Threonine

____________________________________________________________________________________

It should be noted that (2R, 3S)-2-amino-3-hydroxybutanoic acid (D-Threonine) and (2S, 3R)-2-amino-3-

hydroxybutanoic acid (L-Threonine) are molecules of the same molecular formula and structural formula as

the (2S, 3S) & (2R, 3R) forms; thus, are diasteriomers of the (2S, 3S) & (2R, 3R) forms.

------------------ continued on next page ---------------------------

OH

H2N

C1OOH

C4H3

S, C3

R, C2

NH2

C1OOH H

OH

H

C 4H3

OH

H2N

C1OOH

C4H3

R, C3

S, C2

1, OH

4, H

3, C4H3

4, H

1, NH2

2, C1OOH

S, C2

R, C3

NH2

HOOC1 H

OH

H

C 4H3


RULE 8c - Amino acids with two chiral centers: Isoleucine (I)

D-Isoleucine: Furthermore, it can be shown that D-Isoleucine (IUPAC/IUB, p. 603) would have the

following structures and would be named as: (2R, 3 R)-2-amino-3-methylpentanoic acid.

In structure F1, the carboxyl group and methylene group are in the plane of the page.

Chiral centers C2 and C

3 have been analysed through structures F1b & F1a respectively.

D-Isoleucine

F1a

F1

F1b

F1c

F1d

D-Isoleucine

____________________________________

The next section examines L-Isoleucine for its stereochemistry at the two chiral centers.

---------------------- continued on next page ----------------------------

HOOC1

H

H

H2N

C 4H2

C 5H3

CH3

C3

C2

3, C3

4, H

2, HOOC

1, NH2

R

R

2, C4H2CH3

4, H

1, C2

3, CH3

NH2

C1OOH

H

H C H3

C4H2CH3

C2, R C3, R

H NH2

C1OOH

CH3 H

H3C5

C4


L-Isoleucine: Furthermore, it can be shown that L-Isoleucine (IUPAC/IUB, p. 603) would have the

following structures and would be named as: (2S, 3S )-2-amino-3-methylpentanoic acid.

In structure G1, the carboxyl group and methylene group are in the plane of the page.

Chiral centers C2 and C

3 have been analysed through structures G1b & G1a respectively.

L-Isoleucine

G1a

G1

G1b

G1c G1d

L-Isoleucine

In structure G1c, C2 to C

4 are all in the plane of the page. In structure G1d, the view is from C

3 to C

2.

It should be noted that (2S, 3R) & (2R, 3S) forms are molecules of the same molecular formula and

structural formula, but with different stereochemistry, compared to L-Isoleucine {(2S, 3S )-2-amino-3-

methylpentanoic acid} and D- Isoleucine {(2R, 3R)-2-amino-3-methylpentanoic acid.}; thus, are diasteriomers

of L-Isoleucine and D- Isoleucine.

_____________________________________________________________________________________________

RULE 9 - Sulfoxides

Sulfur containing compounds may become oxidized and one oxidized form is the sulfoxide, (Organic

Nomenclature: A Programmed Study Guide for Allinger, Organic Chemistry, 2nd Edition, page 51).

The next two examples show the designation of stereochemistry for 2 oxidzed sulfur atoms.


C2

C3

HOOC1

H

H

NH2

C 4H2

C 5H3

CH3 S

2, C4H2CH3

4, H

1, C2

3, CH3

NH2

C1OOH

H

H C H3

C4H2CH3

3, C3

4, H

2, C1OOH

1, H2N

S

C4

C2, S C3, S

H NH2

C1OOH

CH3 H

H3C5


RULE 9a - Sulfur/oxygen compounds.

A The following compound is provide by Allinger (A Programmed

Study Guide for Allinger, Organic Chemistry, 2nd Edition, page 51).

For the purposes of assigning priority numbers, the sulfoxide can be

considered to be one of the resonance forms with the following

electronic structure, A1 (Allinger, Organic Chemistry, 2nd Edition,

page 221).

A1

The resonance form can be rewritten as two projection formulas, i.e.,

A2 & A3.

A2 A3

From rule 3c, it can be seen that A3 can be represented as A4.

The lone pair of electrons will be assigned the lowest priority, #4. A4

The OCH3 is considered to be:

Similarly, (—O — ) is:

The OCH3 has 2 lone pair of electrons and a carbon

atom; whereas, the (—O —) only has lone pairs of

electrons, thus, the OCH3 will be assigned higher priority, A5

i.e., #1 and (—O —) will be #2. The methyl group will be #3. S

Since the priority numbers 1 → 3 read in a counter clockwise

direction, the configuration of A5 is S.

The next section examines the side chain of Methionine.


O

OCH3 S

●●

CH3

║

O —

OCH3 S+

●●

CH3

O —

O

— O— CH3 ●●

●●

CH3

S+

●●

CH3

S+

●●

CH3 OCH3

O —

CH3

●●

OCH3

O —

— O— CH3 ●●

●●

●● — O

●● ●●

—

3, CH3

●●, 4

OCH3, 1

O —, 2


RULE 9b - Sulfoxides:

The amino acid Methionine (M, Met) has a sulfur atom and it can be oxidized to two different forms, i.e.,

Methionine sulfoxide or methionine sulfone (Darbre, Practical Protein Chemistry, p. 250).

The example in B is the side chain of methionine sulfoxide, only the alpha carbon atom (Cα, atom

connected to the carboxyl & amino group) and the side chain are shown.

side chain: [ —CαCH2CH2—S—CH3 ]

B B1

Since the priority numbers 1 → 3 read in a counter clockwise direction, B2

the configuration of B2 is S. S

______________________________________________________________________________________

Section II

(1) Other Stereochemistry related issues.

An excellent recent review that discusses the types of impurities that can be found in peptide drug

products is cited below:

D'Hondt, Bracke, Taevernier, Gevart, Verbeke, Wynendaele & Spiegelier, Related impurities in

peptide medicines, (2014), Journal of Pharmaceutical and Biomedical Analysis, 101, p. 2-30.

(i) Impurities in synthetic peptide drug products via solid phase peptide synthesis (SPPS)

One topic related to the impurities of synthetic peptides included racemization of amino acids during

the coupling step. Natural amino acids are L-amino acids (S configuration), with the exceptions of

glycine (achiral) and cysteine (R configuration).

During the coupling step, a very small percentage of amino acids can racemize to the D-amino acid (R

configuration), thus leading to an impurity in the drug product. Since naturally occurring peptides are

all L-amino acids, the enantiomer peptide would be all D-amino acids, which is the

non-superimposible mirror image of the all L-amino acid peptide.

If one or more of the amino acids racemize during the synthesis, the peptide will contain one or more

D-amino acids and the peptide impurity would then be termed the diasteriomer of the L-peptide (as

discussed above: diasteriomers are molecules of the same molecular formula and structural formula,

but with different stereochemistry). If only one amino acid has racemized, then the peptide impurity

would be a diasteriomer, but is additionally termed an epimer.

O

CH3 S

●●

║

—Cα CH2CH2

S+

●●

CH3

O —

—Cα CH2CH2

4, ●●

3, CH3

1, O —

2, —Cα CH2CH2


Section II

(1) (i) Impurities in synthetic peptide drug products via solid phase peptide synthesis (SPPS), continued,

An example of this effect is reported in the Proceedings of the 16th American Peptide Symposium,

Large scale synthesis of a 27-mer: Identification and suppression of a diasteriomer contaminant,

p. 131-132. In this case, an Asp residue racemized, resulting in the diasteriomeric impurity. The

paper further discusses ways to reduce the racemization and thus improve the purity of the target

peptide.

A recent report also discuss the improvement of methodologies for suppressing racemization of

amino acids during peptide synthesis: Peptide synthesis beyond DMF:THF & ACN as excellent and

friendlier alternatives, Organic and Biomolecular Chemistry, 2015, 13, 2393.

(1) (ii) Racemization of amino acids during acid hydrolysis of a peptide

During acid hydrolysis of a peptide (for the purposes of analytical determination), a small percentage

of amino acids will racemize, i.e., the L-amino acid will be transformed into the enantiomer, the

D-amino acid. There are methods that can circumvent that issue and allow for determination of the

chiral purity of each amino acid in a batch of peptide, (Goodlett et. al., (1995), Peptide chiral purity

determination, J. Chromatography A, 707, p.213-244).

(1) (iii) Asparagine deamidation and aspartate isomerization

During storage of a peptide pharmaceutical, a peptide may degrade in various ways. One example is

the deamidation of asparagine to form aspartate. The degradation goes through a succinimidyl

intermediate and upon ring opening, can form an L-Asp residue & L-iso-Asp residue (peptide chain

goes through side chain). Furthermore, the succinimidyl intermediate can racemize to form the

D-succnimidyl intermediate and ring opening results in D-Asp & D-iso-Asp (ref: D'Hondt, 2014, J.

Pharm Biomedical Analysis).

Using a model peptide, VTPNGA, the ratio of the four different forms was determined. The following

percentages were reported for each of the different forms:

original peptide 0%

succinimidyl intermediates 0.14%

L-Asp peptide 15.3%

L-iso-Asp peptide 55.8%

D-Asp peptide 7.0%

D-iso-Asp peptide 21.7%

(ref: B.N. Violand & N.R. Siegel, Protein and Peptide Chemical and Physical Stability, in

R.E. Reid, Peptide and Protein Drug Analysis, Drugs and the Pharmaceutical Series, V101,

p. 258).

It is also shown that Asp residues can undergo degradation through the same succinimidyl intermediate

and form the same four different peptide impurities (Reid, p. 261). Also, Wankankar & Borchardt,

(2006), J Pharm Sci, 95, p.2321, reveiw the issue of deamidation of Asn and isomerization of Asp in

bulk pharmaceutical peptides.

Thus, it can be seen that a variety of peptide impurities can form upon degradation of Asn and Asp

residues in a peptide (or protein), resulting in L → D racemization, generating diasteriomeric peptide

impurites or the iso-amino acid versus a normal amino acid.

Thus, one can see the importance of understanding chiral purity of any peptide or protein product, whether

it is produced via SPPS or by recombinant methods.

--------------------------- continued on next page ---------------------------------


Section II

(2) Other chiral molecules.

(i) ephedrine

A A2 S

For A2, the HNCH3, will be assigned the highest

priority and the hydrogen will be the lowest.

Since C1 has oxygen and carbon substituents and the methyl

A3 group has only hydrogen atoms, C1

will be assigned a higher

priority than the methyl group.

Since A2 priority numbers 1 → 3 read in a counter-clockwise

direction, C2 will have the S configuration.

A4 R For A4, the hydroxyl and hydrogen atom will be assigned the

highest (1) and lowest (4) priority respectively.

C2

has the following substituents: C/N/H.

Phenyl has the following substituents: C/C/C.

Thus, C2

will be assigned the higher priority number, i.e., #2

and phenyl will be #3.

Since the priority numbers read clockwise, the configuration

at C1

is R.

Thus, the complete name is (1R, 2S)-ephedrine.

It should be noted that (1R, 2S)-ephedrine is a diasteriomer of (1R, 2R)-ephedrine & (1S, 2S)-ephedrine.

_____________________________

(ii) Methylacetylcholine, Methacholine (use for asthma treatment).

2-(acetyloxy)-N,N,N-trimethylpropan-1-aminium; (structure B).

Three things to notice:

B (i) the carbonyl group at the left of the molecule is a

acetyloxy (or acetoxy), which was described in

section 6e.

(ii) the methyl and acetoxy are in the plane of the

page and hydrogen is behind the page.

(iii) C2 is chiral.

CH3

C2

phenyl

C1

H

H

HO

NHCH3

4, H

3, CH3

1, HNCH3

2, C1

C2

H

HO

phenyl C2

4, H

1, HO

2, C2

H

C

C

, 3 C1

C1

—O

C2

C H3

N+(CH3)3

C1H2—

H CH3 —C

║

O


Section II

(2) (ii) Methylacetylcholine, continued,

The hydrogen atom will be assigned the lowest priority, #4 and the acetoxy substituent is the

highest priority, #1.

C1 is attached to a nitrogen atom and 2 hydrogen atoms, whereas, the methyl group has only

3 hydrogen atoms. Thus, C1 is assigned priority #2 and the methyl is #3.

B B1 S

Since the priority numbers read 1 → 3 in a counter-clockwise direction, the configuration is S.

_____________________________________

(iii) Carboxymethylcysteine (CMC) , p.70-72 Darbre)

Cysteine can be alkylated with iodoacetic acid to yield carboxymethylcysteine. From section

8b, it was shown that L-Cysteine is the R configuration.

As shown below, the alkylation does not change the configuration of the amino acid.

R-Cysteine R-Carboxymethylcysteine

______________________________________

----------------- APPENDIX on next page --------------------------

, 1 C2

C H3

N+(CH3)3

C1H2—

H

O

║

—O CH3 —C

O

║

—O CH3 —C

H, 4

N+(CH3)3; 2

C1H2—

C H3, 3

1, NH2

3, HOOC

H, 4

2, CH2SH

1, NH2

3, HOOC

H, 4

2, CH2SCH2COOH

iodoacetic acid


APPENDIX

Rule 5: Additional material.

If one were to rotate the Ca to Cb single bond in structure A so that the two methyl groups are

pointing downwards, and view the strcture from the Cb to Ca atom, structure , B , the orientations for

the Cl and OH can easily be seen.

A (S, S) @ (a, b) B (S, S) @ (a, b)

If one were to change either one of the chiral centers

, i.e., make the compound (R, S) or (S, R) at (a, b), then

the oreintation of the OH and Cl will change. C (R, S) @ (a, b)

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a

b

CH3

Cl H

OH H

CH3 CH3

CH3

Cl H

H OH

CH3

CH3

Cl H

H OH

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STEREOCHEMISTRY NOTES Table of Contents Stereochemistry Section I · · 2015-10-14STEREOCHEMISTRY NOTES Section I (1) The following post deals with assigning the stereochemistry