chapter 3 stereochemistry 2016

STEREOCHEMISTRY

Learning objectives• Recognize types of isomerism• Recognize and differentiate alkane and cycloalkane conformers• Recognize, draw and analyze open alkane structures using the

different types of projections (Newman, saw-horse and Fischer)• Draw cyclic alkane structures (chair and boat confirmations.)• Analyze and explain conformational stabilities of alkane and

cycloalkanes• Recognize and explain the concept of chirality• Define and explain key terms in optical stereochemistry• Determine R and S sequence in organic molecule• Determine trans and cis, E and Z sequence in organic molecule• Determine other stereoisomerism such as (+), (-); D & L and

meso.

Classification of isomers

4

• Recall that isomers are different compounds with the same molecular formula.

• The two major classes of isomers are constitutional isomers and stereoisomers.

Constitutional/structural isomers have different IUPAC names, the same or different functional groups, different physical properties and different chemical properties.Stereoisomers differ only in the way the atoms are oriented in space. They have identical IUPAC names (except for a prefix like cis or trans). They always have the same functional group(s).

• A particular three-dimensional arrangement is called a configuration. Stereoisomers differ in configuration.

The Two Major Classes of Isomers:

* REFER TO THE SLIDE #3

5

A comparison of constitutional isomers and geometric stereoisomers

Stereochemistry

Stereoisomers may be geometric (cis/trans) or optical.Optical isomers are chiral and exhibit optical activity.

Three main topics to be discussed here:

a) Stereoisomers - isomers that are different to each other only in the way the atoms are oriented in space.

b) Geometric isomers – isomers that owe their existence to hindered rotation about double bond

c) Conformation – different spatial arrangements of a molecule that are generated by rotations about single bonds.

Conformational analysis of straight chain alkanes•Conformational analysis is the study of how conformational factors affect the structure of a molecule and its properties•Rotation about single bond produces isomer that differ in conformation. •Conformers have same connection, interconverts rapidly, thus cannot be isolated.•Can be represented in 2 ways: sawhorse representation or Newman projection

H

H H

H

H H

HH

HH

H

H

rotate

Conformation

Conformation of ethane, CH3CH3

Two conformational isomers or conformers.Eclipsed form = all hydrogen atoms nearest to each other. Staggered form = all hydrogen atoms are furthest apart.

H

H

H H

H

H H

HH

HH

H

H

H

H H

H

H H

HH

HH

H

Eclipsed Staggered

Newman projection

Ball-stick formulasawhorse representation

rotate

Conformational analysis of ethane• Only two conformers.• Torsional strain: resistance to rotation.• For ethane, only 3.0 kcal/mol (or 12 kJ/mol)

• Similar rotation about the C-C single bond as in ethane.• There are two conformational isomer.• The potential energy different are about 3.3 kcal/mol, almost

similar to ethane. Slight increase due to bulkier methyl group.

Conformation of propane

For butane, C4H10, two different structures can be drawn, n-butane and isobutane.The two molecules are different and also having different physical properties:

Physical properties n-butane isobutaneb.p 0 oC -12 oCm.p. -138 oC -159 oCDensity 0.622 0.604Solubility in 100 ml water 1813 ml 1320 ml

Conformation of butane

n-Butane conformation•Consider the two central carbon atoms in the molecule.•6 different conformers can be formed.•60° rotation along central C-C bond

CH3

H H

CH3

HHCH3

H H

H

HH3C

CH3

H H

H

CH3H

Anti conformation (IV)

Gauche conformation (VI)

CH3

H H

H

HH3C

Gauche configuration (II)

Eclipsedconformation (III)

Eclipsedconformation (V)

Eclipsedconformation (I)

CH3

H H

H3C

HH

CH3

H H

H

CH3H

Conformational Analysis of n-Butane

Formed by the linkage of carbon atoms to form a ring.The most common are the five and six-membered rings.

When a carbon atom is bonded to four other atoms, bonding orbital (sp3) are directed towards the corner of tetrahedron.The angle between each pair of orbital is 109.5o with maximum orbital overlapped.

In cyclopropane, this angle cannot be achieved, but instead obtained 60o angle. Thus, orbital overlapped is greatly reduced. The molecule is unstable due to angle strain.

Conformation of cyclic compounds

Conformation of cyclohexaneThe most stable ring structure is cyclohexane which can achieved the maximum angle of 109.5o with maximum overlapped of bonding orbital. Strain free – no angle or torsional strain.There are several conformations of cyclohexane:

Chairconformation

Boatconformation

Twist-boatconformation

Half-chairconformation

Chair conformation – most stable

Conformational analysis of cyclohexane

Chair conformation is the most stable. Why???

H

H

H H

HH

Staggered

H

H HH

HH

HH

HH

1

234

5 6

H

H

H

H

H

HH

H

Looking through C4/C5and C2/C1

The conformation obtainsimilar to staggered form in ethane

H

H

All C-H bond are staggered (similar to staggered form of ethane)

Chair Conformer

Boat Conformer

Axial and equatorial bonds

In the chair conformation, there are two types of C-H bonds: the axial bond and the equatorial bond.

Axial bonds Equatorial bonds Axial and equatorial bonds

This can be visualized by imagining the carbon atoms located on a plane and there will be:6 C-H bonds above the plane6 C-H bonds below the plane

From this, 3 C-H axial bonds pointing upwards 3 C-H axial bonds pointing downwards 3 C-H equatorial pointing upwards 3 C-H equatorial bonds pointing downwards

Cis- when two group on the same face of the ringTrans - when two group on the opposite face of the ring(regardless whether axial or equatorial, or adjacent)

Axial and Equatorial Positions

Conformation of monosubstituted cyclohexaneExample: methylcyclohexane

•Axial and equatorial conformations are not equally stable at room temperature.•Substituents is always more stable in equatorial position•Energy diff. is due to 1,3-diaxial interaction

CH3

H HH

HH

HH

HH

1

234

5 6

H

H

H

H

H

CH3H

H

Looking through C2/C3and C6/C5H

1,3-Diaxial interaction

Hydrogen atoms of the axial methyl group on C1 are too close to the hydrogens three carbon away on C3 and C5

The degree of unstability is about 0.9 kcal/mol for each interaction

1,3-Diaxial Interactions

ring flip

No 1,3-diaxial interaction when CH3 group is equatorial.

Hence, this conformation is more stable since equatorial bonds are pointing towards the outside.

H

H HH

HH

HH3C

HH

1

234

5 6

H

H

H

H

CH3

HH

H

Looking through C2/C3and C6/C5H

At room temperature, most methylcyclohexane conformation (95%) occurs in which the methyl group is at uncrowded equatorial position.

Conformation of di-substituted cyclohexaneExample: 1,2-dimethylcyclohexane

Cis-1,2-dimethylcyclohexane -one CH3 equatorial and one CH3 axial in both chair conformation -each has two 1,3-diaxial interactions and one gauche butane interactionThe two conformations are exactly equal in energy.

CH3

1

234

5 6 CH3

axial-equatorialequatorial-axial

CH3

1

234

5 6CH3

cis-1,2-dimethylcyclohexane

H

H

H

H

CH3

1

234

5 6

CH3

axial-axial equatorial-equatorial

CH3CH3

1

234

5 6

trans-1,2-dimethylcyclohexane

more stable

H

H

H

H

Trans-1,2-dimethylcyclohexane - either both CH3 equatorial (diequatorial) or both CH3 axial (diaxial) - the diequatorial conformation only has one gauche butane interaction - the diaxial has four 1,3-diaxial interactions

Therefore, the diequatorial conformers is more favoured (has lower energy), occurs exclusively.

Conformation of di-substituted cyclohexaneExample: cis-1,3-dimethylcyclohexane

29

• Although everything has a mirror image, mirror images may or may not be superimposable.

• A molecule or object that is superimposable on its mirror image is said to be achiral (lacking-chirality).

• A molecule or object that is not superimposable on its mirror image is said to be chiral.

• Generally, a chiral carbon atom is sp3 with four different attachments.

Chiral and Achiral Molecules:Stereoisomers

30

• Some molecules are like hands. Left and right hands are mirror images, but they are not identical, or superimposable.

Chiral and Achiral Molecules:Stereoisomers

31

• We can now consider several molecules to determine whether or not they are chiral.

Chiral and Achiral Molecules:

Stereoisomers

32

• A carbon atom with four different groups is a chiral center.• The case of 2-butanol. A and its mirror image labeled B are

not superimposable. Thus, 2-butanol is a chiral molecule and A and B are isomers.

• Non-superimposable mirror image stereoisomers like A and B are called enantiomers .

Chiral and Achiral Molecules:Stereoisomers

33

• In general, a molecule with no stereogenic centers will not be chiral.

• With one stereogenic center, a molecule will always be chiral.

• With two or more stereogenic centers, a molecule may or may not be chiral.

• Achiral molecules usually contain a plane of symmetry but chiral molecules do not.

• A plane of symmetry is a mirror plane that cuts the molecule in half, so that one half of the molecule is a reflection of the other half.

Chiral and Achiral Molecules:

Stereoisomers

34

Chiral and Achiral Molecules:

Two identical attachments on an sp3 carbon atom eliminates the possibility of a chiral center.

Stereoisomers

35

Summary of the Basic Principles of Chirality

• Everything has a mirror image. The fundamental question is whether the molecule and its mirror image are superimposable.

• If a molecule and its mirror image are not superimposable, the molecule and its mirror image are chiral.

• The terms stereogenic center and chiral molecule are related but distinct. In general, a chiral molecule must have one or more stereogenic centers.

• The presence of a plane of symmetry makes a molecule achiral.

Stereoisomers

36

• To locate a stereogenic center, examine each tetrahedral carbon atom in a molecule, and look at the four groups—not the four atoms—bonded to it.

• Always omit from consideration all C atoms that cannot be tetrahedral stereogenic centers. These include

CH2 and CH3 groupsAny sp or sp2 hybridized C

Stereogenic Centers:Stereoisomers

37

• Larger organic molecules can have two, three or even hundreds of stereogenic centers.

Identifying of Stereogenic Centers:

Stereoisomers

38

• To draw both enantiomers of a chiral compound such as 2-butanol, use the typical convention for depicting a tetrahedron: place two bonds in the plane, one in front of the plane on a wedge, and one behind the plane on a dash. Then, to form the first enantiomer, arbitrarily place the four groups—H, OH, CH3 and CH2CH3—on any bond to the stereogenic center. Then draw the mirror image.

Drawing Stereogenic Centers - the wedge diagram:Stereoisomers

39

Three-dimensional representations for pairs of enantiomers

Drawing Stereogenic Centers - the wedge diagram:

Stereoisomers

40

• Stereogenic centers may also occur at carbon atoms that are part of a ring.

• To find stereogenic centers on ring carbons, always draw the rings as flat polygons, and look for tetrahedral carbons that are bonded to four different groups.

Contains a plane of symmetry

Identifying of Stereogenic Centers:Stereoisomers

41

• In 3-methylcyclohexene, the CH3 and H substituents that are above and below the plane of the ring are drawn with wedges and dashes as usual.

Drawing Stereogenic Centers - the wedge diagram:

Stereoisomers

42

• Identify the chiral carbons in the compounds below.

Stereochemistry

Identifying of Stereogenic Centers:

43

• In a Fischer projection of a chiral carbon and its mirror image:

horizontal bonds project toward the viewer and vertical bonds project away from the viewer.

• The test for non-superimposability is to slide one on top of the other or rotate 180o and attempt the same.

• Fischer projections of the two enantiomers of 2-butanol:

StereochemistryDrawing Stereogenic Centers – the Fischer Projection:

CH3

CH2CH3

H OH

CH3

CH2CH3

HO HThe chiral carbon atom is at the center of the crossed lines.

44

• Fischer projections of a compound with 2 chiral carbons, (two pairs of enantiomers).

• The maximum number of optical isomers is 2n. (where n = the number of chiral carbon atoms.)

The pairs are diastereomerically related.

Stereochemistry

Drawing Stereogenic Centers – the Fischer Projection:

CH3

OH

OH

COOH

H

H

CH3

H

H

COOH

HO

HO

CH3

OH

COOH

H

H

CH3

OH

COOH

H

HHO

HO

45

• However, there may be severaI different Fischer projections for the same compound depending upon the direction from which is is viewed.

Are these structures the same or different ?

Stereochemistry

Drawing Stereogenic Centers – the Fischer Projection:

CH3

CH3

CH3 CH3

CH2CH3

CH3CH2 CH2CH3

CH2CH3

CH2=CH

CH2=CH CH2=CH

CH2=CH

OH OHOH

HO

a b c d

This is a good place to use your models.

46

• The three dimensional arrangement about a tetrahedral carbon atom is referred to as its configuration.

• Early workers in the late 1800s including Fischer used the terms D and L to label the two molecules in a non-superimposable mirror image pair.

• D and L assignments were chemically related to the structures of glyceraldehyde.

• More recently Cahn, Ingold and Prelog developed the R and S system of assignment which is more convenient.

Labeling Stereogenic Centers:

Stereochemistry

47

• Since enantiomers are two different compounds, they need to be distinguished by name. This is done by adding the prefix R or S to the IUPAC name of the enantiomer.

• Naming enantiomers with the prefixes R or S is called the Cahn-Ingold-Prelog system.

• To designate enantiomers as R or S, priorities must be assigned to each group bonded to the stereogenic center, in order of decreasing atomic number. The atom of highest atomic number gets the highest priority (1).

Labeling Stereogenic Centers with R or S:

Stereochemistry

48

• If two atoms on a stereogenic center are the same, assign priority based on the atomic number of the atoms bonded to these atoms. One atom of higher atomic number determines the higher priority.

Stereochemistry

Labeling Stereogenic Centers with R or S:

49

• If two isotopes are bonded to the stereogenic center, assign priorities in order of decreasing mass number. Thus, in comparing the three isotopes of hydrogen, the order of priorities is:

Stereochemistry

Labeling Stereogenic Centers with R or S:

50

• To assign a priority to an atom that is part of a multiple bond, treat a multiply bonded atom as an equivalent number of singly bonded atoms. For example, the C of a C=O is considered to be bonded to two O atoms.

• Other common multiple bonds are drawn below:

StereochemistryLabeling Stereogenic Centers with R or S:

51

Figure 5.6 Examples of assigning priorities to stereogenic centers

Stereochemistry

Labeling Stereogenic Centers with R or S:

52

Stereochemistry

Labeling Stereogenic Centers with R or S:

53

Stereochemistry

Labeling Stereogenic Centers with R or S:

54

Stereochemistry

Labeling Stereogenic Centers with R or S:

55

Figure 5.7 Examples: Orienting the lowest priority group in back

Stereochemistry

Labeling Stereogenic Centers with R or S:

56

• For a molecule with n stereogenic centers, the maximum number of stereoisomers is 2n. Let us consider the stepwise procedure for finding all the possible stereoisomers of 2,3-dibromopentane.

Stereochemistry

Diastereomers:

57

• If you have drawn the compound and the mirror image in the described manner, you have only to do two operations to see if the atoms align. Place B directly on top of A; and rotate B 180° and place it on top of A to see if the atoms align.

• In this case, the atoms of A and B do not align, making A and B nonsuperimposable mirror images—i.e., enantiomers. Thus, A and B are two of the four possible stereoisomers of 2,3-dibromopentane.

StereochemistryDiastereomers:

58

• Switching the positions of H and Br (or any two groups) on one stereogenic center of either A or B forms a new stereoisomer (labeled C in this example), which is different from A and B. The mirror image of C is labeled D. C and D are enantiomers.

• Stereoisomers that are not mirror images of one another are called diastereomers. For example, A and C are diastereomers.

StereochemistryDiastereomers:

59

Figure 5.8 Summary: The four stereoisomers of 2,3-dibromopentane

Stereochemistry

Diastereomers:

60

• Let us now consider the stereoisomers of 2,3-dibromobutane. Since this molecule has two stereogenic centers, the maximum number of stereoisomers is 4.

Meso Compounds:

• To find all the stereoisomers of 2,3-dibromobutane, arbitrarily add the H, Br, and CH3 groups to the stereogenic centers, forming one stereoisomer A, and then draw its mirror image, B.

Stereochemistry

61

• To find the other two stereoisomers if they exist, switch the position of two groups on one stereogenic center of one enantiomer only. In this case, switching the positions of H and Br on one stereogenic center of A forms C, which is different from both A and B.

• A meso compound is an achiral compound that contains tetrahedral stereogenic centers. C is a meso compound.

StereochemistryMeso Compounds:

62

• Compound C contains a plane of symmetry, and is achiral. • Meso compounds generally contain a plane of symmetry so

that they possess two mirror image halves.

• Because one stereoisomer of 2,3-dibromobutane is superimposable on its mirror image, there are only three stereoisomers, not four.

StereochemistryMeso Compounds:

63

Figure 5.9 Summary: The three stereoisomers 2,3-dibromobutane

StereochemistryMeso Compounds:

64

• When a compound has more than one stereogenic center, R and S configurations must be assigned to each of them.

R and S Assignments in Compounds with Two or More Stereogenic Centers.

One stereoisomer of 2,3-dibromopentaneThe complete name is (2S,3R)-2,3-dibromopentane

Stereochemistry

65

• Consider 1,3-dibromocyclopentane. Since it has two stereogenic centers, it has a maximum of four stereoisomers.

Disubstituted Cycloalkanes:

• Recall that a disubstituted cycloalkane can have two substituents on the same side of the ring (cis isomer, A) or on opposite sides of the ring (trans isomer, B). These compounds are stereoisomers but not mirror images.

Stereochemistry

66

• To find the other two stereoisomers if they exist, draw the mirror images of each compound and determine whether the compound and its mirror image are superimposable.

• The cis isomer is superimposable on its mirror image, making the images identical. Thus, A is an achiral meso compound.

Stereochemistry

Disubstituted Cycloalkanes:

67

• The trans isomer is not superimposable on its mirror image, labeled C, making B and C different compounds. B and C are enantiomers.

• Because one stereoisomer of 1,3-dibromocyclopentane is superimposable on its mirror image, there are only three stereoisomers, not four.

Stereochemistry

Disubstituted Cycloalkanes:

68

Figure 5.10 Summary—Types of isomers

Stereochemistry

69

Figure 5.11 Determining the relationship between two nonidentical molecules

Stereochemistry

70

• The chemical and physical properties of two enantiomers are identical except in their interaction with chiral substances.

• The physical property that differs is the behavior when subjected to plane-polarized light ( this physical property is often called an optical property).

• Plane-polarized (polarized) light is light that has an electric vector that oscillates in a single plane.

• Plane-polarized light arises from passing ordinary light through a polarizer.

Optical Activity

Stereochemistry

71

• Originally a natural polarizer, calcite or iceland spar, was used. Today, polarimeters use a polarized lens similar to that used in some sunglasses.

• A polarizer has a very uniform arrangement of molecules such that only those light rays of white light (which is diffuse) that are in the same plane as the polarizer molecules are able to pass through.

• A polarimeter is an instrument that allows polarized light to travel through a sample tube containing an organic compound and permits measurement of the degree to which the light is rotated.

Optical Activity

Stereochemistry

72

• With achiral compounds, the light that exits the sample tube remains unchanged. A compound that does not change the plane of polarized light is said to be optically inactive.

Optical Activity

Stereochemistry

73

• With chiral compounds, the plane of the polarized light is rotated through an angle . The angle is measured in degrees (°), and is called the observed rotation. A compound that rotates polarized light is said to be optically active.

Optical Activity

Stereochemistry

74

• The rotation of polarized light can be clockwise or counterclockwise.

• If the rotation is clockwise (to the right of the noon position), the compound is called dextrorotatory. The rotation is labeled d or (+).

• If the rotation is counterclockwise, (to the left of noon), the compound is called levorotatory. The rotation is labeled l or (-).

• Two enantiomers rotate plane-polarized light to an equal extent but in opposite directions. Thus, if enantiomer A rotates polarized light +5°, the same concentration of enantiomer B rotates it –5°.

• No relationship exists between R and S prefixes and the (+) and (-) designations that indicate optical rotation.

Optical Activity

Stereochemistry

75

• An equal amount of two enantiomers is called a racemic mixture or a racemate. A racemic mixture is optically inactive. Because two enantiomers rotate plane-polarized light to an equal extent but in opposite directions, the rotations cancel, and no rotation is observed.

Racemic Mixtures

Stereochemistry

76

• Specific rotation is a standardized physical constant for the amount that a chiral compound rotates plane-polarized light. Specific rotation is denoted by the symbol [] and defined using a specific sample tube length (l, in dm), concentration (c in g/mL), temperature (250C) and wavelength (589 nm).

Stereochemistry

Racemic Mixtures

77

• Enantiomeric excess (ee) is a measurement of the excess of one enantiomer over the racemic mixture.

Enantiomeric excess and Optical purity: ee and op

ee = % of one enantiomer - % of the other enantiomer.

• Consider the following example: If a mixture contains 75% of one enantiomer and 25% of the other, the enantiomeric excess is 75% - 25% = 50%. Thus, there is a 50% excess of one enantiomer over the racemic mixture.

• ee is numerically equal to Optical Purity.• The optical purity can be calculated if the specific rotation []

of a mixture and the specific rotation [] of a pure enantiomer are known.

op = ([] mixture/[] pure enantiomer) x 100.

Stereochemistry

78

• Since enantiomers have identical physical properties, they cannot be separated by common physical techniques like distillation.

• Diastereomers and constitutional isomers have different physical properties, and therefore can be separated by common physical techniques.

Physical Properties of Stereoisomers:

Figure 5.12 The physical properties of the three stereoisomers of tartaric

acid.

Stereochemistry

79

• Two enantiomers have exactly the same chemical properties except for their reaction with chiral non-racemic reagents.

• Many drugs are chiral and often must react with a chiral receptor or chiral enzyme to be effective. One enantiomer of a drug may effectively treat a disease whereas its mirror image may be ineffective or toxic.

Chemical Properties of Enantiomers:

Stereochemistry

Geometric IsomerismIsomeration due to hindered rotation about C=C double bonds.A disubstituted alkene can have substituents either on the same side (cis) or opposite side (trans)

Example:The two isomers of 2-butene can have structure I with b.p +4oC and assign as cis configuration and structure II, b.p. +1oC and assign configuration trans.

H

CH3

H

H3C

H

CH3

H3C

H

cis-2-butene trans-2-butene

I II

The two isomers of 2-butene do not interconvert because free roation about the C=C is not possible.

Cis-trans isomerism occurs whenever both double bonded carbons are attached to 2 different groups.

H

H

H3C

H3C

H

CH3

H3C

H3C

Isobutylene 2-Methylbutene

Cl

Cl

H

H3C

1,1-Dichloropropane

No geometric isomerism

CCD

D

A

BCC

D

D

B

A= CC

C

D

A

BCC

C

D

B

A=

same isomer different isomer

(E) And (Z) SystemsDescribes the arrangement of substituents around a double bond

that cannot be described by cis-trans system.

Cahn-Ingold Prelog sequence rule:

1) For each double bond, rank its substituents by atomic number- atoms with high atomic number receive higher priority.

Br (35) > Cl (17) > O (16) > N (14) > C (12) > H (1)H

CH3

H3C

H

trans-2-buteneE-2-butene

Higher

Lower

Lower

Higher

Lower

Higher

Lower

Higher

E double bond(entgegen)higher priority groups on opposite sides

Z double bond(zusammen)higher priority groups on sameside

H

CH3

H

H3C

cis-2-buteneZ-2-butene

2) If a decision cannot be reached by ranking the first atom, look at the second, third or fourth atoms away from the double bond until first point of difference is found

CH

H

H CH

H

C

H

H

H

lower higher

CH

H

CH3

lower

CH

CH3

CH3

higher

O H

lower

O C

H

H

H

higher

CH

CH3

NH2

lower

CH

CH3

Cl

higher

3) Multiple bonded atoms are equivalent to the same number of single-bonded atoms

OCH

this O is bonded to C,C

this C is bonded to O,O,H

= CH O

OC

this O is bonded to C,C

this C is bonded to O,O,H

Geometric isomers in cyclic compoundsMost of the cyclic structures in organic molecules are either five- or six-bonded. Atoms bonded in cyclic structures are also not free to rotate about the single bond. By looking at the relative positions of the substituents on the ring, geometric isomer can also be assigned.

CH3

OH

CH3

OH

above the plane

below the plane

trans-3-Methylcyclohexanol

Other methods representing cyclic structures

CH3

CH3

CH2CH3

H3CH2C

C(CH3)3

C(CH3)3

cis-1,2-Dimethylcyclohexane trans-1,3-Diethylcyclohexane

cis-1,4-di-tert-butylcyclohexane

H3C

H3C

H

H

trans-1,3-Dimethylcyclopentane

Name the following compound. Determine the geometric isomerism of the double bond.

Exercise 3

H3C

CH3

OH

CH3

CH3

H3C

Cl H3C

H3C CH3

CHO

OHCOOH

NH2

CH3

a)

b)

c)

d)

e)

f)