Chem 351: Molecular models [PDF]

MOD.1

MOLECULAR STRUCTURES AND MODELS

Note: There is no need to write a pre-laboratory summary for this experiment, however you may want to start work on the content. The laboratory period will run more like a tutorial, it’s “open book”, you can work in groups and ask your TA about any concepts you don't understand. Your grade for this laboratory is assessed based on an individual quiz at the end of the laboratory period (run under examination conditions).

INTRODUCTION This experiment uses a molecular model kit to help address and clarify the theoretical concepts of covalent bonding and molecular structure. Molecular models are designed to reproduce molecular structures in three dimensions, allowing many subtle features concerning shapes of molecules (such as dipole moment, polarity, bond angle, and symmetry) to become clearer. Learning how to use a model kit correctly can help you to realise how they may be able to help you answer questions about molecular structure. Remember that you can use model kits during examinations to help you answer the questions we might ask. An important fundamental principle is that a molecule tends to position its atoms to give the arrangement with the lowest possible energy. This allows us to predict the shape of a molecule, and the subsequent physical and chemical properties to a very good approximation. In this laboratory period you will learn how to use your model kit to help answer questions and investigate:  the implications of hybridisation on molecular shape  aspects of isomerism  conventions used in 2-D representations of 3-D molecules.  counting types of atoms in a molecule (i.e. looking for equivalent groups, especially H and C)  index of hydrogen deficiency  chirality and chirality centers – R/S nomenclature  Enantiomers and diastereomers – E/Z and cis/trans designation  plane of symmetry, superimposable mirror images, enantiomers  meso compounds st

In preparation you should review the following concepts and terms from 1 year chemistry, 351 lectures and / or tutorial materials:  hybridisation and atomic orbital shape  alkanes, functional groups, constitutional isomers, conformational isomers  Newman projections; eclipsed/staggered, anti/gauche, potential energy diagrams  cis/trans, E/Z and R/S nomenclature  the implications of hybridisation on shape  structural flexibility

MOD.2  conventions used in the two dimensional representation of molecules and add that all important third dimension.  counting types of atoms in a molecule (i.e. looking for equivalent groups)  index of hydrogen deficiency

MOLECULAR MODELS The three dimensional shape of molecules results from the three-dimensional arrangements of their constituent atoms, and as such are often difficult to visualise in terms of a two-dimensional diagram on a page or computer screen. For this reason chemists often make use of molecular structure models (either physical models or computer models). In addition to the qualitative appreciation of molecular structure, scale models can be used to make approximate quantitative measurements. For this experiment you should use your own set of models if you have them. We have a small number of the Molymod Molecular Models that can be borrowed. From the Molymod Models we shall use the following components:

ATOMS Colour

Atom Black White Blue Red Green Orange

No. of Holes C H N O Cl Br

Bond Angle 4 1 4 4 2 1

109 28' 109 28' 109 28'

Atoms are joined together by inserting the appropriate bond into the holes in the atoms. The single short rigid bond should be used to represent sigma () bond. Two curved pieces should be used to represent a double bond and three curved pieces to represent a triple bond. Sometimes more than one sensible structure may be drawn for a particular molecular formula. In this case the arrangement of atoms must be determined experimentally. The different arrangements are said to be "ISOMERS" of each other. Depending upon the relationship of the structures, the pair of structures can be subcategorised as different types of isomers. This is schematically represented by the isomer tree diagram on the following page and is an important part of the materials covered by this experiment. At each branch in the tree, a “yes/no” question is asked in order to decide what path to follow.

The many different possible arrangements of the same set of atoms is the main reason for the enormous number (over a million) of known organic molecules. These different arrangements are possible since carbon has a singular ability to form very strong bonds with itself (as carbon chains or carbon rings), hydrogen atoms, or heteroatoms.

MOD.3

Do the compounds have the same molecular formulae ? NO

Not isomers

YES Isomers Do the compounds have the same connectivity ? YES

NO

Stereoisomers

Constitutional O

Can the compounds be interconverted by rotation about single bonds ?

OH

YES

NO

(skeletal, positional, functional)

Configurational

Conformational H 3C

Is the isomerism at a tetrahedral center ? NO

H

YES

H

H

CH 3

H

H H H 3C

H H CH 3

Optical

Geometric

Are the compounds non-superimposable mirror images ? YES

NO

Diastereomers H 3C

Cl H

H 3C

H Br

H 3C

Enantiomers Cl

H CH 3CH 2

H 3C

Cl

Cl H Br

H H CH 3 H 3C

CH 2CH 3

ISOMER TREE This figure helps you identify the type of isomer between a pair of structures. Note that some classes are not always mutually exclusive (e.g. technically geometric isomers and conformational isomers are also diastereomers). In general, it is usually best to use the more specific term.

EXPERIMENTAL PROCEDURE "Tutorial" work in pairs, open book. Work through the following tutorial questions using your model kit, text book etc. and record your answers, talking to your TA as you work through them. At the end of the laboratory period, you will be given a hands on assessment to be done individually. A.

CARBON:

MOD.4 Tetrahedral - sp3 carbon Since four single bonds are formed, the carbon atom is situated at the centre of a tetrahedron. This is the largest number of -bonds carbon can form and hence the carbon is termed a "saturated carbon". Construct an ethane molecule with the medium straight bonds and confirm that the carbon atoms are both at the centre of a tetrahedron.

H H H C C

H

ethane

H H The molecule is flexible; grasp one carbon atom and view the molecule along the C-C axis. Now rotate the front C atom about the C-C bond for a full 360 rotation. The relative positions of the hydrogen atoms on the different carbon atoms are constantly changing, and every different relative arrangement is called a "CONFORMATION" or they can be described as "CONFORMATIONAL ISOMERS" or "CONFORMERS". There are two extreme conformations, and these have important names.

staggered conformation

eclipsed conformation

It is often useful to inspect interactions between groups on adjacent atoms by viewing along the C-C bond. This particular projection, represented above, is known as the "Newman Projection". (Groups attached to the front carbon intersect at the centre of the circle; those attached to the rear carbon project only as far as the edge of the circle). Another convention very frequently used for the diagrammatic representation of three-dimensional molecules is the wedge-hash diagram.

bond in the plane of the paper bond projecting behind the plane of the paper bond projecting in front of the plane of the paper

Therefore a staggered conformation of ethane could be represented in a wedge-hash diagram as:

H

H H C C H

H H

At room temperature the rotation about C-C bonds takes place many thousands of times per second, however the different conformations do not have identical energies. The staggered and eclipsed conformations are the two extreme energy conformations since the electrostatic repulsion of the pairs of electrons in bonds (or lone pairs) when they are spatially in close proximity destablises the eclipsed conformation (note there are other

MOD.5

explanations of the reasons for the difference in the energies of the two conformations). Thus the staggered conformation is the more stable conformation. This destabilization effect tends to get larger as the groups involved get larger. In real terms, that means that it is at a minimum for two H atoms.

Replace one of the hydrogen atoms on each carbon atom with chlorine to form 1,2-dichloroethane. Starting with the 2 C-Cl bonds lined up with each other in an eclipsed conformation shown below.

Cl

Cl C C H

H

H H

Now for a definition: Torsional or dihedral angle. Let’s look at an example. The torsional angle is the angle between the pair of C-X and C-Y bonds that are part of an X-C-C-Y system when viewed along the middle C-C bond. Rotation about the C-C bond will change this torsional angle. This angle is also known as a dihedral angle. Thinking of “torsional” as “twisting” and this may help you remember and understand the term and differentiate it from a simple “bond angle”.



torsional angle

X

.

Y

There are some specific terms associated with certain torsional angles between a pair of substituents (such as the two Cl atoms in 1,2-dichloroethane): o

Syn

torsional angle = 0

Gauche

torsional angle = 60

Anti

torsional angle = 180

o o

Eclipsed conformations will have the largest amounts of torsional strain due to the electrostatic repulsions between the pairs of electrons in the eclipsing bonds. The conformations that brings atoms (or groups of atoms), especially large atoms, closer together in space than the van der Waals radii allow will have the largest amounts of van der Waals strain. Take your model of 1,2-dichloroethane and rotate the molecule about the middle C-C and observe the equivalent and non-equivalent eclipsed and staggered conformations. 1) Draw the Newman projections for the three eclipsed conformations. Circle the eclipsed conformation(s) with the highest energy.

2) Draw the Newman projections for the three staggered conformations. Circle the staggered conformation(s) with the lowest energy.

MOD.6 3) For the diagrams you drew in qu 1 and qu 2, identify the conformations where the chlorine atoms are syn, gauche or anti. 4) Sketch the following potential energy diagram, and indicate the positions where each of the six conformations (three eclipsed and three anti) of 1,2-dichloroethane would fall:

The connectivity and molecular motion due to bond rotations within a molecule can result in atoms that are considered to be equivalent or non-equivalent types. For example, the six hydrogen atoms in ethane are considered to be chemically equivalent (i.e. of the same type). Each individual hydrogen atom is in an identical environment (attached to a carbon atom that is linked to 2 other hydrogens and one methyl group). The ability to recognise the number of types of H (or indeed other atoms such as C) is a very important and a useful concept. For example, counting types of H is very important in spectroscopy (especially nuclear magnetic resonance) and in reactions (e.g. radical halogenation of alkanes). It will be revisited several times in later questions in this exercise and applied in other components of the course.

There are three methods one can use to establish the number of kinds of H (similar methods can be used for other atoms such as C). We recommend that you start with the first method, and over time as you get better at the task, you will gradually and naturally migrate through the second method to the third:

1.

Substitution method. This method is based on the idea that you replace each H in turn with a "dummy" atom to see if you get a different product (i.e. one that will require a name that differs by more than just E/Z, cis/trans or R/S, e.g. 1-chlorobutane and 2-chlorobutane). If you have a new product, then the H was different to those already considered.

2.

By "verbal" description. The verbal method requires that you describe the position of the H within the molecule. If you need to use different words to describe two H atoms, then they represent different types of H. For example an -OH is different to a -CH (based on what they are attached to), and a -CH3 is different to a -CH2- (because the number of H at that C are different). Other differences could be position on a chain, across a ring or double bond, hybridisation etc.

MOD.7

3.

Symmetry. The symmetry method is the most sophisticated but the quickest method and requires that you look for mirror planes, rotation axes or inversion centers that interchange H atoms. H atoms that can be interchanged are equivalent to each other.

Try replacing different H atoms in 1,2-dichloroethane to determine how many different types of H there are in 1,2-dichloroethane.

CAUTION: Remember that rotation about bonds produces different conformations (conformational isomers or conformers) only, not different molecules.

5) How many different types of H are there in each of the following hydrocarbons: methane, ethane, propane, butane, 2-methylpropane, pentane.

CONSTITUTIONAL ISOMERS are compounds that have the same molecular formula, but differ in the way that the atoms are connected to each other (i.e. due to different branching patterns or functional groups). They cannot be interconverted unless bonds are broken and made. Note that constitutional isomers have different names such as butane and 2-methylpropane (but beyond just differences in stereochemistry such as cis, trans, E/Z or R/S).

6) Using line diagrams, sketch and name all possible constitutional isomers with the molecular formula of C3H7I. 7) How many different types of hydrogen atoms are present in each of the structures that you have drawn?

Trigonal - sp2 carbon Three -bonds are formed with the carbon atom at the centre of a triangle. The other bond formed to carbon is a -bond (remember, carbon is TETRAVALENT). The simplest hydrocarbon with an sp2 carbon is ethene.

H

H C C

H

H

Construct an ethene molecule using the long flexible bonds and satisfy yourself it is flat. Try to rotate the molecule about the C=C bond. It’s only possible if you break the -bond. Replace one of the H atoms with a Cl atom.

8) How many different monochloroethenes can you make? How many different kinds of H are there in ethene?

MOD.8 Replace another H atom by a Cl atom. Because of the lack of rotation about the C=C bond it is possible to construct THREE DIFFERENT dichloroethenes.

H

Cl

Cl

Cl

Cl

H

H

Cl

H

H

H

Cl

1,1-dichloroethene

(Z)-1,2-dichloroethene

(E)-1,2-dichloroethene

The "Z" prefix indicates that the two groups of higher priority according to the Cahn-Ingold-Prelog Rules** (see notes at the end) are situated on the same side (German word Zusammen = together) of the double bond. Conversely, "E" (German word Entgegen = opposite) indicates these groups are across from each other. Note ONLY in the very simplest cases does Z correspond to cis and E to trans, e.g. see Qu 11 below.

9)

How many different kinds of H are there in chloroethene?

Make a model of the Z isomer and then convert this to the E isomer. Note that in order to do this, a chemical bond must be broken, so they are not conformational isomers. The two isomers have the same atoms bonded to each other, but in a different spatial arrangement, so they are called STEREOISOMERS. Note that stereoisomers have the same names except for differences in stereochemistry such as cis, trans, E/Z or R/S, e.g. cis-but-2-ene and trans-but-2-ene, (R)-butan-2-ol and (S)butan-2-ol. The interconversion of stereoisomers requires that bonds are broken. For example, in alkenes, this requires that the -bond be broken. This general kind of isomerism is called CONFIGURATIONAL ISOMERISM and specifically this type is E/Z, or a type of GEOMETRIC ISOMERISM. These molecules are quite different and have different physical and chemical properties.

This is in complete contrast to CONFORMATIONAL

ISOMERS (see above) which are different stereoisomers of the SAME molecule, differing due to rotation about C-C single bonds.

10)

How many different trichloroethenes are there?

Construct, and draw a model of (E)-2-bromobut-2-ene. Replace the hydrogen atom at carbon 3 with a chlorine atom.

11)

Reassign priorities and specify the configuration (E or Z) of each the 2-bromo-3-chlorobut-2-ene(s). The Index of Hydrogen Deficiency (often referred to as the IHD) is a measure of the number of units

(or degree) of unsaturation in a molecule. A saturated hydrocarbon is one that has the maximum number of H atoms for the given number of carbon atoms. This typically means that it only contains single bonds. The IHD is a count of how many molecules of H 2 need to be added to a structure in order to obtain the

MOD.9

corresponding saturated, acyclic species (acyclic means not cyclic, i.e. chain). Remake a model of ethene. The IHD is equal to the number of units of unsaturation, that is the number of rings present (i.e.

 bonds plus the number of

 + r ). IHD can be deduced from a structure by counting these features or it can be

calculated from a molecular formula. Both methods are useful. If we have a molecule with the general molecular formula CcHhNnOoXx, then the following equation can be derived, IHD = 0.5 * [2c+2-h-x+n] see http://www.chem.ucalgary.ca/courses/351/Carey5th/useful/ihd.html 12)

How many hydrogen atoms need to be added to the ethene to turn it into ethane?

How many

molecules of H2 does that correspond to? (This is the IHD for ethene) Linear -sp carbon Two -bonds are formed; the other two bonds to carbon are both -bonds. The angle between the -bonds is 180. Make a model of but-1-yne

H C C CH2CH3 13)

What is the relative spatial relationship of the carbon atoms in the triple bond of but-1-yne?

14)

How many different kinds of H and C atoms are there and what is the index of hydrogen deficiency (IHD) of but-1-yne?

Rearrange these atoms to form but-2-yne

H3C C C CH3 15)

How many different kinds of H and C atoms are there and what is the index of hydrogen deficiency (IHD) of but-2-yne?

B.

NITROGEN and OXYGEN: Many organic molecules contain nitrogen and / or oxygen atoms. The formation of bonds to nitrogen and

oxygen and the arrangements and shapes that result may be considered in terms similar to those discussed for carbon. Nitrogen is often trivalent; the N atom is sp3 hybridised forming 3 -bonds and the fourth group attached to the nitrogen is a lone pair of electrons. These four groups are arranged almost tetrahedrally around the central nitrogen atom. A further bond may be formed to the trivalent nitrogen using the lone pair of electrons from the nitrogen. If this takes place the coordinating group (the term "coordinating" implies that one atom, in this case the nitrogen, supplied both electrons to the bond) occupies the fourth corner of the tetrahedron and the nitrogen becomes positively charged to give a "substituted ammonium cation".

MOD.10 The simplest example of a molecule containing trivalent nitrogen is ammonia; one of the simplest organic nitrogen-containing molecules is methanamine: NH3(ammonia)

CH3-NH2 (methanamine)

Construct models of ammonia and methanamine. Confirm that they both have the same relative arrangements of atoms at the nitrogen atom.

16)

What is the VSEPR description of the shape of the arrangement at nitrogen in both these molecules?

17)

What is the N hybridisation in each structure?

18)

Are ammonia and methanamine saturated compounds?

19)

What is the index of hydrogen deficiency (IHD) for each of these compounds?

From inspection of the models it would seem that the bond angles are the same, as if the nitrogen were carbon. This is only true if there are four identical groups attached to the nitrogen (e.g. as in the NH4 cation). The actual measured bond angles are: NH3 H-N-H

CH3NH2

107.3  0.2

H-N-H H-C-H

105.8 109.5

In fact, deviations from the ideal angle of 109.5 for a regular tetrahedron are often observed. Factors which influence this angle include bond length, size of group or atom attached, and in particular, the presence of a lone pair of electrons. It is found that a lone pair will repel bonding pairs more than will another bonding pair; in the presence of one or more lone pairs the angle between bonding pairs is significantly compressed (VSEPR). An example of a molecule with nitrogen bonded to four groups is the neurotransmitter acetylcholine.

O H3C C O C C H2 H2

CH3  N CH3

OH

CH3

This compound is the chemical mediator which bridges the gap ("synapse") between the endings of two nerve cells. It is by means of this chemical that the nerve impulse is transmitted. Nitrogen is biochemically a very important element. It is found in a large number of biologically active molecules and is often intimately involved in the biological function of the molecule. Oxygen is a commonly occurring element in many organic molecules. The simplest, and most abundant molecule containing oxygen is water. In this molecule the oxygen may be considered sp 3 hybridised with two lone pairs.* Again, the lone pairs compress the angle between the bond pairs as was observed in the ammonia molecule. Methanol, CH3OH, is a simple organic molecule containing an - OH group. If the second H atom of water is also substituted by a methyl (-CH3) group then a molecule of dimethyl ether results.

MOD.11

..O.. H H O H

..O..

H H

104.45o

water

C

H H

H

108.9o 109.3o H C H methanol C O H

CH3

..O..

C O C

CH3 110o

dimethyl ether

* Other treatments of the bonding are also possible. Construct models of molecules of water, methanol, and dimethyl ether.

20)

What is the VSEPR description of the shape of the arrangement at the oxygen in these molecules?

21)

What is the O hybridisation in each structure?

22)

Draw a wedge-hash diagram to show the groups around the O in each structure.

23)

Are water, methanol and dimethyl ether saturated compounds?

24)

What is the index of hydrogen deficiency (IHD) for each of these compounds?

Both nitrogen and oxygen can occur in an sp2 hybridised state and form double bonds, and nitrogen can also form a triple bond when the nitrogen is sp hybridized. (e.g. in an alkyl cyanide (or nitrile) such as CH3-CN, or a ketone such as CH3C=OCH3).

25)

What is the IHD of acetonitrile, CH3-CN?

26)

What is the C-C-N bond angle?

27)

What is the IHD of propan-2-one CH3C=OCH3?

28)

What is the C-C-O bond angle?

C.

CYCLIC SYSTEMS: Carbon atoms, in addition to forming long carbon backbone chains, can also form rings. e.g. cycloalkanes

These rings differ in size and to a limited extent in their chemical properties, particularly in the case of the small-sized rings (e.g. 3 or 4 carbons – which tend to be very reactive) otherwise rings generally have similar reactivity to the analogous acyclic systems. Construct a molecule of cyclopropane using sigma bonds (use the medium, straight pieces)

H

H C

H C H

C H H

29)

How many different kinds of H are there in cyclopropane?

30)

Is cyclopropane saturated?

31)

How many H atoms need to be added to make the related acyclic alkane, propane?

32)

What is the index of hydrogen deficiency (IHD) of cyclopropane?

MOD.12

H2 / Pd

H2 / Ni cyclopropane

propane

propene

You should have concluded that the cyclic system has the same degree of unsaturation as an alkene unit, a fact that is emphasised by the two reactions shown above: C3H6 is the molecular formula for both cyclopropane and propene, which are CONSTITUTIONAL ISOMERS. The ring strain of cyclopropane even makes its reaction resemble an alkene. The index of hydrogen deficiency (IHD) or degree of unsaturation is just a count of the number of  bonds and / or rings. An alternative way to deduce the degree of unsaturation of a cyclic system is to count how many bonds you have to break to make a chain system.

Many other polycyclic ring systems are possible, and you will encounter some during your organic chemistry courses. We will investigate a few examples below:

Build a model of adamantane.

adamantane 33)

C10H16

What in the index of hydrogen deficiency (IHD) of adamantane ? How many types of H are there in adamantane?

Now build a model of cubane.

cubane 34)

C8H8

What in the index of hydrogen deficiency (IHD) of cubane ? (it is probably a good idea to determine this by breaking a bond at a time until you have an acyclic structure). How many types of H are there in cubane?

D.

AROMATIC RINGS: There is a whole branch of organic chemistry based upon structures containing rings of mainly carbon in an

sp2 hybridised state, but also including nitrogen and oxygen in some cases. Compounds comprising rings of sp 2 hybridised atoms where there are 6,10,14,18.. electrons in -orbitals are called AROMATIC and show properties quite different from any other organic structures. The commonest and most familiar of all these compounds is benzene.

MOD.13

H C

H

H

C

C

C

C

H C

H

H

1.39A

benzene

H

C

1.10A

C 120o

H

H

C 120o

C 120o

C

H

C H

H The structure as represented in the left hand diagram has alternating double and single bonds, and it is in this form that you have to construct a benzene ring with your model kit. In reality all the C-C bonds are the same length, 1.39Å, intermediate between double and single, and the ring is completely flat. The angles and bond lengths for benzene are shown in the right-hand diagram. This ring is rigid; there is no flexibility as in the carbon skeleton. The internal angle of a regular hexagon is 120 and so the trigonal sp2 angle of 120 is accommodated without strain. Build a model of benzene. Now determine the degree of unsaturation, by either counting the number of bonds you need to break or taking half the number of H atoms you need to add (equal to counting H 2 molecules) to get a saturated acyclic structure.

35)

What in the index of hydrogen deficiency (IHD) of benzene?

Build a model of chlorobenzene. Now, in turn, replace each of the hydrogens with a second chlorine atom.

36)

How many different dichlorobenzenes are there?

37)

For each dichlorobenzene, how many different types of H and C atoms are there and what is the index of hydrogen deficiency of each compound?

Now take your model of chlorobenzene and replace each of the hydrogens in turn with a bromine atom.

38)

How many different bromochlorobenzenes are there ?

39)

For each bromochlorobenzene, how many different types of H and C atoms are there?

E

OPTICAL ISOMERS : ENANTIOMERS and DIASTEREOMERS There is a further spatial relationship between atoms in molecules that we must consider, and it is a

more subtle than those considered above. There is a type of isomerism called OPTICAL ISOMERISM that can arise as a result of the tetrahedral arrangement around an sp3 hybridised carbon or nitrogen atom. Build two models of CH2ClBr. Position the two molecules of CH2ClBr such that they are ‘reflected’ through an imaginary mirror that runs between them. Try putting one molecule ‘on top’ of the other such that all the matching atoms line up.

MOD.14 40)

Is CH2ClBr superimposable on its mirror image?

41)

Does interchanging any two atoms (Cl, Br, or H’s) create a new molecule?

Looking at only one of the models for now, note the plane of symmetry that bisects the C, Cl and Br atoms. This molecule has an internal plane of symmetry and because of this, it is superimposable on its mirror image. To test this out take a black, tetrahedral C atom and add a white, an orange, a purple and a green piece to the C to make a simple tetrahedral molecule, CHClBrF. Now ignore this one and make as many other models as you can from your model kit (4 or 5 minimum: cooperate with another group if you need to). Now compare them all. Separate them into distinguishable types. You should have only two groups, all those within a group are superimposable on each other and they are all non-superimposable mirror images of all those in the other group. Superimposable means that two models can be placed side by side in such a way that they look identical (i.e. they can be superimposed in each other). Non-superimposable means that when two models are placed side by side, they can always be distinguished. Enantiomers are non-superimposable mirror images of each other. Compare the structures you built and make sure you understand the principle of superimposability. CHClBrF has no internal plane of symmetry, and forms a pair of enantiomers and is said to be chiral (molecules that lack this property are said to be achiral).

42)

What happens when any pair of substituents within these structures are interchanged? (i.e. remove one substituent and switch it with another then see if it belongs to the original group or the other group)

Build each of the following structures and its mirror image, then check for superimposability: 2-chloropropane, 2-chlorobutane, and 2,3-pentadiene.

43)

Which of the structures listed above have non-superimposable mirror images?

The most common scenario that leads to this type of isomerism arises if four different groups are attached to a central tetrahedral atom, then two different molecules can exist depending on the 3D-sequence in which the four groups are attached. The relationship between these two molecules is such that they are non-superimposable mirror images of each other; they are given the name OPTICAL ISOMERS or *

ENANTIOMERS . If the four groups are different there is no element of symmetry (mirror plane, rotation axis, inversion center) in the molecule and the central atom is termed an asymmetric atom. The reason for the term OPTICAL ISOMERS is that most physical and chemical properties of these isomers are identical. However, they have a different effect on a beam of plane polarised light, hence their name. Molecules with no asymmetric atom have no effect - they are optically inactive. One other difference has considerable *

The term enantiomer comes from the Greek enantios = opposite.

MOD.15

biochemical significance - optical isomers typically react at different rates with another optically active compound, e.g. such as an enzyme or a biological receptor. Normally in chemical reactions conducted in the laboratory where a molecule with an asymmetric carbon is generated, equal amounts of the two optical isomers are formed giving a racemic mixture. In natural systems the converse is true. It is a general rule that only one of the pair of enantiomers will be found. Biochemical reactions are so specific that usually the other enantiomer would not give a particular reaction.

Build a model of the isomer of the 1,2-dibromo-1,2-dichloroethane system shown below and its mirror image.

H

H C C

Br

Br Cl

Cl 44)

Are these structures superimposable on each other?

45)

Are there any chirality centers?

This type of compound is a special type of stereoisomer, known as a MESO compound. Note the special relationship of the asymmetric centers. To be considered to be a MESO compound a molecule MUST have two (or more) chiral centers and be superimposable on its mirror image – if there are NO chiral centers (e.g. CH2BrCl) the molecule is NOT considered to be MESO. Keep the last two models and now build the isomer shown below, and its mirror image.

H

H C C

Br Cl 46)

Cl Br

Are these two new models superimposable on each other or either of the other isomers of 1,2-dibromo1,2-dichloroethane you have built?

47)

Are there any chirality centers?

What you have just worked through covers a slightly different type of stereoisomers. Stereoisomers that are non-superimposable mirror images are ENANTIOMERS.

Stereoisomers that are not enantiomers are

DIASTEREOMERS (note that this description is quite broad and therefore includes other types of stereisomers that can be better described by more specific terms). Unlike enantiomers, DIASTEREOMERS typically have different chemical and physical properties, a factor that often makes them much easier to separate and purify.

48)

What is the relationship of last two structures you built to the previous two?

MOD.16 The next optically active molecules we will consider are amino acids. A generic representation is shown below in the Fischer and wedge/hash projections. The (L) differentiates which of the enantiomers we are referring to and is an historical convention that was initially adopted for this purpose.

This has been

superseded by the modern Cahn-Ingold-Prelog Rules (see the end).

All of the amino acids obtained from the hydrolysis of proteins exist as one enantiomer only and those obtained from the animals and the higher plants all have the same arrangement of groups around the asymmetric carbon atom as shown generically below. By the old convention these are the L-amino acids. Based on the R/S convention some are R but most are S. For this reason, the older convention is often retained when describing amino acids particularly (especially by biochemists).

CO2H

CO2H H

H

NH2

NH2 R

R Fischer

wedge/hash

Common representations of amino acids

49)

When R=H in the above formula, the amino acid is called glycine. Can glycine exist as a pair of enantiomers ? If not, why not?

Phenylalanine (R = -CH2C6H5) is an essential amino acid that is not synthesised in the body and so must be ingested in the diet. Whole egg, for instance, contains 5.4% (L)-phenylalanine. It is also one of the two amino acids that make up the artificial sweetener, “Aspartame”. Make a model of naturally occurring phenylalanine. H CH2

O

C OH NH2

(L)- or (S)-phenylalanine

50)

To convert this enantiomer to its mirror image, the (D)- or (R)-enantiomer, which groups can be interchanged ?

(Note:

to convert one enantiomer to the other requires bond breaking and hence these molecules are

configurational isomers).



In a Fischer projection horizontal lines indicate the substituent in front of the plane; vertical lines project backwards.

MOD.17

REFERENCES 1.

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch02/ch2-3.html http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch03/ch3-0.html http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch05/ch5-1.html http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch09/ch9-1.html http://www.chem.ucalgary.ca/courses/351/Carey5th/useful/ihd.html http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-0.html

2.

D.H.R. Barton, Experientia, 6, 316 (1950).

3.

D.H.R. Barton, O. Hassel, K.S. Pitzer and V. Prelog, Science, 119, 49 (1953), Nature, 172, 1096 (1953).

4.

D.H.R. Barton, Science, 169, 539 (1970).

5.

E.L. Eliel, and S.H. Wilen, in "Stereochemistry of Organic Compounds", Wiley, New York, 1994, or E.L. Eliel, "The Stereochemistry of Organic Compounds", McGraw-Hill, New York, 1962.

6.

R.S. Cahn, C.K. Ingold and V. Prelog, Experientia, 12, 81 (1956). http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-6.html http://www.chem.ucalgary.ca/courses/351/WebContent/orgnom/stereo/stereo-03.html http://www.chem.ucalgary.ca/courses/351/WebContent/orgnom/stereo/stereo-02.html