[PDF] Chem 351: Molecular models

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MOD.1

MOLECULAR STRUCTURES AND MODELS

NOTE : There is no pre-laboratory quiz or summary required for this experiment. However, before your

scheduled laboratory session, you should start working through the content. The laboratory period will

run more like a tutorial, you can work in groups and ask your TA about any of the concepts you don't understand or that you need further clarity on. BRING YOUR MODEL KIT and a PRINTED COPY OF THIS DOCUMENT. Remember that model kits are allowed tools during CAL activities, assignments and examinations. This laboratory activity is assessed based on an online Moodle graded activity (30 min. time limit that is to be completed by 6pm two days after your in-person laboratory session.

This activity is essentially the same as one we have successfully used for many years to help students with the

visualization of molecules and to learn how to use their model kits. Model kits are a tool that chemists will use to

visualize and explain aspects of chemical structure. They are very useful for helping students develop the ability

to visualize molecules in 3 dimensions (after all, very few molecules are actually flat). Some students certainly

struggle to grasp and manage this stereochemistry without the use of model kits (like the one shown below which

it the type we typically recommend). Since we regard model kits as a valuable tool, and a tool a chemist might

use, for many years we have allowed model kits to be used during examinations. We also know that many

model kits correctly and effectively so this activity tries to show that while

also exploring some to the topics related to stereochemistry that many students often struggle with. We always

ask UofC Bookstore to stock model kits. While the kits might appear expensive, they are worth the investment to

support your studies and can retain most of their resale value.

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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 and assignments 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

In preparation you should review the following concepts and terms from 1st 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

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

MOD.3

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 No. of Holes Bond Angle

Black C 4 109 28' White H 1 Blue N 4 109 28' Red O 4 109 28' Green Cl 2 Orange Br 1

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

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.4

ISOMER FLOWCHART

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. Start at the top and ask each the

question about the pair of molecules you are trying to define. Based on your YES or NO answer, follow the

corresponding path to the next question, or the end point and the isomer answer.

Isomers

Configurational

StereoisomersConstitutional

OH O

Conformational

H CH3 CH3 H H HCH3 HH CH3 HH

Enantiomers

Cl

CH3CH2H

CH3 Cl

CH2CH3H

CH3

Diastereomers

CH3 CH3H H Cl BrCH3 CH3Cl H H Br

Geometric

Do the compounds have the same molecular formulae ? YES NO

Not isomers

Are the compounds non-superimposable mirror images ? Can the compounds be interconverted by rotation about single bonds ? Do the compounds have the same connectivity ? YES NO YES NO YESNO

Optical

Is the isomerism at a

tetrahedral center ? NO

YES(skeletal, positional, functional)

MOD.5

EXPERIMENTAL PROCEDURE

"Tutorial" work in small groups, 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. After the end of the

laboratory period, you will need to complete an individual online Moodle assessment.

CARBON:

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. HCCH HH HH ethane

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:

MOD.6

CC H H H 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

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 the eclipsed conformation shown below. CC HH ClCl HH

Now for a definition:

Torsional or dihedral angle. . 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

fferentiate it from a simple X Y torsional angle .

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): Syn torsional angle = 0o Gauche torsional angle = 60o Anti torsional angle = 180o

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

MOD.7

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.

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 staggered) 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

try t especially in NMR.

1. Substitution method. This method is based on the idea is 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

MOD.8

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.

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 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.

MOD.9

CC H HH H

Construct an ethene molecule using the long flexible bonds and satisfy yourself it is flat. Try to rotate the

m 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?

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 HCl Cl H Cl H Cl H Cl Cl H

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?

MOD.10

11) Reassign priorities and specify the configuration (E or Z) of each the 2-bromo-3-chloro-2-butene(s).

The Index of Hydrogen Deficiency (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 (or

single bonds) for the given number of carbon atoms. The IHD is a count of how many molecules of H2 need

to be added to a structure in order to obtain the corresponding saturated, acyclic species. Remake a model

of ethene. The IHD is equal to the number of units of unsaturation, that is the number of bonds plus the

number of rings present (i.e. + 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 https://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

CCHCH2CH3

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

CCCH3CH3

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?

MOD.11

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".

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 CH3NH2 H-N-H 107.3 0.2 H-N-H 105.8 H-C-H 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.

CH3COCH2

CH2 NCH3 OCH3 CH3 OH

MOD.12

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 sp3 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. ............

104.45o108.9o

109.3o

110o
watermethanoldimethyl ether HOH H O HH C COH O CH H HH

OCH3CH3

COC H * 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-CN, or a

ketone such as CH3C(=O)CH3).

25) What is the IHD of acetonitrile, CH3-CN?

26) What is the C-C-N bond angle?

27) What is the IHD of propan-2-one CH3C(=O)CH3?

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

MOD.13

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) C C C HH HH HH

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?

H2 / NiH2 / Pd

cyclopropanepropanepropene

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.

adamantaneC10H16

33) 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.

MOD.14

cubaneC8H8

34) 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 sp2

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. 1.39A 1.10A

120o120o

120o
C C C C C CHH H H H H CC C CC C HH HH

HHbenzene

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 H2 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?

MOD.15

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

most commonly arises as a result of the tetrahedral arrangement around an sp3 hybridised carbon. Build two models of CH2ClBr. Position the two molecules of CH2 matching atoms line up.

40) Is CH2ClBr superimposable on its mirror image?

41)

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)

MOD.16

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

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.

CCBrBr

H Cl H 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.

CCClBr

H Cl H Br * The term enantiomer comes from the Greek enantios = opposite.

MOD.17

46) Are these two new models superimposable on each other or either of the other isomers of 1,2-dibromo-

1,2-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?

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). H R NH2 CO2H R NH2 CO2H H

Fischerwedge/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

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

backwards.

MOD.18

a model of naturally occurring phenylalanine. C O OH NH2 H CH2 (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).

REFERENCES

1. https://www.chem.ucalgary.ca/courses/351/Carey5th/Ch02/ch2-3.html

https://www.chem.ucalgary.ca/courses/351/Carey5th/Ch03/ch3-0.html https://www.chem.ucalgary.ca/courses/351/Carey5th/Ch05/ch5-1.html https://www.chem.ucalgary.ca/courses/351/Carey5th/Ch09/ch9-1.html https://www.chem.ucalgary.ca/courses/351/Carey5th/useful/ihd.html https://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).

https://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-6.html https://www.chem.ucalgary.ca/courses/351/WebContent/orgnom/stereo/stereo-03.html https://www.chem.ucalgary.ca/courses/351/WebContent/orgnom/stereo/stereo-02.html

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