organic chemistry experiment 2

8/3/2019 Organic Chemistry Experiment 2

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UNIVERSITI MALAYSIA SABAH

SCHOOL OF SCIENCE AND TECHNOLOGY

ORGANIC CHEMISTRY PRACTICAL I

(SK 10301)

EXPERIMENT 2: MOLECULAR MODELING: ISOMERS, CONFORMERS AND

STEREOISOMERS

Group Members : Daphne William Apin (BS10110079)

Daryl Ng Chuan Yao (BS10110080)

Fatin Binti Sualin (BS10110110)

Ili Shazwani Binti Idris (BS10110151)

Jagadeesh A/L Dewedree (BS10110157)

Title of Experiment : Molecular Modeling: Isomers, Conformers and Stereoisomers

Lecturer : Prof. Madya Dr. Suhaimi Md. Yassir

Demo : Ms. Joyce Kristy Primus

Date of Experiment : 4th October 2011

Date of Submission : 11th October 2011


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Objectives :

To become familiar with the 3-dimensional structures of organic molecules, especially the

tetrahedral structure of alkyl carbon atoms and the planar structures of alkenes.

To be able to construct the 3-dimensional models for organic compounds and draw the

formulas.

Introduction :

Determination of shapes of organic molecules is difficult even through the most powerful

microscopes other than by the usage of Nuclear Magnetic Resonance(NMR) Spectoscropy.

Hence to help in the research of this field, molecular model kit is developed. Molecular models

are very useful in testing hypotheses, in understanding physical and chemical properties, and

also in visualizing the overall structure of an organic molecule.

In this experiment, ball and stick models are used to represent atoms and bonds in molecules

and organic molecules structures are built by utilizing the balls and sticks given. Different

isomers of organic molecules are also able to be constructed by using these models. Isomers

are molecules with the same molecular formulas but different structural formulas. Structural

isomers are compounds with different connections among their atoms. Different

conformations of molecules can be formed by merely rotating groups around a bond without

breaking any bonds, which is also named as conformers.

Materials :

Molecular model kit.


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Elements and bonds represented in the Organic Model Kit

Color Element Number of bonds

Black Carbon 4

White Hydrogen 1

Red Oxygen 2

Yellow Nitrogen 3

Green Chlorine 1

Blue Bromine 1

Method :

1. Models for the diatomic molecules hydrogen (H2), chlorine (Cl2), and hydrogen

chloride (HCl) were constructed. What are the bond between atoms represent?Models for the diatomic molecules hydrogen(), chlorine(), hydrogen

chloride(HCl) are constructed.

HCl

H H Cl Cl H Cl

The bonds between the two atoms are composed of carbon-carbon single bond,

which consists of sigma bond.

2. What is the octet rule and how does it explain the tendency of oxyegen to form

two bonds whereas chlorine, for example, forms only one bond?

The octet rule says that atoms tend to gain, lose or share electrons so as tohave eight electrons in their outer electron shell.

Electronic configuration of oxygen:


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Oxygen tends to form two bonds because it needs another two electrons to

achieve its octet configuration since it has 6 valence electrons and two unpaired

electrons at -orbital as shown above.

Electronic configuration of chlorine:

Chlorine needs another one more electron to achieve octet configuration since it

has 7 valence electrons or one unpaired electron as shown above.

3. Models for methane (CH4) and Propane (CHCl3) were constructed.

Models for Methane() and chloroform() were constructed.

The shape formed is tetrahedral.

4. Models for ethane (C2H6) and propane (C3H8) were constructed.


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There is no other way to construct these molecules since they have carbon

numbers not more than 3, whereby isomerism can only occur with minimum 4

carbon atoms.

C C

H

HH

H

C C C

H

H H

H

H

H

HH

H

H

Ethane Propane

5. Model for butane (C4H10) was constructed. Find another way to construct butane.

Their structures were sketched.

A model for butane() was constructed.

There are two ways of constructing a molecule with the molecular formula .


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CH3 C CH3

H

CH3

CH3 C C CH3

H

H

H

H

Buthane 2-methyl-propane

6. Ethanol and methyl ether both have empirical formula C2H6O but different

structures. The molecules were constructed and the structures were drawnt.

Models for Ethanoland methyl etherare constructed.

Ethanol and methyl ether have the same empirical formula O but different

structural formulas. Ethanol is an alcohol and has anOH group. Ether has a -O-

linkage.

CH3 CH2

OH

CH3 O

CH3

Ethanol Methyl ehter

7. Models for diatomic oxygen and for carbon dioxide were built. How many double

bonds do these molecule have?


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Oxygen molecule has 1 double bond whereas carbon dioxide molecule has 2

double bonds.

O O C OO

Oxygen Carbon dioxide

1 double bond 2 double bonds

8. What formula will you predict for octane? Some up the general expression.

Name Molecular formula

Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane

As shown on the table, every increase of carbon atom by 1, the number of

hydrogen atoms increases by 2. Therefore, the general formula for this kind of

hydrocarbon is .

9. a) A hydrocarbon with the formula consisting of a ring of six C groups,

was constructed using the molecular model kit. Then, another hydrocarbon with

the formula consisting of six CH groups with alternating single and double

bondsbetween the carbons, was constructed. Cyclohexane and benzene were

constructed.


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The benzene ring consists of six carbon atoms bonded in a flat or planar

hexagonal ring. Each carbon is bonded to one hydrogen because of the three

alternating double bonds. This reveals that each carbon is bonded to 3 others

atoms and one of them is a double bond. Hence the molecular geometry at each

carbon is trigonal planar as each carbon is hybridized.

C

C

C

C

C

C

H

H

H H

H

HH

H

HH

H

H

2D plan view of benzene 2D plan view of cyclohexane

Hence the whole ring is flat as compared to the cyclohexane, which is

hybridized, will give rise to a tetrahedral shape for each of the carbon atom.

Each carbon atom is bonded to two carbon atoms and two hydrogen atoms in

carbon-carbon single bond.

Therefore, cyclohexane is not as flat as benzene. In short, benzene has 3

carbon-carbon single bonds and 3 carbon-carbon double bonds while

cyclohexane have 6 carbon-carbon single bonds only.


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b) The smallest unit of units have made without straining the bonds.

Cyclopropane

c) Years ago, chemists tried to create a Cubanemolecule, which consist of 8

carbon atoms with 2 hydrogen atoms attached to each carbon. However, this

molecule are proved to be unstable.

C

C

C

C

CC

C

C

HH H

H

H

H

H

HHH

H

H

H

H

H

H

Cubane in theory Cubane in practical

The carbon atoms in Cubane are hybridized and have an angle of,

where it supposed to be (tetrahedral). Hence, the carbon-carbon single

bonds are too highly strained, which in return, causes more internal energy is

generated renders it becomes very unstable. Therefore, the cubane structure will

easily rupture at most of the conditions.

The cubane in practical also proves that cubane with each carbon bonded to 2

hydrogen atoms are impossible to be made as each carbon by max could only be

bonded to 4 atoms or groups. However, although cubane in practical is able to

be synthesized, researches shown that even though is kinetically stable, its

still remains as highly reactive compound i.e. unstable.

10. a) Model of molecule that has single bond in center and four different atomsattached to it was constructed. It mirror image was constructed.


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The two mirror pairs are not identical since the enantiomers are not

superimposable and remain as distinct compound.

b) The model was placed on the paper. Position of each touching atom wasmarked. The positions of atoms were marked.

The result showed that none of the molecules fitted each other.

c) A model that has a carbon atom in the centre but only three different atoms

attached to it, that is, that has two of one kind of atom instead of four

different atoms or groups.

d) Part b was tried with the model from part c.

The result showed that the molecules fitted each other.


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e) Carbons that have fewer than three different atoms were predicted.

Carbons that have fewer than three different attached atoms would not give

distinct left and right-handed forms.

From the results of part a to part d. A rule on determination on left and right-

handed forms could be formulated:

1. Identify the attached atom or groups of the highest priority.

2. Follow by the second priority to the lowest one.

3. If the arrow pointed clockwise, it is a (R)-handed form. Inversely, if the

arrow pointed anti-clockwise, it is a (S)-handed form.

For example: 2-bromobutane

f) A model of lactic acid was constructed.

Lactic acid


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A 2D plan view of lactic acid is shown as below:

C C OH

CH3O

HO

lactic acid

H

*

The carbon atom with the asterisk(*) has four different groups attached to it.

Hence, it exists as the chiral carbon center that has two forms, (R)-Lactic acid

and (S)-lactic acid.

g) Based on its structure, carvone wold be chiral was predicted.

The structure of carvone:

O

*

Note: * = chiral centre

The asterisk(*) shown that the indicated carbon has four different groups

attached to it. Hence, optical isomerism occurs.


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O

H H

O

(R)-Carvone (S)-Carvone

The two groups and ( are at different side at each of the

isomers. When atom is at the front, ( group is at the back

and vice versa.

h) Left and right-handed forms of thalodimide were constructed.

The structure of thalodimide:

Thalodimide

The different forms of thalodimide will cause different effects. Although theories

stated that (R)-enantiomer is capable to cure morning sickness while (S)-

enantiomer is the one responsible for birth defects, however not much proofs

and evidence are concrete enough to prove the theories.


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The main difference is the position of the group that is attached to the nitrogen

atom on the pentagonal carbon ring. Researches shown that (S)-thalodimide fits

neatly into the major groove at the DNA purine sites. Therefore, it is greatly

believed that (S)-enantiomer will hinder the function of purine during DNA

replication processes. Therefore, causing physiological malfunction or birth

defects.

However, evidence shown that taking either one of the enantiomers at a time

greatly reduced the chance of having birth defects.

N

NH

O

O

O

O

N

NH

O

O

O

O

(S)-thalidomide (R)-thalidomide

Pre-Laboratory Questions-Experiment 2

1. How do you distinguish between molecular formula, empirical formula (simplest),

and structural formula of benzene? (Draw structure)

Molecular formula :

Empirical formula :

Structural formula :

The molecular formula and simplest formula could be distinguished by the

subscripted number at C and H atoms. The empirical formula is a formula only

showing the simplest ratio of carbon to hydrogen atom. Hence, a ratio of 1:1 is


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denoted for empirical formula of benzene as compared to molecular formula,

which shows the actual numbers of carbon and hydrogen bonded together.

Structural formula is different from the two formulas as it shows how the carbon

and hydrogen atoms bonded in benzene.

2. How do you distinguish between geometrical and structural isomers? Give

example.

Geometrical isomers occur only with a presence of carbon-carbon double bond,

which in return will produce cis-and trans-isomers. They both of have the same

molecular formulas but different position of the atoms or groups that attached to

the two double bonded carbons

Structural isomers are isomers that have the same molecular formulas but with

different structural formulas, or, in another words the position of some of the

atoms or groups are different from each of the structural isomers.

For example, butane with molecular formula that has structural isomers

and butane with molecular formula that has geometrical isomers.

H3C C C CH3

H

HH

H

H3C C CH3

CH3

H

Butane

2-methyl-propane

C CH3C CH3

H H

2-butene

C C

H

H3C CH3

H

C C

H

H3

C H

CH3

cis-2-butene trans-2-butene

3. Why should the properties of structural isomers differ?


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This is because of the arrangement of the position of the atoms of the

compound, which results in different physical properties. For example the

melting point of pentane and 2, 2-dimethyl-propane are different as the

arrangement are different.

H3C C C C H3C C CH3

CH3

CH3

H

H

H

H

CH3

H

H

Pentane 2,2-dimethyl-propane

Although both have the same number of carbons, they differ in their molecular

shape. Pentane has a linear shape whereas 2, 2-dimethyl-propane gives a

spherical shape. The melting point of pentane is higher because it has a larger

surface area of contact as compares to 2, 2-dimethyl-propane which is spherical.

Hence, the Van der Waals forces become stronger as Van der Waals forces is

directly proportional to the surface area of contact of the organic molecules.

Therefore, more heat is needed to overcome the intermolecular forces before

any change of physical properties can occur.

4. Draw the skeletal formula for the following compound.

H3C C C

CH3

CH3

CHH

H

CH3H3C


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Answer:

5. Draw all the possible cyclic isomers for C4H8 and name all the isomers.

CH3

Cyclobutane Methyl-cycloprapane

organic chemistry experiment 2

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