the Note

ORGANIC MOLECULES (LIVE)
10 APRIL 2015
Section A: Summary Notes and Examples
Naming and Functional Groups
Important Features of Carbon




There are different allotropes (same element, same phase, different atomic arrangement) of
Carbon: graphite, coal and diamond.
Carbon has a valency of 4 (can form 4 bonds), and it has 4 valence electrons (electrons found in
the outermost energy level).
In the excited state, valence electrons are unpaired i.e. one 2s electron moves to higher 2p
2
1
3
orbital. i.e. 1s 2s 2p
Carbon has the ability to form long chains with other Carbon atoms. This is called Catenation.
Some important definitions



Hydrocarbons – organic molecules that are made up of carbon and hydrogen atoms only.
Homologous series – a family of organic molecules which are identified by the same functional
group and obey the same general formula.
Functional group – a bond, atom or group of atoms which identifies to which homologous series
that molecule belongs and is responsible for the chemical properties of that molecule.
Ways to represent organic molecules
Molecular formula
C6H13OH
Condensed molecular formula
CH3CH2CH2CH2CH2CH2OH
Structural formula
Saturated and unsaturated hydrocarbons
A saturated hydrocarbon only has single bonds between the carbon atoms and unsaturated
hydrocarbons have at a double or triple bond between two carbon atoms.
Functional Groups
Alkanes:
• Single covalent bonds
• SATURATED molecule.
• General formula: CnH2n+2
Alkenes
• have one or more double bonds
• The general formula: CnH2n.
• UNSATURATED molecule
Alkynes
• One or more triple bonds
• General formula: CnH2n-2
• UNSATURATED molecule
Butane: CH3 - CH2-CH2-CH3
but-1-ene: CH2=CH-CH2-CH3
but-1-yne : CHC-CH2-CH3
Alkyl halide (haloalkanes)



Organic compounds containing one or more halogen atoms.
Functional group: C – X where X = F,Cl, Br, I
Numbered such that the position of halogen gets lowest number.
Page 1


F = fluoro, Cl = chloro, Br = bromo, I = iodo
If different halogen atoms in a molecule, then use alphabetical order
Alcohols




–OH (hydroxyl) NOT hydroxide (OH ) functional group.
Named using the ending –ol. –O-H group gets the lowest possible number:
Alcohols are oxidized to carboxylic acids when treated with strong oxidizing agents.
Alcohol molecules have a non-polar hydrocarbon end and a polar –O-H section.

Alcohols are solvents for polar and non-polar solutes.
Polar end
Non –polar
end
Carboxylic Acids

Compounds that have the carboxyl, -COOH functional group

The ending -oic acid denotes that we are dealing with a carboxylic acid.

Are relatively weak acids.
O
CH3 – C – O–H
Esters

These are compounds that have the CO–O–C functional group.

Pleasant smelling substances which are responsible for fragrance of fruit & flowers, perfumes,
food colourants.

Esters arise by the reaction between a carboxylic acid and an alcohol:

Water is also formed as a result of the reaction.

All esters consist of two parts;
o
an alkyl derived from an alcohol,
o
–anoate derived from a carboxylic acid (C=O)

In the example above, ethanoic acid and ethanol react to form the ester, ethyl ethanoate.

NB. Use the oxygen atom as a divider to name the molecule.
Page 2
Ketones
•
Structure:

Select the longest carbon chain containing the carbonyl carbon.

Number the chain starting with the end closest to the ketone

The -e ending of the parent alkane name is replaced by the suffix -one.
(eg. propane becomes propanone)
Aldehydes

Structure:

Select the longest carbon chain containing the carbonyl carbon.

The carbonyl carbon should have the lowest possible number

The -e ending of the parent alkane name is replaced by the suffix –al
(eg propane becomes propanal)
NOTE: In ketones, two carbon groups are attached to the carbonyl carbon, while in aldehydes at
least one hydrogen is attached to the carbon.
ISOMERS
Isomers are compounds which have the same molecular formula, but different structural formulae.
Three types of structural isomers are:

chain isomers– these isomers have different chain lengths
Example: butane and methyl propane
Both have the same molecular formulae C4H10 but with different structural formulae

positional isomers–these isomers have a different position of the same functional group.
Consider the two structural formulae shown below:
CH2=CH-CH2-CH3but-1-ene
CH2-CH=CH2-CH3
but-2-ene
Both have the same molecular formulae C4H8but with different structural formulae

functional isomers–these isomers contain a different functional groups.
Example:propanoic acid and methyl ethanoate
Both have the same molecular formulae C3H6O2 but with different structural formulae
Page 3
Intermolecular Forces

Weak electrostatic forces between molecules.

Holds together molecules (covalently bonded units)

Between non polar molecules – London forces or instantaneous dipoles

Between polar covalent molecules – dipole-dipole van der Waals forces. Dipole-dipole is an
attraction between slightly positive and slightly negative sides (poles) of the molecule.

Hydrogen bonding - between NOF (Nitrogen, Oxygen and Fluorine) bonded to hydrogen. These
are the strongest intermolecular forces.
Inter-molecular forces
Intramolecular force
(covalent bond –
strong)
O
H
H
+
-
Intermolecular
force (weak)
O
H
+
H
Physical Properties
12
Molecules are held together by intermolecular forces. In order to separate these molecules from each
other requires energy. The stronger the intermolecular forces, the more energy that is required to
separate or break these bonds. This will lead to a higher melting point, boiling point, etc.
Boiling and melting points
As the strength of the intermolecular forces increase, the boiling point and melting point will increase.
Thus molecules where there are hydrogen bonds present, will have a higher melting point and boiling
point than those molecules where there are weaker London forces or Van der Waals forces present.
Vapour pressure
Vapour pressure is the amount of pressure that the gaseous molecules exert above the surface of the
liquid phase. The vapour pressure will decrease as the size of the molecule increases (chain length).
Vapour pressure is an indication that there are weak intermolecular forces present in the liquid phase.
Viscosity
A liquid that has a low viscosity will be able to flow more easily. Thus, where hydrogen bonds are
present, there will be a much higher degree of viscosity. The opposite is true for the weak Van der
Waals forces. Viscosity will also increase as the length of the carbon chain increases.
Density
The density of organic molecules will increase as the length of the chain increases.
Surface area
As the surface area of the molecule decreases, there will be lower boiling and melting points as the
intermolecular forces will be weaker.
Phases
As you go down a group, either alkanes, alkenes etc, as the chain length increases, the melting and
boiling points will increase. The smaller molecules will then be gases at room temperature and the
longer chain lengths will be liquids or solids at room temperature
Page 4
Section B: Practice Questions
Naming and Functional Groups
Question 1
(Taken from DoE November Exam 2010)
The chemical properties of organic compounds are determined by their functional groups. The letters
A to F in the table below represent six organic compounds.
1.1
1.2
Write down the LETTER that represents the following:
1.1.1
An alkene
(1)
1.1.2
An aldehyde
(1)
Write down the IUPAC name of the following:
1.2.1
Compound B
(2)
1.2.2
Compound C
(2)
1.3
Write down the structural formula of compound D.
(2)
1.4
Write down the IUPAC name of the carboxylic acid shown in the table.
(2)
1.5
Write down the structural formula of compound F.
(2)
[12]
Intermolecular Forces
Question 2
(Taken from DoE Feb- March 2011)
Knowledge of boiling points can be used to identify chemical compounds. The boiling points of four
organic compounds, represented by the letters A, B, C and D are given in the table below.
Compound
Boiling Point
o
( C)
A
Propane
-42
B
Pentane
36
C
2-methylbutane
27,8
D
Pentan-1-ol
137
Page 5
2.1
Define the term boiling point.
(2)
2.2
Which one of A or B has the higher vapour pressure?
(1)
o
2.3
An unknown straight chain alkane has a boiling point of -0,5 C. Use the information in the
table to identify this alkane and write down its IUPAC name.
(2)
2.4
B and C are structural isomers
2.5
2.4.1
Define the term structural isomer.
(2)
2.4.2
Explain why B has a higher boiling point than C. Refer to structure,
intermolecular forces and energy in your explanation.
(3)
Explain the difference in the boiling points of B and D. Refer to intermolecular
forces and energy in your explanation.
(4)
[14]
Question 3
(Taken from DoE November 2012)
During a practical investigation the boiling points of the first six straight-chain alkanes were
determined and the results were recorded in the table below.
3.1
Alkane
Molecular
formula
Boiling point
o
( C)
Methane
CH4
-164
Ethane
C2H6
-89
Propane
C3H8
-42
Butane
C4H10
-0,5
Pentane
C5H12
36
Hexane
C6H14
69
Write down the general formula of the alkanes.
(1)
Refer to the table to answer Question 3.2 and 3.3.below
3.2
For this investigation, write down the following:
3.2.1
Dependent variable
(1)
3.2.2
Independent variable
(1)
3.2.3
Conclusion that can be drawn from the above results.
o
(2)
3.3
Write down the name of an alkane that is a liquid at 25 C.
(1)
3.4
Will the boiling points of the structural isomers of hexane be higher than, lower than or equal
to that of hexane? Refer to molecular structure, intermolecular forces and energy needed to
explain the answer.
(4)
[10]
Page 6
Section C: Solutions
Naming and Functional Groups
Question 1
(Taken from DoE November Exam 2010)
1.1.1.
1.1.2.
1.2.1.
1.2.2.
1.3.
A 
D
1-bromo-2-methylpropane 
2,4-dimethylhexane 
1.4.
1.5.
Ethanoic acid 
(1)
(1)
(2)
(2)

(2)
(2)

(2)
Intermolecular Forces
Question 2
(Taken from DoE Feb- March 2011)
2.1.
The temperature at which the vapour pressure of a liquid is equal to the external
pressure
(2)
2.2.
A (propane)
(1)
2.3.
Butane
(2)
2.4.1.
Compounds with the same molecular formula but different structural formulae
(2)
2.4.2.
C (2-methylbutane) is more branched / more compact / more spherical / has a smaller
surface area.
Therefore it has weaker intermolecular forces 
And less energy is needed to overcome the intermolecular forces
OR
B (pentane) is less branched / less compact / less spherical / has a larger surface
area.
Therefore it has stronger intermolecular forces
2.5.
And more energy is needed to overcome the intermolecular forces
(3)
The Van der Waals forces in B (pentane)  are weaker , than the hydrogen bonds
in D (penta-1-ol)  and requires less energy to break.
(4)
Page 7
Question 3
(Taken from DoE November 2012)
3.1.
3.2.1.
3.2.1.
3.2.2.
3.3.
3.4.
CnH2n+2
Boiling point 
Chain length / molecular size / molecular mass 
Boiling point increases with an increase in chain length / molecular size / molecular
mass 
Pentane 
Lower than 
 Isomers have more branching / more compact or spherical molecules / smaller
surface areas over which the intermolecular forces act 
 Weaker intermolecular forces 
 Less energy needed to overcome intermolecular forces 
Page 8
(1)
(1)
(1)
(2)
(1)
(4)