TOPIC 10 ORGANIC CHEMISTRY SL
10. 1 Introduction
Homologous series
10.1.1. Describe the features of a homologous series.
10. 2.1 Predict and explain the trends in boiling points of members of a homologous series.
A homologous series is a set of compounds:






which share a general formula (same elements in the same ratio);
members in the same homologous series share the same functional group; a functional group is a group of
atoms which determine the chemical properties of the homologous series;
whose nearest neighbours differ by one repeating unit, most often a methyl group -(CH3 ) or a methylene
group -(CH2)- ;
have similar chemical properties (same functional group);
show a gradual change (gradation) in physical properties as shown by the graphs below:
Examples of homologous series: alkanes, alkenes, alkynes, alcohols, esters, alkanals and amines.
Name
methane
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
undecane
dodecane
eicosane
triacontane
molecular
formula
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
C7H16
C8H18
C9H20
C10H22
C11H24
C12H26
C20H42
C30H62
melting
point
(oC)
-183
-183
-190
-138
-130
-95
-91
-57
-51
-30
-25
-10
37
66
boiling
point
(oC)
-164
-89
-42
-0.5
36
69
98
125
151
174
196
216
343
450
state at
25oC
gas
liquid
solid
The table shows a gradual
increase in boiling point with
increasing number of carbons
and therefore increasing molar
mass.
A trend caused by the fact that
as the number of atoms in the
molecules increases so does
the number of electrons within
the compound which creates
greater polarity during
instantaneous polarisation
(which causes the Van der
Waals’ forces) and therefore
produces greater Van der
Waals’ forces. There is also a
greater surface area over
which instantaneous
polarization can occur.
The graph of boiling points of
alkanes against chain length
gives a steep graph at first but
then flattens out at higher
numbers of carbon atoms.
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Formula of organic compounds
10.1.3 Distinguish between empirical, molecular and structural formulas.
Type of formula
empirical formula
description
Shows most simple ratio
example
molecular formula
Shows the different atoms and how many
of each; no information on how they are
arranged
CH2
structural formula
A structural formula is one that shows
how the atoms are arranged together.
A full structural formula (sometimes
called a graphic formula or displayed
formula) shows every atom and bond.
condensed structural
formula
Structural formula which shows order in
which atoms are arranged but which omits
bonds
C6H14
CH3CH2CH2CH2CH2CH3
or CH3(CH2)4CH3
Naming of organic compounds
10.1.6. Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6.
10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6.
10.1.10. Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following
functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.
When naming an organic compound we want to give a lot of information in its name which is why the name of an
organic compound consists of two parts a shown by the following example of ethane:
eth ane


this part tells us how many carbons atoms
there are in the molecule;
this part could be:
 meth - means it has 1 carbon atom
 eth- = 2 carbon atoms
 prop- = 3 carbon atoms
 but- = 4 carbon atoms
 pent = 5 carbon atoms
 hex = 6 carbon atoms


this part tells us the functional group it has or which
homologous series it belongs to;
the ending could be:
 -ane which means it belongs to the alkanes
(=homologous series)
 -ene = alkene
 -anol = alcohol
 -anal = aldehyde
 -anone = ketone
 -oic acid = carboxylic acid.
Branched compounds
A branch is a carbon atom or group of carbon atoms bonded onto a larger carbon chain.
The name of that branch (or side group) should indicate the number of carbon atoms in it and end with –yl to
indicate it is a branch e.g. methyl, ethyl, propyl.
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You need to identify the longest chain in the branched compound and any carbon atoms not in it must be part of a
branch.
The name of the branched compound should include:
 the name of the branch
 the number of branches if there is two or more
 and, using a number, the position of the branch(es) on the straight chain; the position should be decided
from the end of the longest chain which gives the lowest number.
Examples: 2-methylpentane, 2,2-dimethylpentane, methylpropan-2-ol, 2,2,4-trimethylpentane
Halogenoalkanes
In the case of the halogenoalkanes, the name begins with the name of the halogen and should also indicate the
position and the number of halides if there is more than two.
For example: chloro, bromo, fluoro, iodo, .. in names such as 2-bromopropane, 1,2-dichloroethane.
10.2 Alkanes
10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangement of
atoms
10.1.5. Deduce the structural formulas for the isomers of the non-cyclic alkanes up to C6.
10.2.1 Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity.
10.2.2. Describe, using equations, the complete and incomplete combustion of alkanes.
10.2.3 Describe, using equations, the reactions of methane and ethane with chlorine and bromine.
10. 2.4 Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical
Alkanes mechanism.
are a homologous series in which all non-cyclic compounds have the following general formula: CnH2n + 2.
Structure of alkanes:
There are 3 types of structures but you only need to know two as you do not need to know the cyclic ones:
structure 1
straight chain: all carbon atoms can be
joined by a continuous line.
structure 2
branched chains: molecules have side groups and these are
referred to as alkyl groups.
alkyl group
methyl
ethyl
propyl
butyl
H H H H
   
HCCCCH
   
H H H H
C4H10 called n butane or butane
(n means normal straight chain)
formula
CH3 CH3CH2 CH3CH2CH2 CH3CH2CH2CH3 -
H H H
  
HCCCH


H
H
HCH

H
C4H10 called 2-methylpropane
Structural isomerism
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Structural isomers are compounds with the same molecular formula but the sequence or arrangement in which
the atoms are bonded is different which is why they have a different structural formula. There are many different
types of isomerism but you only need to know structural isomers.
In the space below draw the structural formula for all possible structural isomers and write their names underneath
them.
Number of
carbon atoms
Structural isomers
4
5
6
Physical properties of alkanes: melting/boiling points, viscoscity, density.
Trend: all 4 physical properties increase with increasing number of carbon atoms.
Important factors which affect physical properties:
Chain length: as the carbon chain gets longer, the intermolecular forces become stronger as seen by the
following evidence :
 the higher the melting and boiling point (from colourless gases and liquids to white solids);
 the higher the density (liquid alkanes become more viscous).
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Explanation:
The larger the molecules (greater Mr)



the more sites/surface area there are for intermolecular attraction;
the greater the number of electrons, the greater the polarisation within the molecule;
the greater their mass/inertia and therefore the more energy is needed to increase motion and to melt and
boil.
However, the effect of chain length decreases as the chains get longer. This is the case because in smaller
molecules an additional carbon and its hydrogen causes a much greater % mass increase than in larger
molecules.
Complexity of molecules: has a different effect on the melting point than on the boiling point.

Straight chained alkanes have higher b.p. than branched alkanes of similar molecular mass as there is more
contact/larger surface area between the straight-chained molecules and therefore more sites for induced
polarity and stronger Van der Waals forces; the higher the number of branches, the lower the b.p. As
branching decreases surface area, it increases volatility and decreases density.
name
B.Pt (oC)
36
pentane
CH3CH2CH2CH2CH3
C5H12
2 methylbutane
CH3CHCH2CH3

CH3
C5H12
2,2-dimethylpropane

structural formula
molecular
formula
CH3

CH3C CH3

CH3
28
C5H12
10
However, straight chained alkanes have lower m.p. than branched alkanes. Branching increases m.p. as
branched molecules can fit more closely together and more energy is needed to separate them; the lattice of a
solid straight chained alkane is like wet spaghetti: molecules can easily slide over each other.
Chemical properties of alkanes: unreactive
Alkanes are unreactive because

Large bond enthalpies: the covalent bonds between carbon and hydrogen and carbon and carbon have
large bond enthalpies; this is because both hydrogen and carbon are small atoms so bonding pairs are
attracted strongly by both nuclei.
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
Low or no difference in electronegativity: due to the very small difference in electronegativity between
carbon and hydrogen the covalent between the two atoms very low polarity and does therefore not attract
other reagents.
The only reactions they carry out are combustion and substitution.
Combustion of alkanes: a highly exothermic process
 Complete combustion/oxidation: (in plentiful of oxygen) products are water and carbon dioxide; general
equation :
CxHy + (x + y/4) O2  xCO2 + (y/2)H2O
(multiply by 2 if (x + x/4 produces a 0.5)
Both carbon and hydrogen are oxidised.
Example:
C5H 12 +
8O2
 5CO2 + 6H2O
 Incomplete combustion (carbon is not completely oxidized) : products are water and carbon monoxide or
carbon (=black smoke) depending on the extent of the lack of oxygen (incomplete oxidation); water is always
produced.
Substitution reactions of alkanes
10.2.3 Describe using equations the reactions of methane and ethane with chlorine and bromine
10.2.4. Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical mechanism.
Bond fission
Homolytic fission
When a covalent bond is broken the bonding pair electrons are redistributed between the two atoms; there are two
* occurs
in non-polar
bonds
bonds with a very low polarity when bonding electrons are fairly equally
ways
of redistributing
these
twoorelectrons:
shared in the bond
* each atom gets one of the bonding electrons; each atom has now an unpaired electron and is therefore
unstable and reactive; such a particle with an unpaired electron is called a free radical
* free radicals have a strong tendency to react and usually have a short existence; tend to be intermediates
in reactions
A  B

A
+
B
Heterolytic fission
* Both bonding electrons go to one of the atoms forming a negative and positive ion for example a
carbocation and carbanion
* occurs in polar bonds
* ions are unstable and highly reactive sites
A  B
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
A+
+
B-
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During a substitution reaction, a hydrogen atom on the carbon chain is replaced by a halogen atom. The reaction
needs sunlight/UV as UV/sunlight has the corresponding amount of energy to break the halogen bonds; no
reaction will occur in the dark.
Examples of substitution reactions:
UV
 CH4 (g) +
Cl2 (g)

methane
+
CH3 Cl (g)

chlorine
+
HCl (g)
chloromethane
+
hydrogen chloride
(substitution with chlorine: tetrachloromethane is formed if proportion of chlorine is high compared to the
proportion of methane).

C2H6 (g)
ethane
+
+
Br2 (l)
bromine
UV


C2H5 Br (g)
+
bromoethane
HCl (g)
+
hydrogen chloride
Homolytic free radical substitution reaction
The sequence of events by which substitution takes place is called the reaction mechanism. The mechanism
involves three reaction steps and is called free radical substitution:
step 1:
initiation
The UV light causes homolytic fission of the halogen molecule; each atom takes one of the
electrons in the covalent bond. The two species formed are not atoms but are called free
radicals and each has one electron from the bond. A free radical is the name given to a
species containing an unpaired electron. They are very reactive, because they have an unpaired
electron, and so have a strong tendency to pair up with an electron from another molecule. A
dot is used to represent the unpaired electron.
Using the example of the reaction between methane and chlorine:
Homolytic fission reaction
to form free radicals
step 2:
propagatio
n
Cl - Cl

Cl
+ Cl
free radicals
Chain reaction during which the free radicals react with molecules forming more free
radicals and other molecules:
free radical +
molecule

new free radical
+
new molecule
Free radical knocks an atom off the molecule which itself becomes a free radical while the
radical becomes a molecule.
In the case of the chlorine and methane reaction:
One of the chlorine free radicals reacts with a methane molecule by substituting itself for a
hydrogen atom. A new free radical is formed (or propagated).
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CH4
Cl
+

CH3
+
HCl
The CH3 reacts with the Cl2 to form CH2Cl and the free radical Cl:
CH3
step 3:
termination
+

Cl2
CH3Cl
Cl
+
Occurs when two free radicals react to form a stable molecule.
In the case of the chlorine and methane reaction:
CH3
+
Cl

CH3Cl
Cl
+
Cl

Cl2
CH3
+
CH3

C2H6
10.3 Alkenes
10.1.8 Deduce the structural formulas for the isomers of the straight-chained alkenes up to C6
10.3.1 Describe, using equations, the reactions of alkenes with hydrogen and halogens.
10.3.2. Describe, using equations, the reactions of symmetrical alkenes with hydrogen halides and water.
10.3.3. Distinguish between alkanes and alkenes using bromine water.
10.3.4. Outline the polymerization of alkenes.
10.3.5. Outline the economic importance of the reactions of alkenes.
General:

General molecular formula: - alkenes: CnH2n . This general formula only applies to the straight and branched
alkenes; you should be able to draw the structural formulae of straight-chained alkenes with fewer than 6
carbon atoms + their isomers.

Unsaturated hydrocarbons: because of the double bond, the bonding capacity of carbon is not fully used; the
carbon atom is unsaturated as it is not bonded unto 4 different atoms. Also remember that a double bond is
stronger than single bond but not as strong as two single bonds.
Structure of alkenes:
Structural isomerism:
There are 2 types of structures you need to Draw the structural formula of all possible structural straight-chained
know:
alkenes up to C6 and name each one of them. Remember the
position of the double bond can also change.
 straight chain:
 branched chains:
Physical properties of alkenes:
Same trends and explanations as in
alkanes; however, boiling points/melting
points are generally a little lower than the
boiling points/melting points of the alkanes
as the alkenes have a lower molecular
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mass.
Chemical properties of alkenes:
As they are unsaturated alkenes are
reactive. Also the second bond of the
double bond is weaker than a single
carbon-carbon bond and is broken much
easier.
Alkenes undergo addition reactions; atoms
are added to the carbon chain using the
double bond
Chemical properties: addition reaction of alkenes with hydrogen, hydrogen halides, water and bromine
As they are unsaturated alkenes are reactive. Also the second bond of the double bond is weaker than a single
carbon-carbon bond and is broken much easier.
It is because of this greater reactivity that alkenes, especially ethene, are important starting materials in organic
synthesis of useful chemicals.
It is important to note that alkenes also easily combust and undergo both complete and incomplete combustion.
Alkanes undergo addition reaction which means that atoms are added to the molecule at either side of the double
bond so any addition reaction increases the number of atoms in the molecule. During the addition reaction the
double bond is converted to a single bond. The weaker double bond is replaced by a stronger single bond; this
increases the stability of the molecule.
Reactions:
The reagents (hydrogen, halogen, water and hydrogen halide molecules) are attracted to the double bond; the
double bond breaks open and is replaced by two single bonds to 2 new atoms or groups of atoms. One part of
the reagent bonds to one carbon atom while the other part of the reagent bonds onto the second carbon atom of
the double bond.
Reaction with bromine (bromination):
Occurs under normal conditions (even in the dark); the product is a colourless halogenoalkane.
Equations:
*
molecular:
*
word:
*
structural:
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C2H4 (g)
ethene
+
+
Br2 (l)
bromine

C2H4 Br2 (g)

1,2-dibromoethaneC2H4 Br2 (g)
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Test for unsaturation
The bromination reaction is useful as it can be used to distinguish between an alkane (no discolourisation of
bromine water occurs as it remains yellow/brown) and an alkene (bromine water which is yellow or brown
becomes colourless as 1,2-dibromoethane is a colourless compound).
The amount of bromine water which decolourizes gives an indication of the degree of unsaturation; the greater the
amount of bromine water which needs to be added before it retains its colour means the greater the degree of
unsaturation; the greater the number of double bonds.
Reaction with hydrogen (hydrogenation):
Does not occur under normal conditions but needs a finely divided catalyst (to break the strong H-H bonds), nickel
and some heat, 140 C although it does also occur at room temperature.
This reaction is used to convert unsaturated oils into saturated fats (=hydrogenation of vegetable oils) in the
manufacture of margarine and in the oil industry to convert alkenes into alkanes (= economic importance of
reactions of alkenes).
Reaction with water (=hydration):
Reaction in which H and OH are introduced in the molecule: needs a strong concentrated acid (e.g. H2SO4 or
H3PO4) as catalyst, 300 C and 70 atm. The reaction is used to make industrial ethanol.
Equations:
*
molecular:
*
word:
*
structural:
C2H4 (g)
+
ethene
H2O (l)
+
water


C2H5 OH (l)
ethanol
Other alcohols: propene and steam to form propan-2-ol
Reaction with hydrogen halides:
Concentrated aqueous solutions of hydrogen halide; product is a halogenoalkane.
Examples and equations: but-2-ene reacts with hydrogen chloride.
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Addition polymerisation of alkenes DOUBLE BOND NEEDED!!!
Because they are unsaturated, alkene molecules (or other molecules with double bonds) can be added onto each
other forming longer chains. When this process is allowed to go on for some time a very much longer molecule
called a polymer is formed.
Addition polymerization = when unsaturated monomers combine to form a large molecule or chain called a
polymer.
Conditions: catalyst + heat + high pressure
The monomer which the small alkene is the repeating subunit. The reaction can also be applied to alkenes which
have a hydrogen substituted usually by a halogen (chloroalkenes). This gives a wide variety of addition polymers.
Examples of addition polymers that you need to know very well: polythene and polyvinyl chloride (you should be
able to draw the monomer and general structure of the each of the above polymer).
In addition polymerization, the polymer has the same % carbons as its monomers as no atoms are removed.
Examples:
name
monomer
structural formula
name
polymer
structural formula
repeated unit
ethene
chloroethene
propene
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Economic importance of reactions of alkanes



Hydrogenation of vegetable oils in the manufacture of margarine
Hydration of ethene in the manufacture of ethanol which is used as a solvent, antiseptic or fuel
Polymerization in the manufacture of plastics
10. 4. Alcohols
10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols
10.4.1 Describe, using equations, the complete combustion of alcohols
10.4.2. Describe using equations, the oxidation reactions of alcohols
10.4.3 Determine the products formed by the oxidation of primary and secondary alcohols.
Primary, secondary and tertiary alcohols
The distinction is based on the number of alkyl groups or carbons bonded onto the carbon which carries the –OH
group;
primary
Has one alkyl group or 1 carbon
atom onto the carbon which carries
the –OH group
secondary
Has two alkyl groups or 2 carbon
atoms onto the carbon which carries
the –OH group
tertiary
Has three alkyl groups or 3 carbon
atoms onto the carbon which carries
the –OH group
This distinction is important as the product of the same type of reaction e.g. an oxidation will be different for each
type of alcohol. As such the three different types of alcohols can be considered different homologous series.
Complete combustion of alcohols
Alcohols can be combusted completely (very exothermic reaction) producing carbon dioxide and water as shown
by the symbol equation of the combustion of ethanol.
C2H 5OH + 3O2
 2CO2 + 3H2O
The general equation for the compete combustion of alcohols is
CnH2n+1 OH + (2n-1) O2
 n CO2 +
(n+1) H2O
Oxidation of alcohols using an oxidising agent
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Primary alcohols, such as ethanol, can be oxidised with an oxidising agent e.g. acidified potassium dichromate (V)
which goes from orange (oxidation state +6) to green (oxidation state +3) (potassium permanganate also works
well).
The oxidation of primary alcohols can yield two different products depending on the conditions under which it is
carried out.
1. Partial oxidation to form an aldehyde, e.g. ethanal:
Heating excess alcohol with oxidizing agent and but the aldehyde needs to be distilled as soon as it is formed
as otherwise it will oxidize further to an carboxylic acid.
Equation for the partial oxidation:
 full equation: 3C2H5OH + Cr2O72- + 8H+  3CH3CHO + 2Cr3+ + 7H2O

simpler version:
C2H5OH + [O]  CH3CHO + H2O
2. Full oxidation: if an excess of the oxidizing agent is used and the mixture is heated under reflux, the alcohol
oxidises to a carboxylic acid.
Equation for the further oxidation from ethanol to ethanoic acid:

full equation: 3C2H5OH + 2Cr2O72- + 16H+  3CH3COOH + 4Cr3+ + 11H2O

simpler version:
C2H5OH + 2[O]  CH3COOH + H2O
Secondary alcohols can only be oxidized to ketones whilst tertiary alcohols cannot be oxidized at all.
10. 5. Halogenoalkanes
10.5.1 Describe, using equations, the substitution reations reactions of halogenoalkanes with sodium hydroxide
10.4.2. Explain the substitution reactions of halogenoalkanes with sodium hydroxide in terms of SN1 and SN2.
Primary, secondary and tertiary halogenoalkanes
The distinction is based on the number of alkyl groups or carbons bonded onto the carbon which carries the
halogen;
primary
Has one alkyl group or 1 carbon
atom onto the carbon which carries
the halogen
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secondary
Has two alkyl groups or 2 carbon
atoms onto the carbon which carries
the halogen
tertiary
Has three alkyl groups or 3 carbon
atoms onto the carbon which carries
the halogen
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SN1 and SN2 mechanisms: SN = nucleophilic subsitution
Halogenoalkanes react with dilute sodium hydroxide in nucleophilic substitution reactions of which there are two
types. The type of nucleophilic substitution reaction which is favoured or predominant depends on the structure of
the halogenoalkane i.e. primary, secondary or tertiary. During such a reaction the halogen is substituted by the
hydroxide group to form an alcohol.
Animations of such reactions can be viewed on http://www.btinternet.com/~chemistry.diagrams/Animations.htm
In these reactions the hydroxide ion in sodium hydroxide, which is a negative ion, is called the nucleophile as it
‘seeks’ a positive nucleus.
There are two different ways in which a nucleophilic substitution can occur: SN1 and the SN2.
Whenever a halogenoalkane and a nucleophile react there is a competition between these two different pathways
or mechanisms, the SN1 and the SN2 pathway. Which one is favoured depends on the structure of the
halogenoalkane:

SN1 is favoured with tertiary halogenoalkanes

SN2 is favoured with primary halogenoalkanes

Both mechanisms occur in secondary halogenoalkanes.
SN1
Example: 2-bromo-2-methylpropane with sodium hydroxide to form 2-methyl propan-2-ol and sodium bromide.
Equation:
(CH3)3 CBr + NaOH  (CH3)3 COH + NaBr
Mechanism (multistep):



unimolecular or monomolcular (hence the 1 in SN1 ): rate of reaction depends on 1 molecule only i.e. the
tertiary halogenoalkane;
multistep as it involves 2 steps; involves the formation of an intermediate;
step 1: heterolytic fission results in dissociation or spontaneous ionisation of compound resulting in the
formation of a carbocation - which is the intermediate and can be isolated - and a halide ion; this step
is the slowest and therefore the rate determining step (ionisation caused by positive inductive
effect of alkyl groups
step 2: nucleophilic attack of the carbocation on either side of the carbon carrying the charge as the
halogen has already left (very fast step):
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
the rate of this multistep mechanism depends on the slowest reaction which is the first step and therefore the
rate of the reaction depends on the concentration of the halogenoalkane.
SN2
Example: bromoethane with sodium hydroxide to form ethaonl and sodium bromide.
Equation:
CH3CH2Br + NaOH  CH3CH2OH + NaBr
Mechanism (one step):

bimolecular as both reactants (halogenoalkane and nucleophile) are involved in the same 1 step which is also
the rate determining step; overall reaction is a one step process and involves a transition state which produces
the product instead of an intermediate.

step 1: as the carbon carrying the halogen is positively charged because of the polar CHal bond, the carbon is
attacked by the OH- on the side of the carbon atom opposite to the halogen (halogen affects
approach
by nucleophile) (=slow/rate determining); during this step the carbon-nucleophile bond is formed; the
nucleophile donates its electron pair to form the bond.
step 2: at the same time as step 1, the halogen breaks away (heterolytically) from the molecule
producing a negative halogen ion; the energy needed to break the bond comes from the bond
made between the OH- and the carbon atom; the halogen needs to break away from
the molecule which is a very unstable arrangement as there are 5 groups bonded onto the
carbon atom; the electron pair of the CHal bond is donated to the halogen.

nucleophile donates an electron pair which causes the bonding pair of the halogen to move away; no
intermediate is formed as the transition compound can not be isolated; a transition state is a dynamic process
during which bonds are broken and made; as the halogenoalkane accepts the electron pair it behaves like an
electrophile;
10. 6. Reaction pathways
10.6.1 Deduce reaction pathways given the starting product.
For the reactions in the diagram below you need to be able to identify all starting materials and the conditions to
achieve these reactions.
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alkane
dihalogenoalkane
halogenoalkane
alkene
alcohol
ketone
conversion
alkane  halogenoalkane
tri/tetra halogenoalkane
poly(alkene)
aldehyde
carboxylic acid
starting materials + equation
Free radical substitution
conditions
UV light
Alkane and halogen
Example: methane + chlorine
Equation:
CH4 + Br2  CH3Br + HBr
alkane  alkene
Addition reaction with hydrogen
180 C and nickel catalyst
alkene  halogenoalkane
Addition reaction with hydrogen halide
Normal conditions
alkene  dihalogenoalkane
Addition with halogen
Normal conditions
alkene  poly (alkene)
Addition polymerization
Heat, pressure and catalyst
alkene  alcohol
Catalytic addition with steam
Catalyst e.g. silicon dioxide,
heat, pressure and
concentrated H3PO4 or
H2SO4
UV light
Alkene + water
halogenoalkane 
dihalogenoalkane
Further free radical substitution
Alkane and halogen
Example: bromomethane + bromine
Equation:
CH3Br + Br2  CH2Br2 + HBr
dihalogenoalkane 
trihalogenoalkane
Further free radical substitution
UV light
Alkane and halogen
Example: dibromomethane + bromine
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Equation:
CH2Br2 + Br2  CHBr3 + HBr
dihalogenoalkane 
trihalogenoalkane
Further free radical substitution
UV light
Alkane and halogen
Example: tribromomethane + bromine (product is
tetrabromomethane)
Equation:
CHBr3 + Br2  CBr4 + HBr
halogenoalkane  alcohol
Nucleophilic substitution
Warm mixture
halogenoalkane + dilute sodium hydroxide
Example: 2-bromopronane + sodium hydroxide
alcohol  aldehyde
alcohol  ketone
aldehyde  carboxylic acid
Oxidation using acidified potassium dichromate

Alcohol and oxidising agent
Oxidation using acidified potassium dichromate



Alcohol and oxidising agent

Oxidation using acidified potassium dichromate

Alcohol and oxidising agent

using acidified potassium
dichromate (H+/H2SO4)
heating
distillation
using acidified potassium
dichromate (H+/H2SO4)
heating under reflux
using acidified potassium
dichromate (H+/H2SO4)
heating under reflux
Use the table above to describe how the conversions below can be achieved. For each conversion you must
include the reactions, the stages in the reactions, a balanced symbol equation for each reaction and the reaction
conditions.
a. ethene to ethanoic acid
b. ethene to 1,1,2-tribromoehtane
c. but-2-ene into butanone
d. but-1-ene to butanoic acid; also explain why the product is different from question (c)
e. propane to 1,1,2- trichloropropane
f.
pent-2-ene to 2,3-iodopentane
g. 1,2-dichloroethane to 1,1,2,2-tetrachloroethane
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h. 2-iodopropane to 2-methylpropan-2-ol
Volatility, solubility in water and acid-base behaviour
10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups: alcohol, aldehyde, ketone,
carboxylic acid and halide.
Order of increasing boiling points/volatility:
Comparison of boiling points of compounds of corresponding or very similar mass.
Lowest boiling points:
ALKENES
ALKANES
HALOGENOALKANES
ESTERS
ALKANALS/ALKANONES
AMINES
ALCOHOLS
Least volatile:
NON-POLAR MOLECULES
(Van der Waals’ forces)

INCREASING DEGREE OF POLARITY
(dipole-dipole attraction)

HYDROGEN BONDING
(INCREASED POLARITY OF
HYDROGEN)
ALKANOIC ACIDS
As alkanes and alkenes are non-polar (they are only very slightly polar in case of branched chains) their
intermolecular forces are weak as a result their melting and boiling points are low for their molar mass.
Halogenoalkanes, esters and alkanals and alkanones (less than alkanals) have a greater degree of polarity.
In the other functional groups hydrogen bonds are responsible for a greater attraction between the molecules;
although there are differences in the number and magnitude of the hydrogen bonds. The more polar the
hydrogen atom (which depends on the electronegativity of the other atom it is bonded onto), the stronger the
bond. This explains why acids have stronger hydrogen bonds than amines and alcohols; each molecule can also
make 2 hydrogen bonds.
Order of increasing solubility in water
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Again it is the polarity of the molecules which determines the solubility. Non-polar molecules are not soluble in
water while those that have hydrogen bonds are very soluble in water.
Insoluble
(non-polar molecules)
alkanes/alkenes
esters
halogenoalkanes
Soluble
(dipoles)
aldehydes/ketones
Very soluble
(hydrogen bonding)
alcohols
alkanoic acids
amines
Solubility generally decreases as molecules get longer; this is because the non-polar alkane ends cancel out the
effect of dipole or hydrogen bonds.
Exercises
1. Name and draw the full structural formula of the following organic compounds:
(a) CH3CH2CH(CH3)CH3
(b) CH3CHCHCH2CH3
(c) CH3CH(CH2CHCH2)CH3
(d) CH3CH2CHClCHBrCH3
(e) CH3CH2CH2OH
(f) CH3CH2CH2CH2CO2H
(g) CH3CHClCH2OH
(h) CH3CHBrCH2Cl
(i) (CH3)2CHCH3
(a) heptane
(b) 2-chloro-3-methyl hexane
(c) 3-bromo-2-chloroheptane
(d) pentan-2-ol
(e) hex-2-ene
(f) butanoic acid
2. Write structural formulae for:
3. Draw the formulae of the three structural isomers of formula C3H8O. How many can you find for C4H10O?
4. Draw the formulae of the structural isomers of formula, C4H8Cl2.
5. Draw the structures of
(a) butan-2-ol
(b) 2-amino-1-chlorobutane
(c) 2-chloropropanoic acid
(d) 1,1,2-tribromoethane
(e) ethane-1,2-diol
(f) buta-1,2-dienne
Other functional groups
10.1.13 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring (
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(RCOOR).
) and esters
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Identify the functional groups in the compounds below and draw structural formula of the compounds.
1. CH3COOCH3
2. CH3CH2CH(NH2) CH3
IB Questions
1. (M09)
2. (M09)
3.
(N06) Which of the following products could be formed from the oxidation of ethanol?
I. ethanal
A. I and II only
II. ethanoic acid
B. I and III only
III. ethane
C. II and III only
D. I, II and III
4. (M06) Propane, C3H8, undergoes incomplete combustion in a limited amount of air. Which products are
most likely to be formed during this reaction?
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A. Carbon monoxide and water
B. Carbon monoxide and hydrogen
C. Carbon dioxide and hydrogen
D. Carbon dioxide and water
5. (M06) What is/are the product(s) of the reaction between ethene and hydrogen bromide?
A. CH3CH2Br
B. CH3CH2Br and H2
C. CH2BrCH2Br
D. CH2BrCH2Br and H2
6. (M06) What is the correct name of this compound?
A. 1, -dimethylbutane
B. 2,4-dimethylbutane
C. 2-methylbutane
D. 2-methylpentane
7. (N02) A compound with the empirical formula C2H4O has a relative molecular mass of 88. What
is the formula of the compound?
A. CH3 CH2COCH3
B. CH3COOH
C. HCOOCH3
D. CH3CH2CH2COOH
8. (M02) When one mole of ethene reacts with two moles of oxygen gas
A. H is positive.
B. the oxidation number of carbon is unchanged.
C. an alcohol is formed.
D. carbon monoxide is produced.
9. (M02) What is the product of the reaction between bromine and ethene?
10. (N01) Which of the following statements about single and double bonds between two carbon atoms is
(are) correct?
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I. Double bonds are stronger than single bonds.
bonds.
A. I only
II. Double bonds are more reactive than single
B. II only
C. Both I and II
D. Neither I nor II
11. (N01) Which of the following is an amine?
12. (N01) The boiling points for several bromoalkanes are given below.
CH3Br (4 C)
CH2Br2 (97C)
CHBr3 (150 C)
the increase in boiling points is best attributed to changes in the strengths of
A. covalent bonds.
B. permanent dipole-dipole interactions.
C. hydrogen bonds.
D. van der Waals. forces.
13. (M01) Which compound is not a member of the same homologous series?
A. CH4
B. C2H4
C. C2H6
D. C3H8
14. (M01) Which are the most likely products of the incomplete combustion of a hydrocarbon?
A. Carbon dioxide and water
B. Carbon dioxide and hydrogen
C. Carbon monoxide and water
D. Carbon monoxide and hydrogen
15. (M01) The compound CH3CH2OH is reacted with excess acidified potassium dichromate(VI) solution.
What is the name of the functional group of the final organic product formed?
A. Alkanal
B. Alkanone
C. Alkanoic acid
D. Alkanol
16. (M01) In which pair do both types of compound take part in hydrogen bonding?
A. Alkanals and esters
B. Bromoalkanes and alkanals
C. Alkanes and alkenes
D. Alkanols and amines
17. (N00) How many different structural isomers have the formula C4H9Cl ?
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A. 2
B. 3
C. 4
D. 5
18. (N00) What will be formed when CH2 == CH2 reacts with Br2 in the dark?
A. CH2Br – CH2Br
B. CH3 ---CHBr2
C. CH2 == CHBr + HBr
D. CHBr == CHBr + H2
19. (N00) When the compounds below are listed in order of decreasing boiling point (highest to lowest)
what is the correct order?
1. ethane
A. 4, 3, 1, 2
2. fluoroethane
B. 4, 3, 2, 1
3. ethanol
C. 3, 4, 1, 2
4. ethanoic acid
D. 2, 1, 3, 4
PAPER 2
1. (M09)
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2. (N06)
(a) Butane, C4H10, and but-2-ene, C4H8, are both colourless gases at 0 °C.
(i) Write an equation for the complete combustion of but-2-ene.
[1]
(ii) Describe a chemical test, and its result, to distinguish but-2-ene from butane.
[2]
(iii) Calculate the volume that 0.0200 mol of butane would occupy at 70 °C and 1.10 x105
[3]
(b) CH3COCH3 is the first member of the ketone homologous series. Draw the full structural
formula of the next member of this homologous series and predict how its melting point compares with
that of CH3COCH3 .
[2]
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3. (N02) Organic compounds are arranged as homologous series due to the presence of functional groups.
(a) Explain the meaning of the terms homologous series and functional groups using alcohols as
examples.
[5]
(b) Under certain conditions ethene can be converted to ethanol.
(i) Give a chemical test to identify ethene and state what would be observed.
(ii) Give a balanced equation for the reaction to form ethanol from ethene.
(iii) State the conditions necessary for the reaction in (d) (ii).
[4]
(c) Ethanol may be converted to ethanoic acid.
(i) Identify the reagent needed and state the type of reaction.
(ii) State the colour change observed during the reaction.
[2]
[1]
4. (M01)
(a) The electronegativity value of carbon is 2.5 and that of oxygen is 3.5.
(i) Draw a Lewis (electron dot) structure for the carbon dioxide molecule, state its shape and give the
bond angle.
[3]
(ii) Using the concepts of molecular shape and bond polarity, predict, with an explanation, whether or
not the carbon dioxide molecule is polar.
[3]
(b) Explain at the molecular level why ethanol, C2H5OH, is soluble in water, but cholesterol, C27H45OH,
and ethane, C2H6, are not.
[4]
(c) Give an equation for the complete combustion of methane, CH4. Identify two products formed by the
incomplete combustion of methane, CH4, and identify one harmful effect caused by one of the
products.
[3]
5. (M00)
(a) Discuss the factors which affect the boiling points of covalently bonded compounds by reference to
the following pairs of organic substances, whose boiling points are given:



ethane (184 K) and butane (273 K);
ethane (184 K) and bromoethane (311 K);
bromoethane (311 K) and ethanol (352 K).
[8]
Mark scheme topic 10
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Paper 1
1
2
3
4
5
6
7
A
C
A
A
A
D
D
8
9
10
11
12
13
14
D
D
B
A
B
B
C
15
16
17
18
19
20
21
C
D
A
A
B
22
23
24
25
26
27
28
Paper 2
1. (M09)
2. (N02)
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3.
(b)
(c)
4.
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5.
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