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TOPIC 10 ORGANIC CHEMISTRY
10. 1 Introduction
Homologous series
10.1.1. Describe the features of a homologous series.
10.1.2. Predict and explain the trends in boiling points of members of a homologous series.
A homologous series is a set of compounds which has the following features:






share a general formula (i.e. same elements in the same ratio);
members share the same functional group; a functional group is a group of atoms in a compound with
characteristic chemical properties;
whose nearest neighbours differ by one repeating unit, most often -(CH3 ) group or a -(CH2)- group ;
have similar chemical properties (same functional group);
show a gradual change (gradation) in physical properties as shown by the table below which shows the
melting points and boiling points of some alkanes:
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
The table on the left shows a gradual
increase in boiling point with increasing
number of carbon atoms and therefore
increasing molecular mass.
A trend caused by the fact that as the
number of carbon 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.
solid
A graph of boiling points of alkanes against chain length gives a steep line at first but then flattens out at higher
numbers of carbon atoms suggesting that the size of molecules becomes less influential in affecting boiling point.
Formula of organic compounds
10.1.3 Distinguish between empirical, molecular and structural formulas.
type of formula
empirical formula
description
shows most simple whole number ratio of all atoms.
molecular formula
shows the actual number of the different atoms and how
many of each; no information on how the atoms are
arranged.
structural formula show how atoms are arranged
together in the molecule; a full structural formula
(sometimes called a graphic formula or displayed
formula) shows every atom and bond.
structural formula
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example
CH2
C6H14
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condensed structural
formula
structural formula which shows order in which atoms are
arranged but which omits bonds.
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 at least two parts: one to indicate the number of carbon atoms in the chain and the
other the functional group. Other parts will indicate length and number of branches or if the compound is cyclic.
Example:


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
 benz- = benzene ring


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.
Naming of halogenoalkanes
In the case of the halogenoalkanes, the name begins with the name of the halogen and not the number of carbon
atoms. The name 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.
Alkanes 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:
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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
Naming branched alkanes
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 should end with –yl
to indicate it is a branch e.g. methyl, ethyl, propyl. 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(s).
 the position of the functional group has priority over alkyl.
Examples: 2-methylpentane, 2,2-dimethylpentane, methylpropan-2-ol, 2,2,4-trimethylpentane
Structural isomerism
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.
In your class book draw all isomers for non-cyclic alkanes upto C6.
Physical properties of alkanes: melting/boiling points, viscosity, density
Melting points, boiling points, viscosity (resistance to flow) and density increase with increasing number of carbon
atoms
Explanation:
The larger the molecules (the greater the molar mass)…
 the more sites/surface area there are for intermolecular attraction, …
 or the greater the number of electrons, the greater the polarisation within the molecule, …
 or the greater their mass/inertia …
and therefore the greater the attraction between the molecules, the more energy is needed to increase the motion
of the molecules and cause the alkane to melt or boil; the greater the viscosity and the greater the density.
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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.
See table below: Straight chained alkanes have higher boiling points 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
boiling points. As branching decreases surface area, it increases volatility and decreases density.
molecular
boiling point
structural
formula
name
formula
(oC)
pentane
CH3CH2CH2CH2CH3
C5H12
2 methylbutane
CH3CHCH2CH3

CH3
C5H12
2,2-dimethylpropane
CH3

CH3C CH3

CH3
36
28
C5H12
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However, straight chained alkanes generally have lower melting points than branched alkanes (in particular with
symmetrical structures). 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.
Examples: melting point of octane, C8H8, is - 57 C whilst isomer 2,2,3,3- tetramethylbutane is 121 C.
Chemical properties of alkanes
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.
 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 easily undergo are combustion and substitution reactions.
Combustion of alkanes
 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.
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Examples:
CH 4 +
2O2
 CO2 + 2H2O
and
C3H 8 +
5O2
 3CO2 + 4H2O
 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.
Examples: CH 4 +
1 ½ O2
C3H 8 +
4O2
 CO2 + 2H2O and
 CO2 + 2CO +
CH 4 +
O2
 C + 2H2O
4H2O
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
When a covalent bond is broken the bonding pair electrons are redistributed between the two atoms; there are two
ways of redistributing these two electrons:
Homolytic fission
* occurs in non-polar bonds or bonds with a very low polarity when bonding electrons are fairly equally 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

A+
+
BDuring 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)
 CH3 Cl (g)
methane
+
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
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+
+
Br2 (l)
bromine
UV


C2H5 Br (g)
+
bromoethane
HCl (g)
+
hydrogen chloride
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Homolytic free radical substitution reaction
The sequence of steps/collisions by which substitution reaction takes place is called the reaction mechanism.
The reaction is an example of a free radical substitution reaction.
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).
CH4
Cl
+

CH3
+
HCl
The CH3 reacts with the Cl2 to form CH3Cl 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
CH3
Cl
+
+

CH3
CH3Cl

or
Cl
+
Cl

C2H6
If excess Cl2 is used further substitutions may take place until all the hydrogen atoms are substituted for:
UV
UV
CH3Cl
+
Cl2

CH2Cl2
+
HCl followed by CH2Cl2 +
Cl2  CHCl3
+
CHCl3
+
Cl2
UV

CCl4
+
Cl2 or
HCl
HCl
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.
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10.3.4.
the polymerization of alkenes.
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Structure of alkenes
There are 2 types of structures you need to know:
 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 mass.
Chemical properties of alkenes
As they are unsaturated alkenes are reactive. The second bond of the double bond is weaker than a single
carbon-carbon bond as it has a lower enthalpy change – less energy needed to break it.
Alkenes undergo addition reactions; atoms are added to the carbon chain using the double bond
Structural isomerism
In your classbook, draw the structural formula of all possible structural straight-chained alkenes up to C6 and name
each one of them.
Chemical properties: addition reactions
As they are unsaturated alkenes are reactive. 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 that 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 second bond of the 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.
Example: equations
molecular
word
structural
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ethene
C2H4 (g) +
+
bromine
Br2 (l)

 C2H4 Br2 (g)
1,2-dibromoethane (g)
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Test for unsaturation
The bromination reaction is useful as it can be used to distinguish between an alkane (no decolourization 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, 150 C – 200C, although it does also occur at room temperature if platinum or palladium are
used as catalysts.
Reaction with water (=hydration)
Reaction in which H and OH are introduced in the molecule: needs a strong concentrated acid (e.g. H3PO4 or
H2SO4) as catalyst, 300 C and 70 atm. The reaction is used to make industrial ethanol.
Example: equations
molecular
word
structural
C2H4 (g) +
H2O (l)
ethene +
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 at room temperature; product is a halogenoalkane.
Example: structural equation of reaction between but-2-ene reacts and 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 (each polymerization has own conditions)
The monomer, which is the small alkene molecule, 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 of polymerization reactions
name
monomer
structural formula
polymer
name
repeated unit
ethene
chloroethene
propene
Write structural equations for addition reactions between propene and bromine, water, hydrogen bromide and
hydrogen. Name all species and draw any isomers.
Economic importance of reactions of alkenes



Hydrogenation of vegetable oils in the manufacture of margarine.
Hydration of ethene in the manufacture of industrial ethanol which is used as a solvent, antiseptic or fuel.
Polymerization in the manufacture of plastics.
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

Synthesis of drugs.
Synthesis of ethane-1,2-diol which is used as antifreeze.
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.
Most important alcohol is ethanol which is used as a fuel, solvent, antiseptic, to make esters.
Primary, secondary and tertiary alcohols
Based on the number of alkyl groups or carbons bonded onto the carbon that carries the –OH group.
primary alcohol
Has one alkyl group or 1 carbon
atom onto the carbon which carries
the –OH group
secondary alcohol
Has two alkyl groups or 2 carbon
atoms onto the carbon which carries
the –OH group
tertiary alcohol
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.
The terms can also refer to actual carbon atoms within an alcohol molecule: primary, secondary or tertiary carbon
atom.
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
Oxidation of alcohols using an oxidising agent
Primary alcohols, such as ethanol, can be oxidised by heating them with an oxidising agent e.g. potassium
dichromate (V) which needs to be acidified (e.g. H2SO4) and which goes from orange (oxidation state +6) to
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green (oxidation state +3) (potassium permanganate also works well). The reaction involves the hydrogen atoms
bonded onto the carbon that carries the hydroxyl group.
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 from ethanol:
Heating excess alcohol with oxidizing agent; however, to obtain a high yield of the aldehyde it needs to be
distilled (it has a lower boiling point than ethanoic acid) as soon as it is formed as otherwise it will oxidize
further to a carboxylic acid (full oxidation).
Equation for the partial oxidation of ethanol:


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
Write a full structural equation showing the partial and then full oxidation of propan-1-ol.
Secondary alcohols can only be oxidized to ketones by heating under reflux with acidified potassium
chromate. This is because the ketone does not have any hydrogen atom left on the carbon of the carbonyl group.
Write a structural equation showing the oxidation of propan-2-ol.
Tertiary alcohols cannot be oxidized at all as they have no hydrogen atoms onto the carbon that carries the
hydroxyl group.
10. 5. Halogenoalkanes
10.5.1 Describe, using equations, the substitution 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
As with alcohols, the distinction is again based on the number of alkyl groups or carbons bonded onto the carbon
which carries the halogen.
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primary
Has one alkyl group or 1 carbon
atom onto the carbon which carries
the halogen
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
SN1 and SN2 mechanisms: SN = nucleophilic subsitution
Halogenoalkanes react with warm dilute sodium hydroxide solution 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 (more on this in HL).
During a nucleophilic substitution reaction the halogen is substituted by the hydroxide group to form an alcohol.
Animations of such a reaction 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 i.e. the carbon atoms that carried the halogen atom as the bond C-Hal is polar.
There are two different ways in which a nucleophilic substitution reaction can occur: SN1 and the SN2.
Whenever a halogenoalkane and a nucleophile react there is competition between these two different pathways or
mechanisms, the SN1 and the SN2 pathway. Which one is favoured depends on different factors some of which
are studied in HL.
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 i.e. carbocation;
first order with respect to the halogenoalkane;
molecularity of first step is 1;
rate = k [CHal]
steps:
step 1
(slow step)
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Heterolytic fission results in dissociation or spontaneous ionisation of halogenoalkane
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.
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step 2
(fast)
Nucleophilic attack of the carbocation on either side of the carbon carrying the charge as the
halogen has already left and the ion has a planar shape.
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 ethanol and sodium bromide.
Equation:
CH3CH2Br + NaOH  CH3CH2OH + NaBr
Mechanism (one step) (overall second order):


bimolecular (hence the 2 in SN2 ): as both reactants (halogenoalkane and nucleophile) are involved in the
same single step which is also the rate determining step; overall reaction is a one step process and involves a
transition state/activated complex which produces the product instead of an intermediate.
rate = k [CHal][Nu]

In the single step, a new bond is made at the same time as an old bond is broken:
Bond broken
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
(the halogen atom affects approach by nucleophile); during this process the carbonnucleophile bond is formed; the nucleophile donates its electron pair to the carbon to form
the bond.
Bond made
at the same time, 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.
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In summary, the nucleophile donates an electron pair to the carbon that carries the halogen atom 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.
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.
alkane
dihalogenoalkane
halogenoalkane
alkene
alcohol
tri/tetra halogenoalkane
poly(alkene)
aldehyde
carboxylic acid
starting materials + equation
Free radical substitution
conditions
UV light
ketone
Summary of organic reactions
conversion
alkane  halogenoalkane
Alkane and halogen
Example: methane + chlorine
Equation: CH4 + Br2  CH3Br + HBr
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alkene  alkane
Addition reaction with hydrogen - hydrogenation
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 - hydration
300 C, 70 atm, concentrated
H3PO4 or H2SO4
halogenoalkane 
dihalogenoalkane
Alkene + water
Further free radical substitution
Alkane and halogen
dihalogenoalkane 
trihalogenoalkane
Example: bromomethane + bromine
Equation: CH3Br + Br2  CH2Br2 + HBr
Further free radical substitution
Halogenoalkane and halogen
UV light
UV light
Example: dibromomethane + bromine
Equation: CH2Br2 + Br2  CHBr3 + HBr
trihalogenoalkane 
tetrahalogenoalkane
Further free radical substitution
Halogenoalkane and halogen
UV light
Example: tribromomethane + bromine (product is
tetrabromomethane)
Equation: CHBr3 + Br2  CBr4 + HBr
halogenoalkane  alcohol
Nucleophilic substitution, SN1 or SN2
Warm mixture
Halogenoalkane + dilute sodium hydroxide
Example: 2-bromopropane + sodium hydroxide
alcohol  aldehyde
Oxidation using acidified potassium dichromate

Alcohol and oxidising agent
alcohol  ketone
Oxidation using acidified potassium dichromate



aldehyde  carboxylic acid
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 reaction(s), the stages in the reactions if any, a balanced symbol equation using structural formula for
each reaction and the reaction conditions. Better to answer this in your exercise book.
a. ethene to ethanoic acid
b. ethene to 1,1,2-tribromoethane
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c. but-2-ene into butan-2-one
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
h. 2-iodo-2-methylpropane to 2-methylpropan-2-ol
i.
1-bromopropene to propan-1-ol
j.
1-bromopropane to propanal
k. 3-bromohexane to hexan-3-one
l.
butane to butan-1-ol
m. propene to 1,2,3-trichloropropane
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/decreasing volatility
Comparison of boiling points of compounds of corresponding or very similar mass.
The main factor is the strength of the intermolecular forces BETWEEN the molecules.
Homologous series
Lowest boiling points
ALKENES
ALKANES
HALOGENOALKANES
Least volatile/highest
boiling point
ALKANALS
KETONES
AMINES
ALCOHOLS
CARBOXYLIC ACIDS
Type of intermolecular force
Non-polar molecules
Weak Van der Waals’ forces
Polar molecules
dipole-dipole attraction)
HYDROGEN BONDING
(INCREASED POLARITY OF HYDROGEN)
As alkanes and alkenes are non-polar their intermolecular forces are weak and as a result their melting and boiling
points are low for their molar mass. Halogenoalkanes, esters and aldehydes and ketones (less than aldehydes)
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
soluble
(dipoles)
aldehydes/ketones
very soluble
(hydrogen bonding)
alcohols
carboxylic acids
amines
halogenoalkanes
Solubility generally decreases as molecules get longer; this is because the non-polar alkane ends cancel out the
effect of dipole or hydrogen bonds.
Other functional groups
10.1.13 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring (
and esters (RCOOR).
)
Identify the functional groups in the compounds below and draw structural formula of the compounds.
1. CH3COOCH3
2. CH3CH2CH(NH2) CH3
3. C6H5C2H5
4. C6H5NH2
Exercises
1. Name and draw the full and condensed 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
(g) 3-ethylpentane
(h) 2-methylpropanal
(i) 3-methylbutanoic acid
(j) 2-methylpentan-3-one
(k) 3-chlorobutan-2-ol
(l) 2,3-dihydroxybut-2-ene
(m) butan-2-ol
(n) 2-amino-1-chlorobutane
(o) 2-chloropropanoic acid
(p) 1,1,2-tribromoethane
(q) ethane-1,2-diol
(r) buta-1,2-diene
2. Write the full structural formulae for:
3. Draw the full structural formulae of the structural isomers of (a) C3H8O and (b) C4H10O.
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4. Draw the full structural formulae of the structural isomers of formula, C4H8Cl2.
5. From the list of compounds below select the one which is
(a) an ester
(b) a secondary alcohol
(c) a ketone
(a) CH3CH(CH3)CHO
(b) CH3CH2OCH3
(c) (CH3)3COH
(d) CH3CH2CH(OH)CH3
(e) CH3CH2COCH3
(f) CH3CH2CHCHCH3
(g) CH3CH2COOCH3
(h) (CH3)2CHCOOH
(d) an alkene
6. Classify the following halogenoalkanes as primary, secondary or tertiary
(a) (CH3CH2)3C-Cl
(b) Br-CH2CH2CH2-Br
7. Classify the following amines as primary, secondary or tertiary.
(a)
8.
(CH3CH2)3C-NH2
(b) CH3-NH-CH2-C(CH3)3
Name and draw condensed formula for the products of the following reactions:
a. CH3CH2CH2CH2OH heated with acidified potassium dichromate
b. CH3CHOHCH2CH3 heated with acidified sodium dichromate
c. CH3CH2CH2CH3 reacting with bromine in UV
d. CH3CH2CHCH2 reacting with concentrated hydrogen bromide in normal conditions
e.
CH3CHCH2 reacting with bromine in normal conditions
f.
CH3C(CH3)OHCH3 heated with acidified potassium dichromate
g.
CH3C(CH3) CHCH3 and water; heated to 300 C with concentrated phosphoric acid; pressure of 70 atm
h.
CH3CHO heated with acidified sodium dichromate
Aim 8 : raise awareness of the moral, ethical, social, economic and environmental implications of
using science and technology.
Within the context of this topic the issue of burning fossil fuels can be raised:
There are of course environmental concerns with the burning of fossil fuels:
 release of nitrogen oxides and sulphur oxides to produce acid rain
 production of carbon dioxide which is a greenhouse gas
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
production of carbon monoxide which is toxic when fuels undergo incomplete combustion.
In addition, alkanes and other fuels such as alcohols can now also easily be converted into chemical products
which could be more useful than fuel i.e. dyes, plastics, fibres and so on.
This raises the question as “are some fuels too valuable to burn?”
Factors to be considered whether to continue to use an organic compound as a fuel or process further could be:
 How valuable is it as a fuel? i.e. how much energy does it release per mole/gram of fuel? For instance
small alkanes release a lot of energy per mole/gram; more than for instance alcohols. This can be
checked by using the data booklet (molar enthalpies of combustion) or considering the number of bonds
broken and made, strength of bonds, molar mass, ..
 How much does it pollute? How much CO2/CO does it produce?
 How easy is it to transport/store?
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