Organic Chemistry
 What do we mean by “organic” chemistry?
 What do the following things below have in common?
Organic-dictionary definition
 Oxford English dictionary says:
 “organic” means derived from living matter, i.e. not
produced artificially
 For example, “organic food” refers to the food that is
produced without the use of chemical fertilisers,
pesticides or other artificial chemicals.
Organic- chemical definition
 Of all the elements in the periodic table, one is
much more versatile (stands out) from the rest
 Carbon can form more compounds than any other
element
 Chemical definition: Organic chemistry is the
study of compounds containing carbon (other
than simple binary compounds and salts) and
chiefly or ultimately of biological origin.
 Question: Is carbon itself organic or inorganic?
Carbon-revisited hopefully!
 In group 4 of the periodic table
 Electronic configuration: 2,4 (SL) or 1s2 2s2 2p2
(HL)
 Has a valency (combining power) of 4 with 4
valence (outer shell) electrons
 Achieves the noble gas configuration of neon by
forming 4 covalent bonds
 Take methane for example: carbon is bonded
covalently to 4 hydrogen atoms to achieve its octet
Carbon- continued
 The valency of 4 gives rise to a unique property called catenation
 Catenation: spontaneous linking of atoms of certain chemical

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

elements, such as carbon atoms, to form long chains or stable rings
by forming covalent bonds with itself
C-C bond enthalpy = 348 kJ mol-1
C-H bond enthalpy = 412 kJ mol-1
Catenation allows billions of organic compounds to be formed
Carbon compounds > ∑ all other compounds of all the elements in
the periodic table except hydrogen (since almost all organic
compounds also contain hydrogen)
7 million Organic Compounds
1.5 million Inorganic Compounds
Organic chemistry- link to nature
 Life is based on organic compounds
 The 4 major classes of biomolecules are all organic
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(contain carbon)
Carbohydrates, Proteins, Lipids & Nucleic Acids
Organic chemistry is an excellent foundation for
biochemistry
Biochemistry is just APPLIED organic chemistry!
Just like physics is applied maths!
Hydrocarbons
 A compound which consists of carbon & hydrogen
ONLY
 Are the basis for most organic compounds
 Includes alkanes, alkenes, alkynes (aliphatic
compounds), & arenes (aromatic compounds)
Organic compound familieshomologous series
 Can be grouped into different families with a common
functional group (part of the molecule which gives rise to
common reactivity)
 Homologous series: a group of organic compounds that
follow a regular structural pattern and have the same
general molecular formula, differing only by the addition of
a methylene, -CH2- group, they have almost identical
chemical properties, with physical properties e.g. boiling
point increasing gradually as the number of carbon atoms
increase
Homologous series
 Take the alkanes for example:
 The 1st 4 are methane, ethane, propane, butane
Homologous series- properties
Successive compounds differ from each other by
a -CH2- unit (methylene group)
2) The compounds can all be represented by a
general formula (e.g. alkanes CnH2n+2; if n=3
then the formula is C3H8
3) The compounds have similar chemical
properties
4) Successive compounds have physical properties
that vary in a regular manner as the number of
carbon atoms increases
1)
Homologous series- general
trends in physical properties
 For the alkanes:
As the number of carbon atoms increase the
melting/boiling points gradually increase
 This is caused by an increase in the molar mass of
the molecule and hence there is greater chance of
more temporary dipoles being induced in the case
of hydrocarbons (Van der Waals forces)
Graph of boiling points for the first 10
alkanes
 Curve is initially quite steep, as for small molecules, the addition of an extra
carbon has a proportionally larger effect on the molar mass (eg, from CH4 to C2H6
there is an increase of 97.5%) & hence on the strength of the van der Waals’ forces
 As the length of the carbon chain increases, the percentage change in molar
masses becomes progressively smaller (there is a 10.9% increase in molar mass
from C9H20 to C10H22, hence the curve flattens
 The trend is the same for other physical properties, such as density & viscosity for
the same reasons
Empirical, molecular, structural
(condensed & full) formulae
 Using ethane, C2H6, as an example
 Empirical formula: simplest whole number ratio of atoms e.g. CH3
 Molecular formula: actual number of atoms of each type present in a
molecule e.g. C2H6
 Full structural formula: show the relative positioning of all the atoms
in a molecule & the bonds between them
 Condensed structural formula: omits bonds which can be assumed &
groups atoms together e.g CH3CH3
Shapes of Alkanes

“Straight-chain” alkanes have a zig-zag orientation in 3-D
Formulae- worked example
 Write the empirical, molecular, full structural &
condensed structural formula for ethanol, CH3CH2OH
Answer:
 Empirical: C2H6O
 Molecular: C2H6O
 Full structural:
Further functional groups
 Alkenes: CnH2n e.g. ethene C2H4
 Alkynes: CnHn e.g. ethyne C2H2
 Alcohol: CnH2n+1OH e.g. methanol CH3OH
 Aldehydes: RCHO, e.g. methanal (formaldehyde) HCHO
 Ketones: RCOR’ (R’ can be the same alkyl group as R or different e.g.





propanone (acetone), CH3COCH3
Carboxylic acids: RCOOH, e.g. methanoic (formic) acid, HCOOH
Haloalkanes: RX (X=F,Cl,Br,I) e.g. iodomethane, CH3I
Amines: RNH2 e.g. methylamine, CH3NH2
Esters: RCOOR’ e.g. methyl methanoate HCO2CH3
Arenes: based on the phenyl group (C6H5-) e.g. benzene C6H6
Isomers
 Compounds with the same molecular formula but with
different arrangement of atoms in the molecule
 For the alkanes:
from butane (4 carbon atoms), there is more than 1
structure possible
 C4H10 can refer to either butane, CH3CH2CH2CH3 or
methylpropane, CH3CH(CH3)CH3
Isomers
Structural Isomers
 Different connectivity of atoms
E.g.:
 Butane & methylpropane are structural isomers of each
other

3 types:
1) Chain isomerism
2) Position isomerism
3) Functional group isomerism
Chain isomerism
 Different arrangement of the carbon skeleton
 For example butane & methylpropane are chain
isomers of each other
 Have similar chemical properties but different physical
ones; the branched isomer has a lower melting/boiling
point than the straight-chained one
H
H
H
H
H
C
C
C
C
H
H
H
butane
H
H H
H
CH3
H
C
C
C
H
H
H
methylpropane
H
Position Isomerism
 Have functional group placed at a different position
along the carbon skeleton
 E.g. 1-bromopropane & 2-bromopropane
 Has similar physical & chemical properties
Br
H
H
H
C
C
C
H
H
H
1-bromopropane
H
H
H
Br
H
C
C
C
H
H
H
2-bromopropane
H
Functional group Isomerism
 The molecules have different functional groups (hence
different chemical properties)
 E.g. ethanol & methoxymethane all have the same
molecular formula of C2H6O, however ethanol is an alcohol
while methoxymethane is an ether
ethanol
methoxymethane
Naming organic compounds
 International Union of Pure & Applied Chemistry (IUPAC) devised




nomenclature method
It is irregular for up to 4 carbon atoms, just have the prefixes meth-,
eth-, prop-, butFrom 5 carbon atoms & above the naming becomes systematic, like
those of geometrical shapes, e.g. pent- (pentagon), hex- (hexagon),
hep- (heptagon), oct- (octagon) etc
The name of any organic compound is usually made up of 3 parts:
prefix (substituents), stem (number of carbon atoms in main chain)
& suffix (homologous series of main carbon chain)
e.g. methylpropane
methyl- CH3 substituent
prop- 3 carbon atoms in main chain
-ane- belongs to the alkane homologous series
Mimic for first four prefixes
Monkeys
Eat
Peeled
Bananas
Counting to Ten in Organic
01 = meth
02 = eth
03 = prop
04 = but
05 = pent
06 = hex
07 = hept
hept)
 08 = oct
 09 = non
 10 = dec







Mother
Enjoys
Peanut
BUTter
PENTagon
HEXagon or HEX nut
HEPTember (Roman sept is Greek
OCTober
NONember (Roman nov is Greek non)
DECember
Naming side chains
 If there are alkyl groups (R groups) attached in isomers, the
prefix for the alkyl groups must be used
Prefix
Name of
alkane
Name of Alkyl
group
Structure of
alkyl group
Meth-
Methane
Methyl
CH3-
Eth-
Ethane
Ethyl
CH3CH2-
Prop-
Propane
Propyl
CH3CH2CH2-
But-
Butane
Butyl
CH3CH2CH2C
H2-
Pent-
Pentane
Pentyl
CH3CH2CH2C
H2CH2-
Hex-
Hexane
Hexyl
CH3CH2CH2C
H2CH2CH2-
Steps in naming alkanes- step 1
For the molecule CH3CH(CH3)CH(CH3)CH3
Identify the longest continuous carbon chain (This
may not be the most obvious, straight one), this gives
the stem, given by the 4 blue carbon atoms
Steps in naming alkanes- step 2
CH3CH(CH3)CH(CH3)CH3


Identify & name the side-chains/substituent groups
as the prefix of the name.
Here there are 2 different methyl groups
Steps in naming alkanes- step 3
CH3CH(CH3)CH(CH3)CH3
 Where there is more than 1 side-chain of the same
type, like here, use the prefixes di-, tri-, tetra- and
so on, to indicate this.
 If there are several side-chains within a molecule,
put them in alphabetical order, seperated by
dashes.
 There are 2 methyl groups- hence the prefix is
dimethyl
Steps in naming alkanes- step 4
1CH 2CH(CH )3CH(CH )4CH
3
3
3
3
 Identify the position of the side chains.
 The carbon chain is numbered from the end
which will give the substituent groups the
smallest number.
 Here 1 methyl group is attached to carbon
number 2; the other to carbon number 3.
 The numbers precede the name and each digit is
separated by a comma from the next digit
 Hence the name of this compound is 2,3dimethylbutane
Some exercises
 Name the following compound:
H
H
C
H
H
H
C
H
C
H
C
H
H
C
H
H
H
H
H
C
C
C
H
H
H
H
Step 1- Find the longest carbon chain
H
H
C
H
H
H
C
C
H
H
C
H
H
C
H
H
H
H
H
C
C
C
H
H
H
H
Step 2- Look for any substituents attached
2 methyl
groups
H
H
C
H
H
H
C
C
H
H
C
H
H
C
H
H
H
H
H
C
C
C
H
H
H
H
Steps 3 & 4- Find the location of the substituents to give
the lowest possible overall number
2 methyl
groups
Using the green
numbering gives the
lower substituent
number of 3,3dimethylhexane H
Hence green is
correct
H
H
C
H
3
C
4
C
C
4
H
H
C
2
H
5
C
1
H
6
H
H
H
5
C
3
H
Using the orange
numbering gives the
higher substituent
number of 4,4dimethylhexane
H
H
H
H
2
6
C
H
1
H
Answer
 Hence the name of the molecule is 3,3-dimethylhexane
H
H
C
H
H
H
C
H
C
H
C
H
H
C
H
H
H
H
H
C
C
C
H
H
H
H
Isomers of C4H10- IUPAC vs common name
Methylpropane
methylbutane
dimethylpropane
IUPAC
common
Some well know molecules - IUPAC vs common
name
IUPAC: 2,3,5,4,6-Pentahydroxyhexanal
Common: glucose
IUPAC: 3-carboxy-3-hydroxypentanedioic acid
Common: citric acid
Alkene isomers
 Have the general formula CnH2n
 Have the C=C functional group within the chain
 Simplest alkene is ethene, C2H4
 If molecule is longer than 3 carbon chains, the
double bond can be in more than 1 position
ethene
propene
Naming alkene isomers
 Same 1st 4 steps as alkanes except the name
(suffix) ends in -ene instead of -ane
 Step 5- The position of the double bond C=C is
shown by inserting the numb er of the carbon
atom at which C=C starts
 E.g. for the isomers of C4H8
 CH3CH2CH=CH2 is but-1-ene
 CH3CH=CHCH3 is but-2-ene
 CH2=C(CH3)2 is 2-methylprop-1-ene
Practice problem
 What is the name of the following alkenes:
1) CH3CH=CHCH2CH2CH3
2) CH3CH2CH(CH3)CH=CH2
3) CH2=C(CH3)CH2CH=CH2
Answer:
1) hex-2-ene
2) 3-methylpent-1-ene
3) 2-methylpent-1,4-diene
Naming alcohols (ROH)
 Always end in –ol
 Simplest is methanol
 Like alkenes, the position of the –OH group must be specified after
ethanol
 e.g.
 CH3CH2CH2OH is propan-1-ol
 CH3CH(OH)CH3 is propan-2-ol
ethanol
Propan-1-ol
Propan-2-ol
Practice problems
 Name the following alcohols:
1)
2)
CH3C(OH)(CH3)CH2CH2CH3
Answers
1)2-methylpentan-2-ol
2) propan-1,2,3-triol
Naming aldehydes (RCHO)
 Always end in –al NOT (ol)!
 Simplest is methanal
 -CHO group is always at the end, so this carbon must be
carbon 1 so unnecessary to specify location
 E.g.
methanal
ethanal
propanal
Practice problems
 Name the following aldehydes:
1) CH3CH2CH2CHO
2) HCOCH2
CH2CH3
Answer:
1)butanal
2)butanal
Naming ketone (RCOR’)
 Always ends in suffix –one
 Simplest is propanone (acetone)
 The C=O (carbonyl) group can be inserted anywhere along the
hydrocarbon chain except at the end (why?)
 After butanone, the position of the carbonyl group must be
shown
 E.g pentan-2-one & pentan-3-one
propanone
butanone
Practice problems- ketones
 Name the following ketones:
1) CH3COCH2CH2CH2CH3
2) CH3CH2CH2COCH3
OC
CH3
CH2CH2CH3
Answer:
1) hexan-2-one
2) pentan-2-one
3) pentan-2-one
Naming carboxylic acid (RCOOH)
 End in –oic acid
 Like aldehydes, COOH is always the terminal
group & hence this carbon is always carbon number
1
Methanoic acid
Ethanoic (acetic)
acid
Propanoic acid
Practice problems- carboxylic acids
Name the following carboxylic acids:
1) CH3CH2CH2CO2H
2)HOOCCH2CH2CH3
3) CH3CH2CH(CH3)CH2COOH
Answer:
1) butanoic acid
2)butanoic acid
3) 3-methylpentanoic acid
Naming Haloalkanes (RX)
 Have the prefix halo- e.g fluoro-, chloro-, bromo-. Iodo Involve substituting a halogen atom into an alkane
 Same numbering as alcohols & ketones
 Numbers must be used after 2 carbon (from propane) atoms
 E.g. CH3CHClCH3 is 2-chloropropane
 CH3CH2CHClCH2CH2Br is 1-bromo-3-chloropentane
 Use the prefixes, di-, tri-, tetra- is there are more than 1 of the
same type of halogen atom e.g.
chloromethane
dichloromethane
Trichloromethane
(chloroform)
Tetrachloromethane
(carbon tetrachloride)
Practice problems
 Name the following haloalkanes:
1) CH3CH2FCHCH2CH3
2) CH3CH2CH2CH2CHBrCH3
 Answers:
1) 3-fluoropentane
2)2-bromohexane
Further functional groups
Homologous
series
Amine
Esters
Aromatic
compounds
Functional
group
R-NH2
RCOOR’
C6H5-
Prefix
Amino-
Alkyl
phenyl-
Suffix
Example(s)
Structural
formulae
Methylamine
CH3NH2
2-aminobutane
CH3CH(N
H2)CH2CH3
Ethyl ethanoate
CH3CO2CH
2CH3
Propyl ethanoate
CH3CO2CH
2CH2CH3
Benzene
C6H6
-amine
Alkanoate
-benzene
methylbenzene
C6H5CH3
Primary compounds
 Primary carbon atom attached to a functional
group & also to at least 2 H atoms & 1 alkyl (R)
group e.g.
H
R
C
H
General
structure
H
OH
CH3
C
H
ethanol
OH
Secondary compounds
 Secondary carbon atom attached to a functional
group & just 1 H atoms but 2 alkyl (R) group e.g.
H
R
C
H
OH CH3
R’
General
C
H
OH
CH3
Propan-2-ol
CH3
C
OH
CH2CH3
Butan-2-ol
Tertiary compounds
 Tertiary carbon atom attached to a functional
group & also 3 alkyl (R) groups with no H
attached e.g.
R’’
R
C
R’
General
CH2CH3
CH3
OH CH3
C
OH
CH3
2-methylpropan-2-ol
CH3
C
OH
CH2CH3
3-methylpentan-3-ol
CH2CH2CH3
CH3
C
OH
CH2CH3
3-methylhexan-3-ol
Practice questions
 Are the following molecules primary, secondary or
tertiary?
1) 1-chlorobutane
2) 2-bromobutane
3) 2-chloro-2-methylbutane
1-chlorobutane
H
H
C*
CH2CH2CH3
Cl
 C* attached to 2 H & 1 alkyl group
 Hence it’s primary
2-bromobutane
H
CH3 C*
CH2CH3
Br
 C* attached to 1 H & 2 alkyl groups
 Hence it’s secondary
2-bromo-2-methylbutane
CH3
CH3 C*
CH2CH3
Br
 C* attached to no H & 3 alkyl groups
 Hence it’s tertiary
Volatility
 Measure of how easily a compound evaporates
 Depends on intermolecular forces as kinetic energy
which hold the molecules together must be overcome
 Compounds with stronger intermolecular forces will
evaporate less readily, hence have higher boiling points
 The following factors influence volatility:
1) Molecular size
2) Shape of molecule (branched/linear)
3) Functional group
Volatility 2
 Compounds with longer carbon chains & hence greater
molar mass have higher boiling points & lower volatility
 A molecule with greater molar mass has more electrons
present hence more temporary dipoles can be induced
leading to more Van der Waals’ forces
 The early members of a series e.g. methane to octane are
gases & liquids respectively, at room temperatue
 The latter members are more likely to be solids e.g. C40-
Volatility 3
 Also depends on the number of points of contact
between molecules
 Branched isomers usually have lower boiling
point/higher volatility than straight-chained linear
ones
 Less surface area for attraction in branched
isomers
 Hence methylpropane has lower boiling point than
butane
Diagram of linear & branched
molecules
Volatility 4
 Molecules with functional groups that contain
hydrogen bonds are likely to have the highest
boiling point e.g. ethanol (78C)
 Molecules with functional groups that contain
dipole-dipole attractions are likely to have
intermediate boiling points,e.g. ethanal (20.2 C)
 Molecules which just have van der Waals’ forces
usually have the lowest boiling point e.g propane,
(−42.1 °C)
Diagram of different functional groups vs
volatility
Fair test
 When comparing molecules for boiling
points/volatility in different homologous series, it
is important to compare molecules of similar molar
mass
 This rules out molar mass as being a factor for
different boiling points since it becomes the
“constant” variable
 Hence ethanol (Mr=46) was compared with ethanal
(Mr=44) & propane (Mr=44)
Solubility in water
 Depends on nature of functional group & length of hydrocarbon chain
 Polar functional groups with hydrogen bonding e.g. ethanol are usually the most
soluble = most hydrophilic
 Hydrocarbon chains e.g. alkanes are likely to be least soluble = most hydrophobic
 Depends on relative length of hydrocarbon chain
e.g. very long hydrocarbon chain can overwrite effects of polar functional group
e.g. for the very long chained fatty acids which are solids i.e. butter not oil
Alkanes
 Hydrocarbons
 General formula= CnH2n+2
 Are saturated hydrocarbons
 Saturated- an organic molecule which only has C-C
single bonds & no C=C multiple bonds
 Unreactive in general
Low reactivity of alkanes
 Only have strong C-C & C-H bonds
 Need a large energy input to break these strong bonds
 Hence stable/unreactive under most conditions
 C-C & C-H bonds nonpolar as have similar
electronegativities (EN= 0.4)
 Hence no electron-rich or electron-deficient sites
 2 main reactions:
1) Combustion
2)Halogenation
Combustion of alkanes
 Very exothermic
 Due to high strength of C=O in carbon dioxide & O-H
bond in water
 Burns in excess oxygen to give carbon dioxide & water
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ mol-1
 In a limited supply of oxygen, carbon monoxide & water
are produced:
CH4(g) + 1.5O2(g)  CO(g) + 2H2O(l)
• In an extremely limited supply of oxygen, carbon itself
produced with water
CH4(g) + O2(g)  C(s) + 2H2O(l)
Combustion of alkanes- questions
Write the equation for the reaction of propane in excess
oxygen, limited oxygen & extremely limited oxygen:
Excess oxygen: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Limited oxygen: C3H8(g) + 3.5O2(g)  3CO(g) + 4H2O(l)
Trace amounts of oxygen: C3H8(g) + 2O2(g)  3C(s) + 4H2O(l)
Halogenation of alkanes
 If subjected to UV light, methane reacts with chlorine to
form chloromethane & hydrogen chloride
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
 In excess chlorine, further reaction takes place to produce
dichloromethane, trichloromethane (chloroform) &
tetrachloromethane
Dichloromethane: CH4(g) + 2Cl2(g)  CH2Cl2(g) + 2HCl(g)
Trichloromethane: CH4(g) + 3Cl2(g)  CHCl3(g) + 3HCl(g)
Tetrachloromethane: CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g)
Halogenation of alkanes-problems
Write the equation for the reaction between ethane and
1 mol of bromine:
C2H6(g) + Br2(g)  C2H5Br(l) + HBr(g)
General reaction:
Alkane + halogen  haloalkane + hydrogen halide
RH + X2  RX + HX
R= alkyl group X= halogen atom
Halogenation of alkanes mechanism
 Substitution reaction (chlorine atoms replace
hydrogen atoms in the methane molecule)
 Substitution reaction- a reaction where 1 atom or
group of atoms is replaced by another atom or
functional group
 Mechanism is free radical substitution
 Degree of substitution hard to control, often mixture
of products formed e.g.
 Ethane reacts with excess bromine to form a mixture of
dibromoethanes:
C2H6(g) + 2Br2(g)  C2H4Br2(g) + 2HBr(g)
Free radicals
 Species with unpaired electron e.g. Cl., Br.
 Extremely reactive
 Formed by the homolytic fission of a molecule:
Cl-Cl  2Cl. (produced by action of UV light)
Homolytic fission- the breaking of a covalent bond so
that 1 electron from the bond is left on each atom,
resulting in the formation of 2 free radicals
Free radical substitution
mechanism
• Is a substitution reaction as a halogen atom replaces a
hydrogen/alkyl group.
• 3 steps:
1) Initiation (net increase in radicals)
2) Propagation (chain reaction- amount of radicals
stays the same)
3) Termination (net decrease in radicals)
Mechanism in detail
UV
 Initiation: Cl-Cl(g)  2Cl.(g)
 Propagation: Cl.(g) + CH4(g)  .CH3(g) + HCl(g)
.CH3
(g) + Cl2(g)  CH3Cl(g) + Cl.(g)
 Termination: Cl.(g) + Cl.(g)  Cl2(g)
Cl.(g) + .CH3(g)  CH3Cl(g)
.CH (g) + .CH (g)  C H (g)
3
3
2 6
Overall reaction: CH4(g) + Cl2(g)  CH3Cl(g) + HCl(l)
Exercise
 Write the free radical reaction mechanism for the reaction
of bromine with ethane:
 Initiation: Br-Br(g)  2Br.(g)
 Propagation: Br.(g) + C2H6(g)  .C2H5(g) + HBr(g)
.
C2H5 (g) + Br2(g)  C2H5Br(g) + Br.(g)
 Termination: Br.(g) + Br.(g)  Br2(g)
Br.(g) + . C2H5(g)  C2H5Br(g)
. C H (g) + .C H (g)  C4H (g)
2 5
2 5
10
Overall reaction: :
C2H6(g) + Br2(g)  C2H5Br(l) + HBr(g)
Alkenes
 Unsaturated hydrocarbons with C=C double bond
 Unsaturated molecule- a molecule with 1 or more C=C






double bonds
General formula: CnH2n
C=C double bond stronger & shorter than C-C single bond
Double bond consists of a stronger sigma () bond &
weaker pi () bond
Quite reactive due to double bond
Take part in addition reactions
Addition reaction- a reaction where 2 (or more)molecules
combine together to form a single molecule
Hydrogenation of alkenes
 Alkene + hydrogen  alkane
 E.g.
 C2H4(g) + H2(g)  C2H6(g)
H
H
C
+ H2
C
H
ethene
H
Ni, 180C
H
H
H
C
C
H
H
ethane
H
Hydrogenation of alkenes-exercise
 Write the reactions for the hydrogenation of but-1-ene
& but-2-ene in the presence of a nickel catalyst & high
temperature.





1) With but-1-ene
H2C=CHCH2CH3 + H2  CH3CH2CH2CH3
2) With but-2-ene
CH3CH=CHCH3 + H2  CH3CH2CH2CH3
NB the original position isomerism is lost in this
reaction
Uses of the hydrogenation of
alkenes
Used to convert cooking oils to margarine
unsaturated
saturated
Halogenation of alkenes
 Alkene + Halogen  Dihaloalkane
 Addition reaction
H
Br Br
H
C
H
ethene
+ Br2
C
H
H
C
C
H
H
H
1,2-dibromoethane
Halogenation of alkenes 2
 Alkene + Halogen  Dihaloalkane
H
CH3
C
+ Cl2
C
H
propene
H
H
Cl
Cl
C
C
H
H
CH3
1,2-dichloropropane
Reaction with hydrogen halides
 Alkene + Hydrogen halide  haloalkane
 Addition reaction
H
H
C
H
ethene
+ HCl
C
H
H
Cl
H
C
C
H
H
H
chloroethane
Reaction with hydrogen halides 2
 Reactivity order HI>HBr>HCl
 Weaker strength (longer length) of H-X bond as descending group 7
CH3
C
CH3
+ HCl
C
H
but-2-ene
H
CH3
Br
H
C
C
H
H
CH3
2-bromobutane
Hydration of alkenes
 Reaction where a water molecule reacts with an
unsaturated compound in an addition reaction
 Alkene + water  alcohol
 RC=CR + H2O  ROH
 In Industry:
H
H
C
H
ethene
H
H
C
C
H
H
H3PO4
+ H2O
C
H
H
300C
60 atm
ethanol
OH
Hydration of alkenes- Lab (step 1)
Also addition reaction- H+ & HSO4- added across double bond
H
H
H
H
C
H
C
+ H2SO4
H
ethene
H
C
H
C
OSO3H
H
Ethyl hydrogensulfate
Hydration of alkenes- Lab (step 2)
H
H
H2O
H
C
H
C
OSO3H
H
H
H
C
C
H
H
H
Ethyl
hydrogensulfate
ethanol
OH + H2SO4
Note
sulphuric
acid is
reformedhence it’s a
catalyst
Distinguishing between alkanes &
alkenes
 Use halogenation addition reaction to distinguish
between these 2 functional groups
Used as a test for unsaturation
Addition polymerization of alkenes
 Monomer  Polymer
 Alkene  Polyalkene
Monomer
Ethene
Polymerisation
n can be 1000 or more
Polymer
Poly(e)thene
Everyday examples of poly(e)thene
Polymerisation problem
 Draw the reaction for the polymerisation of
propene:
n(Propene)  (polypropene)n
H
CH3
C
C
H
H
CH3
C
n
H
H
propene
C
H
polypropene
n
Other polymerisations- PVC
H
H
Cl
C
C
H
H
Cl
C
n
H
C
H
Chloroethene
(vinyl chloride)
n
Polychloroethene
(PVC)
Everyday examples of PVC
Other polymerisations- Polystyrene
C6H5
H
C
n
H
C
H
phenylethene
(styrene)
H C6H5
C
C
H
H
n
Polyphenylethene
(Polystyrene)
Everyday uses of Polystyrene
Summary of alkene reactions
Ethene
(from cracking)
React with steam
ethanol
polymerise
React with chlorine
React with benzene
Poly(e)thene
Chloroethene
(vinyl chloride)
Phenylethene
(styrene)
polymerise
polymerise
Polychoroethene
(PVC)
Poly(phenylethene)
(polystyrene)
Alcohols
 General formula: CnH2n+1OH
 Have polar hydroxyl (-OH) group
 Soluble in water due to hydrogen bonding from
hydroxyl group
 2 types of main reactions:
1) Combustion
2) Oxidation
Combustion of alcohols
 Alcohol + oxygen  carbon dioxide + water
 ROH + O2  CO2 + H2O
 Methanol: 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l) small Hc
 Ethanol: 2C2H5OH(l) + 7O2(g)  4CO2(g) + 6H2O(l)
medium Hc
 Propanol: 2C3H7OH(l) + 9O2(g)  6CO2(g) + 8H2O(l) large Hc
Alcohol
Ratio
Alcohol
CO2
Methanol
1
1
Ethanol
1
2
Propanol
1
3
As the alcohol becomes larger the CO2:alcohol ratio is larger, hence
more energy is released (due to the strength of C=O bond)
Combustion of alcohols-exercises
 Which of the following alcohols is likely to release
more energy upon complete combustion, hexanol or
butanol
 Answer:
 Hexanol as has higher molar mass.
Alcohols vs alkanes for fuels
 In general, alkane hydrocarbons release more energy
upon combustion as they are in a more reduced state
(more hydrogen atoms attached)
 Hence petrol (mainly octane) releases more energy
than ethanol upon combustion
 However, alcohols have advantage of being able to be
produced from renewable sources e.g. fermentation
Glucose  ethanol + carbon dioxide
C6H12O6 → 2C2H5OH + 2CO2
Oxidation of alcohols
 Hydroxyl (-OH) group can be oxidised
 Oxidation depends on nature of alcohol (primary,
secondary or tertiary)
 Oxidising agent is potassium dichromate (VI), K2Cr2O7
 Cr2O72- is reduced to Cr3+
 Hence alcohol is the reducing agent
Oxidation of alcohols-primary
alcohols
 Oxidised in 2 stages:
1) Aldehyde (RCHO)
2)Carboxylic Acid (RCOOH)
For ethanol:
CH3CH2OH + [O]  CH3CHO + H2O
H
H
C
H
C
H
OH
H+/Cr2O72-
H
C
[O]
H
H
Ethanol C2H6O
NB 2H atoms removed hence
oxidation
O
C
+ H2O
H
H
Ethanal C2H4O
Oxidation of alcohols-primary
alcohols- stage 2
CH3CHO + [O]  CH3COOH
O
H
H
C
H+/Cr2O72-
C
H
H
O
H
[O]
ethanal
C2H4O
NB: 1 O atom added hence
oxidation
H
C
C
OH
H
Ethanoic acid
C2H4O2
Oxidation of alcohols-primary alcoholscontrolling the degree of oxidation
 How do we control the reaction so that the aldehyde is




formed instead of the carboxylic acid or vice versa?
If the aldehyde is the desired product:
Heat reaction mixture in excess alcohol with distillation
apparatus (to distill off aldehyde as it’s being formed)
If carboxylic acid is the desired product:
Heat reaction mixture under reflux for a longer period of
time with excess oxidizing agent(a vertical condenser to
ensure the aldehyde drops back into reaction vessel for
further oxidation to the carboxylic acid)
Distillation vs Reflux apparatus
Distillation
Reflux
Summary of oxidation of primary
alcohols
 In general:
O
H
R
C
OH
H+/Cr2O72-
R
C
H+/Cr2O72[O]
[O]
H
Primary
alcohol
O
H
aldehyde
R
C
OH
Carboxylic acid
Oxidation of secondary alcohols
Only 1 product formed (ketone) as there is only 1
oxidisable hydrogen attached to the secondary
carbon
e.g. with propan-2-ol:
CH3CH(OH)CH3 + [O]  CH3COCH3 + H2O
O
H
CH3
C
OH
H+/Cr2O72-
CH3
C
CH3
[O]
CH3
Propan-2-ol (C3H8O)
2 H atoms lost
hence oxidation
Propanone (C3H6O)
Oxidation of secondary alcohols general
H
R
C
O
OH
H+/Cr2O72[O]
R
C
R’
Secondary
alcohol
Ketone
R’
Oxidation of tertiary alcohols
 Cannot be oxidised as there is no hydrogen
atom attached to the hydroxyl carbon.
E.g. for 2-methylpropan-2-ol
CH3
H+/Cr
CH3 C
CH3
2O7
2-
OH
[O]
No colour change
Cr2O72- stays orange
No oxidation
possible as no
H atom on C
atom bonded to
alcohol group
NB: tertiary alcohols can only be oxidised under harsh
conditions where the carbon skeleton is broken
Tests to distinguish between aldehydes
& ketones
 All have carbonyl (C=O) group
 To distinguish aldehydes/ketones from other groups:
Aldehydes & ketones both form orange precipitate with 2,4dinitrophenylhydrazine
2)
precipitate can be recrystallised & its melting point found (the melting point
of the crystals can be used to identify the particular aldehyde/ketone)
 To distinguish aldehydes from ketones:
1)Treat the sample with Fehling’s solution (alkaline copper(II) sulfate) or Tollens’
reagent (silver nitrate in ammonia)
2) Only aldehydes react with them; the aldehyde is oxidsed to the carboxylic acid
as an orange-brown precipitate of copper(I) oxide is formed with Fehling’s
solution & a silver mirror is formed with Tollens’ reagent as metallic silver is
deposited on the side of the test tube
3) Nothing happens with ketones
1)
Some pictures
Fehling’s solution- the test
tube on the right has
aldehyde as an orange
precipitate is formed
Tollen’s reagent- the test
tube on the left has
aldehyde as a silver
mirror is formed
Haloalkanes
 General formula: CnH2n+1X (X=halogen atom, F, Cl, Br,
I)
 Are very useful as can be used to synthesize an array of
other organic molecules
 Usually oily liquids
Halogenoalkanes - substitution reactions
• X replaced by another group
• C-X bond reactive due to the polarity difference (X is
more electronegative than C)
• Carbon atom attached to halogen has partial positive
charge (electron deficient) & is prone to attack by
electron-rich species (nucleophiles)
Nucleophiles
 Nucleo + phile
Nucleus loving
 A species (molecule/anion) which has a lone pair of
electrons which can be donated to an electrondeficient centre in an organic molecule to form a
coordinate (dative covalent) bond
 Examples include, -OH, H2O, NH3
Mechanisms for nucleophilic
substitution
 The substitution of an group/group of atoms with a
nucleophile as the attacking species; can occur via an
SN1 (subsitution nucleophic unimolecular) or SN2
(subsitution nucleophic bimolecular) mechanism
 Mechanism depends on nature of haloalkane (primary,
secondary, tertiary)
Primary haloalkanes- mechanisms for
nucleophilic substitution
 Take the following reaction:
 CH3CH2Br(aq) + OH-(aq)  CH3CH2OH(aq) + Br-(aq)
 Rate equation found to be rate = k[CH3CH2Br(aq)] [OH-(aq)]
 Hence 2 species are involved in the rate-determining step, so it’s bimolecular,
hence SN2
 In the transition state, the C-Br bond is broken at the same time as the C-O
bond is being formed
Hydoxide ion
(nucelophile) &
bromoethane
Transition
state
Ethanol &
leaving group
(Br-)
Tertiary haloalkanes- mechanisms for
nucleophilic substitution
 Take the following reaction:
 CH3C(CH3)2Br(aq) + OH-(aq)  CH3C(CH3)2OH(aq) + Br-(aq)
2-bromo-2-methylpropane
Rate = k[CH3C(CH3)2Br(aq)]
Hence it’s unimolecular as only 1 species involved in rate determining step
SN1 mechanism
Different mechanism caused by steric bulk by 3 methyl groups- the
nucelophile (OH-) cannot approach the electron-deficient carbon atom
SN1 mechanism
 Step 1 (rate-determining step): Br- departs as the leaving group;
heterolytic fission
carbocation
Step 2: nucleophile attacks carbocation intermediate
Some definitions
 Carbocation- an organic ion with a positive charge on
an electron-deficient carbon atom
 Heterolytic fission- the breaking of a covalent bond so
that 1 of the atoms/groups takes both of the bonding
electrons & becomes negatively charged, leaving the
other atom/group positively charged
Compare with homolytic fission (met earlier during the
free radical mechanism of alkanes)
Homolytic fission- the breaking of a covalent bond so
that 1 electron from the bond is left on each atom,
resulting in the formation of 2 free radicals
Other factors which favour SN1
 With tertiary haloalkanes, the carbocation
intermediate is stabilised by the positive
inductive effect of the 3 alkyl groups (alkyl
groups are have an electron-donating effect)
which help to reduce the positive charge on the
positive carbon e.g.:
Tertiary > Secondary > Primary
Nuceleophilic substitution of secondary
haloalkanes & relative reactivity
 Go via a mixture of SN1 & SN2 mechanisms, depending
on the reaction conditions, or some intermediate
mechanism
 Relative reactivity of different haloalkanes depends on
the strength of the C-X bond
 Strength: C-F > C-Cl > C-Br > C-I
 Reactivity: C-I > C-Br > C-Cl > C-F
Reaction pathways
 In organic chemistry, usually a desired product cannot
be made from available starting materials (reactants)
in a single step; hence the need for reaction pathways
 The production of new organic compounds from raw
starting materials is called organic synthesis/synthetic
organic chemistry
Summary of reaction pathways
alkane
M1- free radical
subsitution
dihaloalkane
1
haloalkane
1
trihaloalkane
tetrahaloalkane
2
2
alkene
3
Poly(alkene)
M2-nucleophiliic
subsitution
alcohol
4
ketone
4
aldehyde
4
Key:
1- subsitution halogenation
2- addition halogenation
3- polymerisation
4- oxidation
Carboxylic acid
NB M1 & M2
mechanism
required
Reaction pathway puzzles
 How can butanone be synthesised using 2-
bromobutane as one of the starting materials?
Answer:
Reflux with NaOH(aq)
2-bromobutane
Reflux with H+/Cr2O72-
butan-2-ol
butanone
Reaction pathway puzzles 2
 How can ethanoic acid be synthesised using ethene as
one of the starting materials?
Answer:
Reflux with H+/Cr2O72-
Conc. H2SO4/H2O
ethene
ethanol
ethanoic acid