Transcript Organic Chemistry

Organic Chemistry
HL and SL
10.1 Introduction
Important definitions
HOMOLOGOUS SERIES – a family of organic
compounds which all fit the same general formula,
neighbouring members differ from each other by
CH2, have similar chemical properties and show
trends in physical properties.
e.g. the alkanes all fit the general formula CnH2n+2,
members of the family include methane (CH4),
ethane (C2H6) and propane (C3H8).
EMPIRICAL FORMULA – the simplest ratio of
moles of atoms of each element in a compound.
MOLECULAR FORMULA – the actual number of
moles of atoms of each element in a compound.
STRUCTURAL FORMULA (aka displayed formula
or graphical formula) – shows all of the atoms and the
bonds between them. e.g. pentane (C5H12).
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
A condensed structural formula can also be used which
omits the bonds. So for pentane we can use:
CH3CH2CH2CH2CH3
or
CH3(CH2)3CH3
The formulas in the data book are called SKELETAL
formulas. These must NOT be used in the exam.
STRUCTURAL ISOMERS – compounds with the same
molecular formula but a different arrangement of atoms.
e.g. C4H10 represents
H
H
H
H
H
C
C
C
C
H
H
H
H
butane and
H
H
H C
H
H
H
H
C
C
C
H
H
H
H
2-methylpropane
FUNCTIONAL GROUP – atom or group of atoms
which gives an organic compound its characteristic
chemical properties.
C H
Alkanes
suffix -ane
For example
CH3CH3
ethane
C C
Alkenes
suffix -ene
For example
CH2=CH2
ethene
C X
X = Cl, Br or I
Halogenoalkanes
prefix halo-
For example
CH3CH2Cl
chloroethane
C OH
Alcohols
suffix –ol or prefix hydroxy-
For example
CH3CH2OH
ethanol
CH3CH(OH)COOH 2-hydroxypropanoic
acid
O
C
Aldehydes
H
suffix -al
For example
CH3CHO
ethanal
C O
Ketones
suffix –one or prefix oxo-
For example
CH3COCH3
propanone
CH3COCH2COOH 3-oxobutanoic
acid
O
C
Carboxylic acid
OH
suffix –oic acid
For example
CH3COOH
ethanoic acid
C NH2 Amines
suffix –amine or prefix amino-
For example
CH3NH2
methylamine
H2NCH2COOH aminoethanoic
acid
O
C
Esters
OR
suffix –oate
For example
CH3COOC2H5
ethyl ethanoate
Aromatic compounds
Contain the BENZENE ring
H
formula C6H6
C
H C
C H
H C
C H
C
H
delocalised electrons
Some examples CH3CHCH2CH3
nitro group
NO2
(1-methylpropyl)
benzene
nitrobenzene
COOH
1
Br
1
Cl
2
3
4
CH3
Cl
1-bromo-2-chlorobenzene
4-chloro-3-methylbenzenecarboxylic
acid
Sometimes the benzene ring is not regarded as the
key part of the molecule.
In these cases it is referred to as the PHENYL
group.
For example
NH2
phenylamine
CH3CH2CH2C
O
aldehyde
H
butanal
ketone
O
CH3CH2CCH2CH3
CH3
CH3CHCH2C
O
OH
pentan-3-one
carboxylic acid
3-methylbutanoic acid
alkene
alcohol
HO
CH
C
O
C
CH
C
HO
O
ester
ketone
C
O
O
carboxylic acid
carboxylic acid
O
ester
OH
C
O
aldehyde
ether
H2C
O
CH
amine
NH2
C
CH
CH2
O
H2C
O
C
H
C
N
nitrile
Physical Properties of Organic Molecules
Alkanes worksheet
What type of intermolecular force would you
expect to find between alkanes, halogenoalkanes,
aldehydes, ketones, alcohols and carboxylic
acids?
Use this information to deduce the relative boiling
points of these homologous series and their
solubility in water.
10.2 The Alkanes
The alkanes is an homologous series where all
members fit the general formula CnH2n+2.
They have trends in physical properties e.g.
density and m.p. and b.p. all increase with Mr.
They all undergo similar chemical reactions.
Alkanes are SATURATED HYDROCARBONS.
i.e. they contain only single C to C bonds and
are made up of C and H atoms only.
Alkanes are obtained from crude oil by
fractional distillation.
They are mainly used as fuels.
The large Mr alkanes do not ignite easily so
there is little demand for them as fuels so they
are CRACKED to make smaller more useful
alkanes and alkenes.
Apart from combustion alkanes undergo few
chemical reactions. This is for two main
reasons:
1.The bonds in alkanes are relatively strong.
2.The bonds have a relatively low polarity as
the electronegativity of C and H is similar.
As a consequence alkanes can be used as
lubricating oils, although they do degrade
over time.
Reactions of Alkanes:
Combustion:
Alkanes burn exothermically to produce carbon
dioxide and water if there is a plentiful supply
of oxygen. This is known as complete
combustion.
e.g. CH4 + 2O2  CO2 + 2H2O
Write equations for the complete combustion of
butane and octane.
If there is a limited supply of oxygen incomplete
combustion occurs and carbon monoxide or
carbon are formed instead of carbon dioxide.
e.g. CH4 + 1½O2  CO + 2H2O
CH4 + O2  C + 2H2O
What problems do the gases released on
combustion of alkanes cause?
Chlorination
Methane does not react with chlorine in the dark
but in the presence of ultraviolet light reacts to
form hydrogen chloride.
CH4 + Cl2  CH3Cl + HCl
The mechanism for this reaction is known as free
radical substitution.
Substitution = replacement of an atom or group
of atoms by a different atom or group of atoms.
Free radical = species with an unpaired
electron.
Free radicals are formed by homolytic fission of
bonds. In homolytic fission one electron from
the shared pair goes to each atom.
So
Cl
Cl
Cl
or Cl2

+
.
2Cl
Cl
unpaired
electron
Heterolytic fission of Cl – Cl would result in the
formation of Cl+ and Cl-.
There are three steps in the mechanism:
initiation, propagation and termination
Free radical substitution
chlorination of methane
i.e. homolytic breaking of covalent bonds
Overall reaction equation
CH4 + Cl2
CH3Cl + HCl
Conditions
ultra violet light (breaks weakest bond)
excess methane to reduce further substitution
Free radical substitution mechanism
ultraviolet
Cl2
Cl + Cl
initiation step
CH4 + Cl
CH3 + HCl
CH3 + Cl2
CH3Cl + Cl
CH3 + Cl
CH3Cl
two
propagation
steps
termination step
minor
termination step
Also get reverse of initiation step occurring as a
termination step.
CH3 +
CH3
CH3CH3
Further free radical substitutions
Overall reaction equations
CH3Cl + Cl2
CH2Cl2 + HCl
CH2Cl2 + Cl2
CHCl3 + HCl
CHCl3 + Cl2
CCl4 + HCl
Conditions
ultra-violet light
excess chlorine
drag and
drop exercise
Write down two propagation steps to explain the
formation of dichloromethane.
Methane reacts in exactly the same way with
bromine to form hydrogen bromide together with
bromomethane, dibromomethane,
tribromomethane and tetrabromomethane.
Write down the mechanism for the reaction
between chlorine and ethane to form chloroethane.
Use the mechanism to explain why small amounts
of butane are formed. How could the formation of
further substitution products be minimised?
10.3 The Alkenes
All fit the general formula CnH2n.
Are unsaturated hydrocarbons as they contain a
C = C.
Much more reactive than alkanes.
Industrial importance of alkenes:
1. Making polymers (plastics)
2. Hydrogenation of vegetable oils to make
margarine
3. Hydration of ethene to make ethanol.
When naming alkenes have to include position of
double bond, for example:
CH3CH=CHCH3 is but-2ene and
CH3CH2CH=CH2 is but-1-ene
Draw out and name all of the alkenes with the
molecular formula C6H12.
Alkenes undergo ADDITION reactions. Two
substances combine to form one new substance.
Unsaturated molecules are converted to saturated
molecules.
Reactions of Alkenes
1. Addition of hydrogen (hydrogenation)
Alkenes react with hydrogen in the presence of a
nickel catalyst at 180 °C to form an alkane.
H H
e.g. H
H
C=C
H
+ H2
H–C–C–H
H
H H
C2H4 + H2  C2H6
ethene
ethane
Most oils are esters of propane-1,2,3-triol (aka
glycerol) with 3 long chain carboxylic acids (aka
fatty acids). The esters are sometimes called
triglycerides. Hydrogenation of these oils
produces margarine.
The common fatty acids include
• octadeca-9-enoic (oleic) acid – unsaturated acid
found in most fats and olive oil
• octadeca-9,12-dienoic (linoleic) acid –
unsaturated acid found in many vegetable oils
such as soyabean and corn oil
CH2OOC(CH2)7CH=CH(CH2)7CH3
CHOOC(CH2)7CH=CH(CH2)7CH3
CH2OOC(CH2)7CH=CH(CH2)7CH3
Above is the triglyceride formed between
propan-1,2,3-triol and oleic acid.
Hydrogenation using a nickel catalyst and
slight pressure removes some of the C=C.
This enables the chains to pack together
more closely which increases the van der
Waals forces and hence m.p. so the oils are
solidified forming margarine.
2.Addition of halogens (halogenation)
Halogens react with alkenes at room temperature
and pressure in a non-polar solvent to form a
dihalogenoalkane.
e.g. H
C=C
H
H H
H
+ Br2
H–C–C–H
H
Br Br
C2H4 + Br2  C2H4Br2
ethene
1,2-dibromoethane
Bromine water is used as a test for unsaturation.
In the presence of an alkene, bromine water turns
from red brown to colourless. Alkanes do not
react with bromine water.
alkane with
bromine water
alkene with
bromine water
C=C
–C–C–
Br OH
3. Reaction with hydrogen halides
(hydrohalogenation)
Alkenes react with hydrogen halides (HCl, HBr
etc.) to form halogenoalkanes. The reaction occurs
at room temperature and pressure. e.g.
H
C=C
H
H H
H
+ HBr
H–C–C–H
H
H
C2H4 + HBr  CH3CH2Br
ethene
bromoethane
Br
4.Hydration (reaction with water)
This can be done in two ways:
a) Addition of concentrated sulphuric acid to
form an alkyl hydrogensulphate. Water is
then added to hydrolyse the product and
produce an alcohol. The sulphuric acid is
regenerated.
H
C=C
H
H H
H
+ H2SO4
H
H–C–C–H
H
OSO3H
H H
H – C – C – H + H2O
H
OSO3H
H H
H–C–C–H
H
OH
+ H2SO4
C2H4 + H2SO4  CH3CH2OSO3H
CH3CH2OSO3H + H2O C2H5OH + H2SO4
ethanol
b) Alkenes can also undergo direct hydration to
form an alcohol. Ethene can be converted to
ethanol by reaction with steam in the presence of a
phosphoric(V) acid (H3PO4) catalyst at a pressure
of 60 – 70 atm and a temperature of 300 °C.
H
C=C
H
H H
H
+ H2O
H
H–C–C–H
H
OH
What advantages and disadvantages does this method
have over production of ethanol by fermentation?
5. Addition Polymerisation
The formation of polymers involves alkenes reacting
with themselves to form a long chain molecule called
a polymer. The individual molecules used to make
the polymer are called monomers. Ethene is
polymerised to form poly(ethene)
nCH2 = CH2 
CH2 CH2
n
n = about 100 to 10 000
CH2 CH2
is the repeating unit
monomer
repeating unit polymer
CH2=CH2
- CH2 – CH2 - poly(ethene)
polythene
CH2=CHCH3
- CH2 – CH CH3
CH2=CHC6H5
- CH2 – CH C6H5
poly(propene)
polypropylene
typical
uses
film, bags
Moulded
plastic,
fibres
poly(phenylethene) packaging,
polystyrene
insulation
monomer
repeating unit polymer
CH2=CHCl
- CH2 – CH Cl
typical
uses
poly(chloroethene) pipes,
polyvinylchloride flooring
PVC
CF2=CF2
- CF2 – CF2 -
poly(tetrafluoroethene)
PTFE
Non-stick
coating
(Teflon)
10.4 Alcohols
• Alcohols are the homologous series with the
general formula CnH2n+1OH.
• They all contain the functional group, OH,
which is called the hydroxyl group.
• Alcohols can be classified as primary,
secondary or tertiary, depending on the carbon
skeleton to which the hydroxyl group is
attached.
RCH2OH
R2CHOH
R3COH
1 alkyl
group on C
next to OH
so primary
alcohol, 1°
2 alkyl
groups on C
next to OH
so secondary
alcohol, 2°
3 alkyl
groups on C
next to OH
so tertiary
alcohol, 3°
Draw out the structure, name and
classify all the alcohols with the formula
C4H9OH.
H H H H
H C C C C OH
Butan-1-ol, primary
H H H H
H H H H
H C C C C H
H H OH H
Butan-2-ol, secondary
H H H
H C C
C OH
2-methylpropan-1-ol,
primary
H CH3 H
H OH H
H C C
C H
H CH3 H
2-methylpropan-2-ol,
tertiary
Reactions of Alcohols
1.Combustion
In countries such as Brazil, ethanol is mixed with
petrol and used to power cars. Ethanol is less
efficient as a fuel than petrol as it is already
partially oxidised but does make the country less
reliant on supply of petrol. As it can be
produced by fermentation of sugar beet, many
consider ethanol a carbon neutral fuel.
2.Oxidation of Alcohols
Primary alcohols are oxidised first to
aldehydes, such as ethanal. A suitable
oxidising agent is acidified potassium
dichromate(VI)
H H
H C C OH + [O]
H H
ethanol
H
O
H C C
H
ethanal
+ H2O
H
An aldehyde still has one hydrogen atom
attached to the carbonyl carbon, so it can be
oxidised one step further to a carboxylic acid.
H
O
H C C + [O]
H
H
ethanal
H
O
H C C
H
ethanoic acid
OH
In practice, a primary alcohol such as ethanol is dripped
into a warm solution of acidified potassium
dichromate(VI).
The aldehyde, ethanal, is formed and immediately distils
off, thereby preventing further oxidation to ethanoic acid,
because the boiling point of ethanal (23 °C) is much lower
than that of either the original alcohol ethanol (78 °C) or
of ethanoic acid (118 °C). Both the alcohol and the acid
have higher boiling points because of hydrogen bonding.
If oxidation of ethanol to ethanoic acid is required, the
reagents must be heated together under reflux to prevent
escape of the aldehyde before it can be oxidised further.
Secondary alcohols are oxidised to ketones.
These have no hydrogen atoms attached to the
carbonyl carbon and so cannot easily be
oxidised further.
H H H
H C C C H + [O]
H OH H
propan-2-ol
H
H
H C C C H
H O H
propanone
+ H2O
Distinguishing between 1°, 2° and 3° alcohols
When orange acidified potassium dichromate(VI) acts
as an oxidising agent, it is reduced to green
chromium(III) ions.
Primary and secondary alcohols both turn acidified
dichromate(VI) solution from orange to green when
they are oxidised, and this colour change can be used
to distinguish them from tertiary alcohols.
Tertiary alcohols are not oxidised by acidified
dichromate(VI) ions, so they have no effect on its
colour, which remains orange.
10.5 Halogenoalkanes
Named by using the name of the alkane from
which they are derived with the prefix chloro-,
bromo- or iodo-.
For example:
CH3CH2Br is bromoethane
(CH3)2CHCH2Cl is 1-chloro-2-methylpropane
Remember the position of the halogen atom
must be indicated using the appropriate number
so
CH3CH2CH2Cl is 1-chloropropane and
CH3CHClCH3 is 2-chloropropane
Halogenoalkanes can be classified in the same
way as alcohols.
Draw out, name and classify all the isomers with
the formula C5H11Br.
Key feature of halogenoalkanes is
C–X
where X = Cl, Br or I
What is notable about this bond compared with
say, C – C and C – H?
The halogen atom is more
electronegative than C so
the bond is polarised:
d+
d-
C–X
ORDER OF BOND
POLARITIES:
d+
d-
C – Cl
d+
>
d-
C – Br
d+
>
d-
C–I
So is order of reactivity:
chloroalkane > bromoalkanes > iodoalkanes?
Is there another factor that
ought to be considered before
reaching a conclusion?
BOND
ENERGIES
Bond energies:
Bond
Bond energy in
kJmol-1
C - Cl
346
C - Br
C-I
290
234
This suggests that the order of reactivity is:
iodoalkane > bromoalkanes > chloroalkanes
There’s only one way to find out which is best!
Do an experiment, not fight!
But what do halogenoalkanes react with?
The d+ carbon atom is susceptible to attack by
NUCLEOPHILES.
A nucleophile is a species with a lone pair of
electrons.
e.g. OH-, NH3, CN-.
When attack by a nucleophile occurs, the carbon –
halogen bond breaks releasing a halide ion.
A suitable nucleophile for experimentation is OHfrom an aqueous solution of an alkali such as sodium
hydroxide.
CH3CH2X + OH-  CH3CH2OH + XX = Cl, Br or I.
OH has replaced the X so overall we have
NUCLEOPHILIC SUBSTITUTION
How can we follow the rate of this reaction so that
we can determine the order of reactivity?
Halide ions  coloured precipitates when silver nitrate
is added.
So we can measure how long it takes for a precipitate
to form.
See EXPERIMENT SHEET.
Mechanisms for nucleophilic
substitution
SN1 = unimolecular nucleophilic substitution
(only one species in the slow step of the
mechanism, rate determining step)
SN2 = bimolecular nucleophilic substitution (two
species in the slow step of the mechanism, rate
determining step)
Use of curly arrows:
Curly arrows are used in reaction mechanisms to
show the movement of electron pairs.
TAILS come from
either
a bond pair
of electrons
or
a lone pair
of electrons
HEADS point
either
next to an atom
to form
a bond pair
of electrons
X
or
at an atom
to form
a lone pair
of electrons
X
Heterolytic fission
of C – Br bond
H
CH3 C
H
Br
H
Intermediate H
carbocation
H
+
H
H
OH-
- slow
Br
H
CH3 C
CH3 C +
SN1
CH3 C +
OH
fast
H
OH-CH
3
HO
C
H
-
H
Br
C
Br
Transition
state
CH3 H
H
HO
C
H
SN2
CH3
+
-
Br
Which is best? SN1 or SN2?
For primary halogenoalkanes – SN2
For tertiary halogenoalkanes – SN1
3° halogenoalkanes cannot undergo the SN2 mechanism
as 5 bulky groups would not fit around the C in the
transition state - steric hindrance.
1° halogenoalkanes are less likely to undergo SN1 as this
would involve the formation of a primary carbocation as
an intermediate. Alkyl groups push electron density to
the C atom they are attached to (positive inductive effect)
which stabilises the positive charge. More alkyl groups
mean a more stable carbocation.
2° halogenoalkanes react via a mixture of SN1
and SN2. The mechanism predominating
depends upon the nature of the alkyl groups
and the nature of the solvent.