Transcript Organic Chemistry

Organic Chemistry
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 C
H
C
H
C
H
C
H
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
Halogenoalkanes
X = Cl, Br or I
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
H
Aldehydes
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
H2NCH2COOH
methylamine
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-3methylbenzenecarboxylic 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
O
CH3CH2CH2C
H
O
butanal
ketone
CH3CH2CCH2CH3
CH3
aldehyde
O
CH3CHCH2C
OH
pentan-3-one
carboxylic acid
3-methylbutanoic acid
alkene
alcohol HO
CH
C
ketone
O
C
CH
O
O
C
ester
C
HO
O
carboxylic acid
carboxylic acid
O
ester
OH
C
O
H2C
ether
O
C
CH
CH
NH2
amine
aldehyde
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.
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
UV Light
Cl2
Cl + Cl
initiation step
CH4 + Cl
CH3 + HCl
two
propagation
steps
CH3 + Cl2
CH3Cl + Cl
CH3 + Cl
CH3Cl
termination step
CH3 + CH3
CH3CH3
minor
termination step
Also get reverse of initiation step occurring as a
termination step.
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
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?
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 - 2 - ene 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. e.g.
H
C=C
H
H
H
+ H2
H
H
H–C–C–H
H
C2H4 + H2  C2H6
ethene
ethane
H
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
+ Br2
H
H
H–C–C–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.
Bromine Water Test For Alkenes
C = C + Br2(aq)
–C–C–
Br Br
colorless
amber
colorless
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
+ HBr
H
H–C–C–H
H
H Br
C2H4 + HBr  CH3CH2Br
ethene
bromoethane
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
+ H2SO4
H
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
+ H2O
H
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 
n = about 100 to
10 000
CH2
CH2
CH2
CH2
n
is the repeating unit
monomer
repeating
unit
polymer
typical
uses
CH2=CH2
- CH2 – CH2 -
poly(ethene)
polythene
film, bags
- CH2 – CH -
poly(propene)
polypropylene
Moulded
plastic,
fibres
CH2=CHCH3
CH3
CH2=CHC6H5 - CH2 – CH C6H5
poly(phenylethene) packaging,
polystyrene
insulation
monomer repeating
unit
CH2=CHCl
- CH2 – CH Cl
polymer
typical
uses
poly(chloroethene) pipes,
polyvinylchloride flooring
PVC
CF2=CF2
- CF2 – CF2 -
Poly
(tetrafluoroethene)
PTFE
Non-stick
coating
(Teflon)
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
H H H H
Butan-1-ol
primary
H H H H
H C C C C H
H H
H
OH
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
H H
ethanol
Cr2O7
/H+
H
O
H C C
H
+ H2O
H
ethanal
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 Cr O /H+
2 7
H C C
H
H
ethanal
H
O
H C C
H
OH
ethanoic
acid
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
Cr2O7/H+
H
H
H OHH
H C C C H
H O H
propan-2-ol
propanone
H C C C H
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.
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:
+
-
C
X
ORDER OF BOND
POLARITIES:
+
-
C
Cl
>
+
-
C
Br
>
+
-
C
I
So is order of reactivity:
chloroalkane > bromoalkanes > iodoalkanes?
Is there another factor that
BOND
ought to be considered before
ENERGIES
reaching a conclusion?
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 + 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
H
Intermediate
carbocation
H
H
CH3 C OH
CH3 C +
H
H
SN1
CH3 C +
- slow
Br
OH
fast
H
OH - CH
3
-
H
HO
C
C
Br
Br
H
Transitio
n 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° halogenoalknes 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.