Download Reaction with hydrogen halides

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Key facts
Saturated: compounds which contain only single bonds, e.g. alkanes
Unsaturated: compounds which contain double or triple bonds, e.g. alkenes, arenes
Aliphatics: compounds which do not contain a benzene ring; may be saturated or
unsaturated, e.g. alkanes, alkenes
Arenes: compounds which contain a benzene ring; they are all unsaturated
compounds, e.g. benzene, phenol
Electrophile (electron-seeking): an electron-deficient species which is therefore
attracted to parts of molecules which are electron rich, they form bonds by
accepting an electron pair, electrophiles are positive ions or have a partial positive
charge, e.g. NO2+, H+, 𝐵𝑟 𝛿+
Nucleophile (nucleus-seeking): an electron-rich species which is therefore attracted
to parts of molecules which are electron deficient, nucleophiles have a lone pair of
electrons and may also have a negative charge, e.g. Cl-, OH-, NH3
Addition: occurs when two reactants combine to form a single product,
characteristic of unsaturated compounds, e.g. 𝐶2 𝐻4 + 𝐵𝑟2 → 𝐶2 𝐻4 𝐵𝑟2
Substitution: occurs when one atom or group of atoms in a compound is replaced
by a different atom or group, characteristic of saturated compounds and aromatic
compounds, e.g. 𝐶𝐻4 + 𝐶𝑙2 → 𝐶𝐻3 𝐶𝑙 + 𝐻𝐶𝑙
Addition-elimination: occurs when two reactants join together (addition) and in the
process a small molecule such as H2O, HCl or NH3 is lost (elimination), reaction
occurs between a functional group in each reactant, also called condensation
reaction, e.g. 𝑅𝑂𝐻 + 𝑅 ′ 𝐶𝑂𝑂𝐻 → 𝑅 ′ 𝐶𝑂𝑂𝑅 + 𝐻2 𝑂 (alcohol + acid-> ester + water)
Homolytic fission: is when a covalent bond breaks by splitting the shared pair of
electrons between the two products, produces two free radicals, each with an
unpaired electron, e.g. 𝑋: 𝑋 → 𝑋 ∙ +𝑋 ∙
Heterolytic fission: is when a covalent bond breaks with both the shared electrons
going to one of the products, produces two oppositely charged ions, e.g.
𝑋: 𝑋 → 𝑋:− + 𝑋 +
The main features of a homologous series are: successive members of a
homologous series differ by a –CH2- group, can be represented by the same general
formula, show a gradation in their physical properties, have similar chemical
properties
Structural isomers are molecules that have the same molecular formula but
different arrangement of the atoms.
A free radical contains an unpaired electron and so is very reactive.
Alkanes are saturated hydrocarbons and undergo substitution reactions.
Alkenes are unsaturated hydrocarbons and undergo addition reactions.
A carbocation is a positive ion with the charge centered on a carbon atom.
Protic solvents contain –OH or –NH and are able to form hydrogen bonds. Aprotic
solvents are those that are not able to form hydrogen bonds.
The rate of nucleophilic substitution reactions in halogenoalkanes is fastest with
tertiary halogenoalkanes and slowest with primary halogenoalkanes.
The rate of nucleophilic substitution reactions in halogenoalkanes increases down
Group 17 from the fluoroalkane to the iodoalkane.
When an unsymmetrical reagent adds to an unsymmetrical alkene, the electrophilic
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portion of the reagent adds to the carbon that is bonded to the greater number of
hydrogen atoms.
Stereoisomers differ from each other in the spatial arrangement of their atoms.
Enantiomers have opposite configurations at all chiral centers and are mirror
images of each other. Diastereomers have opposite configuration at some but not
all chiral centers, and are not mirror images of each other.
The two enantiomers of a chiral compound rotate plane-polarized light in equal and
opposite directions. A racemic mixture is a 50:50 mix of two enantiomers. A
racemic mixture shows no rotation of plane‐polarized light.
10. 1 Fundamentals of organic chemistry
A homologous series is a series of compounds of the same family that 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 – it is the reactive part;
whose nearest neighbours differ by a common structural unit, most often -(CH3 )
group or a -(CH2)- group ;
have similar chemical properties (same functional group = reactive part of the
molecule);
show a gradual change (gradation) in physical properties as shown by the table
below which shows the boiling points of some alkanes:
examples of homologous series that you need to know: alkanes, alkenes,
alkynes, alcohols, halogenoalkanes esters, ethers, aldehydes, ketones, amines,
carboxylic acids, nitriles, arenes and amines.
The functional group in a homologous series has a great and similar effect on the
size of boiling points of all members in the series.
Trends in boiling points in the same homologous series are then caused by the
carbon chains i.e. their number of carbons and their structure or arrangement
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
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 that 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.
As the molecular mass increases there is also
an increase in inertia and therefore more
energy is needed to increase motion.
solid
Different formula of organic compounds
Type of formula
Empirical formula
Description
Molecular formula
Full structural formula
Example
Shows most simple whole number ratio of all
atoms.
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.
Skeletal structural
formula
Only shows all bonds and atoms of functional
groups; carbon atoms are at lines that are at
angles to each other; also hydrogen atoms
bonded onto carbons not part of a functional
group are not shown.
Condensed structural
formula
Structural formula which shows order in which all atoms
are arranged but which omits bonds.
CH2
C6H14
CH3CH2CH2CH2CH2CH3
or CH3(CH2)4CH3
Different classes of compounds and their functional groups
Name of class
General formula
Functional group
alkanes
CnH2n+2 no functional group
alkenes
CnH2n
double carbon-carbon
bond
Name of
functional
group
Suffix/prefix in
IUPAC name
-ane
alkenyl
-ene
alkynes
halogenoalkanes
alcohols
CnH2n-2 R-Hal
CnH2n+1OH
ethers
aldehydes
ketones
esters
carboxylic acids
amines
amides
nitriles
arenes
R-CHO
R-CO-R’
CnH2n+1COOH
triple carbon-carbon bond
alkynyl
-yne
halogen
halogen
-OH
hydroxyl
prefix: bromo-,
chloro-, iodo-, fluor
-anol
oxygen atom between 2
carbon chains
CHO
CO
ester
-COOH
-NH2
-CONH2
-CN
phenyl ring
-oxyalkane
carbonyl
carbonyl
ester
carboxyl
amine
carboxamide
nitrile
phenyl
-anal
-anone
-anoate
-anoic acid
-anamine
-anamide
-anenitrile
-benzene
Naming of organic compounds: IUPAC rules
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
 -yne = alkyne
 -anol = alcohol
 -anal = aldehyde
 -anone = ketone
 -anoic acid = carboxylic acid.
 -anamine = amine
 -anamide = amide
 -anenitrile = nitrile
Position of functional group
If the functional group is not in a terminal position but in the chain, its position is
indicated by a number (smallest number) between 2 dashes before the functional
group ending e.g. butan-2-one. In some classes the position is indicated at the start
of the name e.g. 1-bromopropane – see below.
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, 1,2dichloroethane.
Naming of ethers
Prefix of alkyl group to the left of the oxygen then -oxy- full name of alkyl group to the right of the
oxygen.
E.g.: CH3CH2OCH3 = ethoxymethane.
Naming of esters
The part attached to the second oxygen atom of the carboxyl group comes from the alcohol used
to make the ester and this is named first as an alkyl group. The carboxylic acid part that contains
the –COO part is given the –anoate suffix.
E.g.: CH3CH2COOCH3 = methylpropanoate.
Branched compounds
See below under the heading naming branched alkanes.
Saturation, unsaturation and aromatic
Saturation = a compound that contains single carbon-carbon bonds only.
Unsaturation = a compound that contains at least one or more double or triple carbon-carbon
bond.
Aromatic = compound contains a phenyl ring; - C6H5.
Structural isomerism
Structural isomers are compounds with the same molecular formula but with different
arrangements of atoms i.e. different structural formula.
For instance, structural isomerism in alkanes
There are 3 types of structures in alkanes but you only need to know two as you do
not need to know the cyclic structures.
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
formula
CH3 CH3CH2 CH3CH2CH2 CH3CH2CH2CH3 -
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)
H H H
  
HCCCH


H
H
HCH
methylpropane

H
C4H10 called 2-
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(s) of the branch (es);
 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) to
all branches;
 the position of the functional group has priority over an alkyl.
Examples: 2-methylpentane, 2,2-dimethylpentane, methylpropan-2-ol, 2,2,4-trimethylpentane.
Primary, secondary and tertiary carbon and nitrogen atoms
Primary, secondary or tertiary carbon atoms in halogenoalkanes and alcohols
Based on the number of alkyl groups or carbons bonded onto the carbon that carries
the –OH group or the halogen.
In alcohols:
primary carbon atom
Has one alkyl group or 1
carbon atom onto the
carbon which carries the –
OH group
secondary carbon atom
Has two alkyl groups or 2
carbon atoms onto the
carbon which carries the –
OH group
tertiary carbon atom
Has three alkyl groups or
3 carbon atoms onto the
carbon which carries the –
OH group
primary carbon atom
Has one alkyl group or 1
carbon atom onto the
carbon which carries the
halogen
secondary carbon atom
Has two alkyl groups or 2
carbon atoms onto the
carbon which carries the
halogen
tertiary carbon atom
Has three alkyl groups or
3 carbon atoms onto the
carbon which carries the
halogen
Nitrogen atoms in amines
primary nitrogen atom
Has one alkyl group on
the nitrogen atom – could
be called a primary amine.
secondary nitrogen atom
tertiary nitrogen atom
Has two alkyl groups on
Has three alkyl groups on
the nitrogen atom.
the nitrogen atom.
Structure and bonding in benzene
Benzene, is the parent molecule of all arenes as all arenes are derived from
benzene.
Structure
Benzene has a regular hexagon structure i.e. a hexagon with
equal sides.
A benzene molecule is planar, cyclic, symmetrical and non-polar
and has 120 bond angles around each carbon atom.
(HL: all carbon atoms are sp2 hybridized).
Bonding
 Each carbon has 3 single (sigma) covalent bonds; two to carbon atoms on either
side and one to a hydrogen atom.
 In addition, all 6 carbon-carbon bonds also have a delocalized π-bond with a high
electron density above and below the ring structure making the 6 π shared
equally between all 6 carbon atoms.
 All carbon-carbon bonds have equal strength.
 All carbon-carbon bonds have a strength between a single and a double bond
which is a bond order of 1.5.
 Resonance hybrid structures between two equivalent structures can be drawn.
Chemical and physical evidence for the structure of benzene:
 Relative stability as compared to other unsaturated compounds - it does not
undergo the reactions typical of unsaturated compounds like oxidation (by
KMnO4) and addition reactions. This is because of the delocalisation of electron
pairs, there are no fixed distinct areas of higher electron density as in alkenes
which means that no electrophiles are attracted strongly enough.
 Thermochemistry:
 The experimental enthalpy of formation is less endothermic than the
enthalpy of formation based on the alternate single-double bond structure
(“cyclohexatriene”); this means that the actual benzene is at a lower energy
level and therefore more stable than the cyclohexatriene structure.
 The heat of combustion of benzene is less exothermic; there is less
difference in terms of energy between the reactants, benzene and oxygen,
and the products, carbon dioxide and water, than if, according to the
calculations the cyclohexatriene structure was burned.
 The heat of hydrogenation of benzene is less exothermic than the heat of
hydrogenation of cyclohexatriene’s; in fact the value is between
cyclohexene and cyclohexadiene ;
cyclohexatriene
energy
Hhydrogenation
benzene
C6 H12
The above diagram shows that benzene is more stable as a result of its structure
than the alternate single-double bond.
Physical evidence for the structure of benzene:
 X-ray crystallography showed that all bonds have the same length and strength;
an intermediate value between the values for C-C and C=C;
10.2 Functional group chemistry
Alkanes
Alkanes are a homologous series in which all non-cyclic compounds have the
following general formula: CnH2n + 2.
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.
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.
name
structural formula
pentane
CH3CH2CH2CH2CH3
2 methylbutane
CH3CHCH2CH3

CH3
molecular
formula
boiling point
(oC)
C5H12
36
28
C5H12
2,2-dimethylpropane
CH3

CH3C CH3

CH3
C5H12
10
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 2,2,3,3tetramethylbutane 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
Both carbon and hydrogen are oxidised.
Examples: CH 4 +
 3CO2 + 4H2O
2O2
 CO2 + 2H2O
and
C3H 8 +
5O2
 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
 C + 2H2O
C3H 8 +
4O2
 CO + 2H2O
 CO2 + 2CO +
and
4H2O
CH 4 +
O2
Substitution reactions of alkanes
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
+
+
Br2 (l)
bromine
UV
 C2H5 Br (g)

+
bromoethane
HCl (g)
+
hydrogen chloride
Photochemical 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. It is also called a photochemical reactions because UV light is needed to
get it started.
step 1:
initiation
The UV light causes homolytic fission of the halogen molecule; each
atom takes one of the electrons of the electron pair in the covalent
bond. The two species, in these case atoms, formed are called free
radicals.
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

Cl - Cl
Cl
+ Cl
f ree radicals
step 2:
Chain reaction during which the free radicals react with molecules
propagation forming more free
radicals and other molecules:
free radical +
new molecule

molecule
new free radical
+
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
Cl
+

Cl

Cl2 or
CH3Cl
or
Cl
+
CH3
+
CH3

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
+
HCl
CHCl3
+
Cl2
UV

CCl4
+
HCl
Exercise
Write the reaction mechanism for a reaction between ethane and fluorine.
Alkenes
Chemical properties of alkenes
As they are unsaturated alkenes are more reactive than alkanes. 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
Addition reactions
It is because of this greater chemical 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 - it is the π bond which is weaker than the sigma bond that breaks.
The π 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 that are added to the molecule.
One part of the reagent bonds to one carbon atom of the double bond 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
(g)
C2H4 (g) +
ethene +
Br2 (l)
 C2H4 Br2 (g)
bromine
 1,2-dibromoethane
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 1,2dibromoethane 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
C2H4 (g) +
H2O (l)
ethene +
water
 C2H5 OH (l)
 ethanol
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 and hydrogen chloride.
Addition polymerization of alkenes DOUBLE BOND NEEDED!!!
Because they are unsaturated, alkene molecules (or other molecules with double
bonds) can also 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.
In addition polymerization, the polymer has the same % carbons as its monomers as
no atoms are removed.
As there are many different monomers that can be used in addition polymerization
there are many different addition polymers which is why the plastic is so versatile.
Alcohols
Most important alcohol is ethanol that is used as a fuel, solvent, antiseptic, to make
esters.
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) or potassium permanganate(VII)
which needs to be acidified (e.g. H2SO4) and which goes from orange (oxidation
state +6) to green (oxidation state +3). 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 distillation needs
to be used to collect the aldehyde (it has a lower boiling point than ethanoic acid)
as soon as it is formed as
otherwise the aldehyde 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 and is then
removed using distillation.
Equation for the further oxidation from ethanol to ethanoic acid:

full equation: 3C2H5OH + 2Cr2O72- + 16H+  3CH3COOH +
4Cr3+ + 11H2O

simpler version:
C2H5OH + 2[O]  CH3COOH + H2O
Secondary alcohols can only be oxidized to ketones 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.
Tertiary alcohols cannot be oxidized at all as they have no hydrogen atoms onto the
carbon that carries the hydroxyl group.
Nucleophilic substitution reactions with acids (esterification or condensation)
Conditions for esterification: heat and concentrated sulphuric acid.
For example when ethanol and ethanoic acid are heated in the presence of
concentrated sulphuric acid an ester called ethyl ethanoate is formed.
CH3CH2OH + CH3COOH  CH3CH2OOCCH3 + H2O
In the above reaction:
 the hydrogen of the hydroxyl functional group in the alcohol is eliminated
 the hydroxyl in the carboxyl group in the carboxylic acid is eliminated on the
acid molecule, It is the carbon-oxygen bond in the acid molecule that breaks.
 the hydrogen and the hydroxyl join to form water whilst the other 2 parts from
the ester.
The above reaction can be started from ethanol only as some ethanol can be
oxidized first using an oxidizing agent and heated under reflux. The ethanoaic acid
can be distilled and then heated with new ethanol in the presence of concentrated
sulphuric acid.
Esters have sweet and fruity smells and are used in: perfumes, flavourings in food,
solvents, pain killers and in the production of fibres e.g. polyester
Halogenoalkanes
Halogenoalkanes are more reactive than alkanes. This is because of the presence of
the halogen atom which is more electronegative than the carbon atom it is attached
to; this makes the carbon atom slightly positive or electron-deficient and provides a
reactive site in the molecule. As a result this electron-deficient carbon atom
halogenoalkanes can undergo (nucleophilic) substitution reactions. A nucleophile is
an electron-rich species containing a lone pair that it donates to an electron-deficient
carbon.
Nucleophilic subsitution
Halogenoalkanes react with warm dilute sodium hydroxide solution in
nucleophilic substitution reactions.
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.
A nucleophile is an electron-rich species containing a lone pair that it donates to an
electron-deficient carbon.
During a nucleophilic substitution reaction the halogen is substituted by the
hydroxide group to form an alcohol.
Polymers
Addition polymers consist of a wide range of monomers and form the basis of the
plastics industry.
Benzene: electrophilic substitution
Because of the stability of the benzene ring as a result of the delocalization of the π
electrons benzene does not easily undergo addition reactions. However, the π
electron cloud (electron-rich) above and below the ring structure is a reactive site
that more readily attracts electrophiles resulting in substitutions of hydrogen atoms
on the ring by the electrophiles.
Electrophiles, like for instance NO+, have a positive charge or a partial positive
charge.
Uses of some of the classes of organic compounds




Alkanes as fuels
Ethene in the ripening of fruit
Alcohols as fuel additives
Esters in perfumes, food flavourings, solvents, nitroglycerin, biofuels and
painkillers.
11.3 Spectroscopic identification of organic compounds
Index of hydrogen deficiency (IHD): information about the degree of
unsaturation
In a hydrocarbon where all the carbon atoms have only single bonds the compound
will have the maximum number of hydrogen atoms; that can be determined using
2n+2 for a n number of carbon atoms. The 2n+2 gives the saturated target for the
molecule of the compound.
If any of the bonds are replaced with double or triple bonds or if rings are involved,
the compound will have a “deficiency” of hydrogen atoms.
By determining the index of hydrogen deficiency (IHD), we can tell from the
molecular formula if there are any multiple bonds or rings in the molecule and how
many. The IHD is the number of hydrogen molecules it would take to make the
structure of the compound saturated.
The IHD or degree of unsaturation can be determined from a molecular formula by:
1. Calculating the maximum number of hydrogen atoms a molecule should have by
applying 2n+2 with n being the number of carbon atoms in the molecular formula
of the compound e.g. the saturated target of C3H6 is C3H8 so the maximum
number of hydrogen atoms C3H6 should have is 8.
2. From the maximum number of hydrogen atoms calculated in step 1 deduct the
number of hydrogen atoms that are in the formula and …..
3. you divide the number in step 2 by 2 because the index is for the deficiency of
hydrogen molecules!!! Example: in C3H6: 8H – 6H = 2H, divide by 2 = 1 so the
IHD is 1
4. What if there are other atoms in the molecule? Before you divide by 2 you should
do the following
a. Subtract any halogen atoms as they are treated as hydrogen atoms –
halogens make 1 bond with a carbon atom just like hydrogen and do not
use up any extra valence electrons/cause any unsaturation.
b. Ignore any oxygen atoms – make 2 bonds/presence has no impact on IHD
c. Add 1 to the number of hydrogen atoms obtained at the end of step 2 for
each nitrogen atom in the molecule – nitrogen atoms add to the
unsaturation of a molecule.
Interpreting the IHD:
 One IHD could be a double bond or a ring structure
 Two IHD is a triple bond
 Three is an aromatic ring
Examples
IHD of C3H6 is:
IHD of C6H12N2Br2 is:
1. Maximum number of hydrogen
1. Maximum number of hydrogen
atoms is 8
atoms is 14
2. There are 6 in the formula so 8 – 6
2. There are 12 in the formula so 14
=2
– 12 = 2
3. Divide 2 by 2 = 1
3. Add 2 for the 2 nitrogen atoms: 2
+2=4
4. Deduct 2 for the 2 bromine atoms:
4–2=2
5. Divide 2 by 2 = 1
Infrared spectroscopy
Spectroscopy is the study of how matter interacts with radiation within the
electromagnetic spectrum.
Infrared radiation occurs at a frequency that causes or increases molecular
vibrations in some covalent bonds – not all but luckily in most of the bonds that occur
in functional groups. There are 3 types of vibrations: symmetric stretch, asymmetric
stretch and symmetric bend. For a vibration to be able to absorb IR it must result in a
change in the molecular dipole.
Infrared spectroscopy is used to identify different functional groups in an organic
compound as each functional group has its own covalent bonds and therefore each
functional group has its own narrow frequency or wavelength range which is shown
by the dips in an infrared spectrum (transmittance mode). The transmittance mode
shows how much of all the wavenumbers in the IR spectrum have been absorbed; a
93% at 4000 cm-1 means that 93% of the radiation at that frequency has passed
through the same sample and 7% has been absorbed.
Each functional group has been given a frequency range and not a precise number
as the actual frequency or wavenumber at which absorption occurs depends on the
other atoms/bonds in the molecule. Often the frequency range is also expressed in
unit of wavenumber which is the number of waves per cm.
The functional groups you need to be able to recognize in an IR spectrum are in
section 26 in the IB Data booklet:
-C-H, O-H, C-Hal, -COOH, C=O, CHOH, C=C, -C C-, C-O and –NH2. The infrared
data concerns stretching vibrations only. Each bond also has own intensity that can
also be used to recognize it.
Extra information:
All molecules have vibrational energy as the COVALENT bonds within the molecule
vibrate permanently (i.e. they stretch and bend). Each type of bond does this at its
own natural frequency. These vibrations are quantized, which means that there are
fixed differences between the vibrational energy levels each bond within a molecule
can occupy (just like electron energy levels in an atom).
The energy needed to make covalent bonds in a polar molecule vibrate more i.e. get
to a higher vibrational energy level can be measured and is always within the
infrared spectrum region of the electromagnetic spectrum (non-polar bonds do not
absorb in the infrared spectrum). The infrared region is a low-energy region so the
energy is not enough to cause the bonds to break so this measurement does not
cause a decomposition of the compound.
Just like the electrons within an atom, a bond can only absorb waves which have a
frequency (corresponding to a fixed amount of energy) which matches the difference
in vibrational energy levels of the actual bond: each polar bond can only absorb a
certain quanta of vibrational energy in the infrared spectrum. The frequencies at
which absorptions take place appear as dips on an infrared spectrum
The vibrational energy levels and therefore the frequency or wavelength absorbed by
a bond depends on:
 The length and strength of the covalent bond which in its turn depends on the
atoms involved in the bond; example a triple bond is stronger than a double bond
and therefore needs to absorb waves of a higher frequency to increase its
vibrations. The stronger the bond the higher the frequency needed to cause that
bond to go to a higher vibrational level.
 The surrounding atoms as they interfere with the vibration of a polar covalent
bond. – this is a reason why
there is a range in the frequencies rather than a precise wavenumber.
Infrared spectroscopy is used to determine functional groups in organic compounds
as each functional group has its own covalent bonds and therefore each functional
group has its own narrow frequency or wavelength range which is shown by the dips.
Each functional group has been given a frequency range and not a precise number;
the actual frequency or wavenumber at which absorption occurs depends on the rest
of the molecule.
Each bond also has own intensity that can be used to recognize it.
Infrared spectra allow the identification of functional groups in an organic
molecule but on their own the IR spectra do not indicate the position of these
functional groups or the chain length.
However, an IR spectrum exists for most organic compounds of which the position of
the functional group is known. The chain length can be derived from mass
spectroscopy.
Mass spectroscopy
Using mass spectrometry it is possible to determine the molecular mass and the
molecular formula of an unknown organic compound. The structure of the molecule
is determined by breaking the molecule into ionised fragments with the aid of an
electron gun and then piecing these fragments (fragmentation pattern) together from
the mass spectrum that was obtained. The electron gun both ionises (removes an
electron) and fragments (causes the rupture of bonds; more easily to break single
bonds as they are weaker).
The identity of the ionized fragments can be determined from their mass. Each ionic
fragment shows as a peak in the mass spectrum and the intensity of the peak
indicates the abundance of that fragment. Usually, the most intense peak in the
mass spectrum is called the base peak and is assigned a value of 100% intensity or
relative abundance.
The peak with the highest mass/charge ratio (m/z) is formed by the loss of one
electron from the molecule is called the molecular ion or parent ion (M) and indicates
the molecular mass of the compound; the other peaks are the ionized fragment
peaks that always have a mass less than the molecular ion. All the peaks on the
mass spectrum and their relative abundance together form the fragmentation pattern
that is used to identify the structure of the compound and determine the molecular
mass.
However, not all fragment ions form with equal ease as double and triple bonds are
difficult to break.
Common fragments which you need to know well – there are more in section 28 in
the data booklet.
fragment
CH3
H2O
CH3O
COOH
CH3CH2
CHO
is identified by the following mass/loss in
mass
15
18
31
45
29
29
Proton nuclear magnetic resonance spectroscopy (1H NMR): informs about the
different hydrogens
1H


NMR detects:
The number of 1 H atoms in a molecule (isotopes other than 1H can also be used
for this technique)
The group of atoms or environment the 1H atoms occur in e.g. the functional
group.
It works on the principle that some atomic nucleii like hydrogen (only atomic nuclei
with odd mass numbers such as 1H), just like electrons, have a magnetic spin which
gives rise to a magnetic field making the nucleus behave like a small magnet. As
there are two directions in which a nucleus can rotate (clockwise and anticlockwise)
there are 2 spin states. When no external magnetic field is applied, both spin states
have equal energy (equivalent).
However, when an external magnetic field is applied, the spin states split in terms of
energy as one of the spin states will now be aligned with the external magnetic field
and the other spin state will not (non-equivalent)
The two spin states within a magnetic field are:


a low spin state/low spin energy level/low energy alignment: the magnetic field of
the rotating nucleus or proton is aligned with the external magnetic field (most
stable!).
a high spin state: the magnetic field of the rotating nucleus is aligned against the
external field.
In a normal sample of a substance exposed to a magnetic field there are always
more nuclei (although not that many more) that are in the lower energy state i.e.
aligned with the external magnetic field. Such nucleii can be made to change their
spin state to the high spin state by radiating these nucleii with radiation of a
frequency that corresponds to the difference in energy between the two spin states.
When this frequency is applied, the nucleus or proton is brought into resonance,
radiation is absorbed and the nucleus flips (= goes to higher spin state). The
frequency of the radiation absorbed is recorded on a chart. The process of
absorption of energy to make the nuclear spin flip to the non-aligned spin state is
called nuclear magnetic resonance. The frequency for most 1H nucleii is within the
radiowaves spectrum.
Resonance (= transfer of energy from the radiation to the nucleus) occurs when the
frequency is correct for the size of the magnetic field applied at the time and the
identity of the nucleus; when this is the case, energy is transferred to the nucleus to
change its magnetic spin hence nuclear magnetic resonance occurs.
So, the actual frequency of the radiowave absorbed in 1H NMR depends on:



The type of hydrogen atom in a compound; equivalent (in terms of environment)
hydrogen atoms need the same frequency (also equivalent in energy).
The electron density of the environment of the hydrogen atom: e.g. a hydrogen
atom part of a CH3 -group absorbs a different frequency - there is a different
energy gap between the low spin state and the high spin state - than a hydrogen
nucleus attached to a carbon for instance in a ring structure or a hydrogen near an
oxygen or a nitrogen atom. This is because:
 The spin of any other hydrogen in the environment affects the spin of the
original hydrogen.
 The shielding effect:
 Electrons in surrounding atoms also have a magnetic spin that affects
the energy level of the spin state of the hydrogen atom.
 Electrons shield the hydrogen nucleus from the effects of the external
magnetic field – lower difference in energy.
 The larger the number and the closer the electrons are to the hydrogen
nucleus the more it is protected or shielded from the external magnetic
field i.e. in the OH group the electronegative oxygen pulls electrons
towards it (away from the hydrogen nucleus) exposing the hydrogen
more to the external magnetic field causing a lower shielding effect.
 Also double bonds can cause a change in shielding effect.
 The lower the shielding effect, the greater the amount of energy needed
to cause a flip.
The strength of the external magnetic field.
The particular position of absorption on the 1H NMR chart/spectrum corresponds to a
hydrogen or proton in a certain chemical environment e.g. in the OH group; the
number of signals recorded at that position indicates the number of hydrogen atoms
that are in that same chemical environment. The number of each type of hydrogen
is obtained from the relative areas underneath each of the peaks in the 1H NMR
spectrum.
In summary, the 1 H NMR tells us the chemical environments the hydrogen nuclei or
protons are in and how many types of protons or hydrogen atoms there are in each
environment.
Chemical shift
A reference compound is added to every sample to be tested, in most cases TMS
(tetramethylsilane) because:
 contains 12 hydrogen atoms that are equivalent so it gives a large single
line/signal which is always at a lower frequency than most other types of organic
protons – the hydrogen atoms in TMS experience greatest shielding due to least
polar bonds. The TMS absorption spectrum (1 line) is used to establish the zero
point or reference point on the scale.
 Signal is also ‘away’ from other signals so easy to identify in spectrum
 non-toxic/unreactive
 volatile – easily removed from sample
The signals from the hydrogen atoms of the compound to be tested are measured in
chemical shift and expressed in ppm. The chemical shift is a value relative to the
signal of the TMS; it indicates the number of units that the signal is shifted from the
signal of the TMS which has a very strong single peak to the right in the 1 H NMR
spectrum. The higher the shielding effect, the lower the chemical shift.
The chemical shift for a number of different protons are in section 27 in the data
booklet.
Interpretation of the 1 H NMR spectrum
The ratio of the areas under all the peaks is the ratio of the number of protons of
each type in the compound; an integration trace is placed on the spectrum in which
the height of each step/type of proton in the trace represents the area of a peak
indicating the relative number of that type of proton.
The ratio is usually expressed with the first number indicating the number of
hydrogen atoms or protons of the peak the nearest to the 1 H TMS.
TOPIC 20 and 21 ORGANIC CHEMISTRY
20.1 Types of organic reactions
Nucleophilic substitution reactions
A nucleophile is an electron rich species. Halogenoalkanes undergo nucleophilic substitution
reactions as the bond between the carbon atoms and the halogen is polar as the halogen is
highly electronegative giving the carbon atoms a partial positive charge; this carbon atom is
the reactive site in the halogenoalkane.
There are two different pathways or reactions mechanisms in which a nucleophilic
substitution reaction involving halogenoalkanes can occur: SN1 and the SN2. In this
substitution reaction where the halogen is substituted, the halogen is referred to as the
leaving group.
Whenever for instance a halogenoalkane and a nucleophile react there is competition
between these two different pathways or mechanisms, the SN1 and the SN2 pathway.
SN1 = Nucleophilic unimolecular substitution
Example: 2-bromo-2-methylpropane with sodium hydroxide to form 2-methyl propan-2-ol
and sodium bromide as SN1 is the predominant mechanism for tertiary halogenoalkanes.
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)
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.
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.
Predominant for primary halogenoalkanes
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]
Sn2 is a concerted reaction as all bond breaking and bond making occurs in a single step.
Only a transition state is formed – no intermediate.
Bond made
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 (1800) to the halogen (the halogen atom affects
approach by the nucleophile); during this process the carbonnucleophile bond is formed; the nucleophile donates its electron pair to
the carbon to form the bond. It is a Lewis base.
Bond broken
at the same time, the halogen breaks away (heterolytic fission) 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.
In the transition state both the nucleophile and halogen carry a partial negative charge; the
(negatively charged) nucleophile (in the case of a hydroxide ion) as it is donating its bonding
pair to the carbon and the halogen as it is taking the covalent bond between itself and the
carbon
The attack by the nucleophile on the opposite side to the halogen causes an inversion of the
configuration around the carbon on to which the halogen is bonded; the H atoms and/or
alkyl groups around the carbon move into the opposite space and this allows space for the
nucleophile to complete its approach to the carbon atom and make its bond. The
tetrahedral arrangement of the bonds on the carbon that carries the halogen is converted
into an inverted tetrahedral arrangement.
If the halogenoalkane is an optical isomer (can only be the case with a secondary
halogenoalkane) than the SN2 mechanism ensures that only one of the optical isomers can
be produced during the reaction as opposed to both enantiomers. The S N2 mechanism is
therefore considered a stereospecific process. The spatial arrangement of the reactant
determines the spatial arrangement of the product. SN1 is not stereospecific.
Secondary halogenoalkanes
Both SN1 and SN2 mechanisms occur for secondary halogenoalkanes.
Factors which affect the rate of nucleophillic reactions in halogenoalkanes
The pathway that is favoured for a particular nucleophilic substitution is the pathway with
the fastest rate (highest k value because of lowest activation energy) for that set of
reactants.
Factor 1: The hydroxide ion is a stronger nucleophile than water because it has a negative
charge and will therefore be attracted more strongly towards the partial positively charged
carbon atom bonded onto the halogen. As a result, the rate of a substitution with hydroxide
(or hydrolysis) is always faster than with water.
Factor 2: class of halogenoalkane: primary, secondary or tertiary.
The first elementary step in the SN1 pathway involves the formation of a carbocation
intermediate as the halogen leaves the molecule as an anion.
The greater the stability of the carbocation, the lower the activation energy of its formation
and the faster it is formed; the more its formation is favoured, the more SN1 is favoured.
The stability of carbocations depends on the number of alkyl groups on the carbon that
carries the positive charge as the electron-releasing alkyl groups make the carbon atom less
positive (delocalises the positive charge more). As a result, the greater the number of alkyl
groups, the more stable the carbocation.
Therefore, in decreasing stability of the carbocation (most stable first):
tertiary carbocation  secondary carbocation  primary carbocation
Tertiary carbocations, which are formed from tertiary halogenoalkanes, are more stable/less
reactive as the electron-releasing alkyl groups make the carbon atom less positive
(delocalises the positive charge more).
With primary halogenoalkanes, the formation of a reactive (unstable) primary carbocation
has an activation energy which is higher than the activation energy of the elementary step
in which the halogenoalkane and the nucleophile collide with each other which is why S N2 is
favoured with primary halogenoalkanes.
Another way of explaining favoured pathways is by considering how likely the transition state
in SN2 can be formed. In the transition state of SN2 carbon needs to accommodate five groups;
this is easier in a primary halogenoalkane as the carbon atom has 2 hydrogen atoms on it which
leaves space for a 5th atom or ion. The alkyl groups on the carbon atom in a tertiary
halogenoalkane allow less space for a fifth particle to bond onto the carbon.
Factor 3: pathway.
primary halogenoalkanes
SN2 - slowest
secondary
halogenoalkanes
SN1 or SN2
tertiary halogenoalkanes
SN1 - fastest
As SN1 reactions generally occur faster (they are unimolecular), nucleophilic substitutions
involving tertiary halogenoalkanes are usually faster than those involving secondary or
primary halogenoalkanes.
Factor 4. Identity of halogen or leaving group:
As the iodine – carbon bond is the weakest bond of the carbon – halogen bonds,
iodoalkanes tend to react at greater rates than other haologenoalkanes in both SN1 and SN2.
Relative rates of reaction of hydrolysis: fastest/highest rate: I  Br  Cl  F.
Factor 5: Nature of solvent: protic and aprotic solvents (both are polar)
SN2 reactions are best carried out in an aprotic solvent such as propanone, CH3COCH3, and
enthanenitrile, CH3CN. An aprotic solvent is a solvent that cannot make any hydrogen bonds
(e.g. with the nucleophile) and is therefore less likely to solvate/surround the nucleophile
ion allowing it to maintain a high energy state and increase the frequency of successful
collisions with the halogenoalkane increasing the reaction rate.
SN1 reactions have the highest reaction rates in protic solvents as protic solvents (protic =
has hydrogen atom onto an oxygen or nitrogen and can donate a H+) either contain an
amine group, –NH, or an hydroxyl group, –OH, and can therefore form hydrogen bonds
between the oxygen or the nitrogen of the solvent molecules and the partially charged
carbon atom on the carbocation solvating the ion – this gives stability to the intermediate
and favouring its formation. Examples of polar protic solvents are water, alcohols, amines,
amides (e.g. methanamide, HCONH2) and carboxylic acids.
Summary:
SN1
2 step process
unimolecular
first order reaction
auto-ionisation of compound
tertiary halogenoalkanes and some secondary
faster
nucleophile (which can be water, alcohol and
amines) can attack from 2 sides
no need for a strong nucleophile
energetically easier to form intermediate
stable intermediate
protic solvent
stereospecific
SN2
1 step process
bimolecular
second order reaction
collision between compound and nucleophile
primary halogenoalkanes and some secondary
slower
nucleophile attacks from opposite side to
halogen
need for a strong nucleophile
energetically easier for nucleophile to collide tha
to form an intermediate
transition state
aprotic solvent
not stereospecifc
Electrophilic addition reactions
An electrophile is an electron-deficient species that can accept electron pairs from a
nucleophile. Electrophiles are Lewis acids. An interhalogen is a compound that has 2 or
more different halogens.
Reaction mechanism in electrophilic addition reaction involving symmetrical alkenes
Step 1: generation of electrophile and the formation of the carbocation
 The double bond is an area of high electron density which induces polarisation (in a nonpolar molecule e.g. a halogen molecule) or further polarisation (in a polar molecule e.g.
hydrogen halide and water) in the other reacting specie; the  electrons can do this
because they are exposed as they are above and below the axis between the two nucleii;
these exposed electrons cause repulsion between themselves and the outer shell
electrons of the attacking halogen or hydrogen halide molecule.
 The  cloud of electrons then attracts the positively charged end of the molecule that
acts like an electrophile as it is attracted to the high electron density of the  bond.
 The polar bond in the attacking halogen molecule breaks (heterolytic fission!) and the
positive end joins the alkene forming an intermediate carbocation; a positive ion as an
electrophile joins a neutral molecule. The  bond is used to bond the halogen onto a
carbon which makes the other carbon atom on the double bond positive as it has lost its
share in the  bond. This carbon only has 6 outer electrons which makes it very unstable;
the partial negative part of the halogen molecule becomes an anion.
Step 2: nucleophilic attack to neutralise carbocation
 Electrostatic attraction (=ionic) between the positively charged carbon atom or
carbocation and the negative ion results in a quick neutralisation forming the end
product;
The addition reaction converts one  bond and one  bond into two  bonds which is
energetically more favourable as the heat needed to break the one  bond and one  bond
is less than the heat released when two  bonds are formed.
examples of electrophilic addition reactions:

Reaction with halogens: test for unsaturation: decolorisation of bromine water; reaction
occurs at room temperature and without sunlight. If reaction is carried out in sunlight


also substitution can occur. As the halogen molecules are non-polar, this reaction needs
polarization caused by the double bond to create the electrophile.
Reaction with interhalogens (e.g. ICl – iodine chloride): Such molecules have permanent
polarity; the least electronegative halogen will act as the electrophile and attack the
alkene.
Reaction with hydrogen halides. The hydrogen ion, H+, becomes the electrophile.
Electrophilic addition reaction with an interhalogen or hydrogen halide involving
unsymmetric alkenes: two products
When a hydrogen halide or an interhalogen (= examples of unsymmetrical electrophiles)
combine with an unsymmetrical alkene, e.g. propene (which is the smallest asymmetric
alkene) two or more intermediate carbocations are formed, each one giving its own
product. The proportion of these products is determined by the rule explained below.
For instance, in the addition of a hydrogen chloride to propene there are two possible
products; one product with the chlorine on the second carbon (2-chloropropane) or one
product with the chlorine on the first carbon (1-chloropropane). When such a reaction
occurs we need to use Markovnikov’s rule to predict the more common or major product:
Markovnikov’s rule:


In case of the addition of a hydrogen halide to an asymmetrical alkene, the major
product is the one in which the hydrogen of the hydrogen halide attaches itself to the
carbon carrying the larger number of hydrogen which in this case is the 2-chloropropane.
In the case of a mixed halide molecule, the least electronegative halogen becomes the
electrophile. The most common product is the one with the least electronegative
electrophile on the carbon with the most hydrogen atoms.
Reason: stability of carbocation produced:
The reason for the above is that during this addition reaction two carbocations are formed;
the carbocation - a secondary carbocation - which leads to the production of 2chloropropane is more stable (in terms of energy it is at a lower level) than the carbocation
– a primary carbocation - which leads to the production of 1-chloropropane. The more
stable ion is formed more readily.
Inductive effect = ability to donate electrons and reduce the charge on an adjacent atom.
decreasing stability
Why? The secondary carbocation (two alkyl groups bonded onto the carbon with the
positive charge) is more stable because each alkyl group donates electrons (any alkyl group
always donates electrons) slightly to the positive carbon atom it is attached to, reducing the
density of that positive charge (= positive inductive effect)
The larger the number of alkyl groups, the more electron density is shifted towards the
positive charge of the carbocation, the more the density of the positive charge is reduced as
the positive charge is now shared by both the carbon atom and the electrons of the alkyl
groups; the positive charge has been delocalised even more.
Again in terms of energy because secondary carbocations are more stable their formation
needs less activation energy than the formation of the primary carbocation which is why it is
produced a lot faster, exists for a longer time and why it dominates and the product it
forms, is formed in larger quantities. The product which is produced by the more stable
carbocation is the major product.
So the stability of a carbocation depends on the number of alkyl groups; tertiary
carbocations (e.g.(CH3)3 C+ ) are even more stable than the secondary ones (CH3)2 CH+; CH3+
is the least stable.
Electrophilic substitution reactions
Benzene is quite unreactive and resistant to chemical change because of its stable electron
structure and bonding with its bond order is 1.5 as a result of the delocalization. However,
benzene participates in substitution reactions with strong electrophiles and with the help of
a catalyst. During such a substitution reaction, the stable ring structure, and therefore its
aromacity, is maintained. During such a reaction a hydrogen atom from the ring is replaced
by another atom or group of atoms referred to as the electrophile. An example of such an
electrophilic substitution is a nitration reaction that uses a mixture of nitric acid and
sulphuric acid.
General mechanism of an electrophilic substitution:
 First, generation of the electrophile which in most cases is formed by the interaction
between the reagent and the catalyst.
 Then, two step substitution: (diagrams from
http://www.chemguide.co.uk/mechanisms/elsub/whatis.html#top)
 Step 1 (rate determining step):
 electron rich  cloud above and below the benzene molecule attracts
electrophiles and donates a  electron pair to it; the electrophile bonds
onto a carbon in the ring ; a  pair moves out of the ring causing the ring
to loose its stability;
 a reactive intermediate cation is formed; the positive charge is distributed
over the ring due to resonance; this restores some stability but not to the
same level as the original benzene molecule.
 Step 2 (fast step): the intermediate breaks up by the loss of a hydrogen ion
(deprotonation), which carries the extra charge away with it; the bonding
electrons between the carbon and the hydrogen move back into the ring as
delocalised electrons; the electrophile has substituted the hydrogen; the
hydrogen is used to regenerate the catalyst which now acts as a Lewis base.

As benzene is symmetrical, monosubstitution gives only 1 product irrespective of the
hydrogen that has been substituted i.e. no isomers.
Nitration of benzene: stepwise mechanism needed!!!
Overall equation: C6H6 + HNO3  C6H5NO2 + H2O
as catalyst.
at 50 oC and with sulphuric acid
Mechanism:
 Generation of the electrophile: the attacking electrophile is the reactive nitryl cation
(NO2+); which is generated from the reaction between concentrated nitric acid and
sulphuric acid after which the two-step substitution occurs;
overall equation: HNO3 + H2SO4  HSO4- + NO2+ + H2O


Main product (if dilute HNO3 and conc H2SO4 at 50 C): nitrobenzene. By-product is water.
Proton released by intermediate is used to regenerate sulphuric acid catalyst.
Reduction reactions
Carboxylic acids and ketones
The reduction of carboxylic acids into aldehydes/alcohols and ketones into alcohols by
heating them in dry ether together with reducing agents such as lithium aluminium hydride,
LiAlH4, and sodium borohydride, NaBH4, is the reverse of the oxidation of alcohols by
oxidizing agents such potassium dichromate.
 Carboxylic acids can be reduced to aldehydes and finally to primary alcohols
 Ketones can be reduced to secondary alcohols.
As lithium aluminium hydride is a stronger reducing agent it reduces carboxylic acids to both
aldehydes and primary alcohols whilst sodium borohydride can only be used to reduce
aldehydes and ketones to alcohols. Both reducing agents produce hydride ions which act as
nucleophiles as they are attracted to the carbon in the carbonyl group that has a partial
positive charge.
Simplified equations of examples of such reductions
Ethanoic acid reduced to ethanol and then reduced to ethanol by heating it dry ether with
LiAlH4
CH3COOH
+ [H]
 CH3CHO
(from LiAlH4)
+
[H]
 CH3CH2OH
(from LiAlH4)
Propanone reduced to ethanol and then reduced to ethanol by heating it dry ether with
NaBH4
CH3COCH3
+ [H]
 CH3CHOHCH3
(from NaBH4)
Conversion of nitrobenzene to phenylamine
This reaction happens in 2 stages:
Step 1: Nitrobenzene is heated under reflux with tin, the reducing agent, and concentrated
hydrochloric acid; the tin is oxidized to tin ions and the nitrogen is reduced from +4 to -2
during the formation of the phenylammonium ion.
C6H5NO2 (l) + 3Sn (s) + 7H+ (aq)  C6H5NH3+ (aq) + 3Sn2+(aq) + 2H2O (l)
Step 2: The addition of sodium hydroxide causes the deprotonation of the
phenylammonium ion to form aniline:
C6H5NH3+ (l) + OH- (aq)  C6H5NH2 (l) + H2O (l)
20.2 Synthetic routes
You need to be able to use the reactions that we have studied above to achieve functional
group interconversions as in many cases in organic chemistry the starting material for a
synthesis reaction is from a different class than the desired product. In many
interconversions this will involve more than one type of reaction i.e. a multistep synthetic
route.
Retro-synthesis
One approach to deducing synthetic routes is the retro-synthesis approach that involves
starting from the desired product (target molecule or lead molecule in the development of a
drug) and working backwards to identify a smaller/simpler starting material that can be
used to make the target molecule, and then identify a even smaller starting material for that
first starting material and so on with the starting materials becoming smaller or simpler and
in greater supply. The starting materials are also referred to as precursors.
Exercise: Deduce a synthetic route to make butanoic acid starting from butene. Show clearly
how you have used the retro-synthesis approach.
The table below shows the reactions that you need to know and can therefore be use to
deduce the synthetic routes.
Functional group
alkane
alkenyl
reaction
free radical substitution
electrophilic addition
hydroxyl


halogeno
carboxyl
nucleophilic substitution
reduction
phenyl
carbonyl
electrophilic substitution
reduction
esterification/condensation
oxidation
products
halogenoalkane
halogenoalkanes,
dihalogenoalkanes, alcohols,
polymers
 ester
 aldehydes, ketones,
carboxylic acids
alcohols
aldehydes, ketones and
alcohols
nitrobenzene
alcohol
20.3 Stereoisomerism
Stereoisomers are compounds with the same molecular formula and same structural
formula but with different arrangements of atoms in space i.e. the order in which atoms
are bonded or connected (same connectivity) along the chain is the same (same
constitution) but the spatial arrangement of atoms around certain (carbon) atoms is
different. The isomers studied in topic 10 were structural isomers as they have their atoms
and/or functional groups connected in different ways.
There are two classes of stereoisomers:
 Conformational isomers are isomers that interconvert spontaneously by free rotation
about a sigma bond without breaking and making any bonds.
 Configurational isomers are isomers that interconvert only by breaking and reforming a
bond. The configurational isomers are further subdivided into a class made up of cistrans isomers and E/Z isomers and a different class called optical isomers.
Configuration refers to the 3D spatial arrangement of atoms or groups of atoms.
Conformational isomers (no bond breaking to interconvert)
Conformations are different shapes a molecule can have as a result of the free rotation
about a sigma bond usually a C-C bond; the different spatial arrangements can be
interconverted by rotating that bond and this can happen spontaneously.
The most simple example of conformational isomerism can be shown using ethane which
has 2 distinct confirmations that can be shown using the diagram below to the right. The 2
different confirmations, eclipsed and staggered, have different stability.
In the eclipsed conformation (atoms in the front block the view of atoms at the back) of a
hydrocarbon such as propane, the hydrogen atoms (or could be different substituent groups
in other molecules) are as close to another as possible; in the staggered conformation they
are as far as possible. In a Newman projection to the left below showing propane, in the
staggered conformation the hydrogen atoms are 600 apart resulting in greater stability of
the molecule.
Configurational isomerism (bond braking to interconvert)
cis-trans isomerism
This type of stereo isomerism occurs because free rotation of a carbon-carbon bond is
prevented either because it is a double bond or because the carbon-carbon bond is part of a
ring structure; this means that both are chemically different compounds! Free rotation is
impossible as there are two regions of overlap within the  bond; above and below the axis
between both nuclei; rotation would cause the overlapping orbitals to separate and
therefore break the bond.
As a result cis-trans isomerism happens in:

Non-cyclic alkenes in which there are 2 different atoms (or groups of atoms) on each
carbon atom on either side of the double bond. The  bond prevents the C=C from
rotating and changing the position of the atoms on either side.

In cycloalkanes when at least two atoms (other than hydrogen) are found on two
different carbons in the ring; again no rotation is possible without breaking the ring.
We only use the cis-trans convention of naming if the different atoms on each carbon on
either side of the double bond are the same - they are called disubstituents. If both sets of
different atoms are not the same (tri- or tetra substituents) than we use a different
convention called the E/Z nomenclature and refer to the isomers as E/Z isomers.
Non-cyclic alkenes
Two possible isomers:
 cis: two identical atoms of groups of atoms (substituents or R) are on the same side of
the double bond or on the same side of the reference plane passing through the double

bond – the plane is perpendicular to all the sigma bonds between the atoms and the
substituents or perpendicular to the trigonal plane of the molecule
trans: two identical groups on opposite side of the reference plane passing through the
double bond.
Example 1:
melting point
(C)
boiling point
(C)
cis-but-2-ene
trans-but-2-ene
- 139
- 106
4
1
cis-1,2-dichloroethene
- 80
trans-1,2-dichloroethene
60
48
Example 2:
melting point
(C)
boiling point
(C)
- 50
Cyclo-alkanes
Cis-trans isomerism also occurs in cyclic alkanes in particular disubstituted cycloalkanes. The
ring forms the reference plane.
Free rotation around a single carbon-carbon bond is also restricted in cyclic alkanes due to
the inflexibility of the ring. Cyclic alkanes have a planar structure that allows cyclo-alkanes
to have other atoms, e.g. halogens, connected on the same side of the plane (cis isomer) or
one halogen on one side and one halogen on the other opposite side of the ring (trans
isomer). In cycloalkanes the substituent groups do not have to be on adjacent carbon
atoms.
Cis-trans isomerism can be recognised by looking for:


a double bond with two different atoms/groups joined to the atoms at either end of the
double bond; but both sets of different groups are the same.
a ring structure with two different atoms or groups of atoms joined to any two carbon
atoms in the ring.
E/Z isomerism
The cis-trans notation is rather limited and only works well when the two different atoms or
groups of atoms on each carbon on the double bond are the same.
E/Z notation is a different set of rules used to describe a double-bond configuration of the
following form R1R2C=CR3R4 whereby R1 ≠R2 and R3 ≠R4; the R represents a substituent
i.e. an atom or group of atoms that has substituted a hydrogen on the carbon chain.
The fact that R1 and R2 could be the same as R3 and R4 means the E/Z notation could also
be used when cis-trans can be used.
As stated earlier E/Z notation tends to be used when there are 3 or 4 different Rs as it is
difficult to apply the cis-trans convention as there could no atoms on each carbon that are
the same. The E/Z convention involves ranking (giving it a relative priority) the R1 and R2 on
the first carbon atom around the double bond and ranking R3 and R4 on the second atom.
The ranking is done using the Cahn-Ingold-Prelog system (CIP) of assigning priority to the
substituent groups which uses the atomic number of the R groups:
Rule 1: Look at the atom (or atom in the substituent group) bonded to the carbon of the
double bond. The atom with the higher atomic number has the higher priority or
precedence. Rule 2: If the atoms are the same, for example if they are both carbon atoms, apply the
same rule to the next bonded atom. This means that longer hydrocarbon chains have
higher priority. When the two groups or substituents (R) with the highest precedence are on the same side
of the reference plane passing through the double bond the molecule is referred to as the Z
isomers (zusammen in German)(equivalent to cis); if the highest groups wit the highest
precedence are on the opposite side of the reference plane the molecule is an E isomer
(equivalent to trans).
Optical isomerism
Optical isomers are isomers that have at least one chiral carbon referred to as an
asymmetrical centre or stereocentre. A chiral carbon or centre is an asymmetric carbon
atom with four different atoms or groups of atoms bonded onto it.
Chiral centres in molecules results in two different spatial arrangements or pairs. There are
two different types of pairs of optical isomers or chiral molecules: enantiomers and
diastereomers.
Enantiomers are a pair of optical isomers that are mirror images from each other but that
cannot be superimposed (non-superimposable). They have no plane of symmetry.
The diagram below is a general diagram of pair of enantiomers.
(from: http://www.docbrown.info/page07/isomerism2.htm)
Similarities between enantiomers

Enantiomers have identical melting and boiling points, density and solubility as the
different spatial arrangement does not affect the strength of intermolecular forces; also
because of this, solubility in a variety of solvents is the same;

Have identical chemical energetic stability and identical reaction paths (same activation
energies) which means that when a chiral molecule is formed during a reaction both
enantiomers are formed in equal amounts (=racemic mixture or racemate);
Different only in the following chemical and physical properties

Chemical: Both enantiomers react differently with other chiral molecules; this explains
why enzymes (which are chiral molecules) only react with a specific protein (other chiral
molecules) to catalyse a reaction.

Physical: Both enantiomers are optically active which means that they rotate planepolarized light; plane-polarised light is light that emerges from a polariser and is
characterised by the fact that the electric field of the light oscillates in one plane only e.g.
perpendicular to the direction in which the light travels. The angle of rotation of the
plane-polarized light can be measured using a polarimeter after a beam of planepolarized light is passed through the sample of the substance to be tested. See diagram
below.
(from http://chemwiki.ucdavis.edu)

Each enantiomer however rotates the plane of polarised light in the opposite direction
(non-chiral centres do not rotate the light at all!) but over the same angle but in the
opposite direction. The enantiomer that rotates it clockwise is the (+) enantiomer whilst
the other (anticlockwise) is the (-) enantiomer.

However, a racemic mixture does not show any optical activity as both the rotation of
one enantiomer is cancelled out by the rotation of the other enantiomer.
Diastereomers are also optical isomers that, just like enantiomers, are non-superimposable
and have at least two or more chiral carbons or stereocentres in their molecule.
However, enantiomers have an opposite configuration at each chiral centre in their
molecule which is why they are mirror images of each other.
This is not the case in diastereomers as they only have an opposite spatial arrangement
around one or two chiral centres in their molecules, but not around all of them.
Diastereomers are stereoisomers that are not mirror images of each other.
Diastereomers with the same molecular formula have both different physical and chemical
properties. Some diastereomers are not chiral i.e. optically active.
21. 1 Measurement and analysis: spectroscopic identification of organic compounds
compound
qualitative analysis
(e.g. with reducing agent)
mass spectrometry
chemical tests
identity of
elements
molecular mass
quantitative analysis
(e.g. combustion analysis)
empirical formula
1H
NMR: - functional groups/type of protons
- number of protons of each type
X-ray crystallography: bond angles/lengths
IR: functional group
MS: - number of carbon atoms
- functional group
molecular formula
(isomerism)
Molecular structure
The structure and formula of a compound can be determined by using information from a
variety of spectroscopic and chemical characterization techniques. However, information
from one technique is usually insufficient to determine or confirm a structure.
High resolution 1H NMR
Splitting pattern
A high resolution 1H NMR spectrometer can show further splitting in some of the single
peaks and this provides us with more information. The splitting of the peaks is caused by
spin-spin coupling. Spin-spin coupling is the consequence of one or more hydrogen atoms in
one environment, as a result of their spinning, shielding (lower difference between the low
and high spin state) or deschielding (greater difference – larger chemical shift) hydrogen
atoms on an adjacent carbon atom. If there is no hydrogen atom on an adjacent carbon
then spin-spin coupling does not occur.
If a hydrogen atom on an adjacent carbon atom to a hydrogen atom, Ha, spins in alignment
with the magnetic field the hydrogen atom on the adjacent carbon atom deshields the
hydrogen atom Ha causing a stronger local magnetic field, a greater difference in energy
between the two spin states and as a result a higher chemical shift.
If its spin is opposed to the magnetic field it shields Ha, lowers its local magnetic field and
the chemical shift will be lower.
Both possible spin states of the adjacent hydrogen will show up as 2 split lines of equal ratio
of intensity, a doublet, instead of a single peak, in a high resolution spectrum. The two lines
correspond to each alignment of the adjacent hydrogen.
Therefore a double peak at a particular chemical shift value means that for the hydrogen
atoms of that chemical shift value there is 1 hydrogen atom on an adjacent atom.
The number of lines or split peaks and their ratio’s allow us to determine the number of
hydrogen atoms that are on adjacent carbon atoms. This is shown by Pascal’s triangle
below.
Splitting pattern
(Pascal triangle)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
ratio
1:1
1:2:1
1:3:3;1
1:4:6 :4 :1
1:5:10:10:5:1
type
singlet
double
triplet
quartet
hydrogen atoms on
adjacent carbon
0
1
2
3
4
5
A singlet means that there is no hydrogen atom on an adjacent carbon atom.
TMS
From topic 10 we remember that a signal on a 1H NMR spectrum corresponds to one type of
hydrogen (or proton) in a certain chemical environment e.g. a hydrogen in an OH group; the
height of a signal or peak recorded at that position in the spectrum indicates the number of
hydrogen atoms that are in that same chemical environment. The number of each type of
hydrogen atom is obtained from the relative areas underneath each of the peaks. As a result
the 1H NMR spectrum tells us the chemical environments the protons are in and how many
types of protons or hydrogen there are in each environment.
The diagram above shows a simplified 1H NMR spectrometer: when the radiation emitted
from the oscillator or radio signal generator is of the correct value to allow protons to move
to higher energy, the energy is absorbed and resonance occurs. Resonance occurs
whenever the radiation and the strength of the magnetic field have specific values. In most
1H NMR spectrometer it is the strength of the magnetic field that is varied and the radiation
kept constant.
The absorption causes a tiny electrical current to flow in an antenna coil surrounding the
sample and which is connected to the detector from which a chart is produced.
The actual absorption frequencies in an experiment depend on several factors like strength
of magnetic field used and the frequency of the radiofrequency oscillator. This can result in
small variations between different 1H NMR spectrometer or even in the same one during an
experiment that makes analysis less accurate. It is for this reason that a standard or
reference is used against which the resonance are frequencies of the sample is compared
and its chemical shift determined.
In most cases that reference chemical is TMS which contains 12 hydrogen that are
equivalent to each other in terms of their environment. The chemical shift of this signal is
given the value of 0 ppm and is the reference point from which all other chemical shifts are
measured.
TMS is used as a reference because:
 It gives a single signal/line because of the large number (12) of equivalent hydrogen
atoms.
 Its line is well removed from other lines or signals.
 It has a strong signal as it has a large number of hydrogen atoms.
 It is chemically inert
 Non-toxic
 Because it has a low boiling point it is easily removed from the sample after the
measurements.
 Its single line which is always at a higher frequency than other types of organic protons –
the hydrogen atoms in TMS experience greatest shielding due to least polar bonds).
The signals from the hydrogen atoms of the compound to be tested are measured in
chemical shift that indicates the number of units that the signal is shifted from the signal of
the TMS. The higher the shielding effect the lower the chemical shift; the chemical shift is
only a relative value.
Interpretation of the spectrum: the ratio of the areas under the peaks is the ratio of the
number of protons of each type; an integration trace is placed on the spectrum in which the
height of each step represents the area of a peak indicating the number of protons. The
greater the shift the less the shielding experienced.
The ratio is usually expressed with the first number indicating the number of hydrogen
atoms or protons of the peak the nearest to the TMS.
X-ray crystallography or diffraction
This technique is used to determine bond angles and bond lengths in crystalline compounds
including organic compounds.
The main principles are:
 X-rays are used as their wavelength is similar to the distances between atoms.
 X-rays are directed at a single crystal (must be a solid) of the substance to be
investigated.
 The interaction between the X-rays and the electrons produce a diffraction pattern.
 The diffraction pattern depends on:
o angle of incidence of the X-rays into the crystal
o wavelength of the X-rays
o distance between the atoms in the crystal
o spatial arrangement or orientation of the atoms in the crystal
 The diffraction pattern is used to produce an electron density map of the crystal.
 The atoms of each element have different electron densities as it depends on the
electron configuration in the atom. Hydrogen atoms do not show up in electron
density maps.
 X-ray crystallography can be used to identify the bond lengths and bond angles of
crystalline compounds.