Download Organic Chemistry: Introduction

Organic Chemistry:
Introduction
IB Topic 10
Fundamentals of Organic
Chemistry
Section 10.1
What is organic chemistry?
Organic Chemistry
• The study of carbon, the compounds it
makes and the reactions it undergoes.
• Over 16 million carbon-containing
compounds are known.
Carbon
• Carbon can form multiple bonds with other carbon and
with atoms of other elements.
• Carbon can make four bonds since it has 4 valence
electrons and most often bonds to H, O, N and S.
• Because the C-C single bond and the C-H bond are
strong, carbon compounds are stable.
• Carbon can form chains and rings.
Hydrocarbons
• Hydrocarbons are organic compounds that
only contain carbon and hydrogen
• Some types of hydrocarbons include
 Alkanes
 CnH2n+2
 Alkenes
 CnH2n
 Alkynes
 CnH2n-2
Hydrocarbons
• Classified as:
• Saturated - all C-C single bonds
• Unsaturated - contain double or triple C-C
bonds
• Aliphatic – straight chains with single C-C
bonds
• Cyclic – ring structures with single C-C
bonds
• Aromatic – ring structures of alternating
single and double C-C bonds (arenes)
Homologous Series
• A homologous series is a series of related
compounds that have the same functional
group. Homologous compounds…
• For example, differ from each other by a –
CH2 – unit (methylene group)
• Can all be represented by a general
formula (alkanes – CnH2n+2)
• Have similar chemical properties
• Have physical properties that vary in a
regular manner as the number of carbon
atoms present increases
Homologous series – Alkanes
#C
Prefix
Alkane (ane)
CnH2n+2
1
meth
CH4
methane
2
eth
C2H6
ethane
3
prop
4
but
5
pent
6
hex
Trends in Boiling Points
What is the trend?
Alkane
Formula
Boiling
Pt./oC
methane CH4
-162.0
ethane
C2H6
-88.6
propane C3H8
-42.2
butane
C4H10
-0.5
Trends in Boiling Points
• Intermolecular forces present:
• Simple alkanes, alkenes, alkynes → van der
Waals’ forces (nonpolar) → lower b.p.
• Aldehydes, ketones, esters & presence of
halogens (polar) → dipole: dipole forces →
slightly higher b.p.
• Alcohol, carboxylic acid & amine → hydrogen
bonding (w/ O, N, F) → even higher b.p.
Formulas
Empirical Formula:
Smallest whole number
ratio of atoms in a
formula
Molecular Formula:
Formula showing the
actual numbers of
atoms
Molecular
Formula
Empirical
Formula
CH4
CH4
C2H6
CH3
C6H12O6
C4H8
C8H16
Formulas
Structural Formula
• Bond angles are drawn as though 90o. The true
shape around C with 4 single bonds is tetrahedral
and the angle is 109.5o.
• Show every atom and every bond. Can use
condensed structural formulas.
• Hexane: CH3CH2CH2CH2CH2CH3 (condensed s.f.)
M.F. = C6H14
E.F. = C3H7
Writing structural formulae
• Write the condensed structural formula for the following
two compounds.
• For branches, write them in parantheses with the C they
branch from.
Isomers
• Isomers: different compounds that have the same
molecular formula
• Structural isomers: an isomer in which the atoms
are joined in a different order so that they have
different structural formulae
• Three kinds: chain isomers, positional isomers,
functional group isomers
• Warning! Some molecules may seem like
isomers, but there is free movement around C-C
single bonds; they can rotate.
Isomers
• Chain isomers: branches are placed in
different spots on the C-C backbone
Isomers
• Positional isomers: important
functional groups are moved around on
the C-C backbone, but the backbone
does not change
Isomers
• Functional group isomers: certain
atoms are rearranged on the C-C
backbone to form different functional
groups
Drawing structural formulae
• Draw out the structural formulas and write the
condensed formula for all isomers that can be
formed by:
• CH4
• C2H6
• C3H8
• C4H10
Naming compounds
1. Determine the longest carbon chain
2. Use the prefix to denote the number carbons in the
chain
3. Use the suffix “-ane” to indicate that the substance
is an alkane
4. If the chain is branched, the name of the side
chain will be written before the main chain and will
end with “–yl”
Naming compounds
Methylpropane
Methylbutane
Dimethylbutane
Naming compounds
1
Meth-
6
Hex-
2
Eth-
7
Hept-
3
Prop-
8
Oct-
4
But-
9
Non-
5
Pent-
10
Dec-
Naming compounds
For chains longer than 4 carbons with side chains:
5. Number the carbons in the chain consecutively,
starting at the end nearest side chains.
6. Designate the location of each substituent group by an
appropriate number and name.
And with 2 or more side chains:
5. Use prefixes di-, tri-, tetra-, to indicate when there are
multiple side chains of the same type.
6. Use commas to separate numbers and hyphens to
separate numbers or letters.
7. Name the side chains in alphabetical order.
Naming compounds
•
•
•
•
Alkenes have one (or more) carbon to carbon double bonds
Suffix changes to “-ene”
When there are 4 or more carbon atoms in a chain, the
location of the double bond is indicated by a number. Begin
counting the carbons closest to the end with the C=C bond
Numbering the location of the double bond(s) takes
precedence over the location of side chains
1-butene
2-butene
24
Functional Groups (p. 243)
Naming Carbon Rings
• To name a molecule with a ring, follow the steps as
previously described for branches and functional
groups.
• The numbering of the base chain/ring begins at the functional
group. Double/triple bonds start the ring, if present.
• For the base chain/ring, insert a “cyclo-” prefix before
the Latin number prefix.
• Example:
Classifying molecules
With reference to the carbon that is directly bonded
to an alcohol or amine group or a halogen:
• Primary = carbon atom is only bonded to one
other carbon
• Secondary = carbon atom is bonded to two other
carbons
• Tertiary = carbon atom is bonded to three other
carbons
Aromatic Hydrocarbons
• Presence of a benzene ring.
• August Kekule proposed a structure with
a ring of C with alternating single and
double bonds.
• Unsymmetrical with different lengths of
bonds.
• Experimentally, it was discovered that
the bonds are, in fact, the same length
(140 pm)
• Single bond – 154 pm; double bond – 134 pm
• Bond order of 1.5
• Actually symmetrical!
Aromatic Hydrocarbons
Delocalized electrons from the ½ bond – resonance!
Aromatic Hydrocarbons
Benzene is uncharacteristically stable.
When adding H to this molecule, you would expect the
enthalpy to change 3 times that of adding H to
cyclohexene, but it isn’t – it’s much less.
This difference in energy (expected vs. actual) is known
as resonance energy or delocalization energy.
Functional Group Chemistry
Section 10.2
Alkanes
• Simplest hydrocarbons
• Low bond polarity
• Strong covalent bonds
• C-C (346 kJ mol-1)
• C-H (414 kJ mol-1)
• Relatively inert
• Important reactions:
• Combustion
• Halogenation
Combustion of Alkanes
• Used as fuels (propane, butane, octane, etc) for the
large energy released
• Volatility decreases as the length of C-chain increases –
short chains used as fuel
• Undergoes complete combustion in presence of excess
O2 to produce CO2 and H2O.
Combustion of Alkanes
• Undergoes incomplete combustion in presence of
limiting O2 to produce CO and H2O.
• CO irreversibly binds hemoglobin the blood thus
reducing its oxygen-carrying capacity.
• Suffocation results
Quick Question
Deduce the balanced equations for the
complete combustion of:
1. Propane
2. Pentane
3. Hexane
Types of Reactions
• Substitution: replacement of individual atoms
with other single atoms or with a small group
of atoms.
• Addition: two molecules are added together
to produce a single molecule.
• Elimination: the removal of two substituents
from the molecule.
Halogenation of Alkanes
• Halogenating alkanes increases reactivity.
• Free-radical substitution and elimination
• Free-radical refers to a species that is formed when a
molecule undergoes homolytic fission: two electrons
of a covalent bond are split evenly between two atoms
resulting in two atoms with a single electron.
• Heterolytic fission: both electrons in the bond are
transferred to one atom resulting in cation and anion
Halogenation of Alkanes
Example:
methane reacts with chlorine in the presence of UV light:
Halogenation of Alkanes
3 stages to free-radical substitution:
1. Initiation
2. Propagation
3. Termination
Halogenation of Alkanes
Initiation: homolytic fission of the chlorine
molecule in presence of UV light produces 2 freeradicals
Halogenation of Alkanes
Propagation: first stage is reaction of methane
and chlorine free-radical to produce methyl
radical.
Halogenation of Alkanes
Propagation: second stage is the reaction of
methyl radical with chlorine to produce
chloromethane and chlorine radical
Halogenation of Alkanes
Termination: reduces the concentration of
radicals. Radicals begin to react with other
radicals.
Alkenes
• Unsaturated hydrocarbons – contain at least
one C-C double bond.
• Double bond makes them more reactive than
corresponding alkane
• Undergoes addition reactions.
• Test for unsaturation:
• bromine water
• Addition of alkene to
bromine water adds Br to
molecule, thus rendering
it colorless.
Hydrogenation
• Addition of hydrogen
• Important in food industry – removing the doublebond increases melting point thus making a
substance that is solid rather liquid at room
temperature
• Partial hydrogenation of fats and oils can be
harmful to health – saturated vs unsaturated fats.
Halogenation of Alkenes
• Electrophilic halogenation of symmetrical alkenes
involves addition of elemental halogens resulting in
dihalogenated alkane:
Halogenation of Alkenes
• Example: but-2-ene and bromine
Halogenation of Alkenes
Addition of hydrogen halide, HX, to a symmetrical alkane
results in mono-halogenated alkane:
Polymerization of Alkenes
• Plastics industry utilizes addition polymerization
• Reaction of many smaller monomers with a C=C linking
together to form a polymer.
• Monomer ethane supplied by petrochemical industry
undergoes addition polymerization to form polyethene.
• Any monomer with a C-C double bond can undergo
polymerization wherever the double bond is located.
Polymerization of Alkenes
Alcohols
Can undergo complete combustion reactions to
form CO2 and H2O.
Oxidation of Alcohols
Oxidation of Alcohols
• Oxidation of primary alcohols occurs in two steps. First
step produces an aldehyde. Second step produces a
carboxylic acid.
Oxidation of Alcohols
• Aldehydes can be
recovered using
distillation.
• Distillation involves
the gentle evaporation
of liquids utilizing the
difference in boiling
points. The gas is
collected and cooled
into a pure distillate.
Oxidation of Alcohols
• Oxidation of secondary alcohols produces a ketone.
Oxidation of Alcohols
• To get a carboxylic acid, the
aldehyde has to remain in the
solution with the oxidizing agent for
a longer amount of time. Instead of
distillation, a reflux column is used.
• Refluxing is a technique that
involves the cyclic evaporation and
condensation of a volatile reaction
mixture, preserving the solvent as it
is does not evaporate.
Condensation reaction of
alcohol and carboxylic acid
• Esters are derived from carboxylic acids
• Applications include flavoring agents, medications,
solvents, and explosives.
• Esterification – a reversible reaction that occurs when a
carboxylic acid and an alcohol are heated in the
presence of a catalyst (e.g. H2SO4)
Nucleophilic substitution
• A halogenoalkane can undergo other reactions.
• The polar carbon-halogen bond, C-X, creates
an electron deficient carbon making it open to
‘attack’ by electron-rich species known as
nucleophiles – species that contain a lone pair
of electrons and sometimes a full negative
charge.
Electrophilic substitution
• Benzene does not readily undergo addition
reactions but will undergo electrophilic
substitution reactions.
• Electrophiles – electron-poor substance
capable of accepting an electron pair.
• The double-bond attracts the electrophile but
the stability of benzene leads to substitution
NOT addition (like with alkenes).
Nucleophilic substitution
Electrophilic substitution