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Chapter 2Chapter 2Alkanes and Cycloalkanes: Alkanes and Cycloalkanes:
Introduction to Introduction to HydrocarbonsHydrocarbons
Hydrocarbons only contain carbon and hydrogen atoms.
Hydrocarbons are either classed as aliphatic or aromatic.
Aliphatic hydrocarbons contain three main groups: alkanes which only have carbon-carbon single bonds,
alkenes which have a carbon-carbon double bond, or alkynes which have a carbon-carbon triple bond.
Classes of HydrocarbonsClasses of Hydrocarbons
Aromatic hydrocarbons are more complex but the simplest aromatic hydrocarbon is benzene. Aromatic hydrocarbons are called arenes.
Classes of HydrocarbonsClasses of Hydrocarbons
The Lewis model of chemical bonding predates the idea
that electrons have wave properties.
Two widely used theories of bonding based on the wave
nature of an electron are:
Valence Bond Theory, and
Molecular Orbital Theory
Models for Chemical BondingModels for Chemical Bonding
Which electrostatic forces are involved as two hydrogen
atoms approach each other and form a H-H bond.
These electrostatic forces are:• attractions between the electrons and the nuclei• repulsions between the two nuclei • repulsions between the two electrons
+ e– + e–
Formation of HFormation of H22 from Two Hydrogen Atoms from Two Hydrogen Atoms
Potential
energyH• + H•
Internuclear distance
H H
weak net attraction at
long distances
Potential Energy vs Distance Potential Energy vs Distance Between Two Hydrogen Atoms Between Two Hydrogen Atoms
Potential
energyH• + H•
Internuclear distance
H H
H H
H H
attractive forces increase
faster than repulsive forces
as atoms approach each other
Potential Energy vs Distance Potential Energy vs Distance Between Two Hydrogen Atoms Between Two Hydrogen Atoms
Potential
energyH• + H•
H2
Internuclear distance
74 pm
H H
H H
H H-436 kJ/mol
maximum net attraction
(minimum potential energy)
at 74 pm internuclear distance
Potential Energy vs Distance Potential Energy vs Distance Between Two Hydrogen Atoms Between Two Hydrogen Atoms
Potential
energyH• + H•
H2
Internuclear distance
74 pm
H H
H H
H H-436 kJ/mol
repulsive forces increase
faster than attractive forces
at distances closer than 74 pm
Potential Energy vs Distance Potential Energy vs Distance Between Two Hydrogen Atoms Between Two Hydrogen Atoms
Valence Bond Theory
constructive interference between two half-filled atomic
orbitals is basis of shared-electron bond
Molecular Orbital Theory
derive wave functions of molecules
by combining wave functions of atoms
Models for Chemical BondingModels for Chemical Bonding
Waves interactions include:
Constructive interference when the waves are in phase
and reinforce each other
Destructive interference when the waves are out of
phase and oppose each other
Behavior of WavesBehavior of Waves
Electron pair can be shared when half-filled orbital of
one atom overlaps in phase with half-filled orbital of
another. For example with overlap of two 1s orbitals
of two hydrogen atoms shown below:
Valence Bond Model for Valence Bond Model for Bonding in HydrogenBonding in Hydrogen
The approach of the two hydrogen atoms can be modeled
showing electrostatic potential maps. The high electron
density between the nuclei is apparent.
Valence Bond ModelValence Bond Model
Electrons feel the
attractive force of the
protons
Orbitals begin
to overlap
Optimal distance
between nucleiHigh electron density
between the nuclei
The Sigma (The Sigma () Bond) Bond
A bond in which the orbitals overlap along a line
connecting the atoms is called a sigma (s) bond.
Two perpendicular views are shown below.
Electrons in molecules occupy molecular orbitals (MOs)
just as electrons in an atom occupy atomic orbitals (AOs).
MOs are combinations of AOs.
Two electrons per MO.
The additive combination of two atomic orbitals
generates one bonding orbital.
The subtractive combination of the two atomic orbitals
generates an antibonding orbital.
Bonding in HBonding in H22::
The Molecular Orbital ModelThe Molecular Orbital Model
Addition of the AOs to form the bonding MO ()
Molecular Orbital Model for HMolecular Orbital Model for H22
Subtraction of the AOs to form the antibonding MO (*)
Format is AOs on the sides and MOs in the middle.
Combination of n AOs results in n MOs.
Bonding MOs lower in energy than antibonding MOs.
Fill electrons in MOs the same as for AOs – lowest first.
Molecular Orbital DigramsMolecular Orbital Digrams
Introduction to Alkanes:Introduction to Alkanes:Methane, Ethane, and PropaneMethane, Ethane, and Propane
Small AlkanesSmall Alkanes
General formula for alkanes is CnH2n+2.
Smallest alkane is methane CH4 - also the most abundant.
Ethane (C2H6) and propane (C3H8) are the next alkanes.
Natural gas is 75% methane 10% ethane and 5% propane.
These alkanes have the lowest boiling points.
Structures of AlkanesStructures of Alkanes
All carbons in methane, ethane and propane have four
bonds.
Bond angles (which are close to 109.5o) and bond lengths
are:
The dilemma:
Methane has tetrahedral geometry.
This is inconsistent with electron configuration of carbon of1s2, 2s2, 2px
1,2py1 with only two unfilled orbitals.
Structure and Bonding TheoryStructure and Bonding Theory
spsp33 Hybrid Orbitals Hybrid Orbitals
Linus Pauling proposed a mixing or hybridization of the s
and three p orbitals to create 4 equal unfilled orbitals called
sp3 orbitals.
Properties of Properties of spsp33 Hybrid Orbitals Hybrid Orbitals
All four sp3 orbitals are of equal energy.
The axes of the sp3 orbitals point toward the corners of a
tetrahedron.
σ Bonds involving sp3 hybrid orbitals of carbon are
stronger than those involving unhybridized
2s or 2p orbitals.
Bonding with Bonding with spsp33 Hybrid Orbitals Hybrid Orbitals
Bonding in methane involves orbital overlap between each
partially filled carbon sp3 orbital and a partially filled s
orbital of the hydrogen atom.
Bonding and Structure of EthaneBonding and Structure of Ethane
Ethane also has tetrahedral geometry about the
carbon
atoms.
Hybridization can be used to rationalize the bonding.
The C-H bonds are formed as described for methane.
The C-C bond is formed by overlap of sp3 orbitals on
each of the carbon atoms.
C-C Bond Formation in EthaneC-C Bond Formation in Ethane
Two half-filled sp3
orbitals on each C
Electrons with
opposite spin
Overlap of orbitals
to form a bonding
orbital.
Structure of Ethylene and sp2 HybridizationStructure of Ethylene and sp2 Hybridization
Ethylene is planar with bond angles close to 120o.
sp3 Hybridization cannot be used to explain this
bonding.
Three atoms are bonded to each carbon so three
hybrid
orbitals are formed. Called sp2 orbitals.
One p orbital is not hybridized.
spsp22 Hybrid Orbitals Hybrid Orbitals
The 2s and two of the 2p orbitals are mixed to form
three sp2 orbitals with a trigonal planar arrangement.The 2pz orbital remains half filled.
Sigma (Sigma () Bonding in Ethylene) Bonding in Ethylene
Form C-H bonds
by overlap of sp2
and s orbitals
These are all sigma () bonds. An unfilled p
orbital remains on each carbon atom.
Form C-C bond
by overlap of sp2
orbitals on each
carbon
Pi (Pi () Bonding in Ethylene) Bonding in Ethylene
This called a pi () bond and the electrons in the
bond are called electrons.
Form second C-C
bond by overlap
of p orbitals on
each carbon
Structure of Acetylene and sp HybridizationStructure of Acetylene and sp Hybridization
Acetylene is linear with bond angles of 180o.
sp3 and sp2 Hybridization cannot explain this bonding.
sp Hybridization explains this. There are two half filled
p orbitals no hybridized.
spsp Hybrid Orbitals Hybrid Orbitals
The 2s and one of the 2p orbitals are mixed to form
two sp orbitals with a linear arrangement. The 2py and 2pz orbitals remain half filled.
Sigma (Sigma () Bonding in Acetylene) Bonding in Acetylene
Form C-H bonds
by overlap of sp
and s orbitals
These are all sigma () bonds. Two unfilled p
orbitals remain on each carbon atom.
Form C-C bond
by overlap of sp
orbitals on each
carbon
Pi (Pi () Bonding in Acetylene) Bonding in Acetylene
There are two pi () bonds and a total of 4 electrons.
Form one bond by overlap of py orbitals
on each carbon
Form second
bond by overlap of pz orbitals on each
carbon
Hybridization of CarbonHybridization of Carbon
Carbons bonded to four atoms are sp3 hybridized with
bond angles of approximately 109.5o.
Carbons bonded to three atoms are sp2 hybridized with
bond angles of approximately 1200 and one C-C -bond.
Carbons bonded to two atoms are sp hybridized with
bond angles of approximately 1800 and two C-C -bonds.
Theories of Chemical BondingTheories of Chemical Bonding
Approaches to chemical bonding:
1.Lewis model;
2.Orbital hybridization model;
3.Molecular orbital model.
Considerations of Chemical BondingConsiderations of Chemical Bonding
Lewis and Orbital hybridization models work together
and success in organic depends on writing correct Lewis
structures.
Molecular orbital theory provides insights into structure
and reactivity lacking in the other models. This model
requires higher level theory which will not be presented.
The results of MO theory will be used – for example
electrostatic potential maps.
Isomers of ButaneIsomers of Butane
There is only one isomer for each of the molecular formulas CH4, C2H6 and C3H8.
For C4H10 there are two distinct connectivities of the
carbon atoms. They are constitutional isomers.
H
H
C
H
C
H
C
H
C
HH HH
HH
H
CH
C
H
CH
C
H HH
H
H
Bondline
formulas
Isomers of ButaneIsomers of Butane
The isomers have different physical properties.
All carbon atoms are sp3 hybridized.
Higher n-AlkanesHigher n-Alkanes
n-Alkanes are straight-chain alkanes with general formula CH3(CH2)nCH3. n-Pentane is CH3CH2CH2CH2CH3 and
n-hexane is CH3CH2CH2CH2CH2CH3. These formulas can be
abbreviated as CH3(CH2)3CH3 or CH3(CH2)4CH3.
Isomers of CIsomers of C55HH1212
There are three isomers C5H12.
It is important to realize that these are all representations
of isopentane.
Isomers of higher n-alkanesIsomers of higher n-alkanes
For higher n-alkanes there are many isomers and it is not
possible to easily predict how many isomers can be
formed.
IUPAC Naming IUPAC Naming
Alkane names are the basis of the IUPAC system of
nomenclature. The –ane suffix is specific to alkanes.
The IUPAC Rules for Branched AlkanesThe IUPAC Rules for Branched Alkanes
Rules for naming branched alkanes:
1.Find the longest continuous carbon chain and its IUPAC name. This is the parent alkane.
2.Identify the substituents on this chain.
substituent
longest chain
(5 carbons)
The IUPAC Rules for Branched AlkanesThe IUPAC Rules for Branched Alkanes
Rules for naming branched alkanes:
3. Number the longest continuous chain in the direction that gives the lowest number to the first substituent.
4. Write the name of the compound. The parent alkane is the last part of the name and is preceded by the names of the substituents and their numerical locations (locants). Hyphens separate the locants from the words.
2-methylpentane
The IUPAC Rules for Branched AlkanesThe IUPAC Rules for Branched Alkanes
Rules for naming branched alkanes:
5.When the same substituent appears more than once,
use the multiplying prefixes di-, tri-, tetra-, and so on. A separate locant for each substituent. Locants are separated from each other by commas and from the words by hyphens.
2,2-dimethylbutane 2,3-dimethylbutane
Alkyl GroupsAlkyl Groups
Alkyl groups are substituents derived from alkanes.
They lack one hydrogen at the point of attachment.
The alkyl group is named from the alkane by replacing the
-ane suffix with –yl.
For example a CH3CH2CH2CH2- substituent is a butyl group.
Classification of Carbon AtomsClassification of Carbon Atoms
Carbon atoms are defined as primary, secondary, tertiary
or quaternary.
A primary carbon is directly attached to one other carbon.
A secondary carbon is directly attached to two other
carbons. A tertiary carbon to 3 and a quaternary carbon
to 4.
Complex Alkyl Groups (Substituents)Complex Alkyl Groups (Substituents)
Secondary and tertiary groups may have common names
and IUPAC names.
The base name of these groups is the longest chain
including the attachment carbon form and the
substituents are located on this chain.
Naming Highly Branched AlkanesNaming Highly Branched Alkanes
When two or more different substituents are present
number from the end closest to the first point of difference.
When two or more different substituents are present, they
are listed in alphabetical order in the name. Prefixes such
as di-, tri-, and tetra- are used but ignored when alphabetizing.
tert-Butyl precedes isobutyl. sec-Butyl precedes tert-butyl.
4-ethyl-3,5-dimethyloctane
Naming Highly Branched AlkanesNaming Highly Branched Alkanes
When two or more different substituents are present
number from the end closest to the first point of difference.
If the first substituent is located an equal distance from each
end then the second substituent becomes the first potential
point of difference and so on.
Naming CycloalkanesNaming Cycloalkanes
Cycloalkanes contain a ring of carbons and have general formula CnH2n.
Add the prefix cyclo- to the name of the corresponding
alkane.
Naming CycloalkanesNaming Cycloalkanes
Identify and name substituents as before.
For one substituent no numbers are used.
Naming CycloalkanesNaming Cycloalkanes
For multiple substituents the locations must be specified.
Number the carbon atoms of the ring in the direction that
gives the lowest number to the substituents at the first
point of difference.
First substituent is on C1 by default.
Naming CycloalkanesNaming Cycloalkanes
If the ring has fewer carbons than the alkyl group attached
to it then the ring is the substituent.
Sources of Alkanes and CycloalkanesSources of Alkanes and Cycloalkanes
Natural is mainly methane with ethane and propane.
Petroleum is a liquid mixture containing approximately 150
hydrocarbons. Half of these are alkanes or cycloalkanes.
Distillation of crude oil gives fractions based on boiling point.
Petroleum RefiningPetroleum Refining
The yield of the more useful petroleum fraction used as
automotive fuel is increased by two processes:
Cracking. Cracking is the cleavage of carbon–carbon bonds
in high molecular weight alkanes induced by heat (thermal
cracking) or with catalysts (catalytic cracking).
Reforming. Reforming converts the hydrocarbons in
petroleum to aromatic hydrocarbons and highly branched
alkanes, both of which are better automotive fuels than
unbranched alkanes and cycloalkanes.
Other Natural Sources of AlkanesOther Natural Sources of Alkanes
Solid n-alkanes are waxy and coat the outer surface of many
living things to prevent loss of water. Examples include:Pentacosane (CH3(CH2)23CH3 is found in the waxy outer
layer of many insects.
Hentriacontane is a component of beeswax and the outer
layer of leaves of tobacco, peach trees and others.
Hopanes are found in petroleum
and geologic sediments.
Boiling PointBoiling Point
Boiling points of n-alkanes increase with increasing
molecular weight (number of carbons).
Branched alkanes generally have lower boiling points than
unbranched alkanes with the same number of carbons.
Intermolecular Forces and Boiling PointIntermolecular Forces and Boiling Point
Attractive forces between molecules in the liquid phase
affect the boiling point of the liquid.
These Intermolecular forces are van der Waals forces and
may be divided into three types:
Dipole-dipole (including hydrogen bonding);
Induced dipole-dipole; or
Induced dipole-induced dipole.
Intermolecular Forces and AlkanesIntermolecular Forces and Alkanes
Alkanes have no dipole so the van der Waals forces are the
temporary induced dipole-induced dipole.
This interaction is dynamic and fluctuates.
Intermolecular Forces and AlkanesIntermolecular Forces and Alkanes
Long chain alkanes have more induced dipole-induced
dipole interactions so the boiling point increases with
increasing chain length.
Intermolecular Forces and AlkanesIntermolecular Forces and Alkanes
Branched alkanes have lower surface area than isomeric n-
alkanes and therefore have lower boiling points.
Melting PointMelting Point
Solid n-alkanes are soft low melting solids. The same
intermolecular forces hold the molecules together in the
solid state.
Solubility of Alkanes in WaterSolubility of Alkanes in Water
Alkanes (and all hydrocarbons) are virtually insoluble in
water and are said to be hydrophobic.
The densities of most alkanes are in the range 0.6-0.8
g/mL therefore alkanes float on the surface of water.
Acidity of HydrocarbonsAcidity of Hydrocarbons
Hydrocarbons are very weak acids. Alkynes have the Hydrocarbons are very weak acids. Alkynes have the
lowest pKa.lowest pKa.
Hydrocarbons are very weak acids. Alkynes have the Hydrocarbons are very weak acids. Alkynes have the
lowest pKa.lowest pKa.
Combustion of HydrocarbonsCombustion of Hydrocarbons
Combustion of hydrocarbons is exothermic generating Combustion of hydrocarbons is exothermic generating COCO22 and water. and water.Combustion of hydrocarbons is exothermic generating Combustion of hydrocarbons is exothermic generating COCO22 and water. and water.
Combustion of Relative StabilityCombustion of Relative Stability
All isomers of CAll isomers of C88HH1818 generate 8 molecules of CO generate 8 molecules of CO2 2 and 9 and 9
of Hof H22O yet different amounts of energy. This energy O yet different amounts of energy. This energy
difference must be directly related to the relative energies difference must be directly related to the relative energies
of the isomers.of the isomers.
All isomers of CAll isomers of C88HH1818 generate 8 molecules of CO generate 8 molecules of CO2 2 and 9 and 9
of Hof H22O yet different amounts of energy. This energy O yet different amounts of energy. This energy
difference must be directly related to the relative energies difference must be directly related to the relative energies
of the isomers.of the isomers.
Most stable
isomer
Most stable
isomer
Least stable
isomer
Least stable
isomer
Least energy
released
Oxidation and Reduction in Organic ChemistryOxidation and Reduction in Organic Chemistry
Assuming the oxidation state of H is +1 and O is -2 it is Assuming the oxidation state of H is +1 and O is -2 it is
possible to calculate the oxidation state of C in possible to calculate the oxidation state of C in
compounds containing C, H and O.compounds containing C, H and O.
Assuming the oxidation state of H is +1 and O is -2 it is Assuming the oxidation state of H is +1 and O is -2 it is
possible to calculate the oxidation state of C in possible to calculate the oxidation state of C in
compounds containing C, H and O.compounds containing C, H and O.
Oxidation and Reduction in Organic ChemistryOxidation and Reduction in Organic Chemistry
Oxidation of carbon corresponds to an increase in the Oxidation of carbon corresponds to an increase in the
number of bonds between carbon and oxygen or to a number of bonds between carbon and oxygen or to a
decrease in the number of carbon–hydrogen bonds. decrease in the number of carbon–hydrogen bonds.
Reduction corresponds to an increase in the number of Reduction corresponds to an increase in the number of
carbon–hydrogen bonds or to a decrease in the number carbon–hydrogen bonds or to a decrease in the number
of carbon–oxygen bonds.of carbon–oxygen bonds.
Oxidation of carbon corresponds to an increase in the Oxidation of carbon corresponds to an increase in the
number of bonds between carbon and oxygen or to a number of bonds between carbon and oxygen or to a
decrease in the number of carbon–hydrogen bonds. decrease in the number of carbon–hydrogen bonds.
Reduction corresponds to an increase in the number of Reduction corresponds to an increase in the number of
carbon–hydrogen bonds or to a decrease in the number carbon–hydrogen bonds or to a decrease in the number
of carbon–oxygen bonds.of carbon–oxygen bonds.
Oxidation and Reduction in Organic ChemistryOxidation and Reduction in Organic Chemistry
Any element more electronegative than C has the same Any element more electronegative than C has the same
effect as O on the oxidation state of C. effect as O on the oxidation state of C.
Oxidation state of C is +2 in CHOxidation state of C is +2 in CH33Cl and CHCl and CH33OH.OH.
Any element less electronegative than C has the same Any element less electronegative than C has the same
effect as H on the oxidation state of C.effect as H on the oxidation state of C.
Oxidation state of C is -4 in CHOxidation state of C is -4 in CH44 and CH and CH33Li.Li.
Any element more electronegative than C has the same Any element more electronegative than C has the same
effect as O on the oxidation state of C. effect as O on the oxidation state of C.
Oxidation state of C is +2 in CHOxidation state of C is +2 in CH33Cl and CHCl and CH33OH.OH.
Any element less electronegative than C has the same Any element less electronegative than C has the same
effect as H on the oxidation state of C.effect as H on the oxidation state of C.
Oxidation state of C is -4 in CHOxidation state of C is -4 in CH44 and CH and CH33Li.Li.