Download HYDROCARBONS HYDROCARBONS Types of Hydrocarbons

HYDROCARBONS
Chapter 2
• Compounds composed of only carbon and hydrogen atoms
(C, H). Each carbon has 4 bonds.
Introduction to
Hydrocabons
• They represent a “backbone” when other “heteroatoms”
(O, N, S, .....) are substituted for H. (The heteroatoms give
function to the molecule.)
Carbon Backbone,
Nomenclature, Physical &
Chemical Properties
• Acyclic (without rings); Cyclic (with rings); Saturated:
only carbon-carbon single bonds; Unsaturated: contains
one or more carbon-carbon double and/or triple bond
Types of Hydrocarbons
HYDROCARBONS
• Alkanes contain only single (σ ) bonds and have the
generic molecular formula: [CnH2n+2]
• Alkenes also contain double (σ + π) bonds and have the
generic molecular formula: [CnH2n]
• Alkynes contain triple (σ + 2π) bonds and have the
generic molecular formula: [CnH2n-2]
Each C atom is tetrahedral with sp3 hybridized orbitals. They only have single bonds.
• Aromatics are planar, ring structures with alternating
single and double bonds: eg. C6 H 6
Each C atom is trigonal planar with sp2 hybridized orbitals.
There is no rotation about the C=C bond in alkenes.
Question 2.1
Question 2.2
• What is the hybridization of the starred
carbon in humulene (shown)?
• What is the hybridization of the starred
carbon of geraniol?
•
A)
sp
•
A)
sp
•
B)
sp2
•
B)
sp2
•
C)
sp3
•
C)
sp3
•
D)
1s2 2s2 2p2
•
D)
1s2 2s2 2p2
1
Types of Hydrocarbons
Propane
It is easy to rotate about the C-C bond in alkanes.
Each C atom is linear with sp hybridized orbitals.
Each C--C bond is the same length; shorter than a C-C bond: longer than a C=C bond.
The concept of resonance is used to explain this phenomena.
Naming Alkanes
C1 - C10 : the number of C atoms present in the chain.
Each member C 3 - C 10 differs by one CH2 unit. This is called a homologous series.
Methane to butane are gases at normal pressures.
Pentane to decane are liquids at normal pressures.
Nomenclature of Alkyl Substituents
Examples of Alkyl Substituents
2
Constitutional or structural isomers have the same
molecular formula, but their atoms are linked
differently. Naming has to account for them.
Question 2.3
• How many hydrogens are in a molecule of
isobutane?
•
A)
6
•
B)
8
•
C)
10
•
D)
12
A compound can have more than one name, but a
name must unambiguously specify only one compound
C7 H16 can be any one of the following:
Question 2.4
• How many isomeric hexanes exist?
•
A)
2
•
B)
3
•
C)
5
•
D)
6
Question 2.5
• The carbon skeleton shown at the bottom right
accounts for 9 carbon atoms. How many other
isomers of C10H 22 that have 7 carbons in their
longest continuous chain can be generated
by adding a single carbon to various positions
on this skeleton?
•
A)
2
•
B)
3
•
C)
4
•
D)
5
3
Alkanes
(Different types of sp3 carbon atoms)
Different Kinds of sp3 Carbons and
Hydrogens
• Primary, 1o, a carbon atom with 3 hydrogen atoms: [RCH 3 ]
• Secondary, 2o, a carbon atom with 2 hydrogen atoms: [RCH 2 -R]
• Tertiary, 3o, a carbon atom with 1 hydrogen atom: [R-CHR]
R
• Quaternary, 4o, a carbon atom with 0 hydrogen atoms:
CR4
Question 2.6
• In 3-ethyl-2-methylpentane, carbon #3
(marked by a star) is classified as:
•
•
•
•
A)
B)
C)
D)
primary (1°)
secondary (2°)
tertiary (3°)
quaternary (4°)
Nomenclature of Alkanes
Question 2.7
• How many primary carbons are in the
molecule shown at the bottom right?
•
A)
2
•
B)
3
•
C)
4
•
D)
5
3. Number the substituents to yield the lowest possible number
in the number of the compound
1. Determine the number of carbons in the parent hydrocarbon
(substituents are listed in alphabetical order)
2. Number the chain so that the substituent gets the lowest
possible number
4. Assign the lowest possible numbers to all of the substituents
4
5. When both directions lead to the same lowest number for one
of the substituents, the direction is chosen that gives the lowest
possible number to one of the remaining substituents
6. If the same number is obtained in both directions, the first
group receives the lowest number
7. In the case of two hydrocarbon chains with the same number of
carbons, choose the one with the most substituents
8. Certain common nomenclatures are used in the IUPAC system
Question 2.7
• The correct structure of 3-ethyl-2methylpentane is:
•
A)
B)
•
C)
D)
Cycloalkanes
CnH2n
Cycloalkane Nomenclature
•
Cycloalkanes are alkanes that contain a ring
of three or more carbons.
•
Count the number of carbons in the ring,
and add the prefix cyclo to the IUPAC name of
the unbranched alkane that has that number of
carbons.
Cyclopentane
Cyclohexane
5
Cycloalkanes
•
Name any alkyl groups on the ring in the
usual way. A number is not needed for a single
substituent.
CH2CH3
Cycloalkanes
•
Name any alkyl groups on the ring in the
usual way. A number is not needed for a single
substituent.
•
List substituents in alphabetical order and
count in the direction that gives the lowest
numerical locant at the first point of difference.
H3 C CH3
Ethylcyclopentane
CH2CH3
3-Ethyl-1,1-dimethylcyclohexane
Question 2.8
For more than two substituents,
• Which one contains the greatest number
of tertiary carbons?
•
A)
2,2-dimethylpropane
•
B)
3-ethylpentane
•
C)
sec-butylcyclohexane
•
D)
2,2,5-trimethylhexane
Naphtha
(bp 95-150 °C)
2.17
Physical Properties of Alkanes
and Cycloalkanes
Kerosene
(bp: 150-230 °C)
C5 -C12
Light gasoline
(bp: 25-95 °C)
C12-C15
Crude oil
Gas oil
(bp: 230-340 °C)
Refinery gas
C1 -C4
C15-C25
Residue
6
Fig. 2.15
Question 2.9
• Arrange octane, 2,2,3,3-tetramethylbutane
and 2-methylheptane in order of increasing
boiling point.
•
A)
2,2,3,3-tetramethylbutane < octane
< 2methylheptane
•
B)
octane < 2-methylheptane <
2,2,3,3- tetramethylbutane
•
C)
2,2,3,3-tetramethylbutane < 2methylheptane < octane
•
D)
2-methylheptane < 2,2,3,3tetramethylbutane < octane
van der Waals Forces
Weak Intermolecular Attractive Forces
Example of Intramolecular Forces:
Protein Folding
10-40kJ/mol
Ion-dipole
(Dissolving)
40-600kJ/mol
150-1000kJ/mol
700-4,000kJ/mol
0.05-40kJ/mol
Intermolecular Forces
Ion-Dipole Forces (40-600 kJ/mol)
• Interaction between an ion and a dipole (e.g. NaOH and
water = 44 kJ/mol)
• Strongest of all intermolecular forces.
The boiling point of a compound increases with
the increase in van der Waals force…and a
Gecko uses them to walk!
Ion-Dipole & Dipole-Dipole Interactions:
like dissolves like
• Polar compounds dissolve in polar solvents
& non-polar in non-polar
7
Intermolecular Forces
Dipole-Dipole Forces
(permanent dipoles)
Intermolecular Forces
Dipole-Dipole Forces
5-25 kJ/mol
Boiling Points &
Hydrogen Bonding
Hydrogen Bonding
• Hydrogen bonds, a
unique dipole-dipole (1040 kJ/mol).
Intermolecular Forces
Hydrogen Bonding
8
DNA: Size, Shape & Self Assembly
Intermolecular Forces
http://www.umass.edu/microbio/chime/beta/pe_alpha/atlas/atlas.htm
Views & Algorithms
10.85 Å
10.85 Å
London or Dispersion Forces
• An instantaneous dipole can induce another dipole in an
adjacent molecule (or atom).
• The forces between instantaneous dipoles are called
London or Dispersion forces ( 0.05-40 kJ/mol).
Gecko: toe, setae, spatulae
Boiling Points of Alkanes
6000x Magnification
•
Full et. al., Nature (2000)
5,000 setae / mm2
600x frictional force; 10-7
Newtons per seta
Geim, Nature Materials
(2003)
Glue-free Adhesive
100 x 10 6 hairs/cm2
governed by strength of intermolecular
attractive forces
•
alkanes are nonpolar, so dipole-dipole and
dipole-induced dipole forces are absent
•
only forces of intermolecular attraction are
induced dipole-induced dipole forces
http://micro.magnet.fsu.edu/primer/java/electronmicroscopy/magnify1/index.html
Induced dipole-Induced dipole
Attractive Forces
+–
•
•
Induced dipole-Induced dipole
Attractive Forces
+–
+–
two nonpolar molecules
center of positive charge and center of
negative charge coincide in each
•
+–
movement of electrons creates an
instantaneous dipole in one molecule (left)
9
Induced dipole-Induced dipole
Attractive Forces
+
•
–
Induced dipole-Induced dipole
Attractive Forces
+
+–
temporary dipole in one molecule (left)
induces a complementary dipole in other
molecule (right)
•
•
–
+
Boiling Points
•Increase with increasing number of carbons
•
more atoms, more electrons, more
opportunities for induced dipole-induced
dipole forces
•Decrease with chain branching
•
Induced dipole-Induced dipole
Attractive Forces
–
–
the result is a small attractive force
between the two molecules
branched molecules are more compact with
smaller surface area—fewer points of contact
with other molecules
–
temporary dipole in one molecule (left)
induces a complementary dipole in other
molecule (right)
Induced dipole-Induced dipole
Attractive Forces
+
+
–
•
+
–
+
the result is a small attractive force
between the two molecules
Intermolecular Forces
London Dispersion
Forces
Which has the higher
attractive force?
10
Boiling Points
Question 2.10
• Which alkane has the highest boiling
point?
•
A)
hexane
•
B)
2,2-dimethylbutane
•
C)
2-methylpentane
•
D)
2,3-dimethylbutane
•Increase with increasing number of carbons
• more atoms, more electrons, more
opportunities for induced dipole-induced
dipole forces
Heptane
bp 98°C
Octane
bp 125°C
Nonane
bp 150°C
Boiling Points
•Decrease with chain branching
• branched molecules are more compact with
smaller surface area—fewer points of contact
with other molecules
Octane: bp 125°C
2-Methylheptane: bp 118°C
2.18
Chemical Properties:
Combustion of Alkanes
•All alkanes burn in air to give
carbon dioxide and water.
2,2,3,3-Tetramethylbutane: bp 107°C
Heats of Combustion
Heptane
Heats of Combustion
•Increase with increasing number of carbons
4817 kJ/mol
654 kJ/mol
Octane
5471 kJ/mol
• more moles of O2 consumed, more moles
of CO2 and H2O formed
654 kJ/mol
Nonane
6125 kJ/mol
What pattern is noticed in this case?
11
Figure 2.17
Heats of Combustion
5471 kJ/mol
5471 kJ/mol
5 kJ/mol
5466 kJ/mol
+
8 kJ/mol
5466 kJ/mol
25
O2
2
5458 kJ/mol
5458 kJ/mol
+
6 kJ/mol
5452 kJ/mol
25
O2
2
25
+
O2
2
+
25
O2
2
5452 kJ/mol
8CO2 + 9H2O
What pattern is noticed in this case?
Heat of Combustion
Patterns
Important Point
•Increase with increasing number of carbons
•Isomers can differ in respect to their stability.
• more moles of O2 consumed, more moles
of CO2 and H2O formed
•Equivalent statement:
–Isomers differ in respect to their potential energy.
•Decrease with chain branching
• branched molecules are more stable
(have less potential energy) than their
unbranched isomers
Differences in potential energy can
be measured by comparing heats of
combustion. (Worksheet problems)
O
2.19
Oxidation-Reduction in Organic Chemistry
Oxidation of a carbon atom corresponds
to an increase in the number of bonds to
the carbon atom and/or a decrease in the
number of hydrogens bonded to the
carbon atom. See examples on the board.
O
increasing oxidation
state of carbon
O
H
H
H
C
H
-4
H
H
C
H
C
H
C
HO
C
OH
OH
H
OH
H
-2
0
+2
+4
12
HC
increasing oxidation
state of carbon
H
H
H
H
C
C
H
H
H
C
H
C
H
-3
• How to calculate the oxidation state
of each carbon in a molecule that contains
carbons in different oxidation states?
CH
CH3CH2 OH
H
-2
-1
Table 2.5 How to Calculate
Oxidation Numbers
• 1. Write the
Lewis structure
and include
unshared electron
pairs.
H
H
H
C
C
H
H
••
O
••
H
Table 2.5 How to Calculate
Oxidation Numbers
• 3. For a bond
between two
atoms of the
same element,
assign the
electrons in the
bond equally.
C 2 H 6O
H
H
••
••
••
••
H •• C
H
C
H
••
•• O ••
••
H
Table 2.5 How to Calculate
Oxidation Numbers
• 2. Assign the
electrons in a
covalent bond
between two
atoms to the more
electronegative
partner.
H
H
••
••
••
••
H •• C
H
C
••
•• O ••
••
H
H
Table 2.5 How to Calculate
Oxidation Numbers
• 3. For a bond
between two
atoms of the
same element,
assign the
electrons in the
bond equally.
H
H
••
••
••
••
H
H
H •• C• • C
••
•• O ••
••
H
13
Table 2.5 How to Calculate
Oxidation Numbers
• 4. Count the number
of electrons
assigned to each
atom and subtract
that number from
the number of
valence electrons in
the neutral atom;
the result is the
oxidation number.
H
H
••
••
••
••
H
H
H •• C• • C
••
•• O ••
••
Each H
C of CH3
C of CH2O
O
=
=
=
=
H
+1
-3
-1
-2
Generalization
Oxidation of carbon occurs when a bond between
carbon and an atom which is less electronegative
than carbon is replaced by a bond to an atom that
is more electronegative than carbon. The reverse
process is reduction.
oxidation
C X
C
Y
reduction
X less electronegative than carbon
Y more electronegative than carbon
Examples
Question 2.11
Oxidation
CH4 + Cl2
CH3Cl + HCl
Reduction
CH3Cl + 2Li
CH3Li + LiCl
• To carry out the reaction shown below we
need:
•
•
CH3OH → H2C=O
•
•
A)
an oxidizing agent
•
B)
a reducing agent
14