CHEM 112
Table of Contents
CHAPTER 1 .............................................................................................................................................. 1
INTRODUCTION TO ORGANIC CHEMISTRY ............................................................................................. 1
1.1 The Origins of Organic Chemistry ................................................................................................. 1
1.2 Bond Formation: The Octet Rule .................................................................................................. 1
Covalent Bonds................................................................................................................................ 1
Lewis Diagrams for Multiple covalent Bonds .................................................................................. 2
1.3 Bond Polarity and Electronegativity ............................................................................................. 5
1.4 Polar Covalent Bonds: Dipole Moments ....................................................................................... 6
1.5 VSEPR Theory ................................................................................................................................ 6
1.6 Resonance structures.................................................................................................................. 10
1.7 Acid and Base Strength ............................................................................................................... 14
1.8 Acids and Bases: The Lewis Definition ........................................................................................ 16
Summary ....................................................................................................................................... 19
1.9 How to represent Lewis Acids and Bases in a reaction............................................................... 20
CHAPTER 2 ............................................................................................................................................ 21
STRUCTURE AND STEREOCHEMISTRY OF ALKANES .............................................................................. 21
2.1 Sigma Bonding............................................................................................................................. 21
2.2 The alkanes ................................................................................................................................. 21
2.3 The Nomenclature of Alkanes..................................................................................................... 27
2.4 Physical properties of alkanes..................................................................................................... 30
2.5 Reactions of Alkanes ................................................................................................................... 32
2.6 Structure and conformations of alkanes .................................................................................... 32
2.7 Cis-Trans Isomerism .................................................................................................................... 37
CHAPTER 3 ............................................................................................................................................ 44
THE STUDY OF CHEMICAL REACTIONS .................................................................................................. 44
3.1 Chlorination of Methane............................................................................................................. 44
3.2 Chlorination of Propane .............................................................................................................. 46
3.3 Bromination of Propane.............................................................................................................. 51
CHAPTER 4 ............................................................................................................................................ 56
ALKYL HALIDES: NUCLEOPHILIC SUBSTITUTION AND ELIMINATION .................................................... 56
4.1 Stereochemistry, Substitution and Elimination Reactions ......................................................... 56
i
4.2 IUPAC Nomenclature of alkyl halides/Haloalkanes .................................................................... 57
4.3 Preparation of alkyl halides ........................................................................................................ 59
4.4 Reactions of Alkyl Halides: Substitution and Elimination ........................................................... 61
4.5 Second-Order Nucleophilic Substitution: The SN2 reaction Mechanism .................................... 62
4.6 SN1 AND E1 REACTIONS .............................................................................................................. 67
4.7 Rearrangements (hydride and methyl shifts) ............................................................................. 71
CHAPTER 5 ............................................................................................................................................ 82
STRUCTURE AND SYNTHESIS OF ALKENES ............................................................................................ 82
5.1 Isomerism in the alkenes ............................................................................................................ 82
5.2 Physical properties of the alkenes .............................................................................................. 99
5.3 IUPAC Nomenclature of alkenes ............................................................................................... 101
5.4 Structure and Preparation of Alkenes....................................................................................... 104
CHAPTER 6 .......................................................................................................................................... 111
REACTIONS OF ALKENES: ADDITION REACTIONS................................................................................ 111
6.1 Reactions of Alkenes: Addition Reactions ................................................................................ 111
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CHAPTER 1
INTRODUCTION TO ORGANIC CHEMISTRY
1.1 The Origins of Organic Chemistry
Organic chemistry is the chemistry of carbon compounds. Carbon is quite special in this
study;
C
- forms strong C- C bonds and to other elements
- can form C-C chains in a process called catenation.
This diversity of carbon provides the basis for life on the earth. The term “organic” means
“derived from living organism”. Although organic compounds are largely naturally available,
some can be synthesized from inorganic compounds. Friedrich Wohler in 1828 converted
ammonium cyanate, made from ammonia and other inorganic chemicals, to urea simply by
heating it in the absence of oxygen.
O
NH4+-OCN
heat
H2N
inorganic
C
NH2
organic
The synthesis of this first ever organic compound from inorganic sources weakens the above
definition. We may define organic chemistry as a study of the chemistry hydrocarbon
compounds and their derivatives.
1.2 Bond Formation: The Octet Rule
G. N. Lewis is a scientist who in 1915 invented an illustration on how atoms bond together to
form molecules. He stated in his principle that a filled shell (energy level) of electrons is
stable, and atoms transfer or share electrons in such a way as to attain a filled shell of
electrons (noble configuration). This is the octet rule.
Ionic bonding
Ionic bond: bond in which one or more electrons from one atom are removed and attached to
another atom, resulting in positive and negative ions which attract each other.
Li
F
Li+
He conf
+
F
Ne conf
Li + F
ionic bond
Covalent Bonds
Covalent chemical bonds involve the sharing of a pair of valence electrons by two atoms, in
contrast to the transfer of electrons in ionic bonds. Such bonds lead to stable molecules if
they share electrons in such a way as to create a noble gas configuration for each atom.
1
Hydrogen gas forms the simplest covalent bond in the diatomic hydrogen molecule. The
halogens such as chlorine also exist as diatomic gases by forming covalent bonds. The
nitrogen and oxygen which makes up the bulk of the atmosphere also exhibits covalent
bonding in forming diatomic molecules.
Covalent bonding can be visualized with the aid
of Lewis diagrams.
Lewis Diagrams for Multiple covalent Bonds
Lewis symbols and Lewis diagrams can be used to describe multiple bonds. For multiple
single bonds, the procedure is similar to that for a single bond.
A single bond can be represented by the two
dots of the bonding pair, or by a single line
which represents that pair. The single line
representation for a bond is commonly used in
drawing Lewis structures for molecules.
The Lewis structures are useful for
visualization, but do not reveal the bent
structure for water (105°), the pyramidal shape
for ammonia, or the tetrahedral geometry of the
methane molecule. The Lewis diagrams can
also help visualize double and triple bonds, see
below.
Lewis Structures
Symbolizes the bonding in a valent molecule, each valence electron is symbolized by a dot as
we saw above or a bonding pair of electrons is symbolized by a dash.
Example
H
H
CH4
(methane)
-
-
H
H
H
C2H6 (ethane)
C
H
H
C
C
H
H
H
2
The ethane structure shows the most important characteristics of carbon – its ability to form
strong C-C bonds. We will encounter many structures with nonbonding electrons too in the
valence shell. These are called lone pairs. These lone pairs of nonbonding electrons help to
determine the reactivity of their parent compounds. Can you identify the lone pairs of
electrons in the following compounds?
H
H
C
N
H
H
H
H
methyleamine
H
H
H
C
C
O
H
H
H
H
C
Cl
H
chloromethane
ethanol
Sometimes we may assume to draw the lone pairs of electrons as long as we all understand
that they still exist.
Exercise
Draw Lewis structures for,
a) Ammonia, NH3
b) hydronium ion, H3O+
c) propane C3H8
d) fluoroethane, CH3CH2F
e) boron trifluoride, BF3
Multiple Bonding
Sharing of one pair of electrons between two atoms is called a single bond, we also have
double and triple bonds between atoms.
Example
H
H
C
C2H4
C
H
ethylene; double bond between C to C atoms
H
H
C
H2CO
O
formaldehyde; double bond between C to O atoms
H
H
C
H2CNH
C2H2
N
H
H
H
C
C
formaldimine; double bond between C to N atoms
H
acetylene; triple bond between C to C atoms
IMPORTANT: Please, note that C normally forms four bonds in neutral organic compounds
– tetravalent, N three (trivalent), O two (divalent) and H one (monovalent). It is wrong to
draw any of these atoms with less or more bonds attached to other atoms in neutral
molecules.
3
Bond Polarity
A bond with the electrons shared equally between the two identical atoms is called a nonpolar bond.
Example
The H-H bond and the C-C bond in ethane.
These are non-polar covalent bonds
 Bonding electrons attracted more strongly by one atom than by the other
 Electron distribution between atoms in not symmetrical
Covalent bonds can have ionic character. In two different elements, electrons in a bond are
unequally shared resulting to a polar bond.
Example
H
H
C
Cl
H
These are polar covalent bonds. Bonding electrons attracted more strongly by one atom than
by the other. Electron distribution between atoms in not symmetrical.
Another example is water.
Water forms a polar covalent molecule. The figure above shows that oxygen has 6 electrons
in the outer shell. Hydrogen has one electron in its outer energy shell. Since 8 electrons are
needed for an octet, they share the electrons.
However, oxygen gets an unequal share of the two electrons from both hydrogen atoms.
Again, the electrons are still shared (not transferred as in ionic bonding), the sharing is
unequal. Since the electrons are more close to the oxygen then it acquires a "partial" negative
charge. At the same time, since hydrogen loses the electron most - but not all of the time, it
acquires a "partial" positive charge. The partial charge is denoted with a small Greek symbol
for delta.
4
1.3 Bond Polarity and Electronegativity

Electronegativity (EN) refers to the intrinsic ability of an atom to attract the shared
electrons in a covalent bond
 Differences in EN produce bond polarity
 Arbitrary scale. As shown in Figure below electronegativities are based on an
arbitrary scale
 F is most electronegative (EN = 4.0), Cs is least (EN = 0.7)
 Metals on left side of periodic table attract electrons weakly, thus they have a lower
EN
 Halogens and other reactive nonmetals on right side of periodic table attract electrons
strongly, higher electronegativities
EN of C = 2.5
The Periodic Table and Electronegativity
Bond Polarity and Inductive Effect
 Non-polar Covalent Bonds exist between atoms with similar EN
 Polar Covalent Bonds exist between difference in EN of atoms < 2
 Ionic Bonds exist between difference in EN > 2
 C–H bonds, relatively nonpolar C-O, C-X bonds (more electronegative
elements) are polar
 Bonding electrons toward electronegative atom
 C acquires partial positive charge, +
 Electronegative atom acquires partial negative charge,  Inductive effect: shifting of electrons in a bond in response to EN of nearby atoms
5
1.4 Polar Covalent Bonds: Dipole Moments



Molecules as a whole are often polar from vector summation of individual bond
polarities and lone-pair contributions
Strongly polar substances soluble in polar solvents like water; nonpolar substances are
insoluble in water.
Dipole moment - Net molecular polarity, due to difference in summed charges
o symbol , unit D (debye)
1.5 VSEPR Theory




Valence Shell Electron Pair Repulsion:
The VSEPR theory assumes that each atom in a molecule will achieve a geometry
that minimizes the repulsion between electrons in the valence shell of that atom
Prediction 1: Electron pairs repel one another – attain maximum distance
Prediction 2: Non-bonding electrons repel more than bonding ones
Dipole Moments in Water and Ammonia
 Large dipole moments
o EN of O and N > H
o Both O and N have lone-pair electrons oriented away from all nuclei
6
Absence of Dipole Moments
 In symmetrical molecules, the dipole moments of each bond has one in the opposite
direction
 The effects of the local dipoles cancel each other
7
PROBLEM
When you ingest aspirin, it passes through your stomach, which has an acidic pH, before
traveling through the basic environment of your intestine. Provide the correct structure of
aspirin a) as it exists in the stomach and b) as it exists in the intestine.
Formal Charges
 Sometimes it is necessary to have structures with formal charges on individual atoms
 We compare the bonding of the atom in the molecule to the valence electron structure
8



If the atom has one more electron in the molecule, it is shown with a “-” charge
If the atom has one less electron, it is shown with a “+” charge
Neutral molecules with both a “+” and a “-” are dipolar
Exercise: Use Electronegativities of elements to predict the direction of the dipole moments
of the following bonds.
a) C-Cl b) C-O,
c) C-N
d) B-Cl
Ionic structures
Some organic compounds contain ionic bonds.
Example
Methylammonium chloride (CH3NH3Cl)
9
H
H
H
C
C
N
H
H
= C
N
H
H
H
H
N
H
H
H
resonance hybrid
resonance structures
H
H
-
Na+
O
C
or
H
Na
O
C
H
H
H
H
H
H
C
N
H
H
Cl
H
Some molecules can be drawn either covalently or ionically.
Example NaOCH3
H
H
-
Na+
O
C
H
or
Na
O
C
H
H
H
1.6 Resonance structures
Some compounds are not adequately represented by a single Lewis structures. Because two
or more valent bond structures are possible.
Example
H
H
H
C
C
N
H
H
= C
N
H
H
H
H
N
H
H
H
resonance hybrid
resonance structures
The spreading of the positive charge makes the molecule more stable – resonance stabilized
cation, there is a partial +ve charge in C and N atoms which we normally symbolize as δ+.
Resonance stabilization plays a crucial role in organic chemistry, especially in the chemistry
of compounds having double bonds.
Example
H
H
H
C
O
+
C
H2O
H
C
H
H
O
H
O
O
H
C
O
H
acetic acid
10
C
H
+
C
O
H3O+
We use double headed arrow between resonance structures and often enclose the structures in
block brackets. Some uncharged molecules actually have resonance-stabilized, equal positive
and negative formal charges.
Example
Nitromethane, (CH3NO2) (again)
H
H
H
H
N+
C
H
O-
H
C
H
Od-
O-
O
H
N+
O
C
H
N+
O d-
Other resonance structures
C
N+
C
major
+
N
minor
In drawing resonance structures, we try to draw structures that are as low in energy as
possible. Only electrons can be delocalized, unlike electrons nuclei can not be delocalized.
They must remain in the same places, with the same bond distances and angles.
Example
Draw resonance structures for,
H
O
C
H
C
H
H2C=CH-NO2
Resonance Hybrids
 A structure with resonance forms does not alternate between the forms
 Instead, it is a hybrid of the two resonance forms, so the structure is called a
resonance hybrid
 For example, benzene (C6H6) has two resonance forms with alternating double and
single bonds
o In the resonance hybrid, the actual structure, all its C-C bonds equivalent,
midway between double and single
11
Rules for Resonance Forms
 Individual resonance forms are imaginary - the real structure is a hybrid (only by
knowing the contributors can you visualize the actual structure)
 Resonance forms differ only in the placement of their  or nonbonding electrons
 Different resonance forms of a substance don’t have to be equivalent
 Resonance forms must be valid Lewis structures: the octet rule applies
 The resonance hybrid is more stable than any individual resonance form would be
Drawing Resonance Forms
 Any three-atom grouping with a multiple bond has two resonance forms
Different Atoms in Resonance Forms
 Sometimes resonance forms involve different atom types as well as locations
 The resulting resonance hybrid has properties associated with both types of
contributors
 The types may contribute unequally
 The “enolate” derived from acetone is a good illustration, with delocalization between
carbon and oxygen
12
2,4-Pentanedione
 The anion derived from 2,4-pentanedione
o Lone pair of electrons and a formal negative charge on the central carbon
atom, next to a C=O bond on the left and on the right
o Three resonance structures result
Acids and Bases: The Brønsted–Lowry Definition
 The terms “acid” and “base” can have different meanings in different contexts
 For that reason, we specify the usage with more complete terminology
 The idea that acids are solutions containing a lot of “H+” and bases are solutions
containing a lot of “OH-” is not very useful in organic chemistry
 Instead, Brønsted–Lowry theory defines acids and bases by their role in reactions that
transfer protons (H+) between donors and acceptors
Brønsted Acids and Bases
 “Brønsted-Lowry” is usually shortened to “Brønsted”
 A Brønsted acid is a substance that donates a hydrogen ion (H+)
 A Brønsted base is a substance that accepts the H+ - “proton” is a synonym for H+ loss of an electron from H leaving the bare nucleus—a proton
The Reaction of HCl with H2O
 When HCl gas dissolves in water, a Brønsted acid–base reaction occurs
 HCl donates a proton to water molecule, yielding hydronium ion (H3O+) and Cl
 The reverse is also a Brønsted acid–base reaction of the conjugate acid and conjugate
base
Acids are shown in red, bases in blue. Curved arrows go from bases to acids
Ka – the Acidity Constant
 The concentration of water as a solvent does not change significantly when it is
protonated
 The molecular weight of H2O is 18 and one liter weighs 1000 grams, so the
concentration is ~ 55.6 M at 25°
13


The acidity constant, Ka for HA Keq times 55.6 M (leaving [water] out of the
expression)
Ka ranges from 1015 for the strongest acids to very small values (10-60) for the weakest
1.7 Acid and Base Strength

The “ability” of a Brønsted acid to donate a proton to is sometimes referred to as the
strength of the acid (imagine that it is throwing the proton – stronger acids throw it
harder)
 The strength of the acid is measured with respect to the Brønsted base that receives
the proton
 Water is used as a common base for the purpose of creating a scale of Brønsted acid
strength
pKa – the Acid Strength Scale
 pKa = -log Ka
 The free energy in an equilibrium is related to –log of Keq (DG = -RT log Keq)
 A smaller value of pKa indicates a stronger acid and is proportional to the energy
difference between products and reactants
 The pKa of water is 15.74
14
Organic Acids
Those that lose a proton from O–H, such as methanol and acetic acid
Those that lose a proton from C–H, usually from a carbon atom next to a C=O double bond
(O=C–C–H
Organic Bases
 Have an atom with a lone pair of electrons that can bond to H+
 Nitrogen-containing compounds derived from ammonia are the most common organic
bases
 Oxygen-containing compounds can react as bases when with a strong acid or as acids
with strong bases
15
1.8 Acids and Bases: The Lewis Definition


Lewis acids are electron pair acceptors and Lewis bases are electron pair donors
The Lewis definition leads to a general description of many reaction patterns but there
is no scale of strengths as in the Brønsted definition of pKa
Lewis Acids and the Curved Arrow Formalism
 Group 3A elements, such as BF3 and AlCl3, are Lewis acids because they have
unfilled valence orbitals and can accept electron pairs from Lewis bases
 Transition-metal compounds, such as TiCl4, FeCl3, ZnCl2, and SnCl4, are Lewis acids
 Organic compounds that undergo addition reactions with Lewis bases (discussed
later) are called electrophiles and therefore Lewis Acids
 The combination of a Lewis acid and a Lewis base can shown with a curved arrow
from base to acid
Illustration of Curved Arrows in Following Lewis Acid-Base Reactions
16
Lewis Bases
 Lewis bases can accept protons as well as Lewis acids, therefore the definition
encompasses that for Brønsted bases
 Most oxygen- and nitrogen-containing organic compounds are Lewis bases because
they have lone pairs of electrons
 Some compounds can act as both acids and bases, depending on the reaction
Structural Formulas
We as organic chemists, use several kinds of formulas. One of them is condensed structural
formulas – they don’t show all the individual bonds.
Example
CH3CH3
CH3(CH2)4CH3
-CHO, -COOH, CH3CN, CH3COOCH3, CH3COOH or CH3CO2H
Drawing Chemical Structures
Practice drawing the above compounds in expanded forms.
 Chemists use shorthand ways for writing formulas: Kekule, sum, condensed, skeletal
(bond-line), wedge
 Condensed structures: C-H and C-C and single bonds aren't shown but understood
o If C has 3 H’s bonded to it, write CH3
o If C has 2 H’s bonded to it, write CH2; and so on. The compound called 2methylbutane, for example, is written as follows:
 Horizontal bonds between carbons aren't shown in condensed structures—the CH3,
CH2, and CH units are simply but vertical bonds are added for clarity
17
Re-visiting Skeletal (Bond-Line) Structures
 Minimum amount of information but unambiguous
 C’s not shown, assumed to be at each intersection of two lines (bonds) and at end of
each line
 H’s bonded to C’s aren't shown – whatever number is needed will be there
 All atoms other than C and H are shown
Line-angle formulas
This type of formula represents a quick and convenient way of representing compounds. In
each compound, there is a methyl (-CH3) group at the beginning and end of the structure, and
a C atom at each point where two of three lines meet. Remember hydrogen atoms are not
shown on any of the carbon atoms therefore, the missing bonds must be linking to hydrogen.
18
Butane
iso-butane
3-hexane
2-hexane
2-cyclohexanone
O
Give a Lewis structure for;
HN
N
H
OH
O
O
O
O
Wedge Structures – the wedge like bonds imply a three dimensional orientation in space
wither towards or away from you.
Summary



Organic molecules often have polar covalent bonds as a result of unsymmetrical
electron sharing caused by differences in the electronegativity of atoms
The polarity of a molecule is measured by its dipole moment, .
(+) and () indicate formal charges on atoms in molecules to keep track of valence
electrons around an atom
19

Some substances must be shown as a resonance hybrid of two or more resonance
forms that differ by the location of electrons.
A Brønsted(–Lowry) acid donatea a proton
A Brønsted(–Lowry) base accepts a proton
The strength Brønsted acid is related to the -1 times the logarithm of the acidity
constant, pKa. Weaker acids have higher pKa’s
A Lewis acid has an empty orbital that can accept an electron pair
A Lewis base can donate an unshared electron pair
In condensed structures C-C and C-H are implied
Skeletal structures show bonds and not C or H (C is shown as a junction of two
lines) – other atoms are shown
Molecular models are useful for representing structures for study








1.9 How to represent Lewis Acids and Bases in a reaction
Remember the Bronsted-Lowry definition of acids and bases depends on the transfer of a
proton from the acid to the base. The base uses a pair of nonbonding electrons to form a bond
to the proton. G.N. Lewis reasoned that this kind of reaction does not need a proton: A base
could use its lone pair of electrons to bond to some other electron-deficient atom. This way
we look at an acid-base reaction from the view point of the bonds that are being formed and
broken rather than a proton that is transferred.
-
B:H
H:A
B:
-
+
:A
Here a base is a species with nonbonding electrons that can be donated to form new bonds.
Lewis acids are species that can accept these electron pairs to form new bonds. Since a Lewis
acid accepts a pair of electrons it is called an electrophile (electron loving), a Lewis base is
called a nucleophile (nucleus loving). Try to identify a nucleophile and electrophile in the
reactions below.
H+
B:-
H
H
F
N:
B
H
F
B
H
bond formed
H
F
H
N+
H
F
-
B
F
F
20
CHAPTER 2
STRUCTURE AND STEREOCHEMISTRY OF ALKANES
2.1 Sigma Bonding
Two atoms with s and p orbitals can overlap to form a sigma bond. Remember the sp, sp2 and
sp3 hybridization story.
Example
H2: s-s overlap
Cl2: p-p overlap
HCl: s-p overlap
A sigma bond is the first bond that forms between two atoms. Sigma bonds can also form as
a result of the hybridization of atomic orbitals.
Example
H
H
C
H
H
and
H
H
H
C
C
H
H
H
We shall discuss more about hybridization elsewhere later.
Structure of alkanes
Organic chemistry is studied using families of compounds to organize the material. The
properties and reactions of the compounds in a family are similar, just as their structures are
similar. Organic molecules are classified according to their reactive parts, called functional
groups.
Classification of hydrocarbons
We have four classes of hydrocarbons;
 Alkanes
 Alkenes
 Alkynes
 Aromatic hydrocarbons
2.2 The alkanes
There are four carbon or hydrogen atoms bonded to each carbon atom and all have C-C single
bonds. The general formula for the n-alkanes is a chain of –CH2- groups (methylene groups)
terminated at each end by a hydrogen atom. The general formula is;
21
CnH2n+2
Note that there is always an even number of hydrogen atoms in a hydrocarbon. Alkane
homologs refer to each member in an alkane series that differs from the next member by a CH2- group.
Lower membered alkanes have common or trivial names that are easy to memorize, higher
alkanes starting with C4H10 have structural isomers and we need a way of systematically
naming them.
n-utane and iso-butane are the two isomers of butane. Isomers are different compounds with
the same molecular formula. The simplest alkane is methane: CH4. The Lewis structure of
methane can be generated by combining the four electrons in the valence shell of a neutral
carbon atom with four hydrogen atoms to form a compound in which the carbon atom shares
a total of eight valence electrons with the four hydrogen atoms.
Methane is an example of a general rule that carbon is tetravalent; it forms a total of four
bonds in almost all of its compounds. To minimize the repulsion between pairs of electrons in
the four C H bonds, the geometry around the carbon atom is tetrahedral, as shown in the
figure below.
H
C
H
H
H
Practice Problem:
Use the fact that carbon is usually tetravalent to predict the formula of ethane, the alkane
that contains two carbon atoms.
The alkane that contains three carbon atoms is known as propane, which has the formula
C3H8 and the following skeleton structure.
22
The four-carbon alkane is butane, with the formula C4H10.
The names, formulas, and physical properties for a variety of alkanes with the generic
formula CnH2n+2 are given in the table below. The boiling points of the alkanes gradually
increase with the molecular weight of these compounds. At room temperature, the lighter
alkanes are gases; the midweight alkanes are liquids; and the heavier alkanes are solids, or
tars.
The Saturated Hydrocarbons, or Alkanes
Melting
Point (oC)
Boiling
Point (oC)
State
at 25oC
methane CH4
-182.5
-164
gas
ethane
C2H6
-183.3
-88.6
gas
propane C3H8
-189.7
-42.1
gas
butane C4H10
-138.4
-0.5
gas
pentane C5H12
-129.7
36.1
liquid
hexane C6H14
-95
68.9
liquid
heptane C7H16
-90.6
98.4
liquid
octane
-56.8
124.7
liquid
nonane C9H20
-51
150.8
liquid
decane C10H22
-29.7
174.1
liquid
Name
Molecular
Formula
C8H18
The alkanes in the table above are all straight-chain hydrocarbons, in which the carbon
atoms form a chain that runs from one end of the molecule to the other. The generic formula
for these compounds can be understood by assuming that they contain chains of -CH2- groups
with an additional hydrogen atom capping either end of the chain. Thus, for every n carbon
atoms there must be 2n + 2 hydrogen atoms: CnH2n+2. Because two points define a line, the
carbon skeleton of the ethane molecule is linear, as shown in the figure below.
H
H
H
C
C
H
H
H
Because the bond angle in a tetrahedron is 109.5, alkane molecules that contain three or four
carbon atoms can no longer be thought of as "linear," as shown below.
H2
C
H3C
H2
C
C
H2
CH 3
In addition to the straight-chain examples considered so far, alkanes also form branched
structures. The smallest hydrocarbon in which a branch can occur has four carbon atoms.
This compound has the same formula as butane (C4H10), but a different structure. Compounds
with the same formula and different structures are known as isomers (from the Greek isos,
23
"equal," and meros, "parts"). When it was first discovered, the branched isomer with the
formula C4H10 was therefore given the name isobutane.
Isobutane
The best way to understand the difference between the structures of butane and isobutane is
to compare the ball-and-stick models of these compounds shown in the figure below.
Butane
Isobutane
Butane and isobutane are called constitutional isomers because they literally differ in their
constitution. One contains two CH3 groups and two CH2 groups; the other contains three CH3
groups and one CH group. There are three constitutional isomers of pentane, C5H12. The first
is "normal" pentane, or n-pentane.
A branched isomer is also possible, which was originally named isopentane. When a more
highly branched isomer was discovered, it was named neopentane (the new isomer of
pentane).
Practice Problem 2:
The following structures all have the same molecular formula: C6H14. Which of these
structures represent the same molecule?
Practice Problem 3:
Determine the number of constitutional isomers of hexane, C6H14.
By the way, there are two constitutional isomers with the formula C4H10, three isomers of
C5H12, and five isomers of C6H14. The number of isomers of a compound increases rapidly
with additional carbon atoms. There are over 4 billion isomers for C30H62, for example.
The Cycloalkanes
If the carbon chain that forms the backbone of a straight-chain hydrocarbon is long enough,
we can envision the two ends coming together to form a cycloalkane. One hydrogen atom
has to be removed from each end of the hydrocarbon chain to form the C C bond that
closes the ring. Cycloalkanes therefore have two less hydrogen atoms than the parent alkane
and a generic formula of CnH2n.
24
The smallest alkane that can form a ring is cyclopropane, C3H6.
Any attempt to force the four carbons that form a cyclobutane ring into a plane of atoms
would produce the structure shown in the figure below, in which the angle between adjacent
C C bonds would be 90.
The angle between adjacent C
C bonds in a planar cyclopentane molecule would be 108.
By the time we get to the six-membered ring in cyclohexane, a puckered structure can be
formed by displacing a pair of carbon atoms at either end of the ring from the plane of the
other four members of the ring. One of these carbon atoms is tilted up, out of the ring,
whereas the other is tilted down to form the "chair" structure shown in the figure below.
Conformation of ethane - Rotation around C C Bond
As one looks at the structure of the ethane molecule, it is easy to fall into the trap of thinking
about this molecule as if it was static. Nothing could be further from the truth. At room
temperature, the average velocity of an ethane molecule is about 500 m/s
more than twice
the speed of a Boeing 747. While it moves through space, the molecule is tumbling around its
center of gravity like an airplane out of control. At the same time, the C H and C C
bonds are vibrating like a spring at rates as fast as 9 x 1013 s-1.
There is another way in which the ethane molecule can move. The CH3 groups at either end
of the molecule can rotate with respect to each around the C C bond. When this happens,
the molecule passes through an infinite number of conformations that have slightly different
energies. The highest energy conformation corresponds to a structure in which the hydrogen
atoms are "eclipsed." If we view the molecule along the C C bond, the hydrogen atoms on
one CH3 group would obscure those on the other, as shown in the figure below.
The lowest energy conformation is a structure in which the hydrogen atoms are "staggered,"
as shown in the figure below.
25
The difference between the eclipsed and staggered conformations of ethane are best
illustrated by viewing these molecules along the C C bond, as shown in the figure below.
Eclipsed
Staggered
The difference between the energies of these conformations is relatively small, only about 12
kJ/mol. But it is large enough that rotation around the C C bond is not smooth. Although
the frequency of this rotation is on the order of 1010 revolutions per second, the ethane
molecule spends a slightly larger percentage of the time in the staggered conformation.
The different conformations of a molecule are often described in terms of Newman
projections. These line drawings show the six substituents on the C C bond as if the
structure of the molecule was projected onto a piece of paper by shining a bright light along
the C C bond in a ball-and-stick model of the molecule.
Newman projections for the different staggered conformations of butane are shown in the
figure below.
Because of the ease of rotation around C C bonds, there are several conformations of some
of the cycloalkanes described in the previous section. Cyclohexane, for example, forms both
the "chair" and "boat" conformations shown in the figure below.
Chair
Boat
Try to draw the boat here without the hydrogen atoms
The difference between the energies of the chair conformation, in which the hydrogen atoms
are staggered, and the boat conformation, in which they are eclipsed, is about 30 kJ/mol. As a
result, even though the rate at which these two conformations interchange is about 1 x 10 5 s-1,
we can assume that most cyclohexane molecules at any moment in time are in the chair
conformation.
26
The Nomenclature of Alkanes
How to name organic compounds using the IUPAC rules
In order to name organic compounds you must first memorize a few basic names. These
names are listed within the discussion of naming alkanes. In general, the base part of the
name reflects the number of carbons in what you have assigned to be the parent chain. The
suffix of the name reflects the type(s) of functional group(s) present on (or within) the parent
chain. Other groups which are attached to the parent chain are called substituents.
 Alkanes - saturated hydrocarbons. The names of the straight chain saturated hydrocarbons
for up to a 12 carbon chain are shown below. The names of the substituents formed by the
removal of one hydrogen from the end of the chain is obtained by changing the suffix ane to -yl.
Number of Carbons
Name
Number of Carbons
Name
1
methane
6
hexane
2
ethane
7
heptane
3
propane
8
octane
4
butane
9
nonane
5
pentane
10
decane
Naming Alkyl Groups
There are a few common branched substituents which you should memorize. These are
shown below.
Name the substituent groups attached to the longest chain as alkyl groups. Give the location
of each alkyl by the number of the main chain carbon atom to which it is attached. See below.
 CH3-, methyl
 CH3CH2-, ethyl
 CH3CH2CH2-, n-propyl
 CH3CH2CH2CH2-, n-butyl
Here is a simple list of rules to follow. Some examples are given at the end of the list.
1. Identify the longest carbon chain. This chain is called the parent chain.
27
2. Identify all of the substituents (groups appending from the parent chain).
3. Number the carbons of the parent chain from the end that gives the substituents
the lowest numbers. When comparing a series of numbers, the series that is the
"lowest" is the one which contains the lowest number at the occasion of the first
difference. If two or more side chains are in equivalent positions, assign the
lowest number to the one which will come first in the name.
4. If the same substituent occurs more than once, the location of each point on which
the substituent occurs is given. In addition, the number of times the substituent
group occurs is indicated by a prefix (di, tri, tetra, etc.).
5. If there are two or more different substituents they are listed in alphabetical order
using the base name (ignore the prefixes). The only prefix which is used when
putting the substituents in alphabetical order is iso as in isopropyl or isobutyl. The
prefixes sec- and tert- are not used in determining alphabetical order except when
compared with each other.
6. If chains of equal length are competing for selection as the parent chain, then the
choice goes in series to:
a) the chain which has the greatest number of side chains
b) the chain whose substituents have the lowest- numbers
c) the chain having the greatest number of carbon atoms in the smaller side
chain
d) the chain having the least branched side chains
7. A cyclic (ring) hydrocarbon is designated by the prefix cyclo- which appears
directly in front of the base name.
In summary, the name of the compound is written out with the substituents in alphabetical
order followed by the base name (derived from the number of carbons in the parent chain).
Commas are used between numbers and dashes are used between letters and numbers. There
are NO spaces in the name.
Here are some examples:
Alkyl halides
The halogen is treated as a substituent on an alkane chain. The halo- substituent is considered
of equal rank with an alkyl substituent in the numbering of the parent chain. The halogens are
represented
as
follows:
F-
28
fluoro-, Cl - chloro-,
Br - bromo-, I- iodoHere are some examples:
Practice Problem
Name the following compound.
Practice Problem
Name the following compound.
Complex Substituents
 If the branch has a branch, number the carbons from the point of attachment.
 Name the branch off the branch using a locator number.
 Parentheses are used around the complex branch name.
1-methyl-3-(1,2-dimethylpropyl)cyclohexane
Give name for the compound
29
Problem: Write structures for; 3-ethyl-methylpentane, 3-methyl-5-propylnonane, 4-t-butyl-2methlheptane.
Name the following
NB: there should be no spaces in between the name.
Physical properties of alkanes
Solubility:
Alkanes are nonpolar, so they dissolve in nonpolar or weakly polar organic solvents. Alkanes
are hydrophobic – water hating so they form good lubricants and preservatives for metals by
keeping water away from the surface of metals.
Density:
Less than 1 g/mL, this makes them float on water.
Boiling points
Boiling points increase smoothly with increasing numbers of carbons and increasing
molecular weights. Larger molecules have larger surface areas, resulting in icreased
intermolecular van der waals attractions. These increased attractions must be overcome for
vaporization and boiling to occur. A larger molecule, with greater surface area and greater
van der waals attractions, therefore boils at a higher temperature.
A graph of n-alkane boiling points versus the number of carbon atoms shows the increase in
boiling points with increasing molecular weight. Each additional CH2 grup increases the
boiling point by about 30oC up to about ten carbons, and by about 20oC in the higher
alkanes. The other curve represents the boiling points for some branched alkanes. In general,
a branched alkane boils at lower temperature than the n-alkane with the same number of
carbon atoms. This difference in boiling is also explained by the intermolecular van der
waals forces. Branched alkanes are more compact, with less surface area for London force
interactions.
30
Melting points
Melting points increase with increasing carbons (less for odd-number of carbons). Branched
alkanes pack more efficiently into a crystalline structure, so have higher m.p.
 Lower b.p. with increased branching



 Higher m.p. with increased branching
Examples:
Major uses of alkanes
 C1-C2: gases (natural gas)
 C3-C4: liquified petroleum (LPG)
31




C5-C8: gasoline (petrol)
C9-C16: diesel, kerosene, jet fuel
C17-up: lubricating oils, heating oil
Origin: petroleum refining
Reactions of Alkanes
Alkanes are least reactive class of organic compounds – thus the name paraffin - don’t react
with strong acids or bases nor with most other reagents
Most useful reactions take place under energetic or high-temp conditions and we rarely see
them in the lab but in kitchen, engines and industry.
Combustion
Rapid oxidation takes place at high temperatures, converting alkanes to carbon dioxide and
water. Little control over the reactions is possible, except for moderating the temperature and
controlling the fuel/air ratio to achieve efficient burning.
CnH(2n+2) + excess O2  n CO2 + (n+1)H2O
2 CH3CH2CH2CH3
+
13 O2
heat
8 CO2
+ 10 H2O
Cracking and hydrocracking (industrial)
The catalytic cracking of large hydrocarbons at high temperatures gives smaller
hydrocarbons. This produces high yields of gasoline. Hydrogen is added to give saturated
hydrocarbons; cracking without hydrogen gives mixtures of alkenes and alkanes.
long-chain alkanes
catalyst
shorter-chain alkanes
H2, heat
catalyst
Halogenations
Alkanes react with halogens (F2, Cl2, Br2, I2) to form alkyl halides. For example methane
reacts with chlorine (Cl2) to form chloromethanes (methyl chloride), dichloromethane
(methylene chloride), trichloromethane and tetrachloromethane.
CH4 + Cl2
heat or light
CH3Cl + CH2Cl2 + CHCl3 + CCl4
Heat or light is needed to initiate the halogenations. We discuss this section in detail later.
2.6 Structure and conformations of alkanes
Although alkanes are not as reactive as other organic compounds, they have many of the
same structural characteristics. We will use simple alkanes as examples to study some of the
properties of organic compounds, including the structure of the sp3 hybridized carbon atom
and properties of C-C and C-H single bonds.
Structure of methane
32
The simplest alkane is methane, CH4. Methane is perfectly tetrahedral, with the 109.5o bond
angles predicted for a sp3 hybrid carbon. The four hydrogen atoms are covalently bonded to
the central carbon atom.
H
109.5O
H
H
H
Ethane Conformers
Ethane, the two-carbon alkane, is composed of two methyl groups with overlapping sp3
hybrid orbitals forming a sigma bond between them. Structures resulting from the free
rotation of a C-C single bond result to conformations – same molecule with different energies
because of bonding electron pair repulsions.
The two methyl groups are not fixed in a single position but are relatively free to rotate about
the sigma bond connecting two carbon atoms. The different arrangements formed by
rotations about a single bond are called conformations, and a specific conformation is called
a conformer (conformational isomer). Pure conformers cannot be isolated easily because the
molecules are constantly rotating through all the possible conformations. The lowest-energy
conformer is most prevalent.
In drawing conformations, we will use Newman projections, a way of drawing a molecule
looking straight down the bond connecting two carbon atoms. The front carbon atom is
represented by three lines (three bonds) coming together in a Y shape. The back carbon is
represented by a circle with three bonds pointing out from it. Until you become familiar with
the Newman projection, you should practice with models of each example and compare with
the drawings. An infinite number of conformers are possible for ethane because the angle
between the hydrogen atoms on the front and back carbon atoms can take on an infinite
number of values from 0o to 360o.
33
Any of these conformations can be specified by its dihedral angle, the angle between the C-H
bonds on the front carbon atom and the C-H bonds on the back carbon in the Newman
projection. Two of the conformations have special names. The conformation with () = 0o is
called the eclipsed conformation because the Newman projection shows the hydrogen atoms
on the back carbon to be hidden by those on the front carbon. The staggered conformation,
with () = 60o, has the hydrogen atoms on the back carbon staggered halfway between the
hydrogens on the front carbon. Any other intermediate conformation is called a skew
conformation.
Staggered conformer has lowest energy with electron clouds in the C-H bonds separated as
much as possible. The eclipsed conformation places the C-H electron clouds closer together;
torsional strain is higher than staggered conformation.
Atoms and bonds remain the same on the molecule, the only variation is the angles in which
certain parts of molecule are bent or twisted. Molecules have conformations with high strain
as well as conformations with low strain. High strain conformations are when certain parts of
the molecule that repel are forced to be close to one another. Molecules, therefore, want to
have a low strain structure.
Conformational study is the study of the energetics of different conformations. Many
reactions depend on a molecule’s ability to twist into a particular conformation;
conformational analysis can help to predict which conformations are favored, and which
reactions are more likely to take place. We will apply conformational analysis to ethane and
propane and butane first, and later to some interesting cycloalkanes.
Conformational Analysis
For ethane, only 3.0 kcal/mol
34
Propane Conformers
Note slight increase in torsional strain due to the more bulky methyl group.
Butane Conformers C2-C3 rotation
 Highest energy has methyl groups eclipsed.
 Steric hindrance is causes high torsional strain
 Dihedral angle = 0 degrees


Lowest energy has methyl groups anti.
Dihedral angle = 180 degrees between the methyl groups
35



Methyl groups eclipsed with hydrogens
Higher energy than staggered conformer but lower than methyl-methyl eclipse
Dihedral angle = 120 degrees



Gauche, staggered conformer
Methyls closer (60 degrees) than in anti conformer
Dihedral angle = 60 degrees
Higher Alkanes
 Anti conformation is lowest in energy.
36

“Straight chain” actually is zigzag.
CH3CH2CH2CH2CH3
H H H H H
C
C
C
C
C
H
H
H H H H
H
Cycloalkanes
Cyclohexanes are rings of carbon atoms with -CH2- units. The general molecular formula is
CnH2n. They are nonpolar, insoluble in water with a compact shape. Melting and boiling
points similar to branched alkanes with same number of carbons.
Naming Cycloalkanes
 Cycloalkane is usually the base compound
 Number the carbons in the ring if there is more than 1 substituent, with the first
substituent in alphabet getting the lowest number.
 A ring may be attached to a chain (cycloalkyl) this will be a substituent.
CH2CH3
CH2CH3
CH3
2.7 Cis-Trans Isomerism


Cis: like groups on same side of ring
Trans: like groups on opposite sides of ring
Stability of Cycloalkane
 5- and 6-membered rings most stable
 Bond angle closest to 109.5
 Angle (Baeyer) strain
Measured by heats of combustion per -CH2 –
Heats of Combustion Alkane + O2  CO2 + H2O
37
Cyclopropane
 Large ring strain due to angle compression
 Very reactive, weak bonds
Torsional strain because of eclipsed hydrogens
Cyclobutane
 Angle strain due to compression
 Torsional strain partially relieved by ring-puckering
38
Cyclopentane
 If planar, angles would be 108, but all hydrogens would be eclipsed.
 Puckered conformer reduces torsional strain.
Cyclohexane
 Combustion data shows it’s unstrained.
 Angles would be 120, if planar.
 The chair conformer has 109.5 bond angles and all hydrogens are staggered.
No angle strain and no torsional strain.
Chair Conformer
39
Boat Conformer
Conformational Energy
Axial and Equatorial Positions
40
Monosubstituted Cyclohexanes
1,3-Diaxial Interactions
Disubstituted Cyclohexanes
Cis-Trans Isomers
41
Bonds that are cis, alternate axial-equatorial around the ring.
CH3
CH3
Bulky Groups
 Groups like t-butyl cause a large energy difference between the axial and equatorial
conformer.
 Most stable conformer puts t-butyl equatorial regardless of other substituents.
Bicyclic Alkanes
 Fused rings share two adjacent carbons.
 Bridged rings share two nonadjacent C’s.
bicyclo[3.1.0]hexane
bicyclo[3.1.0]hexane
bicyclo[2.2.1]heptanes
bicyclo[2.2.1]heptane
Cisand Trans-Decalin
 Fused cyclohexane chair conformers
 Bridgehead H’s cis, structure more flexible
 Bridgehead H’s trans, no ring flip possible.
42
43
CHAPTER 3
THE STUDY OF CHEMICAL REACTIONS
Tools for Study
 To determine a reaction’s mechanism, look at:
 Equilibrium constant
 Free energy change
 Enthalpy
 Entropy
 Bond dissociation energy
 Kinetics
 Activation energy
3.1 Chlorination of Methane
Alkanes are not very reactive molecules. Most reactions require some energy input to initiate
a reaction e.g. high temperature and catalyst for cracking, uv light for chlorination or a spark
to ignite them (initiating free radical reactions). A combination of two main reasons account
for this lack of reactivity compared to most other homologous groups of organic molecules.
Bond Strength:
The single covalent C-C (bond enthalpy 348 kJ mol-1) and C-H (bond enthalpy 412 kJ mol-1)
bonds are very strong so bond fission does not readily happen. The carbon atom radius is
small, giving a short and strong bond with other small atoms. Therefore the reactions will
tend to have high activation energies resulting in slow/no reaction.
1.
Nature of bonding:
Carbon and hydrogen have similar electronegativities, so there is no polar bond giving a
slightly positive carbon (Cδ+) which can be attacked by electron pair donating nucleophiles.
All the C-C and C-H bonds are single covalent and no region of particularly high electron
density susceptible to attack by electron pair accepting electrophiles.
Reaction of methane with chlorine produces a mixture of chlorinated products, depending on
the amount of Cl2 added and reaction conditions such as;
 Requires heat or light for initiation.
 The most effective wavelength is blue, which is absorbed by chlorine gas.
Lots of products can be formed from absorption of only one photon of light - (chain reaction).
2.
44
H
H
C
H
+
Cl2
H
heat or light
H
H

C
Cl
+
HCl
H
The above reaction may continue; heat or light is needed for each step:
H
Cl
Cl
Cl
2
H C Cl
H C Cl Cl2
Cl C Cl
H

H
Cl2
H
Cl
Cl
C Cl
+ HCl
Cl
Chlorination does not occur at r.t on absence of light
Free-Radical Chain Reaction
 A chain reaction mechanism can explain chlorination in methane – it consists 3 steps
 Initiation: generates a reactive intermediate
 Propagation: the intermediate reacts with a stable molecule to produce another
reactive intermediate (and a product molecule).
 Termination: side reactions that destroy the reactive intermediate.
Initiation Step
A chlorine molecule splits homolytically into chlorine atoms - (free radicals). The chlorine
molecule is split into two chlorine atoms/radicals by homolytic bond fission by the impactabsorption of the ultraviolet photon. Its quantum of energy, E=hv, must be great enough to
break the Cl-Cl bond. These are highly reactive chlorine atoms that react as soon as they are
formed; they are cable of attacking an actet molecule.
Cl




Cl
+
photon ( h)
Cl
+
Cl
Homolytic bond fission means the original pair of (Cl-Cl) bonding electrons is split
between the two radicals formed.
Did you notice the half arrows? Step (1) illustrates how to use half-arrows to show a
homolytic bond fission step. The single dots represent the unpaired electron on the
free radical and the half-arrows show the individual electron 'shifts'.
The breaking of the Cl-Cl bond in the chlorine molecules begins the reaction because
it is the weakest of the bonds of any reactant molecule involved. Bond
enthalpies/kJmol-1: Cl-Cl = 242, C-C = 348, C-H = 412, and even the new bond
formed, C-Cl, is 338.
Free radicals are highly reactive species with an unpaired electron and tend to form a
new bond as soon as is possible e.g. in this case by abstracting another atom from
another molecule e.g. step (2) H abstracted, and step (3) chlorine abstracted or by
pairing up with another radical e.g. steps (4) to (6). See below!
45
Propagation Step (1)
The chlorine atom collides with a methane molecule and abstracts (removes) a H, forming
another free radical and one of the products (HCl).
H
H
H
C
H
+
H
Cl
H
H
+
C
Cl
H
Propagation Step (2)
The methyl free radical collides with another chlorine molecule, producing the other product
(methyl chloride) and regenerating the chlorine radical.
Overall Reaction
Cl
Cl
+
photon ( h)
H
H
C
H
+
H
Cl
+
C
H
Cl
H
H
C
Cl
H
H
H
+
Cl
H
H
+
Cl
Cl
H
H
C
Cl
+
H
Cl
H
Termination Steps
 Collision of any two free radicals depletes the reacting species until none is left for
further reaction.
 Combination of any free radical with contaminant or collision with wall.
H
H
C
H
+ Cl
H
H
C
H
Quiz: Can you suggest others?
3.2 Chlorination of Propane
46
Cl
CH3
CH2
CH3
+ Cl2
h
Cl
CH2
Cl
CH2
CH3
+ CH3
CH
CH3
There are six 1 H’s and two 2 H’s. We expect 3:1 product mix, or 75% 1-chloropropane
and 25% 2-chloropropane. Typical product mix however is 40% 1-chloropropane and 60% 2chloropropane. Therefore, not all H’s are equally reactive, more reactive ones result to higher
yields associated with their position.
Reactivity of Hydrogens
 To compare hydrogen reactivity, find amount of product formed per hydrogen: 40%
1-chloropropane from 6 hydrogens and 60% 2-chloropropane from 2 hydrogens.
 40%  6 = 6.67% per primary H and 60%  2 = 30% per secondary H
Secondary H’s are 30%  6.67% = 4.5 times more reactive toward chlorination than primary
H’s.
Predict the Product Mix
Given that secondary H’s are 4.5 times as reactive as primary H’s, predict the percentage of
each monochlorinated product of n-butane + chlorine.
Problems
1. Write the propagation steps leading to the formation of dichloromethane (CH2Cl2)
from chloromethane
2. Explain why free-radical halogenation usually gives mixtures of products
3. Free radical chlorination of hexane gives very poor yields of 1-chlorohexane, while
cyclohexane can be converted to chlorocyclohexane in good yields – how do you
account for this difference? What ratio of reactants (cyclohexane and chlorine) would
you use for the synthesis of chlorocyclohexane?
Equilibrium constant - Keq
Now that we have determined a mechanism for the chlorination of methane, we can consider
the energetics of the individual steps. Let’s begin by reviewing some of the principles needed
for this discussion.
Thermodynamics is the chemistry that deals with the energy changes accompanying chemical
and physical transformations. The equilibrium concentrations of reactants and products are
governed by the equilibrium constant of the reaction. Equilibrium constants can be used to
determine the extent to which reactions take place. The concentration of the species present at
the end of the reaction can be used to calculate Keq.
For Example, if A and B react to give C and D, then the equilibrium constant Keq is defined
by the following equation;
A+B→C+D
47
The equilibrium constant for chlorination Keq = 1.1 x 1019 is so large that the remaining
amounts of the reactants are close to zero at equilibrium. Large value indicates reaction “goes
to completion” because they are energetically favored.
Bond Dissociation Energy
 Bond breaking requires energy (+BDE)
 Bond formation releases energy (-BDE)
Table below gives BDE for homolytic cleavage of bonds in a gaseous molecule.
A B
A
B
+
We can use BDE to estimate H for a reaction.
Which is more likely?
Estimate DH for each step using BDE.
CH4 +
Cl
+ Cl2
CH3
+
CH3
HCl
CH3Cl
+ Cl
CH3Cl
+ H
or
CH4 +
Cl
+ Cl2
H
HCl
+ Cl
Kinetics
 Answers question, “How fast?”
 Rate is proportional to the concentration of reactants raised to a power.
Rate law is experimentally determined.
Reaction Order
 For A + B  C + D, rate = k[A]a[B]b
a is the order with respect to A
a + b is the overall order
 Order is the number of molecules of that reactant which is present in the ratedetermining step of the mechanism.
The value of k depends on temperature as given by Arrhenius: ln k = -Ea + lnA
RT
Activation Energy
 Minimum energy required to reach the transition state.
H
H
C
H
Cl
H

At higher temperatures, more molecules have the required energy.
48
Reaction-Energy Diagrams
 For a one-step reaction:
reactants  transition state  products
 A catalyst lowers the energy of the transition state.
Energy Diagram for a Two-Step Reaction
 Reactants  transition state  intermediate
 Intermediate  transition state  product
Rate-Determining Step
 Reaction intermediates are stable as long as they don’t collide with another molecule
or atom, but they are very reactive.
 Transition states are at energy maximums.
 Intermediates are at energy minimums.
49
The reaction step with highest Ea will be the slowest, therefore rate-determining for the entire
reaction.
Rate, Ea, and Temperature
X
+
CH4
HX
+
CH3
X
Ea
Rate @ 300K
Rate @ 500K
F
1.2 kcal
140,000
300,000
Cl
4 kcal
1300
18,000
Br
18 kcal
9 x 10-8
0.015
I
34 kcal
2 x 10-19
2 x 10-9
Conclusions
 With increasing Ea, rate decreases.
 With increasing temperature, rate increases.
 Fluorine reacts explosively.
 Chlorine reacts at a moderate rate.
 Bromine must be heated to react.
Iodine does not react (detectably).
Free Radical Stabilities
Energy required to break a C-H bond decreases as substitution on the carbon increases.
Stability:
3
>
2
>
1
>
methyl
DH(kcal) 91, 95, 98, 104
Chlorination Energy Diagram
Lower Ea, faster rate, so more stable intermediate is formed faster.
50
3.3 Bromination of Propane
Br
Br
CH3
•
•
CH2
CH3
+ Br2
heat
CH2
CH2
CH3 + CH3
CH
CH3
There are six 1 H’s and two 2 H’s. We expect 3:1 product mix, or 75% 1bromopropane and 25% 2-bromopropane.
Typical product mix: 3% 1-bromopropane and 97% 2-bromopropane!!!
Bromination is more selective than chlorination.
Reactivity of Hydrogens
• To compare hydrogen reactivity, find amount of product formed per hydrogen: 3% 1bromopropane from 6 hydrogens and 97% 2-bromopropane from 2 hydrogens.
• 3%

6
=
0.5%
per
primary
H
and
97%  2 = 48.5% per secondary H
Secondary H’s are 48.5%  0.5% = 97 times more reactive toward bromination than primary
H’s.
Bromination Energy Diagram
 Note larger difference in Ea
 Why endothermic?
Bromination vs. Chlorination
51
Endothermic and Exothermic Diagrams
52
Hammond Postulate
 Related species that are similar in energy are also similar in structure. The structure
of a transition state resembles the structure of the closest stable species.
 Transition state structure for endothermic reactions resemble the product.
Transition state structure for exothermic reactions resemble the reactants.
Radical Inhibitors
 Often added to food to retard spoilage.
 Without an inhibitor, each initiation step will cause a chain reaction so that many
molecules will react.
 An inhibitor combines with the free radical to form a stable molecule.
Vitamin E and vitamin C are thought to protect living cells from free radicals.
Reactive Intermediates
 Carbocations (or carbonium ions)
 Free radicals
 Carbanions
 Carbene
Carbocation Structure
 Carbon has 6 electrons and a positive charge. Carbon is sp2 hybridized with vacant p
orbital.


Stabilized by alkyl substituents 2 ways:
(1) Inductive effect: donation of electron density along the sigma bonds. (2)
Hyperconjugation: overlap of sigma bonding orbitals with empty p orbital.
53
Carbocations
The general stability order of simple alkyl carbocations is: (most stable) 3o > 2o > 1o > methyl
(least stable)
This is because alkyl groups are weakly electron donating due to hyperconjugation and
inductive effects. Resonance effects can further stabilise carbocations when present.
Free Radicals
 Also electron-deficient
 Stabilized by alkyl substituents
Order of stability: 3 > 2 > 1 > methyl
Carbanions
 Eight electrons on C: 6 bonding + lone pair
 Carbon has a negative charge.
 Destabilized by alkyl substituents.
Methyl >1 > 2  > 3 
Carbenes
 Carbon is neutral and has vacant p orbital, therefore it can be electrophilic.
 Its lone pair of electron can act as nucleophilic.
54
55
CHAPTER 4
ALKYL HALIDES: NUCLEOPHILIC SUBSTITUTION AND ELIMINATION
Our study of organic chemistry is organized into families of compounds according to their
functional groups. Alkyl halides contain halogen atoms as their functional groups. We have
already studied that alkyl halides may be formed by free-radical halogenations of alkanes.
In this chapter we consider the physical properties and reactions of alkyl halides. We use their
reactions to introduce substitution and elimination, two of the most important types of
reactions in organic chemistry.
There are three major classes of organohalogen compounds: the alkyl halides, the vinyl
halides, and the aryl halides. An alkyl halide simply has a halogen atom bonded to one of the
sp3 hybrid carbon atoms of an alkyl group. A vinyl halide has halogen atom bonded to one of
the sp2 hybrid carbon atoms of an alkene. An aryl halide has a halogen atom bonded to one
of the sp2 hybrid carbon atoms of an aromatic ring. The structures of some representative
alkyl, vinyl and aryl halides are shown below.
Alkyl halide
CHCl3 Chloroform
CF2Cl2 Freon-12
CCl3-CH3
1,1,1- CF3-CHClBr
trichloroethane
Halothane
Vinyl halides
Cl
H
C
C
Cl
H
H
Aryl halides
I
I
NH 2
HO
O
CH 2
CH
COOH
Cl
I
I
4.1 Stereochemistry, Substitution and Elimination Reactions
Polarity and Reactivity
Carbon-halogen bond is polar, because halogen atoms are more electronegative than carbon
atoms so carbon has partial positive charge. Most reactions of alkyl halides result from
breaking this polarized bond. The carbon atom has a partial positive charge, making it
56
somewhat electrophilic. A nucleophile can attack groups. This versatility allows alkyl his
electrophilic carbon, or the halogen atom can leave as a halide ion, taking the bonding pair of
electrons with it. By serving as a leaving group, the halogen can be eliminated from the alkyl
halide, or it can be replaced (substituted for) by a wide variety of functional groups. This
versatility allows alkyl halides to serve as intermediates in the synthesis of many other
functional groups.

Halogen can leave with the electron pair.
H + H C Br
H
Classes of Alkyl Halides
 Methyl halides: only one C, CH3X
 Primary: C to which X is bonded has only one C-C bond.
 Secondary: C to which X is bonded has two C-C bonds.
 Tertiary: C to which X is bonded has three C-C bonds.
Classify These:
CH3
CH
CH3
CH3CH2F
(CH3)3CBr
CH3I
Cl
Dihalides
 Geminal dihalide: two halogen atoms are bonded to the same carbon
 Vicinal dihalide: two halogen atoms are bonded to adjacent carbons.
H
H
H
C
C
H
Br
H
H
C
C
Br
Br H
vicinal dihalide
H Br
geminal dihalide
4.2 IUPAC Nomenclature of alkyl halides/Haloalkanes
Functional group suffix = halide (i.e. fluoride, chloride, bromide, iodide)
Substituent name = halo- (i.e. fluoro, chloro, bromo, iodo)
Structural unit : haloalkanes contain R-X where X = F, Cl, Br, I etc.
Notes :


Haloalkanes can also be named as alkyl halides despite the fact that the halogens are
higher priority than alkanes.
The alkyl halide nomenclature is most common when the alkyl group is simple.
57
Haloalkane style:
 The root name is based on the longest chain containing the halogen.
 This root give the alkane part of the name.
 The type of halogen defines the halo prefix, e.g. chloro The chain is numbered so as to give the halogen the lowest possible number
Alkyl halide style:
 The root name is based on the longest chain containing the halogen.
 This root give the alkyl part of the name.
 The type of halogen defines the halide suffix, e.g. chloride
 The chain is numbered so as to give the halogen the lowest possible number.
Haloalkane style:
 Functional group is an alkane, therefore suffix = -ane
 The longest continuous chain is C3 therefore root = prop
 The substituent is a chlorine, therefore prefix = chloro
 The first point of difference rule requires numbering from the right as drawn, the
substituent locant is 11-chloropropane
Cl
CH3CH2CH2Cl
Alkyl halide style:
 The alkyl group is C4, it's a tert-butyl
 The halogen is a bromine, therefore suffix = bromide
tert-butyl bromide
Haloalkane style:
 Functional group is an alkane, therefore suffix = -ane
 The longest continuous chain is C3 therefore root = prop
 The substituent is a bromine, therefore prefix = bromo
 There is a C1 substituent = methyl
 The substituent locants are both 22-bromo-2-methylpropane
Br
(CH3)3CBr
Haloalkane style:
 Functional group is an alkene, therefore suffix = -ene
 The longest continuous chain is C4 therefore root = but
 The substituent is a bromine, therefore prefix = bromo
 Since bromine is named as a substituent, the alkene gets priority
58

The first point of difference rule requires numbering from the left as drawn to make
the alkene group locant 1 Therefore the bromine locant 4CH2=CHCH2CH2Br
4-bromobut-1-ene
CH3
CH CH2CH3
Cl
4-(2-bromoethyl)heptanes
CH2CH2Br
CH3(CH2)2CH(CH2)2CH3
2-chlorobutane
Systematic Common Names
 Name as alkyl halide.
 Useful only for small alkyl groups.
Name these:
CH3
CH CH2CH3
Cl
(CH3)3CBr
CH3
CH3
CH CH2F
Uses of Alkyl Halides
 Solvents - degreasers and dry cleaning fluid
 Reagents for synthesis of other compounds
 Anesthetic for example Halothane CF3CHClBr, but CHCl3 used originally
is toxic and carcinogenic, hot in use now.
 Freons, chlorofluorocarbons or CFC’s. Freon 12, CF2Cl2, now replaced with Freon22, CF2CHCl, not as harmful to ozone layer.
 Pesticides - DDT banned in many countries
4.3 Preparation of alkyl halides
Many synthesis of alkyl halides use the chemistry of functional groups we have not yet
covered. For now, we review free-radical halogenations and summarize the other, often more
useful, synthesis of alkyl halides.
Free-Radical Halogenation of Alkanes
Free-radical halogenations is rarely an effective method for the synthesis of alkyl halides. It
usually produces mixtures of products, because there are different kinds of hydrogen atoms
59
that can be abstracted. Also, more than one halogen atom may react, giving multiple
substitution. For example, the chlorination of propane can give a messy mixture of products.
CH3CH2CH3 + Cl2 → CH3CH2CH2Cl, CH3CHClCH3 + CH3CHClCH2Cl + CH3CCl2CH3 +
CH3CH2CHCl2 + others
Laboratory syntheses using free-radical halogenation are generally limited to specialized
compounds that give a single major product, such as the following examples. All H’s in the
cyclohexane are equivalent, and free-radical halogenations give good yields of
chlorocyclohexane. Restrict amount of halogen to prevent di- or trihalide formation
H
H
H
+
Br2
h
Br
+
HBr
Formation of dichlorides and trichlorides is possible, but these side reactions are controlled
by using only a small amount of chloride and an excess of cyclohexane. Below, highly
selective bromination of 3 C, as that is the most easy H to be abstracted.
CH3
CH3
C
H
CH3
+ Br2
h
CH3
CH3
C
Br +
HBr
CH3
90%
Allylic Halogenation
Allylic radical is resonance stabilized. Bromination of cyclohexene gives a good yield of 3bromocyclohexene, where bromine has substituted for an allylic hydrogen on the carbon
atom next to the double bond (sp3 C next to C=C).
Reaction Mechanism
Starts with free radical chain reaction (initiation, propagation, termination).
h
Br2
2 Br
Quiz: Write the overall reaction for this mechanism.
Avoid a large excess of Br2 by using N-bromosuccinimide (NBS) to generate Br2 as product
HBr is formed.
60
O
N
O
Br
N
+ HBr
O
H
+
Br2
O
PROBLEMS
Show how free-radical halogenations might be used to synthesize each of the following
compounds. In each case, explain why we expect to get a single major product.
a) 1-chloro-2,2-dimethylpropane (neopentyl chloride)
b) 2-bromo-2-methlybutane
Br
CH
CH 2CH 2CH 3
c)
1-bromo-1-phenylbutane
The light-catalyzed reaction of 2,3-dimethyl-2-butene with a small concentration of bromine
gives two products:
CH 3
H3C
CH 3
C
Br2, hv
C
H3C
H3C
CH 3
CH 2
C
C
H3C
CH 2
Br
Br
+
C
CH 3
CH 3
C
CH 3
2,3-dimethyl-2-butane
4.4 Reactions of Alkyl Halides: Substitution and Elimination
Alkyl halides are easily converted to most other functional groups. The halogen atom can
leave with its bonding pair of electrons to form a stable halide ion; we say that a halide is a
good leaving group (good LG). when another atom replaces the halide ion, the reaction is a
substitution. When the halide ion leaves with another atom or ion (often H+), the reaction is
an elimination. In many eliminations, a molecule of dehydrohalogens to indicate that a
hydrogen halide has been removed from the alkyl halide. Substitution and elimination
reactions often occur in competition with each other.
In a nucleophilic substation, a nucleophile (Nuc:-) replaces a group (X-) from a carbon atom,
using its lone pair of electrons to form a new bond to the carbon atom.
C
C
H
X
+
Nuc:
-


C
C
H
Nuc
+
X:
-
The halogen atom on the alkyl halide is replaced with another group.
Since the halogen is more electronegative than carbon, the C-X bond breaks
heterolytically and X- leaves.
The group replacing X- is a nucleophile.
61
In elimination, both the halide ion and another substituent are lost. A new pi bond is formed.
C
C
H
X
-
+
B:
C
C
+
X:
-
+ HB
In the elimination the reagent (B:-) reacts as a base, abstracting a proton from the alkyl halide.
Most nucleophiles are also basic and can engage in either substitution or elimination
depending on the alkyl halide and the reaction conditions. Note that the alkyl halide loses
halogen as a halide ion, and also loses H+ on the adjacent carbon to a base. A pi bond is
formed. Product is alkene. Also called dehydrohalogenation (-HX).
Problem: Classify each reaction as a substitution, elimination, or neither
H
H
+ NaBr
Na+ -OCH3
OCH 3
Br
H
+ H3O+ + HSO4-
H2SO4
OH
H
Br
+ IBr + KBr
KI
H
Br
Give the structures of the substitution products expected when 1-bromohexane reacts with
a) Na+ -OCH2CH3
b) NaCN
c) NaOH
4.5 Second-Order Nucleophilic Substitution: The SN2 reaction Mechanism
A nucloephilic substitution has the general form
Nuc:- +
C
X
Nuc
+
C
X-
H
-
Where Nuc: is the nucleophile and X- is the leaving halide ion. An example is the reaction of
iodomethane (CH3I) with potassium hydroxide. The product is methanol.
H
H-O
+ H
C
H
H
I
HO
C
H
+
I-
H
Hydroxide ion is a good nucleophile (donor of an electron pair) because the oxygen atom has
unshared pairs of electrons and a negative charge. The carbon atom of methyl iodide is
62
electrophilic because it is bonded to an electronegative iodine atom. Electron density is drawn
away from the carbon atom by the halogen atom, giving carbon atom a partial positive
charge. The negative charge of hydroxide ion is attracted to this partial positive charge.
H
H
O-
H
C
I
H
electrophile
Hydroxide ion attacks the back side of the electrophilic carbon atom, donating a pair of
electrons to form a new bond. Notice that arrows are used to show movement of electron
pairs, from the electron-rich nucleophile to the electron-poor carbon atom of the electrophile.
(remember the reverse does not happen). Carbon must accommodate only 8 electrons in its
valelnce shell, so the carbon-iodine bond must begin to break as the carbon-oxygen bond
begins to form. Iodine is the leaving group; it leaves with the pair of electrons that once
bonded it to the carbon atom. The following mechanism shows attack by the nucleophile
(hydroxide), the transition state, and the departure of the leaving group (bromine).
H
H
H O
H
C Br
HO C Br
H H
H
H
HO C
H
+
-
Br
H
The reaction of methyl iodine with hydroxide ion is a concerted reaction, taking place in a
single step with bonds breaking and forming at the same time. The middle structure is a
transition state, a point of maximum energy, rather than an intermediate. In this transition
state, the bond to the nucleophile (OH) is partially formed, and the bond to the leaving group
is partially broken. The one-step mechanism shown for this reaction is called Bimolecular
nucleophilic substitution (SN2). The abbreviation SN2 stands for Substitution, Nucleophilic,
bimolecular. The term bimolecular means that the transition state of the rate-determining step
(the only step in this reaction) involves the collision of two molecules.
Uses for SN2 Reactions
 Synthesis of other classes of compounds.
 Halogen exchange reaction.
Nucleophile
R-X + I
-
R-X +
R-OH
alcohol

R-OR'
ether
-

R-SH
thiol
-

R-SR'
thioether

R-NH3+X
amine salt

R- N3
azide
-

R-CC-R'
alkyne
-

R-CN
nitrile

R-COO-R'
ester
OR'
SH
-
R-X + N3
R-X +
Class of Product
akyl halide

R-X +
SR'
R-X + NH3
R-X +
Product
R-I
-
R-X + OH
R-X +
-

CC-R'
CN
R-X + R-COO
-
Strength of the Nucleophile in SN2 Reaction
63
The nature of the nucleophile strongly affects the rate of the SN2 reactions. A good
nucleophile is much more effective than a weak one in attacking an electrohilic carbon atom.
For example, both methanol (CH3OH) and methoxide ion (CH3O-) are nucleophilic; but
methoxide ion reacts with electrophiles in the SN2 reaction about a million times faster than
methanol. It is generally true that species with a negative charge is a stronger nuclleophile
than a similar species without a negative charge.
Methoxide ion has nonbonding electrons that are readily available for bonding. In the
transition state, the negative charge is shared by the oxygen of the methoxide ion and by the
halide leaving group. Methanol, however has no negative charge; the transition state has a
partial negative charge on the halide but a partial positive charge on the methanol oxygen
atom. As in the case of methanol and the methoxide ion, a base is always a stronger
nucleophile than its conjugate acid.
Summary: Trends in Nucleophilic Strength
 Of a conjugate acid-base pair, the base is stronger: OH- > H2O, NH2- > NH3
 Decreases left to right on Periodic Table. More electronegative atoms less likely to
form new bond: OH- > F-, NH3 > H2O
Increases down Periodic Table, as size and polarizability increase: I- > Br- > ClPolarizability Effect
64
Bulky Nucleophiles
Sterically hindered for attack on carbon, so weaker nucleophiles.
CH3
CH2
O
ethoxide (unhindered)
weaker base, but stronger nucleophile
Solvent Effects (1)
Polar protic solvents (O-H or N-H) reduce the strength of the nucleophile. Hydrogen bonds
must be broken before nucleophile can attack the carbon.
Solvent Effects (2)
65


Polar aprotic solvents (no O-H or N-H) do not form hydrogen bonds with nucleophile
Examples:
CH3
C N
acetonitrile
O
H
C
CH3
N
CH3
dimethylformamide
(DMF)
O
C
H3C
CH3
acetone
SN2: Reactivity of Substrate
 Carbon must be partially positive.
 Must have a good leaving group
 Carbon must not be sterically hindered.
Leaving Group Ability
 Electron-withdrawing
 Stable once it has left (not a strong base)
 Polarizable to stabilize the transition state.
66
Structure of Substrate
Relative rates for SN2:
CH3X > 1° > 2° >> 3°
Tertiary halides do not react via the SN2 mechanism, due to steric hindrance.
Stereochemistry of SN2
Walden inversion
4.6 SN1 AND E1 REACTIONS
The terms SN1 and E1 mean "substitution, nucleophilic, unimolecular" and "elimination,
unimolecular," respectively. These two reaction types are being considered together for two
reasons:
1. They often occur simultaneously and competitively with one another, under the same
reaction conditions.
2. They each involve the formation of a carbocation as the crucial intermediate in the ratedetermining step; these reactions exhibit unimolecular (or "first-order") kinetics, because
only one molecule -- the immediate precursor of the carbocation -- is involved in the ratedetermining step.
SN1 mechanism/E1 mechanism
SN1 indicates a substitution, nucleophilic, unimolecular reaction, described by the
expression rate = k [R-LG]. This implies that the rate determining step of the mechanism
67
depends on the decomposition of a single molecular species. This pathway is a multi-step
process with the following characteristics:
step 1: slow loss of the leaving
group, LG, to generate a
carbocation intermediate, then
step 2 : rapid attack of a
nucleophile
on
the
electrophilic carbocation to
form a new sigma bond
Example
(CH3)3C
Br
+
(CH3)3C
+
-
Br
The generalized mechanism for each of these reaction types has been depicted below, using
tert-butyl chloride as the starting material:
Multi-step reactions have intermediates and a several transition states (TS).
In an SN1 there is loss of the leaving group generates an intermediate carbocation which is
then undergoes a rapid reaction with the nucleophile..
68
Notice that the product of an SN1 substitution reaction has simply replaced the chlorine atom
with the new substituent, "Nu" in this case. The alkyl group (t -butyl) present in the starting
material is still intact, and the hybridization of the substituent-bearing carbon has not changed
(it's still sp3). Conversely, the product of the elimination reaction is an alkene: the starting
material has lost the elements of HCl, and the hybridization of the carbon originally bearing
the chlorine atom has changed from sp3 to sp2.
Note also that the nucleophile in an SN1 reaction does not have to bear a negative charge. In
fact, these reactions are typically performed under "solvolysis" conditions, i.e., simply
heating the starting material in a protic solvent (an alcohol or carboxylic acid) that can also
act as a nucleophile. In these cases, of course, the product of carbocation capture by the
solvent will bear a "+" charge, and it will have to lose H+ in order to form a neutral product.
Similarly, the base in an E1 reaction does not have to be strong. In fact, the base must not be
strong, otherwise the E2 mechanism will be followed. It is common for the solvent to act as
the base in an E1 reaction, just as it acted as the nucleophile in an SN1 process.
Lets look at how the various components of the reaction influence the reaction pathway:
RReactivity order : (CH3)3C- > (CH3)2CH- > CH3CH2- > CH3In an SN1 reaction, the rate determining step is the loss of the leaving group to form the
intermediate carbocation. The more stable the carbocation is, the easier it is to form, and the
faster the SN1 reaction will be. Some students fall into the trap of thinking that the system
with the less stable carbocation will react fastest, but they are forgetting that it is the
generation of the carbocation that is rate determining.
The following images show a selection of alkyl bromides and their relative rates of reaction
in an SN1 hydrolysis. Try to correlate the structure of the alkyl bromide with the type of
carbocation that will be formed. If you need help, click the L button to show you where the
carbocation will be formed.
Self-test question
The following products are formed when tert-butyl bromide is heated in ethanol:
What three (organic) products would you expect to be formed if t -pentyl bromide were
heated in ethanol? Which of these products is (are) formed by substitution and which is (are)
formed by elimination pathways?
69
Self-test question #2
Can you explain why heating either enantiomer of 2-bromo-2-phenylbutane in ethanol leads
to the same substitution product, i.e., a racemic mixture of 2-ethoxy-2-phenylbutane?
Effect of Substrate Structure
Because the mechanisms of SN1 and E1 reactions each involve a carbocation intermediate,
only those substrates that ionize to produce particularly stable carbocations will be able to
react via these pathways. Typically this means tertiary alkyl halides (or alcohols, in acidic
media; see "Self-test question #3"), or substrates that can ionize to form carbocations
stabilized by resonance. SN1 and E1 reactions are much rarer for secondary alkyl halides (or
alcohols), and these mechanistic pathways are never followed for simple primary or methyl
alkyl halides (or alcohols).
Effect of Reaction Medium
SN1 and E1 reactions are most favorable in protic solvents, such as carboxylic acids or
alcohols. Neutral or acidic conditions are most common, but sometimes the media are slightly
basic. Recall, however, that strongly basic conditions will promote other modes of reactivity,
such as the E2 elimination, even in substrates that otherwise might have been susceptible to
SN1 or E1 reactions.
You should have found that the carbocations get more stable as you go left to right in the
table. As the carbocation gets easier to form, so the rate of reaction increases.
-LG
The only event in the rate determining step of the SN1 is breaking the C-LG bond. Therefore,
there is a very strong dependence on the nature of the leaving group, the better the leaving,
the faster the SN1 reaction will be.
Nu
Since the nucleophile is not involved in the rate determining step, the nature of the
nucleophile is unimportant in an SN1 reaction. However, the more reactive the nucleophile,
the more likely an SN2 reaction becomes.
70
Stereochemistry
of
SN1
(Racemization:
inversion
and
retention)
In an SN1, the nucleophile attacks the planar carbocation. Since
there is an equally probability of attack on each face there will be a
loss of stereochemistry at the reactive center as both products will
be observed.
4.7 Rearrangements (hydride and methyl shifts)
Since a carbocation intermediate is formed, there is the possibility of rearrangements (e.g.
1,2-hydride or 1,2-alkyl shifts) to generate a more stable carbocation. This is usually
indicated by a change in the position of the alkene or a change in the carbon skeleton of the
product when compared to the starting material.
This pathway is most common for systems with good leaving groups, stable carbocations and
weaker nucleophiles. A typical example is the reaction of HBr with a tertiary alcohol.
 Carbocations can rearrange to form a more stable carbocation.
 Hydride shift: H- on adjacent carbon bonds with C+.
Methyl shift: CH3- moves from adjacent carbon if no H’s are available.
Hydride Shift
71
H
Br H
CH3
C
C
H
CH3
CH3
CH3
C
H
CH3
C
C
H
CH3
CH3
CH3
C
C
H
CH3
H
CH3
CH3
H
H
CH3
C
C
C
CH3
H
CH3
Nuc
CH3
CH3
H
Nuc
C
C
H
CH3
CH3
Methyl Shift
CH3
Br CH3
CH3
C
C
H
CH3
CH3
CH3
C
C
H
CH3
CH3
CH3
CH3
C
C
H
CH3
CH3
CH3
CH3
CH3
C
C
CH3
H
CH3
CH3
Nuc
CH3
C
C
H
CH3
CH3 Nuc
C C CH3
H
72
CH3
CH3
SN1 MECHANISM FOR REACTION OF ALCOHOLS WITH HBr
Step 1:
An acid/base reaction. Protonation of the alcoholic oxygen to make a better
leaving group. This step is very fast and reversible. The lone pairs on the
oxygen make it a Lewis base.
Step 2:
Cleavage of the C-O bond allows the loss of the good leaving group, a
neutral water molecule, to give a carbocation intermediate. This is the rate
determining
step
(bond
breaking
is
endothermic)
Step 3:
Attack of the nucleophilic bromide ion on the electrophilic carbocation
creates the alkyl bromide.
73
SN1 MECHANISM FOR REACTION OF ALKYL HALIDES
WITH H2O
Step 1:
Cleavage of the already polar C-Br bond allows the loss of the
good leaving group, a halide ion, to give a carbocation
intermediate. This is the rate determining step (bond breaking is
endothermic)
Step 2:
Attack of the nucleophile, the lone pairs on the O atom of the water
molecule, on the electrophilic carbocation creates an oxonium
species.
Step 3:
Deprotonation by a base yields the alcohol as the product.
Note that this is the reverse of the reaction of an alcohol with HBr.
In principle, the nucleophile here, H2O, could be replaced with any
nucleophile, in which case the final deprotonation may not always
be necessary.
E1 mechanism (cont’d)
E1 indicates a elimination, unimolecular reaction, where rate = k [R-LG].
This implies that the rate determining step of the mechanism depends on the decomposition
of a single molecular species.
Overall, this pathway is a multi-step process with the following two critical steps:
loss of the leaving
group, LG, to generate a
carbocation
74
intermediate, then
loss of a proton, H+,
from the carbocation to
form the π-bond
Example
H
H
H
Br
C
C
H
CH3
CH3
H
C
C
CH3
H
CH3
H
H
O
H
H
CH3
H
C
C
H
CH3
C
CH3
+
+
C
H3O
CH3
H
 Halide ion leaves, forming carbocation.
 Base removes H+ from adjacent carbon.
Pi bond forms.
H
H
O
H
H
CH3
H
C
C
H
CH3
C
CH3
H
+
C
+
H3O
CH3
Let's look at how the various components of the reaction influence the reaction pathway:
RReactivity order : (CH3)3C- > (CH3)2CH- > CH3CH2- > CH3In an E1 reaction, the rate determining step is the loss of the leaving group to form the
intermediate carbocation. The more stable the carbocation is, the easier it is to form, and the
faster the E1 reaction will be. Some students fall into the trap of thinking that the system with
the less stable carbocation will react fastest, but they are forgetting that it is the generation of
the carbocation that is rate determining. Since carbocation intermediates are formed during an
E1, there is always the possibility of rearrangements (e.g. 1,2-hydride or 1,2-alkyl shifts) to
generate a more stable carbocation. This is usually indicated by a change in the position of
75
the alkene or a change in the carbon skeleton of the product when compared to the starting
material.
-LG
The only event in the rate determining step of the E1 is breaking the C-LG bond. Therefore,
there is a very strong dependence on the nature of the leaving group, the better the leaving
group, the faster the E1 reaction will be. In the acid catalysed reactions of alcohols, the -OH
is protonated first to give an oxonium ion, providing the much better leaving group, a water
molecule (see scheme below).
B
Since the base is not involved in the rate determining step, the nature of the base is
unimportant in an E1 reaction. However, the more reactive the base, the more likely an E2
reaction becomes.
Selectivity
E1 reactions usually favour the more stable alkene as the major product : i.e. more highly
alkyl substituted and trans- > cisThis E1 mechanistic pathway is most common with: good leaving groups, stable
carbocations, weak bases. A typical example is the acid catalysed dehydration of 2o or 3o
alcohols.
E1 Mechanism for Alcohols
76
Step 1:
An acid/base reaction. Protonation of the alcoholic oxygen to
make a better leaving group. This step is very fast and
reversible. The lone pairs on the oxygen make it a Lewis base.
Step 2:
Cleavage of the C-O bond allows the loss of the good leaving
group, a neutral water molecule, to give a carbocation
intermediate. This is the rate determining step (bond breaking is
endothermic)
Step 3:
An acid/base reaction. Deprotonation by a base (a water
molecule) from a C atom adjacent to the carbocation center
leads to the creation of the C=C
E1 Mechanism For Alkyl Halides
77
Step 1:
Cleavage of the polarised C-X bond allows the loss of the
good leaving group, a halide ion, to give a carbocation
intermediate. This is the rate determining step (bond
breaking is endothermic)
Step 2:
An acid/base reaction. Deprotonation by a base (here an
alkoxide ion) from a C atom adjacent to the carbocation
center
leads
to
the
creation
of
the
C=C
E1 Energy Diagram
Note: first step is same as SN1
E2 Reaction
E2 indicates an elimination, bimolecular reaction, where rate = k [B][R-LG].
This implies that the rate determining step involves an interaction between these two species,
the base and the organic substrate.
This pathway is a concerted process with the following characteristics:
E2 Mechanism
78
H
H
H
O
Br
CH3
H
C
C
H
CH3
C
CH3
H
-
+ H2O + Br
C
CH3
Simultaneous removal of the proton, H+, by the base, loss of the leaving group, LG, and
formation of the pi-bond.
Let's look at how the various components of the reaction influence the reaction pathway:
 Bimolecular elimination
 Requires a strong base
Halide leaving and proton abstraction happens simultaneously - no intermediate.
Effects of RIn an E2 reaction, the reaction transforms 2 sp3 C atoms into sp2 C atoms. This moves the
substituents further apart decreasing any steric interactions. So more highly substituted
systems undergo E2 eliminations more rapidly. This is the same reactivity trend as seen in E1
reactions.
-LG
The C-LG bond is broken during the rate determining step, so the rate does depend on the
nature of the leaving group. However, if a leaving group is too good, then an E1 reaction may
result.
B
Since the base is involved in the rate determining step, the nature of the base is very
important in an E2 reaction. More reactive bases will favour an E2 reaction.
Stereochemistry
E2 reactions occur most rapidly when the H-C bond and C-LG bonds involved are co-planar,
most often at 180o or antiperiplanar. This sets up the σ-bonds that are broken in the correct
alignment to become the π-bond.
79
Selectivity
The outcome of E2 reactions is controlled by the stereochemical requirements described
above. Where there is a choice, the more stable alkene will be the major product.
The E2 pathway is most common with:

high concentration of a strong base
 poorer leaving groups
 R-LG that would not lead to stable carbocations (when the E1 mechanism will
occur).
A typical example is the dehydrohalogenation of alkyl halides using KOtBu / tBuOH.
Saytzeff’s Rule
Many compounds
examples shown
eliminate in two
predominate. For
hydroxide:
can eliminate in more than one way, to give mixtures of products. In the
above, both 2-bromopentane and 1-bromo-1-methylcylcohexane can
ways. In most cases, we can predict which elimination product will
example, consider the E2 reaction of 2-bromobutane with potassium
The rule states: If more than one elimination product is possible, the most-substituted alkene
is the major product (most stable).The first product has a monosubstituted double bond, with
80
one substituted on the double bonded carbons. It has the general formula R-CH=CH2. The
second product has a disubstituted double bond, with general formula R-CH=CH-R (or
R2C=CH2). In most E1 and E2 eliminations where there are two or more possible elimination
products, the product with the most highly substituted double bond will predominate. The
general principle is called Sytzeff rule, and reactions that give the most highly substituted
alkene are said to follow Saytzeff orientation.
R2C=CR2>R2C=CHR>RHC=CHR>H2C=CHR
tetra >
tri
>
di > mono
PROBLEM:
The reaction of 2-bromobutane with potassium hydroxide can give elimination and
substitution. Show the substitution products and give the mechanisms of its formation.
81
CHAPTER 5
STRUCTURE AND SYNTHESIS OF ALKENES
Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only)
containing a carbon-carbon double bond. The first two are:
ethene
C2H4
propene
C3H6
You can work out the formula of any of them using: CnH2n
The table is limited to the first two, because after that there are isomers which affect the
names.
5.1 Isomerism in the alkenes
Structural isomerism
All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means
that there are two or more different structural formulae that you can draw for each molecular
formula. For example, with C4H8, it isn't too difficult to come up with these three structural
isomers: There is, however, another isomer. But-2-ene also exhibits geometric isomerism.
Geometric (cis-trans) isomerism
The carbon-carbon double bond doesn't allow any rotation about it. That means that it is
possible to have the CH3 groups on either end of the molecule locked either on one side of the
molecule or opposite each other. These are called cis-but-2-ene (where the groups are on the
same side) or trans-but-2-ene (where they are on opposite sides).
82
How geometric isomers arise
These isomers occur where you have restricted rotation
somewhere in a molecule. At an introductory level in organic
chemistry, examples usually just involve the carbon-carbon
double bond - and that's what this page will concentrate on.
Think about what happens in molecules where there is
unrestricted rotation about carbon bonds - in other words where
the carbon-carbon bonds are all single. The next diagram shows
two possible configurations of 1,2-dichloroethane.
These two models represent exactly the same molecule. You
83
can get from one to the other just by twisting around the carboncarbon single bond. These molecules are not isomers.
If you draw a structural formula instead of using models, you
have to bear in mind the possibility of this free rotation about
single bonds. You must accept that these two structures
represent the same molecule:
But what happens if you have a carbon-carbon double bond - as
in 1,2-dichloroethene?
These two molecules aren't the same. The carbon-carbon
double bond won't rotate and so you would have to take the
models to pieces in order to convert one structure into the other
one. That is a simple test for isomers. If you have to take a
model to pieces to convert it into another one, then you've got
isomers. If you merely have to twist it a bit, then you haven't!
Note: In the model, the reason that you can't rotate a carbon-carbon
double bond is that there are two links joining the carbons together. In
reality, the reason is that you would have to break the pi bond. Pi
bonds are formed by the sideways overlap between p orbitals. If you
tried to rotate the carbon-carbon bond, the p orbitals won't line up any
more and so the pi bond is disrupted. This costs energy and only
happens if the compound is heated strongly.
If you are interested in the bonding in carbon-carbon double bonds,
follow this link. Be warned, though, that you might have to read several
pages of background material and it could all take a long time. It isn't
necessary for understanding the rest of this page.
84
Drawing structural formulae for the last pair of models gives two
possible isomers.
In one, the two chlorine atoms are locked on opposite sides of
the double bond. This is known as the trans isomer. (trans :
from latin meaning "across" - as in transatlantic).
In the other, the two chlorine atoms are locked on the same side
of the double bond. This is know as the cis isomer. (cis : from
latin meaning "on this side")
The most likely example of geometric isomerism you will meet at
an introductory level is but-2-ene. In one case, the CH3 groups
are on opposite sides of the double bond, and in the other case
they are on the same side.
The importance of drawing geometric isomers properly
It's very easy to miss geometric isomers in exams if you take
short-cuts in drawing the structural formulae. For example, it is
very tempting to draw but-2-ene as
CH3CH=CHCH3
If you write it like this, you will almost certainly miss the fact that
there are geometric isomers. If there is even the slightest hint in
a question that isomers might be involved, always draw
compounds containing carbon-carbon double bonds showing
the correct bond angles (120°) around the carbon atoms at the
ends of the bond. In other words, use the format shown in the
last diagrams above.
85
How to recognise the possibility of geometric isomerism
You obviously need to have restricted rotation somewhere in the
molecule. Compounds containing a carbon-carbon double bond
have this restricted rotation. (Other sorts of compounds may
have restricted rotation as well, but we are concentrating on the
case you are most likely to meet when you first come across
geometric isomers.) If you have a carbon-carbon double bond,
you need to think carefully about the possibility of geometric
isomers.
What needs to be attached to the carbon-carbon double
bond?
Note: This is much easier to understand if you have actually got some
models to play with. If your school or college hasn't given you the
opportunity to play around with molecular models in the early stages of
your organic chemistry course, you might consider getting hold of a
cheap set. The models made by Molymod are both cheap and easy to
use. An introductory organic set is more than adequate. Google
molymod to find a supplier and more about them.
Alternatively , get hold of some coloured Plasticene and some used
matches and make your own.
Think about this case:
Although we've swapped the right-hand groups around, these
are still the same molecule. To get from one to the other, all you
would have to do is to turn the whole model over.
You won't have geometric isomers if there are two groups the
same on one end of the bond - in this case, the two pink groups
on the left-hand end.
So . . . there must be two different groups on the left-hand
86
carbon and two different groups on the right-hand one. The
cases we've been exploring earlier are like this:
But you could make things even more different and still have
geometric isomers:
Here, the blue and green groups are either on the same side of
the bond or the opposite side.
Or you could go the whole hog and make everything different.
You still get geometric isomers, but by now the words cis and
trans are meaningless. This is where the more sophisticated E-Z
notation comes in.
Summary
To get geometric isomers you must have:


restricted rotation (often involving a carbon-carbon double
bond for introductory purposes);
two different groups on the left-hand end of the bond and
two different groups on the right-hand end. It doesn't
matter whether the left-hand groups are the same as the
right-hand ones or not.
87
Note: The rest of this page looks at how geometric isomerism affects
the melting and boiling points of compounds. If you are meeting
geometric isomerism for the first time, you may not need this at the
moment.
If you need to know about E-Z notation, you could follow this link at
once to the next page. (But be sure that you understand what you have
already read on this page first!)
Alternatively, read to the bottom of this page where you will find this
link repeated.
The effect of geometric isomerism on physical properties
The table shows the melting point and boiling point of the cis
and trans isomers of 1,2-dichloroethene.
melting point (°C)
boiling point (°C)
cis
-80
60
trans
-50
48
isomer
In each case, the higher melting or boiling point is shown in red.
You will notice that:


the trans isomer has the higher melting point;
the cis isomer has the higher boiling point.
This is common. You can see the same effect with the cis and
trans isomers of but-2-ene:
isomer
cis-but-2-ene
melting point (°C)
boiling point (°C)
-139
4
88
trans-but-2-ene
-106
1
Why is the boiling point of the cis isomers higher?
There must be stronger intermolecular forces between the
molecules of the cis isomers than between trans isomers.
Taking 1,2-dichloroethene as an example:
Both of the isomers have exactly the same atoms joined up in
exactly the same order. That means that the van der Waals
dispersion forces between the molecules will be identical in both
cases.
The difference between the two is that the cis isomer is a polar
molecule whereas the trans isomer is non-polar.
Note: If you aren't sure about intermolecular forces (and also about
bond polarity), it is essential that you follow this link before you go on.
You need to know about van der Waals dispersion forces and dipoledipole interactions, and to follow the link on that page to another about
bond polarity if you need to.
Use the BACK button on your browser to return to this page.
Both molecules contain polar chlorine-carbon bonds, but in the
cis isomer they are both on the same side of the molecule. That
means that one side of the molecule will have a slight negative
charge while the other is slightly positive. The molecule is
therefore polar.
Because of this, there will be dipole-dipole interactions as well
as dispersion forces - needing extra energy to break. That will
raise the boiling point.
89
A similar thing happens where there are CH3 groups attached to
the carbon-carbon double bond, as in cis-but-2-ene.
Alkyl groups like methyl groups tend to "push" electrons away
from themselves. You again get a polar molecule, although with
a reversed polarity from the first example.
Note: The term "electron pushing" is only to help remember what
happens. The alkyl group doesn't literally "push" the electrons away the other end of the bond attracts them more strongly. The arrows with
the cross on (representing the more positive end of the bond) are a
conventional way of showing this electron pushing effect.
By contrast, although there will still be polar bonds in the trans
isomers, overall the molecules are non-polar.
The slight charge on the top of the molecule (as drawn) is
exactly balanced by an equivalent charge on the bottom. The
slight charge on the left of the molecule is exactly balanced by
the same charge on the right.
This lack of overall polarity means that the only intermolecular
attractions these molecules experience are van der Waals
dispersion forces. Less energy is needed to separate them, and
so their boiling points are lower.
90
Why is the melting point of the cis isomers lower?
You might have thought that the same argument would lead to a
higher melting point for cis isomers as well, but there is another
important factor operating.
In order for the intermolecular forces to work well, the molecules
must be able to pack together efficiently in the solid.
Trans isomers pack better than cis isomers. The "U" shape of
the cis isomer doesn't pack as well as the straighter shape of the
trans isomer.
The poorer packing in the cis isomers means that the
intermolecular forces aren't as effective as they should be and
so less energy is needed to melt the molecule - a lower melting
point.
E-Z NOTATION FOR GEOMETRIC ISOMERISM
This page explains the E-Z system for naming geometric isomers.
Important! If you have come straight here via a search engine
(including the Google site search on the Main Menu of Chemguide),
you should be aware that this page follows on from an introductory
page about geometric isomerism. Unless you are already confident
about how geometric isomers arise, and the cis-trans system for
naming them, you should follow this link first. You will find links
back to this current page at suitable points on that page.
The E-Z system
The problem with the cis-trans system for naming geometric
isomers
Consider a simple case of geometric isomerism which we've already
discussed on the previous page.
You can tell which is the cis and which the trans form just by looking
at them. All you really have to remember is that trans means "across"
91
(as in transatlantic or transcontinental) and that cis is the opposite. It is
a simple and visual way of telling the two isomers apart. So why do we
need another system?
There are problems as compounds get more complicated. For example,
could you name these two isomers using cis and trans?
Because everything attached to the carbon-carbon double bond is
different, there aren't any obvious things which you can think of as
being "cis" or "trans" to each other. The E-Z system gets around this
problem completely - but unfortunately makes things slightly more
difficult for the simple examples you usually meet in introductory
courses.
How the E-Z system works
We'll use the last two compounds as an example to explain how the
system works.
You look at what is attached to each end of the double bond in turn,
and give the two groups a "priority" according to a set of rules which
we'll explore in a minute.
In the example above, at the left-hand end of the bond, it turns out that
bromine has a higher priority than fluorine. And on the right-hand end,
it turns out that chlorine has a higher priority than hydrogen.
If the two groups with the higher priorities are on the same side of the
double bond, that is described as the (Z)- isomer. So you would write it
as (Z)-name of compound. The symbol Z comes from a German word
(zusammen) which means together.
Note: I'm not getting bogged down in the names of these more
complex compounds. As soon as I put the proper full names in, the
whole thing suddenly looks much more complicated than it really is,
and you will start to focus on where the whole name comes from
rather than on if it is a (Z)- or (E)- isomer.
92
If the two groups with the higher priorities are on opposite sides of the
double bond, then this is the (E)- isomer. E comes from the German
entgegen which means opposite.
So the two isomers are:
Summary
 (E)- : the higher priority groups are on opposite sides of the
double bond.
 (Z)- : the higher priority groups are on the same side of the
double bond.
Note: Two possible suggestions for remembering this:
 E is for "Enemies", which are on opposite sides.
You don't, of course, need a way of remembering the Z as well - it's
just the other way around from E.
 In Z isomers, the higher priority groups are on zee zame zide.
That works best if you imagine you are an American
speaking with a stage German accent!
I would really welcome a suggestion which would work for anybody
- preferably something more visual so that it isn't languagedependent. If you have a way of remembering this which might help
other students, please get in touch with me via the address on the
about this site page.
Rules for determining priorities
These are known as Cahn-Ingold-Prelog (CIP) rules after the people
who developed the system.
The first rule for very simple cases
You look first at the atoms attached directly to the carbon atoms at
each end of the double bond - thinking about the two ends separately.
 The atom which has the higher atomic number is given the
higher priority.
Let's look at the example we've been talking about.
93
Just consider the first isomer - and look separately at the left-hand and
then the right-hand carbon atom. Compare the atomic numbers of the
attached atoms to work out the various priorities.
Notice that the atoms with the higher priorities are both on the same
side of the double bond. That counts as the (Z)- isomer.
The second isomer obviously still has the same atoms at each end, but
this time the higher priority atoms are on opposite sides of the double
bond. That's the (E)- isomer.
What about the more familiar examples like 1,2-dichloroethene or but2-ene? Here's 1,2-dichloroethene.
Think about the priority of the two groups on the first carbon of the
left-hand isomer.
Chlorine has a higher atomic number than hydrogen, and so has the
higher priority. That, of course, is equally true of all the other carbon
atoms in these two isomers.
In the first isomer, the higher priority groups are on opposite sides of
the bond. That must be the (E)- isomer. The other one, with the higher
priority groups on the same side, is the (Z)- isomer.
And now but-2-ene . . .
This adds the slight complication that you haven't got a single atom
attached to the double bond, but a group of atoms.
That isn't a problem. Concentrate on the atom directly attached to the
double bond - in this case the carbon in the CH3 group. For this simple
case, you can ignore the hydrogen atoms in the CH3 group entirely.
94
However, with more complicated groups you may have to worry about
atoms not directly attached to the double bond. We'll look at that
problem in a moment.
Here is one of the isomers of but-2-ene:
The CH3 group has the higher priority because its carbon atom has an
atomic number of 6 compared with an atomic number of 1 for the
hydrogen also attached to the carbon-carbon double bond.
The isomer drawn above has the two higher priority groups on
opposite sides of the double bond. The compound is (E)-but-2-ene.
A minor addition to the rule to allow for isotopes of, for example,
hydrogen
Deuterium is an isotope of hydrogen having a relative atomic mass of
2. It still has only 1 proton, and so still has an atomic number of 1.
However, it isn't the same as an atom of "ordinary" hydrogen, and so
these two compounds are geometric isomers:
The hydrogen and deuterium have the same atomic number - so on that
basis, they would have the same priority. In a case like that, the one
with the higher relative atomic mass has the higher priority. So in these
isomers, the deuterium and chlorine are the higher priority groups on
each end of the double bond.
That means that the left-hand isomer in the last diagram is the (E)form, and the right-hand one the (Z)-.
Extending the rules to more complicated molecules
If you are reading this because you are doing a course for 16 - 18 year
olds such as UK A level, you may well not need to know much about
this section, but it really isn't very difficult!
Let's illustrate this by taking a fairly scary-looking molecule, and
seeing how easy it is to find out whether it is a (Z)- or (E)- isomer by
applying an extra rule.
Focus on the left-hand end of the molecule. What is attached directly
95
to the carbon-carbon double bond?
In both of the attached groups, a carbon atom is attached directly to the
bond. Those two atoms obviously have the same atomic number and
therefore the same priority. So that doesn't help.
In this sort of case, you now look at what is attached directly to those
two carbons (but without counting the carbon of the double bond) and
compare the priorities of these next lot of atoms.
You can do this in your head in simple cases, but it is sometimes useful
to write the attached atoms down, listing them with the highest priority
atom first. It makes them easier to compare. Like this . . .
In the CH3 group:
The atoms attached to the carbon are H H H.
In the CH3CH2 group:
The atoms attached directly to the carbon of the CH2 group are C H H.
In the second list, the C is written first because it has the highest
atomic number.
Now compare the two lists atom by atom. The first atom in each list is
an H in the CH3 group and a C in the CH3CH2 group. The carbon has
the higher priority because it has the higher atomic number. So that
gives the CH3CH2 group a higher priority than the CH3 group.
Now look at the other end of the double bond. The extra thing that this
illustrates is that if you have a double bond, you count the attached
atom twice. Here is the structure again.
So, again, the atoms attached directly to the carbon-carbon double
bond are both carbons. We therefore need to look at what is attached to
those carbons.
In the CH2OH group:
The atoms attached directly to the carbon are O H H.
In the CHO group:
The atoms attached directly to the carbon of the CH2 group are O O H.
In both lists, the oxygens are written first because they have a higher
atomic number than hydrogen. In the CHO group list, the oxygen is
written twice because of the C=O double bond.
So, what is the priority of the two groups? The first atom in both lists is
an oxygen - that doesn't help. Look at the next atom in both lists. In the
CH2OH group, that's a hydrogen; in the CHO list, it's an oxygen.
The oxygen has the higher priority - and that gives the CHO group a
higher priority than the CH2OH group.
The isomer is therefore a (Z)- form, because the two higher priority
96
groups (the CH3CH2 group and the CHO group) are both on the same
side of the bond.
That's been a fairly long-winded explanation just to make clear how it
works. With a bit of practice, it takes a few seconds to work out in any
but the most complex cases.
One more example to make a couple of additional minor points . . .
Here's an even more complicated molecule!
Before you read on, have a go at working out the relative priorities of
the two groups on the left-hand end of the double bond, and the two on
the right-hand end. There's another bit of rule that I haven't specifically
told you yet, but it isn't hard to guess what it might be when you start
to look at the problem. If you can work this out, then you won't have
any difficulty with any problem you are likely to come across at this
level.
Look first at the left-hand groups.
In both the top and bottom groups, you have a CH2 group attached
directly to the carbon-carbon double bond, and the carbon in that CH2
group is also attached to another carbon atom. In each case, the list will
read C H H.
There is no difference between the priorities of those groups, so what
are you going to do about it? The answer is to move out along the
chain to the next group. And if necessary, continue to do this until you
have found a difference.
Next along the chain at the top left of the molecule is another CH2
group attached to a further carbon atom. The list for this group is again
C H H.
But the next group along the chain at the bottom left is a CH group
attached to two more carbon atoms. Its list is therefore C C H.
Comparing these lists atom by atom, leads you to the fact that the
bottom group has the higher priority.
Now look at the right-hand groups. Here is the molecule again:
97
The top right group has C H H attached to the first carbon in the chain.
The bottom right one has Cl H H.
The chlorine has a higher atomic number than carbon, and so the
bottom right group has the higher priority of these two groups.
The extra point I am trying to make with this bit of the example is that
you must just focus on one bit of a chain at a time. We never get
around to considering the bromine at the extreme top right of the
molecule. We don't need to go out that far along the chain - you work
out one link at a time until you find a difference. Anything beyond that
is irrelevant.
For the record, this molecule is a (Z)- isomer because the higher
priority groups at each end are on the same side of the double bond.
Can you easily translate cis- and trans- into (Z)- and (E)-?
You might think that for simple cases, cis- will just convert into (Z)and trans- into (E)-.
Look for example at the 1,2-dichloroethene and but-2-ene cases.
But it doesn't always work! Think about this relatively uncomplicated
molecule . . .
This is clearly a cis- isomer. It has two CH3 groups on the same side of
the double bond. But work out the priorities on the right-hand end of
the double bond.
The two directly attached atoms are carbon and bromine. Bromine has
the higher atomic number and so has the higher priority on that end. At
the other end, the CH3 group has the higher priority.
98
That means that the two higher priority groups are on opposite sides of
the double bond, and so this is an (E)- isomer - NOT a (Z)-.
Never assume that you can convert directly from one of these systems
into the other. The only safe thing to do is to start from scratch in each
case.
Does it matter that the two systems will sometimes give different
results? No! The purpose of both systems is to enable you to decode a
name and write a correct formula. Properly used, both systems will do
this for you - although the cis-trans system will only work for very
straightforward molecules.
Where would you like to go now?
To the isomerism menu. . .
To menu of basic organic chemistry. . .
To Main Menu . . .
© Jim Clark 2007 (last modified October 2009)
5.2 Physical properties of the alkenes
Boiling Points
The boiling point of each alkene is very similar to that of the alkane with the same number of
carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the
rest that you are likely to come across are liquids.
In each case, the alkene has a boiling point which is a small number of degrees lower than the
corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and
these depend on the shape of the molecule and the number of electrons it contains. Each
alkene has 2 fewer electrons than the alkane with the same number of carbons.
Solubility
Alkenes are virtually insoluble in water, but dissolve in organic solvents.
Chemical Reactivity: Bonding in the alkenes
We just need to look at ethene, because what is true of C=C in ethene will be equally true of
C=C in more complicated alkenes.
Ethene is often modelled like this:
99
The double bond between the carbon atoms is, of course, two pairs of shared electrons. What
the diagram doesn't show is that the two pairs aren't the same as each other.
One of the pairs of electrons is held on the line between the two carbon nuclei as you would
expect, but the other is held in a molecular orbital above and below the plane of the molecule.
A molecular orbital is a region of space within the molecule where there is a high probability
of finding a particular pair of electrons.
In this diagram, the line between the two carbon atoms represents a normal bond - the pair of
shared electrons lies in a molecular orbital on the line between the two nuclei where you
would expect them to be. This sort of bond is called a sigma bond.
The other pair of electrons is found somewhere in the shaded part above and below the plane
of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move
around anywhere in this shaded region and can move freely from one half to the other. Note:
This diagram shows a side view of an ethene molecule. The dotted lines to two of the
hydrogens show bonds going back into the screen or paper away from you. The wedge shapes
show bonds coming out towards you.
The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the
sigma bond and, because they lie exposed above and below the rest of the molecule, they are
relatively open to attack by other things.
The sigma Bond Framework
In CHEM 200 we saw how visualize the sigma bonds of organic molecules using hybridized
orbitals. In ethylene each carbon atom is bonded to three other atoms, and there are no
nonbonding electrons. Three hybrid orbitals are needed, implying sp2 hybridization. We saw
that sp2 hybridization corresponds to bond angles of about 120o giving optimum separation
of three atoms bonded to the carbon atom.
100
Each of the carbon-hydrogen bonds is formed by overlap of an sp2 hybrid orbital on carbon
with the 1s orbital of a hydrogen atom. The C-H bond length in ethylene is slightly shorter
than the C-H bond in ethane, because the sp2 orbital in ethylene has more s character than an
sp3 orbital in ethane.
Two electrons must go into the carbon-carbon bonding region. Each carbon atoms still has an
additional unhybridized p orbital, and these overlap to form a pi bonding molecular orbital.
The two electrons in this orbital, form the second bond between the double-bonded carbon
atoms. No rotation is possible without breaking the pi bond (63 kcal/mole). Cis isomer cannot
become trans without a chemical reaction occurring.
5.3 IUPAC Nomenclature of alkenes
Many of the same rules for alkanes apply to alkenes
1. Name the parent hydrocarbon by locating the longest carbon chain that contains the
double bond and name it according to the number of carbons with the suffix -ene.
2. a) Number the carbons of the parent chain so the double bond carbons have the lowest
possible numbers. In a ring, the double bond is assumed to be between carbon 1 and
carbon 2.
b) If the double bond is equidistant from each end, number so the first substituent has the
lowest number.
101
3. Write out the full name, numbering the substituents according to their position in the
chain and list them in alphabetical order.
4. Indicate the double bond by the number of the first alkene carbon.
5. If more than one double bond is present, indicate their position by using the number
of the first carbon of each double bond and use the suffix -diene (for 2 double bonds),
-triene (for 3 double bonds), -tetraene (for 4 double bonds), etc.
6. a) Cycloalkenes are named in a similar way. Number the cycloalkene so the double
bond carbons get numbers 1 and 2, and the first substituent is the lowest possible
number.
b. If there is a substituent on one of the double bond carbons, it gets number 1.
102
SOLVED PROBLEM:
Name These Alkenes
Alkene Substituents
Cis-trans Isomerism
 Similar groups on same side of double bond, alkene is cis.
 Similar groups on opposite sides of double bond, alkene is trans.
103
 Cycloalkenes are assumed to be cis.
Trans cycloalkenes are not stable unless the ring has at least 8 carbons.
Name these:
trans-2-pentene
cis-1,2-dibromoethene
Stability of alkenes:
There are 3 factors that influence alkene stability:
1. Degree of substitution: more highly alkylated alkenes are more stable, so tetra > tri > di >
mono-substituted.
2. Stereochemistry: trans > cis due to reduced steric interactions when R groups are on
opposite sides of the double bond.
3. Conjugated alkenes are more stable than isolated alkenes.
trans-2-butene
more stable than
cis-2-butene
5.4 Structure and Preparation of Alkenes
Reactivity of alkenes


A pi bond is a region of high electron density
(red) so alkenes are typically nucleophiles.
Alkenes typically undergo addition reactions in
which the pi bond is converted to two new sigma
bonds. See example below
Dehydration of Alcohols to get alkenes is an addition reaction
Elimination Reactions
104
Elimination reactions are important as a method for the preparation of alkenes.
The term "elimination" describes the fact that a small molecule is lost during the process.
A 1,2-elimination indicates that the atoms that are lost come from adjacent C atoms.
The two most important methods are:

Dehydration (-H2O) of alcohols, and
 Dehydrohalogenation (-HX) of alkyl halides.
There are three fundamental events in these elimination reactions:
1. removal of a proton
2. formation of the CC pi-bond
3. breaking of the bond to the leaving group
Depending on the relative timing of these events, different mechanisms are possible:
Loss of the LG to form a carbocation, removal of H+ and formation of C=C bond :
E1 reaction
 Simultaneous H+ removal, C=C bond formation and loss of the LG : E2 reaction
 Removal of H+ to form a carbanion, loss of the LG and formation of C=C bond (E1cb
reaction)
In many cases the elimination reaction may proceed to alkenes that are constitutional isomers
with one formed in excess of the other. This is described as regioselectivity.

Zaitsev's rule, based on the dehydration of alcohols, describes the preference for
eliminations to give the highly substituted (more stable) alkene, which may also be described
as the Zaitsev product. The rule is not always obeyed, some reactions give the anti-Zaitsev
product.
Similarly, eliminations often favour the more stable trans-product over the cis-product
(stereoselectivity)
E2 mechanism
E2 indicates an elimination, bimolecular reaction, where rate = k [B][R-LG].
This implies that the rate determining step involves an interaction between these two species,
the base and the organic substrate.
105
B:H
C
C
+
C
B
H
+
:X-
C
X
This pathway is a concerted process with the following characteristics: Simultaneous removal
of the proton, H+, by the base, loss of the leaving group, LG, and formation of the pi-bond.
The E2 dehydrogenation gives excellent yields with bulky secondary and tertiary halides that
are poor SN2 substrates. A strong base forces second-order elimination by abstracting a
proton. The molecule’s bulkiness hinders second-order substitution, and a relatively pure
elimination product results. Tertiary halides are the best E2 substrates because they are prone
to elimination and cannot undergo SN2 substitution.
CH3
CH3
H3C
Br
+ OH-
H2O
H2C
+ Br-
CH3
CH3
(>90%)
Let's look at how the various components of the reaction influence the reaction pathway:
Effects of R (alkyl group)
In an E2 reaction, the reaction transforms 2 sp3 C atoms into sp2 C atoms. This moves the
substituents further apart decreasing any steric interactions. So more highly substituted
systems undergo E2 eliminations more rapidly. This is the same reactivity trend as seen in E1
reactions.
Effect of –LG (leaving group)
The C-LG bond is broken during the rate determining step, so the rate does depend on the
nature of the leaving group. However, if a leaving group is too good, then an E1 reaction may
result.
Effect of B (base)
Since the base is involved in the rate determining step, the nature of the base is very
important in an E2 reaction. More reactive bases will favour an E2 reaction.
The E2 pathway is most common with:

high concentration of a strong base
106

poorer leaving groups
 R-LG that would not lead to stable carbocations (when the E1 mechanism will
occur).
A typical example is the dehydrohalogenation of alkyl halides using KOtBu / tBuOH.
Selectivity
In many cases elimination reactions may proceed to alkenes that are isomeric but with one
formed in excess of the other.
Regioselectivity (products are constitutional isomers):
Zaitsev's rule, based on experiment observations of the dehydration of alcohols, expresses
the preference for eliminations to give the highly substituted (more stable) alkene, which
may also be described as the Zaitsev product.
The rule is not always obeyed, some reactions give the anti-Zaitsev product which is
sometimes described as the Hoffman product. (Hoffman studied the elimination of
ammonium salts). Care is needed with E2 eliminations of cyclic systems since the
antiperiplanar alignment of the C-H and C-LG bonds can dictate that the anti-Zaitsev
products dominate.
Stereoselectivity (products are stereoisomers)
Similarly, eliminations often favour the more stable trans-product over the cis-product.
107
Elimination Questions
What are the major products produced by heating the following isomeric alcohols with
Qu 1:
H2SO4 ?
Qu 2:
What are the major products produced by heating the following isomeric alkyl
bromides with NaOEt ?
Qu 3:
Draw Newman projections of product forming steps for cis- and trans-butene from the
reactions of:
(a) 2-bromo-butane with NaOEt / EtOH / heat
(b) 2-butanol with H2SO4 / heat
Qu 4:
Explain the following experimental observation for stereoisomers of menthyl chloride:
Elimination Answers
Qu1:
When heated with H2SO4, 2o alcohols undergo dehydration via an E1 mechanism.
The major product is the tri-substituted alkene, methylcyclohexene
Qu 2:
When heated with strong bases such as NaOEt, alkyl bromides undergo E2
elimination.
108
The outcome of E2 reactions is dependent on the antiperiplanar arrangement of the CH and C-LG bonds. For substituted cyclohexanes this requires that the LG be axial.
For the cis-isomer with the -Br axial, the more highly susbtituted alkene can be
formed by removal of the H adjacent to the methyl group.
For the trans-isomer, when the -Br is axial the methyl group is also axial. Therefore
the elimination must occur from the C3-H bond giving the anti-Zaitsev product.
The reactive conformation is an unfavourable diaxial conformer, therefore the
reaction will be slower than that of the cis-isomer.
Qu3: These are elimination reactions :
(a) First the E2 reaction of an alkyl halide with a strong base. There two possible H atoms at C3
that can be removed to give 2-butene, look at each in turn:
In this staggered conformation with the With the H anti to the Br, the methyl groups are a
methyl groups anti, the H is anti to the Br less stable gauche relationship. This leads to the
leaving group giving trans-2-butene.
cis-alkene
(b) Now the E1 reaction of an alcohol with a strong acid. Again there are two possible H atoms at
C3 to consider:
109
With H+ lined up as shown to allow with H+ lined up as shown to allow formation
formation of the pi bond, the methyl of the pi bond, the methyl groups are in close
groups end up trans in the alkene.
proximity and end up cis in the alkene.
Implications: The steric interactions in the product forming steps control the stereoselectivity
favouring trans-2-butene.
Qu4:
The lowest energy conformation of menthyl chloride has the chlorine atom in a
equatorial postion.
In this position there is no antiperiplanar H , ring flip is difficult as it would require to
formation
of
a
triaxial
conformer.
In contrast, in neomenthyl chloride, the lowest energy confromation has the chlorine
atom axial with 2 H in the correct orientation to give the products. The major product is
the more highly substituted alkene.
110
CHAPTER 6
REACTIONS OF ALKENES: ADDITION REACTIONS
Alkenes contain the unsaturated C=C functional group which characteristically undergo
addition reactions. This is driven by the conversion of the weaker pi-bond into 2 new,
stronger sigma bonds. The reactions of alkenes can seem a little bewildering as a wide variety
of reagents undergo this type of reaction providing access to products with various
regioselectivities and stereoselectivities depending on the reagent and / or reaction conditions,
but ultimately on the mechanism by which the reaction occurs. Once again, the concept of the
functional group helps to organize and simplify the study of chemical reactions. By studying
the characteristic reactions of the double bond, we can predict the reactions of alkenes we
have never seen before.
There are potentially 3 factors associated with each of the addition reactions of alkenes:

overall transformation (i.e. C=C to what functional group)
regioselectivity (provided the two atoms that add are different)
stereoselectivity (most easily revealed with cyclic alkenes)


Study Tip: For each reagent, identify the electrophilic atom as it will be involved in the first
step of adding to the pi-bond. Remember the E1 and E2 mechanism.
6.1 Reactions of Alkenes: Addition Reactions
Because single bonds (sigma bonds) are more stable than pi bonds, we might expect the
double bond to react and transform the pi bond into a sigma bond. In fact, this is the most
common reaction of double bonds. See the reaction below;
C
C
+
H
H
catalyst
C
C
H
H
Hydrogenation of alkene is an example of an addition reaction, one of the three major
reaction types we have studied, addition, elimination, and substitution. In an addition, two
molecules combine to form one product molecule. When an alkene undergoes addition, two
groups add to the carbon of the double bond, and the carbons become saturated. In many
ways, addition is the reverse of elimination, in which one molecule is split into two fragment
molecules. In a substitution, one fragment replaces another fragment in a molecule.
111
addition
Elinination
Substitution
C
+
C
X
C
C
X
Y
C
X +Y
Y
catalyst
C
-
C
C
X
Y
C
C
Y
+
X
+
X
Y
Addition is the most common reaction of alkenes, and we will consider additions in detail. A
wide variety of functional groups can be formed by addition of suitable reagents to the double
bonds of alkenes.
Structure of alkenes:





The alkene functional group consists of two sp2 hybridised
C atoms bonded to each other via a sigma and a pi-bond.
The pi-bond is produced by the side-to-side overlap of the
p-orbitals not utilized in the hybrids (see left).
The electrons in the pi bond are spread farther from the
carbon nuclei than the sigma electrons, and they are
loosely held.
The substituents are attached to the C=C unit via sigma
bonds.
The 2 C of the C=C and the 4 atoms attached directly to
the C=C are all in the same plane.
Reactivity:




A pi bond is a region of high electron density (red) so
alkenes are typically nucleophiles.
Alkenes react with electrophiles (e.g. H+, X+)
Alkenes typically undergo addition reactions in which the
pi bond is converted to two new stronger sigma bonds.
Overall reaction : Electrophilic addition
Reactions of Alkenes: Addition Reaction
112
A strong electrophile pulls the electrons out of the pi bond to form a new sigma bond. A
carbocation results. The curved arrow shows the movement of electrons, from the electronrich pi bond to the electron-poor electrophile.
E+
E
C
C+
C
C
Electrophilic Addition reactions occur in two steps;
Step 1: attack of the pi bond on the electrophile
C
C
+
E+
C
C+
+
Nuc:-
E
C
C
E
Nuc
This type of reaction requires a strong electrophile to attract the electrons of the pi bond and
generate a carbocation in the rate-determining step. Most alkene reactions fall this large class
of electrophilic additions to alkenes.
Electrophilic addition reactions are an important class of reactions that allow the
interconversion of double and triple carbon triple bonds into a range of important
functional groups. Conceptually, addition is the reverse of elimination. What does the term
"electrophilic addition" imply?
An electrophile, E+, is an electron poor species that will react with an electron rich species
(the C=C).
An addition implies that two systems combine to a single entity. Depending on the relative
timing of these events, slightly different mechanisms are possible:

Reaction of the C=C with E+ to give a carbocation (or another cationic intermediate)
that then reacts with a Nu Simultaneous formation of the two sigma bonds
The following pointers may aid your understanding of these reactions:





Recognise the electrophile present in the reagent combination
The electrophile adds first to the alkene, dictating the regioselectivity.
If the reaction proceeds via a planar carbocation, the reaction is not stereoselective
If the two new bonds form at the same time from the same species, then syn addition
is observed
If the two new bonds form at different times from different species, then anti
addition is observed
113
Hydrogenation of Alkenes
Reaction Type: Electrophilic Addition
Summary




Alkenes can be reduced to alkanes with H2 in the presence of metal catalysts such as
Pt, Pd, Ni or Rh.
The two new C-H bonds are formed simultaneously from H atoms absorbed into the
metal surface.
The reaction is stereospecific giving only the syn addition product.
This reaction forms the basis of experimental "heats of hydrogenation" which can be
used to establish the stability of isomeric alkenes.
Reactions of Alkenes: Addition Reactions
Reaction of Alkenes with Hydrogen Halides
Reaction type: Electrophilic Addition
Summary






When treated with HX alkenes form alkyl halides.
Hydrogen halide reactivity order : HI > HBr > HCl > HF (paralleling acidity order).
Regioselectivity predicted by Markovnikov's rule: "For addition of hydrogen halides
to alkenes, the H atom adds to the C with the most H atoms already present"
Reaction proceeds via protonation to give the more stable carbocation intermediate.
Not stereoselective since reaction proceeds via planar carbocation.
For HBr, care must be taken to avoid the formation of radicals when an alternate
mechanism occurs.
114
Mechanism for reaction of alkenes with HBr
Step 1:
An acid/base reaction. Protonation of the alkene to generate the more
stable carbocation. The pi electrons act as Lewis base.
Step 2:
Attack of the nucleophilic bromide ion on the electrophilic
carbocation creates the alkyl bromide.
Hydration of Alkenes (addition of water)
Reaction type: Electrophilic Addition
Summary




When treated with aq. acid, most commonly H2SO4, alkenes form alcohols.
Regioselectivity predicted by Markovnikov's rule
Reaction proceeds via protonation to give the more stable carbocation intermediate.
Not stereoselective since reactions proceeds via planar carbocation.
115
Mechanism For Reaction Of Alkenes With H3O+
Step 1:
An acid/base reaction. Protonation of the alkene to
generate the more stable carbocation. The pi electrons
pairs act as a Lewis base.
Step 2:
Attack of the nucleophilic water molecule on the
electrophilic carbocation creates an oxonium ion.
Step 3:
An acid / base reaction. Deprotonation by a base
generates the alcohol and regenerates the acid catalyst.
Hydroboration / Oxidation of Alkenes
Reaction type: Electrophilic Addition
Summary





Overall transformation : C=C to H-C-C-OH
Reagents (two steps) 1. BH3 or B2H6 then 2) NaOH/ H2O2
Regioselectivity : Anti-Markovnikov, since the B is the electrophile.
Stereoselectivity : Syn since the C-B and C-H bonds form simultaneously from the
BH3.
The alcohol is formed over a series of steps involving the B center (see below), with
retention of configuration at the C.
116

Compliments simple hydration with opposite regiochemistry and stereospecificity.
Reactivity of Borane
The image shows the electrostatic potential for borane, BH3.
The more red an area is, the higher the electron density and the more
blue an area is, the lower the electron density.

The low electron density (blue) is on B atom, after all, it has
an incomplete octet.
 The high electron density (red) is on the H atoms.
 Electronegativities: B = 2.0, H = 2.1
How many electrons are there around the B in borane ?



Borane reacts with alkenes via a concerted mechanism
Simultaneous making of C-B and C-H bonds as C=C and B-H break.
Electrophilic B atom adds at the least substituted end of the alkene.
Although a carbocation is not involved, the reaction occurs as if the electrophilic B atom
were making the more stable carbocation.
117
MECHANISM FOR REACTION OF ALKENES WITH BH3
Step 1:
A concerted reaction. The pi-electrons act as the nucleophile with
the electrophilic B and the H is transferred to the C with syn
stereochemistry.
Step 2:
First step repeats twice more so that all of the B-H bonds react
with C=C
Step 3:
Peroxide ion reacts as the nucleophile with the electrophilic B
atom.
Step 4:
Migration of C-B bond to form a C-O bond and displace
hydroxide. Stereochemistry of C center is retained.
Step 5:
Attack of hydroxide as a nucleophile with the electrophilic B
displacing the alkoxide.
Step 6:
An acid / base reaction to form the alcohol.
Halogenation of Alkenes
Reaction type: Electrophilic Addition
118
Summary





Overall transformation : C=C to X-C-C-X
Reagent: normally the halogen (e.g. Br2) in an inert solvent like methylene chloride,
CH2Cl2.
Regioselectivity: not relevant since both new bonds are the same, C-X.
Reaction proceeds via cyclic halonium ion.
Stereoselectivity: anti since the two C-X bonds form in separate steps one from X2
the other X-.
Mechanism for reaction of alkenes with halogens
Step 1:
The pi electrons act as a nucleophile, attacking the bromine,
displacing a bromide ion but forming a cationic cyclic
bromonium ion as an intermediate.
Step 2:
Attack of the nucleophilic bromide from the side away from the
bromonium center in an SN2 like fashion opens the cyclic
bromonium ion to give overall trans addition.
Halohydration of Alkenes
Reaction type: Electrophilic Addition
Summary






Overall transformation: C=C to HO-C-C-X
Reagents: X2 / H2O or HOX (hypohalous acid)
Reaction proceeds via cyclic halonium ion (compare with halogenation)
The alternative nucleophile, water opens the halonium ion (instead of the halide ion)
Regioselectivity: X reacts as the electrophile so the C-O bond forms at the more
stable cation center.
Stereoselectivity: anti since the two new bonds form in separate steps.
119
Mechanism for reaction of alkenes with Br2 / H2O
Step 1:
Same first step as for the reaction of Br2/CH2Cl2. The electrons act
as a nucleophile, attacking the bromine, displacing a bromide ion
but forming a cationic cyclic bromonium ion as an intermediate.
Step 2:
Attack of the nucleophilic water molecule from the side away
from the bromonium center in an SN2 like fashion opens the
cyclic bromonium ion to give overall trans addition.
Step 3:
An acid / base reaction converts the oxonium into the alcohol.
Epoxidation of Alkenes
Reaction type: Electrophilic Addition
Summary

Overall transformation : C=C to epoxide
 Reagent : a peracid, RCO3H
 Regioselectivity: not relevant since both new bonds are the same, C-O
 Stereoselectivity: syn since the two new C-O sigma bonds form at the same time from
the peracid.
What other functional group description could be given to an epoxide? What other addition
reactions convert an alkene into a three membered ring?
Mechanism For Reaction Of Alkenes With Peracid
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A single step reaction involving several changes. Start at the C=C as
the nucleophile, make a bond to the slightly electrophilic O, break the
weak O-O bond and form a C=O, break the original C=O to make a
new O-H bond, break the original O-H to form the new C-O bond!
(phew! Am sure that was a bit of a task! Get encouraged anyway)
Ozonolysis of Alkenes
Reaction type: Electrophilic Addition
Summary





Overall transformation : C=C to 2 x C=O
Ozonolysis implies that ozone causes the alkene to break (-lysis)
Reagents : ozone followed by a reducing work-up, usually Zn in acetic acid.
It is convenient to view the process as cleaving the alkene into two carbonyls:
The substituents on the C=O depend on the substituents on the C=C.
What would be the products of the ozonolysis
reactions
of: (c) cis or
(a) ethene? (b) 1-butene?
methylpropene?
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trans-2-butene?
(d)
2-
Mechanism For Reaction Of Alkenes With O3
Step 1:
The pi-electrons act as the nucleophile, attacking the ozone at the
electrophilic terminal O. A second C-O is formed by the nucleophilic
O attacking the other end of the C=C.
Step 2:
The cyclic species called the malozonide rearranges to the ozonide.
Step 3:
On work-up (usually Zn / acetic acid) the ozonide decomposes to give
two carbonyl groups.
Addition Questions
Qu 1:
Show the major products, with stereochemistry where applicable, for the
reactions of :
(a) propene and (b) methylcyclohexene with each of the following :
(i)
H2
/
Pd (ii) HBr / dark
(iii)
HBr
/
H2O2 atmosphere
(v) BH3 then NaOH / H2O2 (iv)
aq.
(vii)
Cl2
/
H2O (vi)
(ix) O3 then Zn / CH3CO2H
(viii) CH3CO3H
Qu 2:
/
N2
H2SO4
Cl2
Using diagrams, mechanisms with curly arrows, and / or short paragraphs,
explain the following observation:
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Qu 3:
The following paragaraph describes a series of reactions on a series of unknown
but related compounds:
An achiral hydrocarbon A was analysed by combustion analysis. 0.01g of A
gave 0.03217g of CO2 and 0.011g H2O.
A was added to conc. H2SO4 and the non-polymeric portion of the organic
extract was found to contain small amounts of the starting material A and three
other isomeric hydrocarbons, B, C and D. All four compounds were further
investigated chemically by their reactions with H2 / Pt and with HBr (dark).
Reaction of either A or D with H2 / Pt gave ethylcyclobutane.
Reaction of A with HBr (dark) gave E, a secondary bromide whereas reaction of
D
with
HBr
(dark)
gave
F,
a
tertiary
isomer
of
E.
Reaction of either B or C with H2 / Pt gave methylcyclopentane.
Reaction of B or C with HBr (dark) gave G.
What are A - G? Suggest why and how the A converts to B and C.
Addition Reactions Answers
Qu1:
Pay attention to the regiochemistry and stereochemistry in each case.
(a) propene
(b) methylcyclohexene
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Qu2: This question deals with the Markovnikov and anti-Markovnikov hydration of alkenes.
Aq. H2SO4 reacts by protonation of the alkene to yield the more stable 3o carbocation
followed by attack of H2O as the nucleophile and finally loss of H+ to get the
Markovnikov alcohol. In contrast, borane (BH3) undergoes a concerted addition, so
there is no carbocation intermediate. The reaction is controlled by a combination
electronic and steric factors.
- in this process (check the
electronegativities of B and H). Thus, the B adds at the least hindered end generating
cation character at the more stable carbocation site. This generates an alkyl borane.
During the oxidation step, an O atom is inserted into the C-B bond to give the C-O-B
system that generates the alcohol on reaction with NaOH.
Qu3:
The following flow chart is constructed from the problem and gradually built up to
arrive at a solution.
This type of problem is virtually impossible to slove by "guessing" A then trying to
work forwards from there. You need to extract and sift through the chemical
information to derive the solution. Information in black comes straight from the
problem, with deductions from that indicated by red arrows. Green arrows indicate
critical or key pieces of information. In this question, the most likely route to solve it is
probably starting from G......
124
A reacts with the acid to form a 2o carbocation, as we would expect is the first step of
an acid addition reaction. This carbocation can rearrange to give a more stable
carbocation by an alkyl shift that makes the 5 membered ring (less strain) which can
give a 3o carbocation via a hydride shift. Loss of a proton can then generate two
different alkenes.
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