SOME FUNCTIONAL GROUPS
H
R
H
H
C
C
H
R
R
H
H
alkane
R
H
C
C
R
O
phenol
N
N
aldehyde
R
nitro
OH
R
carboxylic acid
O
R
R
Cl
C
O
O
C
R
C
acid anhydride
O
O
C
O
O
C
R
ketone
O
R
N
O
C
N
O
+
N
R
3º amine
O
R
C
nitrile
R
2º amine
H
R
R
R
H
O
R
O
ether
R
R
H
C
R
H
1º amine
H
H
H
arene
..
O
.. H
H
R
H
R
C
alkyne
alcohol
alkyl halide
C
R
alkene
X
R
R
H
N
C
R
O
H
R
R
N
C
H
H
ester
O
R
1º amide
acid chloride
R
N
R
S
R
R
H
R
thiol
: O:
S
S
O
..
..OH
R
N:
R
H
: OCHEMISTRY
ORGANIC
INTRO
:
1º sulfonamide
S
R
..
O
..
S
sulfone
..
..OH
R
..
O
..
dialkylperoxide
R
R
..
..
O
..
..O
O
S
..
OH
..
O
O
sulfonic acid
sulfinic acid
H
R
R
R
sulfoxide
sulfide
O
S
..S
O
3º amide
R
O
O
R
C
2º amide
bisulfate
..
..
O
_
H
hydroperoxide
R : MgX+
Grignard
1
O
_
C O:
R
O
_
O:
R
carboxylate
R
Ar = aryl group
e.g., phenyl
CH2
H3C
CH
CH
H3C
CH
CH2
CH3
CH
H3C
CH2
CH2
CH2
CH3
C
H3C
C
CH2
CH3
CH3
t-butyl group
sec-butyl group
CH
isopentyl group
isobutyl group
H3C
CH3
CH
vinyl group
CH3
H3C
CH2
CH2
CH3
isopropyl group
benzyl group
carbonyl group
allyl group
naphthyl
CH3
CH2
C
acyl group
alkoxide
R = alkyl group
e.g., methyl CH3
ethyl CH3CH2
: O:
C
neopentyl group
Structures above this line must be memorized. Those below are for reference.
..
R
N
..
..O
C
..
R
..
N
N
R
R
.. _
N
+
azo compound
isocyanate
_ ..
R
N
..
+
N
..
N
diazo compound
: O:
..
..
..
C
C
N
N
urea
carbonyl
O
C
..
N
+
O
H
R
C
imine
H
C
..
N
..
N
:O:
C
oxime
hydroxylamine
_HO
2
..
NH2
C
H
semicarazide
H
R
ORGANIC CHEMISTRY INTRO
..
N
N
: O:
C
..
NH2
C
O R'
H+ + R'OH
H
O R'
aldehyde
..
H
semicarbazone
OH
H+ + R'OH
.. ..
N OH
..
..
OH
..
N
hemiacetal
R
C
H
+
H2O
O R'
acetal
2
WHAT IS ORGANIC CHEMISTRY?:
Organic chemistry is the study of the chemistry of carbon (C) compounds. Exceptions
include CO, CO2 , carbonates, bicarbonates, cyanates (CNO-), thiocyanates (CNS-), and
elemental C compounds like diamond and graphite. Although these compounds contain
carbon, their chemistry is typical of inorganic chemicals and they are generally classed as
inorganic.
 Most organic compounds contain H.
 Many also contain N, O, P, S, Cl, Br and I.
 Considering only C, H, O, and N, there are over 18  106 known C-containing
compounds and this number increases by about 500,000 each year.
 C is unique among the elements in its ability to bond with itself forming long chains of
compounds from simple to immense, i.e., from methane (CH4 ) to DNA containing 10’s
of billions of C atoms.
 Synthetic organic compounds include medicines, dyes, paints, polymers, food additives,
pesticides, fibers, etc.
 Natural organic compounds include the matter contained in all living and once-living
organisms, e.g., hair, skin, muscles, genes, food, etc. Aside from water, living organisms
are made up primarily of organic compounds.
HISTORY:
 In 1807, J. J. Berzelius coined the name ‘organic’ chemistry for materials derived from
living organisms. (‘Inorganic’ refers to materials derived from minerals).
 Up until ~ 1800, the only source of organic chemicals was from living organisms (hence
the name ‘organic’). In 1826, Friedrich Wohler converted an inorganic C-containing
chemical, ammonium cyanate into urea, a previously known ‘organic’ substance isolated
from urine .
O
heat
+ -
NH4
OCN
ammonium
cyanate
H2N
C
NH2
urea
Wohler was intrigued that chemicals of the same elemental composition could be
different and invented the term ‘isomerism’ to describe this.
 Organic chemistry began with chemists synthesizing natural organic compounds. For
example, aspirin, which was originally obtained as an extract from the bark of the willow
tree, is now produced synthetically in millions of tons per year.
 The availability of large, inexpensive sources of raw materials, i.e., petroleum, coal, and
natural gas, has caused synthetic organic chemistry to flourish.
 Unfortunately, < 10% of the fossil fuels consumed are used to make chemicals; > 90% is
burned to supply energy.
ORGANIC CHEMISTRY INTRO
3
Some Basics:

Carbon atoms have 4 bonds or less, never 5 bonds!

The 4 bonds on carbon may exist in any of 4 arrangements.
C
C
4 single () bonds
 = 'sigma'
C
C
2 single bonds + 1 double bond
(3  bonds + 1  bond)
A double bond is made of one  and one  bond.
2  bonds +2  bonds
one single and
one triple bond
or
two double bonds

Note that each line (---) in the structures represents a covalent bond, i.e., a pair of
electrons that is shared between two atoms.

When carbon has 4 bonds it is neutral (as above), but when carbon has only three bonds
it exists in one of 3 possible forms, a cation (‘carbocation’ or ‘carbonium ion’), an anion
(‘carbanion’) or a neutral radical.
..
+
C
C
carbocation (+ 1 charge)
carbanion (-1 charge)
.
C
radical (no charge)
Electron Configuration of the Elements:
In order to understand organic chemistry we must learn the electron configuration of the first
20 elements (H to Ca) plus Br and I.
Electrons are continuously buzzing around the nucleus at mind-boggling speeds (ca. 1/10 the
speed of light). We don’t know the exact position of electrons from one moment to the next
(Heizenburg uncertainty principle) but we do know that their movement is not entirely
random.
Electrons fly within well-defined flight paths (orbitals) around the nucleus. Each orbital can
hold a maximum of 2 electrons. Think about the heavier elements on the periodic table, with
100+ electrons flying around the nucleus in 50+ different flight paths. Inevitably, some of
the orbitals overlap. Just imagine how busy their flight controllers must be while trying to
prevent all those flying electrons from colliding.
The orbitals lowest to ground zero (the nucleus) are lowest in energy and are occupied by
electrons before the outer, high-energy orbitals. The 50+ orbitals around all atoms are
grouped into 7 different ‘energy levels’ (also called layers or ‘shells’) with n = 1 being the
shell closest to the nucleus (lowest energy) and n = 7 being the farthest from the nucleus
(highest energy). Study the order of orbitals and shells in the planetary model of the atom.
ORGANIC CHEMISTRY INTRO
4
PLANETARY MODEL OF THE ATOM SHOWING ENERGY LEVELS n = 1 to 6
P shell (n = 6)
6s  1, 6p  3, 6d  5, 6f  7, 6g  9, 6h  11
O shell (n = 5)
5s  1, 5p  3, 5d  5, 5f  7, 5g  9
N shell (n = 4)
4s  1, 4p  3, 4d  5, 4f  7
M shell (n = 3)
3s  1, 3p  3, 3d  5
L shell (n = 2)
2s  1, 2p  3
K shell (n = 1)
1s  1
Nucleus
This model is not spatially correct. There is some overlap of orbitals in the 3rd shell and
higher.

Within each shell, there exist subshells or types of orbitals. The types of orbitals are
named s, p, d, f, g, h, etc.
 The 1st shell has only an s orbital, named 1s.
 The 2nd shell has both s and p-type orbitals, named 2s and 2p.
 The 3rd shell has s, p and d-type orbitals, named 3s, 3p and 3d.
 The 4th shell has s, p, d and f-type orbitals, named 4s, 4p, 4d and 4f.
 etc.
ORGANIC CHEMISTRY INTRO
5
SHAPES OF ATOMIC ORBITALS
An s orbital
(l = 0, ml = 0)
z
z
z
z
x
x
y
x
x
y
y
y
A px orbital
A py orbital
A pz orbital
(on the x axis)
(l = 1, ml = -1)
(on the y axis)
(l = 1, ml = 0)
(on the z axis)
(l = 1, ml = +1)
z
z
Three p orbitals
z
z
z
y
x
x
y
y
y
A dxy orbital
A dxz orbital
(between the x and y axes)
(l = 2, ml = -2)
(between the x and z axes)
(l = 2, ml = -1)
ORGANIC CHEMISTRY INTRO
x
x
x
A dyz orbital
(between the y and z axes)
(l = 2, ml = 0)
y
A dx2-y2 orbital
A dz2 orbital
(on the x and y axes)
(l = 2, ml = +1)
(on the z axis)
(l = 2, ml = +2)
6
The number of each type of orbital, their shape and orientation are listed below.
 There is only one s-orbital in each shell and it is spherical.
 There are three p-orbitals in the 2nd and all higher shells. The three p-orbitals
are propeller shaped and are oriented along an x, y or z axis in space. They are
named px, py and pz, respectively.
 There are five d-orbitals in the 3rd and all higher shells. Four of the five
d-orbitals look like four-leaf clovers each oriented differently around the
nucleus. The fifth d-orbital looks like a propeller inside a donut. They are
named dxy, dxz, dyz, dx2-y2 and dz2.
 There are seven f-orbitals in the 4th and all higher shells. All but one have six
lobes. Each one is oriented differently around the nucleus.
In writing the electron configuration of the elements we fill lowest energy orbitals first
(Aufbau principle), with a maximum of 2 electrons per orbital –with opposite spins (Pauli
Exclusion principle). Orbitals of the same energy level (‘degenerate orbitals’) are all singly
filled (half-filled) before electrons pair up. This occurs, for example in the 2px, 2py, and 2pz
orbitals.
The filling order (increasing energy level) of the various orbitals is shown in the following
chart.
ns
(n-2)f
(n-1)d
np
H
1s
He
Li
2s
2p
Ne
Na
3s
3p
Ar
K
4s
3d
4p
Kr
Rb
5s
4d
5p
Xe
Cs
6s
4f
5d
6p
Rn
Fr
7s
5f
6d
7p
Uuo
Compare this table with a periodic
table. The filling order is the same as
the layout of the s-, p-, d- and fblocks on the periodic table, i.e., this
is the filling order. Note the patterns.
The orbital filling order is:
ns, (n-2)f, (n-1)d, np.
The p-orbitals begin filling after the
2s orbital.
The d-orbitals begin filling after the
4s orbital.
The f-orbitals begin filling after the
6s orbital.
Reading the table left to right and top to bottom, the orbital filling order is as follows:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p
ORGANIC CHEMISTRY INTRO
7
For the purpose of learning organic chemistry, we need only study the electron configuration
of the first 20 elements (H to Ca) plus Br and I, i.e., in the following orbitals:
1s 2s 2p 3s 3p 4s
Br and I have electrons in the 4p and 5p orbitals respectively, but their electron arrangement
is analogous to F and Cl.
The electron configuration of atoms is shown using a notation in which the number of
electrons in each orbital is written as a superscript. The orbital is shown as a line, _ or as a
circle, O. Each electron in the orbital is written as an arrow, . The direction of the arrow is
either up, , (indicating clockwise rotation) or down,  , (indicating counterclockwise
rotation). Complete the following table.
Full Orbital Notation
1s
2px
2py
2pz
__
1s1
He __
1s2
1
2
2s
Simplified
Notation
H
3
Li
4
Be
5
B
__
__
__
__
__
__
__
__
__
6
1s
__
__
2s
__
__
2px
__
__
2py
__
__
2pz
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
__
C
7
N
8
O
9
F
10
Ne
11
Na
1s
Write out the full electron
configuration of carbon:
6C
1s2
1s2
1s2
2s1
2s2
2s2
2p1
1s2
1s2
2s2
2s2
2p2
2p3
1s2
1s2
1s2
1s2
2s2
2s2
2s2
2s2
2p4
2p5
2p6
2p6
3s
__
2s
2px 2py 2pz
12
ENERGY
The 2px, 2py, & 2pz
orbitals are equal in
energy ('degenerate').
outer, valence electrons
bonding electrons
Problem: Write out the electron configuration for Mg through Ca in both 'Full Orbital
Notation' and 'Simplified Orbital Notation'. Recall that the 4s orbital is filled before the
3d orbital.
ORGANIC CHEMISTRY INTRO
8
Full Orbital Notation
3s 3px 3py 3pz
12
M
g
13

[Ne]
3s2
Si
15
P
16
S
17
Cl
Ar
19
[
4s
Al
14
18
[Ne]
Simplified Orbital Notation
K
Ca
The outermost occupied shell is referred to as the ‘valence’ shell. Orbitals of the valence
shell are thus ‘valence orbitals’ and electrons in the valence orbitals are ‘valence electrons’.
The outer (valence) electrons are transferred or shared in chemical reactions. Chemistry is
understood in terms of valence electron arrangement.
The number of valence electrons determines the ability of an atom to combine with other atoms
 The number of covalent bonds an atom forms to become isoelectronic with its nearest
noble gas is called its ‘covalence’. (Isoelectronic means ‘having the same valence
electronic configuration’.)
 The number of valence electrons in an atom is shown with a Lewis Symbol. One dot is
drawn for each valence electron. The dots are placed into four positions (one for each of
the one s plus three p orbitals) around the symbol of the element, i.e., north, south, east or
west. Once all four positions are singly filled, electrons (dots) are paired up until a
maximum of 8 valence electrons (dots) have been drawn.
some Lewis symbols
.
.
B
.
.
.C.
.
..
. N.
.
Although Lewis symbols do not always show the lowest energy electron arrangement of an
unbonded atom, they are a good depiction of the electron arrangement just prior to bonding.
When dots (electrons) are drawn, each orbital is first half-filled before electrons are paired up
in orbitals.
The arrangement of elements in the periodic table is based on the number of valence
electrons. For example, elements in Group IVA have 4 valence electrons.
For all representative elements (A-group elements), the number of valence electrons equals
the group number.
ORGANIC CHEMISTRY INTRO
9
 Complete the following table. Note that He is an exception. Although it has 2 valence
electrons like Be and Mg, it is unreactive (like other noble gases) and therefore has a
valence of 0.
Group #
1A
2A
3A
4A
5A
6A
7A
8A
# valence
electrons
1
2
3
4
5
6
7
8
covalence
(# bonds)
Period
1
H
Period
2
Li
Be
B
C
N
O
F
Ne
Period
3
Na
Mg
Al
Si
P
S
Cl
Ar
Period
4
K
Ca
Br
Kr
electron
config.
s1
s2
show lone
pairs of
electrons
after
bonding
H
He
Be
s2 p1
B
s2 p2
C
s2 p3
N
s2 p4
O
I
Xe
s2 p5
s2 p6
F
 Note that covalence is the same as the group number for groups 1A to 4A, but covalence
is equal to [8 - (group number)] for groups 5A to 8A.
 It is very important to appreciate the relationship between the Lewis symbols and the
number of covalent bonds formed by an atom (its covalence). Usually, in most stable
organic compounds, the atoms form a covalent bond for each unpaired electron in the
Lewis symbol of the atom.
Study the bonding arrangements of the neutral atoms shown below. Note that all single
(unpaired) electrons in a Lewis structure will bond (as shown in the bonded structure). The non
bonded electron pairs (‘lone pairs’) may either remain unbonded or form two additional bonds
per electron pair. The two additional bonds may be two single bonds or one double bond
Octet Rule:
ORGANIC CHEMISTRY INTRO
10
Note however that 2nd period elements (B, C, N, O and F) will never have more than 4 bonds
(8 electrons) around themselves as they can only use four orbitals - their 2s and 2p orbitals
for bonding (‘octet rule’).
Hypervalent Atoms:
3rd period elements and higher (Si, P, S, Cl and Br) can form more than 4 bonds (more than 8
electrons) by using their d-orbitals. Examples include PCl5, SF6, ClF7, and BrF7. Such
elements that exceed the octet rule are called ‘hypervalent’.
Group #
Lewis
Symbol
3A
.
.
B
.
4A
.
.C.
.
5A
.
B
.
N
..
.
Al .
.
. Si .
.
..
..
: F.
..
..
:F
N
C
.
..
P
.
..
P
Bonded
Structure
..
..
..
O
..
N
.
..
. ..O .
O
..
C
C
Lewis
Symbol
N.
7A
..
C
Bonded
Structure
..
6A
.
..
..
..
. ..S .
: Cl .
..
..
: Cl
..
..
S
Si
: Cl
Al
..
P
S
..
Si
Cl
..
P
S
Cl
Carbon and nitrogen are the only two elements that can form a triple bond.
ORGANIC CHEMISTRY INTRO
11
ELECTRONEGATIVITY:
One way to estimate the degree of ionic or covalent character in a chemical bond is to
compare electronegativities of atoms involved. Electronegativity is a measure of the force of
an atom’s attraction for electrons that it shares in a chemical bond with other atoms.
 In the 1930’s, Linus Pauling assigned electronegativity values to all elements relative to F
(the most electronegative element), which he gave a value of 4.0 .
Linus Pauling's Table of Electronegativities
H
2.1
Li
1.0
Na
1.0
K
0.9
Rb
0.9
Cs
0.8
Fr
0.8
Be
1.5
Mg
1.2
Ca Sc Ti V Cr Mn Fe
1.0 1.3 1.4 1.5 1.6 1.6 1.7
Sr Y Zr Nb Mo Tc Ru
1.0 1.2 1.3 1.5 1.6 1.7 1.8
Ba La Hf Ta W Re Os
1.0 1.1 1.3 1.4 1.5 1.7 1.9
Ra Ac
1.0 1.1
Co Ni Cu Zn
1.7 1.8 1.8 1.6
Rh Pd Ag Cd
1.8 1.8 1.6 1.6
Ir Pt Au Hg
1.9 1.8 1.9 1.7
B
2.0
Al
1.5
Ga
1.7
In
1.6
Tl
1.6
C
2.5
Si
1.8
Ge
1.9
Sn
1.8
Pb
1.7
N
3.0
P
2.1
As
2.1
Sb
1.9
Bi
1.8
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
1.9
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.1

Bear in mind that these values can vary slightly depending upon the chemical
environment and so the values are average values.
Note that EN increases across any period and decreases down any group (in most cases).

Pure covalent bonds involve equal sharing of the bonding electron pairs. Pure covalent
bonds occur when both atoms involved have equal EN (i.e., EN = 0) ...
For example, H2, N2, O2, F2, Cl2, Br2, I2, Cx, S8, PH3, and CS2 are all pure covalent.



Nonpolar covalent bonds are those in which EN  0.4. Examples include all the pure
covalent compounds listed above as well as compounds, e.g., CH4 (EN = 0.4) and BH3
(calculate EN)
Polar covalent bonds are those in which the bonding pair of electrons is unequally
shared (0.5  EN  1.7). Examples include HBr, HCl, etc. In polar covalent
compounds, the more electronegative atom has a partial negative charge ( -) and the less
electronegative atom has a partial positive charge ( +).
Ionic bonds are those in which EN  1.8 and are generally considered to have complete
charge separation, i.e., considered to be made up of cations and anions. Examples include
ORGANIC CHEMISTRY INTRO
12
NaCl, Li3N, and CaO. Exceptions are HF and alkali metal iodides. EN in HF is 1.9 but
this compound behaves as a polar covalent compound. EN in LiI is only 1.5 but LiI
behaves ionically. BF3 is also anomalous. It is covalent although EN = 2.0.
 A scale of bond type versus EN follows...
E
0
1.8
0.4
n o n p o la r
0.5
covalent
p o la r c o va le n t
3.2
io n ic
1.7
p u re c o va le n t
In polar covalent bonds, the electron distribution is said to be polarized, i.e., not equally
distributed but closer to the more electronegative atom. For example, in HCl....
EN = (3.0 – 2.1) = 0.9 i.e., polar covalent
A separation of '+' and '-' charge is called a dipole. Dipoles are sometimes illustrated with an
arrow pointing toward the more electronegative atom.
+
The tail of the arrow is crossed to look like a + sign.
The head of the arrow points in the direction of electron shift.
H  Cl
The shifting of electron density through sigma bonds due to EN differences between atoms is
called an 'inductive effect'. Electropositive elements (metals) such as Zn and Hg, inductively
donate electrons through sigma (single) bonds with carbon. Electronegative elements
(nonmetals) such as oxygen and chlorine inductively withdraw electrons from carbon through
their sigma bonds with carbon.
Problem: Calculate EN values and show bond dipoles and dipole arrows.
H
H
C
H
:O:
C
H
..
O
..
H
H
H
C
C
H
H
..
H
N
H
acetic acid
ethyl amine
ORGANIC CHEMISTRY INTRO
13
BONDING (INTRAMOLECULAR FORCES):
Atoms bond because the resulting compound is more stable (lower energy). For the
representative elements, filled valence orbitals (isoelectronic with the nearest noble gas) is a
stable arrangement.
Note the # of electrons in each PEL for the noble gases and the # of valence electrons..
period
noble gas
PEL = 1
PEL = 2
PEL = 3
PEL = 4
1
He
2
2
Ne
2
8
3
Ar
2
8
8
4
Kr
2
8
18
8
5
Xe
2
8
18
18
PEL = 5
8
 For main group elements, bonds form in which the combining atoms obtain a noble gas
electron configuration by either transferring electrons (ionic) or sharing electrons
(covalent). A-Group element bond to become isoelectronic with their nearest noble gas.
 Ionization energy (Ei) is the amount of energy added to remove an electron form an
isolated atom (endothermic process). Metals have low Ei whereas metalloids and non
metals have high Ei. Ei decreases down all groups (due to increased shielding and
distance from the nucleus) and from right to left across all periods (due to increased
distance from the nucleus).
 Write the full electron configuration of the reactants and products in the reaction ...
Na + Cl  [Na+ + Cl-]
Na (1s2 2s2 2p6 3s1) + Cl (1s2 2s2 2p6 3s2 3p5)  Na+ (1s2 2s2 2p6) + Cl- (1s2 2s2 2p6 3s2 3p6)
isoelectronic with Neon
isoelectronic with Argon
 Ionic bonds result when metals lose electrons to nonmetals forming cations (+) and
anions (-). Electrostatic attraction holds the solid together, e.g. ...
Na (g)  Na+ (g) + 1eEi = + 119 kcal/mol
-85 kcal
overall
Cl (g) + 1e-  Cl- (g)
Eea = - 83 kcal/mol
Na+ (g) + Cl- (g)  NaCl (g)
Elattice  -121 kcal/mol
The lattice energy (Elattice) is the energy released due to coulombic (electrostatic)
attraction as cations and anions combine in a crystallatice.
 Ionic bonds are common in inorganic compounds, but are less common in organic
compounds. The Ei of C is too high (+261 kcal/mol). C shows little tendency to act as a
source of a cation or anion in an ionic bond so most of its compounds are characterized
by covalent bonds.
ORGANIC CHEMISTRY INTRO
14
Covalent bonding is readily shown using Lewis diagrams or Lewis structures.
Using Lewis structures show the bonding in CH4, CH3OH, and H3O+ (hydronium ion)
CH4
4H.
+
H.
.
. C.
.
H
.
H . . C. . . H
.
C
H
H
H
H
Lewis structure
Lewis structures show a bonding electron pair as a line (  ) and a nonbonded pair as
two dots (.. or :).
CH3OH
3H.
+
.
. C.
.
+
..
.O.
..
H
+
.H
H
C
..
O
..
H
H
H2O
..
.O.
..
O2
+
..
.O.
..
..
O
..
..
O
..
Draw Lewis structures for H2, and N2 and include lone pair electrons.

Most of the 2nd and 3rd period representative elements we encounter in organic chemistry
(C, N, O, F, Na, Mg, Al, Si, P, S, Cl, etc.) obey the ‘octet rule’, i.e., in forming
compounds, atoms will gain, lose or share electrons to obtain 8 valence electrons. This
produces a stable valence electron configuration (stable octet) like that of Ne and Ar.

H and Li also react to obtain a noble gas electron configuration, i.e., that of Helium with 2
valence electrons.
Reaction Mechanisms: We show the transfer of electrons with curved arrows. A oneelectron transfer is shown with a half arrow head, while a two-electron transfer is shown with
a full (double) arrowhead. Arrows point away from the donor atom and toward the acceptor
atom. Be careful to draw covalent bonds as a line (a shared pair) but recall that an ionic bond
is shown by a non bonded pair and electric charges on the anion and cation. .
.
Li
..
+
. F:
..
.
.
H
H
+
ORGANIC CHEMISTRY INTRO
..
+
: F:
..
EN = (4.0 - 1.0) = 3.0 (ionic bond)
H
H
EN = (2.1 - 2.1) = 0.0 (covalent bond)
Li
15
EXCEPTIONS TO THE OCTET RULE:
The octet rule indicates that atoms bond to share enough electrons so that each atom has 8
valence electrons after bonding. Most main group elements (Group 1A to 7A) obey this but
there are a few exceptions, divided into 2 kinds, i.e., molecules in which atoms contain fewer
than 8 valence electrons and molecules in which atoms contain more than 8 valence
electrons.
 Molecules in which atoms contain less than 8 valence electrons include BeCl2, AlCl3 and
BF3. Be has only 4 valence electrons around it while both Al and B have only 6 valence
electrons around them. These compounds thus behave as Lewis acids or electrophiles
(electron acceptors). Learn the Lewis structures for these compounds.
..
..
: Cl :
: F:
..
..
: Cl
Be
..
..
Cl :
..
..
:F
B
..
F:
..
..
..
: Cl
Al
..
..
Cl :
 Molecules in which atoms contain more than 8 valence electrons.....
Atoms of the 2nd period use 2s and 2p orbitals for bonding and these orbitals can contain
only 8 valence electrons, hence, the octet rule.
Atoms of the 3rd, 4th, and 5th period have ns, np, nd, etc. orbitals and can accommodate
more than 8 electrons in their valence shells.
Phosphorus and sulfur have 3s, 3p, and 3d orbitals and in some compounds have > 8
valence electrons. Study the Lewis structures for the following .....
PI3
..
P
..
:
PI5
I
.. : I :
..
..
I:
..
..
:I:
..
:I
..
P
:I:
..
..
:
..I
:I :
..
..
: F:
SF6
..
..
..
:F
..
:F
S
..
:
..F
..
F:
..
: F:
..
ORGANIC CHEMISTRY INTRO
16
DRAWING LEWIS STRUCTURES OF COMPOUNDS:
1. The least electronegative atom is central, e.g., in CO2
O=C=Onot
O=O=C
2. H is never central (max 1 bond), e.g., in H2O
H-O-H
not
H-H-O
3. In oxyacids (‘ternary’ acids), e.g., HNO3, H2SO4, HClO3, etc., acidic H’s are bonded to O and O’s
are bonded to the central atom,
e.g., in HClO (hypochlorous acid)
H-O-Cl
not
H-Cl=O
4. Join all atoms with single bonds. Add additional bonds and non-bonded electron pairs consistent
with the Lewis symbols and normal covalence of atoms. Non-bonded e pairs may remain
unbonded but single electrons always form bonds. Make each atom isoelectronic with its nearest
noble gas but ‘3rd period plus’ atoms may be hypervalent. Check for normal bonding as shown
on page 10.
5. Count (check) the number of valence electrons in the structure. Add 1 e- for each negative
charge on an ion and subtract 1 e- for each positive charge on an ion.
Note: Oxygen does not normally bond to itself, except in peroxides, e.g.,
hydrogen
peroxide,
H-O-O-H
Alternate method:
1. Add up the number of valence electrons in each atom to find the total number of valence
electrons in the compound. Add 1 e- for each negative charge on an ion and subtract 1 e- for each
positive charge on an ion.
2. Initially, connect all atoms by single bonds (2 electrons per bond) and then add electron pairs to
each atom to complete their octet. Place any leftover electrons on the central atom. If there are
not enough electrons to give the central atom an octet, try multiple bonds. Use one or more of
the unshared pairs of electrons on the atoms bonded to the central atom to form double or triple
bonds. Note that some atoms often have one or more lone pairs of electrons, i.e., group 5A
(1 lone pair), groups 6A (2 lone pair) and group 7A (3 lone pairs).
3. Check that B, C, N, O, and F do not have more than 8 electrons around them after bonding.
carbonic acid
H2CO3
(24 valence electrons)
sulfuric acid
H2SO4
( 32 valence electrons)
phosphoric acid
H3PO4
(32 valence electrons)
H
:O :
..
O
..
C
..
O
..
H
H
(26 valence electrons)
..
:O
..
..
H
O
..
S
..
O
..
P
..
O
..
:O
..
H
H
S
..
O
..
..
O
..
:O
..
H
H
acetic acid
CH3COOH
nitrous acid
(24 valence electrons)
(18 valence electrons)
sulfurous acid
H2SO3
:O
..
:O :
:O :
H
H
C
HNO2
:O :
C
..
O
..
H
H
..
O
..
..
N
..
O
..
H
H
ORGANIC CHEMISTRY INTRO
17
Draw Lewis structures for thionyl chloride (SOCl2), phosphorus oxychloride (POCl3), and
orthosilicic acid [Si(OH)4.
SOCl2
POCl3
thionyl chloride
: O:
..
: Cl
..
S
..
orthosilicic acid
phosphorus oxychoride
H4SiO4
..
: O:
..
: Cl
..
..
..
Cl :
P
: OH
..
Cl :
..
..
HO
....
HO
..
:
: Cl
..
Si
..
..
OH
Thus far we have only drawn structures of neutral molecules. Charged polyatomic ions are
also encountered in organic chemistry. Consider the following reaction.
H2O
+
H3O+
H+
hydronium ion
Lewis structures and mechanism:
H
..
O
..
H
+
+
..
O
H+
H
H
H
We see that hydronium ion, a charged polyatomic cation, forms when water is ‘protonated’
(bonds with H+). In hydronium ion it is the oxygen atom (no longer the hydrogen ion) that
has most of the positive charge. This can be seen by determining the ‘formal charge’ of each
atom in the complex.
FORMAL CHARGE:
Formal charge allows us to assign charges to all atoms in a polyatomic ion or molecule. A
charged atom is usually a site of activity. The charge of an atom in a polyatomic ion or
molecule is called its formal charge.
To derive the formal charge, follow these steps....
1. Write a correct Lewis structure for the molecule or ion.
2. Assign to each atom, all of its unshared (non bonding) electrons and ½ of its shared
electrons (half of each bonding pair).
3. Compare this with the number of electrons in the neutral, unbonded atom. The difference
is the formal charge.
4. The total charge on a molecule or ion is the sum of its formal charges.
ORGANIC CHEMISTRY INTRO
18
Formal Charge = the theoretical charge on a bonded atom assuming the atom owns all its
non bonded electrons and half of its bonded electrons. Compare this to the Lewis symbol of
the neutral atom.
+
Oxygen 'owns' 5 valence e's, i.e.,
2 non bonded and 3 bonded (half
of each of the 3 sigma bonds).
This is one less than the Lewis
symbol of neutral oxygen.
Oxygen has a formal charge
equal to +1
..
O
H
H
H
..
. O.
..
Each hydrogen 'owns' 1 valence
electron - the same as the Lewis
symbol of neutral hydrogen.
All hydrogens have a formal charge
equal to zero.
.H
Lewis symbol of a neutral hygrogen atom.
Lewis symbol of a neutral oxygen atom.
Although oxygen has 8 valence electrons (1 non bonded pair + 3 shared pairs), it only ‘owns’
5 valence electrons, 1 less than that seen in the Lewis symbol of neutral oxygen. Hence
oxygen has a formal charge of +1 hydronium ion.
BH3 (borane) has an incomplete octet (an empty 2pz orbital). As a result it has a tendency to
accept a pair of electrons from an electron donor, e.g., sodium hydride (NaH). The resulting
product is a polyatomic anion in which boron has a formal charge of –1. Study the following
Lewis structures.
H
B
H
H
H
+
Na
H
sodium hydride is a polar covalent
compound (not ionic). EN = 1.1
Na+
B
H
H
H
sodium borohydride
BH4- is a polyatomic anion
in which boron has a formal charge
equal to -1
. B.
.
The Lewis symbol of a neutral
boron atom has only 3 valence e's.
Exercises:
1. Draw Lewis structures (with a curved arrow) to show the reaction of lithium hydride
(LiH) with aluminum hydride (AlH3). Lithium aluminum hydride (LiAlH4) forms.
2. Draw the Lewis structure of sodium hydroxide (NaOH). Note that the Na to O bond
is ionic, not covalent. Do not draw a line (covalent bond) showing a shared electron
pair between Na and O. Rather the pair of electrons will appear as an extra non
bonded pair on oxygen. Oxygen will have a formal charge of –1, while sodium will
have a formal charge of +1. Hydrogen has a formal charge equal to 0.
3. Draw Lewis structures (with a curved arrow) to show the reaction of ammonia (NH3)
with anhydrous hydrogen chloride gas (HCl). Ammonium chloride (NH4Cl) forms.
ORGANIC CHEMISTRY INTRO
19
4. Draw the Lewis structure of sodium amide (NaNH2). As with NaOH, the Na to N
bond is ionic, not covalent. The amide ion (NH2-) is a polyatomic anion in which N
has formal charge of –1.
Note that only valence electrons are considered when determining formal charge. Inner shell
electrons are not involved in bonding and need not be considered. Even if inner shell
electrons are counted, the result is the same so we use the simpler method of considering only
outer valence electrons when calculating formal charge.
Bonding electron pairs are not necessarily shared equally so formal charges do not
necessarily represent actual charges. However, in most cases, if the Lewis structure shows
that an atom has a formal charge, it actually bears at least part of that charge. This partial
transfer of '+' and '-' charge is called polarization.
SOME ACID BASE REACTION MECHANISMS:
Neutralization of H2SO4 (a diprotic acid) with KOH (a monoprotic base) occurs in two steps.
First potassium bisulfate (NaHSO4) forms and finally sodium sulfate (Na2SO4). The reaction
mechanism can be shown by drawing Lewis structures and curved arrows.
H
K+
:O :
H
..
O
..
..
O
..
S
..
:O
..
H
..
O
..
H
:O :
:O :
H
H
..
..
O
S
..
O:
..
: O:
: O:
sulfuric acid
K+
K+
potassium bisulfate
-
K+
..
..
:O
..
..
:O
..
O:
..
: O:
H
H
S
..
O
..
potassiuim sulfate
H
+
Note that OH donates a pair of non bonded electrons to remove H from the acid.
H simultaneously leaves its covalently bonded electrons with oxygen in the acid (producing
H+) and accepts the pair of electrons from OH- forming H2O. This leads us to two useful
definitions of acids and bases:
A Lewis base, such as OH-, is an electron (pair) donor.
A Lewis acid, such as H2SO4 is an electron (pair) acceptor.
A Bronsted base, such as OH-, is a proton (H+) acceptor.
A Bronsted acid, such as H2SO4, is a proton (H+) donor.
K+ ion proceeds throughout the reaction unchanged and is thus called a ‘spectator ion’.
ORGANIC CHEMISTRY INTRO
20
Study the mechanism of Na2CO3 (a diprotic base) reacting with aqueous HCl. Note that
when HCl (and other very strong acids) dissolves in water it reacts completely forming
hydronium chloride (H3O+ Cl-). No HCl remains in dilute aqueous solutions of the acid.
..
O+
:O :
Na+
..
..
:O
C
H
..
..
..
: Cl :
..
H
..
O
..
:O :
H
H
O : Na+
sodium carbonate
H
Na+
..
..
:O
C
:O :
..
..
O
sodium bicarbonate
..
..
H
H
O+
H
H
H
..
: Cl :
..
H
C
..O
..
O
..
..
..
O
carbonic acid
H
Recall from the rules for writing Lewis structures that acidic hydrogens are attached to the
oxygen, not the central atom in oxyacids.
Exercises:
1. Explain why Na2CO3 is a Lewis base.
2. Explain why H3O+ is a Lewis acid.
3. Explain why Na2CO3 is a Bronsted base.
4. Explain why H3O+ is a Bronsted acid.
5. Determine the formal charge on the carbon and oxygen atoms in HCO3(Carbon is 0. One oxygen is –1, the other two have a formal charge of 0).
6. Draw Lewis structures and show the mechanism (with curved arrows) of the
reaction of perchloric acid (HClO4) plus H2O. Perchloric acid is a stronger acid
than HCl and it reacts completely with water producing perchlorate ion (ClO4-) and
hydronium ion (H3O+).
7. Sodium amide (NaNH2) is a stronger base than sodium hydroxide. Draw Lewis
structures to show the mechanism of its reaction with acetic acid (CH3COOH)
forming sodium acetate (Na+ CH3COO-) and ammonia (NH3).
Having become familiar with Lewis structures, now is good time to learn the Lewis
structures of the generic functional groups shown on page 1 in this section. Note that ‘R’
stand for ‘the Rest of the molecule’ and is usually interpreted as some alkyl group such as
methyl (CH3-), ethyl (CH3CH2-), etc.
Look closely at the Lewis structure of nitro compounds. Notice that the overall charge of
a nitro compound is zero but N is +1 while one O is –1.
Exercise: Draw two Lewis structures of nitric acid (HNO3)..
ORGANIC CHEMISTRY INTRO
21
H
..
:O :
..
..
:O
N
+
:O :
..
O
..
..
O
..
H
N
+
..
O
..
H
Two Lewis structures can be drawn for nitric acid.
One oxygen must have only one bond because
N (a 2nd period element) cannot have more than 4 bonds.
Either of two oxgyens can share the negative charge.
The two structures are called 'resonance hybrids'.
A double headed arrow is drawn between resonance hybrids
and the hybrids are enclosed in square braces.
The Lewis structure of nitric acid shows permanent charge separation within the molecule.
Two resonance structures are possible (above). In resonance structures, the positions of all
atoms are unchanged but electron arrangements differ.
We will most commonly see resonance when we study reaction mechanisms. Study the
energy diagram for the combustion of methane.
REACTION ENERGY DIAGRAM
Transition
state
ENERGY
Activation
Energy (G*)
CH4 + 2 O2
Energy released
(Exotherm, Go <0)
CO2 + 2 H2O
Reactants
Products
H
H
C.
H
+ .H +
H
C
H
+
2
..
O
..
H
Reactants
..
..
. O. + . O.
..
..
The free radicals formed
represent a highly reactive
transition state.
Transition state
Energy must be applied (activation energy)
to break C to H bonds and O to O bonds.

This is the activation energy.
H
2
Energy is given off (exothermic)
as the free radical species in the
transition state combine to form
the products.
..
O
..
..
O
..
C
..
O
..
+ 2
.. ..
O
H
H
Products
Free radicals (unpaired electrons) are very reactive (high energy). If resonance could occur
such that the free electrons could move between several atoms (instead of just one), then the
free radical transition state would be less reactive and the activation energy would be lowered
and the reaction would proceed more quickly.
ORGANIC CHEMISTRY INTRO
22
RESONANCE:
As with free radicals, electric charges in an organic molecule indicates a high energy and
high instability in the compound. Reaction intermediates (or transition states) are often charged and
thus represent the highest energy level of a chemical reaction. The higher the energy of a transition
state, the greater is the activation energy of a reaction and hence the slower the reaction proceeds.
Resonance is a means by which electric charge on an atom is spread (delocalized) over several
atoms in a molecule. The effect of charge delocalization in a reaction's transition state is to decrease
the activation energy (G*) and speed up a reaction. The chemical reactivity of many conjugated
unsaturated compounds is enhanced by their ability to delocalized charge via resonance in a transition
state. More resonance structures indicates greater charge delocalization and lower G*.
The student must become skilled at drawing resonance structures of conjugated organic
molecules as this is often a way of understanding a reaction as well as a way of successfully
predicting reaction products.
Rules for Resonance:
1. Electron(s) may move from an atom to an adjacent -bond
(forming a -bond)
2. Electron(s) may move from a -bond to an adjacent atom
(eliminating a  bond)
..
..
X
+
X
+
X
..
..
X
X
3. A  bond may flop from one side of an atom to another
side, i.e., from atop one  bond to atop an adjacent  bond on the same atom.
X
Use a full arrowhead for showing movement of a pair of electrons and a half-arrowhead for
movement of a single electron (a free radical):
Benzyl Cation:
+
C
C
C
C
C
C
C
C
C
C
Benzyl Anion:
C
Benzyl Free Radical:
C
ORGANIC CHEMISTRY INTRO
23
VALENCE BOND THEORY & VSEPR:
Valence bond theory and Valence Shell Electron Pair Repulsion theory (VSEPR) are simple
but effective models for understanding the bonding that occurs in organic compounds.
 Covalent bonds are usually formed between two atoms by the overlap of 2 half-filled
(singly-occupied) orbitals of the bonding atoms. For example, two hydrogen atoms form
a covalent bond by the overlap of their half-filled 1s atomic orbitals.
H
1s1
+
H
H
H
H
H
1s1
Valence Bond theory predicts that
two, half-filled 1s atomic orbitals
of hydrogen overlap to form a single
(sigma) bond.
This H 2 molecule is stable (low
in energy) because the electrons
spend most of their time between
H nuclei-drawing them together.
Each H atom in
H2 is isoelectronic
with Helium
 Carbon is in group IVA of the periodic table and therefore has 4 valence (bonding)
electrons, i.e., 4 electrons in its highest Principal Energy Level (PEL). Its ground state
electron configuration is given as …
1s2 2s2 2px1 2py1 2pz0
4 valence electrons
 Carbon normally forms 4 covalent bonds using its 4 valence shell electrons, however, its
ground state electron configuration is not suitable for this because the 2s orbital is full and
the 2pz orbital is empty. In a process called ‘hybridization’, one of the 2s electrons is
promoted to the vacant 2pz orbital producing 4 half-filled orbitals. These orbitals alter
their shape in order to achieve the best possible overlap with orbitals from other atoms
and also to move as far away from each other as possible.
1s2 2s2 2px1 2py1 2pz0
hybridization
1s2 2s1 2px1 2py1 2pz1
 Only the outer, valence electrons are involved in bonding and thus the inner shells of
electrons will not be considered further.
 The vast majority of organic compounds are formed from just a few elements, i.e., C, N,
O, P, S and the halogens. We will look at their hybridization states in the following pages.
ORGANIC CHEMISTRY INTRO
24
BONDING IN CARBON COMPOUNDS
6
C
orbital shape
Group 4A
4 valence e’s.
Covalence = 4
2py1
2px1
2s2
ground state
valence electron
configuration
A change (hybridization) occurs
to the orbital shape and electronic
configuration to facilitate bonding.
The 3 hybridizations occurring in
cabon are shown below.
2pz0
sp 2 hybridized
(triangular planar)
sp hybridized
(linear)
Carbon forms a double bond
Carbon forms 1 triple or 2 double bonds
sp 3 hybridized (tetrahedral)
Carbon forms 4 single bonds
The arrangement of the 4 atomic
orbitals and the electronic
configuration in the carbon atom
are not suitable to form 4 bonds.
180º
90º
109.5º
or
C
120º
C
C
C
orbital
shape
(sp2 orbitals x 3) + (p x 1)
hybridized
orbitals
(sp orbitals x 2) + (p x 2)
(sp3 orbitals x 4)
electron
configuration
(3  bonds) + (1  bond)
(forms 4  bonds)
alkanes
e.g., ethane (C 2H6)
bonds
C
C
H
H
H
H
H
C
C
H
H
H
C
H
C
H
H
H
H
H
ORGANIC CHEMISTRY INTRO
H
H
C
H
H
H
H
C
C
H
H
H
H
H
H
H
H
H
C
alkynes
e.g., acetylene (C 2H2)
H
H
H
H
(2  bonds) + (2  bonds)
alkenes and arenes
e.g., ethene and benzene
(C 2H4)
(C 6H6)
H
H
C
C
H
also CO2
O C O
25
BONDING IN NITROGEN COMPOUNDS
Group 5A
7
The shape and orientation of the
4 atomic orbitals in the nitrogen
atom are not suitable for forming
3 (or 4) bonds.
N
orbital shape
5 valence e’s.
Covalence = 3
2s2
ground state
valence electron
configuration
2py1
2px1
A change (hybridization) occurs
to the orbital shape to facilitate
bonding. The 3 hybridizations
occurring in nitrogen are shown
below.
2pz1
sp 3 hybridized
sp 2 hybridized
sp hybridized
Nitrogen forms 3 or 4 single bonds
Nitrogen forms a double bond
Nitrogen forms 1 triple bond
......
..
..
90º
or
.:.
N
N
107º
orbital
shape
hybridized
orbitals
180º
N
120º
..
....
..
(sp 2 orbitals x 3) + (p x 1)
(sp3 orbitals x 4)
(sp orbitals x 2) + (p x 2)
electron
configuration
bonds
N
..
1 lone pair
+
3  bonds
....
or
H
N
N
H
H
(1 lone pair) + (2  bonds) + (1  bond)
(4  bonds)
Cl H
ammonium
chloride
NH3
NH4 Cl
+ -
nitriles
e.g., ethanenitrile
N
CH3
C
..:
N
(1 lone pair) + (1  bond) + (2  bonds)
..
N
amines
ammonia
ORGANIC CHEMISTRY INTRO
..
+
H
H
H
azo compounds
e.g., (trans azobenzene)
.:.
N
....
N
:N
C
CH3
26
BONDING IN OXYGEN COMPOUNDS
8
The shape and orientation of the
4 atomic orbitals in the oxygen
atom are not optimal for forming
2 (or 3) bonds.
O
orbital shape
Group 6A
6 valence e’s.
Covalence = 2
2s2
ground state
valence electron
configuration
2px2
2py1
A change (hybridization) occurs
to the orbital shapes to facilitate
bonding. The 2 hybridizations
occurring in oxygen are shown
below.
2pz1
sp 3 hybridized
sp 2 hybridized
Oxygen forms 2 single bonds
Oxygen forms a double bond
O
..
..
hybridized
orbitals
.:.
.... O
....
..
orbital
shape
90º
105º
or
..
..
O
:
120º
..
(sp 2 orbitals x 3) + (p x 1)
(sp3 orbitals x 4)
electron
configuration
bonds
2 lone pair
+
2  bonds
or
e.g., methanol (CH 3OH)
CH3
.... O
....
e.g., acetone (CH3CCH3)
+
hydronium ion (H3O )
H
.. O
H
ORGANIC CHEMISTRY INTRO
(2 lone pair) + (1  bonds) + (1  bond)
O
1 lone pair
+
3  bonds
H
..:
O
..:
C
+
H
H3C
: O:
C
H3C
CH3
CH3
27
 Note that a double bond is made of one  and one  bond.
 Note that a triple bond is made of one  and two  bonds.
 Halogens (groups VIIA elements) generally form only one  bond in organic compounds.
They do not reshape their orbitals (hybridize) when they bond.
The shape and orientation of the
4 atomic orbitals in the halogens
are adequate for forming one
single bond.
17
Cl
orbital shape
ground state
3s2
valence electron
configuration
3px2
3py2
(3 lone pairs and 1  bond)
H
Cl
Hybridization does not occur
when halogens form single
bonds.
3pz1
C
H
H
..
: Cl
..
CH3
methyl chloride
(chloromethane)

Hydrogen, like the halogens, does not hybridize its 1s orbital when bonding.

Silicon, like carbon, is a group 4A element with 4 valence electrons. As expected, silicon forms
sp3 hybridized tetrahedral compounds with 4 substituents. Simple examples include silicon
tetrabromide (SiBr4) and tetramethylsilane [(CH3)4Si]. Silicon forms a few compounds in which
it has double bonds, e.g., H2Si=CH2. However, silicon's large size makes p-orbital overlap for 
bonds less effective than in carbon compounds. Unlike 2nd period elements which cannot
accommodate more than 8 electrons in their valence orbitals, Si, a 3rd period element can expand
its valence shell to accommodate 10 electrons (sp3d hybridized – 5 substituents, e.g., SiF5-) and
even 12 electrons (sp3d2 hybridized – 6 substituents, e.g., fluorosilicic acid, H2SiF6).

Phosphorus, like nitrogen, is a group 5A element with 5 valence electrons. As expected,
phosphorus forms sp3 hybridized compounds with 3 substituents. Simple examples include
phosphorus tribromide (PBr3) and trimethylphosphine [(CH3)3P]. Phosphorus forms some
compounds in which it has double bonds to oxygen, e.g., phosphoric acid (H3PO4). However,
phosphorus’ large size makes p-orbital overlap for  bonds less effective than in C or N
compounds. Like other 3rd period elements, phosphorus can be bonded to 4, 5, and 6 atoms.
e.g., phosphorus oxychloride (Cl3P=O), phosphorus dibromide trichloride (PBr2Cl3), and
phosphorus hexafluoride anion (PF6-).

Sulfur, like oxygen, is a group 6A element with 6 valence electrons. As expected, sulfur forms
sp3 hybridized compounds with 2 substituents. Simple examples include dimethyl sulfide
[(CH3)2S] and methyl mercaptan (methane thiol) (CH3SH). Sulfur can form bonds to three
(H2SO3), four (H2SO4), five (SOF4) and six atoms (SF6).
 A methyl cation has an sp2 hybridized carbocation with a vacant p orbital. A methyl radical has
an sp2 hybridized carbon atom with a ½-filled 2p orbital. A methyl anion contains an sp3
hybridized carbanion with a lone pair in one of its sp3 orbitals. Draw them.
ORGANIC CHEMISTRY INTRO
27
In saturated compounds, all atoms have only  bonds, whereas in unsaturated compounds,
one or more  bonds are present.
 Conjugated unsaturation occurs when alternating  and  bonds are present. In such
compounds, all p-orbitals in conjugated  bonds overlap.
1,3-butadiene
CH2
CH
H
CH2
H
H
O
C
C
C
CH
H
C
H
H
H
H
H
C
C
C
2-cylcopentenone
H
C
H
H
 Isolated unsaturation occurs when  bonds are separated by more than one  bond.
In such compounds, p-orbitals of one  bond cannot overlap with p-orbitals of other 
bonds.
CH2
1,4-pentadiene
CH
CH2
H
O
C
H
C
C
H
C
H
H
H
H
H
H
3-cylcopentenone
C
C
H
C
C
H
CH2
H
H
C
CH
H
C
H
H
 Cumulated unsaturation describes immediately adjacent unsaturation. Cumulated
carbon-to-carbon compounds are not very stable and are rarely encountered.
1,2-butadiene
ORGANIC CHEMISTRY INTRO
H2C
C
CH
CH3
28
HYBRIDIZATION STATE IS BASED ON THE NUMBER OF REGIONS OF ELECTRON DENSITY
Be
sp
H
Be
C
B
..
O
..
H
N
..
O
..
C
C
H
C
H
H
C
H
H
C
H
..
HO
..
..
O
..
..
O
..
..
O H
..
-
H
+
N
C
B
H
H
..
OH
..
..
O
H
H
H
C
H
H
H
..
O
N
H
H
H
H
H
NH4+Cl-
Si
+
+
H
Al
:
H
H
H
Na+BH4-
Mg
:
..
H
O
H
H
sp3
:
-: .. :
O
N
-
H
..
F
:
..
N
..
N
+
sodium azide
H
..
O.. H
..
O
..
Na+
N
N
H
H
C
..
N
..
C
B
sp2
H
O
hydronium ion
P
Cl
S
sp
H
H
H
Al
sp2
..
H
P
C Si
H
H
H
CH2
:O :
:O :
..
S
H3C
S
:O :
:O:
:O :
:O :
:O :
S
:O :
-
H
OH
sp3
HO
HO
Al
H
H
H
+
Li
AlH4-
HO
HO
Si
HO
HO
P
Cl
..
..O
HO
:O :
..
Cl
S
P
Cl
..
O
..
:O :
OH
OH
Cl
..
Cl :
..
HO
HO
Cl
Cl
ORGANIC CHEMISTRY INTRO
..
.O
.
..
..O
:
:
HO
:O :
..
Cl
S
:
:
HO
29
:
Study the following table. In the last 3 columns Lewis structures are drawn as if the atoms
were bonded. Learn these names and structures and identify their hybridization states.
Lewis
Symbol
#
valence
#
#
bonds
unshared
e- 's
e- 's
+ 1 F.C.
neutral
- 1 F.C.
B
B
3
3
0
-
B
none
boride ion
..
C
+
C-
C
C
4
4
0
carbonium
ion
N
N
5
3
2
carbide ion
2
4
N
N
:
..
..
O
:
F
7
1
6
Cl +
:
7
1
6
..
.. F
..
..
F
..
:
..
fluoronium ion
Cl
..
an oxide ion
..
F+
O-
:
oxonium
ion
:
-
nitride ion
O+
6
..
nitronium
ion
..
O
..
+
..
chloronium ion
unhybridized
..
..
.. Cl
unhybridized
unhybridized
fluoride ion
.. ..
Cl :
..
unhybridized
chloride ion
Bromine and iodine are analogous to fluorine and chlorine. Draw the structures of
bromonium and iodinium cations, bromide and iodide anions, and bromine and iodine.
ORGANIC CHEMISTRY INTRO
30
HOMOLYTIC AND HETEROLYTIC PROCESSES:
The amount of energy required to dissociate a hydrogen molecule into 2 H atoms is called its
‘bond (dissociation) energy and is quite large (104 kcal/mol)
Bond breaking: There are 2 ways in which a covalent, 2-electron bond can break: in an
electronically symmetrical way so that one electron remains with each product fragment or
in an electronically unsymmetrical way so that both electrons remain in one product
fragment, leaving the other fragment with an empty orbital.
Symmetrical cleavage is called a homolytic process and the unsymmetrical cleavage is
called a heterolytic process.
A:B  A + B
homolytic cleavage (radical reaction)
A:B  A+ + B:-
heterolytic cleavage (polar reaction)
Bond making: Conversely, there are 2 ways in which a covalent 2-electron bond can form:
in an electronically symmetrical (homogenic) way when 1 electron is donated to the new
bond by each reactant, or in an electronically unsymmetrical (heterogenic) way when both
bonding electrons are donated to the new bond by one reactant.
A + B  A:B
homogenic bond making (radical reaction)
A+ + B:-  A:B
heterogenic bond making (polar reaction)
Processes that involve symmetrical bond breaking and making are called radical reactions. A
radical is an atom or group of atoms that contains one or more unpaired valence electrons and
thus has one or more orbitals with only 1 electron...
Cl2  Cl  + Cl 
(2 chlorine radicals/atoms)
O2   O  +  O 
(2 oxygen diradicals/atoms)
CH4  CH3  + H 
Radicals are very reactive species and generally only exist for a short time.
ORGANIC CHEMISTRY INTRO
31
HOMOLYTIC BOND DISSOCIATION ENERGY:
Bond dissociation energy (D) is the amount of energy consumed when a bond is broken
(cleaved). Homolytic cleavage produces free radicals:
AB  A. + B.
Homolytic bond breaking is always endothermic but varies widely from I2 (36 kcal/mol) to HF
(135 kcal/mol). Some values are listed in the following tables ....
Homolytic Bond dissociation Energies (D) of Various A  B Bonds in kcal/mol
B in A  B
A in A  B
group
H
F
 Cl
 Br
I
 OH
 NH2
 CH3
H
---
104
135
103
87
71
119
107
104
CH3 
methyl
105
110
85
71
57
93
80
90
CH3CH2 
1
98
107
81
69
53
92
77
89
CH3CH2CH2 
1
98
107
81
69
53
91
78
85
(CH3)2CH 
2
94.5
106
81
68
53
92
93
84
(CH3)3C 
3
92
110
79
63
52
93
93
80
C6H5 CH2
benzyl
88
72
55
45
70
75
CH2 CHCH2 
allyl
86
69
55
41
72
72
CH2 CH 
vinyl
108
84
C6H5 
phenyl
111
116
96
92
81
65
103
91
101
Homolytic Bond Dissociation Energies (D) of Miscellaneous Bonds
compound
D (kcal/mol)
compound
D (kcal/mol)
CH3  OCH3
80
CH2 CH2 (both bonds)
165
CH3  OH
93
CH2 CH2 ( bond only)
62
CH3O  H
104
C N (both bonds)
143
HO  H
119
C  C (all bonds)
~ 200
CH3O OCH3
36
C  N (all bonds)
~ 213
N  C  H
130
C  C  H
132
C  H (aldehydes)
87
C O (aldehydes & ketones)
176-179
 Others: H2 =104, F2 = 38, Cl2 = 58, Br2 = 46, I2 = 36, RS  H  80, Si  Si  53


Bond dissociation energies are often measured in the vapor phase, which ignores other effects
such as salvation, by the solvent. Despite these limitations, bond dissociation energies provide
a set of relative bond strengths that are helpful in understanding many reactions.
Note particularly the trend in the dissociation energies for the C  H bond. There is a
decrease with the progression from methyl to 1 to 2 to 3 C's. The same trend is seen in the
dissociation of C  X and C  C bonds. This indicates an increasing stability of more
substituted free radicals. In addition, conjugation stabilizes radicals due to resonance, i.e.,
allylic and benzylic radicals are unusually stable.
ORGANIC CHEMISTRY INTRO
32
HETEROLYTIC BOND DISSOCIATION ENERGY:
Heterolytic cleavage produces ions:
A  B  A+ + B:Heterolytic bond breaking is more endothermic than homolytic. The separation of
opposite electric charges requires additional energy. Some values are listed ....
Heterolytic Bond dissociation Energies of Various A  B Bonds in kcal/mol
B in A  B
A in A  B
group
H
F
 Cl
 Br
I
 OH
H
---
401
370
334
324
315
390
CH3 
methyl
218
256
227
219
212
274
CH3CH2 
1
182
191
184
176
242
CH3CH2CH2 
1
185
178
171
235
(CH3)2CH 
2
163
170
164
156
222
(CH3)3C 
3
148
157
149
140
208
C6H5CH2
benzyl
166
157
149
215
CH2 CHCH2 
allyl
173
165
159
223
CH2 CH 
vinyl
207
200
194
C6H5 
phenyl
219
210
202
275
 Note that the dissociation energies for C  X bonds generally decrease with the
progression from methyl to 1 to 2 to 3. The stability of carbocations is increased by
alkyl substitution since alkyl groups are inductively electron donating.
 In addition, carbocation stability is increased by resonance. Thus allylic and benzylic
carbocations are unusually stable.
 Vinyl and phenyl compounds dissociate less readily. No resonance occurs in vinyl
and phenyl carbocations. Furthermore, the terminal carbon in vinyl and phenyl
compounds are sp2 hybridized (more electronegative – holds the bonded electrons
more strongly) compared to all others listed, in which the terminal carbon is sp3
hybridized (less electronegative – holds the bonded electrons less strongly).
 The relative stability of carbocations is as follows...
+
CH2
b enzyl
R
R
C+
>
2º
R
R
3º
R
R
C+
>
H2C
C+
>
H2C
CH + >
H
H
allyl
H
H
CH
CH2 +
1º
+
>
H
C+
H
vinyl
phenyl
methyl
increasing C+ stab ility
ORGANIC CHEMISTRY INTRO
33
 In the case of heterolytic dissociation, removal of a halogen atom produces a carbocation
(C+). The relative stabilities of some carbocations is shown.
increasing carbocation stability
more
stable
CH3
CH3
H3C
C+
CH3
tertiary
3º
CH3
C+
H3C
H
C+
H
H
>
primary
1º
C+
H
H
secondary
2º
>
less
stable
>
H
methyl
 Two reasons proposed for the increased stability of more highly substituted carbocations
are inductive effects and hyperconjugation.
 Inductive effects refer to the shift of electron density through a sigma bond. Methyl
groups (and other alkyl groups) which are directly bonded to electron-deficient atoms
(e.g., C+) shift electron density to that site and hence are found to be weak electron
donors. Alkyl groups stabilize C+’s by inductive effects.
CH3
H3C
Inductive effects:
Alkyl groups donate (shift) electron
density through sigma bonds to
electron deficient atoms.
This stabilizes the carbocation.
C+
CH3

Hyperconjugation refers to the shift of electron density by the partial overlap of
C-H sigma orbitals with a vacant p orbital. A carbocation is sp2 hybridized. Its
vacant p-orbital extends above and below the plane of the sigma bonds at right angles.
vacant p orbital
of a carbocation
overlap (hyperconjugation)
sp 2
hybridized
carbocation
..
C
+
C
H
Csp 3-Hs
sigma bond
orbital
H
H
HYPERCONJUGATION

Note that the order of stability of free radicals is also 3º > 2º > 1º > methyl due to
hyperconjugation with neighboring alkyl groups.
ORGANIC CHEMISTRY INTRO
34
ACIDS and BASES:
Arrhenius Theory: In 1884, Svante Arrhenius defined acids and bases as follows...
 An Arrhenius acid contains H and produces H+ (or H3O+) in water (e.g., HCl).
 An Arrhenius base contains the OH group and produces OH- in water (e.g., NaOH).
 Neutralization is the combination of H+ (or H3O+) and OH- ions to form HOH.
The Arrhenius theory explains reactions of protonic acids with metal hydroxides
(hydroxy bases), e.g., HCl (aq) + NaOH  HOH + NaCl
Bronsted-Lowry Theory: In 1923, J. N. Bronsted and T. M. Lowry defined acids and bases as
 A Bronsted acid is a proton (H+) donor
 A Bronsted base is a proton acceptor.
 Neutralization was defined as the transfer of a proton from a proton donor (acid) to a proton
acceptor (base).
The Bronsted Theory is broader than the Arrhenius and includes non aqueous reactions,
e.g., HCl (g) + :NH3 (g)  NH4Cl (s)
e.g., H-CC-H + Na+NH2-  H-CC: - Na+ + :NH3
Conjugate acid / base pairs are defined as species that differ by a proton.
Na+ OH- + CH3COOH
base

acid
HOH +
Na+ CH3COO-
conj. acid
conj. base
Strength of Acids and Bases:
 A very strong acid (HCl) has a very weak conjugate base (Cl-), i.e., if the acid releases a
proton readily, its conjugate base will not attract a proton strongly.
 A very strong base (NaOH) has a very weak conjugate acid (H2O), i.e., if the base accepts a
proton readily, its conjugate acid will not readily give up the proton.
 Acid strength is quantified by Ka and pKa ....
HA + HOH  H3O+ + A-
and the degree of dissociation is given by...
[H 3 O + ] [A - ]
but in dilute soln., [HOH] is a constant (55.5 M) and this is
[ HA] [HOH]
combined with the equilibrium constant (Keq) giving another constant, Ka, which is a
[ H 3O  ] [A - ]
measure of the strength of an acid... Ka = Keq  [HOH] = Keq  55.5 =
[ HA ]
K eq =
and note that:
pKa = -log10 Ka

A strong acid (H3O+) has a very weak conjugate base (H2O)

Similarly, a strong base (OH-) has a very weak conjugate acid (H2O).
ORGANIC CHEMISTRY INTRO
35
 Similarly for bases: B + HOH  BH+ + OH- and the degree of dissociation is ...
[BH + ] [OH - ]
but in dilute soln., [HOH] is a constant (55.5 M) and this is combined with
Keq =
[ B] [HOH]
the equilibrium constant (Keq) giving another constant, Kb, which is a measure of the strength
[ BH  ] [OH - ]
of a base... Kb = Keq  [HOH] = Keq  55.5 =
[ B]
and pKb = -log10 Kb
 The relationships that apply to conjugate acid base pairs are ...
Ka  Kb = 10-14
or
pKa + pKb = 14
For example, CH3COOH has pKa = 4.7. Its conjugate base, CH3COO- has pKb = (14 – 4.7) = 9.3

It is important that students of organic chemistry be familiar with pK values of acids and
bases. Not only do they provide a quantitative measure of the strength of an acid or base but,
as we will soon see, these values can also combined to predict the extent of reaction of
acid/base reactions

In the following tables, the approximate strengths versus pK value is given.
pKa
<1
1-5
5 - 15
> 15
acid strength
strong
moderate
weak
very weak
example
H2SO4
H3PO4
H2CO3
HOH
pKb
<1
1-5
5 - 15
> 15
base strength
strong
moderate
weak
very weak
example
NaOH
Na3PO4
NaHCO3
HOH
Consider Acids Stronger Than H3O+: H3O+ has a pKa = -1.74. Stronger acids such as HCl,
with pKa's lower than -1.74, protonate HOH producing H3O+ (hydronium ion).
HCl
pKa = -7
+
HOH

pKb = 15.74
H3O+
pKa = -1.74
Cl-
+
pKb = 21
Thus H3O+ is the strongest acid which can exist in water. Any stronger acid will be 'leveled'
(reduced) in strength to pKa = -1.74 by the solvent water. Water is said to be a leveling solvent.
Consider Bases Stronger Than OH-: OH- has a pKb = -1.74. Stronger bases such as NaNH2,
with pKb's lower than -1.74, are protonated by HOH producing OH-.
NaNH2
pKb = -21
+
HOH
pKa = 15.74

NH3
pKa = 35
+
NaOH
pKb = -1.74
Thus OH- is the strongest base that can exist in water. Any stronger base will be 'leveled'
(reduced) in strength to pKb = -1.74 by the solvent water.
ORGANIC CHEMISTRY INTRO
36
ACIDITY & BASICITY OF WATER
Water is both weakly acidic and weakly basic, i.e., ‘amphoteric’ or ‘amphiprotic’.

Water acts as a base by accepting protons from acids...
H-O-H

+
H-Cl
 H3O
+ Cl
-
Water acts as an acid by donating protons to bases ...
H-O-H + :NH3  NH4

+
+
-
+ OH
+
Water undergoes self-ionization, or ‘autopyrolysis’ to form equal concentrations of H3O and
OH- ions ...
H-O-H + H-O-H  H3O
+
-
+ OH
The extent of self-ionization of water is very small. At room temperature (25C) the
+
_
-7
+
_
-7
concentrations of H3O and OH are 1.0  10 moles/L each, i.e., [H3O ]= [OH ] = 10
M

The equilibrium constant, Keq, for this reaction is given by ...
Keq =

[H 3O+ ] [OH - ] [107 ] [10-7 ]
1014


[HOH] [HOH] [55.5] [55.5] [101.74 ] [101.74 ]
-14
where 10 is ‘KW’, the ion product
constant for water
As for all other acids and bases, Ka and Kb for water is calculated as follows ...
Ka = Kb = Keq  [10
1.74
]=
1014
 1015.74
1.74
10
and thus pKa = pKb = 15.74 for water.
THREE CLASSES OF SOLVENTS
1. Amphiprotic solvents: Other ionizable solvents are also amphiprotic, e.g., methanol, ethanol,
acetic acid, and ammonia. The autopyrolysis constants for some are listed.
Solvent
- log Ks or (pKw)
water
14
acetic acid
14.5
ethylenediamine
15.3
methanol
16.7
ethanol
19.1
2. Non-polar Aprotic solvents: These have no protons and are not polar. They are nonionizable
and inert, i.e., neither acidic nor basic, e.g., benzene, CCl4.
3. Polar Aprotic (Basic but not Acidic): These are also nonionizable solvents but can accept a
proton because of the presence of atoms such as O or N which have lone pairs of electrons,
e.g., ether, dioxane, ketones, and pyridine. The are no known examples of solvents that are
acidic but not basic.
ORGANIC CHEMISTRY INTRO
37
Predicting the extent of a reaction using the equation:
(pKeq = pKa + pKb - 14)
Consider an acid in HOH:
Consider a base in HOH:
HA + HOH  H3O+ + AB + HOH  BH+ + OH[H 3O + ] [A - ]
[BH + ] [OH - ]
Ka = Keq  [55.5] =
Kb = Keq  [55.5] =
[HA]
[B]
Now consider an acid (HA) reacting with a base (B):
HA + B  BH+ + A-
[BH + ] [A - ]
Keq =
[HA] [B]
By manipulation of these expressions we derive an expression for predicting the extent of
reaction ...
 [H O + ] [A - ]  [BH + ] [OH - ]  [A - ] [BH + ]
+
Ka  Kb =  3
 
 
 [H 3 O ] [OH ]
[HA]
[ B]
  [ HA] [B] 

 


or
Ka  Kb =
( Keq )

( Kw )
or
Ka  Kb =
( Keq )

( 10-14 )
Taking the log of both sides and then multiplying the equation by (-1) yields ...
pKa + pKb =
pKeq
+
14
Finally rearranging this we have a simple expression for calculating the extent of an acid-base
reaction ...
pKeq = pKa + pKb - 14
The equilibrium constant (Keq) can then be determined (if desired) by taking the negative antilog
of pKeq calculated above, however it is easier to convert from pKeq to extent of reaction
graphically (following).
One can readily get a sense for extent of reaction from basic
thermodynamics …
Keq = 1 (pKeq = 0) means that the extent of a reaction = 50%
A reaction is said to go to completion when Keq  103 or pKeq  -3
These calculations assume that reagents are 100% pure (not dilute) and yet they work
quite well as estimates even in dilute aqueous solutions. For example, Pasto, et. al. state
(p. 274) that "compounds containing acidic functional groups with pKa's of less than ca.
12 will dissolve in dilute aqueous (5%) NaOH".
HA + NaOH  Na+ A- + HOH
We can understand this by calculating pKeq, i.e., [12 + (-1.74) - 14] = -3.74 = pKeq or
Keq  5500 (quite favorable even in dilute aqueous solution)
On p.275, the same author states that "only acids with pKa's < 6 will dissolve in dilute (5%)
NaHCO3" and we could also have predicted this because pKeq = [ 6 + 7.6 - 14] = -0.4 or
Keq = 2.5 Using this equation, determine if 5% NaHCO3 will dissolve a) acetic acid, b) phenol.
Note: in order to quantitatively titrate an acid base pair in aqueous media, pKeq must be  -8.
ORGANIC CHEMISTRY INTRO
38
Determine pKeq (the
pKeq = pKa + pKb –14.
negative
logarithm
of
the
equilibrium
constant,
Keq)
using
the
formula:
Read the graph (pKeq vs. Extent of Reaction in %).
Note: This only works for Acid/Base reactions.
pKeq vs. Extent of Reaction (%)
[pKeq = pKa + pKb - 14]
pKeq
20
10
0
-10 0
10
20
30
40
50
60
70
80
90 100
-20
Extent of Reaction (%)
ORGANIC CHEMISTRY INTRO
39
More on Predicting Acid / Base Reactions: In addition to calculating pKeq we can also predict
whether an acid/base reaction will occur by making use of a few simple principles. Let's begin
with a familiar example which we know reacts essentially to completion ...
CH3COOH
+
moderate
acid
4.7
NaOH 
HOH
strong
base
-1.74
weak
acid
15.74
+
Na+ CH3COOweak
base
9.3
 (pKa or pKb)
Note the pK values. In general, an acid and base will spontaneously react only if the reaction
products are a weaker acid and a weaker base than the reagent acid and reagent base.
 A stronger acid will donate H+ to a base whose conjugate acid is weaker (higher pKa) or

An acid with a lower pKa will donate H+ to a base whose conjugate acid has a higher pKa.
Assign pK values to all species in the following reactions and predict which of the following
reactions will proceed as shown ...
CH3CH2O-
+
CH3COOH

CH3CH2OH
H2N-
+
CH3CH2OH

NH3
+
CH3CH2O-
H-CC-H
+
NaNH2

NH3
+
H-CC:-Na+
C6H5OH
+
OH-

C6H5O-
+
HOH
C5H5N
+
H3O+

C5H5NH+
+
HOH
ORGANIC CHEMISTRY INTRO
+
CH3COO-
40
acid
pKa
conjugate base
pKb
CH4
ca. 55
:CH3-
ca. -41
C2H4
44
-30
C6H6
NH3
43
35
:C2H3C6H5:NH2-
H2
C2H2
35
25
:H:C2H-
-21
-11
(CH3)2CO
(CH3)3COH
20
18
CH3COCH2(CH3)3CO-
-6
-4
CH3CHO
C2H5OH
17
16
CH2- CHO
C2H5O-
-3
-2
H2O
15.74
-1.74
HCO3HPO4-2
10.3
12.3
OHCO3-2
PO4-3
CH3NH3+
10.6
CH3NH2
3.4
C6H5OH
9.9
C6H5O-
4.1
HCN
NH4+
9.3
9.2
CNNH3
4.7
4.8
CH3SH
H2PO4-
8
7.2
CH3SHPO4-2
6
6.8
H2S
7.1
HS-
6.9
6.4
HCO3-
7.6
C6H5SH
C5H5NH+
6
5.3
C6H5SC5H5N
8
8.7
HN3
CH3COOH
4.7
4.7
N3CH3COO-
9.3
9.3
C6H5NH3+
4.6
C6H5NH2
9.4
C6H5COOH
4.0
COO-
10.0
HF
H3PO4
CF3COOH
3.2
2.1
0.2
CF3COO-
10.8
11.9
13.8
C2H5COH+NH2
HNO3
-1
-1.4
C2H5CONH2
NO3-
15
15.4
H3O+
-1.74
HOH
15.74
CH3CH2OH2+
-3.6
CH3CH2OH
17.6
H2SO4
-5
HSO4-
19
HCl
HBr
HI
-7
-8
-9
-10
-10
-12
ClBrIC2H5CN:
ClO4SbF6-
21
22
23
24
24
26
weakest
acids
Cannot
protonate
water
(pKa + pKb) = 14
for conjugate
acid-base pairs
Determined
in water.
Protonate
water
increasingly.
H2CO3
Protonate
water
completely
strongest
acids
C2H5CNH+
HClO4
HSbF6
ORGANIC CHEMISTRY INTRO
C6H5
FH2PO4-
-29
-21
strongest
bases
Completely
protonated
by water
3.7
1.7
Determined
in water.
Protonated
by water
increasingly.
Not
protonated
by water
weakest
bases
41
Determine the extent of the following reactions
(pKeq)
Base
(pKb)
Acid
(pKa)
Conjugate
Base (pKb)
Conjugate
Acid (pKa)
NaOH
CH3COOH

CH3COO-
HOH
NaHCO3
HCl (aq)

HOH
H2CO3
CH3NH2
H3 O+

HOH
CH3NH3+
C5H5N
H3 O+

HOH
C5H5NH+
C6H5NH2
H3 O+

HOH
C6H5NH3+
NaOH
C2H5OH

C2H5O-
HOH
C2H5OH
H3 O+

HOH
C2H5OH2+
NO2--NH2
H3 O+

HOH
NO2--NH3+
NaHCO3
HA
(6.0)
HA
(12.0)

A-
H2CO3

A-
HOH
NaHCO3
-OH

-O-
H2CO3
I-
CH4

CH3-
HI
B
7
HA
7

A-
BH+
B
8
H3 O+

HOH
BH+
NaOH
HA
8

A-
HOH
NaOH
-OH

-O- Na+
HOH
NaOH
Note: pKeq [forward rxn.] = -pKeq [reverse rxn.]
ORGANIC CHEMISTRY INTRO
and
(pKeq)
reverse rxn
(% forward rxn. + % reverse rxn = 100%)
42
Typical pKa values of functional groups
Typical pKb values of functional groups
pKa
group
example
pKb
group
example
55
alkane
CH4
24
nitrile
CH3CN:
45
alkene
CH2=CH2
23
acid chloride
ethanoyl
chloride
35
ammonia
NH3
22
aldehyde
ethanal
25
alkyne
CH CH
21.2
ketone
acetone
20
ketone
acetone
20.5
ester
ethyl acetate
17
amide
methanamide
20
carboxylic acid
acetic acid
17
aldehyde
ethanal
17.6
ether
C2H5OC2H5
15 - 18
alcohol
ethanol
16 - 18
alcohols
CH3CH2OH
15.74
water
H2O
15.74
water
H2O
10
phenol
-OH
15
aliphatic amides
ethanamide
9.3
hydrocyanic acid
HCN
9.3
acetate
CH3COO-
8
aliphatic thiol
CH3SH
9
aromatic amine
-NH2
7
hydrogen sulfide
H2S
7.6
bicarbonate
NaHCO3
6
aromatic thiol
-SH
6.0
aliphatic sulfide
CH3S-
5
carboxylic acid
CH3COOH
4.7
cyanide
CN-
-1
sulfonic acid
-SO3H
4.6
ammonia
NH3
-1
bisulfate
R-OSO3H
4
phenoxide
-O-
-1.74
hydronium ion
H3O+
3
aliphatic amine
CH3NH2
-2
protonated alcohol
CH3CH2OH2+
-1.74
hydroxide
OH-
-7
mineral acid
HCl
-2
alkoxide
CH3CH2O-
borohydride
NaBH4
aluminum hydride
LiAlH4
Grignard
CH3- +MgBr
-21
hydride
Li+ :H-
-21
sodium amide
Na+ :NH2-
-41
1 carbanion
Li+ :CH3-
sodium
Na0
ORGANIC CHEMISTRY INTRO
43
LEWIS ACIDS AND BASES:
A Lewis acid (electrophile, E+) is a substance that accepts an electron pair.
A Lewis base (nucleophile, Nu: -) is a substance that donates an electron pair.
As a result of this electron donation from a base to an acid, a bond is formed.
Lewis Acids:
+
 Lewis acids must have vacant, low energy orbitals, or a polar bond to H, so H can be lost.
 Lewis acids include, but are much broader than Bronsted-Lowry and Arrhenius acids. For
+3
example, metal cations, such as Al , are Lewis acids because they can accept a pair of
electrons when they form a bond to a base.
 In the same way, compounds of 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.
 Similarly, many transition metals salts, such as TiCl4, FeCl3, ZnCl2, and SnCl4 are Lewis
acids.
By means of Lewis structures, show the following acid-base reactions ...
1. HCl(g) + H2O
..
.. O
H
+
H
H
Lewis base
..
Cl
.. :
H
.. O +
H
H
+
.. _
: Cl :
..
Lewis acid
Chemical reactions invovle the transfer of electrons from electron donors to electron acceptors
Arrows show the electron transfer (from nucleophile to electrophile) (from Lewis base to Lewis acid).
Arrows do not show movement of molecules, atoms or ions!!!!!!!
The large arrow shows that one of the lone pairs of electrons on oxygen (nucleophile) is used to form a
covalent bond with the H atom in HCl (the electrophile). Notice that the oxygen atom in the hydronium ion
product has one less pair of electrons than the oxygen in HOH and notice that the oxygen atom has a + charge.
The small arrow shows that the hydrogen to chlorine bond breaks as the shared (bonding) pair of electrons
moves to the chlorine atom which thus becomes a chloride anion. Notice that the chloride anion has one more
lone pair of electrons than the chlorine atom in HCl.
2. BF3 + (CH3)2O (dimethyl ether)
3. AlCl3 + (CH3)3N (trimethylamine)
ORGANIC CHEMISTRY INTRO
44
Some Lewis Acids: (proton donors or electron pair acceptors):

strong acids:
H3O+, HCl, HBr, HNO3, H2SO4

weak acids
CH3COOH, CH3CH2OH, C6H5OH (phenol), H2O


cations
Cl+, Br+
compounds with vacant orbitals
AlCl3, BF3, TiCl4, FeCl3, ZnCl2
Lewis Bases:
Lewis bases have nonbonding electron pairs that can be donated to Lewis acids. Most nitrogencontaining and oxygen-containing organic compounds have 1 or 2 lone pairs of electrons,
respectively.
Examples of Bases: (proton acceptors or electron pair donors)
CH3CH2
CH3
..
O
..
CH3CH2
H
..
O
..
alcohol
ether
: O:
: O:
C
Cl
CH3
acid chloride
C
:O:
CH2CH3
amine


N
: O:
..
O
..
CH3
H
CH3
CH3
H
C
: O:
..
O
..
CH3
S
CH3
CH3
CH3
sulfide
C
..
NH2
amide
..
CH3
C
ketone
ester
carboxylic acid
CH2CH3
C
aldehyde
..
CH3CH2
CH3
: O:
H
..
..
O
H
water
Note that some compounds can act as both Lewis acids or Lewis bases, depending upon the
reaction conditions.
Alcohols and carboxylic acids act as acids by donating a proton but can also act as bases
when their oxygen atoms donate an electron pair (to very strong acids), and accept a proton.
Draw Lewis structures showing the following acid-base reactions.....
1. CH3OH + HBr
2. (CH3)2O + H2SO4

Note that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one
atom with a lone pair of electrons and can therefore react at more than one site. For example,
acetic acid can be protonated on the doubly-bonded or singly-bonded O atoms ...
ORGANIC CHEMISTRY INTRO
45
Some Factors Affecting Acidity:
HI > HBr > HCl > HF
       
decreasing acidity
Decreasing halogen size decreases acid strength, or conversely, increasing halogen size increases
acid strength. In the same vertical group of the periodic table, the larger atoms (higher molecular
weight), can disperse '-' charge over a larger region and thus add stability to the conjugate base.
As the stability of the conjugate base increases (i.e., weaker conjugate base), the greater is the
strength of the acid. I- is a weaker base than F- so HI is stronger than HF
Consider EN also: In the same row of the periodic table, the EN of the atom bonded to H
increases from left to right across the table. Since more EN atoms can carry a negative charge
more readily than a less EN atom, the acidity increases as shown....
(CH3)3C ---- H
   

(CH3)2N ---- H
CH3O ---- H
        
increasing acidity

EN
stability
C
- CH
3
<
<
N
- NH
2
<
<
O
OH -
<
<
F
F-
acidity
H ----CH3
<
H ----NH2
<
H ----OH
<
H ----F
basicity
- CH
>
- NH
>
OH -
>
F-
3
2
For organic acids, i.e., carboxylic acids, the proximity of an electronegative atom to an acidic
H affects the acidity of the compound ...
(CH2Cl)CH2CH2COOH
   

F ---- H
>

CH3(CHCl)CH2COOH
     
increasing acidity
> CH3CH2(CHCl)COOH
    
As previously mentioned in the section on resonance, delocalization of electrons in a
conjugate base increases the stability of the anion, therefore increasing the acidity of its
conjugate acid.....
CH3SOOOH
 
>
 
ORGANIC CHEMISTRY INTRO
CH3COOH
>
CH3CH2OH
        
decreasing acidity

46
Group 5A
Group 6A
Group 7A
CH4
NH3
H2O
HF
55
35
15.74
3.2
SiH4
PH3
H2S
HCl
35
27
7.1
-7
GeH4
AsH3
H2Se
HBr
25
23
3.8
-8
H2Te
HI
2.6
-9
EN
Group 4A
acidity, size
acidity, EN
pKa’s of Binary Acids

Note that the arrows point in the direction of an increase.

Increasing EN for an acid (H-B) increases its acidity. Increasing size of H-B also increases acidity.

Acidity of binary acids increases left to right across each period as the EN increases.
Acidity of binary acids increases down each group as the size of the heteroatom increases greatly, even though
EN is increasing up the group.
Basicity trends of conjugate bases are opposite those of acidity trends of acids. See below.


basicity
Group 4A
Group 5A
Group 6A
Group 7A
CH3-
NH2-
OH-
F-
-41
-21
-1.74
10.8
HS-
Cl-
6.9
21
Br
-
-
SiH3
PH2
-21
-13
-
-
GeH3
AsH2
HSe-
-11
-9
10.2
22
HTe-
I-
11.4
23
basicity
pKb’s of Conjugate Bases of Binary Acids
-
pKa’s of Ternary (Oxy-) Acids
Ternary acids (oxyacids) contain hydrogen-oxygen-hereroatom links and conform to the general formula HmXOn.
These compounds are acidic when X is a nonmetal or metalloid but are basic when X is metallic. With few
exceptions (such as phosphorus oxyacids), H-atoms are bonded exclusively to O-atoms. Several factors affect the
strength of an oxyacid.
1.
If there is more than one ionizable hydrogen, Ka1 > Ka2 > Ka3. The successive acidity constants differ by ca. 5
powers of 10, i.e., 105, e.g., H3PO4, pKa1 = 2.1, pKa2 = 7.2, pKa3 = 12.4
2.
Inductive and resonance effects are important. Acidity is greatest when the heteroatom is highly electronegative
and is increased by the presence of electron-withdrawing groups (e.g., -F or –CF3). Thus the acidity increases
left to right in the following series:
H3PO4 < H2SO4 < HClO4
HIO3 < HBrO3 < HClO3
CH3CO2H < FCH2CO2H < F2CHCO2H < CF3CO2H
3.
Acidity increases with increasing number of oxygens. This is also an inductive (and resonance) effect. The
Ka’s increase successively in the following series by factors of ca.10 5, i.e., pka’s decrease by ca. 5:
HClO < HClO2 < HClO3 < HClO4
ORGANIC CHEMISTRY INTRO
47

Electrophiles (E+) and Lewis acids are both electron pair acceptors (proton donors in the
binary acids shown below). They have similar periodic trends as seen in the following table.
Note that the arrows point in the direction of an increase.
acidity, E+

Group 4A
Group 5A
Group 6A
Group 7A
CH4
55
NH3
H2O
HF
35
15.74
3.2
SiH4
PH3
H2S
HCl
35
27
7.1
-7
GeH4
AsH3
H2Se
HBr
25
23
3.8
-8
H2Te
HI
2.6
-9
acidity, E+
pKa’s of Binary Acids
Nucleophiles (Nu:-) and Lewis bases are both electron pair donors, however their periodic
trends are opposite in vertical columns (as seen in the following table). Note that the arrows
point in the direction of an increase.
basicity, Nu:-
Group 4A
Group 6A
Group 7A
NH2
OH-
F-
-41
-21
-1.74
10.8
SiH3-
PH2-
HS-
Cl-
6.9
21
Br
-
-21
-13
CH3
-

Group 5A
-
-
GeH3
AsH2
HSe-
-11
-9
10.2
22
HTe-
I-
11.4
23
Nu: basicity
pKb’s of Conjugate Bases of Binary Acids
-
The size (polarizability) of nucleophiles has the largest effect on nucleophilicity. Larger,
more polarizable nucleophiles are more able to donate an electron pair (through distortion of
their large valence electron clouds).
ORGANIC CHEMISTRY INTRO
48
STRENGTH OF NUCLEOPHILES (NUCLEOPHILICITY)
The relative rate at which a nucleophile (Nu:-) reacts to displace (substitute for) a leaving group
is called ‘nucleophilicity’. Consider the following nucleophilic substitution reactions:
CH3OH + HI  CH3I + HOH
CH3OH + HCl  CH3Cl + HOH
The first reaction is much faster than the second because I- is a much better Nu:- than Cl-. The
leaving group (HOH) was the same in both cases. The nucleophilicity (relative reactivity) of
various Nu:-’s is listed in the following table ...
Nu:-
Reactivity
-
very weak
HSO4 ,
weak
H2PO4-,
Relative Reactivity
RCOOH
< 0.01
ROH
1
-
HOH, NO3
100
-
fair
F
500
Cl-, RCOO-
20  103
NH3, CH3SCH3
~ 300  103
N3-, Br-
~ 600  103
OH-, CH3O-
2  106
CN-, HS-, RS-, R3P:, NH2- , I-, H-, R-
> 100  106
good
very good
Note that Nu:-’s are electron donors as are Lewis bases and reducing agents. Nu:-’s are either
uncharged (with nonbonded electrons) or they are anions, but they are never cations. Nu:-’s are
basic, neutral, or sometimes weakly acidic, but not strongly acidic. Strong acids (HCl, H2SO4)
and Lewis acids (AlCl3, SnCl2) are electrophiles (E+’s), i.e., electron acceptors as are oxidizing
agents.
1. Within any given row of the periodic table, nucleophilicity decreases from left to right as
basicity & polarizability decrease (because electronegativity of the central atom is increasing)
CH3-
>
>
NH2NH3
PH2PH3
>
>
>
>
OHOH2
SHSH2
>
>
>
>
FHF
ClHCl
2. For nucleophiles with the same attacking atom, the anion is more basic and more
nucleophilic than the neutral compound.
Cl > HCl
OH- > HOH RO- > ROH NH2- > NH3 CH3CO2- > CH3CO2 H CN- > HCN
3. Nucleophilicity increases down any column of the periodic table; as the polarizability of
atoms increases (despite decreasing basicity – Polarizability has greater influence).
NH2H2PH2AsH2Sb-
ORGANIC CHEMISTRY INTRO
OHHSHSeHTe-
FClBrI-
increasing
nucleophilicity
increasing
polarizability
49
Note the similarities and differences of nucleophiles and bases ...
 Nu:-’s and bases are both electron donors
 Basicity deals with equilibrium position (Keq). At equilibrium, a stronger base holds a
greater proportion of H+. Nucleophilicity deals with kinetics. A stronger Nu:- attacks
faster than a weaker one.
 Basicity deals with interaction with H+ while nucleophilicity is broader and also deals
with interaction with other atoms, especially, but not only C atom.
Polarizability (‘squashiness’) of Nucleophiles:
 A polarizable nucleophile, e.g., I-, is large and soft (‘squashy’) because its valence
(donor) electrons are far from the nucleus (in the 5th period). The electron cloud is
readily distorted during bond making and breaking which reduces the energy maximum in
the transition state and thus speeds up reactions.
 A non-polarizable nucleophile, e.g., F- is small and hard (rigid). Its outer valence
electrons are close to the nucleus (in the 2nd period) and tightly held. F- forms strong
bonds but its electron cloud is not easily distorted during bond formation and breaking so
its transition states are high energy (slow reaction).
 It is often true that good nucleophiles are also good leaving groups for the same reasons,
i.e., they are polarizable and stabilize a negative charge (which leaving groups often
have).
Efficiency of Leaving Groups:
 Groups which best stabilize a '-' charge are the best leaving groups, i.e., the weakest bases
are most stable as anions and are the best leaving groups. These are salts of strong acids
(conjugate bases of strong acids), e.g., HI + H2O  I- + H3O+
 In the following table, the relative rate at which various groups will ‘leave’ in a
substitution reaction are listed. Note that the weakest bases (which are least reactive –
most stable) are the best leaving groups.
pKb = 23
pKb = 22
pKb = 21
pKb = 11
pKb = -1.7
pKb = -2
pKb = -21
TosO-
I-
Br -
Cl-
F-
HO-
RO-
H2N-
60,000
30,000
10,000
200
1
0
0
0
 Note: F-, OH-, RO-, & NH2- are not easily displaced by nucleophiles, i.e., they are lousy
leaving groups.

Based on pKb values, where would the following leaving groups be placed in the
preceding table? H2O ? ROH? carboxylate, RCOO- ?
ORGANIC CHEMISTRY INTRO
50
Trends in Strength of Nucleophiles and Electrophiles
Nucleophiles (Nu:-) are electron donors. Nucleophiles are anions or neutral but never cations.
Electrophiles (E+) are electron acceptors. Electrophiles are usually cations or neutral but
occasionally are anions.
Study the following tables and note that
 for nucleophiles with the same attacking atom (in the same column), the anion is a much
better nucleophile than the neutral atom
+
 for electrophiles of the same kind (e.g., H2O and H3O ), the cation is a much better electrophile
than the neutral species.
+
 some neutral molecules (like H2O) can be nucleophilic in the presence of a strong E
(e.g., H2O + HBr) and can be electrophilic in the presence of a strong Nu:- (e.g., H2O + NH2-).
BH3
CH3+
NH4+
H3O+
R-OH2+
HF
BH4-
:CH3-
:NH3
H2O
R-OH
F-
NH2-
OH-
R-O-
E+ strength
Nu:- strength
O-2
PH4+
H2S
R-SH
HCl
:PH3
HS-
R-S-
Cl-
-
H3BO3
H2CO3
-
-
H2BO3
HCO3
PH2
S-2
HNO3
H3O+
HBr
H2O
Br-
-
NO3
CO3-2
OH-
CN-
H3PO4
H2SO4
-
HSO4
-
-2
-2
H2PO4
HPO4
-3
PO4




Nu:- strength
E+ strength
Nu:- strength
O-2
HCN
E+ strength
HClO3
ClO3-
E+ strength
SO4
Nu:- strength
Nucleophiles donate loosely held electrons. These are  electrons (e.g., CH2=CH2) or non
bonded electrons (e.g., :NH3) but rarely are they  electrons such as in Na-H. Other
examples include metals such as Mg: and Na.
Some electrophiles have empty orbitals such as BH3 or AlCl3.
Other electrophiles produce an empty orbital in order to accept electrons from the Nu:-.
Strong Bronsted acids (protic acids such as HCl  H+) are good examples.
Atoms with + charge are potential electrophiles, such as the C in a carbonyl group. It is +
and its weak  bond to O can cleave to form an empty orbital (C+) . Similarly, a + C
attached to a good leaving group (e.g., -Br) can cleave a  bond to form an empty orbital
(C+).
ORGANIC CHEMISTRY INTRO
51
Study the following reactions. Write products where they are not shown and identify
nucleophiles and electrophiles.

..
Cl :
..
H
..
: Cl : - + H+
..
+
Na
+
- ..
:O
..
+
+ -
H
acetone
lithium hydride
Li
CH3
C
..
O
..
H
.. - +
: O : Li
: O:
+
H
H
+
Na Cl
CH3
CH3
C
CH3
H
..
Br :
..
H
CH3
+
H3C
..
: Br: - + H+
..
C
..
O
..
CH3
CH3
H
H3C
C
CH3
..
O
+
H
C+
CH3
H
CH3
H3C
CH3
H3C
..
: Br: ..
C
Br
CH3
: O:
CH3
..
O
..
C
H
sodium amide
: O:
CH3
C
.. :
O
..
H
+
..
: NH 2
..
Cl
.. :
H
H
H
C
H
Na+
+
+ H+
+
C
..
: Cl : - + H+
..
H
ORGANIC CHEMISTRY INTRO
H
C
H
+
C
H
H
52
REDOX REACTIONS IN ORGANIC CHEMISTRY
There are a variety of ways for calculating oxidation numbers for organic chemistry. In single C
compounds, we use the same (rigorous) procedure used for inorganic chemistry, i.e., assign
oxidation numbers (ox. #) to all atoms other than C based on their EN values and calculate the
ox. # of C from these with the aid of a couple rules ...
1. The ox. # of a compound equals the charge on the compound
2. Certain atoms have fixed ox. #’s when bonded to C, i.e., H= +1, O = -2, halogens (X) = -1
Calculate the ox. # of C in each of the following compounds ...
Ox. # -4
CH4
-2
CH3OH
CH3Cl
0
+2
+4
O
O
O
H-C-H
CH2Cl2
H-C-OH
HO-C-OH
+4
CO2
CCl4
CHCl3
Oxidation of an organic compound causes an increase in oxidation number and is also defined
as:
 any reaction that increases the number of atoms which are more EN than C, i.e., O, N, X or
 any reaction that decreases the number of atoms which are less EN than C, i.e., H, Li, Na, Mg
Reduction is the opposite, i.e., a decrease in oxidation number and a reaction that decreases the
number of EN atoms atoms.
Identify the following reactions as oxidation or reduction...
CH 3-CH 3
CH 2=CH2
O
CH3CH2OH
CH 3-C-H
CH
CH
O
CH 3-C-OH
+H2O
CH 2=CH2
CH 3-CH 2OH
- H 2O
ORGANIC CHEMISTRY INTRO
53
Oxidizing Reagents for Organic Chemistry
Oxidant (media)
Reduced Form
Oxidant
Reduced Form
(purple) MnO4 (OH or neutral)
MnO2 (brown ppte.)
X2, e.g., Cl2
X-, e.g., Cl-
(purple) MnO4- (H+)
Mn+2 (colorless, aq.)
H2O2
2 O-2
hot conc. HNO3
NO2 (brown gas)
O2
2 O-2
(orange Cr+6 ) H2CrO4 (aq. H2SO4)
Cr+3 (blue-green)
Pb(OAc)4
Pb(OAc)2
PCC (still Cr+6) (in CH2Cl2)
Cr+3 (blue-green)
Hg(OAc)2
HgOAc
HIO4 (colorless)
HIO3 (colorless)
NaOCl (bleach)
Cl-
-
-
H2CrO4 is prepared by dissolving CrO3 (chromia or chromic anhydride) or K2Cr2O7 in aq.
H2SO4.
Jones reagent:
CrO3 + H2SO4 (aq.) + acetone  H2CrO4
Pyridinium Chlorochromate (PCC): CrO3 + C5H5N: + HCl  C5H6NCrO3Cl
Collins Reagent: CrO3 + 2 C5H5N:  (C5H5N)2CrO3
Cr+6 in aqueous acid soln. are strong oxidants. Cr+6 in anhydrous solvent are mild oxidants.
Reducing Agents for Organic Chemistry
Reducing Reagent (conditions)
Oxidized form
H2 (g) (Pt, Pd, Ni, etc. catalyst in EtOH)
H+
LiAlH4 (in ether) liberates :H-
H2
NaBH4 (in EtOH) liberates :H-
H2
BH3 (g)
H2
Li, Na, K, Zn, Hg, Mg, etc.
Li+, Na+, K+, Zn+2, etc.
SnCl2 + 2HCl = H2SnCl4
H2SnCl6
Zn[Hg] + HCl
ZnCl2 + H2
ORGANIC CHEMISTRY INTRO
54
Balancing Redox Equations:
In order to calculate the theoretical and actual yield from a reaction, a balanced chemical
equation is necessary. The following rules describe the Ion-Electron Half-Reaction Method for
balancing redox equations ...
1. Break the equation into 2 half-reactions, i.e., oxidation and reduction
2. Balance all atoms in each half-reaction other than H and O
3. Balance O and H as follows ...
 add H2O to balance O’s first, then balance H’s by adding H+
 add sufficient electrons to balance the charges
Note: e-'s are always added to the right side of the oxidation ½-reaction because ox. = loss of e-'s
Note: e-'s are always added to the left side of the reduction ½-reaction because red. = gain of e-'s
 if the reaction is carried out in alkaline media, add enough OH- to neutralize all H+
(combine them to make HOH) and be sure to add the same quantity of OH- to both sides to
keep mass and charges balanced
4. Multiply each ½-reaction by a least common multiple so the number of e-'s transferred in each
½-reaction is equal.
5. Add the ½-reactions and cancel common terms from each side
6. Check for charge and mass balance
Balance the following redox equations for practice ...
1. Cyclopentanol is cleaved by strong oxidants like hot, conc. HNO3 producing pentanedioic
acid.
2. Cyclohexene is cleaved by strong oxidants like hot, acidic KMnO4 producing hexanedioic
acid.
3. Cyclohexene is oxidized to a diol by cold, neutral or alkaline KMnO4 producing
1,2-cyclohexanediol.
ORGANIC CHEMISTRY INTRO
55
Elimination Often Competes with Substitution:
 Strong dehydrating acids (H2SO4, H3PO4) favor elimination (dehydration) in alcohols. Because they
are strong acids, they readily protonate the alcohol thereby converting a poor leaving group (OH -)
into a good leaving group (HOH), however, the anions produced after protonation of the alcohol
(HSO4- or H2PO4-) are very poor nucleophiles and can’t replace the leaving group.
+
H2SO4 (catalyst)

CH2=CH2
H2O
(elimination)
(CH3)2CH-OH +
H2SO4 (catalyst)

CH3CH=CH2 + H2O
(elimination)
(CH3)3C-OH
H2SO4 (catalyst)

(CH3)2C=CH2 + H2O
(elimination)
CH3CH2-OH

+
Strong non-dehydrating acids (like HI, HBr and HCl) also readily protonate an alcohol creating a
good leaving group (HOH) but with the difference that the resulting Nu: -’s (like I-, Br-, and Cl-), are
much better Nu:-’s and readily replace the leaving group which results in substitution.
+
HBr

CH3CH2-Br
(CH3)2CH-OH +
HBr

(CH3)2CH-Br
+ H2O
(CH3)3C-OH
HBr

(CH3)3C-Br
+ H2O
CH3CH2-OH

+
+
+ H2O
(substitution)
(substitution)
(substitution)
Very strong bases can cause elimination reactions with alkyl halides because strong hydrohalic acids
(HX) produced by elimination react rapidly and completely with the excess strong base in a
neutralization reaction, as per Le Chatalier. This is especially true with 2 and 3 alkyl halides which
are bulky (hindered) and the Nu:- has difficulty contacting the reactive C atom.
CH3CH2-Cl
1
+
CH3O Na+  [CH3CH2-OCH3 + NaCl] & [CH2=CH2 + CH3O-H + NaCl]
(v. strong base)
> 90 % sub.
<10% elim.
-
(CH3)2CH-Cl
2
+
CH3O Na+  [(CH3)2CH-OCH3 + NaCl] & [CH3CH=CH2 + CH3OH + NaCl]
~ 20 % sub.
~80% elim.
(CH3)3C-Cl
3
+
CH3O Na+  [(CH3)3C-OCH3 + NaCl] & [(CH3)2C=CH2+ CH3OH + NaCl]
~3 % sub.
~ 97% elim.
(CH3)3C-Cl
+
Na+CN-
3
-
-
(v. good Nu:-)
ORGANIC CHEMISTRY INTRO

(CH3)3C-CN + NaCl
(almost all sub.)
56