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CHEM 3013
ORGANIC CHEMISTRY I
LECTURE NOTES
CHAPTER 2
1.
Formal Charge
The Lewis structures we have drawn thus far have all been neutral covalent
molecules. However, some covalently bonded molecules may contain charged atomic
components. Furthermore, many ionic species also contain covalent bonds. To manage
the electronic bookkeeping we assign a formal charge to each of the atoms of a structure.
The formula , Formal Charge = Valence Electrons - 1/2( Bonding Electrons) - N onBonding Electrons will be used for this purpose.
+
H
H
N
For Nitrogen
H
FC = VE-(1/2BE + NBE)
FC = 5 - ( 8/2 + 0) = 1
For Hydrogen
FC = VE-(1/2BE + NBE)
FC = 1 - (2/2 + 0) = 0
H
Nitrogen bears the positive charge
Calculation of Formal Charge
2.
Polar Covalent Bonds
Like most things in nature there are "gray" areas between the ionic and
covalent bond. A polar covalent bond is a covalent bond between two atoms of differing
electronegativity. Electronegativity increases as you move to the right and the top of the
periodic table. Electronegative elements need more electron density near them. We can
predict the nature of a chemical bond by using the difference in Pauling Electronegativities,
(∆ E.Neg.). For example, acetonitrile is a compound containing a triple bond where the
electronegativity of nitrogen is greater than that of carbon. Thus the electron density on the
nitrogen is greater than on the carbon.
δ+
δ−
C
N
Pauling
2.5
Electroneg.
3.0
H 3C
E.NEG = 0.5
3.
POLAR COVALENT BOND
Dipole Moments
The polarization of a covalent bond (separation of charge densities) gives rise to a
dipole along the internuclear axis. This dipole is a measurable quantity which is exerted
in a localized direction ( a vector). The dipole moment, µ is a product of the charge
separation in electrostatic units, q,( esu), the bond length, r, (A) and a constant term (1010
debye/esu . A ).
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F
H
H
F
µ = 2.76 D
No Net Dipole
C
F
F
H
C
F
µ = q x r x 1010 debye/esu A
Y
A vector can be resolved into it's
three component directions.
A combination of dipole moments
may result in a net cancellation.
X
Z
Bond Dipoles
4.
Resonance Structures
Resonance structures are groups of Lewis-type representations which differ only in
the position of localized electron density. A resonance structure is a formula which can be
written which involves the movements of bonds, unshared electron pairs, single electrons
or charges. It is important to understand that atoms stay fixed in space in resonance
structures, only electrons are shared over one atom. Structures which involve the
movement of atoms are not valid resonance forms. The standard rules for writing valid
Lewis structures (octet rule, no pentavalent carbons, etc.) apply for resonance forms.
Two possible Lewis Structures for the acetate ion: CH3COO
O a
H
C
C
H H
O b
For Oxygen a: FC = 0
For Oxygen b: FC = -1
a
O
C
H
C
H H
O b
FC = -1
FC = 0
RESONANCE STRUCTURES
O
H
- 1/2
C
C
H H
O - 1/2
Each Oxygen has a -1/2
Formal Charge. The one
double bond is shared.
RESONANCE HYBRID
DO NOT USE TWO SINGLE
ARROWS, THIS DEPICTS
AN EQUILIBRIUM
Use of Double-Headed Arrows to indicate Resonance Structures
The formalism used to depict resonance structures is the double-headed arrow.
This must be well-understood to be distinct from the pair of single headed arrows used to
describe an equilibrium reaction. It must be remembered that while we may draw several
separate structural formulae to depict the various resonance contributors, these are not
individual molecules in equilibrium with one another, but a composite depiction of the
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chemical species being discussed. Resonance and equilibrium ARE NOT THE
SAME THING .
A molecule is a weighted average of all its contributing resonance structures. The
structural depiction of the weighted average is called the Resonance Hybrid . The best
real world analogy for a resonance hybrid would be to consider a biological hybrid formed
when two related species form offspring. A good example of this is the mule, a hybrid
formed in the crossing of a horse and a donkey. A mule is always a mule, it show some
characteristics of horses and some of donkeys, but it is always a mule. It never spends part
of its time as a horse and part of its time as a donkey ( this would be like equilibrium).\
While a molecule is an average of its contributing resonance structures, it need not
be true that all resonance forms contribute equally. To assess the weighting of the
contributions we examine the various structures and consider their stability as if they were
real individual molecules ( which , of course, they are not). The forms that have the
greatest stability are the forms which have the greatest contribution to the resonance hybrid.
1. Identical structures contribute equally.
H
C
H
C
CH2
CH2
CH2
CH2
allyl anion
2. Structures with the greater number of bonds are more important
H2C
H2C
CH2
H2C
CH2
CH2
BEST
3. Structures that delocalize charge or unshared electrons are important
H2C
H2C
O
CH3
O
CH3
4. When considering structures with separated charges, atoms with
appropriate electronegativity will best accomodate the charges.
H3C
O
H3C
O
C
C
H3C
H3C
BEST
Rules for Writing Resonance Structures
In General, the more important resonance structures one can write
for a compound, the more stable will be that compound . Resonance structures imply
additional orbital overlap, and hence, stability.
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5.
Bronstead Acidity and Basicity
A general acid HA, in aqueous solution, can dissociate into its component parts,
H+ (the proton) and A- (the conjugate base of HA). The larger the equilibrium constant
for this ionization reaction, the stronger the acid will be. Because of the great range of
organic acids, chemists usually discuss this subject using a log scale, where pKa is defined
as the negative log of the equilibrium constant Ka. Most compounds fall within the pKa
range of 55 (exceedingly weak organic acids like methane) to -7 (strong inorganic acids like
H2SO4). Chemical reactivity and reaction mechanisms are often intimately related to acid
strength. Factors related to acid strength will be of critical importance throughout this
course.
Ka
HA
H+
+
A-
Ka = H+
A-
1/Ka
HA
pKa = - log Ka
HOH
H+ + HO
Ka = 10-16
pKa = 16
Equilibrium lies to the left,
water is only slightly ionized.
Water is a relatively weak acid.
Smaller, more negative pKa............Stronger acid
Larger, more positive pKa..............Weaker acid
Acids, Bases and pKa
6.
Acid, Base Reactions
Being able to predict whether an acid-base reaction will proceed or not is very
important. A reaction will proceed to the right (towards products) only if the
conjugate acid is a weaker acid.
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HA
acid
+
NH3 +
pKa = 35
B:base
H-B
+
conjugate
acid
A:conjugate
base
HO:-
HOH
+
pKa = 16
:NH2-
Equilibrium favors reactants, little products formed.
Keq = ?
To solve : combine the Ka for each of the individual
species.
Keq = Ka(NH3) x 1/Ka(HOH) = 10-35/10-16 = 10-19
The reaction is extremely unfavorable!
Calculation of acid-base reaction using pKa data.
7.
Lewis Acids and Bases
A Lewis acid is defined as an electron acceptor. They are species which are at least one
electron pair short of a filled outer shell configuration(either octet or duet), because of this they are
very reactive toward electron sources. Another term for Lewis acids is electrophiles (electron
loving species).Typically, a Lewis acid is a cation , (e.g. H+, Li+, (CH 3)3C+) or a metal atom in
a salt.A Bronstead acid (proton source) is also a Lewis acid
A Lewis base is defined as an electron donor. They are species that react with Lewis acids
by supplying an electron pair. Another name for Lewis bases is nucleophiles (nucleus loving
species). A typical Lewis base is an anion (e.g. OH-) or a neutral heteroatom with at least one pair
of non-bonding electrons (e.g. :NH 3).
A Lewis acid must have available a low energy empty orbital to accept an
electron pair from the Lewis base.
empty orbital
F
F
+
F
B F
F
Flouride anion
Lewis base
Boron triflouride
Lewis acid
F
B F
F
Tetrafluoroborate anion
Reaction of Lewis Acid with a Lewis Base
7.
Combustion Analysis
Converting organic compounds to carbon dioxide and water by combustion provides
an
analytical means for determination of the % composition of the
molecule in question. This
allows an organic chemist to assign an empirical
formula to the compound in question.
The equations:
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Weight H in sample = weight H2O x (2.016g H / 18.016 g H2O)
Weight C in Sample = weight CO2 x (12.01 g C / 44.01 g CO2)
% H in Sample = Weight H in sample / sample weight
% C in Sample = Weight C in Sample / sample weight