Transcript Document

Chapter 2: Polar Bonds and Their Consequences
Coverage:
1. Polar Covalent and Dipole Moment
2. Intermolecular Forces
3. Resonance
4. Acids and Bases – Bronsted-Lowry and Lewis
5. Prediction of Acid/Base Reactions
6. Drawing Organic Structures
Goals:
1.
Be able to calculate the charge separation on two bonded atoms given the bond length
and the measured dipole moment.
2.
Be able to calculate formal charges on atoms.
3.
Be able to draw resonance forms for simple organic molecules.
4.
Know the definitions of Bronsted-Lowry and Lewis acids and bases.
5.
Understanding the meaning of pKa and use it to predict the direction of an acid-base
reaction.
6.
Know how to used curved arrows to indicate flow of electrons in a chemical reaction.
7.
Know the structures of common organic bases and acids.
8.
Know the nature of intermolecular forces, including hydrogen bonding, dipole-dipole
interactions, and London dispersion forces.
9.
Know how to draw condensed and skeletal structures for simple hydrocarbons.
2-1
Bond Dipoles
A bond is polar if there is a difference in electronegativities of the two atoms.
If the difference, DEN, is greater than or equal to 0.9 (Pauling scale), the bond
is ionic.
DEN > 2.0
DEN  2.0 and  0.5
DEN < 0.5
Ionic Bond
Polar Bond
Nonpolar Bond
Calculation of Bond Dipole Moment
= Q x r x 4.8
where Q is magnitude of the charge separation (electrons)
r is internuclear distance (Angstroms)
 is the bond dipole moment (debyes, D)
2-2
2-3
Example: The carbonyl group in a ketone
O
Carbonyl Group of Ketone
CH3CCH3
The dipole moment of the group is measured at 2.4 D.
The bond distance, r, is 1.21 Angstroms.
Calculate the charge separation
Q =

r
 - 0.413
= _____2.4_____ = 0.413 electrons O
1.21 x 10-10
CH3CCH3
The symbol  (delta) indicates a partial charge
 +0.413
on the atom, either negative or positive.
2-4
Some typical Bond Dipoles:
Bond
 Debyes
C-N
C-O
C-F
C-Cl
C-Br
C-I
H-C
H-N
H-O
C=O
C=N
-
0.22
0.86
1.51
1.56
1.48
1.29
0.3
1.31
1.53
2.4
3.6
Q: Why is the C-F bond less polar than the C-Cl, since F is
more electronegative than Cl?
2-5
Molecular Dipole Moments
Molecular Dipole Moments are the vector sum of the individual bond dipole
moments. They depend on the magnitude and direction of the bond dipoles.
NH3
H2O
H
H N
H
H
:
 = 1.5 D
H
CO2
:
O
:
1.9 D
..
..
:O=C=O:
0.0 D
CH3Cl
H
H
H
C Cl
1.87 D
2-6
2-7
Intermolecular Forces
Intermolecular Forces (IMFs) determine the physical properties (e.g.
melting point, boiling point) of organic molecules. There are three
major types of IMFs; hydrogen bonding, dipole – dipole and London
dispersion forces
1.
Hydrogen Bonding. This force is dominant if present. It is not a true
bond. Organic compounds must possess an O-H or N-H group to hydrogen
bond to itself.
Example: Methanol
:
H3C
O
:
O
CH3
- H
H
+
Hydrogen Bond
2-8
Hydrogen Bonding raises the boiling point of organic molecules.
Isomers of C3H9N
..
H3C N CH3
..
N CH3
CH3CH2
CH3
Boiling Point
No. of NH
Groups
CH3CH2CH2
H
3.50C
0
370C
1
.. H
N
H
490C
2
2-9
2. London Dispersion Forces - the next most dominant force. It arises from
temporary dipoles induced by nearby molecules. This force depends on the
surface area and, hence, molecular weight.
Nonpolar
Nonpolar
+
-
+
-
Coordinated Temporary Dipole caused by distortion of electron density around molecule.
When these molecules diffuse away from each other, the coordinated dipole disappears.
2-10
London Dispersion forces are proportional to surface area, and therefore molecular weight.
Which hydrocarbon among these straight-chain alkanes has the highest boiling point?
CH3CH2CH2CH3
CH4
Lowest
CH3CH2CH2CH2CH2CH3
Highest
If hydrocarbons have the same molecular weight, the molecule with the least branching
will possess the highest boiling point.
Consider isomers of C6H14
Boiling Point
60oC
69oC
58oC
2-11
3. Dipole- Dipole Interaction. These forces result from permanent dipole moments
in polar molecules.
+
-
+
-
Rank these molecules, which have similar molecular weights, according to boiling point.
OH
O
Lowest
Highest
Summary
1.
2.
3.
Hydrogen bonding raises boiling point.
Increased molecular weight raises boiling point
Polar molecules possess higher boiling point than nonpolar molecules
if same molecular weight.
2-12
Geckos climb vertical and even inverted
surfaces with ease using millions of micronscale adhesive foot-hairs on each toe (setae).
Each foot-hair splits into hundreds of tips only
200 nanometers in diameter, permitting
intimate contact with rough and smooth
surfaces alike. All of this occurs due to van
der Waal forces (sum total of intermolecular
forces), not sticky substances.
2-13
Resonance Structures
Some structures are not adequately expressed by a single structure. If two or more valence
bond, structures are possible, the molecule shows characteristics of both – termed resonance
structures.
H
+
C
H
H
H
H
C N
N
+
H
H
H
Resonance forms differ in the placement of p and nonbonding valence electrons. The atoms
themselves occupy the same positions in both resonance forms. The molecule is not changing,
it is just represented by two or more different Lewis structures.
..
:O:
:O:
H3C C
-
H3C C
:O:
..
-
:O:
The charged is shared by the two oxygen atoms.
2-14
Rules for Resonance
1.
The real structure is a composite of the resonance forms. The structure does not
change, but is a resonance hybrid of the individual structures.
:O:
H
H
C C
H
.. :O:
C C
H
:O:
.. -
:O:
-1/2
:O:
H
C C
H
:O:
-1/2
2.
Resonance forms differ only in the placement of electrons, not the placement of atoms.
3.
Different resonance forms are not necessarily equivalent.
4. Resonance forms must be valid Lewis structures and obey normal rules of valency.
Usually the octet rule must be obeyed, but in some cases a valid structure shows only
six electrons around an atom. (See previous example).
2-15
To become proficient in drawing resonance structures, we need to learn to recognize
five patterns. I will illustrate all of these on the chalkboard.
1. Allylic lone pair (two curved arrows):
2. Allylic positive charge (one curved arrow):
3. Lone pair adjacent to positive charge (one or two curved arrows):
4. A π bond between two atoms of differing electronegativity. (one arrow):
5. Conjugated π bonds enclosed in a ring: (three curved arrows):
2-16
Assessing Resonance Structures; Three Rules
1. Minimize charges. Resonance forms with no charges are more significant.
Example on board.
2. Electronegative atoms, such as N, O, and Cl, can bear a positive charge, but only if
they possess an octet of electrons.
Example on board
3. Avoid drawing resonance structures in which two carbon atoms bear opposite charges.
Example on board.
2-17
Acids and Bases
Bronsted-Lowry Acid – substance that donates a hydrogen ion (proton).
Bronsted-Lowry Base – substance that accepts a hydrogen ion (proton).
HA +
H2 O
Ka = [H2O] [A-]
[HA] [H2O]
A- + H3O+
pKa = -log Ka
The lower the pKa, the stronger the acid
Acid-Base Reaction
Conjugate
Acids
Conjugate
Bases
Ka
pKa
HBr + H2O
H3O(+) + Br(-)
HBr
H3O(+)
Br(-)
H2O
105
-5
CH3CO2H + H2O
H3O(+) + CH3CO2(-)
CH3CO2H
H3O(+)
CH3CO2(-)
H2O
1.77*10-5
4.75
C2H5OH + H2O
H3O(+) + C2H5O(-)
C2H5OH
H3O(+)
C2H5O(-)
H2O
10-16
16
NH3 + H2O
H3O(+) + NH2(-)
NH3
H3O(+)
NH2(-)
H2O
10-34
34
2-18
Common Organic and Inorganic Acids
Weakest
Strongest
Acid
ethane
ammonia
ethanol
water
methylammonium ion
bicarbonate ion
phenol
ammonium ion
carbonic acid
acetic acid
benzoic acid
hydrogen fluoride
phosphoric acid
hydronium ion
oxonium ion
sulfuric acid
hydrogen chloride
hydrogen bromide
hydrogen iodide
Formula
CH3CH3
NH3
CH3CH2OH
H2 O
CH3NH3+
HCO3–
C6H5OH
NH4+
H2CO3
CH3CO2H
C6H5CO2H
HF
H3PO4
H3 O+
R2OH+
H2SO4
HCl
HBr
HI
pKa
48
33
15.9
15.7
10.64
10.33
9.95
9.24
6.26
4.76
4.19
3.5
2.1
– 1.74
– 1.75
– 5.2
–7
–8
–9
Conjugate Base
CH3CH2–
NH2–
CH3CH2O–
OH –
CH3NH2
CO32–
C6H5O–
NH3
HCO3–
CH3CO2–
C6H5CO2–
F–
H2PO4–
H2O
R2O
HSO4–
Cl –
Br –
I–
2-19
Predicting Acid –Base Reactions
In general, an acid with a lower pKa will react with a conjugate base of an acid with a
higher pKa.
O
O
CH3COH
Acid
pKa
+
Na+OHBase
+
CH3CO- Na
Conjugate Base
4.76
+
H2O
Conjugate Acid
15.7
Keq ≈ Ka reactant/ Ka product ≈ 10-5/10-16 = 1011 Very Large.
This means that the above reaction is quantitative.
In general, if there is at least 5 pKa units difference, then the reaction will be quantitative,
that is, the reaction will be nearly complete.*
2-20
Will the following acid and base reaction be quantitative, i.e go to completion?
CH3CH2O
Acid
pKa
H
+
Na+ OH-
Base
+
CH3CH2O- Na+
Conjugate Base
H2O
Conjugate Acid
15.9
15.7
Answer: the reaction will take place, but it will not be quantitative.
Acetone, a ketone, possesses an acidic hydrogen on the methyl group. Will acetone
quantitatively react with sodium amide (NaNH2)?
Answer: Yes
H O H
H C C C H
H
Na+NH2-
H
Acid
pKa
+
19
H O -..
H C C C H
H
Base
Na
+
+
NH3
H
Conjugate Base
Conjugate Acid
33
2-21
Irving Langmuir
His interest in fundamentals involved him in the theory of chemical
bonding in terms of electrons, and he elaborated on ideas first expressed
by Gilbert Lewis. Langmuir proposed that octets could be filled by
sharing pairs between two atoms—the "covalent" bond. His studies of
surface chemistry—the study of chemical forces at the contact surfaces
(interfaces) between different substances, where so many biologically
and technologically important reactions occur—earned him the Nobel
Prize in chemistry in 1932.
Gilbert Newton Lewis
1875-1946. An American chemist at the University of California,
Berkeley who revolutionized concepts of bonding. The Lewis octet rule is
named after him. He also generalized the concept of acids and bases and
we now think of Lewis acids and Lewis bases. He had a summer home in
Marin County, Calif.
2-22
Lewis Acids and Bases
Lewis Acid – substance that accepts an electron pair
Lewis Base – substance that donates an electron pair
B
:
Base – electron rich
Nucleophile
+
A
B
:
A
Acid – electron poor
Electrophile
Curved Arrow Formalism - a curved arrow always indicates the direction of
electron flow. The electron flow is from the Lewis base (electron rich) to
the Lewis acid (electron poor) or from the nucleophile to the electrophile.
2-23
F3 B
+
:NH3
F3B---NH3
Boron trifluoride
:
Acid
Acid-Base Adduct
Base
2-24
Drawing Structures
Focus on Hydrocarbons
Alkanes
1.
Complete Lewis Structures
H
H
H
H H C H H H H
H C C C C C C C H
H H H H H H H
CH3
2.
Condensed Formula Structure
3.
Skeletel or Line-angle Formulas
CH3CH2CHCH2CH2CH2CH3
2-25
Alkenes
H
H
H
H H C
H H
H C C C C C C C H
H H H H H H H
CH3
CH3CH2CHCH=CHCH2CH3
2-26