heat of reaction estimation
TRANSCRIPT
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Chapter 1 Concepts and
Models in Organic Chemistry
General discussion of some overall concepts
in organic chemistry.
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Chapter 1 Problems
Problems each worth 5 pts.
1, 4, 6, 14, 16, 21
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What is a model?
A representation of the real
Details of the model are only as useful as
they are helpful to explain observations.
When a model does not explain
observations it must be modified or thrown
out.
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Modern Models
How do we arrive at Models
Imagination
Creativity
Desire to explain what we observe
Often quantitative in order to explain
observations from instruments.
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Examples of Models.
Lewis dot diagrams
Tinker Toy Models
Dreiding Models
CPK space filling models
Quantum Mechanical Models
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How do we know what amolecule look like?
Imagination
Not very quantitative
Helps sharpen understanding
Graphic representation of quantitative
models.
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Ethane Model
Different views
Each has a use.
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Molecular dimensions.
How do we get the molecular dimensions of
molecules?
X-ray
Electron and Neutron diffraction
Only on solids
Give nuclei positions of heavy atoms
Microwave
Usually low MW molecules
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Bond Lengths and Angles
Represent the inter-nuclei distances as
points in space
The electron density extends out from these
points.Molecule rC-H rC-X
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Covalent Radii
Functions of charge.
C4+ 0.15A while C4- 2.60A
Depends on bonding
C bonded to four groups 0.772A
C double bonded 0.67A
Effective Radii for groups of atoms. CH3 2.0A
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Atomic Volume and Surface
Atomic volumes can be calculated from
values in Table 1.3 (p.8)
Butane
2xCH3 = 2x17.12 = 34.24
2x CH2 = 2x10.23 = 20.46
total=54.7cm3/mole
Theory cal = 99.91 A3 = 60.14 cm3/mole
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Heats of Formation
o is the difference in enthalpy between
the compound and the elements in their
standard state.
Obtained from heats of combustion data.
m C(graphite) + n/2 H2(gas ) CmHn Hof
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Heats of Combustion
Ho when the compound is burned with
excess O2
CmHn + (m+ n/4) O2 m CO2 + n/2 H2O Ho
= heat of compustio
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General caseC(graphite) + O2(gas) CO2(gas)
Ho
r= Ho
f(CO2)
H2(gas) + 1/2 O2(gas) H2O (liquid)
Ho
r= Ho
f(H2O)
Combining these quations.
Ho
f(CmHn) = mHo
f(CO2) + n/2 Ho
f(H2O) - Ho
(combustion) (CmHn)
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Corrections
Change of phase
heat of vaporization for liquids
heat of sublimation for solids
Temperature corrections
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Example From BookO
O
+ 8 O2 6 CO2 + 4 H2O Ho
= -735.9 Kcal/mol
Ho
f(cyclohexanedione) = 4 Hof(H2O) + 6 H
of(CO2) - H
ocombustion (cyclohexanedio
Hof(cyclohexanedione) = 4 (68.32) + 6 (94.05) - (-735.9) = -101.68 Kcal/mol
Correction of solid to gas Ho
sub = +21.46 Kc al/mol
Hof(cyclohexanedione) = -101.68 Kcal/mol + 21.46 = -80.22 Kcal/mol
Hof(H2O) = -68.32 Kcal/mol H
of(CO2) = -94.05 Kc al/mol
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Differences BetweenCompounds
You can use heats of reactions if you go to a
common product.
Tran more stable than cis by 1 Kcal/mol
CH3
H
CH3
H
CH3CH2CH2CH3
H
CH3
CH3
H
-27.6 Kcal/mol-28.6 Kcal/mol
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Principle Additivity
The property of a molecule can be estimated
from the the parts (bonds or groups).
If you know the contribution from a group
in one molecule you can use it in another
calculation.
Hof(CH2) = -35.1 - -30.4 Kcal/mol = -4.7 Kcal/mol
CH3CH2CH2CH3 CH3CH2CH2CH2CH3
Ho
f= -30.4 Kcal/mol Ho
f= -35.1 Kcal/mol
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Table 1.5 Bond Increments toHof
How well does this method work?
CH4 cal -15.32 found -17.89 Kcal/mol
Ethane
6 C-H and 1 C-C
6 (-3.83) + 1 (2.73) = -20.25 Kcal/mol
Found -20.24 (not bad)
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Problem
Calculated values
They are the same
Obs. -32.4 Kcal/mol and -30.4 Kcal/molrespectively.
CH3CHCH3
CH3
CH3CH2CH2CH3
10 (-3.83)
3 (2.73)= -30.11 Kcal/mol
10 (-3.83)
3 (2.73)= -30.11 Kcal/mol
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Group Increments(Table 1.7)
Group contributions allow for different
bond types.
Obs. -32.4 Kcal/mol and -30.4 Kcal/mol
respectively (better?)
CH3CHCH3
CH3
CH3CH2CH2CH3
3 CH3 = 3 (-10.8)
1 CH = -1.9= -34.3 Kcal/mol
2 CH3 = 2 (-10.8)
2 CH2 = 2 (-4.5)
= -29.6 Kcal/mol
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Limitation
Group increments do not consider strain
contribution
CH3CHCH2CH2CH3
CH3
3 CH3 = 3 (-10.8)
2 CH2 = 2 (-4.5)
1 CH = 1 (-1.9)=-42.04 Kcal/mol (exp. -41.66)
CH2CH3
HH
CH3
CH3 H
1 gauscheinteratction 0.8
Correction -42.04 + 0.8 = -41.24 (better)
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Bond Dissociation Energies
The energy needed to break the bond.
Homolytic
Hetrolytic
A B A B Hor
A B A B Hor
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Using Bond Dissociation
Energies
Example
This calculation does not say anything
about how fast the reaction occurs.
Ho
= 2(-104. Kcal/mol) + 104.2 + 88 = -15.8 Kcal/mol
A checkHor= 2 H
of (CH4) - H
of(CH3CH3) - H
of(H2) = 2(-17.9) - (-20.2) - 0 = -15.6 Kcal/mol
CH3CH3 + H2 2 CH4
H2 2 H DHo
= 104.2 Kcal/mol
DHo
= 88 Kcal/molCH3CH3 2 CH3
2 CH3 + 2 H 2 CH4 2 DHo
= 2 (-104 Kcal/mol)
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Bonding Models
What holds molecules together?
Groups of atoms find they are a lower energy
when they are associated with each other. They
are bonded.
Electrostatic attractions.
Covalent bond
Na+
Cl- NaCl energy
H H H H
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Models
Lewis Model
For some reason atoms (first row) are in a low
energy state if the have 8 electrons around them.
Why?
Electron transfer between atoms of different
electronegativity
F-1
O-2
N-3Ne
F Cs Cs+1
F-1
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Covalent Bond
Both atoms have the same EN
Bonding only involves outer shell electrons
(valence shell electron)
F F F F
P 1s2 2s2 2p6 3s2 3p3
core valence electrons
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Dipole Moment
Vector sum of the dipole moments of the
bonds.
Dipole moments useful in understandingproperties.
= charge (esu) X distance (cm) = Debye (D)
= 0.1 (4.8 x 10-10
esu) 1.5 A(1 x 10-8
cm/A) = 0.72 x 10-18
esu cm = 0.72 Debye (D)
A B+0.1 esu -0.1 esu
1.5A
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Which Molecules Have DipoleMoments?
Cl HCH3
Cl
H
H
CH3
CH3
H
H
CH3CH2CH2OH
Cl
Cl
Cl
OO
O
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Bond Polarity
Bond polarity comes from unequal sharing
of bonding electrons.
Polar Bond = Covalent part + ionic part
molecule =covalent + ionic
% Ionic character = 2
(1 + 2)
(100)
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Electronegativity
Unequal sharing of electrons between atoms
of a bond.
Pauling Electronegativity Scale
Compares bond dissociation energies of like
atoms with those of unlike atoms.
H2 and Cl2 (104.2 + 58)/2 = 81.1 Kcal/mol HCl 103.2 (larger due to ionic character)
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Table 1.8
Other EN scales
Pauling,p Mullikin,m (I+A)/2
Allen,spec Nagel, Benson,
x Which EN scale to use depends on
application.
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Bonding
Valence Bond Theory (VB)
Bring together of atoms which share electrons.
Molecular Orbital Theory (MO)
Bring together nuclei which are held together
by electrons.
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H2 Molecule
Two Hydrogen atoms
We can write the equations for a 1s orbital.
Wave functions
H1s H1s
1 = c a(1) b(2)
2 = c a(2) b(1)
VB = c a(2) b(1) + c a(2) b(1)
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LCAO Method
LinearCombinations ofAtomic Orbitals
1 = c1 a(1) + c2 b(1)
2 = c1 a(2) + c2 b(2)
MO = 12 = c12 a(1) a(2) + c2
2 b(1) b(2) + c1c2[a(1)b(2) + a(2)b(1)
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Comparison
VB is simpler and is incorporated into the
MO wave function
LCAO method has ionic contribution
As terms are added to either model they
both give better answers and they become
very much alike.
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Energy Levels
The MO model generates the orbital and the
energies of the orbitals.
Energy
H H-H H
AO AOMO
s s
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Methane
Atomic Orbital view.
1s2 2s22p2
This model predicts
CH2 with a bond angle
of 90 o
Energy
s
s
p
AO
Carbon
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Observation
Methane is CH4.
Need new model with 4 unpaired electrons.
Two possible solutions
unpair one of the s electrons and put it in a p
orbital.
Completely reorganize the orbital.
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Hybridization
A model that can be used to explain what
we observe.
Based on AOs
Linear Combination of an s and three p orbitals
Higher in energy than an s but lower than a p.
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Energy
s
sp3
Energy p
Carbon
s
Hybridized
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LCAO
Four AOs must yield 4 MOs
There is an error in the book
sp3 = 1/2( C2s + C2px + C2py + C2pz)
sp3 = 1/2( C2s - C2px - C2py + C2pz)
sp3 = 1/2( C2s + C2px - C2py - C2pz)
sp3 = 1/2( C2s - C2px + C2py - C2pz)
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Shape and Orientation
The electron distribution must have the
characteristics of both s (spherical) and p
(node) orbitals.
The four new electron distributions
(orbitals) will orient as far away from each
other as possible. Toward the corners of atetrahedron.
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Photoelectron Spectroscopy
Measures the energy of electrons ejected
from a molecule when hit by a photon.
Binding energy = photon energy - emitted
photon energy
h = T + EB
Methane shows >290 ev (C1s) and two morebands, 23.0 ev and 12.7 ev. (Two types of
orbitals)
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Orbital Symmetry
The MOs must be symmetric or anti-
symmetric with respect to the AOs (basis
set).
The sigma bond resulting from the C sp3
and H 1s do not have the symmetry of the s
an p orbital.
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New Approach
Consider all the AOs of C and 4 Hs.
Select symmetric or anti-symmetric linear
combination.
If we ignore the C1s electrons there should
be 4 bonding and 4 non-bonding orbitals.
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Orbitals and Energies
-13.30708 Kcal/mol
-28.87874 Kcal/mol
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VSEPR Theory
Valence Shell Electron PairRepulsion
Theory
Electrons in orbitals will take on a geometry
that has the electrons as far away from each
other as possible.
2 sp 180 deg.
3 sp2 120 deg.
4 sp3 109.5 deg.
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Methyl Chloride
Predicted bond angles
H-C-H 109.5o
H-C-Cl 109.5o
Experimental
H-C-H 110o 52
H-C-Cl 108o 0
Why?
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VSEPR Explanation
Cl more EN than Carbon. Draws electon
density toward Cl.
C-H bond shorter.
The electron density is shifted toward the Cl
so the Hs repel each other more than in
methane making H-C-H bond angle greater. The H are attracted to the Cl so the H-C-Cl
bond angle is smaller
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MO Explanation
The HOMOs show significant bonding
between the Cl and C-H bond
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Variable Hybridization
Because of the EN of chlorine the methyl
chloride molecule will be less symmetric.
The C-Cl bond will have more p character
which results in the C-H bond having more
s character.
C-H shorter and H-C-H angle larger.
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Hybridization Parameter
For a bond spn is the hybridization
parameter.
2 = n
For an sp3 the fractional s and p character
are1
1+ 2i% s = % p =2
i1+ 2i
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Interorbital Angles
Based on hybridization model
Angle A-C-B
Angle A-C-A
1 + a b cos ab = 0
1 + a2 cos aa = 0
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Methyl Chloride (again)
Two bond angles H-C-H and H-C-Cl
Relationship between these two angles
If we know one we can calculate the other.
If H-C-Cl is 108 o then
3 sin2ab = 2 (1 - cos a a
cos aa = 1 -3 sin
2ab
2
cos aa = 1 -3 (sin (108))
2
2
aa = 110o
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Carbon Orbital
Orbital to H.
sp2.86
The other C orbital to Cl
sp3.5
a2 =
- 1
cos aa
= 2.86
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1 + 2.86+
1
1 + b2
= 1
b2 = 3.50
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Curved Bonds
Consider CH2Cl2 Measured Cl-C-Cl 111o 47; H-C-H 112 o 0
Cal. C to Cl orbital sp2.69; C to H sp3.37 or a
bond angle of 107.
The bond angle does not fit experimental.
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Are Bonds Straight Lines
Between Nuclei?
Electron density may lie outside the line
between two nuclei.
Internuclear bond angle - defined by nuclear
positions
Interorbital bond angle - defind by the
direction the orbital leaves the atom.
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Bent Bonds
Best example cyclopropane
C-H hybridization sp2.36 which would make
the C-C sp2.69 which would predict an
interorbital angle of 111.8 o; > 60 o.
C
C
C
H
H
Interatomic 115.1
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Measurable Properties
Hybridization parameter can not be
measured directly but it can be correlated to
some measurable parameters.
J13C-H (cps) = 500/ (1 - 2)
rC-H = 1.1597 - 10-4 J13C-H
log k rel = 0.129 J13C-H - 15.9
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, Model for Ethene
Carbon is sp2 hybridized
H-C bond is shorter than normal because
bond has more s character
C-C bond has two parts; from overlap of
two C sp2 hybrids and a from overlap of
two p orbital. H-C-H 120 o and the same for H-C-C.
C-C shorter, more s character
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Energy Diagram
PEsp
2
Hybrid Orbitalsof Carbon
p
Hybrid Orbitalsof Carbon
sp2
p
MO's
to H 1s to H 1s
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Bent Bond Model
Pauling
Carbons are sp3 with two sp3 from each
carbon overlapping to for two bent bonds.
Predicts planar but bond angles of 109.5 o
C
H
H
C
H
H
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Acidity of Hydrocarbons
Hydrocarbons are weak acids.
Ethane 10-42, ethene 10-36.5, ethyne 10-25
R H + B R -
+ HB+
Ka =[R-] [HB+]
[RH] [B]
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Why?
The anion stability follows the s character
of the orbital that the resultant anion exist
in.
sp3 anion less stable than sp2 which is less
stable than sp. Therefore, alkanes less acidic
than alkenes which are less acidic thanalkynes
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Bent Bond Rationalization
The bent bond pulls the electron pair of the
anion closer to the center of the internuclear
axis of the C-C bond.
Less electron pair -- electron pair repulsion
HHH
H H
H
H H
H
0O
60O?
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Which Model is Correct?
Neither!
Both models are useful and must be used
with care.
Usually they converge on the same solution
when considered in great detail.