Principles of Organic Chemistry
lecture 9, page 1
ORGANIC STURCTURE
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Enolate of acetone in terms of MOs
o Last time we were in the process of connecting the MO phenomena of (.)H3,
to allyl radical to allyl anion to the enolate of acetone.
o The bell rang
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Variable Hybridization and Molecular Geometry
o Felix states that an incisive window into molecular geometry is had by the
concept of variable hybridization.
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o Author continually refers to the hybridization of orbitals when the inputs
are only bond angles.
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When a set of orbitals is manipulated mathematically to establish a
hybridization scheme, we can refer to the result as orbitals
hybridized in a certain way.
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When the hybridization is calculated from data (bond angles) the
result tells us the s versus p orbital character in the bond. All the
orbitals contribute to this bond.
o My undergrads see this concept in Organic Chemistry I (che230)
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http://www.chem.uky.edu/courses/che230/ACG/230.cammers.200
3/lecture/05.2003.09.08.pdf
Hybridization describes bonds or orbitals.
o What exactly is the difference between a bond and an orbital?
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When a molecule breaks into two can we say that a particular orbital
has been broken, destroyed or disrupted?
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Remember that orbitals are mathematical descriptions used to
model the reality of electronic populations.
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When a molecule becomes two molecules electronic
populations around the nuclei are certainly redistributed.
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However the stability of more than one orbital contributes to
the stability of the bond.
Referring to the hybridization of a bond and the hybridization of an
orbital are two different matters entirely.
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When we consider hybridization changes as a function of
molecular geometry, we are talking about bond
hybridization.
You may have learned that atoms and or molecules can be assigned hybridization.
Principles of Organic Chemistry
lecture 9, page 2
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This is not true.
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In the reading and this lecture you learned that hybridization applied
to orbitals generates localised orbitals.
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Strictly, hybridization is a description of either atomic orbitals or
bonds derived from these orbitals at a particular atom.
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When we talked about methane in terms of its molecular orbitals,
we described the sp3 hybridized bonds in the methane molecule in
terms of orbitals derived from carbon s and p atomic orbitals.
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Hybridization in the delocalized description of methane
does not apply to the orbitals, they were certainly NOT
hybridized.
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Hybridization in the delocalized description of methane
does apply to the bonds.
o All four borrowed equally from C atom’s s and p
atomic orbitals.
o All four had equal s character.
o All four were sp3 hydridized.
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Geometry and hybridization/ think about the reactivity of the terminal
alkyne, alkene, and alkane!
o Stability of electrons, anions, radicals etc. Stability of s versus p
orbitals.
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s orbitals stabilize bonds and electrons better.
pKa
~50
~35
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
o
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~25
H
H
H
H
H
H
pKa is an energy parameter!
---------------------------------------------------------------------------PKa of alkanes, alkenes and alkynes
- think about the creation of anions by deprotonation of
- ethane, ethylene, and acetylene
a proton abstracted by a base is generalized in the following reaction
Principles of Organic Chemistry
AH + B(-)
lecture 9, page 3
A(-) + BH
the half reaction in water
AH + H2O
A(-) + H3O(+)
has an equilibrium constant associated with it
Keq = [A][H3O]/[AH][H2O]
The terms in this equation are concentration. The molarity of water, [H2O] is 55 M or 55
moles/Liter
The acidity constant is
Ka = [A][H3O]/[AH]
Thus,
Ka = Keq[H2O]
Taking the negative log of both sides and multiplying
by RT gives:
∆G
-RTlogKa = -RTlog(Keq[H2O])= -RTlogKeq RTlog[H2O]
-2.30RTlogKa = -RTLnKeq - RTLn[H2O]
anions
Rxn Coordinate
but –logKa = pKa
and –RTLnKeq = ∆G
2.30RTpKa + RTLn[H2O]= ∆G
---------------------------------------------------------------------------o The last equation above says that there is a linear relationship between the
stability of the anions in the deprotonation reaction and the pKa of the
conjugated carbon acid.
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hybridization as a function of molecular Geometry
o The fractional s character of the ith bond at a particular atom in a molecule is
given by:
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1/(1+λi2) and ∑i (1/(1+λi2) = 1
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Why? Do you recognize the equation above as a restatement
of something you already know?
o The fractional p character is given by:
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λi2/(1+λi2) and ∑i λi2/(1+λi2) = 3
Principles of Organic Chemistry
lecture 9, page 4
o If spn is the hybridization of said bond at a given atom.
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λi2 = n of the ith spn hybridized bond.
o Hybridization is related to molecular geometry as we discussed last time in
the inversion of ammonia and phosphine.
o Example: sp3, in spn
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λ i2 = n = 3
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Fraction of s character:
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1/(1+λi2 )= 0.25 or 25%
o The bond angle between two spn hybridized orbitals is in the case of
unequal orbital hybridization.
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1 + λaλbcosθ = 0
o for equal hybridization
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1 + λa2cosθ = 0
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in sp3,
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cosθ = −1/3
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θ = 109.47°
o as shown before from a vectorial argument.
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On the variable hybridization of H3C-X
o Given that Cl-C-H = 108° and H-C-H = 110.9°
o λC-H2 = -1/cos(110.9°) = 2.80
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thus sp2.8
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since 1 + λaλbcosθ = 0
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1+√2.8* λC-Clcos(Cl-C-H) = 1+√2.8* λC-Clcos(108°) = 0
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-1/√2.8/ cos(108°) = λC-Cl = 1.93
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and λC-Cl2 = 3.74
o This is not how your author does it but it works just as well.
o We can rationalize this result using the following resonance structures.
Cl
Cl
H
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H
H
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H
H
H
Cl atom is more electronegative than the C atom.
Principles of Organic Chemistry
lecture 9, page 5
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The carbon atom in a cation is flat with an empty p orbital.
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If the charge separated resonance structure contributes to the
real structure, the C-Cl bond should be rich in p character as
indicated by sp3.74.
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Likewise the C-H bonds in CH3Cl give up p character in an
sp2.8 hybridization scheme that is decidedly less p character
than sp3, plain vanilla tetrahedral.
o ∑i λi2/(1+λi2) = 3
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3.74/(1+3.74) + 3 (x/(1+x) = 3
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0.7890 + 3x/(1+x) = 3
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3x/(1+x) = 2.211
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3x= 2.211+2.211x
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0.789x = 2.211
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x = λi2 = 2.8
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CH3Cl, HCH = 110° 0', CH3Br, HCH = 110° 15', CH3I,
HCH = 110° 58 I (Walter Gordy, James W. Simmons, and
A. G. Smith, Department of Physics, Duke University,
Durham, North Carolina, Phys. Rev. 1948, 74 243–249.
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Thus the hybridization of the CH bonds at C in the methyl
halides are X = Cl, Br, I are sp2.92, sp2.89, sp2.79.
o The p character appears to decrease.
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The author is surprised that F breaks the trend.
o The C-F bond is very strong compared to the other
halides: 108 kcal/mol for CH3F vs. 83 kcal/mol for
CH3Cl vs. 70 kcal/mol for CH3Br vs. 56 kcal/mol for
CH3I.
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The CF bond is short and strong.
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The shorter the bond gets the more
pyramidal the CH3 fragment should
get.
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The more polar the bond the flatter
the CH3 fragment should get.
Bent Bonds
o The author jumps into an apology for the hybridization / geometry
argument by suggesting that the bonds in CH3F may not be straight.
Principles of Organic Chemistry
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lecture 9, page 6
He sights strained rings and their attendant bent-bond descriptions
offered to explain their reactivity and electron density.
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However it is strain that drives these anomalies.
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INSTRUCTOR shows similar reactivity of alkenes and
cyclopropanes.
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Electron density outside the bond vectors.
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My problem with bent bonds. Think about benzene!
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What will you have to do to get a symmetrical structure?
o Author makes an important point at the end of Chapter 1.
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We can’t judge the relative validity of a bent bond description of
‘double bonds’ versus a description of double bonds as a sigma and
a pi molecular orbital.
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Each description has it strong and weak points.
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The electrons are not the mathematicians we are.