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Lecture 2
Simple Molecular Orbitals - Sigma and Pi Bonds in Molecules
bond order =
(number of bonding electrons) - (number of antibonding electrons)
amount of
bonding
1sa
hydrogen molecule = H 2
LUMO
HOMO
σ = 1s (^) a + 1sb = bonding MO =
potential energy
higher, less stable
lower, more stable
LUMO = lowest unoccupied molecular orbital HOMO = highest occupied molecular orbital
Similar phase of electron density (no node) adds together constructively.
energy of isolated atoms
bond order (H 2 molecule) =
2 = 1 bond
1sb
H H
H H
σ∗ = 1s (^) a - 1s (^) b = antibonding MO = H^ H
LCAO = linear combination of atomic orbitals node = zero electron density because of opposite phases
∆E = bond energy
There is a big energy advantage for a hydrogen molecule over two hydrogen atoms.
Sigma ( σ ) bonding molecular orbital - Shared electron density is directly between the bonding atoms,
along the bonding axis. A sigma bonds is always the first bond formed between two atoms.
Sigma star ( σ *) antibonding molecular orbital – Normally this orbital is empty, but if it should be
occupied, the wave nature of electron density (when present) is out of phase (destructive interference) and
canceling in nature. There is a node between the bonding atoms (zero electron density).
Problem 1 – What would the MO pictures of He 2 , H 2
, H 2
and He 2
look like? Would you expect that
these species could exist? What would be their bond orders?
node = zero electron density because of opposite phases
2pa
π bond
LUMO
HOMO π = 2p (^) a + 2p (^) b = bonding MO =
potential energy
higher, less stable
lower, more stable
LUMO = lowest unoccupied molecular orbital HOMO = highest occupied molecular orbital
Similar phase of electron density (no node) adds together constructively.
energy of isolated p orbitals
bond order of a pi bond =
(2) - (0) 2 = 1 bond
2pb
π∗ = 2pa - 2pb = antibonding MO =
LCAO = linear combination of atomic orbitals
∆E = bond energy
There is a big energy advantage for a pi bond over two isolated p orbitals.
Overlap is above and below the bond axis, not directly between the bonded atoms.
Lecture 2
Pi bond ( π ): bonding molecular orbital –The bonding electron density lies above and below, or in
front and in back of the bonding axis, with no electron directly on the bonding axis, since 2p orbitals
do not have any electron density at the nucleus. These are always second or third bonds overlapping
a sigma bond formed first. The HOMO of a pi system is especially important.
Pi star ( π *): antibonding molecular orbital – Normally this orbital is empty, but if it should be
occupied, the wave nature of electron density is out of phase (destructive interference) and canceling
in nature. This produces repulsion between the two interacting atoms, when electrons are present.
Atoms gain a lot by forming molecular orbitals. They have more stable arrangement for their electrons
and the new bonds help them attain the nearest Noble gas configuration.
The Hybridization Model for Atoms in Molecules
The following molecules provide examples of all three basic shapes found in organic chemistry.
C
H
H H
C
H
H
H
H C C H Ca Cb
H
H
Ca
ethane tetrahedral carbon atoms
HCH bond angles ≈ 109 o HCC bond angles ≈ 109 o
ethene trigonal planar carbon atoms
HCH bond angles ≈ 120 o^ (116 o^ ) CCH bond angles ≈ 120 o^ (122 o)
ethyne linear carbon atoms
HCC bond angles = 180 o
allene trigonal planar carbon atoms at the ends and a linear carbon atom in the middle
HCa H bond angles ≈ 120 o HCa Cb bond angles ≈ 120 o Ca Cb Ca bond angles = 180 o
C C
H H
H H
H
H
To understand these shapes we need to understand hybridization or atomic orbitals. In organic
chemistry our orbital mixtures will be simple combinations of the valence electrons in the 2s and 2p
orbitals on a single carbon atom. Though not exactly applicable in the same way for nitrogen, oxygen and
the halogens, this model will work fine for our purposes in beginning organic chemistry. We will mix
these orbitals three ways to generate the three common shapes of organic chemistry: linear (2s+2p),
trigonal planar (2s+2p+2p) and tetrahedral (2s+2p+2p+2p).
1. sp hybridization – carbon and other atoms of organic chemistry
2p's
2s isolated carbon atom (not typicalin our world)
promote a 2s electron to a 2p orbital
mix (2s and 2p), two ways (2s + 2p) and (2s - 2p) to create two sp hybrid orbitals
2p's
2s
2p
sp
sp arrangement for carbon atom bonded to other atoms, two p orbitals remain to become part of pi bonds
2p
sp
Overall, this would be a favorable trade.
cost = promotion energy ≈ 100 kcal/mole gain = electron/electron repulsion in s ortibal is removed ≈ 20-40 kcal/mole gain = two additional bonds are possible ≈ 150-200 kcal/mole gain = more directional orbitals form, that have better overlap of electron density between the bonding atoms, thus forming stronger bonds
potential energy
higher, less stable
lower, more stable
Lecture 2
We rarely draw our 3D structures like this, preferring simpler ways of representing the details.
H 2 C (^) C H
3D ethyne drawn with p orbitals as lobes (p orbitals with phase shown in the left structure and without phase in the right structure.
Alternative ways of drawing 3D structures that are simpler than the above drawing at showing the 3D details.
H H C C H
H C C H
3D ethyne drawn with p orbitals as lines and pi electrons explicitly drawn in, in a manner similar to showing lone pair electrons. In this book I will usually draw pi bonds this way in 3D structures.
p orbital lobes are in the plane of the paper.
p orbital lobe is in back of the paper.
p orbital lobe is in front of the paper.
H C C H
Each line represents a bond. While the three simple lines of the triple bond appear equivalent, we know that the first bond formed is a sigma bond of overlapping sp hybrid orbitals. The second and third bonds are overlapping 2p orbitals, above and below and in front and in back. Since the C-H bonds are single bonds, we know that they are sigma bonds too, using hybrid orbitals. This is how you will determine the hybridization of any atom in a structure. Knowing how many pi bonds are present will tell you how many 2p orbitals are being used in those pi bonds. The remaining s and 2p orbitals must be mixed together in hybrid orbitals (in this example, only an s and a 2p remain to form two sp hybrid orbitals).
HCCH
The connections of the atoms are implied by the linear way the formula is drawn. You have to fill in the details about the number of bonds and where they are from your understanding of each atom's bonding patterns. A C-H bond can only be a single bond so there must be three bonds between the carbon atoms to total carbon's normal number of four bonds. This means, of course, that the second and third bonds are pi bonds, using 2p orbitals, leaving an s and p orbitals to mix, forming two sp hybrid orbitals.
A bond line formula only shows lines connecting the carbon atoms and leaves off the hydrogen atoms. Every end of a line is a carbon (two in this drawing) and every bend in a line is a carbon (none in this drawing). You have to figure out how many hydrogen atoms are present by substracting the number of lines shown (bonds to non-hydrogen atoms) from four, the total number of bonds of a neutral carbon (4 - 3 = 1H in this drawing). The shape of the carbon atoms must be linear, because we know the hybridization is sp.
C 2 H (^2)
All of the details in this group go together. If you have any one of them, you should be able to fill in the remaining details.
This is the ultimate in condensing a structure. Merely writing the atoms that are present and how many of them there are provides no details about the connectivity of the atoms. It only works for extremely simple molecules that have only one way that they can be drawn. Ethyne is an example of such molecule. Other formulas may have several, hundreds, thousands, millions, or more ways for drawing structures. Formulas written in this manner are usually not very helpful.
carbon atom shape = linear hybridization = sp bond angles about sp carbon = 180o number of sigma bonds = 2 number of pi bonds = 2
Lecture 2
2. sp
2
hybridization
2p's
2s isolated carbon atom (not typicalin our world)
promote a 2s electron to a 2p orbital
mix (2s and two 2p's), three ways to create 2p's three sp^2 hybrid orbitals
2s
sp^2
sp^2 arrangement for a carbon atom bonded to other atoms, one p orbital remains to become part of a pi bond
2p
Overall, this would be a favorable trade.
cost = promotion energy ≈ 100 kcal/mole gain = electron/electron repulsion in s ortibal is removed ≈ 20-40 kcal/mole gain = two additional bonds are possible ≈ 150-200 kcal/mole gain = more directional orbitals form, that have better overlap of electron density between the bonding atoms, thus forming stronger bonds
sp^2 sp^2
potential energy
higher, less stable
lower, more stable
Creating sp
2
hybrid orbitals
2s sp^2 a 2px and 2py (^) mixing 2s+2p+2pone example of
mathematically, mix three ways
The "mixing" process symbolized here is repeated two additional ways, creating three sp 2 hybrid orbitals.
Similar phase mixes constructively in the right front quadrant
sp^2 a
sp^2 b
sp^2 c
All three sp^2 hybrid orbitals lie in a plane and divide a circle into three equal pie wedges of 120o^. The descriptive term for the shape is trigonal planar. Usually we draw the sp^2 hybrid orbitals without their small backside lobes and no p orbital is shown. These hybrid orbitals will form sigma bonds.
120 o
120 o
120 o
top-down view
sp2 hybrid orbitals
sp^2 a
sp^2 b
These represent sp^2 a hybrid orbitals. The small, opposite phase lobe on the backside has been left off to simplify the picture.
C
An isolated sp^2 hybridized carbon atom for viewing. A bonded carbon atom would need orbital overlap for each orbital present, sp^2 a, sp^2 b , sp^2 c and 2pz.
There remains one 2p orbital perpendicular to the three sp^2 hybrid orbitals. The 2p orbital extends along the entire axis with opposite phase in each lobe.
sp^2 2pz c
side-on view
The Complete Picture of an sp2 Hybridized Carbon Atom
C C (^) C
sp^2 hybridized carbon atom sp^2 hybridized carbon atom^ sp^2 hybridized carbon atom We will use this approach.
Examples of alternative methods of drawing a three dimensional sp carbon atom, using simple lines, dashed lines and wedged lines.
Lecture 2
3. sp
3
hybridization
2p's
2s isolated carbon atom (not typicalin our world)
promote a 2s electron to a 2p orbital
mix (2s and three 2p's), four ways to create 2p's four sp 3 hybrid orbitals
2s
sp^3
sp^3 arrangement for carbon atom bonded to other atoms, no 2p orbitals remain so no pi bonds can form
Overall, this would be a favorable trade.
cost = promotion energy ≈ 100 kcal/mole gain = electron/electron repulsion in s ortibal is removed ≈ 20-40 kcal/mole gain = two additional bonds are possible ≈ 150-200 kcal/mole gain = more directional orbitals form, that have better overlap of electron density between the bonding atoms, thus forming stronger bonds
sp^3 sp^3 sp^3
potential energy
higher, less stable
lower, more stable
Creating sp3 hybrid orbitals
sp (^2) a 2s 2px , 2py , 2pz One example is shown of mixing 2s+2p+2p+2p to form an sp^3 hybrid orbital.
mathematically, mix four ways
The "mixing" process symbolized here is repeated three additional ways, creating four sp 3 hybrid orbitals.
z
x
y
Similar phases interact constructively in the front, right, upper octant where the large lobe will be located.
C
A tetrahedron has four equivalent triangular sides. The atoms at the ends of the bonds with carbon define the vertices of the tetrahedron. The carbon is sitting in the middle of the tetrahedron.
sp^3 orbitals minus the small backside lobes
sp 3 orbitals drawn using our 3D conventions
C
= C
One sp3 carbon atom bonded in methane and two sp3 carbon atoms bonded in ethane
H C
H
H
H
All of the details in this group go together. If you have any one of them, you should be able to fill in the remaining details.
carbon atom shape = tetrahedral hybridization = sp 3 bond angles about sp carbon = 109o number of sigma bonds = 4 number of pi bonds = 0
C C
H
H
H
H
H
H
A single bond allows rotation to occur about the carbon-carbon bond, which alters the shape and the energy of the molecule.
C C
H
H
H
H
H
H
C-C single bond rotation
Lecture 2
Each line represents a bond. Since there are only single bonds, we know that they must be sigma bonds. There cannot be any pi bonds becasue there are no second or third bonds between the same two atoms. The 2s and all three 2p orbitals must all be mixed, meaning that the hybridization has to be sp 3 and all of the terms that go along with sp^3 hybridization.
H 3 CCH 3
The connections of the atoms are implied by the linear way the formula is drawn. You have to fill in the details about the number of bonds and where they are located from your understanding of each atom's bonding patterns. A CH 3 has three single bonds between carbon and hydrogen, so there can only be one additional bond between the carbon atoms to total carbon's normal number of four bonds. This means, of course, that there is no pi bond, using a 2p orbital, leaving the 2s and all three 2p orbitals to mix, forming four sp^3 hybrid orbitals.
A bond line formula only shows lines connecting the carbon atoms and leaves off the hydrogen atoms. Every end of a line is a carbon (two in this drawing) and every bend in a line is a carbon (none in this drawing). You have to figure out how many hydrogens are present by substracting the number of lines shown (bonds to non-hydrogen atoms) from four (the total number of bonds of a neutral carbon (4 - 1 = 3H on each carbon atom in this drawing).
C 2 H 6
This is the ultimate in condensing a structure. Merely writing the atoms that are present and how many of them there are provides no details about the connectivity of the atoms. A structure can be generated only for extremely simple molecules that have only one way that they can be drawn. Ethane is an example of such molecule.
C
H
H
H
C
H
H
H
or
CH 3 CH 3
Problem 2 – Draw a 3D representation or hydrogen cyanide, HCN. Show lines for the sigma bond
skeleton and the lone pair of electrons. Show two dots for the lone pair. Also show pi bonds represented
in a manner similar to above. What is different about this structure compared with ethyne?
Show lines for the sigma bond skeleton and the lone pairs of electrons with two dots for each lone pair.
Also show pi bonds represented in a manner similar to above. What is different about this structure
compared with ethene?
Problem 3 – Draw a 3D representation of methanal (common name = formaldehyde), H 2 C=O and
methanimine, H 2 C=NH. Show lines for the sigma bond skeleton and the lone pairs of electrons with two
dots for each lone pair. Also show pi bonds represented in a manner similar to above. What is different
about these structures compared with ethene above?
Problem 4 – Draw a 3D representation or hydrogen methanol, H 3 COH and methanamine, H 3 CNH 2. Show
lines for the sigma bond skeleton and a line with two dots for lone pairs, in a manner similar to above.
What is different about this structure compared with ethane above?
This represents all of our bonding pictures that you need to understand. Almost every example of
hybridization used in this book, whether carbon, nitrogen, oxygen or the halogens will use one of these
three shapes. I hope you can see how important this is to your organic career (…and your biochemistry
career).
Lecture 2
Molecular Orbital Diagrams
Ethyne MOs
First we need to form the sigma and sigma-star MOs between the two carbon atoms using their sp hybrid orbitals ( cc and *cc). The vertical scale represents relative potential energy among the various orbitals. Lower is more stable.
potential energy
higher, less stable
lower, more stable
C C
spa sp (^) b
σCC = spa + sp (^) b = bonding MO
σCC∗ = spa - spb = antibonding MO
C C
C C
sigma bond
sigma-star antibond, has a node
bond order =
= 1 bond
Next we need to form sigma and sigma-star MOs between each carbon atom and a hydrogen atom. We'll just show one MO diagram and you can imagine doing it twice.
C
sp
σCΗ = sp + 1s = bonding MO
σCΗ ∗ = sp - 1s = antibonding MO
C
C
sigma bond
bond order =
= 1 bond
H
1s
H
H
potential energy
higher, less stable
lower, more stable
Two C-H sigma/sigma-star MOs form. This scheme shows one of them. The other would look just like it.
sigma-star antibond, has a node
Finally we need to form two pi and pi-star MOs between the carbon atoms using carbon 2p orbitals ( CC and *CC). We'll just show one MO diagram and you can imagine doing it a second time.
πCC = 2p + 2p = bonding MO
πCC∗ = 2p - 2p = antibonding MO
pi bond
pi-star antibond
bond order =
= 1 bond
2p
C C
2p
C C
C C
potential energy
higher, less stable
lower, more stable
node
Lecture 2
If we put all of the molecular orbitals of ethyne together, in a single energy diagram, it would look as
follows. The pi MOs determine the highest occupied molecular orbital (HOMO) and lowest unoccupied
molecular orbital (LUMO). The 2p orbital overlap is the least bonding and the least antibonding. The
HOMO electrons are the easiest place to donate electrons from (least tightly held) and the LUMO orbital
is the best place to accept electrons, if accepted into the molecular orbitals (lowest potential energy empty
orbital = most stable of the empty orbitals). The pi molecular orbitals determine much of the chemistry of
alkynes.
potential energy
higher, less stable
lower, more stable
= 3 bonds
two πCC bonding MOs
two πCC∗ antibonding MOs
two σ∗CH antibonding MOs
one σ∗CC antibonding MOs
two σCH bonding MOs
one σCC bonding MOs
energy of orbitals on isolated atoms
LUMOs
HOMOs
Best place to take electrons from.
Best place to donate electrons to.
MO diagram for ethyne
bond order between the = C & C atoms, use σCC & two πCC σ CC
σCH σCH
πCC πCC
πCC∗ πCC∗
σ∗CH σ∗CH
σ∗CC
Problem 5 – Use ethyne (H-CC-H) as a model to draw an MO diagram for hydrogen cyanide (H-CN) and
propanenitrile (CH 3 CH 2 CN). Lone pairs of electrons belong to a single atom and are found at middle
energies (they do not form bonding and antibonding orbitals). Label lone pair orbitals with the letter “n”
for nonbonding electrons. Notice there is one fewer bond for each lone pair in the structures above.
Ethanenitrile (common name = acetonitrile),is shown below as an example.
potential energy
higher, less stable
lower, more stable
= 3 bonds
two πCN bonding MOs
two πCN∗ antibonding MOs
three σ∗CH antibonding MOs
one σ∗CC antibonding MO and one σ∗CN antibonding MO
three σCH bonding MOs
one σCC bonding MO and one σCN bonding MO, arbitrarily shown at same energy for simplicity
LUMOs
HOMO
Best place to take electrons from.
Best place to donate electrons to.
MO diagram for ethanenitrile
H 3 C C N
energy of nonbonding orbitals (lone pair electrons), same as orbitals on isolated atoms
n 1 MO (nitrogen)
σ^ σCN CC
σ∗CC σ∗CN
σCH σCH σCH
πCN πCN
πCN∗ πCN∗
σ∗CH σ∗CH σ∗CH
bond order between the = C & N atoms, use σCN & two πCN
Lecture 2
Ethane molecular orbitals. There are no pi and pi-star MOs in ethane. There are now seven sigma bonds (six
C-H and one C-C) and zero pi bonds. The important HOMO / LUMO orbitals are the sigma and sigma-star
MOs in this example (our designation of relative sigma energies is arbitrary). Because there are no pi / pi-
star HOMO / LUMO molecular orbitals, ethane is much less reactive than ethyne and ethene.
potential energy
higher, less stable
lower, more stable
= 1 bond
six σ∗CH antibonding MOs
one σ∗CC antibonding MO
six σCH bonding MOs
one σCC bonding MO
energy of orbitals on isolated atoms
LUMO
HOMO
MO diagram for ethane
σCC
σCH σCH σCH σCH σCH σCH
σ∗CH σ∗CH σ∗CH σ∗CH σ∗CH σ∗CH
πCC∗
bond order between the = C & N atoms, use σCC
Problem 7 – Use ethane (CH 3 -CH 3 ) as a model to draw an MO diagram for methyl amine (CH 3 NH 2 ) and
methanol (CH 3 -OH). Lone pairs of electrons belong to a single atom and are found at middle energies
(they do not form bonding and antibonding orbitals). Label lone pair orbitals with the letter “n” for
nonbonding electrons. If there is more than one lone pair label them as n 1 and n 2. And show them at the
same energy. Notice there is one fewer bond for each lone pair in the structures above. Dimethyl ether is
shown below as an example.
potential energy
higher, less stable
lower, more stable
= 1 bond
six σ∗CH antibonding MOs
two σ∗CO antibonding MOs
six σCH bonding MOs
two σCO bonding MOs
energy of nonbonding orbitals (lone pair electrons), same as orbitals on isolated atoms
LUMO
HOMOs
MO diagram for dimethyl ether
H 3 C O CH 3
n 2 MO (oxygen)
n 1 MO (oxygen)
bond order between the = C & O atoms, use σCO σCO^ σCO
σCH σCH σCH σCH σCH σCH
σ∗CH σ∗CH σ∗CH σ∗CH σ∗CH σ∗CH
σ∗CO σ∗CO
