CLAYDEN. Organic Chemistry. 2ª edição. Oxford. 2012.
1261 pág.

CLAYDEN. Organic Chemistry. 2ª edição. Oxford. 2012.


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molecule is made up from 
isolated atoms. Take ethane, for example. Each carbon uses three sp3 AOs orientated 
towards the three hydrogen atoms, leaving one sp3 orbital on each carbon atom for the 
C\u2013C bond.
In the MO energy level diagram we now have both C\u2013H bonding \u3c3 and antibonding \u3c3* 
orbitals (made from combining sp3 orbitals on C with 1s orbitals on H) and also a C\u2013C 
bonding \u3c3 and antibonding \u3c3* orbital, made from two sp3 orbitals on C. The diagram 
below just shows the C\u2013C bond.
molecular orbitals of ethane (just C\u2013C bond shown)
e
 n
 e
 r
 g
 y
2p
2s
filled bonding C\u2013C \u3c3 orbital
mix to make 4 x sp3 orbitals
C
2p
2s
C
3 x sp3 used in
bonds to 3 x H
3 x sp3 used in
bonds to 3 x H
mix to make 4 x sp3 orbitals
empty antibonding C\u2013C \u3c3\u2217 orbital
For ethene (ethylene), the simplest alkene, we need a new set of hybrid orbitals. Ethene is a 
planar molecule with bond angles close to 120°. Our approach will be to hybridize all the 
H
H
H
H
C
methane
 Interactive bonding orbitals in 
methane
H
H
H
C
H
H
H
C
C\u2013H \u3c3 bonds made from
overlap of sp3 orbital on
C and 1s orbital on H
C\u2013C \u3c3 bond made from
overlap of sp3 orbital on
each C atom
ethane
H
H
H
H
ethene (ethylene)
117.8°
C\u2013H bonds
108 pm
C\u2013C bond
133 pm
CHAPTER 4   STRUCTURE OF MOLECULES100
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orbitals needed for the C\u2013H framework and see what is left over. In this case we need three 
equivalent bonds from each carbon atom (one to make a C\u2013C bond and two to make C\u2013H 
bonds). Therefore we need to combine the 2s orbital on each carbon atom with two p orbitals 
to make the three bonds. We could hybridize the 2s, 2py, and 2pz orbitals (that is, all the AOs 
in the plane) to form three equal sp2 orbitals, leaving the 2pz orbital unchanged. These sp2 
hybrid orbitals will have one-third s character and only two-thirds p character.
2s
2px
2py 2pz
+ +
add together then divide into
three equivalent hybrid orbitals:
s + p + p
3
3 x (1/3s + 2/3p)
sp2 orbital
plus remaining 2px orbital
The three sp2 hybrid AOs on each carbon atom can overlap with three other orbitals (two 
hydrogen 1s AOs and one sp2 AO from the other carbon) to form three \u3c3 MOs. This leaves the 
two 2px orbitals, one on each carbon, which combine to form the \u3c0 MO. The skeleton of the 
molecule has \ufb01 ve \u3c3 bonds (one C\u2013C and four C\u2013H) in the plane and the central \u3c0 bond is 
formed by two 2px orbitals above and below the plane.
+ CC
each C atom has three
sp2 orbitals plus a p orbital
H
H
H
H
CC
H
H
H
H
C\u2013C \u3c3 bonds made from
overlap of sp2 orbital on each C atom
C\u2013C \u3c0 bond made from overlap 
of p orbital on each C atom
This is the \ufb01 rst MO picture we have constructed with a C=C double bond, and it is worth 
taking the time to think about the energies of the orbitals involved. We\u2019ll again ignore the 
C\u2013H bonds, which involve two of the sp2 orbitals of each C atom. Remember, we mixed two 
of the three 2p orbitals in with the 2s orbital to make 3 × sp2 orbitals on each C atom, leaving 
behind one unhybridized 2p orbital.
Now, \ufb01 rst we need to generate the \u3c3 and \u3c3* orbitals by interacting one sp2 orbital on each atom. 
Then we need to deal with the two p orbitals, one on each C, which interact side-on. The unhy-
bridized p orbitals are a bit higher in energy than the sp2 orbitals, but they interact less well (we 
discussed this on p. 93) so they give a \u3c0 orbital and a \u3c0* orbital whose energies are in between the 
\u3c3 and \u3c3* orbitals. Each C atom donates two electrons to these orbitals (the other two electrons are 
involved in the two bonds to H), so the overall picture looks like this. Two AOs give two MOs.
e
 n
 e
 r
 g
 y
2p
2s
filled bonding C\u2013C \u3c3 orbital
mix to make
3 x sp2 orbitals
C C
2 x sp2 used in
bonds to 2 x H
keep one
p orbital
unhybridized
2 x sp2 used in
bonds to 2 x H
2p
2s
2p 2p
keep one
p orbital
unhybridized
mix to make
3 x sp2 orbitals
sp2 sp2
filled bonding C\u2013C \u3c0 orbital
\u3c3\u2217
\u3c0\u2217
unfilled antibonding orbitals
molecular orbitals of ethene (ethylene)
 Interactive bonding orbitals in 
ethene
HYBRID IZAT ION OF ATOMIC ORB ITALS 101
2069_Book.indb 101 12/12/2011 8:24:37 PM
The fact that the sideways overlap of the p orbitals to form a \u3c0 bond is not as effective as the head-on overlap of the 
orbitals to form a \u3c3 bond means that it takes less energy to break a C\u2013C \u3c0 bond than a C\u2013C \u3c3 bond (about 260 kJ 
mol\u22121 compared to about 350 kJ mol\u22121).
CC CC
overlap of orbitals is more efficient 
in a \u3c3 bond than in a \u3c0 bond
Ethyne (acetylene) has a C\u2261C triple bond. Each carbon bonds to only two other atoms to 
form a linear CH skeleton. Only the carbon 2s and 2px have the right symmetry to bond to the 
two atoms at once so we can hybridize these to form two sp hybrids on each carbon atom, 
leaving the 2py and 2pz to form \u3c0 MOs with the 2p orbitals on the other carbon atom. These 
sp hybrids have 50% each s and p character and form a linear carbon skeleton.
2s
2px 2py
2pz
+
add together then divide into
two equivalent hybrid orbitals:
s + p
2
2 x (1/2s + 1/2p)
sp orbital
plus two remaining 2p orbitals
We could then form the MOs as shown below. Each sp hybrid AO overlaps with either a 
hydrogen 1s AO or with the sp orbital from the other carbon. The two sets of p orbitals com-
bine to give two mutually perpendicular \u3c0 MOs.
+
each C atom has two
sp orbitals plus two p orbitals
H H CC
C\u2013C \u3c3 bond made 
from overlap of sp 
orbital on each
C atom
C\u2013C \u3c0 bonds each 
made from overlap of 
one p orbital on each 
C atom
H HCC
Hydrocarbon skeletons are built up from tetrahedral (sp3), trigonal planar (sp2), or linear (sp) 
hybridized carbon atoms. Deciding what sort of hybridization any carbon atom has, and 
hence what sort of orbitals it will use to make bonds, is easy. All you have to do is count up the 
atoms bonded to each carbon atom. If there are two, that carbon atom is linear (sp hybrid-
ized), if there are three, that carbon atom is trigonal (sp2 hybridized), and if there are four, that 
carbon atom is tetrahedral (sp3 hybridized). Since the remaining unhybridized p orbitals are 
used to make the \u3c0 orbitals of double or triple bonds, you can also work out hybridization state 
just by counting up the number of \u3c0 bonds at each carbon. Carbon atoms with no \u3c0 bonds 
are tetrahedral (sp3 hybridized), those with one \u3c0 bond are trigonal (sp2 hybridized), and those 
with two \u3c0 bonds are linear (sp hybridized).
There\u2019s a representative example on the left. This hydrocarbon (hex-5-en-2-yne) has two linear 
sp carbon atoms (C2 and C3), two trigonal sp2 carbon atoms (C5 and C6), a tetrahedral sp3 CH2 
group in the middle of the chain (C4), and a tetrahedral sp3 methyl group (C1) at the end of the 
chain. We had no need to look at any AOs to deduce this\u2014we needed only to count the bonds.
We can hybridize any atoms
We can use the same ideas with any sort of atom. The three molecules shown on the next page 
all have a tetrahedral structure, with four equivalent \u3c3 bonds from the central tetrahedral sp3 
ethyne (acetylene)
HH
 Interactive bonding orbitals in 
ethyne
CH3
H H
H
H
H
1
2
3
45
6
hex-5-en-2-yne
CHAPTER 4   STRUCTURE OF MOLECULES102
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atom, whether this is B, C, or N, and the same total number of bonding electrons\u2014the mol-
ecules are said to be isoelectronic. The atoms contribute different numbers of electrons so to 
get the eight bonding electrons we need we have to add one to BH4 and subtract one from 
NH4\u2014hence the charges in BH4
\u2212 and NH4
+. In each case the central atom can be considered to 
be sp3 hybridized, using an sp3 orbital