3. Alkenes and Alk ynes.
3.1 H ybridiza tion in Alkenes.
When C is bound to four groups, as we saw with methane, it is
sp3-hybridized, which means mathematically that the s orbital
and all three p orbitals are added up and divided by four to
make four new orbitals. We can hybridize orbitals in other
ways, too. If we combine the s orbital with two p orbitals and
leave the third one unaltered, we have sp2 hybridization. An
sp2-hybridized atom has three equivalent sp2 orbitals, each with
1/3 s and 2/3 p character, and one unadulterated p orbital.
The sp2 orbitals are 120° apart. If we combine the s orbital with
one p orbital and leave the other two unaltered, we have sp
hybridization. An sp-hybridized atom has two equivalent sp
orbitals, each with 1/2 s and 1/2 p character, and two
unadulterated p orbitals. The sp orbitals are 180° apart. The
two p orbitals are perpendicular to each other and to the line
containing the sp orbitals.
Under what circumstances do we find C(sp3 ), C(sp2 ), and
C(sp ) hybridization, and how does C "choose" a particular
hybridization? Suppose CH4 were sp2-hybridized. Three H atoms
would be coplanar, and the fourth, whose s orbital overlapped
with the p orbital, would have a 90° angle with respect to the
other three. Moreover, half the p orbital's bonding density
would be wasted on the side of the C atom opposite the unique
H atom. So we can see that sp3 hybridization is best for a C
atom that is bound to four different atoms.
However, when a C atom is bound to three different atoms, the
situation changes. Consider ethane (ethylene, H2C=CH2 ). If the
C atoms were sp3-hybridized, we would need to use two hybrid
orbitals to construct the double bond. They would need to be
canted with respect to the C–C axis, thus putting only a small
amount of electron density between the two C nuclei. Not a
good situation!
To put maximum electron density between the two C nuclei,
the C atom must place an orbital in the same plane as the C–H
bonds. This arrangement is achieved by sp2 hybridization. The
three C(sp2 ) orbitals on one C atom are used to form two σ
bonds to H atoms and one σ bond to the other C. The p orbital
is used to form the second bond to the other C atom. The C(p)
orbitals overlap with each other lengthwise, giving a bond that
does not look like the σ bonds we have seen already. This bond
is a π bond. It has a node, i.e., a plane in which no electron
density resides. The C(p) orbital has a higher energy than the
C(sp2 ) orbital used to make the σ bond, and the C(p)–C(p)
overlap is poorer than the C(sp2 )–C(sp2 ) overlap, so the energy
of the C–C π bond is higher than the energy of the C–C σ bond.
Whenever you have a double bond between two atoms, there
is one σ MO and one π MO. The energy of the C(sp2 )–C(sp2 ) σ
bond is lower than the energy of the C(sp3 )–C(sp3 ) bond of
ethane, because the C(sp2 ) orbitals are lower in energy than a
C(sp3 ) orbital.
Just like there is a σ* orbital associated with a σ orbital in the
H–H bond, so there are anti-bonding orbitals associated with
the C(sp2 )–C(sp2 ) σ bond and the C(p)–C(p) π bond. A C atom
that makes one double bond and two single bonds is sp2hybridized.
3.2 ( N o ) Ro t a tion in Alkenes.
We saw that the two ends of ethane can rotate with respect to
one another with only a small barrier to rotation due to the loss
of hyperconjugative stabilization in the eclipsed isomer. Is the
same true in ethylene? If you look down the C–C axis in
ethylene, you will see that each C–H bond on one C has a
180° dihedral angle with a C–H bond on the other C. A 90°
rotation about the C–C bond gives a new conformational
diastereomer of ethylene, in which no C–H bond has a 180°
dihedral angle with another C–H bond. As a result, you might
expect that ethylene would be predominantly planar, with a
small barrier to rotation. In fact, the barrier to rotation in
ethylene is very high, about 66 kcal/mol. Why? In the 90°
isomer of ethylene, the two p orbitals are perpendicular to each
other, so they don't overlap. No overlap, no bond. In other
words, rotation about the C=C bond in ethylene breaks the π
bond, increasing the energy of the electrons associated with
that bond, so it is a very high energy process. Since room
t emp era ture supplies abou t 1 5 kcal/mol of energ y,
t h er e is no ro t a tion abou t C=C π bonds a t room
t emp era ture!
3.3 Subs ti tu t ed and Higher Alkenes. Configura tional
Dias t er eomers.
The four C–H bonds in alkenes are identical. Replacing an H with
another group results in a vinyl compound. E.g., vinyl chloride is
CH2=CHCl. The "vinyl" you are familiar with is polyvinyl chloride
(PVC), a polymer that is made by linking multiple vinyl chloride
units together. (The electrons linking the units together come
from the π bond of vinyl chloride, so PVC has no π bonds.)
Another name for vinyl chloride is chloroethylene. In ethene,
"eth" means "two carbons", and "ene" means that there's a
double bond. If we replace an H in ethene with CH3 to get a
three-carbon compound with a double bond, we call it propene.
Propene has three different kinds of C atoms. Two are sp2hybridized, and the third is sp3-hybridized. The energy of the
C(sp3 )–H bond is higher than the energy of the C(sp2 )–H bond
because the energy of the C(sp3 ) orbital is higher than the
energy of the C(sp2 ) orbital. We number the C atoms in
propene from one end to another. The ends of propene are
different, so we can number it in two different ways. We do it
so that the atoms of the π bond have the lowest numbers
possible.
We can replace one of the H atoms of propene with Cl, to get
the skeletal isomers 1-chloropropene, 2-chloropropene, or 3chloropropene. Actually, there are two kinds of 1chloropropene. There is the kind where the Cl atom is near the
CH3 group attached to C(2), and there is the kind where the Cl
atom is near the H atom attached to C(2). The relationship
between these two compounds is that they have the same
atom-to-atom connections, but different shapes, so they are
stereoisomers. They are stereoisomers that have different
internal dimensions, so they are diastereomers. Because
rotation to interconvert one diastereomer into the other is a
high-energy process, requiring cleavage of a π bond, they are
configurational diastereomers. (In the past, isomers such as
these have been called double-bond isomers or geometric
isomers.)
Any alkene in which one C of the C=C bond is attached to two
different groups, and the other C is also attached to two
different groups, will have two diastereomeric forms. For
alkenes in which each C is attached to one H and one other
group, we call these two forms cis and trans. So cis-1chloropropene is the diastereomer in which the Cl and the CH3
group are on the same side of the π bond, and trans-1chloropropene is the diastereomer in which the Cl and the CH3
group are on opposite sides of the π bond.
Remember "cis — same side". In a moment we will see another,
more general way of naming the two diastereomers of alkenes.
If we replace a terminal H in propene with a CH3 group, we
generate either 1-butene or 2-butene. (If one replaces the
internal H in propene with CH3, we generate 2-methylpropene,
also known as isobutylene.) The number designates the
position of the C=C bond in the chain. Not only do 1-butene
and 2-butene have their π bond in different locations, but, by
necessity, the C–H bonds have also been moved around, so 1butene and 2-butene are skeletal isomers. Just as there are in
1-chloropropene, there are two diastereomers of 2-butene: cis2-butene and trans-2-butene. They cannot interconvert by
rotation, so they are configurational diastereomers.
The interconversion of cis and trans alkenes requires cleavage
of the C=C π bond. This can happen when light of the proper
wavelength is absorbed by an alkene, which promotes one
electron from the π to the π* orbital. Because 1/2 bond + 1/2
anti-bond = no bond, there is now no π bond and the molecule
is free to rotate about what used to be the π bond.
Nomenclature
Higher alkenes are named as you would expect. The root
designates the number of C atoms in the longest chain
containing the double bond, the suffix "ene" designates an
alkene, the number preceding the root designates the position
of the double bond in the chain, and any prefixes designate
substituents. An alkene with two double bonds is called a
diene, one with three is called a triene, etc., and more than one
number is used before the root to designate the position of the
double bonds: e.g., 1,3-butadiene or 1,4-cyclohexadiene. A
cycloalkene is named so that the C's of the π bond are 1 and 2
and so that the first substituent has as low a number as
possible. All cycloalkenes up to 8-membered rings must be cis,
so we don't need to indicate whether they are cis or trans in
their name. (Try to make a model of transcyclohexene!) Larger
cycloalkenes, however, must be so designated: e.g., trans- or
cis-cyclodecene.
A compound XHC=CHY may be classified as cis or trans.
But what does one do for a compound like, say, 1-iodo-1bromo-2-chloro-1-propene? This compound can also exist as
two diastereomers, but it's not clear which one is cis and which
one is trans. In these cases we use the E/Z nomencla ture.
First we need to assign priorities to the four groups attached
to the double bond. Look at the atomic numbers of the two
atoms attached to each C of the double bond. One C has I and
Br attached: I has higher priority. The other C has C and Cl
attached: Cl has higher priority. (It is helpful to circle the
higher-priority group attached to each C.) If the two highpriority groups are on the same side of the double bond, the
compound is "zusammen" (German for together), or Z; if they
are on opposite sides, the compound is "entgegen" (German
for against), or E. In the name of the compound, the E or Z
follows immediately after the number indicating the position of
the double bond.
If the two atoms directly attached to a C are the same, one
looks for the first difference between the groups attached to
each of those atoms to determine which is higher priority. For
example, in the compound below, C2 is attached to C1 and C6:
no difference. C1 is attached to O, O, and O (the
double bond to O is counted as two O atoms); C6 is attached
to S, H, and H. The heaviest atom attached to C6 is heavier
than the heaviest atom attached to C1, so C6 has higher
priority. Similarly, C3 is attached to C4 and C7: no difference.
C4 is attached to C5, C9, and H; C7 is attached to C8, H, and
H. The heaviest atom attached to C4 is the same as the
heaviest atom attached to C7, but the second-heaviest atom
attached to C4 is heavier than the second-heaviest atom
attached to C7, so C4 has higher priority. The two high-priority
groups, C4 and C6, are against one another, so the isomer is E.
We assign priority to groups in compounds so that we can
name the compounds. It is purely a matter of language. It has
no chemical reality whatsoever.
3.4 S t abilit y o f Alkenes.
There are two rules involved in evaluating the relative stability
of two alkenes:
1. Trans alkenes are lower in energy than cis alkenes (except
when the alkene is in a ring with 10 or fewer members).
2. More substituted alkenes (alkenes with more non-H groups
attached to the C=C bond) are lower in energy than less
substituted alkenes.
The reason that cis alkenes are higher in energy than
trans alkenes is simple. In a cis alkene, the two large groups are
both on the same side of the π bond, so they bump into one
another, creating a steric interaction. This interaction is not
present in the trans isomer. The steric interaction can cost as
little as 0.9 kcal/mol for 2-butene, but it can be larger if larger
groups are attached to the C=C bond.
The reason that more substituted alkenes are lower in
energy than less ubstituted ones is more subtle, and it is still a
matter of controversy. Jones gives one explanation; I will give
another one, again involving hyperconjugation. The C=C π bond
has both a bonding and antibonding orbital associated with it.
The π* orbital is not very high in energy, and it can interact
with C–H σ bonds of the type C=C–C–H. (In other words, the C–
H bonds must involve a C that is adjacent to but not part of
the C=C bond.) The more C–H σ bonds that can hyperconjugate
with the π bond, the lower in energy the molecule is. Other
bonds, e.g. C–C bonds, can also interact hyperconjugatively
with the C=C π* orbital, as long as they involve a C that is
adjacent to but not part of the C=C bond.
We mentioned earlier that all cycloalkenes up to 8-membered
rings must be cis. The trans isomers are much too high in
energy to exist at room temperature. In a trans alkene, the two
allylic C atoms are so far apart that one needs at least four C
atoms to bridge them.
In bridged bicyclic compounds, the C atoms at the bridging
points (those C atoms attached to three C atoms of the ring
system) cannot participate in a π bond. The reason is that the
p orbital of the bridgehead C atom cannot overlap with the p
orbital on a neighboring C atom. This phenomenon is called
Br ed t 's rule. Bredt's rule applies only to smaller bicyclic
compounds. In large rings, Bredt's rule can be violated (and is
violated regularly).
3.5 H ybridiza tion in Alk ynes.
The same logic that told us that the C atoms in ethylene
cannot be sp3-hybridized tells us that the C atoms in ethyne
(acetylene, HC≡CH) cannot be sp3- or sp2-hybridized. Instead,
each C atom is sp-hybridized. The s orbital is averaged with
just one p orbital to make two new hybrid sp orbitals that are
pointing 180° from one another. The energy of the sp orbital is
halfway between that of an s orbital and that of a p orbital.
One of the sp orbitals is used to make a σ bond with the other
C, and the other is used to make a σ bond with H. The two
remaining, unhybridized p orbitals are used to overlap with two
p orbitals on the other C atoms to make two π bonds.
Whenever you have a triple bond between two atoms, there
are one filled σ MO and two filled π MOs (and the corresponding
empty anti-bonding MOs). Again, there are anti-bonding
orbitals associated with the C(sp)–C(sp) σ bond and the C(p)–
C(p) π bonds. A C atom that makes one triple bond and one
single bond is sp-hybridized.
Even though it's hard to tell, the π electrons around the C≡C
bond form a hollow cylinder of electron density.
3.6 H ybridiza tion in He t eroa t o ms.
Heteroatoms also hybridize. We treat most lone pairs as if they
were groups to which the heteroatom was bound; usually they
reside in hybrid orbitals. If a heteroatom has only σ bonds to its
neighbors, it is sp3-hybridized. (Examples: dimethyl ether,
ammonia.) If a heteroatom has one π bond to one neighbor, it
is sp2-hybridized. (Examples: pyridine, acetone, protonated
acetone.) If a heteroatom has two π bonds to its neighbors, it
is sp-hybridized. (Acetonitrile.) These hybridizations have
consequences for both structure (bond angles and lengths)
and reactivity (lone pairs with certain energies).
To summarize: Hybrid orbitals are used to form σ bonds and to
contain lone pairs. Unhybridized p orbitals are used to form π
bonds or are empty. To de t ermine th e hybridiza tion of an
a tom, coun t th e number of σ bonds and lone pairs no t
used in resonanc e; you need tha t man y hybrid
orbitals.
Problems. For each of the following compounds: (1) Fill in any
non-bonding pairs of electrons. (2) Indicate the kinds of
orbitals (π and σ) that are made. (3) Indicate the hybridization
of each non-H atom. (4) Indicate the approximate bond angles.
(a) Formaldehyde (methanal), H2C=O. (b) 2-butynal Nmethylimine,
CH3–C≡C–CH=N–CH3.
(c)
Methyl
acetate,
CH3CO2CH3. (d) Nitrilamine, H2N–CN. (e) Trimethyl borate,
(CH3O) 3 B.
We said that lone pairs are put in hybrid orbitals. If the lone
pair can be used in resonance, however, it must be in a p
orbital for maximum overlap to occur. Therefore: Hybrid
orbitals are used to form σ bonds and to contain lone pairs not
used in resonance. Unhybridized p orbitals are used to form π
bonds, hold lone pairs used in resonance, or are empty.
Be careful when determining hybridization of heteroatoms in
cyclic compounds. Pyridine, pyrrole, furan all have sp2hybridized heteroatoms. In pyrrole and furan one lone pair is
involved in resonance; in pyridine the lone pair is not involved in
resonance.
3.7 Higher alk ynes
Replacement of an H atom in acetylene with a CH3 group gives
propyne, "prop" because there are three carbons, and "yne"
because there is a triple bond. Propyne has two kinds of H
atoms, one attached to C(sp) and one attached to C(sp3).
Replacement of the latter with another CH3 group gives 1butyne, and replacement of the former gives 2-butyne, where
the number is associated with the position of the triple bond.
Higher alkynes are called pentyne, hexyne, heptyne, octyne,
etc., with associated numbers.
Isopropylacetylene is also called 3-methyl-1-butyne. A
compound with both a double and a triple bond is called an
"enyne", and the positions of the bonds are given numbers this
way: 2-octen-6-yne has the double bond in position 2 and the
triple bond in position 6. (No t e t he double bond is giv en
t h e smaller number.)
The linear nature of the alkyne makes it very difficult to make
cycloalkynes. Cyclooctyne is the smallest cyclic alkyne isolable
at room temperature.
3.8 Alk yn e a cidi t y
The C(sp) orbital is much lower in energy than the C(sp2 ) or
the C(sp3 ) orbital. As a result, a pair of electrons in such an
orbital is quite strongly stabilized, and a carbanion in which the
lone pair is in a C(sp) orbital is quite low in energy. This means
that RC CH alkynes are much more acidic at the terminal C(sp)
than most hydrocarbons. On the scale we use to measure
acidity, the pKa scale, ethyne (acetylene) falls at 25. It is less
acidic than H2O (15) but much more acidic than ethene (37) or
ethane (~45). Even NH3, an acid with a central atom more
electronegative than C, has a pKa of 37, much less acidic than
acetylene.
3.9 Degrees o f unsa tura tion
Alkanes have the general formula CnH2n+2. Alkenes have two
fewer H atoms than alkanes, so they have the formula CnH2n, as
do cycloalkanes. Compounds with two π bonds (alkynes,
alkadienes), one π bond and one ring (cycloalkenes), and two
rings (spiro, fused, or bridged alkanes) have the formula CnH2n–
2. In fact, for each π bond or ring in a compound, two H atoms
are removed from the formula. Conversely, if we want to draw
a hydrocarbon with a particular formula, CxHy, we can calculate
how many rings or π bonds it has by the following formula: #
rings or π bonds = [ 2x + 2 – y ] / 2
The number of rings or π bonds in a compound is called its
degrees of unsaturation (a bit of a misnomer, actually, because
an unsaturated compound refers to one with a π bond, not a
ring, but there you go).
Calculating the number of degrees of unsaturation is a quick
way of making sure that you have drawn a compound with the
right number of H atoms. For example, if you were asked to
draw one example of C30H54, the easy way to do it would be to
calculate the number of degrees of unsaturation as follows: #
rings or π bonds = [ 2(30) + 2 – 54 ] / 2 = 4
It has four degrees of unsaturation, so any example of C30H54
would have to have 30 C atoms and four rings, three rings and
one π bond, two rings and two π bonds, one ring and three π
bonds, or four π bonds.
Suppose your compound has halogen atoms, CxHyXz ? When you
replace a H atom with a halogen atom, you don't change the
number of rings or π bonds, so halogens can be treated like H
atoms. Thus:
# rings or π bonds = [ 2x + 2 – (y + z) ] / 2
Since we're on the subject, let's learn how to deal with
chalcogens (O, S) and pnictogens (N, P). If we look at EtOH and
acetone, CH3C(=O)CH3, we can see that the number of degrees
of unsaturation of each compound CxHyOw is the same as CxHy
(none in the case of EtOH, one in the case of acetone). So, in
calculating the number of degrees of unsaturation in a
compound with O or S, CxHyOwXz, we ignore the O atoms.
# rings or π bonds = [ 2x + 2 – (y + z) ] / 2
W h a t abou t N a toms ? Since nitrogen is trivalent, an
organonitrogen compound has one more hydrogen than an
equivalent hydrocarbon has, and we therefore subtract the
number of nitrogens from the number of hydrogens to arrive at
the equivalent hydrocarbon number. Thus, the N-containing
compound, CxHyNv, has the same number of degrees of
unsaturation as CxH( y–v ).
H
H
H
N
H
H
H
H
H
H
N
= C5H8
C5H9N
two unsaturations: one ring
and one double bond
removed
H
For example, Trimethylamine, NMe3 (C3H9N), has no degrees of
unsaturation, like propane with the formula C3H8.
Acetone hydrazone, CH3C(=NNH2 )CH3 (C3H8N2 ), has one degree
of unsaturation like an hydrocarbon with the formula C3H6 (e.g.
CH3CH=CH2 )
Acetonitrile, CH3C≡N, (C2H3N) has two degrees of unsaturation
like the hydrocarbon with the formula C2H2 (HC≡CH)
So, for CxHyNvOwXz, # rings or π bonds = [ 2x + 2 – (y + z – v) ]
/2
No t e tha t Si, P, and S are trea t ed like C, N, and O,
respec ti v el y.