Jones M., Fleming S.A. Organic Chemistry

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3.2 Alkenes: Structure and Bonding 99

3.2 Alkenes: Structure and Bonding

The simplest alkene is properly called ethene but is almost universally known by its

common name, ethylene.

Several spectroscopic measurements and chemical reactions show that ethylene

has the formula C

and is a symmetrical compound composed of a pair of meth-

ylene groups: H

CCH

. Earlier, when we encountered methane (CH

), we con-

structed a bonding scheme designed to reproduce the way four hydrogens were

attached to the central carbon.In ethylene,we have a slightly different arrangement.

Each carbon atom in ethylene is attached not to four other atoms, as is the carbon

in methane or ethane,but to three.It seems we will need a different bonding ration-

ale with which to describe Nature now. Our strategy will be to develop a bonding

scheme for the simplest trivalent compound of carbon, methyl (CH

), and then

extend it to ethylene, in which each carbon is attached not to three hydrogens as in

methyl, but to two hydrogens and the other methylene (CH

) group (Fig. 3.2).

We like the old names and will use many of them. The newer, systematic names are fine when complexity

develops—we obviously could not have a common, or “trivial” name for every compound. Yet some of the fla-

vor of organic chemistry is lost when the system is applied too universally. The trivial names connect to his-

tory, to the quite correct image of the bearded geezer slaving over the boiling retort, and we like that.

FIGURE 3.2 Replacement of one

hydrogen in methyl (CH

), with a

methylene (CH

) group leads to the

framework of ethylene (H

CCH

Note that each carbon so far has only

three bonds. In this drawing the full

bonding scheme for ethylene is not

yet in place.

Three bonds are needed to make attachments to the three hydrogens in CH

. Using

the atomic orbitals of carbon to overlap with the three hydrogen 1s orbitals leads to

problems, just as it did in our earlier construction of methane (see Section 2.2), as

Worked Problem 3.1 shows.

WORKED PROBLEM 3.1 Produce a bonding model for neutral methyl ( CH

)

using the unhybridized atomic orbitals of carbon and the hydrogen 1s orbitals.

Critically discuss the shortcomings of this model. What’s wrong with it?

ANSWER There is more than one way to construct such a model. Start by deter-

mining how many electrons are available. Carbon (

C) supplies four (six less the

two low-energy 1s electrons), and each hydrogen supplies one, for a total of seven.

We might form three carbon–hydrogen bonds by overlap of the carbon 2p

,2p

and 2p

orbitals with three hydrogen 1s orbitals, for example.This process uses six

electrons (two in each of the three carbon–hydrogen bonds), and leaves the car-

bon 2s orbital to hold the remaining electron.

C H

2s Orbital

containing

one electron

(continued)

100 CHAPTER 3 Alkenes and Alkynes

C (1s

) 2s

Combine 2s, 2p

, 2p

three sp

hybrid orbitals; 2p

is not used

FIGURE 3.3 The combination of

three atomic orbitals of carbon.

n sp

hybrid H

C H

Bonding

σ orbital

FIGURE 3.4 The bonding σ

orbital.

However, this model gives a “methyl”with carbon–hydrogen bonds 90° apart. It is

possible for a central carbon to be surrounded by three hydrogens 120° apart, and

so we might suspect that this unhybridized model with its 90° angles will be desta-

bilized by electron–electron repulsion. Moreover, overlap of a 1s orbital with a 2p

orbital is inefficient because the rear lobes of the 2p orbitals are wasted. Stronger

bonds can be constructed from directed, hybrid orbitals.

Other possible answers include making two carbon–hydrogen bonds from 2p/1s

overlap and the third from 2s/1s overlap. This model leaves the last electron in a

2p orbital and consequently has the same problems as noted previously.

C H Bonds

close together

Wasted lobe

90⬚

The new sp

hybrids are generally shaped like the sp

hybrids constructed

earlier. They are directed orbitals—there is a fat lobe and a thin lobe, so they lead

to quite efficient overlap with the hydrogen 1s orbitals (Fig. 3.4). The directed sp

hybrid orbital overlaps with the spherical 1s orbital on hydrogen to produce a

carbon–hydrogen σ bond.The antibonding orbital (σ*) is not shown in Figure 3.4.

What do you guess is the angle between the three hybrid orbitals? Remember

that one of the problems with using unhybridized carbon atomic orbitals is that this

process produces bonds that are too close together, and repulsions between filled

orbitals are not minimized. Hybrid orbitals produce strong bonds (good overlap) that

As in our earlier discussion of sp

hybridization (p. 55), we will create three new

hybrid orbitals that will do a better job of attaching the carbon atom to the three

hydrogens than do the pure atomic orbitals. Three hybrid orbitals are needed, and

our mathematical operations will therefore involve combining three wave functions

(atomic orbitals) to produce the three new hybrid orbitals. Recall that in such quan-

tum mechanical calculations the number of orbitals created always equals the num-

ber of orbitals combined in the calculation, here three. So, let’s combine the carbon

2s,2p

, and 2p

orbitals to produce three new, sp

hybrids.The 2p

orbital unused in

our calculation remains, unhybridized and waiting to be incorporated in our picture

(Fig. 3.3). Remember that the choice of 2p orbitals to be combined is arbitrary. Any

pair of p orbitals can be used, leaving the third left over.

3.2 Alkenes: Structure and Bonding 101

120⬚

FIGURE 3.5 In planar CH

, the angle

between the carbon–hydrogen bonds

is 120°. Repulsive overlap between

the filled carbon–hydrogen bonds is

minimized in this arrangement.

Energy

␴*

– 1s (antibonding

molecular orbital)

␴

+ 1s (bonding

molecular orbital)

FIGURE 3.6 The overlap of sp

and 1s orbitals can occur in a

bonding, stabilizing way (sp

 1s),

or an antibonding, destabilizing way

(sp

 1s).

are directed so as to minimize repulsions. As Figure 3.5 shows, the best way to

arrange three things in space (the hydrogens in this case) surrounding a central object

(here, the carbon) so they are as far from one another as possible, is to put them at

the corners of an equilateral triangle. And indeed, the angles between the three sp

hybrids are exactly 120°. Recall the discussion of BH

in Section 2.2c.

Each of the three sp

/1s overlapping systems produces a bonding (σ  sp

 1s)

and an antibonding (σ*  sp

 1s) molecular orbital. Stabilization is maximized

if we form two-electron bonds, as the stabilized bonding molecular orbital

will be filled and the antibonding, high-energy molecular orbital is empty

(Fig. 3.6).

Each hydrogen supplies a single electron.Therefore,in order to form three two-

electron bonds we need a single electron from carbon for each of the three sp

hybrids

(Fig. 3.7).

FIGURE 3.7 Three two-electron

carbon–hydrogen bonds can be

formed using one electron from each

hydrogen and three electrons from

carbon (all electrons are shown in

red).This leaves one electron on

carbon (shown in red) left over. In

this figure, two of the sp

orbitals are

shown schematically as wedges.

102 CHAPTER 3 Alkenes and Alkynes

/1s

Bond

/1s

Bond

/1s

Bond

Singly occupied

orbital

FIGURE 3.8 A view of sp

hybridized

Remember that we did not use the carbon 2p

orbital in creating the sp

hybrids.

That’s where the last electron of carbon goes. So our sp

bonding scheme leads to the

picture in Figure 3.8: a carbon atom surrounded by a planar array of three hydrogen

atoms, with the unhybridized 2p

orbital extending above and below the plane of the

four atoms.The carbon–hydrogen bonds are familiar two-electron bonds (don’t forget

the empty antibonding orbitals,though) and there is a single electron in the 2p

orbital.

Don’t be confused by the electron shown in only one lobe of the 2p orbital. The two

lobes are not separate, and the electron occupies the whole orbital, not just one lobe.

We have seen this molecule, CH

, before! It is nothing more than a methyl radical.

In the text so far, we have built up a picture that emphasizes localized bonds. In

methyl, for example, electrons occupy three bonding orbitals made up of overlap-

ping sp

and 1s orbitals. In Problem 3.2, we develop a picture of methyl in which

the delocalization of electrons throughout the molecule is emphasized. Delocalizing

electrons,spreading them out over several atoms rather than localizing them between

two atoms,is almost always energetically favorable.But there are advantages to both

schemes. In one sense the delocalized picture is probably more “real,” because elec-

trons are not limited to the regions of space the hybridization scheme suggests.

However, we do not make horrible energetic mistakes if we ignore the delocaliza-

tion that the orbitals you develop in Problem 3.2 show so clearly, and the hybridiza-

tion scheme is excellent for “bookkeeping” purposes. It helps us to keep track of

electrons in the chemical reactions that follow in later chapters, for example.

Molecular orbitals for

Atomic orbitals for carbon

–

It is important that we do not view the molecular orbital and hybridization schemes

as being in conflict or as giving substantially different pictures of the bonding in CH

Note, for example, that the geometry is exactly the same in the two schemes. We are

humans, stuck with our inability to apprehend the properties of electrons easily, and

needing approximations in order to represent Nature. Different approximate bonding

schemes have been developed that emphasize different properties of the molecules.The

molecular orbital picture does an excellent job of showing the distribution of electrons

throughout the molecule.The hybridization picture sacrifices an ability to show this delo-

calization for the advantages of clarity, and ease of following the course of chemical reac-

tions. We need to keep both representations in mind as we study chemical reactions.

PROBLEM 3.2 Construct the bonding molecular orbitals for planar methyl (CH

Use the molecular orbitals for cyclic H

(A, B, and C) given below. The molecu-

lar orbitals for H

are shown in the drawing, but if they are unfamilar, take a look

now at Problems 1.62 and 1.63. Allow these orbitals to interact with the appro-

priate atomic orbitals of a carbon atom placed at the center of the triangle of

hydrogens. Next, place the bonding molecular orbitals in order of energy, lowest

first. The dot in B shows the position of the third hydrogen atom.

3.2 Alkenes: Structure and Bonding 103

C C

FIGURE 3.9 Our first orbital picture

of ethylene.

We could easily imagine exchanging one of the three hydrogens of CH

for another

atom or group. In constructing ethylene (H

CCH

), this other group would be another

methylene (CH

; Fig. 3.9). This transformation gives us our first orbital picture of

ethylene although the carbon–carbon double bond is not yet complete-

ly drawn in Figure 3.9. So our first picture of ethylene is derived from joining a pair of

methylene groups.The two sp

hybrids not involved in the carbon–hydrogen bonds over-

lap to make an sp

/sp

σ bond joining the two carbons of ethylene.The two 2p

orbitals

remain, and extend above and below the plane defined by the six atoms (Fig. 3.10).

C CCC

/sp

σ Bond sp

/sp

σ Bond

FIGURE 3.10 Another orbital picture

of ethylene.

The structure of ethylene shown in Figure 3.10 is quite analogous to that pro-

duced in the construction of ethane from a pair of sp

hybridized carbons (Section

2.5). In ethane, two sp

hybrids overlapped to form a bonding σ orbital and an anti-

bonding σ* orbital (p. 68). Construction of the carbon–carbon sigma bond system

in ethylene begins with the similar overlap of a pair of sp

hybrids. Once more, a

bonding σ orbital and antibonding σ* orbital are produced, this time by the con-

structive and destructive overlap of a pair of sp

hybrids (Fig. 3.11). Because there

are only two electrons, only the low-energy, bonding molecular orbital is filled.

Energy

σ*

+ sp

(bonding molecular orbital)

– sp

(antibonding molecular orbital)

FIGURE 3.11 Overlap of the two sp

hybrids to form a bonding and

antibonding pair of molecular

orbitals.

104 CHAPTER 3 Alkenes and Alkynes

So far, we have ignored the 2p

orbital on each carbon. It is time to take them

into account. These two carbon 2p

orbitals are only the distance of a single bond

apart and will interact quite strongly. Moreover, they are of equal energy and that

too contributes to a strong interaction. Orbitals of equal energy interact most

strongly (p. 34). We know what happens when atomic orbitals overlap—bonding

and antibonding molecular orbitals are created. The new molecular orbitals

formed from overlap of a pair of 2p orbitals are shown schematically in Figure 3.12.

Energy

2p – 2p = π* (antibonding)

2p + 2p = π (bonding)

Symmetry plane

contains all four

hydrogens and

both carbons

66 kcal/mol

Stabilization

C C

2p/2p Overlap gives a π bond

/sp

Overlap gives a σ bond

FIGURE 3.12 Overlap of a pair of 2p orbitals results in

a lower-energy, bonding π orbital and a higher-energy,

antibonding π* orbital.The stabilization amounts to

about 66 kcal/mol when the π orbital is filled.

WEB 3D

The coded representation for a double

bond; note that there is no distinction

made between the

σ and π bonds; a

typical C

C bond distance is 1.33 A

⬚

FIGURE 3.13 The π and σ bonds

of a carbon–carbon double bond are

not differentiated in the coded

representation of an alkene.

These new orbitals are not σ orbitals because they do not have cylindrical symme-

try.There is a plane of symmetry instead, and this makes them pi (␲) orbitals, here

called π (bonding) and π* (antibonding).There are only two electrons to be put into

the new system of two molecular orbitals, one from each carbon 2p

orbital. These

are accommodated very nicely in the highly stabilized bonding π molecular orbital.

So the two carbons of ethylene are held together by two bonds. One is made up

of sp

/sp

σ overlap and the other of 2p/2p π overlap.There is a double bond between

the two carbons made up of the σ and π bonds. The convention is to simply draw

two lines between the atoms, making no distinction between the two bonds,

although, as you have just seen, this approximation is far from justified (Fig. 3.13).

3.2 Alkenes: Structure and Bonding 105

The overall bonding scheme for a carbon–carbon double bond includes both σ and

π bonds (and their empty antibonding counterparts) and is also shown in Figure 3.13.

Carbon–carbon double bonds are quite short, with a typical bond distance

being 1.33 Å.

In ethane, there is nearly free rotation about the carbon–carbon bond

(⬃3 kcal/mol, p. 65). Is the same true for ethylene? Actually, the argument is

a bit more subtle. There isn’t really free rotation about the carbon–carbon

bond in ethane. There is a 3-kcal/mol barrier, produced by the need to pass

through a structure in which the carbon–hydrogen bonds are eclipsed. Our real

question should be, How high is the barrier to rotation about the carbon–carbon

double bond in ethylene? To begin the answer we construct Newman projections

by sighting along the carbon–carbon bond (Fig. 3.14).

In this form, the two 2p

orbitals do not overlap

90⬚ Form of ethylene

Eclipsing

(torsional strain)

minimized

Eclipsing

(torsional strain)

minimized

90⬚ rotation

0⬚ Form of ethylene Newman projections

90⬚ rotation

EclipsedEclipsed

FIGURE 3.14 Although rotation

about the carbon–carbon bond

relieves torsional strain, it also

destroys the overlap between the two

orbitals on the adjacent carbons.

This loss of orbital overlap leads to a

high barrier to rotation. When one

2p orbital is rotated 90°, orbital

overlap is completely lost.

Note that in the 0° conformation of ethylene there is torsional strain

induced by the eclipsing of the carbon–hydrogen bonds. This strain is removed

in the 90° rotated arrangement because now the electrons are as far apart

as possible. In ethane, each pair of eclipsed hydrogens induces approximately

1 kcal/mol of destabilization. Therefore our initial guess might be that the

0° form of ethylene would be about 2 kcal/mol higher in energy than the 90°

form. That estimate would be approximately right if torsional strain were

the whole story. But it is not, as we have ignored the pair of 2p

orbitals,

and their interaction completely overwhelms the small amount of torsional

strain.

In the 0° form, the two p orbitals overlap; in the 90° form they do not

(Fig. 3.14).When rotation occurs, the 2p/2p overlap declines and so does the sta-

bilization derived from occupying the new bonding molecular orbital with two

electrons.There is an enormous stabilizing effect in the 0° arrangement that over-

comes the relatively minor effects of torsional strain.

ETHYLENE: A PLANT HORMONE

106 CHAPTER 3 Alkenes and Alkynes

Figure 3.15 plots the energy change as rotation occurs. The barrier to rota-

tion here is no paltry 3 kcal/mol as it is in ethane but 65.9 kcal/mol, an amount

far too high to be available under normal conditions. Alkenes are “locked” in a

planar conformation by this amount of energy. We will soon see some conse-

quences of this locking.

Energy

Rotation angle

0⬚ 180⬚90⬚

90⬚

65.9 kcal/mol

0⬚

180⬚

FIGURE 3.15 A plot of potential

energy versus rotation angle in

an alkene.

to the discovery of ethylene as a plant hormone. Trees sur-

rounding gas lamps were often defoliated when gas leaks

occurred. A principal component of the illuminating gas was

ethylene, and a little investigation showed that this chemical

was the cause of the defoliation.

It’s amazing, but the simplest of all

alkenes, ethylene, is an important plant

hormone. Among other functions, ethyl-

ene acts to promote ripening of fruit. Moreover, produc-

tion of ethylene is autocatalytic; that is, a little ethylene

induces the formation of more from the amino acid

methionine, and its effects are magnified. Accordingly,

fruits such as tomatoes and bananas are now typically

shipped green in well-ventilated containers so they will

arrive unspoiled. Ripening can then be started by exposure

to ethylene.

Ethylene also induces abscission; the falling of leaves or

flowers, by promoting the formation of the enzyme cellulase

that weakens cell walls by destroying the cellulose from

which they are made. A weakened abscission layer forms at

the base of the leaf or flower, and wind or rain then breaks

the stem. Indeed, it was this phenomenon that indirectly led

PROBLEM 3.3 Show that in the 90° rotated form of an alkene there is zero over-

lap between the pair of carbon 2p

orbitals.

3.3 Derivatives and Isomers of Alkenes 107

Summary

Two carbons joined by a double bond are held together by a σ bond made up of

overlapping sp

orbitals and a π bond made from 2p orbitals overlapping side to

side.There is no free rotation about a carbon–carbon double bond, because such

rotation diminishes overlap between the 2p orbitals making up the π bond and

costs energy. A full 90° rotation requires about 66 kcal/mol of energy. Therefore,

alkenes are planar.

In the hybridization model, we have constructed alkenes with both relatively

strong σ bonds and weaker π bonds. The weaker, higher-energy π bonds will be

more reactive than the stronger, lower-energy σ bonds.This point is critical and is

often a source of confusion. A bond with lower energy is more stable and less reac-

tive than a higher-energy bond. The weaker π bonds are likely to be the locus of

reactivity of alkenes. In fact, alkenes and alkynes undergo all sorts of reactions that

are unknown for alkanes with their exclusive set of strong, unreactive σ bonds. As

soon as we work a bit more on nuts and bolts (nomenclature, isomers, and some

thermochemistry), we’ll get to what is usually the most fun in organic chemistry—

reactions—and they will be triggered by the relatively weak π bonds in alkenes.

3.3 Derivatives and Isomers of Alkenes

After we developed a structure for ethane (p. 68), we examined substituted

alkanes.Here we do the same thing for ethene,and discover a new kind of isomerism.

There is but one ethylene (or ethene, if you insist). We can make derivatives of

this molecule by mentally replacing a hydrogen with some “X” group. Figure 3.16

shows a few such compounds. These are not called “ethenyl” or even “ethylenyl”

compounds but instead bear the delightfully trivial name, so far untampered with,

of vinyl.

If we now let X  CH

(methyl), we get a compound, C

(propene or, com-

monly, propylene, Fig. 3.17). Notice how the root prefix prop- has been retained for

this three-carbon species, and only the vowel changed to note the difference between

saturated propa

ne and unsaturated propene.

X =

SiH

Vinyl fluoride

Vinyl bromide

Vinyl chloride

Vinyl iodide

Vinyl alcohol

Vinyl cyanide

Vinyl silane

Vinylamine

Ethylene Vinyl X

WEB 3D

FIGURE 3.16 Some substituted

alkenes called vinyl compounds.

WEB 3D

X = CH

Propene

FIGURE 3.17 When X  CH

(methyl), we get propene (C

Note the change of vowel from

propane to propene.

108 CHAPTER 3 Alkenes and Alkynes

replace H

1-Chloropropene

2-Chloropropene

3-Chloropropene

or allyl chloride

CCH

Terminal

position

HC CH

CHX

HC CH

Propene

CHCl

CH X

CH Cl

replace H

FIGURE 3.18 Substitution of three different hydrogens in propene

appears to give three different compounds.propenyl

We can now imagine replacing a hydrogen in propene with an “X”atom or group,

but the situation is not as simple as it was with ethylene. Although there is only one

kind of hydrogen in ethylene, there are several different hydrogens in propene that

might be replaced. A quick count of the different hydrogens in propene is mislead-

ing. Care must be taken from now on to draw out the three-dimensional structures

of the alkenes on which we are doing our mental replacements.In practice, this task

is quite simple, because alkenes are planar, and most structures are easily drawn. If we

just write a shorthand structure,we predict only three new kinds of compounds when

a hydrogen is replaced with an X atom or group. As shown in Figure 3.18, these

compounds are all known.Note the use of the frequently encountered common name

allyl for the group. Figure 3.19 shows some allyl compounds.

There is one severe problem with the approach of Figure 3.18.In reality, there are not

three kinds of substituted propenes, but four.The more detailed drawing in Figure 3.20

shows why. The schematic structure for propene used in Figure 3.18 is inadequate

The allyl group

CCH

ClCH

Allyl chloride

Allyl bromide

Allylamine

Allyl alcohol

Allyl cyanide

WEB 3D

FIGURE 3.19 Some allyl compounds.

CCH

replace H

The two H

atoms are really

different! H

is on the same

side as the methyl group,

but H

is on the opposite side

This schematic drawing hides too much detail

replace H

These compounds

are not the same!

FIGURE 3.20 A detailed drawing of

propene reveals four different

potential substitution sites.