1

Dienes

Building Bridges to Knowledge

A railway track entering Aquas Calientes, the pathway to Machu Picchu

Dienes have the general formula CnH2n. Double bonds in dienes can be isolated as in 1,4-pentadiene, H2C=CH2-CH2-CH=CH22

; cumulated as in allene, H2C=C=CH2; or conjugated as in 1,3-butadiene, H2C=CH-CH=CH2, the simplest conjugated diene.

If the carbon skeleton of 1,3-butadiene is written in the following manner,

2

then the C2-C3 bond distance should approximate 1.54 Å (1.54 x 10-10 m). Actual measurements have shown that the C2-C3 bond distance is 1.48 Å (1.48 x 10-10 m); therefore, the C2-C3 bond must possess some double bond character. The C2-C3 partial double bond character can be accounted for by resonance.

Resonance is the phenomenon in which the actual electron configuration is a hybrid between two or more electron configurations. Consequently, the resonance structures do not adequately represent the “true” structure of butadiene, because the actual structure is a hybrid of all three structures. The molecular

3

orbital picture would be a more accurate representation of butadiene.

As in the case of the alkene carbon skeleton, the carbons of dienes participate in hybridization. In the case of butadiene, two 2sp2 hybridized atomic orbitals of the C1 carbon atom linearly combine with two 1s atomic orbitals of two hydrogen atoms to form the terminal CH22

molecular bonds. A third 2sp2 hybridized atomic orbital of the C1 carbon atom linearly combines with a 2sp2 hybridized atomic orbital of the C2 carbon atom. C2 has three 2sp2 hybridized atomic orbitals; therefore, one of the remaining two 2sp2 hybridized atomic orbitals can linearly combine with a 1s atomic orbital of hydrogen, and the other can linearly combine with a 2sp2 hybridized atomic orbital of the C3 carbon atom. Since the C3 hybridized atomic orbitals has three 2sp2 hybridized orbitals, one of the remaining 2sp2 hybridized atomic orbitals can linearly combine with a 1s atomic orbital of hydrogen, and the other can linearly combine with a 2sp2 hybridized atomic orbital of carbon atom 4. Since carbon atom 4 has three 2sp2 hybridized atomic orbitals, then the remaining two can participate in linear combinations analogous to carbon atom 1, i.e., two 2sp2 hybridized atomic orbitals can linearly combine with two 1s atomic orbitals of two hydrogen atoms. The results of these linear combinations to form bonding molecular orbitals can be represented by the following picture:

4

The linear combination of atomic orbitals for the C-H sigma, σ, bonding molecular orbitals would be (2sp2 + 1s) and linear combination of atomic orbitals for the C-C sigma, σ, bonding molecular orbitals would be (2sp2 + 2sp2).

The above description takes care of the sigma bonding molecular orbitals. The molecule also includes pi, π, bonds. Each carbon atom has an electron in a 2p atomic orbital that is perpendicular to the plane containing the carbon atoms, the hydrogen atoms, and the sigma bonding molecular orbitals. The 2p atomic orbitals can linearly combine to form two pi, π, bonding molecular orbitals.

Linear combinations of atomic orbitals result in orbital symmetry; therefore, the linear combination of n atomic orbitals would result in the formation of n molecular orbitals. Therefore, if a 2sp2 hybridized atomic orbital linearly combined with a 1s atomic orbital of hydrogen, two molecular orbitals are formed. One molecular orbital would be a

5

bonding molecular orbital and the other would be an antibonding molecular orbital. In our previous discussions, we have only looked at the bonding molecular orbital, but it must be noted that all linear combinations of molecular orbitals will lead to bonding molecular orbitals and antibonding molecular orbitals.

The four 2p atomic orbitals can linearly combine with each other to form four molecular orbitals as described in Figure 3.1

Figure 3.1 π bonding and antibonding molecular orbitals of 1,3-butadiene.

The bonding molecular orbitals of 1,3-butadiene are indicated without asterisks and the antibonding molecular orbitals are indicated with asterisks. The four electrons that were in the 2p atomic orbitals of the four carbon atoms now reside in the two bonding molecular orbitals of 1,3-butadiene. The bonding molecular orbitals are filled and the antibonding molecular orbitals are

6

unoccupied. This suggests that 1,3-butadiene should be a stable molecule. The bonding molecular orbital of lowest energy, 𝜋

1, has a

lower potential energy than the bonding molecular orbital in ethene. This suggests that the conjugated double bond system is more stable than the isolated double bond system.

The pi frame of the 1,3-butadiene system can be illustrated in figure 3.2.

Figure 3.2 linear combination of the four 2p orbitals of 1,3-butadiene

Figure 3.2 is an illustration of the linear combination of the four 2p orbitals that are perpendicular to the sigma frame as described in Figure 3.1. Each 2p atomic orbital contains one electron. After the four 2p orbitals, each containing one electron, linearly combine, the resulting π

1 molecular orbital contains two paired electrons and π

2

molecular orbital contains two electrons. The antibonding molecular orbitals, π3

* and π4* have no electrons.

7

Experimental evidence suggests that conjugated double bonds are more stable than isolated double bonds. Table 3.1 lists heats of hydrogenation for selected alkenes and dienes.

Compound Enthalpy*

propene 126 1-pentene 127

1,4-pentadiene 254 1,3-butadiene 238

* Enthalpy of hydrogenation, kJ mol-1 Table 3.1 ∆H (hydrogenation) for selected alkenes and dienes

The structural formulas corresponding to the names in Table 3.1 are:

propene 1-pentene

1,4-pentadiene 1,3-butadiene

Table 3.1 suggests that the energy released upon hydrogenating a double bond is approximately 126 kJ mol-

-1, The heat of hydrogenation of the isolated diene, 1,4-pentadiene, is 254 kJ/mol about twice that for propene. Therefore, everything being equal, one would suspect that the heat of hydrogenation of 1,3-butadiene would be about 254 kJ mol-1, but the actual heat of hydrogenation of the diene is 238 kJ mol-1 . The difference of 16 kJ mol-1 is attributed to the stabilization of the diene resulting from the overlap of the four 2p atomic orbitals forming two bonding molecular orbitals. This results

8

in delocalization of the four electrons that reside in the π1 and π

2

bonding molecular orbitals. Spectroscopic evidence supports the stabilization argument and the double bond character between carbon atoms 2 and 3 in 1,3-butadiene.

There is a small barrier of rotation about carbon atoms 2 and 3. The small barrier to rotation is approximately 21 kJ/mol (approximately 8.4 kJ mol-1 higher than the C-C barrier to rotation in ethane). Rotation about the C2 and C3 atoms of 1,3-butadiene is interesting. It can lead to the single, “s”, bond cis structure (I) or the single, “s”, bond trans structure (II)

The energy barrier to this rotation is represented in Figure 3.3

9

Figure 3.3: The rotational barrier around the C2-C3 bond of 1,3-butadiene converting s-cis to s-trans and vice versa

Reactions of Dienes

Conjugated dienes will react with excess halogens, Br2 and Cl2, to yield tetrahaloalkanes. The halogenation reaction can be represented by the following chemical equation.

H2C=CH-CH=CH2 + X2 → X-CH2CHX-CHX-CH2X

If the reaction is conducted under conditions favorable to the addition of one mole of halogen, then the products would be 3,4-dihalo-1-butene (the 1,2-addition product) and 1,4-dihalo-2-butene

10

(the 1,4-addition product). If the reaction is carried out at low temperature (e.g., 193K), then the major product is the 1,2-addition product. If the reaction is carried out at 313K, then the 1,4-addition product predominates.

At 313 K

11

At 193 K

A mixture of 80% of the1,2-addition product and 20% of the 1,4-addition product at 193K when raised to a temperature 313K will equilibrate to a mixture of 20% of the 1,2-addition product and 80% of the 1,4-addition product. The reverse does not occur when a mixture of 20% of the 1,2-addition product and 80% of 1,4-addition product at 313K is lowered to 193K.

These observations are explained by realizing that initially halogen reacts with the double bond on the diene in an analogous manner as halogen reacts with an alkene, but instead of forming a cyclic halonium ion, a resonance stabilized allylic carbocation is formed.

12

At 193K, the reaction is kinetically controlled, i.e., the 1,2 addition

13

product is formed more rapidly than the 1,4 addition product since at this temperature the halide ion is in proximity to carbocation 1 and, therefore, can immediately bond to the carbon atom to form compound III, 3,4-dibromo-1-butene.

At 313K, the formation of the 1,2-addition product is accelerated, but the allylic halide formed can ionize to produce a resonance stabilized carbocation.

resonance stabilized carbocation

14

Two processes are operating here- (1) the acceleration of the formation of the 1,2-addition product, and (2) the ionization of the 1,2-addition product to form the 1,4-addition product. However, after the 1,4-addition product has been formed, it is more difficult to ionize it since an internal double bond is more stable than a terminal double bond. Thus 1,4-addition product is described as being the more stable product, and the addition reaction is said to be thermodynamically controlled.

1,2-addition product

15

1,4-addition product

Figure 3.4 is an energy profile diagram that represents the products resulting from the addition of bromine to the resonance stabilized carbocation intermediate. The carbocation intermediate is formed from the addition of bromine to 1,3-butadiene.

16

Figure 3.4: kinetically controlled versus thermodynamically controlled ionic addition of bromine to 1,3-butadiene where ∆Hreation≈ ∆G and ∆G = -RT lnK

If the diene is unsymmetrical, then the yield of the 1,4-addition thermodynamically controlled product would be increased.

17

Free radical bromination of 1,3-butadiene produces a stable allylic radical. The allylic free radical can undergo a propagation step with Br2 to form either 3,4-dibromo-2-butene, compound I, or 1,4-dibromo-2-butene, compound II.

Initiation step

propagation steps

18

propagation steps

19

Termination steps

20

In this reaction, compound II predominates. The radical mechanism is not subject to a rapid equilibrium process since 192 kJ/mol of energy would be required to homolytically cleave Br2 while 280 kJ/mol would be required to homolytically cleave C-Br. The allylic radical must be formed for a finite period and the kinetic advantage of the proximity which exists in the ionic addition does not exist in the free radical addition reaction.

Polymerization of Dienes

Both ionic and radical polymerization of dienes favor the 1,4-addition product.

21

Polymerization of substituted dienes follows a tail-to head addition process. For example, isoprene (2-methyl-1,3-butadiene), compound III polymerizes in the following manner:

The following is a shorter representation of the tail-to-head polymerization of isoprene:

Cis-trans isomerism is possible in diene isomerism. If the polymerization process leads to an all trans isoprene polymer, then the product is referred to as gutta percha (http://mw1.m-w.com/medical/gutta-percha).

22

If the polymerization process leads to an all cis isoprene polymer, then the product is referred to as natural rubber or latex.

In 1859, Charles Goodyear discovered that treating latex with sulfur resulted in the production of an elastic substance. This process became known as the vulcanization of rubber.

23

http://www.google.com/imgres?imgurl=http://www.chemistrydaily.com/chemistry/upload/5/58/Vulcanization.png&imgrefurl=http://www.chemistrydaily.com/chemistry/Vulcanization&h=317&w=598&sz=5&t

24

bnid=XnpaJ3u66XeXkM:&tbnh=72&tbnw=135&prev=/images%3Fq%3Dvulcanization%2Bof%2Brubber&hl=en&usg=__W5rRanDEiA7Q0VvSeJbElHTrzkk=&ei=fgRRStfaCIHWsgPWqbyqDQ&sa=X&oi=image_result&resnum=7&ct=image

The process of synthesizing an all cis isoprene polymer was eluded for many years; however, in 1955, a polymerization process using a special catalyst, the Ziegler-Natta Catalyst, led to the successful polymerization of an all cis polymer.

http://chem.chem.rochester.edu/~chem421/zn.htm

Dimerization of Acetylene using Copper (I) Chloride

The dimerization of acetylene using copper (I) chloride produces vinyl acetylene, 3-butyn-1-ene. Copper (I) chloride functions as a catalyst in this reaction. The double bond receives preference over a triple bond when naming alkenynes.

The mechanism for this reaction would probably follow the following sequence of steps.

(1) Cu+ - 1e → Cu2+

25

(2)

(3)

(4)

26

(5)

The addition of HCl to 3-butyn-1-ene in the presence of Cu+ produces 2-chloro-1,3-butadiene, compound IV. Compound IV is also referred to as chloroprene, because it is structurally similar to isoprene.

Chloroprene can polymerize by a free radical reaction to produce neoprene, the first synthetic rubber developed by DuPont in 1932.

http://www.wisegeek.com/what-is-neoprene-rubber.htm

27

Neoprene can be vulcanized by heating it with ZnO or MgO.

Isoprene polymerization appears in biological systems using isopentenyl pyrophospate, compound V, as a precursor.

Isopentenyl pyrophosphate can polymerize, in vivo, by a stepwise process to form cholesterol.

http://community.middlebury.edu/~chem/chemistry/students/ho/cholesterol.html

28

29

Problems

Dienes

1. In 1,3-butadiene, four 2p orbitals linearly combine. Three sets of

2p orbitals could linearly combine in the following manner: C1 to C2;

C2 to C3 and C3 to C4.

The following four bonding molecular orbitals are possible in 1,3-

butadiene:

π11; π22

; π33; and π44

If the lowest energy level for the molecular orbitals formed from the

linear combination of 2p orbitals is represented as:

How would you represent π2 2, π33

and π44?

2. Complete the following reactions:

(a) H2C=CHCH=CH2 + Br22 →

(b) H2C=CHCH=CH2 + Br2 + (CH3)3COOC(CH33

)3 →

1

2 34

CC

C C++

++

--

--

Π1

30

(c)

(d) H2C=CHCH=CH2 + HBr

3. Show how the heat of hydrogenation can be used to calculate the

resonance energy of stabilization of a conjugated diene.

4. Draw and give an explanation of the energy profile diagram of the

following reaction:

5. Calculate the equilibrium constant for the following reaction, if ∆G

= -3.608 x 103 J/mol

BrCH2CH(Br)CH=CH2 → BrCH2CH=CHCH

2Br

What does the value of the equilibrium constant suggest about the

direction of the equilibrium reaction?

C CH H + Cu2Cl22

313K⎯⎯⎯→

CH3CH=CHCH=CHCH3Br2

CH3CHCHCH=CHCH3

Br Br

CH3CHCH=CHCHCH3

Br Br

193K

313K

31

6. Write a mechanism to account for the following transformation:

7. Show how isopentenyl pyrophosphate could be the precursor to

the biosynthesis of vitamin A, compound VI.

8. Draw structures for the following compounds:

(a) 2,2,5,5-tetramethyl-3-hexene

(b) 2,5-heptadiene

(c) 2,4,6,8-decatetraene

(d) 1,4-cyclohexadiene

(e) bicyclo[2.2.1]-2,5-heptadiene or bicyclo[2.2.1]hepta-2,5-diene

OH

VI