Minggu, 11 Oktober 2009

Free Radical Chemistry

Free Radical Chemistry

Introduction

The cleavage of a covalent heteronuclear chemical bond generally results in the formation of two oppositely charged ions, with the negative charge developing on the more electronegative atom of the bonded pair and the positive charge on the less electronegative atom. As the electronegativity difference between two bonded atoms becomes smaller and smaller, the tendency of the bond between them to break homolytically becomes larger and larger; each atom retains one of the electrons from the bond. Figure 1 presents several examples of reactions that involve both heterolytic and homolytic bond cleavage.

Figure 1

Divvy 'em up

Species in which an atom has an unpaired electron are called free radicals. The three free radicals shown in the bottom panel of Figure 1 are called chlorine atoms, hydroxyl radicals, and triphenylmethyl radicals. Chlorine atoms and hydroxyl radicals are extremely reactive materials. The triphenylmethyl radical, on the other hand, is very stable. In fact, no one has ever isolated hexaphenylethane, presumably because it decomposes spontaneously into two triphenylmethyl radicals.


Exercise 1 Suggest a reason that hexaphenylmethane is not stable.


Free Radical Shape

According to VSEPR theory carbocations, R3C+, should be trigonal planar, while the carbanions, R3C:-, should be pyramidal. These predictions have been verified experimentally. You might expect that the geometry of a free radical, R3C., would be intermediate between that of a carbocation and a carbanion. Experimental measurements indicate, however, that carbon free radicals are essentially planar. Figure 2 compares the structures of these three species.

Figure 2

Three Forms of Trivalent Carbon


Exercise 2 What is the hybridization of the central carbon in the carbocation shown in Figure 2? In the free radical? In the carbanion?

Exercise 3 How many valence electrons does the central carbon "own" in a carbocation? In a free radical? In a carbanion ?


Free Radical Stability

The stability of carbocations increases in the order CH3+ <>3CH2+ < (CH3)2CH+ < (CH3)3C+. In other words, methyl <>o <>o <>0. The structural similarity between carbocations and carbon free radicals illustrated in Figure 2 suggests that these species should display a similar increase in stability as a function of increasing substitution at the central carbon. This expectation is borne out by experimental measurements. Presumably replacement of a hydrogen atom by an alkyl group creates the possibility for hyperconjugative stabilization of the free radical in the same way it does for the carbocation. Figure 3 demonstrates this partial obrital overlap between a C-H sigma bond of a methyl group and a p orbital on the central carbon atom of a free radical.

Figure 3

Hyperconjugation to the Rescue

When a free radical center is flanked by a pi system, a resonance interaction between the p orbital on the central carbon and the p orbitals of the pi bond(s) is possible. Figure 4 presents two views of this type of interaction for one common structure, the allylic radical. Note in this figure that the unpaired electron is originally on the carbon atom bearing the R groups. Consideration of the picture in the left hand panel reveals that overlap of the p orbital on the middle carbon with either of the flanking p orbitals is, to a first approximation, equally likely. This view is reenforced by the familiar electron pushing scheme depicted in the right hand panel. Note the use of a single barbed arrow to denote resonance delocalization of a single electron.

Figure 4

A Resonance Stabilized Free Radical


Exercise 4 The benzylic radical, C6H5CH2., is stabilized by resonance interactions similar to those shown in Figure 4 for the allylic radical. Draw three resonance structures to demonstrate how the unpaired electron can be delocalized into the aromatic ring of the benzylic radical.

Exercise 5 Rank free radicals 1-4 in order of increasing stability.

1 <>

2 <>

3 <>

4 <>

3 <>


Free Radical Reactions

Polymerization

In our discussion of addition polymers we described the polymerization of alkenes in terms of an acid-catalysed process. In practice, polymerization of simple alkenes is more often initiated by a free radical than it is by a proton. Regardless of the initiator, the basic chain reaction mechanism applies. Figure 5 reviews the process which most often begins with the homolytic cleavage of the O-O bond of an organic peroxide.

Figure 5

Free Radical Polymerization


Exercise 6 Write out all the steps involved in the formation of a trimer (n=3) of vinyl chloride, H2C=CHCl. Use t-butylperoxide as your free radical source.


Halogenation

At room temperature, in the absence of light, a mixture of methane and dichlorine is stable indefinitely. However, upon exposure to visible light, the components of the mixture react rapidly as shown in Equation 1. The reaction is called a free radical subsitituion.

The composition of the product depends upon the molar ratio of methane to dichlorine. The exclusive formation of chloromethane would require a very large excess of methane. However, the most interesting feature about reaction 1 is that a single photon causes the formation of thousands of molecules of chloromethane. In other words, one photon sets off a chain reaction. This and other observations led to the formulation of the 4-step mechanism shown in Figure 6.

Figure 6

Workin' on the Chain Gang

The process begins with the photochemically induced homolytic cleavage of the Cl-Cl bond to produce two chlorine atoms. This step is called initiation. In the second step, one of the chlorine atoms collides with a molecule of methane; the collision results in the transfer of a hydrogen atom, from the carbon to the chlorine. The hydrogen atom transfer results in the formation of a molecule of hydrogen chloride and a methyl radical. Since the second reaction creates a new radical, it is called a propagation reaction. A second propagation step follows as the methyl radical collides with a dichlorine molecule, an encounter that yields a molecule of chloromethane and a chlorine atom. Collision of the chlorine atom generated in the second propagation step with another molecule of methane forms a second molecule of HCl and a second methyl radical. The two propagation steps continue until the concentrations of Cl2 and/or CH4 are reduced to the point where the probability of a collision between two radicals becomes significant. Such a collision constitutes the termination step of the process.


Exercise 7 Termination reactions involve the combination of two radicals. Write an equation for another combination that would terminate the radical chain reaction involved in the chlroination of methane.

Exercise 8 Write out the steps involved in the formation of methylene chloride, CH2Cl2, by the free radical chlorination of chloromethane.


Free radical halogenation of alkanes is a general reaction. The chlorination of 2,3-dimethylbutane provides an informative example. This simple alkane contains two types of hydrogen atoms, twelve primary and two tertiary. As shown in Equation 2, chlorination of this compound produces a mixture of 2-chloro-2,3-dimethylbutane and 1-chloro-2,3-dimethylbutane. The composition of this mixture is 82.5% 2-chloro-2,3-dimethylbutane and 17.5% 1-chloro-2,3-dimethylbutane. This product distribution is different than would be expected if the only factor governing the ratio of the two isomers were the ratio of primary and tertiary hydrogens in the starting material.


Exercise 9 Following the format outlined in Figure 2, write equations for each step involved in the formation of each of the products shown in Equation 2.

Exercise 10 Which free radical is more stable A or B ? Which C-H bond would you expect to be broken more readily C2-H or C1-H ?

Exercise 11 Calculate the composition of the product mixture you would expect theoretically from reaction 2 if the only factor governing the ratio of the two isomers were the ratio of primary and tertiary hydrogens in the starting material. Express you answer to the nearest tenth of a percent.

2-chloro-2,3-dimethylbutane = %

1-chloro-2,3-dimethylbutane = %

Exercise 12 (Enter your answer to 3 significant figures.)


The product distribution in reaction 2 suggests that a 3o hydrogen atom is nearly six time more reactive than a 1o hydrogen atom.This increased reactivity has been attributed to the greater stability of the 3o free radical in comparison to the 1o alternative. The more stable free radical is formed more readily, i.e. faster. Figure 7 presents a reaction coordinate diagram that compares the relative energies of the reactants, intermediates, and products involved in reaction 2. The key feature of the figure is not the relative energies of the products, but rather the relative energies of the intermediates, and more importantly, the relative activation energies for their formation.

Figure 7

One Reactant, Two Products

Follow the reaction pathway highlighted in red in the figure: When a chlorine atom collides with a molecule of 2,3-dimethylbutane at one of the 3o hydrogen atoms, the C-H bond begins to break as the H-Cl bond begins to form. At the point labeled T3,1 the configuration of the C---H---Cl fragment has its maximum energy. As the H-Cl bond becomes stronger, the C-H bond weakens. As the Cl atom rebounds from the collision, it takes the H atom with it, leaving the intemediate tertiary free radical, I3, behind. Collision of this intermediate with a molecule of dichlorine produces further bonding changes that lead to the transition state labeled T3,2. From that point the path to the product is energetically downhill. A similar series of events occurs along the pathway highlighted in blue. The key difference is that E3act is less than E1act.


Exercise 13 Draw structures for the species labeled T3,1, T3,2, T1,1, and T1,2. Use dotted lines to indicate bonds that are partially made or partially broken.
While we now interpret experimental outcomes such as those seen for reaction 2 in tems of the relative stabilities of the intermediates formed along the alternative reaction pathways, it is important to remember that our understanding of relative stabilities evolved from the analysis of product distribution data from many reactions similar to reaction 2.

Combustion

As shown in Equation 3, the chlorination of methane will produce carbon tetrachloride if there is sufficient dichlorine in the reaction mixture.

The carbon atom in this reaction undergoes a change in oxidation level from -4 to +4; it is oxidized. A similar oxidation occurs when methane is treated with dioxygen:

In this case the reaction is called combustion. Like chlorination, combustion of methane, Equation 4, involves free radical reactions.


Exercise 14 What is the oxidation level of the carbon atom in CO2? What is the oxidation level of oxygen in dioxygen ? What is the oxidation level of oxygen in water ? ?


In our introduction to MO theory we saw that dioxygen is a diradical; there is an unpaired electron on each oxygen atom. Dioxygen is a very reactive molecule. While its reactivity is essential for our existence, it can be a bane to that existence, and it has even been implicated in our demise. Dioxygen-promoted free radical reactions are definitely involved in the spoilage of food and have been implicated as contributors to the aging process.

Free Radical Inhibitors

Since many free radical reactions are chain reactions, a simple strategy for breaking the chain involves addition of a compound that will form stable, i.e. non-reactive, free radicals. For example, the free radical produced by the reaction of 2,6-di-t-butyl-4-methylphenol with dioxygen, Equation 5, has very low reactivity. Not only is the free radical center sterically hindered by the two bulky t-butyl groups, it is also stabilized by resonance interaction with the aromatic ring.


Exercise 15 Draw resonance structures that depict the delocalization of the unpaired electron into the aromatic ring of the 2,6-di-t-butyl-4-methylphenol radical. Use single-barb arrows to indicate the motion of electrons.
The abbreviation for 2,6-di-t-butyl-4-methylphenol is BHT, which stands for butylated hydroxytoluene. This compound is added to many packaged foods as a preservative to prevent the food from becoming stale, a process that involves oxygen- promoted free radical reactions. Another common preservative is BHA, which stands for butylated hydroxyanisole, a compound in which the CH3 group of BHT is replaced by an OCH3 group.

Free radical inhibitors are also used to prevent spoilage of food products that contain unsaturated fats and oils. The allylic C-H bonds in these structures are easily attacked by free radicals, which in turn react with dioxygen in a radical chain reaction that is called autooxidation. Figure 8 summarizes the essential features of the process using the polyunsaturated fatty acid linolenic acid as an example.

Figure 8

Autooxidation of An Unsaturated Fatty Acid

In this scheme In. represents any species that initiates the chain reaction. Note that the free radical formed in step 1 is doubly allylic. Therefore it is formed readily. Once formed, it reacts with dioxygen in step 2 to produce a peroxy radical which abstracts a hydrogen atom from another linolenic acid molecule in step 3. This produces an organic peroxide and a new free radical, thereby propagating the chain reaction. Food additives such as BHT inhibit autooxidation and retard spoilage.

Fats present in cells within our bodies are subject to similar autooxidation, which may lead to cellular damage. Vitamin E is a natural antioxidant that has received considerable attention in the poplular press recently for its "anti-aging" capabilities. Consideration of the structure of vitamin E would suggest that, to the extent that free radical reactions contribute to aging, there is some basis for these claims.


Exercise 16 Draw the structure of the free radical derived from vitamin E. Then draw the three cyclohexadienyl resonance contributors that are involved in the stabilization of this free radical.

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