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Organic Chemistry Second Edition. David Klein. Chapter 11 Radical Reactions. 11.1 Free Radicals. Free radicals form when bonds break homo lytically Note the single-barbed or fishhook arrow used to show the electron movement. 11.1 Free Radicals.
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Organic Chemistry Second Edition David Klein Chapter 11 Radical Reactions Klein, Organic Chemistry 2e
11.1 Free Radicals • Free radicals form when bonds break homolytically • Note the single-barbed or fishhook arrow used to show the electron movement Klein, Organic Chemistry 2e
11.1 Free Radicals • Recall the orbital hybridization in carbocations and carbanions Klein, Organic Chemistry 2e
11.1 Free Radicals • Free radicals can be thought of as sp2 hybridized or quickly interconverting sp3 hybridized sp2 hybridized Klein, Organic Chemistry 2e
11.1 Free Radical Stability • Free radicals do not have a formal charge but are unstable because of an incomplete octet • Groups that can push (donate) electrons toward the free radical will help to stabilize it. WHY? HOW? Consider hyperconjugation Klein, Organic Chemistry 2e
11.1 Free Radical Stability • Use arguments that involve hyperconjugationand an energy diagram to explain the differences in bond dissociation energy below • Practice with conceptual checkpoint 11.1 Klein, Organic Chemistry 2e
11.1 Free Radical Resonance • Drawing resonance for free radicals using fishhook arrows to show electron movement • Remember, for resonance, the arrows don’t ACTUALLY show electron movement. WHY? • Draw the resonance hybrid for an allyl radical • HOW and WHY does resonance affect the stability of the free radical? Klein, Organic Chemistry 2e
11.1 Free Radical Resonance • The benzylic radical is a hybrid that consists of 4 contributors • Draw the remaining contributors • Draw the hybrid Klein, Organic Chemistry 2e
11.1 Resonance Stabilization • How does resonance affect the stability of a radical? WHY? • What is a bigger factor, hyperconjugation or resonance? • Practice with SkillBuilder 11.1 Klein, Organic Chemistry 2e
11.1 Resonance Stabilization • Vinylic free radicals are especially unstable • What type of orbital is the vinylic free radical located in, and how does that affect stability? • Practice with SkillBuilder 11.2 • No resonance stabilization Klein, Organic Chemistry 2e
11.2 Radical Electron Movement • Free radical electron movement is quite different from electron movement in ionic reactions • For example, free radicals don’t undergo rearrangement • There are SIX key arrow-pushing patterns that we will discuss Klein, Organic Chemistry 2e
11.2 Radical Electron Movement • Homolytic Cleavage: initiated by light or heat • Addition to a pi bond • Hydrogen abstraction: NOT the same as proton transfer • Halogen abstraction Klein, Organic Chemistry 2e
11.2 Radical Electron Movement • Elimination: the radical from the α carbon is pushed toward the β carbon to eliminate a group on the β carbon (reverse of addition to a pi bond) • The –X group is NOT a leaving group. WHY? • Coupling: the reverse of homolytic cleavage • Note that radical electron movement generally involves 2 or 3 fishhook arrows Klein, Organic Chemistry 2e
11.2 Radical Electron Movement • Note the reversibility of radical processes • Practice with SkillBuilder 11.3 Klein, Organic Chemistry 2e
11.2 Radical Electron Movement • Radical electron movement is generally classified as either initiation, termination, or propagation • Initiation occurs when radicals are created • Termination occurs when radicals are destroyed Initiation Termination Klein, Organic Chemistry 2e
11.2 Radical Electron Movement • Propagation occurs when radicals are moved from one location to another Propagation Klein, Organic Chemistry 2e
11.3 Chlorination of Methane • Let’s apply our electron-pushing skills to a reaction • We must consider each pattern for any free radical that forms during the reaction • Is homolytic cleavage also likely for CH4? Klein, Organic Chemistry 2e
11.3 Chlorination of Methane Klein, Organic Chemistry 2e
11.3 Chlorination of Methane • Why do termination reactions happen less often than propagation reactions? Klein, Organic Chemistry 2e
11.3 Chlorination of Methane • The propagation steps give the net reaction • Initiation produces a small amount Cl• radical • H abstraction consumes the Cl• radical • Cl abstraction generates a Cl• radical, which can go on to start another H abstraction • Propagation steps are self-sustaining Klein, Organic Chemistry 2e
11.3 Chlorination of Methane • Reactions that have self-sustaining propagation steps are called chain reactions • Chain reaction: the products from one step are reactants for a different step in the mechanism • Polychlorination is difficult to prevent, especially when an excess of Cl2 is present. WHY? • Practice with SkillBuilder 11.4 Klein, Organic Chemistry 2e
11.3 Chlorination of Methane • Draw a reasonable mechanism that shows how each of the products might form in the following reaction • Which of the products above are major and which are minor? Are other products also possible? Klein, Organic Chemistry 2e
11.3 Radical Initiators • An initiator starts a free radical chain reaction • Which initiator above initiates reactions most readily? WHY? • The acyl peroxide will be effective at 80 °C Klein, Organic Chemistry 2e
11.3 Radical Inhibitors • Inhibitors act in a reaction to scavenge free radicals to stop chain reaction processes • Oxygen molecules can exist in the form of a diradical, which reacts readily with other radicals. Use arrows to show the process • How can reaction conditions be modified to stop oxygen from inhibiting a desired chain reaction? Klein, Organic Chemistry 2e
11.3 Radical Inhibitors • Hydroquinone is also often used as a radical inhibitor • Draw in necessary arrows in the reactions above Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics • If we want to determine whether a process is product favored, we must determine the sign (+/-) for ΔG • In a halogenation reaction, will ΔH or -TΔS have a bigger effect on the sign of ΔG? WHY? Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics • Because the -TΔS term should be close to zero, we can simplify the free energy equation • Considering the bonds that break and form in the reaction, estimate ΔHhalogenation for each of the halogens in the table below Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics • Fluorination is so exothermic that it is impractical. WHY does that make it impractical? • Iodination is nonspontaneous, so it is not product favored Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics • Consider chlorination and bromination in more detail • Chlorination is more product favored than bromination Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics • Which step in the mechanism is the slow step? Which reaction has a faster rate? Which is more product favored? Both steps are exothermic Second step is exothermic First step is endothermic Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • With substrates more complex than ethane, multiple monohalogenation products are possible • If the halogen were indiscriminant, predict the product ratio? Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • For the CHLORINATION process, the actual product distribution favors 2-chloropropane over 1-chloropropane • Which step in the mechanism determines the regioselectivity? • Show the arrow pushing for that key mechanistic step Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • In one reaction, a 1° free radical forms, and in the other a 2° radical forms • Is the chlorination process thermodynamically or kinetically controlled? • WHY? Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • For the BROMINATION process, the product distribution vastly favors 2-bromopropane over 1-bromopropane • Which step in the mechanism determines regioselectivity? • Show the arrow pushing for that key mechanistic step Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics • Is the key step in the bromination mechanism kinetically or thermodynamically controlled? WHY? Both steps are exothermic Second step is exothermic First step is endothermic Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • Focus on the H abstraction step, and consider the Hammond postulate: species on the energy diagram that are similar in energy are similar in structure Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • When the bromine radical abstracts the hydrogen, the carbon must be able to stabilize a large partial radical • When the chlorine radical abstracts the hydrogen, the carbon does not carry as much of a partial radical Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • Which process is more regioselective? WHY? Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • Bromination at the 3° position happens 1600 times more often than at the 1° position Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • Which process is least regioselective? WHY? • What is the general relationship between reactivity and selectivity? WHY? • Practice with SkillBuilder 11.5 Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity • Ignoring possible addition products for now, draw the structure for EVERY possible monobromination product for the reaction below • Rank the products in order from most major to most minor Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry • The halogenation of butane or more complex alkanes forms a new chirality center • 2-chlorobutane will form as a racemic mixture • Which step in the mechanism is responsible for the stereochemical outcome? Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry • Whether the free radical carbon is sp2 or a rapidly interconverting sp3, the halogen abstraction will occur on either side of the plane with equal probability sp2 hybridized Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry • Three monosubstituted products form in the halogenation of butane • Draw all of the monosubstituted products that would form in the halogenation of 2-methylbutane including all stereoisomers Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry • In the halogenation of (S)-3-methylhexane, the chirality center is the most reactive carbon in the molecule. WHY? • Name the product and predict the stereochemical outcome • Practice with SkillBuilder 11.6 Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry • Draw all of the monosubstituted products that would form in the halogenation below including all stereoisomers • Classify any stereoisomer pairs as either enantiomers or diasteriomers Klein, Organic Chemistry 2e
11.7 Allylic Halogenation • When an C=C double bond is present it affects the regioselectivity of the halogenation reaction • Given the bond dissociation energies below, which position of cyclohexene will be most reactive toward halogenation? Klein, Organic Chemistry 2e
11.7 Allylic Halogenation • When an allylic hydrogen is abstracted, it leaves behind an allylic free radical that is stabilized by resonance • Based on the high selectivity of bromination that we discussed, you might expect bromination to occur as shown below • What other set of side-products is likely to form in this reaction? Hint: addition reaction Klein, Organic Chemistry 2e
