Stability 11.1 Free Radicals –Resonance

Home , Hyperconjugation, Radical (chemistry)

11.1 Free Radicals 11.1 Free Radicals

• Free radicals form when bonds break HOMOLYTICALLY. • Recall the orbital hybridization in carbocations and carbanions.

• Note the single‐barbed or fishhook arrow used to show the electron movement.

11.1 Free Radicals 11.1 Free Radicals – Stability

• Free radicals can be thought of as sp2 hybridized or • Free radicals do not have a formal charge but are quickly interconverting sp3 hybridized. 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.

11.1 Free Radicals – Stability 11.1 Free Radicals –Resonance

• Use arguments that involve hyperconjugation and an • Drawing resonance for free radicals using fishhook energy diagram to explain the differences in bond arrows to show electron movement. dissociation energy (BDE) below. • 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? • Practice with CONCEPTUAL CHECKPOINT 11.1. Copyright 2012 John Wiley & Sons, Inc. 11 -5 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -6 Klein, Organic Chemistry 1e 11.1 Free Radicals – Resonance 11.1 Free Radicals –Resonance Stabilization • The benzylic radical is a hybrid that consists of four • How does resonance affect the stability of a radical? contributors. WHY?

• Draw the remaining contributors. • Draw the hybrid. • What is a bigger factor, hyperconjugation or resonance? • Practice with SKILLBUILDER 11.1. Copyright 2012 John Wiley & Sons, Inc. 11 -7 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -8 Klein, Organic Chemistry 1e

11.1 Free Radicals – Resonance 11.2 Radical Electron Movement Stabilization • Vinylic free radicals are especially unstable. • Free radical electron movement is quite different from • No resonance stabilization. electron movement in ionic reactions. – For example, free radicals don’t undergo rearrangement.

• What type of orbital is the vinylic free radical located in, • There are SIX key arrow‐pushing patterns that we will and how does that affect stability? discuss. • Practice with SKILLBUILDER 11.2. Copyright 2012 John Wiley & Sons, Inc. 11 -9 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -10 Klein, Organic Chemistry 1e

11.2 Radical Electron Movement 11.2 Radical Electron Movement

1. Homolytic cleavage, initiated by light or heat: 5. 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): 2. Addition to a pi bond:

– The –X group is NOT a leaving group. WHY? 3. Hydrogen abstraction (NOT the same as proton transfer): 6. Coupling, the reverse of homolytic cleavage:

4. Halogen abstraction: • Note that radical electron movement generally involves two or three fishhook arrows. Copyright 2012 John Wiley & Sons, Inc. 11 -11 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -12 Klein, Organic Chemistry 1e 11.2 Radical Electron Movement 11.2 Radical Electron Movement

• Note the reversibility of radical processes: • Radical electron movement is generally classified as either initiation, termination, or propagation.

INITIATION TERMINATION occurs when occurs when radicals are radicals are created. destroyed.

11.3 Radical Reactions – 11.2 Radical Electron Movement Chlorination of Methane • PROPAGATION occurs when radicals are moved from • Let’s apply our electron‐pushing skills to a reaction. one location to another.

• We must consider each pattern for any free radical that forms during the reaction.

11.3 Radical Reactions – 11.3 Radical Reactions – Chlorination of Methane Chlorination of Methane • Why are termination reactions less common than propagation reactions?

Copyright 2012 John Wiley & Sons, Inc. 11 -17 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -18 Klein, Organic Chemistry 1e 11.3 Radical Reactions – 11.3 Radical Reactions – Chlorination of Methane Chlorination of Methane • The propagation steps give the net reaction: • Reactions that have self‐sustaining propagation steps are called CHAIN REACTIONS. • In a 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 Cl is present. WHY? 1. Initiation produces a small amount Cl• radical. 2 2. H abstraction consumes the Cl• radical. 3. Cl abstraction generates a Cl• radical, which can go on to start another H abstraction. • Propagation steps are self‐sustaining. • Practice with SKILLBUILDER 11.4. Copyright 2012 John Wiley & Sons, Inc. 11 -19 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -20 Klein, Organic Chemistry 1e

11.3 Radical Reactions – 11.3 Radical Reactions – Radical Chlorination of Methane Initiators • Draw a reasonable mechanism that shows how each of • An initiator starts a free radical chain reaction the products might form in the following reaction. ∆Hº = +159 kJ/mol Cl ∆Hº = +243 kJ/mol Cl2

heat Cl Cl ∆Hº = +121 kJ/mol Cl

• Which of the products above are major, and which are • Which initiator above initiates reactions most readily? minor? WHY? • The acyl peroxide will be effective at 80 °C. Copyright 2012 John Wiley & Sons, Inc. 11 -21 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -22 Klein, Organic Chemistry 1e

11.3 Radical Reactions – Radical 11.3 Radical Reactions – Radical Inhibitors Inhibitors • Inhibitors act in a reaction to scavenge free radicals to • Hydroquinone is also often used as a radical inhibitor. 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? Draw in necessary arrows in the reactions above.

Copyright 2012 John Wiley & Sons, Inc. 11 -23 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -24 Klein, Organic Chemistry 1e 11.4 Halogenation Thermodynamics 11.4 Halogenation Thermodynamics

• If we want to determine whether a process is product • Because the ‐TΔS term should be close to zero, we can favored, we must determine the sign (+/‐) for ΔG. 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. • In a halogenation reaction, will ΔH or ‐TΔS have a bigger effect on the sign of ΔG? WHY?

11.4 Halogenation Thermodynamics 11.4 Halogenation Thermodynamics

• Fluorination is so exothermic that it is impractical. WHY does that make it impractical?

• Iodination is nonspontaneous, so does not favor products. The reaction is not PRODUCT‐FAVORED.

11.4 Halogenation Thermodynamics 11.4 Halogenation Thermodynamics

• Consider chlorination and bromination in more detail. • Which step in the mechanism is the slow step? Which reaction has a faster rate? Which is more product favored? –Both steps are – First step is endothermic exothermic – Second step is exothermic

• Chlorination is more product‐favored than bromination. Copyright 2012 John Wiley & Sons, Inc. 11 -29 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -30 Klein, Organic Chemistry 1e 11.5 Halogenation Regioselectivity 11.5 Halogenation Regioselectivity

• With substrates more complex than ethane, multiple • For the CHLORINATION process, the actual product MONOHALOGENATION products are possible. distribution favors 2‐chloropropane over 1‐ chloropropane.

• If the halogen were indiscriminant, predict the product • Which step in the mechanism determines the ratio? regioselectivity? • Show the arrow pushing for that key mechanistic step.

11.5 Halogenation Regioselectivity 11.5 Halogenation Regioselectivity • In one reaction, a 1° free radical forms, and in the other, a 2° radical forms. • For the BROMINATION process, the product distribution • Is the chlorination process thermodynamically or vastly favors 2‐bromopropane over 1‐bromopropane. kinetically controlled? • WHY?

• Which step in the mechanism determines regioselectivity? • Show the arrow pushing for that key mechanistic step.

11.5 Halogenation Regioselectivity – 11.5 Halogenation Regioselectivity Kinetic vs. Thermodynamic • Is the key step in the bromination mechanism • Focus on the H abstraction step, and consider the kinetically or thermodynamically controlled? WHY? Hammond postulate—species on the energy diagram that are similar in energy are similar in structure.

Copyright 2012 John Wiley & Sons, Inc. 11 -35 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -36 Klein, Organic Chemistry 1e 11.5 Halogenation Regioselectivity 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.

11.5 Halogenation Regioselectivity 11.5 Halogenation Regioselectivity

• Which process is more regioselective? WHY?

• Bromination at the 3° position happens 1600 times more often than at the 1° position .

11.5 Halogenation Regioselectivity 11.5 Halogenation Regioselectivity

• Which process is least regioselective? WHY? • For the reaction below, draw the structure for EVERY possible monobromination product.

• What is the general relationship between reactivity and selectivity? WHY? • Rank the products in order from most major to most minor. • Practice with SKILLBUILDER 11.5. Copyright 2012 John Wiley & Sons, Inc. 11 -41 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -42 Klein, Organic Chemistry 1e 11.6 Halogenation Stereochemistry 11.6 Halogenation Stereochemistry

• The halogenation of butane or more complex alkanes • Whether the free radical carbon is sp2 or a rapidly forms a new chirality center. interconverting sp3, the halogen abstraction will occur on either side of the plane with equal probability.

• 2‐chlorobutane will form as a racemic mixture. • Which step in the mechanism is responsible for the stereochemical outcome?

11.6 Halogenation Stereochemistry 11.6 Halogenation Stereochemistry

• Three monosubstituted products form in the • In the halogenation of (S)‐3‐methylhexane, the chirality halogenation of butane. center is the most reactive carbon in the molecule. WHY?

• Draw all of the monosubstituted products that would form in the halogenation of 2‐methylbutane including • Name the product and predict the stereochemical all stereoisomers. outcome. • Practice with SKILLBUILDER 11.6. Copyright 2012 John Wiley & Sons, Inc. 11 -45 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -46 Klein, Organic Chemistry 1e

11.6 Halogenation Stereochemistry 11.7 Allylic Halogenation

• Draw all of the monosubstituted products that would • When an C=C double bond is present, it affects the form in the halogenation below including all regioselectivity of the halogenation reaction. stereoisomers. • Given the bond dissociation energies below, which position of cyclohexene will be most reactive toward halogenation? Br2

light

• Classify any stereoisomer pairs as either enantiomers or diastereomers.

Copyright 2012 John Wiley & Sons, Inc. 11 -47 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -48 Klein, Organic Chemistry 1e 11.7 Allylic Halogenation 11.7 Allylic Halogenation with NBS

• When an allylic hydrogen is abstracted, it leaves behind • To avoid the competing halogen ADDITION reaction, an allylic free radical that is stabilized by resonance. NBS can be used to supply Br• radicals.

• Based on the high selectivity of bromination that we discussed, you might expect bromination to occur as shown:

• Show how resonance stabilizes the succinimide radical. • What other set of side‐products is likely to form in this reaction? Hint: addition reaction. • Heat or light INITIATES the process. Copyright 2012 John Wiley & Sons, Inc. 11 -49 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -50 Klein, Organic Chemistry 1e

11.7 Allylic Halogenation with NBS 11.7 Allylic Halogenation with NBS

• The succinimide radical that is produced in the initiation step can also undergo propagation when it collides with an H–Br molecule. Show a mechanism.

• Propagation produces new Br• radicals to continue the + H-Br chain reaction. • Where does the Br–Br above come from? The amount of Br–Br in solution is minimal, so the competing • Give some examples of some termination steps that addition reaction is minimized. might occur.

11.8 Applications – 11.7 Allylic Halogenation with NBS Atmospheric Chemistry and O3 • Ozone is both created and destroyed in the upper atmosphere. • Give a mechanism that explains the following product • O molecules absorb harmful UV radiation. distribution. Hint: resonance. 3 • O3 molecules are recycled as heat energy is released.

• Practice with SKILLBUILDER 11.7.

Atmospheric Chemistry and O3 Atmospheric Chemistry and O3 • For this process to be spontaneous, the entropy of the • O depletion (about 6% each year) remains a serious universe must increase. 3 health and environmental issue. • Heat is a more disordered form of energy than light. WHY? • Compounds that are most destructive to the ozone layer: 1. Are stable enough to reach the upper atmosphere .

2. Form free radicals that interfere with the O3 recycling process.

– CHLOROFLUOROCARBONs (CFCs) fit both criteria.

• Atmosphere O3 is vital for protection, but what effect does O3 have at the earth’s surface? Copyright 2012 John Wiley & Sons, Inc. 11 -55 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -56 Klein, Organic Chemistry 1e

11.8 Applications – 11.8 Applications –

Atmospheric Chemistry and O3 Combustion and Firefighting • CFC substitutes that generally decompose before • Like most reactions, combustion involves breaking

reaching the O3 layer include hydrochlorofluorocarbons. bonds and forming new bonds:

– A fuel is heated with the necessary Eact to break bonds (C‐C, C‐ H, and O=O) homolytically. – The resulting free radicals join together to form new O–H and C=O bonds. • Hydrofluorocarbons don’t even form the chlorine • Why does the process release energy overall? radicals that interfere with the O cycle. 3 • What types of chemicals might be used in fire extinguishers to inhibit the process?

11.8 Applications – 11.8 Applications – Combustion and Firefighting Combustion and Firefighting

• Water deprives the fire of the Eact needed by absorbing • Halons suppress combustion in three main ways:

the energy . 1. As a gas, they can smother the fire and deprive it of O2. 2. They absorb some of the E for the fire by undergoing • CO2 and argon gas deprive the fire of needed oxygen. act homolytic cleavage. • Halons are very effective fire‐suppression agents.

3. The free radicals produced can combine with C• and H• free radicals to terminate the combustion chain reaction. • Halons suppress combustion in three main ways.

Copyright 2012 John Wiley & Sons, Inc. 11 -59 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -60 Klein, Organic Chemistry 1e 11.8 Applications – 11.9 Autooxidation vs. Antioxidation – Combustion and Firefighting Autooxidation • How do you think the use of halons to fight fires affects • Autooxidation is the process by which compounds react the ozone layer? with molecular oxygen.

• FM‐200 is an alternative firefighting agent.

• The process is generally very slow. • Practice with CONCEPTUAL CHECKPOINT 11.18.

11.9 Autooxidation vs. Antioxidation – 11.9 Autooxidation vs. Antioxidation – Autooxidation Mechanism Autooxidation Mechanism • The mechanism • Propagation can be more precisely defined as the steps illustrates the that add together to give the NET chemical equation. need for a more refined definition of initiation and propagation. HOW? • Steps in the mechanism that are not part of the NET equation must be either initiation or termination.

11.9 Autooxidation vs. Antioxidation – 11.9 Autooxidation vs. Antioxidation – Autooxidation Autooxidation • Some compounds such as ethers are particularly • Light accelerates the autooxidation process. susceptible to autooxidation. • Dark containers are often used to store many chemicals such as vitamins. • In the absence of light, autooxidation is usually a slow process. • Compounds that can form a relatively stable C• radical upon H abstraction are especially • Because hydroperoxides can be explosive, ethers like susceptible to autooxidation. WHY? diethyl ether must not be stored for long periods of • Consider the autooxidation of compounds time. with allylic or benzylic hydrogen. • They should be dated and used in a timely fashion. Copyright 2012 John Wiley & Sons, Inc. 11 -65 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -66 Klein, Organic Chemistry 1e 11.9 Autooxidation vs. Antioxidation – 11.9 Autooxidation vs. Antioxidation – Antioxidants Antioxidants • Triglycerides are important to a healthy diet. • Foods with unsaturated fatty acids have a short shelf life unless preservatives are used.

• Autooxidation can occur at the allylic positions causing the food to become rancid and toxic. Copyright 2012 John Wiley & Sons, Inc. 11 -67 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -68 Klein, Organic Chemistry 1e

11.9 Autooxidation vs. Antioxidation – 11.9 Autooxidation vs. Antioxidation – Antioxidants Natural Antioxidants • Preservatives can undergo H abstraction to quench the • Vitamins C is hydrophilic. C• radicals that form in the first step of autooxidation. • Vitamin E is hydrophobic. • What parts of the body do these vitamins protect? • For each vitamin, show its oxidation mechanism, and explain how that protects the body from autooxidation. • One molecule of BHT can prevent thousands of autooxidation reactions by stopping the chain reaction. • How does BHT’s structure make it good at taking on a free radical? Consider resonance and sterics.

11.10 Addition – Anti‐Markovnikov 11.10 Addition – Anti‐Markovnikov Addition Addition • We learned in Chapter 9 that H–X will add across a C=C • The O–O single bond can break homolytically with a

double bond with anti‐Markovnikov regioselectivity relatively low Eact. when peroxides are present.

• Now that we have discussed free radicals, we can explain the mechanism for the anti‐Markovnikov addition. Copyright 2012 John Wiley & Sons, Inc. 11 -71 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -72 Klein, Organic Chemistry 1e 11.10 Addition – Anti‐Markovnikov 11.10 Addition – Anti‐Markovnikov Addition Addition • The Br• radical reacts to give the more stable C• radical. • Most of the radicals in solution at any given moment will be Br• radical. WHY?

• Give some examples of other termination steps that might occur but that would be less common.

• A 3° radical forms through a lower transition state than • Go back through each step in the mechanism to explain a 2° radical. the ΔH value for each step.

11.10 Addition –Thermodynamics 11.10 Addition –Thermodynamics

• Anti‐Markovnikov radical addition of HBr is generally • Explain the sign (+/–) for the entropy term in each spontaneous (product favored). process. • Yet, radical addition of HCl and HI are generally nonspontaneous (reactant favored).

• Will the entropy change favor products or reactants? • We will examine the thermodynamics of each WHY? propagation step. • The enthalpy change is affected most by the relative stability of the two radical species. Copyright 2012 John Wiley & Sons, Inc. 11 -75 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -76 Klein, Organic Chemistry 1e

11.10 Addition –Thermodynamics 11.10 Addition –Thermodynamics

• For the second propagation step, why is the entropy • The HI reaction will never favor products. WHY? term approximately zero?

• What small differences in entropy between products • How can the temperature be adjusted to favor products and reactants might account for entropy changes being for the HCl and HBr reaction? slightly + or –?

Copyright 2012 John Wiley & Sons, Inc. 11 -77 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -78 Klein, Organic Chemistry 1e 11.10 Addition –Thermodynamics 11.10 Addition –Stereochemistry

• Addition reactions often form a new chiral center. • `

` • Recall that at least one reagent in the reaction must be • For HCl and HI, the second propagation step will not be chiral for the reaction to be stereoselective. product‐favored. WHY? • Predict the product distribution for the reaction below – HBr is the only reactant that favors product formation for both propagation steps. and explain the stereochemical outcome. • Might the overall propagation involving HCl be product‐ favored if the –ΔG of step 1 outweighs the +ΔG of step 2 ? • Practice with SKILLBUILDER 11.8. Copyright 2012 John Wiley & Sons, Inc. 11 -79 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -80 Klein, Organic Chemistry 1e

11.11 Radical Polymerization 11.11 Radical Polymerization • Radical polymerizations generally proceed through a • In Chapter 9, we learned how some ionic chain reaction mechanism. polymerizations occur. • Free radical conditions are also frequently used to form polymers. • Recall that a polymerization process joins together many small units called monomers in a long chain.

11.11 Radical Polymerization 11.11 Radical Polymerization • Radical polymerizations generally proceed through a • Radical polymerizations generally proceed through a chain reaction mechanism. chain reaction mechanism.

• Note how the sum of the propagation steps yields the overall reaction. Copyright 2012 John Wiley & Sons, Inc. 11 -83 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -84 Klein, Organic Chemistry 1e 11.11 Radical Polymerization 11.11 Radical Polymerization

• Radical polymerizations generally produce chains of • Branching is common in some radical polymerizations. monomers with a wide distribution of lengths. • How might experimental conditions be optimized to control the average length of the chains? • Because the mechanism we just learned proceeds • Branching makes polymer materials through 1° carbon free radical intermediates, it is more flexible, such as a polyethylene usually not facile. used for squeeze bottles. • In Chapter 27, we will discuss specialized catalysts that • When catalysts are used to minimize can be used to control such polymerization reactions. branching, more rigid materials are produced, such as the material used for squeeze bottle caps. • Why does branching affect rigidity? Copyright 2012 John Wiley & Sons, Inc. 11 -85 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -86 Klein, Organic Chemistry 1e

11.11 Radical Polymerization 11.11 Radical Polymerization

• Many derivatives of ethylene are also polymerized.

11.13 Synthetic Utility of 11.12 Radical Petroleum Processes Halogenation • Recall the CRACKING and REFORMING processes from • Radical chlorination and bromination are both useful chapters 4 and 8. processes. • Crude oil is CRACKED to produce smaller alkanes and • Recall that bromination is more selective. WHY? alkenes.

• Temperature can be used to help avoid polysubstitution. HOW? • Reforming increases branching. Propose a mechanism.

Copyright 2012 John Wiley & Sons, Inc. 11 -89 Klein, Organic Chemistry 1e Copyright 2012 John Wiley & Sons, Inc. 11 -90 Klein, Organic Chemistry 1e 11.13 Synthetic Utility of 11.13 Synthetic Utility of Halogenation Halogenation • Chlorination can be useful with highly symmetrical • Synthesizing a target molecule from an alkane is substrates. challenging because of its limited reactivity. • Often halogenation is the best option.

• It is difficult to avoid polysubstitution. WHY? • The synthetic utility of halogenation is limited: – Chlorination is difficult to control. – Bromination requires a substrate with one site that is significantly more reactive than all others.