11 -1 Klein, Organic Chemistry 1e Copyright 11 -2 Klein...

• Free radicals form when bonds break HOMOLYTICALLY.

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

11.1 Free Radicals

Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e11 -1

• Recall the orbital hybridization in carbocations and carbanions.

11.1 Free Radicals


• Free radicals can be thought of as sp2 hybridized or quickly interconverting sp3 hybridized.

11.1 Free Radicals


• 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.

11.1 Free Radicals – Stability


• Use arguments that involve hyperconjugation and an energy diagram to explain the differences in bond dissociation energy (BDE) below.

• Practice with CONCEPTUAL CHECKPOINT 11.1.

11.1 Free Radicals – Stability


• 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?

11.1 Free Radicals – Resonance


• The benzylic radical is a hybrid that consists of four contributors.

• Draw the remaining contributors.

• Draw the hybrid.

11.1 Free Radicals – Resonance


• How does resonance affect the stability of a radical? WHY?

• What is a bigger factor, hyperconjugation or resonance?

• Practice with SKILLBUILDER 11.1.

11.1 Free Radicals – Resonance Stabilization


• Vinylic free radicals are especially unstable.

• No resonance stabilization.

• What type of orbital is the vinylic free radical located in, and how does that affect stability?


11.1 Free Radicals – Resonance Stabilization


• 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.

11.2 Radical Electron Movement


1. Homolytic cleavage, initiated by light or heat:

2. Addition to a pi bond:

3. Hydrogen abstraction (NOT the same as proton transfer):

4. Halogen abstraction:



5. Elimination: the radical from the α carbon is pushed toward the β carbon to eliminate a group on the βcarbon (reverse of addition to a pi bond):

– The –X group is NOT a leaving group. WHY?

6. Coupling, the reverse of homolytic cleavage:

• Note that radical electron movement generally involves two or three fishhook arrows.



• Note the reversibility of radical processes:




• Radical electron movement is generally classified as either initiation, termination, or propagation.



INITIATION occurs when radicals are created.

TERMINATION occurs when radicals are destroyed.

• PROPAGATION occurs when radicals are moved from one location to another.



• 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?

11.3 Radical Reactions –Chlorination of Methane




• Why are termination reactions less common than propagation reactions?



• The propagation steps give the net reaction:

1. Initiation produces a small amount Cl• radical.

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.



• 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 Cl2 is present. WHY?




• 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?



Cl2

heatCl Cl

Cl

Cl

• 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.

11.3 Radical Reactions – Radical Initiators


∆Hº = +159 kJ/mol



• 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?

11.3 Radical Reactions – Radical Inhibitors


• Hydroquinone is also often used as a radical inhibitor.

• Draw in necessary arrows in the reactions above.

11.3 Radical Reactions – Radical Inhibitors


• 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?

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.





• 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.



• Consider chlorination and bromination in more detail.

• Chlorination is more product‐favored than bromination.



• Which step in the mechanism is the slow step? Which reaction has a faster rate? Which is more product favored?



–Both steps are exothermic

– First step is endothermic

– Second step is exothermic

• With substrates more complex than ethane, multiple MONOHALOGENATION products are possible.

• If the halogen were indiscriminant, predict the product ratio?

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.



• 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?



• 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.



• Is the key step in the bromination mechanism kinetically or thermodynamically controlled? WHY?

11.5 Halogenation Regioselectivity –Kinetic vs. Thermodynamic


• 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.





• 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.



• Which process is more regioselective? WHY?



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



• Which process is least regioselective? WHY?

• What is the general relationship between reactivity and selectivity? WHY?




• For the reaction below, draw the structure for EVERY possible monobromination product.

• Rank the products in order from most major to most minor.



• The halogenation of butane or more complex alkanesforms a new chirality center.

• 2‐chlorobutane will form as a racemic mixture.

• Which step in the mechanism is responsible for the stereochemical outcome?

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.



• 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.



• In the halogenation of (S)‐3‐methylhexane, the chiralitycenter is the most reactive carbon in the molecule. WHY?

• Name the product and predict the stereochemicaloutcome.




• Draw all of the monosubstituted products that would form in the halogenation below including all stereoisomers.

• Classify any stereoisomer pairs as either enantiomers or diastereomers.



Br2

light

• 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?

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:

• What other set of side‐products is likely to form in this reaction? Hint: addition reaction.

11.7 Allylic Halogenation


• To avoid the competing halogen ADDITION reaction, NBS can be used to supply Br• radicals.

• Show how resonance stabilizes the succinimide radical.

• Heat or light INITIATES the process.

11.7 Allylic Halogenation with NBS


• Propagation produces new Br• radicals to continue the chain reaction.

• Where does the Br–Br above come from? The amount of Br–Br in solution is minimal, so the competing addition reaction is minimized.



• 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.

• Give some examples of some termination steps that might occur.



+ H-Br

• Give a mechanism that explains the following product distribution. Hint: resonance.




• Ozone is both created and destroyed in the upper atmosphere.

• O3 molecules absorb harmful UV radiation.

• O3 molecules are recycled as heat energy is released.

11.8 Applications –Atmospheric Chemistry and O3


• For this process to be spontaneous, the entropy of the universe must increase.

• Heat is a more disordered form of energy than light. WHY?



• O3 depletion (about 6% each year) remains a serious health and environmental issue.

• 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?



• CFC substitutes that generally decompose before reaching the O3 layer include hydrochlorofluorocarbons.

• Hydrofluorocarbons don’t even form the chlorine radicals that interfere with the O3 cycle.



• Like most reactions, combustion involves breaking 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.

• Why does the process release energy overall?

• What types of chemicals might be used in fire extinguishers to inhibit the process?

11.8 Applications –Combustion and Firefighting


• Water deprives the fire of the Eact needed by absorbing the energy .

• CO2 and argon gas deprive the fire of needed oxygen.

• Halons are very effective fire‐suppression agents.

• Halons suppress combustion in three main ways.



• Halons suppress combustion in three main ways:1. As a gas, they can smother the fire and deprive it of O2.

2. They absorb some of the Eact for the fire by undergoing homolytic cleavage.

3. The free radicals produced can combine with C• and H• free radicals to terminate the combustion chain reaction.



• How do you think the use of halons to fight fires affects the ozone layer?

• FM‐200 is an alternative firefighting agent.

• Practice with CONCEPTUAL CHECKPOINT 11.18.



• Autooxidation is the process by which compounds react with molecular oxygen.

• The process is generally very slow.

11.9 Autooxidation vs. Antioxidation –Autooxidation


• The mechanism illustrates the need for a more refined definition of initiation and propagation. HOW?

11.9 Autooxidation vs. Antioxidation –Autooxidation Mechanism


• Propagation can be more precisely defined as the steps that add together to give the NET chemical equation.

• Steps in the mechanism that are not part of the NET equation must be either initiation or termination.

11.9 Autooxidation vs. Antioxidation –Autooxidation Mechanism


• Some compounds such as ethers are particularly susceptible to autooxidation.

• Because hydroperoxides can be explosive, ethers like diethyl ether must not be stored for long periods of time.

• They should be dated and used in a timely fashion.



• Light accelerates the autooxidation process.

• 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 susceptible to autooxidation. WHY?

• Consider the autooxidation of compounds with allylic or benzylic hydrogen.



• Triglycerides are important to a healthy diet.

• Autooxidation can occur at the allylic positions causing the food to become rancid and toxic.

11.9 Autooxidation vs. Antioxidation –Antioxidants


• Foods with unsaturated fatty acids have a short shelf life unless preservatives are used.



• Preservatives can undergo H abstraction to quench the C• radicals that form in the first step of 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.



• Vitamins C is hydrophilic.

• 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.

11.9 Autooxidation vs. Antioxidation –Natural Antioxidants


• We learned in Chapter 9 that H–X will add across a C=C double bond with anti‐Markovnikov regioselectivitywhen peroxides are present.

• Now that we have discussed free radicals, we can explain the mechanism for the anti‐Markovnikov addition.

11.10 Addition – Anti‐MarkovnikovAddition


• The O–O single bond can break homolytically with a relatively low Eact.



• The Br• radical reacts to give the more stable C• radical.

• A 3° radical forms through a lower transition state than a 2° 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.

• Go back through each step in the mechanism to explain the ΔH value for each step.



• Anti‐Markovnikov radical addition of HBr is generally spontaneous (product favored).

• Yet, radical addition of HCl and HI are generally nonspontaneous (reactant favored).

• We will examine the thermodynamics of each propagation step.

11.10 Addition –Thermodynamics


• Explain the sign (+/–) for the entropy term in each process.

• Will the entropy change favor products or reactants? WHY?

• The enthalpy change is affected most by the relative stability of the two radical species.

11.10 Addition – Thermodynamics


• The HI reaction will never favor products. WHY?

• How can the temperature be adjusted to favor products for the HCl and HBr reaction?

11.10 Addition –Thermodynamics


• For the second propagation step, why is the entropy term approximately zero?

• What small differences in entropy between products and reactants might account for entropy changes being slightly + or –?



• `

`

• For HCl and HI, the second propagation step will not be product‐favored. WHY?– HBr is the only reactant that favors product formation for

both propagation steps.

• Might the overall propagation involving HCl be product‐favored if the –ΔG of step 1 outweighs the +ΔG of step 2 ?



• Addition reactions often form a new chiral center.

• Recall that at least one reagent in the reaction must be chiral for the reaction to be stereoselective.

• Predict the product distribution for the reaction below and explain the stereochemical outcome.


11.10 Addition – Stereochemistry


• In Chapter 9, we learned how some ionic 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


• Radical polymerizations generally proceed through a chain reaction mechanism.




• Note how the sum of the propagation steps yields the overall reaction.






• Radical polymerizations generally produce chains of 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 through 1° carbon free radical intermediates, it is usually not facile.

• In Chapter 27, we will discuss specialized catalysts that can be used to control such polymerization reactions.



• Branching is common in some radical polymerizations.

• Branching makes polymer materials more flexible, such as a polyethylene used for squeeze bottles.

• When catalysts are used to minimize branching, more rigid materials are produced, such as the material used for squeeze bottle caps.

• Why does branching affect rigidity?



• Many derivatives of ethylene are also polymerized.





• Recall the CRACKING and REFORMING processes from chapters 4 and 8.

• Crude oil is CRACKED to produce smaller alkanes and alkenes.

• Reforming increases branching. Propose a mechanism.

11.12 Radical Petroleum Processes


• Radical chlorination and bromination are both useful processes.

• Recall that bromination is more selective. WHY?

• Temperature can be used to help avoid polysubstitution. HOW?

11.13 Synthetic Utility of Halogenation


• Chlorination can be useful with highly symmetrical substrates.

• 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.



• Synthesizing a target molecule from an alkane is challenging because of its limited reactivity.

• Often halogenation is the best option.



11 -1 Klein, Organic Chemistry 1e Copyright 11 -2 Klein...

Documents

Transcript of 11 -1 Klein, Organic Chemistry 1e Copyright 11 -2 Klein...