Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

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Pricing

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Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

Up to 40-60% more affordable

Lifetime access on any device

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

$219

Hardcover print text only

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

$301

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David R. Klein, “Organic Chemistry”, 3rd Edition

$301

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Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

About this textbook

Lead Authors

Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

Contributing Authors

Felix NgassaGrand Valley State University

Neil GargUCLA

Jennifer ChaytorSaginaw Valley State University

Greg DomskiAugustana College

Christian E. MaduCollin Community College

Christopher NicholsonUniversity of West Florida

Franklin OwEast Los Angeles College, UCLA

Robert S. PhillipsUniversity of Georgia

Grigoriy SeredaUniversity of South Dakota

Simon E. LopezUniversity of Florida

Brannon McCulloughNorthern Arizona University

Jason JonesKennesaw State University

José BoquinAugustana College

Stephanie BrouetSaginaw Valley State University

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Chapter 18: Radicals

The photodissociation of nitrogen dioxide produces two radicals: nitrogen oxide and an oxygen atom. Nitrogen oxide is a key player in the production of smog. [1]


Contents

Learning Objectives

  • Use bond dissociation tables to estimate bond strength and relative stability of radicals.
  • Comprehend the difference between highly reactive free radicals and very stable free radicals.
  • Evaluate a series of radicals and rank their relative stability from most to least stable.
  • Understand the chain reaction and predict the products formed in initiation, propagation and termination reactions.
  • Understand the reactivity and selectivity of the reagents I2, Br2, Cl2, and F2 in the radical halogenation of alkanes.
  • Predict the major and minor products formed in the radical halogenation of alkanes with Br2 and Cl2.
  • Understand the stereochemistry of radical reactions.
  • Predict the products formed in the allylic bromination reactions using N-bromosuccinimide (NBS).
  • Understand the mechanism of the addition of HBr with peroxides to an alkene and predict the products formed.

18.1 Introduction

Free radical reactions are involved in many chemical processes, such as the combustion of fuels, biochemical reactions, aging and diseases, and the industrial manufacturing of fabrics and plastics.

Radicals are molecules that possess an unpaired electron. Radicals are neutral, but because they do not have a full octet they are very reactive. Most radicals have very short lifespans. Some examples of radicals are given in Figure 18.1.

Figure 18.1. Examples of radicals

The shape of carbon radical orbitals is hard to determine, but experiments and theory suggest that their shapes lie between a planar sphybridized carbocation shape and a pyramidal sphybridized carbanion shape. Thus, carbon radicals can be regarded as being sphybridized with the single electron occupying an unhybridized p orbital.

Figure 18.2. Orbital structure comparison between a carbocation, carbon radical, and carbanion

18.2 Homolysis and Bond Dissociation Energies

Bond dissociation energies (BDEs) are defined as the change in enthalpy (∆Ho) for the homolytic bond cleavage process shown in Figure 18.3.

Figure 18.3. Sigma (σ) bond homolysis and free radical formation

The BDE values obtained for the above general reaction are valuable thermodynamic quantities because they correlate directly to bond strength. These values are generally determined for the gas phase reaction with all species in their standard states (at the standard pressure of 1 bar). However, like all thermodynamic measurements, the values obtained are temperature-dependent and depend on whether they are measured in the liquid or gas phase.

BDEs describe the homolytic cleavage of a bond: In a homolytic process, each of the atoms whose bond is broken receives one electron from the σ bond being cleaved to produce two neutral radicals. This is different from heterolytic cleavage in which the two electrons from the σ bond move onto one atom to produce a cation and an anion. Remember from previous chapters that a double-headed arrow is used to show the movement of two electrons and a single-headed arrow is used to show the movement of one electron. Can you identify the correct use of arrows for homolytic and heterolytic bond cleavage in the reaction below?

Q18.1 - Level 1

Match heterolytic bond cleavage and homolytic bond cleavage to the correct mechanism.

question description
Premise
Response
1

Heterolytic bond cleavage

A

Mechanism 4

2

Homolytic bond cleavage

B

Mechanism 5

C

Mechanism 1

D

Mechanism 2

E

Mechanism 3


18.2.1 Calculation of Heats of Reaction

We can use BDEs, which correlate to bond strengths, to determine the relative stability and reactivity of different molecules. Let’s review thermochemistry and molecular stability.

Why is one molecule more stable than another? All molecules possess internal strain caused by conformational distortion or non-optimal bonding with respect to a reference compound. The internal strain is an internal energy stored within the molecule. Thus, molecules have a potential energy (PE) or stored energy that can be released when given the right reaction conditions. The most common analogy is a compressed spring (under stress) that has stored, or potential energy (PE). The stressed spring has a higher potential energy than the relaxed state of the spring and has the potential to do work, such as moving a block when released. Similarly, chemicals can do work via a chemical reaction, such as the force created by the decomposition of sodium azide (NaN3) with an electrical spark to produce sodium metal and nitrogen gas in car airbags.

In previous chapters, the concept of Gibbs free energy (∆G°) was introduced. Gibbs free energy has two components, enthalpy (∆H°) and entropy (∆S°), given by the Gibbs free energy equation:

∆G° = ∆H°-T∆S°

The entropy of a system is a measure of the disorder of a system. Entropy describes a molecule’s ability to move through, rotate, or tumble within the system (translational and rotational energy), and the internal vibrational motion of the atoms in the molecule (vibrational energy). Entropy changes are very important when looking at changes in chemical structure, such as cyclization reactions, conformational changes, or mixing of solutions. However, for many reactions ∆S° is small. Therefore, when focusing on bond strengths and reactivities, the “-T∆S°” component of the Gibbs free energy equation can be considered negligible.

Enthalpy change (∆H°) is the heat absorbed or released at a constant pressure and temperature, and relates mainly to the differences between both intramolecular and intermolecular bond strengths of the products and reactants. For any reaction, the heat of reaction ∆H° indicates the difference in bond energies between the reactants and products.

∆H° = sum of ∆H° bonds broken – sum of ∆H° bonds formed

When energy is released by the system in the form of heat during a reaction, the reaction is exothermic (∆H° is negative). If energy is absorbed by the system, then the reaction is endothermic (∆H° is positive).

Since bond dissociation energies are a measure of the change of enthalpy during homolysis, we can use BDE to predict the change in enthalpy between the reactants and products or the heat of reaction ∆H°react. If the reactants have stronger bonds than the products, then the reaction will be endothermic (∆H° is positive). If the products are more stable with strong bonds, then the reaction will be exothermic (∆H° is negative).

This allows us to predict whether a reaction is endothermic or exothermic just by determining if the products or reactants have stronger or more stable bonds. Always remember that thermodynamics only tells us the relative stabilities between the products and reactants and nothing about how fast the reaction will proceed; for that, we need to consider reaction rates or kinetics.

Table 18.1

Table 18.1


Click here to go to Question 18.2; click here to go to Question 18.7; click here to go to Question 18.21

Example: Using the values in Table 18.1, determine whether the following reaction is exothermic or endothermic.

CH4 + Cl2 → CH3Cl + HCl

For the purpose of this calculation, we can assume the following pathway:

Figure 18.4. Bond formation and breaking steps in the reaction of methane with chlorine

This is not the mechanism; however, because the heat of reaction is a thermodynamic quantity and is not dependent on the pathway, and only depends on the initial and final steps of the reaction, we need to look at only the bonds being broken and the bonds formed.

Bond breaking: The values in Table 18.1 are all positive because to break a bond, energy must be absorbed. The gas phase potential energy reaction coordinate diagram is given below for the homolytic cleavage of Cl2. Notice that there is no transition state. This is because bond breaking is the only process and for a gas phase reaction the activation energy (Ea) is usually equal to the change in enthalpy.

Figure 18.5. Potential energy diagram for the homolytic cleavage of Cl2

Bond formation: When two radicals combine, a new σ bond is formed and energy is released. Thus, the energy given off during bond formation is equal to the negative values of the BDEs given in Table 18.1. The potential energy diagram is given below and, for small radicals in the gas phase, the activation energy is usually zero.

Figure 18.6. Potential energy diagram for bond formation between two radicals

To determine the ∆H° of this reaction (CH4 + Cl2 → CH3Cl + HCl), we need to look at the bonds being broken and formed, which were given in the proposed pathway above. One C–H bond in methane (∆H° = 439 kJ mol-1) and one Cl–Cl bond (∆H° = 243 kJ mol-1) are broken. The C–Cl bond of chloromethane (∆H° = 350 kJ mol-1) and the H–Cl bond (∆H° = 431 kJ mol-1) are being formed.

∆H°      = energy input – energy output

               = ∆H° (bonds broken) –∆H° (bonds formed)

               = (439 + 243) – (350 + 431)

               = - 99 kJ mol-1

From the calculation, we see that this is an exothermic reaction. This calculation tells us that the products formed possess stronger bonds than the reactants; therefore, the forward reaction is favored. Remember, thermodynamics only tells us the relative stability of the reactants and products and does not tell us how fast the reaction will proceed.

18.2.2 Bond Strength and Radical Stability

Homolytic bond dissociation energies can be used to estimate bond strength and the relative stability of radicals. What is meant by stability? Stability is an intrinsic property of one molecule with respect to another molecule. For example, a tertiary carbocation is more thermodynamically stable than a methyl cation. However, the tertiary carbocation is still very reactive and is not expected to persist in the reaction media for a long period of time. Thus, stability is a thermodynamic property and not a kinetic property; that is, stability will not tell us the rate at which the radical, once formed, will react.

BDE can be used to predict bond strength. There are some trends seen in the BDEs given in Table 18.1:

a) Decreasing electronegativity decreases bond strength (C–X)

CH3–F > CH3–OH > CH3–NH2
(460, 385, 356 kJ mol-1, respectively)

b) Decreasing electronegativity decreases bond strength (X–H)

O–H > N–H > C–H
(497, 450, 439 kJ mol-1, respectively)

c) Decreasing electronegativity and increase in size decreases bond strength (C–X)

CH3–F > CH3–Cl > CH3–Br > CH3–I
(460, 350, 297, 232 kJ mol-1, respectively)

There are two reasons for trends a–c: electronegativity and bond overlap. The larger the electronegativity difference between two atoms covalently bonded together, the shorter and stronger the bond is (trends a and b). The same electronegativity trend is seen going down the periodic table (trend c). In addition, as you go down the periodic table, the progressively larger halogens lead to size mismatch with the carbon valence orbitals, therefore, weaker bonds are formed.

d) Decreasing s character decreases bond strength (C–H)

C(sp)–H > C(sp2)–H > C(sp3)–H
(557, 464, 439 kJ mol-1, respectively)

Trend d is due to the hybridization of the carbon atom. The more s character in the hybrid orbital, the shorter the bond length and the greater the bond strength.

e) Increasing hyperconjugation decreases bond strength (C–H)

CH3–H > CH3CH2–H > (CH3)2CH–H > (CH3)3C–H
(439, 421, 411, 400 kJ mol-1, respectively)

Trend e shows that the bond strength decreases from methyl > primary > secondary > tertiary. In this case, the trend is not due to the hydrogen being bonded to different atoms with different electronegativity, size, or hybridization. We now must shift our attention to the stability of the radical produced. Since BDE gives us an idea of bond strength or the ease of forming a radical, BDE can also give us an indication of the stability of the radical once formed.

In the homolytic cleavage of the methane, ethane, isopropane, and 2-methyl propane, a carbon radical and a hydrogen radical are formed.

Figure 18.7. Potential energy diagram of the homolytic cleavage of methyl, 1°, 2°, and 3° C–H bonds​

In all of the above cases, a C–H bond is being homolytically cleaved; therefore, the differences in BDEs are related to the differences in the stabilities of the radicals being produced during the homolysis reaction.

Thus, the trend parallels the stability of the radicals produced. Why is a tertiary radical more stable than a methyl radical? Radicals are electron-deficient and, like carbocations, adjacent sp3 orbitals help stabilize the radical through hyperconjugation.

Figure 18.8. Radical stabilization by hyperconjugation

To answer the following question, click here to refer to Table 18.1.

Question 18.2

Q18.2 - Level 2

Using the BDE values in Table 18.1, match the homolytic bond dissociation reactions with the potential energy curves.

question description
Premise
Response
1

A)

A

Curve 2

2

B)

B

Curve 1


f) Resonance increases the stability of radical

(CH3)3C–H > CH2=CHCH3 > C6H5CH3
(400, 369, 376 kJ mol-1, respectively)

Trend f also results from increasing radical stability: bond strength decreases as tertiary > allylic > benzylic.

The greater stability of benzylic and allylic radicals is due to resonance stabilization. This resonance stabilization has an even stronger stabilizing effect than hyperconjugation as can be seen by the smaller values of BDEs.

Figure 18.9. Resonance stabilization of allylic and benzylic radicals


Q18.3 - Level 2

Rank the stability of the following radicals from most to least stable (most stable at the top).

question description
A

2)

B

3)

C

1)


Q18.4 - Level 2

Sort the H-X bonds from strongest to weakest (strongest at the top).

question description
A

2)

B

1)

C

4)

D

3)


Q18.5 - Level 2

Sort the C-N, C-F, and C-O bonds from strongest to weakest (strongest at the top).

question description
A

3)

B

1)

C

2)


18.2.3 Radicals Stabilized by Functional Groups

It is interesting to see that adjacent functional groups have an effect on neighboring C–H BDEs; whether they are electron donating (EDG) or withdrawing groups (EWG), the C–H bond next to the functional group is weakened. The radicals formed after homolytic cleavage are similar to or are more stable than tertiary radicals. A simplified explanation of this stabilization effect is shown with the resonance structures below. Fundamentally, it can be stated that anything that stabilizes an anion or cation will stabilize a radical. For a more detailed explanation, please consult other resources since a more in depth explanation is beyond the scope of this textbook.

Figure 18.10. Bond Dissociation Energies of C–H bonds adjacent to electron donating and withdrawing groups

In summary, electron withdrawing groups and electron donating groups, including alkyl groups and conjugated groups, stabilize the radical that is formed as can be seen in the smaller BDE values. Thus, as seen in BDEs given below, the weakest C–H bond is adjacent to the alcohol functional group.

Figure 18.11. Carbon–hydrogen Bond Dissociation Energies of propanol


Q18.6 - Level 3

Which numbered C-H bond would be the weakest among the following molecules?

question description
A

1)

B

2)

C

3)


18.2.4 Very Stable Radicals

Some radicals are far more stable and persist far longer than would be expected if only electronic factors are considered.

Examples are given below.

Figure 18.12. Examples of stable radicals

The tri-tert-butylphenoxyl radical is very stable and tri-tert-butylphenol (BHT) is a very important antioxidant additive used in plastics, elastomers, petroleum products, and food preservatives. TEMPO, DPPH, and Galinoxyl are all very stable free radicals used in radical trapping (scavengers) for mechanistic studies. They are also used as inhibitors to prevent unwanted competitive side radical reactions. The use of BHT as a food preservative is a potential health risk. A very reputable website for environmental concerns is the David Suzuki Foundation.

There are two reasons for the stability of the given radicals: steric hindrance and electronic stabilization. As seen in the structures above, the radicals are produced adjacent to electron donating groups and electron withdrawing groups and, therefore, can become resonance-stabilized. The radicals are also adjacent to large bulky groups, which sterically shield the radical from reacting with other molecules.

18.2.5 Radical Initiators

Industrial polymerization reactions are very important in the production of plastics, thermoplastics, hydrogels, and countless numbers of consumer products. Initiators are needed to start the polymerization reactions to produce these materials. Most σ bonds can be cleaved above 200°C; however, heating the reaction mixture above this temperature creates unwanted radical species and side products and also wastes energy. Thus, initiators are compounds that have weak bonds with small BDEs and can be homolytically cleaved at lower energies and/or by using ultraviolet (UV) radiation.

In Table 18.1, you can see that some of the weakest bonds occur between O–O peroxy σ bonds.  Di-tert-butyl peroxide and dibenzoyl peroxide are common reagents used for initiating free radical reactions. 

Figure 18.13. Common peroxide initiators

The homolytic bond cleavage of dibenzoyl peroxide occurs at lower temperatures than dialkyl peroxide (di-tert-butyl peroxide) because of the weaker peroxide bond, which is due to resonance stabilization with the carbonyl group. Benzoyl peroxide then further decomposes to produce a phenyl radical and neutral carbon dioxide.

Another important industrial and synthetic radical initiator is azobisisobutyronitrile (AIBN). AIBN can be decomposed readily with either heat or UV light to produce a radical plus nitrogen gas (N2).

18.14.png
Figure 18.14. Azobisisobutyronitrile initiator


18.3 Radical Halogenation of Alkanes

18.3.1 Chlorination of Methane and the Chain Reaction

Chlorination of methane takes places in the presence of light (hv) or heat. Energy in the form of heat or light is needed to homolytically cleave the Cl–Cl bond to generate chlorine radicals (BDE = 243 kJmol-1). The chlorine radical is not very selective for C–H bonds and a mixture of chlorinated methanes is produced as shown below. It should be noted that when irradiation occurs, the concentration of chlorine radicals produced is only a small fraction of the Cl2 present in the reaction mixture. However, only small concentrations of chlorine radicals are needed to produce many chlorine radical reactions.

Figure 18.15. Chlorination of methane

Industrially, these products are separated by fractional distillation and sold separately. Using methane in excess of the chlorine can increase the proportion of CH3Cl. This then leaves unreacted methane at the end of the reaction that can be easily separated and then recycled. To increase the proportion of the more highly chlorinated products, excess chlorine is used in the reaction mixture. The chlorinated products produced from this reaction are used synthetically and industrially as solvents and degreasing agents. However, industrial and scientific researchers try to avoid using these chemicals because of their toxic and carcinogenic properties.

Chain reaction mechanism

The chlorination of methane follows a chain reaction mechanism. In a chain reaction process, the products from one step are the reactants for the next step in the mechanism. A chain reaction mechanism involves three steps:

  • Initiation: generation of reactive radicals
  • Propagation: reactive radical intermediates react with stable molecules to generate new radical intermediates in a self-perpetuating reaction called a chain reaction
  • Termination: any reaction that stops the chain reaction from proceeding.

Initiation: 
The initiation step in the halogenation of methane involves the absorption of energy in the form of light () or heat by chlorine to cause homolytic cleavage of the Cl–Cl bond.

Figure 18.16. Initiation reaction for the chlorination of methane

Propagation: 

The chlorine radical generated from the initiation reaction is a highly reactive intermediate and when it collides with a methane molecule it abstracts a hydrogen atom to form HCl and a methyl radical. The methyl radical then reacts with chlorine to form chloromethane and regenerates a chlorine radical. The chlorine radical could react with another methane molecule or could also react with chloromethane to form HCl and a haloalkyl radical, which in turn will react with chlorine to form dichloromethane and another chlorine radical. This chain reaction process continues until all of the reagents are quenched or some other reaction consumes the radical intermediates.

Figure 18.17. Propagation steps for the chlorination of methane

Termination: 

A few of the many possible termination reactions are given below. A termination reaction is any reaction that does not generate new radical and stops radical propagation. Termination reactions slow down the overall process because they eliminate the number of free radicals in the reaction mixture. The number of termination reactions is actually quite small because radical intermediates are very unstable and short-lived. Thus, the probability of two radicals encountering each other is low. Termination reactions do become important at the end of a reaction process as the concentration of reagents decreases.

Figure 18.18. Termination coupling reactions for the chlorination of methane​

Termination reactions can also occur with the walls of the reaction vessels so the design and materials used for the reaction vessel are very important. Industrially, another very important termination step that can occur is with impurities in the reaction mixture. Impurities can change the kinetics of the radical reaction with explosive consequences. Impurities destroy radicals and can cause an induction period in which the rate of reaction is very slow. Once the inhibiting impurities have been reacted with, a rapid rate of reaction can then occur, resulting in generation of considerable amount of heat and pressure. If the heat is not dissipated or the pressure released, then an explosion can occur.

To answer the following question, click here to refer to Table 18.1.

Question 18.7

Q18.7 - Level 2

Look at the BDEs in Table 18.1. Which of the following is most reactive to the abstraction of a hydrogen by a chlorine radical?

A

Methane

B

Chloromethane (CH3ClCH_3Cl)

C

Dichloromethane (CH2Cl2CH_2Cl_2)

D

Chloroform (CHCl3CHCl_3)

E

Ethane


18.3.2 Selectivity of Halogenation Reactions

The halogenation of alkanes with bromine and chlorine is a very important industrial reaction because alkanes are very unreactive and the halogenation reaction is one of the few reactions that can add a functional group to an alkane.

The mechanism for the halogenation of alkanes is exactly the same as for the chlorination of methane: initiation, propagation, and termination. However, the different radical halogenation reactions are not equal in either rate of reaction or the products formed. Given below are the product ratio results for the mono-halogenation of propane.

Figure 18.19. Halogenation of propane

Propane has two 2° hydrogens and six 1° hydrogens; thus, statistically, if each hydrogen abstracted were equal in energy and equal in the rate of reaction, then the product ratio would be 25:75 for 2° and 1° hydrogens, respectively. This is not what we find. Fluorination initially produces slightly more than statistically predicted 2-fluoropropane and then rapidly reacts further to form multiple halogenated and cleaved products. Reaction conditions used to minimize multiple fluorination (large excess alkane compared to the halogen) are not successful for this extremely exothermic reaction. Chlorination produces slightly more of the 2° alkyl halide, whereas bromination produces significantly more of the 2° alkyl halide and iodide does not react at all.

Before we go into the explanation, intuitively you may already know why we obtain the observed product ratios for these reactions. From acid–base chemistry we know that the stability of a base increases with the size of the anion. Halogen radicals follow the same trend in stability as their anions.

Q18.8 - Level 2

Rank the following halogen radicals from the most reactive to the least reactive.

question description
A

Br•

B

F•

C

I•

D

Cl•


Question 18.9 - Level 1

Which radical is more stable?

question description
A

1)

B

2)


Looking back at the product ratios, we can see that as the reactivity decreases from fluorine to bromine, the ability to abstract a proton that forms a less stable radical decreases. For example, the highly reactive fluorine molecule is almost equally able to abstract a secondary or primary hydrogen to form 2° and 1° fluoroalkanes, whereas the less reactive bromine radical is only able to abstract predominantly secondary hydrogens to produce 2-bromopropane in a 92% yield. Chlorine radical reactivity is in between fluorine and bromine and the iodide radical is so stable and unreactive that it cannot abstract a hydrogen.

Now, let us take a closer look at the thermodynamics. What bond breaking and forming reactions cause the difference in the reactivity and the ratio of products? Carefully look at the BDEs of the two reactions given below.

Figure 18.20. Initiation and propagation steps for the halogenation of propane


Figure 18.21. BDEs and ∆H° values for the propagation steps for the synthesis of 1° alkyl halides


Figure 18.22. BDEs and ∆H° values for the propagation steps for the synthesis of 2° alkyl halides​



Q18.10 - Level 1

Using the BDEs given above, order the H-X bonds from strongest to weakest (strongest at the top).

A

H-I

B

H-Br

C

H-F

D

H-Cl


Q18.11 - Level 1

Using the BDEs given above, order the C-X bonds from strongest to weakest (strongest at the top).

A

C-Cl

B

C-Br

C

C-F

D

C-I


Q18.12 - Level 3

Using the ΔHo react\Delta H^o\ _{react} values given above, match the reaction with the thermodynamic description.

Premise
Response
1

CH3CH2CH3+F2CH_3CH_2CH_3 + F_2

A

Slightly exothermic

2

CH3CH2CH3+Cl2CH_3CH_2CH_3 + Cl_2

B

Exothermic

3

CH3CH2CH3+B2CH_3CH_2CH_3 + B_2

C

Highly exothermic

4

CH3CH2CH3+I2CH_3CH_2CH_3 + I_2

D

Endothermic


Observations from BDE and ∆H° values:

Fluorine versus iodine:

  • The fluorine radical creates very strong F–H and C–F bonds in the two propagation steps. Both propagation steps are exothermic and the overall reaction is highly exothermic.
  • In contrast, I–H and C–I bonds are weak and both propagation steps with iodine are endothermic. The iodine radical is quite stable and will not abstract a hydrogen atom from an alkane.

Abstraction of primary and secondary hydrogens:

  • The abstraction of a hydrogen from a carbon by fluorine or chlorine radicals is an exothermic reaction.
  • The abstraction of a secondary hydrogen is slightly more exothermic than the abstraction of a primary hydrogen for both fluorination and chlorination reactions.
  • There is a very large difference in the bond energies involved in fluorination and chlorination radical reactions. Chlorination is far less exothermic in both propagation steps.
  • Bromine radicals are more stable and form weaker bonds with carbon and hydrogen.
  • The abstraction of a hydrogen by a bromine radical is an endothermic process.
  • The abstraction of a secondary hydrogen by a bromine radical is slightly less endothermic than the abstraction of a primary hydrogen.

The ∆H° react values provide the overall changes in energy for each of these reactions and the calculations show us that in all cases, either exothermically or endothermically, the abstraction of a secondary hydrogen is more favorable. However, these calculations do not tell us how fast the reaction will proceed or give us insight into why we see different product ratios for the different halogenation reactions. For this, we need to consider reaction kinetics.

To explain product ratios we need to look at the transition states. The Hammond postulate was studied comprehensively in the chapters on substitution and elimination reactions. The Hammond postulate states: The transition state for an exothermic reaction looks very much like the starting materials. If a bond is forming in the reaction, then that bond is less than half-formed and if a bond is breaking, then it is less than half-broken in the transition state.

The halogen radical reactivities are directly related to the activation energies (Ea). Thus, we must relate thermodynamics to kinetics. Generally, if product A is more stable than product B, then product A will form at a faster rate than product B and in greater yield. This statement combines kinetics with thermodynamics; the more stable product (thermodynamics–equilibrium) forms at a faster rate (kinetics–energy barrier). 

This statement assumes that the stability of the products is directly related to the energies of the transition state. It assumes that the more stable product has a more stable transition state (smaller Ea) and will form at a faster rate. We are using thermodynamics to predict the rate of a reaction without knowing anything about the magnitude of the activation energy. Fortunately, for radical reactions, there is parallelism between the energies of the transition state and the stability of the products formed.

What do we see experimentally? Fluorination of an alkane is a highly exothermic, extremely rapid reaction. In fact, only a small amount of the fluorine radical is needed to produce thousands of fluorine reactions. This dramatic rapid release of heat can have explosive consequences. Experimentally, chlorination reactions are much slower than fluorination reactions and bromination reactions are considerably slower and require heating. However, this still does not explain product ratios.

The fluorine radical reaction has a negligible activation barrier in the propagation steps. In the transition state, the fluorine atom is relatively far from the hydrogen and possesses most of the radical character. The hydrogen is still very close to the carbon atom and the carbon does not possess much of a radical character.

Chlorination is also an exothermic reaction, but far less so than fluorination. In this case, because the chlorine radical is less reactive, the chlorine atom is much closer to the hydrogen and the carbon possesses a more radical character than in the fluorination reaction.

Abstraction of hydrogen by a bromine radical is endothermic. The transition state for an endothermic reaction looks very much like the product. Any bonds that are forming are more than half-formed, and any bonds that are breaking are more than half-broken. Thus, in the transition state, the carbon has a large amount of radical character.


Q18.13 - Level 3

Order the halogenation reactions from greatest to least amount of radical character on the carbon in the transition state (most radical character on carbon at the top).

question description
A

Bromination

B

Fluorination

C

Chlorination


Now, to finally answer the question about product ratios. Given in Figure 18.23 are the potential energy reaction coordinate diagrams for the abstraction of a hydrogen atom. As this is the slowest step for all of the halogenation reactions, this is also the rate-determining step in the propagation steps. The diagrams are not to scale; however, the diagrams depict the relative differences in energies between the three reactions. Look closely at the differences in Ea among the three reactions and answer the following questions.

Figure 18.23a. Abstraction of a hydrogen by a halogen radical in the halogenation of propane


Figure 18.23b. Abstraction of a hydrogen by a halogen radical in the halogenation of propane


Figure 18.23c. Abstraction of a hydrogen by a halogen radical in the halogenation of propane


Q18.14 - Level 2

Sort the halogenation reactions from greatest to smallest difference in Ea secondaryE_a\ _{secondary } and Ea primaryE_a\ _{primary} (largest difference in EaE_a at the top).

A

Fluorination

B

Chlorination

C

Bromination


Q18.15

Since EaE_a is proportional to the rate of the reaction, which reaction would produce an almost equal amount of primary and secondary radicals due to small differences in activation energies?

A

Bromination

B

Chlorination

C

Fluorination