(BQ) Part 2 book Organic chemistry has contents: Radical reactions, alcohols and ethers synthesis and reactions; alcohols from carbonyl compounds oxidation–reduction and organometallic compounds; conjugated unsaturated systems, aromatic compounds... and other contents.
solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 459 Radical Reactions 10 Unpaired electrons lead to many burning questions about radical types of reactivity In fact, species with unpaired electrons are called radicals, and they are involved in the chemistry of burning, aging, disease, as well as in reactions related to destruction of the ozone layer and the synthesis of products that enhance our everyday lives For example, polyethylene, which can have a molecular weight from the thousands to the millions, and practical uses ranging from plastic films and wraps to water bottles, bulletproof vests, and hip and knee replacements, is made by a reaction involving radicals Oxygen that we breathe and nitric oxide that serves as a chemical signaling agent for some fundamental biological processes are both molecules with unpaired electrons Highly colored natural compounds like those found in blueberries and carrots react with radicals and may protect us from undesirable biological radical reactions Large portions of the economy hinge on radicals, as well, from reactions used to make polymers like polyethylene, to the target action of pharmaceuticals like Cialis, Levitra, and Viagra, which act on a nitric oxide biological signaling pathway Reactions with radicals also play a role in organic synthesis In this chapter we study the properties and reactivity of species with unpaired electrons, and we shall find that they are radically important to chemistry and life 459 Openmirrors.com solom_c10_459-501hr.qxd 28-09-2009 460 15:09 Page 460 Chapter 10 Radical Reactions 10.1 Introduction: How Radicals Form and How They React So far almost all of the reactions whose mechanisms we have studied have been ionic reactions Ionic reactions are those in which covalent bonds break heterolytically and in which ions are involved as reactants, intermediates, or products Another broad category of reactions has mechanisms that involve homolysis of covalent bonds with the production of intermediates possessing unpaired electrons called radicals (or free radicals): Each atom takes one electron from the covalent bond that joined them A B homolysis A ϩ B Radicals Helpful Hint A single-barbed curved arrow shows movement of one electron This simple example illustrates the way we use single-barbed curved arrows to show the movement of a single electron (not of an electron pair as we have done earlier) In this instance, each group, A and B, comes away with one of the electrons of the covalent bond that joined them 10.1A Production of Radicals ᭹ Energy in the form of heat or light must be supplied to cause homolysis of covalent bonds (Section 10.2) For example, compounds with an oxygen–oxygen single bond, called peroxides, undergo homolysis readily when heated, because the oxygen–oxygen bond is weak The products are two radicals, called alkoxyl radicals: R O O R Dialkyl peroxide heat 2R Homolysis of a dialkyl peroxide O Alkoxyl radicals Halogen molecules (X2) also contain a relatively weak bond As we shall soon see, halogens undergo homolysis readily when heated or when irradiated with light of a wavelength that can be absorbed by the halogen molecule: X X homolysis heat or light (hn) X Homolysis of a halogen molecule The products of this homolysis are halogen atoms, and because halogen atoms contain an unpaired electron, they are radicals 10.1B Reactions of Radicals ᭹ Almost all small radicals are short-lived, highly reactive species When radicals collide with other molecules, they tend to react in a way that leads to pairing of their unpaired electron One way they can this is by abstracting an atom from another molecule For example, a halogen atom may abstract a hydrogen atom from an alkane This hydrogen abstraction gives the halogen atom an electron (from the hydrogen atom) to pair with its unpaired electron Notice, however, that the other product of this abstraction is another radical intermediate, in this case, an alkyl radical, R·, which goes on to react further, as we shall see in this chapter solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 461 10.2 Homolytic Bond Dissociation Energies (DH°) 461 A MECHANISM FOR THE REACTION Hydrogen Atom Abstraction General Reaction ϩ X X H ϩ R H R Reactive Alkane radical intermediate Alkyl radical intermediate (reacts further) Specific Example Cl ϩ H CH3 Cl H ϩ CH3 Chlorine Methane atom (a radical) Methyl radical intermediate (reacts further) This behavior is characteristic of radical reactions Consider another example, one that shows another way in which radicals can react: They can combine with a compound containing a multiple bond to produce a new radical, which goes on to react further (We shall study reactions of this type in Section 10.10.) A MECHANISM FOR THE REACTION Radical Addition to a p Bond R R + C C Reactive Alkene alkyl radical intermediate C C Further reaction (Section 10.10) New radical intermediate 10.2 Homolytic Bond Dissociation Energies (DH°) When atoms combine to form molecules, energy is released as covalent bonds form The molecules of the products have lower enthalpy than the separate atoms When hydrogen atoms combine to form hydrogen molecules, for example, the reaction is exothermic; it evolves 436 kJ of heat for every mole of hydrogen that is produced Similarly, when chlorine atoms combine to form chlorine molecules, the reaction evolves 243 kJ molϪ1 of chlorine produced: H ϩ H H H ⌬ H ° ϭ Ϫ 436 kJ molϪ1 Cl ϩ Cl Cl Cl ⌬ H ° ϭ Ϫ 243 kJ molϪ1 Bond formation is an exothermic process Reactions in which only bond breaking occurs are always endothermic The energy required to break the covalent bonds of hydrogen or chlorine homolytically is exactly equal Openmirrors.com solom_c10_459-501hr.qxd 462 28-09-2009 15:09 Page 462 Chapter 10 Radical Reactions to that evolved when the separate atoms combine to form molecules In the bond cleavage reaction, however, ⌬H° is positive: H Cl H H ϩ H ⌬ H ° ϭ ϩ 436 kJ molϪ1 Cl Cl ϩ Cl ⌬ H ° ϭ ϩ 243 kJ molϪ1 Bond breaking is an endothermic process ᭹ Energy must be supplied to break covalent bonds ᭹ The energies required to break covalent bonds homolytically are called homolytic bond dissociation energies, and they are usually abbreviated by the symbol DH° The homolytic bond dissociation energies of hydrogen and chlorine, for example, can be written in the following way: H9H Cl Cl (DH ° ؍436 kJ mol؊1) (DH ° ؍243 kJ mol؊1) The homolytic bond dissociation energies of a variety of covalent bonds have been determined experimentally or calculated from related data Some of these DH° values are listed in Table 10.1 TABLE 10.1 Single-Bond Homolytic Dissociation Energies (DH°) at 25°Ca ACB 9: Aؒ ؉ Bؒ Bond Broken (shown in red) kJ mol؊1 Bond Broken (shown in red) kJ mol؊1 HoH DoD FoF Cl o Cl Br o Br IoI HoF H o Cl H o Br HoI CH3 o H CH3 o F CH3 o Cl CH3 o Br CH3 o I CH3 o OH CH3 o OCH3 CH3CH2 o H CH3CH2 o F CH3CH2 o Cl CH3CH2 o Br CH3CH2 o I CH3CH2 o OH 436 443 159 243 193 151 570 432 366 298 440 461 352 293 240 387 348 421 444 353 295 233 393 CH3CH2 o OCH3 CH3CH2CH2 o H CH3CH2CH2 o F CH3CH2CH2 o Cl CH3CH2CH2 o Br CH3CH2CH2 o I CH3CH2CH2 o OH CH3CH2CH2 o OCH3 (CH3)2CH o H (CH3)2CH o F (CH3)2CH o Cl (CH3)2CH o Br (CH3)2CH o I (CH3)2CH o OH (CH3)2CH o OCH3 (CH3)2CHCH2 o H (CH3)3C o H (CH3)3C o Cl (CH3)3C o Br (CH3)3C o I (CH3)3C o OH (CH3)3C o OCH3 C6H5CH2 o H 352 423 444 354 294 239 395 355 413 439 355 298 222 402 359 422 400 349 292 227 400 348 375 Bond Broken (shown in red) kJ mol؊1 CH2" CHCH2 o H CH2" CH o H C6H5 o H HC# C o H CH3 o CH3 CH3CH2 o CH3 CH3CH2CH2 o CH3 CH3CH2 o CH2CH3 (CH3)2CH o CH3 (CH3)3C o CH3 HO o H HOO o H HO o OH (CH3)3CO o OC(CH3)3 369 465 474 547 378 371 374 343 371 363 499 356 214 157 O O ' ' C6H5COoOCC6H5 139 CH3CH2O o OCH3 CH3CH2O o H O ' CH3CoH 184 431 364 a Data compiled from the National Institute of Standards (NIST) Standard Reference Database Number 69, July 2001 Release, accessed via NIST Chemistry WebBook (http://webbook.nist.gov/chemistry/) Copyright 2000 From CRC Handbook of Chemistry and Physics, Updated 3rd Electronic Edition; Lide, David R., ed Reproduced by permission of Routledge/Taylor & Francis Group, LLC DH° values were obtained directly or calculated from heat of formation (Hf) data using the equation DH°[A B] ϭ Hf [A·] ϩ Hf [B·] Ϫ Hf [A B] 10.2A How to Use Homolytic Bond Dissociation Energies to Calculate Heats of Reaction Bond dissociation energies have, as we shall see, a variety of uses They can be used, for example, to calculate the enthalpy change (⌬H°) for a reaction To make such a calcula- solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 463 463 10.2 Homolytic Bond Dissociation Energies (DH°) tion (see following reaction), we must remember that for bond breaking ⌬H° is positive and for bond formation ⌬H° is negative ⌬H ° ϭ Ϫ(net DH °products) ϩ (net DH °reactants) Negative sign because energy is released in bond formation ⌬H ° ϭ Ϫ͚ DH °products ϩ ͚ DH °reactants or is the mathematical ͚symbol for summation Let us consider, for example, the reaction of hydrogen and chlorine to produce mol of hydrogen chloride From Table 10.1 we get the following values of DH°: HoH ϩ Cl o Cl 9: (DH ° ؍436 kJ mol؊1) (DH ° ؍243 kJ mol؊1) ϩ679 kJ is required to cleave mol of H2 bonds and mol of Cl2 bonds H o Cl (DH ° ؍432 kJ mol؊1) ؋ Ϫ864 kJ is evolved in formation of the bonds in mol of HCl Overall, the reaction of mol of H2 and mol of Cl2 to form mol of HCl is exothermic: Two moles of product formed ⌬H ° ϭ Ϫ2 (432 kJ molϪ1) ϩ (436 kJ molϪ1 ϩ 243 kJ molϪ1) Bond forming (exothermic; negative sign) Bond breaking (endothermic; positive sign) ϭ Ϫ864 kJ molϪ1 ϩ 679 kJ molϪ1 ϭ Ϫ185 kJ molϪ1 Overall ⌬H ° for mol HCl produced from H2 ؉ Cl2 For the purpose of our calculation, we have assumed a particular pathway, which amounts to and then H H 9: H· Cl Cl 9: Cl· H· ϩ Cl· 9: H Cl This is not the way the reaction actually occurs Nevertheless, the heat of reaction, ⌬H°, is a thermodynamic quantity that is dependent only on the initial and final states of the reacting molecules Here, ⌬H° is independent of the path followed (Hess’s law), and, for this reason, our calculation is valid Calculate the heat of reaction, ⌬H°, for the following reactions: (a) H2 ϩ F2 9: HF (b) CH4 ϩ F2 9: CH3F ϩ HF (c) CH4 ϩ Cl2 9: CH3Cl ϩ HCl (d) CH4 ϩ Br2 9: CH3Br ϩ HBr (e) CH4 ϩ I2 9: CH3I ϩ HI Review Problem 10.1 (f) CH3CH3 ϩ Cl2 9: CH3CH2Cl ϩ HCl (g) ϩ Cl2 9: ϩ HCl Cl (h) ϩ Cl2 9: Cl ϩ HCl 10.2B How to Use Homolytic Bond Dissociation Energies to Determine the Relative Stabilities of Radicals Homolytic bond dissociation energies also provide us with a convenient way to estimate the relative stabilities of radicals If we examine the data given in Table 10.1, we find the following values of DH° for the primary and secondary C H bonds of propane: Openmirrors.com H H (DH° = 423 kJ mol–1) (DH° = 413 kJ mol–1) solom_c10_459-501hr.qxd 28-09-2009 464 15:09 Page 464 Chapter 10 Radical Reactions This means that for the reaction in which the designated C H bonds are broken homolytically, the values of ⌬H° are those given here H ϩ Hи ⌬H ° ϭ ϩ423 kJ molϪ1 ϩ Hи ⌬H ° ϭ ϩ413 kJ molϪ1 Propyl radical (a 1° radical) Isopropyl radical (a 2° radical) H These reactions resemble each other in two respects: They both begin with the same alkane (propane), and they both produce an alkyl radical and a hydrogen atom They differ, however, in the amount of energy required and in the type of carbon radical produced These two differences are related to each other ᭹ Alkyl radicals are classified as being 1°, 2°, or 3° based on the carbon atom that has the unpaired electron, the same way that we classify carbocations based on the carbon atom with the positive charge More energy must be supplied to produce a primary alkyl radical (the propyl radical) from propane than is required to produce a secondary carbon radical (the isopropyl radical) from the same compound This must mean that the primary radical has absorbed more energy and thus has greater potential energy Because the relative stability of a chemical species is inversely related to its potential energy, the secondary radical must be the more stable radical (Fig 10.1a) In fact, the secondary isopropyl radical is more stable than the primary propyl radical by 10 kJ molϪ1 We can use the data in Table 10.1 to make a similar comparison of the tert-butyl radical (a 3° radical) and the isobutyl radical (a 1° radical) relative to isobutane: ϩ Hؒ ⌬H° ϭ ϩ400 kJ molϪ1 Hؒ ⌬H° ϭ ϩ422 kJ molϪ1 tert-Butyl radical (a 3° radical) Helpful Hint H Knowing the relative stability of radicals is important for predicting reactions H ϩ Isobutyl radical (a 1° radical) Here we find (Fig 10.1b) that the difference in stability of the two radicals is even larger The tertiary radical is more stable than the primary radical by 22 kJ molϪ1 The kind of pattern that we find in these examples is found with alkyl radicals generally ᭹ Overall, the relative stabilities of radicals are 3° Ͼ 2° Ͼ 1° Ͼ methyl Tertiary (3°) Ͼ Secondary (2°) Ͼ C Ͼ C9Cи C ᭹ Primary (1°) Ͼ C Ͼ C9Cи H Methyl H H Ͼ C9Cи H9 Cи H H The order of stability of alkyl radicals is the same as for carbocations (Section 6.11B) Although alkyl radicals are uncharged, the carbon that bears the odd electron is electron deficient Therefore, alkyl groups attached to this carbon provide a stabilizing effect through hyperconjugation, and the more alkyl groups bonded to it, the more stable the radical is Thus, the reasons for the relative stabilities of radicals and carbocations are similar solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 465 465 10.3 Reactions of Alkanes with Halogens 1Њ radical 1Њ radical 2Њ radical 3Њ radical +H +H +H 22 kJ mol–1 +H Potential energy Potential energy 10 kJ mol–1 ∆H ° = +423 kJ mol–1 ∆H ° = +413 kJ mol–1 Figure 10.1 (a) Comparison of the potential energies of the propyl radical (ϩH·) and the isopropyl radical (ϩH·) relative to propane The isopropyl radical (a 2° radical) is more stable than the 1° radical by 10 kJ molϪ1 (b) Comparison of the potential energies of the tert-butyl radical (ϩH·) and the isobutyl radical (ϩH·) relative to isobutane The 3° radical is more stable than the 1° radical by 22 kJ molϪ1 ∆H° = +400 kJ mol–1 ∆H ° = +422 kJ mol–1 (b) (a) Solved Problem 10.1 Classify each of the following radicals as being 1°, 2°, or 3°, and rank them in order of decreasing stability A B C STRATEGY AND ANSWER We examine the carbon bearing the unpaired electron in each radical to classify the radical as to its type B is a tertiary radical (the carbon bearing the unpaired electron is tertiary) and is, therefore, most stable C is a primary radical and is least stable A, being a secondary radical, falls in between The order of stability is B Ͼ A Ͼ C List the following radicals in order of decreasing stability: Review Problem 10.2 CH3и 10.3 Reactions of Alkanes with Halogens ᭹ Alkanes react with molecular halogens to produce alkyl halides by a substitution reaction called radical halogenation A general reaction showing formation of a monohaloalkane by radical halogenation is shown below It is called radical halogenation because, as we shall see, the mechanism involves species with unpaired electrons called radicals This reaction is not a nucleophilic substitution reaction R H ϩ X2 9: R X ϩ HX Openmirrors.com solom_c10_459-501hr.qxd 28-09-2009 466 15:09 Page 466 Chapter 10 Radical Reactions In these reactions a halogen atom replaces one or more of the hydrogen atoms of the alkane and the corresponding hydrogen halide is formed as a by-product Only fluorine, chlorine, and bromine react this way with alkanes Iodine is essentially unreactive, for a reason that we shall explain later 10.3A Multiple Halogen Substitution One complicating factor of alkane halogenations is that multiple substitutions almost always occur The following example illustrates this phenomenon If we mix an equimolar ratio of methane and chlorine (both substances are gases at room temperature) and then either heat the mixture or irradiate it with light of the appropriate wavelength, a reaction begins to occur vigorously and ultimately produces the following mixture of products H H H C 9H ϩ Cl2 H Methane heat or light Chlorine H 9C 9Cl Cl ϩ Cl ϩ H 9C 9Cl H Cl H 9C 9Cl H ϩ Cl9C 9Cl Cl ϩ H Cl Cl Chloromethane Dichloromethane Trichloromethane Tetrachloromethane Hydrogen chloride (The sum of the number of moles of each chlorinated methane produced equals the number of moles of methane that reacted.) To understand the formation of this mixture, we need to consider how the concentration of reactants and products changes as the reaction proceeds At the outset, the only compounds that are present in the mixture are chlorine and methane, and the only reaction that can take place is one that produces chloromethane and hydrogen chloride: H H 9C9H H ϩ Cl2 H H 9C9C l ϩ H9 Cl H As the reaction progresses, however, the concentration of chloromethane in the mixture increases, and a second substitution reaction begins to occur Chloromethane reacts with chlorine to produce dichloromethane: H H 9C9C l H Cl ϩ Cl2 H9C 9C l ϩ H9 Cl H The dichloromethane produced can then react to form trichloromethane, and trichloromethane, as it accumulates in the mixture, can react with chlorine to produce tetrachloromethane Each time a substitution of Cl for H takes place, a molecule of H Cl is produced Solved Problem 10.2 If the goal of a synthesis is to prepare chloromethane (CH3Cl), its formation can be maximized and the formation of CH2Cl2, CHCl3, and CCl4 minimized by using a large excess of methane in the reaction mixture Explain why this is possible ANSWER The use of a large excess of methane maximizes the probability that chlorine will attack methane mol- ecules because the concentration of methane in the mixture will always be relatively large It also minimizes the probability that chlorine will attack molecules of CH3Cl, CH2Cl2, and CHCl3, because their concentrations will always be relatively small After the reaction is over, the unreacted excess methane can be recovered and recycled solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 467 467 10.4 Chlorination of Methane: Mechanism of Reaction 10.3B Lack of Chlorine Selectivity Chlorination of most higher alkanes gives a mixture of isomeric monochloro products as well as more highly halogenated compounds ᭹ Chlorine is relatively unselective; it does not discriminate greatly among the different types of hydrogen atoms (primary, secondary, and tertiary) in an alkane An example is the light-promoted chlorination of isobutane: Cl2 Cl light ϩ ϩ Cl Isobutane ᭹ ᭹ Isobutyl chloride (48%) Polychlorinated products (23%) ϩ HCl tert-Butyl chloride (29%) Because alkane chlorinations usually yield a complex mixture of products, they are not useful as synthetic methods when the goal is preparation of a specific alkyl chloride Helpful Hint Chlorination is unselective An exception is the halogenation of an alkane (or cycloalkane) whose hydrogen atoms are all equivalent [Equivalent hydrogen atoms are defined as those which on replacement by some other group (e.g., chlorine) yield the same compound.] Neopentane, for example, can form only one monohalogenation product, and the use of a large excess of neopentane minimizes polychlorination: ϩ Cl2 heat or light Neopentane (excess) ᭹ Cl ϩ HCl Neopentyl chloride Bromine is generally less reactive toward alkanes than chlorine, and bromine is more selective in the site of attack when it does react We shall examine the selectivity of bromination further in Section 10.6A 10.4 Chlorination of Methane: Mechanism of Reaction The reaction of methane with chlorine (in the gas phase) provides a good example for studying the mechanism of radical halogenation CH4 ϩ Cl2 9: CH3Cl ϩ HCl (ϩ CH2Cl2, CHCl3, and CCl4) Several experimental observations help in understanding the mechanism of this reaction: The reaction is promoted by heat or light At room temperature methane and chlorine not react at a perceptible rate as long as the mixture is kept away from light Methane and chlorine react, however, at room temperature if the gaseous reaction mixture is irradiated with UV light at a wavelength absorbed by Cl2, and they react in the dark if the gaseous mixture is heated to temperatures greater than 100°C The light-promoted reaction is highly efficient A relatively small number of light photons permits the formation of relatively large amounts of chlorinated product A mechanism that is consistent with these observations has several steps, shown below The first step involves the dissociation of a chlorine molecule, by heat or light, into two chlorine atoms The second step involves hydrogen abstraction by a chlorine atom Openmirrors.com solom_c10_459-501hr.qxd 28-09-2009 468 15:09 Page 468 Chapter 10 Radical Reactions A MECHANISM FOR THE REACTION Radical Chlorination of Methane REACTION heat CH4 ϩ Cl2 : CH3Cl ϩ HCl or light MECHANISM Cl Chain Initiation Step 1: Halogen dissociation heat or light Cl Under the influence of heat or light a molecule of chlorine dissociates; each atom takes one of the bonding electrons Cl ϩ Cl This step produces two highly reactive chlorine atoms H Chain Propagation S Cl ϩ H C Step 2: Hydrogen abstraction H Cl H ϩ H This step produces a molecule of hydrogen chloride and a methyl radical H H H ϩ C Cl Cl H H C Cl Cl This step produces a molecule of methyl chloride and a chlorine atom The chlorine atom can now cause a repetition of step ϩ C H Helpful Hint Remember: These conventions are used in illustrating reaction mechanisms in this text Curved arrows or always show the direction of movement of electrons Single-barbed arrows show the attack (or movement) of an unpaired electron Double-barbed arrows show the attack (or movement) of an electron pair H H H ϩ H A methyl radical abstracts a chlorine atom from a chlorine molecule Chain Termination H H H A chlorine atom abstracts a hydrogen atom from a methane molecule Step 3: Halogen abstraction C Cl H C Cl H Coupling of any two radicals depletes the supply of reactive intermediates and terminates the chain Several pairings are possible for radical coupling termination steps (see text) In step the highly reactive methyl radical reacts with a chlorine molecule by abstracting a chlorine atom This results in the formation of a molecule of chloromethane (one of the ultimate products of the reaction) and a chlorine atom The latter product is particularly significant, for the chlorine atom formed in step can attack another methane molecule and cause solom_STG_G01-G18hr1.qxd 14-10-2009 G-7 16:45 Page 1006 Special Topic G Carbon–Carbon Bond–Forming and Other Reactions TBS O O O O N OTBS Grubbs 1999 second generation catalyst 75% S R R = CH2OMOM TBS O O O O O OH O O N N OTBS S OH S O R R = CH2OMOM [mixture of (Z) and (E, Z) dienes] Epothilone B Another example is ring-opening olefin metathesis polymerization (ROMP), as can be used for synthesis of polybutadiene from 1,5-cyclooctadiene N Cl Review Problem G.9 H Ru Cl n N PCy3 Ph ROMP n What products would form when each of the following compounds is treated with (PCy3)2Cl2Ru " CHPh, one of Grubbs’ catalysts? (a) O (b) OTBDMS O C6H5 H N N O (c) O O (d) O O O N C6H5 Openmirrors.com OH solom_STG_G01-G18hr1.qxd 14-10-2009 16:45 Page 1007 G.4 Some Background on Transition Metal Elements and Complexes G-8 G.3 The Corey–Posner, Whitesides–House Reaction: Use of Lithium Dialkyl Cuprates (Gilman Reagents) in Coupling Reactions The Corey–Posner, Whitesides–House reaction involves the coupling of a lithium dialkylcuprate (called a Gilman reagent) with an alkyl, alkenyl, or aryl halide The alkyl group of the lithium dialkylcuprate reagent may be primary, secondary, or tertiary However, the halide with which the Gilman reagent couples must be a primary or cyclic secondary alkyl halide if it is not alkenyl or aryl General Reaction R2CuLi ϩ A lithium dialkyl cuprate (a Gilman reagent) RЈ9 X 9999999: R9 R ϩ RCu ϩ LiX Alkenyl, aryl, or 1° or cyclic 2° alkyl halide Specific Example I CH3 (CH3)2CuLi ϩ ϩ CH3Cu Lithium dimethylcuprate ϩ LiI 75% The required lithium dimethylcuprate (Gilman) reagent must be synthesized by a two-step process from the corresponding alkyl halide, as follows Synthesis of an organolithium compound Synthesis of the lithium dialkylcuprate (Gilman) reagent Li R9X 999999: R9 Li 999999: Cul R9 Li R2CuLi ϩ ϩ LiX Lil All of the reagents in a Corey–Posner, Whitesides–House reaction are consumed stoichiometrically The mechanism does not involve a catalyst, as in the other reactions of transition metals that we have studied Show how 1-bromobutane could be converted to the Gilman reagent lithium dibutylcuprate, and how you could use it to synthesize each of the following compounds (a) Review Problem G.10 (b) G.4 Some Background on Transition Metal Elements and Complexes Now that we have seen examples of some important reactions involving transition metals, we consider aspects of the electronic structure of the metals and their complexes Transition metals are defined as those elements that have partly filled d (or f ) shells, either in the elemental state or in their important compounds The transition metals that are of most concern to organic chemists are those shown in the green and yellow portion of the periodic table given in Fig G.1, which include those whose reactions we have just discussed Transition metals react with a variety of molecules or groups, called ligands, to form transition metal complexes In forming a complex, the ligands donate electrons to vacant solom_STG_G01-G18hr1.qxd 14-10-2009 G-9 16:45 Page 1008 Special Topic G Carbon–Carbon Bond–Forming and Other Reactions 1/IA 1 Periods H 1.00797 2/IIA Li Be 6.941 9.01218 11 12 Na Mg 22.98977 24.305 19 20 3/IIIB 4/IVB 5/VB 6/VIB 7/VIIB 21 22 23 24 25 K Ca Sc Ti V 8/VIIIB 9/VIIIB 10/VIIIB 11/IB 26 27 28 29 12/IIB 30 Cr Mn Fe Co Ni Cu Zn 39.098 40.08 44.9559 47.90 50.9414 51.996 54.9380 55.847 58.9332 58.71 63.546 65.38 37 38 39 40 41 42 43 44 45 46 47 48 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd 85.4678 87.62 89.9059 91.22 92.9064 95.94 98.9062 55 56 57 72 73 74 75 101.07 102.9055 76 186.2 190.2 192.22 Valence electrons 107.868 112.40 78 79 80 77 Cs Ba La Hf Ta W Re Os Ir 132.9054 137.34 138.9055 178.49 180.9479 183.85 106.4 Pt Au Hg 195.09 196.9665 200.59 10 11 12 Figure G.1 Important transition elements are shown in the green and yellow portion of the periodic table Given across the bottom is the total number of valence electrons (s and d) of each element orbitals of the metal The bonds between the ligand and the metal range from very weak to very strong The bonds are covalent but often have considerable polar character Transition metal complexes can assume a variety of geometries depending on the metal and on the number of ligands around it Rhodium, for example, can form complexes with four ligands in a configuration called square planar On the other hand, rhodium can form complexes with five or six ligands that are trigonal bipyramidal or octahedral These typical shapes are shown below, with the letter L used to indicate a ligand L L L L L Rh L L Rh Square planar rhodium complex L L L Trigonal bipyramidal rhodium complex L Rh L L L L Octahedral rhodium complex G.5 Electron Counting in Metal Complexes Transition metals are like the elements that we have studied earlier in that they are most stable when they have the electronic configuration of a noble gas In addition to s and p orbitals, transition metals have five d orbitals (which can hold a total of 10 electrons) Therefore, the noble gas configuration for a transition metal is 18 electrons, not as with carbon, nitrogen, oxygen, and so on When the metal of a transition metal complex has 18 valence electrons, it is said to be coordinatively saturated.* *We not usually show the unshared electron pairs of a metal complex in our structures, because to so would make the structure unnecessarily complicated Openmirrors.com solom_STG_G01-G18hr2.qxd 15-10-2009 12:55 Page 1009 G.5 Electron Counting in Metal Complexes To determine the valence electron count of a transition metal in a complex, we take the total number of valence electrons of the metal in the elemental state (see Fig G.1) and subtract from this number the oxidation state of the metal in the complex This gives us what is called the d electron count, d n The oxidation state of the metal is the charge that would be left on the metal if all the ligands (Table G.1) were removed oxidation state of d n ϭ total number of valence electrons Ϫ the metal in the complex of the elemental metal Then to get the total valence electron count of the metal in the complex, we add to d n the number of electrons donated by all of the ligands Table G.1 gives the number of electrons donated by several of the most common ligands total number of valence electrons ϭ d n ϩ electrons donated by ligands of the metal in the complex Let us now work out the valence electron count of two examples TABLE G.1 Common Ligands in Transition Metal Complexesa Ligand Number of Electrons Donated Count as Negatively charged ligands Hydride, H Alkanide, R Halide, X Allyl anion H:Ϫ R:Ϫ X:Ϫ Ϫ Cyclopentadienyl anion, Cp Ϫ Electrically neutral ligands Carbonyl (carbon monoxide) :C # O: R3P: or Ph3P: Phosphine Alkene C 2 2 C Diene Benzene a Used with permission from the Journal of Chemical Education, Vol 57, No 1, 1980, pp 170-175, copyright © 1980, Division of Chemical Education Example A Consider iron pentacarbonyl, Fe(CO)5, a toxic liquid that forms when finely divided iron reacts with carbon monoxide OC Fe ϩ CO Fe(CO)5 CO Fe or OC CO CO Iron pentacarbonyl From Fig G.1 we find that an iron atom in the elemental state has valence electrons We arrive at the oxidation state of iron in iron pentacarbonyl by noting that the charge on the complex as a whole is zero (it is not an ion), and that the charge on each CO ligand is also zero Therefore, the iron is in the zero oxidation state G-10 solom_STG_G01-G18hr1.qxd G-11 14-10-2009 16:45 Page 1010 Special Topic G Carbon–Carbon Bond–Forming and Other Reactions Using these numbers, we can now calculate d n and, from it, the total number of valence electrons of the iron in the complex dn ϭ Ϫ ϭ total number of ϭ d n ϩ 5(CO) ϭ ϩ 5(2) ϭ 18 valence electrons We find that the iron of Fe(CO)5 has 18 valence electrons and is, therefore, coordinatively saturated Example B Consider the rhodium complex Rh[(C6H5)3P]3H2Cl, a complex that, as we shall see later, is an intermediate in certain alkene hydrogenations L H L Rh L Cl L ؍Ph3P [i.e., (C6H5)3P] H The oxidation state of rhodium in the complex is ϩ3 [The two hydrogen atoms and the chlorine are each counted as Ϫ1 (hydride and chloride, respectively), and the charge on each of the triphenylphosphine ligands is zero Removing all the ligands would leave a Rh3ϩ ion.] From Fig G.1 we find that, in the elemental state, rhodium has valence electrons We can now calculate d n for the rhodium of the complex dn ϭ Ϫ ϭ Each of the six ligands of the complex donates two electrons to the rhodium in the complex, and, therefore, the total number of valence electrons of the rhodium is 18 The rhodium of Rh[(C6H5)3P]3H2Cl is coordinatively saturated total number of valence ϭ dn ϩ 6(2) ϭ ϩ 12 ϭ 18 electrons rhodium G.6 Mechanistic Steps in the Reactions of Some Transition Metal Complexes Much of the chemistry of organic transition metal compounds becomes more understandable if we are able to follow the mechanisms of the reactions that occur These mechanisms, in most cases, amount to nothing more than a sequence of reactions, each of which represents a fundamental reaction type that is characteristic of a transition metal complex Let us examine three of the fundamental reaction types now In each instance we shall use steps that occur when an alkene is hydrogenated using a catalyst called Wilkinson’s catalyst In Section G.7 we shall examine the entire hydrogenation mechanism In Section G.8 we shall see how similar types of steps are involved in the Heck–Mizokori reaction Ligand Dissociation–Association (Ligand Exchange) A transition metal complex can lose a ligand (by dissociation) and combine with another ligand (by association) In the process it undergoes ligand exchange For example, the rhodium complex that we encountered in Example B above can react with an alkene (in this example, with ethene) as follows: L H H2 C L Cl L L ϩ H2 C Rh H C H2 CH2 H L H L ؍Ph3P [i.e., (C6H5)3P] Openmirrors.com ϩ Rh Cl L solom_STG_G01-G18hr1.qxd 14-10-2009 16:45 Page 1011 G.6 Mechanistic Steps in the Reactions of Some Transition Metal Complexes Two steps are actually involved In the first step, one of the triphenylphosphine ligands dissociates This leads to a complex in which the rhodium has only 16 electrons and is, therefore, coordinatively unsaturated H L L L H L H Rh L H ϩL Rh Cl (18 electrons) Cl (16 electrons) L ؍Ph3P In the second step, the rhodium associates with the alkene to become coordinatively saturated again H2 C H L L H ϩ H2 C Rh L C H2 H Rh CH2 L Cl (16 electrons) H Cl (18 electrons) The complex between the rhodium and the alkene is called a p complex In it, two electrons are donated by the alkene to the rhodium Alkenes are often called p donors to distinguish them from s donors such as Ph3P:, ClϪ, and so on In a p complex such as the one just given, there is also a donation of electrons from a populated d orbital of the metal back to the vacant p* orbital of the alkene This kind of donation is called “back-bonding.” Insertion–Deinsertion An unsaturated ligand such as an alkene can undergo insertion into a bond between the metal of a complex and a hydrogen or a carbon These reactions are reversible, and the reverse reaction is called deinsertion The following is an example of insertion–deinsertion H2 C C H2 L H Rh L H Cl insertion deinsertion (18 electrons) Cl L C H3 Rh L CH H (16 electrons) In this process, a p bond (between the rhodium and the alkene) and a s bond (between the rhodium and the hydrogen) are exchanged for two new s bonds (between rhodium and carbon, and between carbon and hydrogen) The valence electron count of the rhodium decreases from 18 to 16 This insertion–deinsertion occurs in a stereospecific way, as a syn addition of the M H unit to the alkene C C M C H C M H Oxidative Addition–Reductive Elimination Coordinatively unsaturated metal complexes can undergo oxidative addition of a variety of substrates in the following way.* A M ϩ A B oxidative addition M B *Coordinatively saturated complexes also undergo oxidative addition G-12 solom_STG_G01-G18hr1.qxd G-13 14-10-2009 16:45 Page 1012 Special Topic G Carbon–Carbon Bond–Forming and Other Reactions The substrate, A B, can be H H, H X, R X, RCO H, RCO X, and a number of other compounds In this type of oxidative addition, the metal of the complex undergoes an increase in the number of its valence electrons and in its oxidation state Consider, as an example, the oxidative addition of hydrogen to the rhodium complex that follows (L ϭ Ph3P) H L L ϩ Rh H oxidative addition H reductive elimination Cl L L L Rh L Cl (16 electrons) Rh is in ؉1 oxidation state H (18 electrons) Rh is in ؉3 oxidation state Reductive elimination is the reverse of oxidative addition With this background, we are now in a position to examine the mechanisms of two applications of transition metal complexes in organic synthesis G.7 The Mechanism for a Homogeneous Hydrogenation: Wilkinson’s Catalyst The catalytic hydrogenations that we have examined in prior chapters have been heterogeneous processes Two phases were involved: the solid phase of the catalyst (Pt, Pd, Ni, etc.), containing the adsorbed hydrogen, and the liquid phase of the solution, containing the unsaturated compound In homogeneous hydrogenation using a transition metal complex such as Rh[(C6H5)3P]3Cl (Wilkinson’s catalyst), hydrogenation takes place in a single phase, i.e., in solution When Wilkinson’s catalyst is used to carry out the hydrogenation of an alkene, the following steps take place (L ϭ Ph3P) Step L L L ϩ H Rh L Rh H Cl L H L 16 valence electrons Cl Oxidative addition H 18 valence electrons Step H H L L L Rh Rh L Cl H 18 valence electrons Openmirrors.com H ϩ L L Cl 16 valence electrons Ligand dissociation solom_STG_G01-G18hr1.qxd 14-10-2009 16:45 Page 1013 G.7 The Mechanism for a Homogeneous Hydrogenation: Wilkinson’s Catalyst G-14 Step H2 C H L L H CH2 ∆ H ϩ H2C Rh L CH2 L Cl 16 valence electrons Ligand association Rh Cl H 18 valence electrons Step H2C C H2 L L H Rh L ∆ C H3 Rh L Cl H Cl Insertion CH H 18 valence electrons 16 valence electrons Step Cl CH3 L Rh L ∆ CH2 Cl ϩ H3C Rh Reductive elimination CH L L H 16 valence electrons 14 valence electrons Step H L L Rh Cl ϩ H2 L ∆ Rh Oxidative addition H L Cl 14 valence electrons 16 valence electrons (Cycle repeats from step 3.) Step regenerates the hydrogen-bearing rhodium complex and reaction with another molecule of the alkene begins at step Because the insertion step and the reductive elimination step are stereospecific, the net result of the hydrogenation using Wilkinson’s catalyst is a syn addition of hydrogen to the alkene The following example (with D2 in place of H2) illustrates this aspect H H H ϩ D2 EtO2C CO2Et A cis-alkene (diethyl maleate) Rh(Ph3P)3Cl H CO2Et EtO2C D D A meso compound What product (or products) would be formed if the trans-alkene corresponding to the cisalkene (see the previous reaction) had been hydrogenated with D2 and Wilkinson’s catalyst? Review Problem G.11 solom_STG_G01-G18hr1.qxd 14-10-2009 G-15 16:45 Page 1014 Special Topic G Carbon–Carbon Bond–Forming and Other Reactions THE CHEMISTRY OF Homogeneous Asymmetric Catalytic Hydrogenation: Examples Involving L-DOPA, (S)-Naproxen, and Aspartame University) (The other half of the 2001 prize was awarded to K B Sharpless for asymmetric oxidation reactions See Chapter 8.) Knowles, Noyori, and others developed chiral catalysts for homogeneous hydrogenation that have proved extraordinarily useful for enantioselective syntheses ranging from small laboratory-scale reactions to industrial- (ton-) scale reactions An important example is the method developed by Knowles and co-workers at Monsanto Corporation for synthesis of L-DOPA, a compound used in the treatment of Parkinson’s disease: Development by Geoffrey Wilkinson of a soluble catalyst for hydrogenation [tris(triphenylphosphine)rhodium chloride, Section 7.13 and Special Topic G] led to Wilkinson’s earning a share of the 1973 Nobel Prize in Chemistry His initial discovery, while at Imperial College, University of London, inspired many other researchers to create novel catalysts based on the Wilkinson catalyst Some of these researchers were themselves recognized by the 2001 Nobel Prize in Chemistry, 50% of which was awarded to William S Knowles (Monsanto Corporation, retired) and Ryoji Noyori (Nagoya Asymmetric Synthesis of L-DOPA H3CO AcO COOH H3CO H2 (100%) NHAc COOH [(Rh(R,R)-DIPAMP)COD]ϩBF4Ϫ(cat.) H3Oϩ HO COOH H NHAc AcO H NH2 HO (100% yield, 95% ee [enantiomeric excess]) O ‘ Ac=CH3C ¬ L-DOPA OCH3 P P H3CO COD ؍ 1,5-Cyclooctadiene (R,R)-DIPAMP (Chiral ligand for rhodium) Another example is synthesis of the over-the-counter analgesic naproxen using a BINAP rhodium catalyst developed by Noyori (Sections 5.11 and 5.18) Asymmetric Synthesis of (S)-Naproxen CH2 H COOH (S)-BINAP-Ru(OCOCH3)2 (0.5 mol%) ϩ H2 CH3 COOH MeOH H3CO H3CO (S)-Naproxen (an anti-inflammatory agent) (92% yield, 97% ee) P(Ph)2 (Ph)2P P(Ph)2 (Ph)2P (S)-BINAP (R)-BINAP (S)-BINAP and (R)-BINAP are chiral atropisomers (see Section 5.18) Openmirrors.com solom_STG_G01-G18hr1.qxd 14-10-2009 16:45 Page 1015 G-16 G.8 The Mechanism for an Example of Cross-Coupling: The Heck–Mizokori Reaction Catalysts like these are important for asymmetric chemical synthesis of amino acids (Section 24.3D), as well A final example is the synthesis of (S)-phenylalanine methyl ester, a compound used in the synthesis of the artificial sweetener aspartame This preparation employs yet a different chiral ligand for the rhodium catalyst Asymmetric Synthesis of Aspartame COOH (1) (R,R)-PNNP-Rh(I) (cat.), H2 (83% ee) (catalytic asymmetric hydrogenation) (2) MeOH, HA NHAc Ph H3C COOCH3 H NH2 (S)-phenylalanine methyl ester (97% ee after recrystallization) Ph N N (Ph)2P P(Ph)2 CH3 (R,R)-PNNP (Chiral ligand for rhodium) HOOC COOH H2N H (S)-aspartic acid H NH2 NH COOH H O COOCH3 Aspartame The mechanism of homogeneous catalytic hydrogenation involves reactions characteristic of transition metal organometallic compounds A general scheme for hydro- genation using Wilkinson’s catalyst is shown here We have seen structural details of the mechanism in Section G.7 Cl[(C6H5)3P]2Rh Cl[(C6H5)3P]3Rh Ϫ(C6H5)3P Cl[(C6H5)3P]2Rh(H)2 Cl[(C6H5)3P]2Rh H A general mechanism for the Wilkinson catalytic hydrogenation method, adapted with permission of John Wiley & Sons, Inc from Noyori, Asymmetric Catalysis in Organic Synthesis, p 17 Copyright 1994 H2 H Cl[(C6H5)3P]2RhH H G.8 The Mechanism for an Example of Cross-Coupling: The Heck–Mizokori Reaction Having seen steps such as oxidative addition, insertion, and reductive elimination in the context of transition metal–catalyzed hydrogenation using Wilkinson’s catalyst, we can now see how these same types of mechanistic steps are involved in a mechanism proposed for the Heck–Mizokori reaction Aspects of the Heck–Mizokori mechanism are similar to steps proposed for other cross-coupling reactions as well, although there are variations and certain steps that are specific to each, and not all of the steps below are involved or serve the same purpose in other cross-coupling reactions solom_STG_G01-G18hr1.qxd 14-10-2009 G-17 16:45 Page 1016 Special Topic G Carbon–Carbon Bond–Forming and Other Reactions A MECHANISM FOR THE REACTION The Heck–Mizokori Reaction Using an Aryl Halide Substrate GENERAL REACTION Ar ¬ X + R Pd catalyst Ar Base (an amine) R MECHANISM Pd(L)4 –2L (L = ligand, e.g., Ph3P) Ar¬ X base¬HX Pd(L)2 Reductive elimination (regenerates catalyst) Oxidative addition (incorporates halide reactant) Coordinatively unsaturated catalyst base H¬ Pd(L)2¬ X Ar¬ Pd(L)2¬ X Ar R Alkene insertion (incorporates alkenyl reactant, forms new C¬C bond) R 1,2-syn elimination (forms the product as a trans alkene) H Ar H Pd(L)2X H R C¬C bond rotation Ar H H Pd(L)2X H R G.9 Vitamin B12: A Transition Metal Biomolecule The discovery (in 1926) that pernicious anemia can be overcome by the ingestion of large amounts of liver led ultimately to the isolation (in 1948) of the curative factor, called vitamin B12 The complete three-dimensional structure of vitamin B12 [Fig G.2(a)] was elucidated in 1956 through the X-ray studies of Dorothy Hodgkin (Nobel Prize, 1964), and in 1972 the synthesis of this complicated molecule was announced by R B Woodward (Harvard University) and A Eschenmoser (Swiss Federal Institute of Technology) The synthesis took 11 years and involved more than 90 separate reactions One hundred coworkers took part in the project Vitamin B12 is the only known biomolecule that possesses a carbon–metal bond In the stable commercial form of the vitamin, a cyano group is bonded to the cobalt, and the cobalt is in the ϩ3 oxidation state The core of the vitamin B12 molecule is a corrin ring [Fig G.2(b)] with various attached side groups The corrin ring consists of four pyrrole Openmirrors.com solom_STG_G01-G18hr1.qxd 14-10-2009 16:45 Page 1017 G-18 G.9 Vitamin B12: A Transition Metal Biomolecule subunits, the nitrogen of each of which is coordinated to the central cobalt The sixth ligand [(below the corrin ring in Fig G.2(a)] is a nitrogen of a heterocyclic group derived from 5,6-dimethylbenzimidazole The cobalt of vitamin B12 can be reduced to a ϩ2 or a ϩ1 oxidation state When the cobalt is in the ϩ1 oxidation state, vitamin B12 (called B12s) becomes one of the most powerful nucleophiles known, being more nucleophilic than methanol by a factor of 1014 Acting as a nucleophile, vitamin B12s reacts with adenosine triphosphate (Fig 22.2) to yield the biologically active form of the vitamin [Fig G.2(c)] A carbon–cobalt s bond O H2N R O CH3 H2N O CH3 CH3 N ϩ N Co H2N N CH3 O P CH3 O O HO N NH2 O CH3 O H3C N OϪ O O CH3 N H3C HN NH2 CH3 N N NH2 H OH OH H H H H CH3 N N CH2 O Adenine Co O HOCH2 (a) (b) (c) Figure G.2 (a) The structure of vitamin B12 In the commercial form of the vitamin (cyanocobalamin), R " CN (b) The corrin ring system (c) In the biologically active form of the vitamin (5Ј-deoxyadenosylcobalamin), the 5Ј carbon atom of 5Ј-deoxyadenosine is coordinated to the cobalt atom For the structure of adenine, see Section 25.2 See Special Topic H in WileyPLUS Openmirrors.com 3.0 OϭC–H 3.5 m, broad O–H associated 2000 4.0 CϵN –––m(1–2) –––w–m CϵC w–m ␦Ar–H 2400 (Microns) CϭO s ␦N–H w–overtones CϭC 1800 m 1400 s sk s ␦XC–H2 m–s m C–N ␦O–H m ␦O–H w–m s 1200 ␦–C–H m–s(2) m–s ␦ϭC–H 1600 Typical IR absorption frequencies for common functional groups Absorptions are as follows: = stretching; ␦ = bending; w = weak; m = medium; s = strong; sk = skeletal From Multiscale Organic Chemistry: A Problem-Solving Approach by John W Lehman © 2002 Reprinted by permission of Pearson Education, Inc., Upper Saddle River, NJ 2.5 s w–m O–H m N–H O–H free 2800 –C–H s ϭC–H m–s ϵC–H m Ar–H ––w–m 3200 10 s 600 16 m–s C–Br m–s m–s s ␦ϵC–H m–s 800 11 12 13 14 C–Cl ␦N–H C–O ␦Ar–H m sk ␦ϭC–H 1000 15:13 Alkane Alkene Alkyne Aromatic 1Њ alcohol 2Њ alcohol 3Њ alcohol Phenol Ether Ester Carboxylic acid Ketone Aldehyde Amide 1Њ amine 2Њ amine 3Њ amine Alkyl chloride Alkyl bromide Nitrile 3600 Frequency (cm–1) 2-10-2009 4000 See Table 2.7 for a Table of IR frequencies solom_ep_B01-B03v1.qxd Page SOLOMONS solom_ep_B01-B03v1.qxd 2-10-2009 15:13 Page SOLOMONS 13C NMR Approximate Chemical Shift Ranges C Cl, Br C N C OR C OH C O C N C N O CH C CH2 C OR 220 O O C R,H C OH 200 180 C C 160 140 C C 120 100 ␦C (ppm) 80 60 40 Type of Carbon Atom Chemical Shift (dd, ppm) 1° Alkyl, RCH3 2° Alkyl, RCH2R 3° Alkyl, RCHR2 0–40 10–50 15–50 Alkyl halide or amine, C Alcohol or ether, O Alkyne, C X qX=Cl, Br, or N r 10–65 50–90 60–90 C Alkene, C 100–170 C Aryl, C 100–170 N 120–130 N 150–180 O Amide, 20 Approximate Carbon-13 Chemical Shifts TABLE 9.2 Nitrile, CH3 C O Carboxylic acid or ester, C O 160–185 O Aldehyde or ketone, C 182–215 solom_ep_B01-B03v1.qxd 2-10-2009 15:13 Page SOLOMONS H NMR Approximate Chemical Shift Ranges C OH, NHn C C H O OH C CH O C C C OH Ar CH H O H X,O,N CH C H 12 TABLE 9.1 11 10 dH (ppm) 3Њ,2Њ,1Њ C C CH CH Approximate Proton Chemical Shifts Type of Proton Chemical Shift (dd, ppm) Type of Proton Chemical Shift (dd, ppm) 1° Alkyl, RCH3 2° Alkyl, RCH2R 3° Alkyl, R3CH Allylic, R2C “ C ¬ CH3 0.8–1.2 1.2–1.5 1.4–1.8 1.6–1.9 Alkyl bromide, RCH2Br Alkyl chloride, RCH2Cl Vinylic, R2C " CH2 Vinylic, R2C “ CH 3.4–3.6 3.6–3.8 4.6–5.0 5.2–5.7 ƒ ƒ R Ketone, RCCH3 R 2.1–2.6 ‘ 2.2–2.5 2.5–3.1 3.1–3.3 3.3–3.9 3.3–4.0 O Alcohol hydroxyl, ROH Amino, R NH2 Phenolic, ArOH Carboxylic, RCOH ‘ O a The chemical shifts of these protons vary in different solvents and with temperature and concentration Openmirrors.com 6.0–8.5 9.5–10.5 ‘ O Benzylic, ArCH3 Acetylenic, RC # CH Alkyl iodide, RCH2I Ether, ROCH2R Alcohol, HOCH2R Aromatic, ArH Aldehyde, RCH 0.5–6.0a 1.0–5.0a 4.5–7.7a 10–13a ... CH3CH2CH2 o F CH3CH2CH2 o Cl CH3CH2CH2 o Br CH3CH2CH2 o I CH3CH2CH2 o OH CH3CH2CH2 o OCH3 (CH3)2CH o H (CH3)2CH o F (CH3)2CH o Cl (CH3)2CH o Br (CH3)2CH o I (CH3)2CH o OH (CH3)2CH o OCH3 (CH3)2CHCH2... CH3CH2 o H CH3CH2 o F CH3CH2 o Cl CH3CH2 o Br CH3CH2 o I CH3CH2 o OH 436 443 159 24 3 193 151 570 4 32 366 29 8 440 461 3 52 293 24 0 387 348 421 444 353 29 5 23 3 393 CH3CH2 o OCH3 CH3CH2CH2 o H CH3CH2CH2... (CH3)3C o OCH3 C6H5CH2 o H 3 52 423 444 354 29 4 23 9 395 355 413 439 355 29 8 22 2 4 02 359 422 400 349 29 2 22 7 400 348 375 Bond Broken (shown in red) kJ mol؊1 CH2" CHCH2 o H CH2" CH o H C6H5 o H HC#