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18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 831

18.5 Which of the two compounds in each of the following sets should react more rapidly in a nu-

cleophilic aromatic substitution reaction with CH3O_ in CH3OH? Explain your answers. (a) F F (b) F F NO NO " 2 " 2 " " M or M or r r r r %NO2 "C "NO2 O OCH ( % 3

INTRODUCTION TO TRANSITION-METAL 18.5 CATALYZED REACTIONS

We’ve just learned that SN1 and SN2 reactions cannot be carried out on either aryl or vinylic halides. However, reactions that look very much like nucleophilic substitutions can be carried out using certain transition-metal catalysts. Here are some examples.

H3C $ Pd P L Lc 34 CH3 (catalyst) CH3 (CH3CH2)3N % H2C A CH2 % HBr (18.17) CH C' N (solvent) i + 3 i + neutralized by Br 18 h, 125 °C CH A CH2 % % the (CH3CH2)3N o-bromotoluene o-methylstyrene (86% yield) This reaction, called the Heck reaction, has become very important in organic synthesis. We’ll revisit this reaction in Sec. 18.6A. Notice the formation of the carbon–carbon bond and the re- lease of bromide as HBr. Superficially, it looks as if the conjugate-base anion of ethylene dis- places bromide ion from the aromatic ring. However, this reaction occurs by a very different mechanism and does not happen without the palladium catalyst. (Only about 1 mole % of the catalyst is required.) In the following reaction, we see the substitution of a vinylic bromide by a thiolate anion.

Ph Br Ph SCH2CH3 Pd(PPh3)4 (catalyst) CCA Li SCH CH CCA Li Br (18.18) $ ) | _ 2 3 benzene $ ) | _ + + HH) $ HH) $ (93% yield) This reaction, too, looks superficially like a nucleophilic substitution reaction. But this reac- tion also proceeds by a different mechanism and does not take place without the catalyst, which is present in only 1 mole %. Notice also the retention of alkene stereochemistry, a very

different result from that expected in an SN2 reaction. These are but two examples of thousands now known in which transition-metal catalysts bring about seemingly “impossible” reactions. The field of transition-metal catalysis has ex- ploded in the last four decades, and it has become very important in both laboratory and indus- trial chemistry, as well as in some areas of biology. This field is part of the larger field of 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 832

832 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

organometallic chemistry: the chemistry of carbon–metal bonds. (Grignard reagents and lithium dialkylcuprate reagents are examples of organometallic compounds that you encoun- tered in Secs. 8.8 and 11.4C and will encounter again in subsequent chapters.) Our goal in this section is to understand some of the basic ideas of transition-metal catalysis. Then, in Sec. 18.6, we’ll examine a few important transition-metal catalyzed reactions in the light of these principles.

A. Transition Metals and Their Complexes Recall from general chemistry that transition metals are the elements in the “d block” or “B” groups of the periodic table (groups 3–12 in the IUPAC numbering). These elements are shown in Fig. 18.3. In a given period n, elements are characterized by the progressive filling of d orbitals in quantum level n 1 and the s orbital in quantum level n. Thus, in the fourth period, the transition elements are- characterized by the filling of the one 4s and the five 3d or- bitals. Because the 4s and 3d orbitals have very similar energies, it is usually convenient to think of the electrons in both types of orbitals together as valence electrons. For example, Ni has the electronic configuration [Ar]4s23d8, but we classify Ni as a 10-valence electron atom. Central to transition-metal chemistry are a wide variety of compounds containing transition metals surrounded by several groups, called ligands. Such compounds are called coordination compounds or transition-metal complexes. These can be neutral molecules, as in the first of the following examples, or complex ions, as in the second example.

3 NH3 | Cl NH H3N NH3 Pt 3 Co" Cl NH 3 H3N NH3 "NH3 cis-diamminedichloroplatinum(II) hexamminecobalt(III) ion (cis-platin, an antitumor drug) a complex ion a neutral complex To deal systematically with transition-metal complexes, we must be aware of, and be able to apply, certain conventions: 1. how to classify ligands 2. how to specify formal charge on the metal 3. how to calculate the oxidation state of the metal 4. how to count electrons around the metal In transition-metal chemistry, all ligands are Lewis bases. That is, ligands interact with tran- sition metals by donating electron pairs. There are two types of ligands. The first we’ll call an

Group number 3B 4B 5B 6B 7B 8B 1B 2B Valence electrons in the neutral atom 3 4 5 6 7 8 9 10 11 12

Period 4 Sc Ti V Cr Mn Fe Co Ni Cu Zn

Period 5 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd

Period 6 La Hf Ta W Re Os Ir Pt Au Hg

Figure 18.3 The transition metals.The red numbers indicate the number of valence electrons (outer shell s and d electrons) in the neutral atoms. (These are the same as the IUPAC group numbers in the periodic table; see the inside of the rear cover.) 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 833

18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 833

L-type ligand. If you imagine that if a ligand dissociates from the metal with its bonding elec- tron pair and thus becomes a neutral molecule, the ligand is an L-type ligand. For example, any

one of the NH3 ligands in the two preceeding complexes is an L-type ligand because if we re- move it with its bonding electron pair, we get NH3, the neutral molecule ammonia. The second type of ligand is termed an X-type3 ligand. If you imagine that if a ligand disso- ciates from the metal with its bonding electron pair and thus becomes a negative ion, the lig- and is an X-type ligand. Thus, Cl in cis-platin (the first example) is an X-type ligand, because

removing it with its bonding pair of electrons gives the chloride ion, Cl_. The classification of ligands has implications for computing formal charge. From a formal- charge perspective, the bonding electrons on L-type ligands are considered to “belong” com- pletely to the ligand. Let’s see how this differs from the way we treat bonds in main-group chemistry. We know that a nitrogen with four bonds in main-group chemistry, for example, the

ammonium ion, |NH4,hasapositiveformalcharge.Ifweweretotakeasimilarviewwithcis- platin, the nitrogens would each have a positive charge; and, because the complex is neutral, the Pt would have a charge of 2. A neutral transition-metal complex bearing six L-type ligands would thus have a charge of- 6 on the metal and a positive charge on each ligand. It is incon- venient to draw out all of these- charges; moreover, a formal charge of 6 on a metal is highly unrealistic. Instead, we adopt the convention that the electron pair in- an L-type ligand is as- signed completely to the ligand. Sometimes this point is emphasized by leaving the bonding electron pair on the ligand and depicting the ligand–metal bond as an arrow from these elec- trons to the metal. This is called a dative bond.

Cl NH3 3 L Pt dative bonds

Cl L NH3 3 Because electrons on an L-type ligand belong to the ligand, removal of the ligand does not change the formal charge on either the ligand or the metal:

Cl NH3 Cl NH3 LL L Pt Pt NH3 (18.19) + 3 Cl LLNH3 Cl LL In contrast, electrons in the bonds to X-type ligands are assigned in the same way that we assign electrons in main-group chemistry: one electron is assigned to the ligand and one to the metal. This means that if we remove an X-type ligand, it takes on an additional negative charge and the metal takes on a compensating positive charge:

Cl NH3 Cl NH3 LL LL | Pt Pt L Cl _ (18.20) + 332 NH Cl LL 3 NH3 2 3 2 Differentiating between2 X-type and L-type bonds is a very convenient bookkeeping device, but we should bear in mind that both types of bonds are covalent bonds, and the degree to which electrons (and charge) are transferred to the metal varies widely in both types of bonds, depending on the metal and the ligand. Table 18.1, on p. 834, lists some of the common ligands used in transition-metal chemistry. These are classified as L-type or X-type ligands. It is worth noting two things about this table. First, alkenes or aromatic rings can act as ligands by donating their p electrons to a metal. Sec- ond, allyl and cyclopentadienyl (Cp) are classified as both L-type and X-type ligands. Let’s 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 834

834 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

TABLE 18.1 Some Typical Ligands Used in Transition-Metal Chemistry

Ligand Name Abbreviation Type Electron count*

H3N ammine L 2 1 3 H2O aquo L 2† 3

R3P (R alkyl, aryl) trialkylphosphino, L 2

3 = 1 triarylphosphino

C A O carbonyl CO L 2‡ 33 § H2CACH2 ethylene L 2 §§ CH3C'N acetonitrile MeCN L 2 3

benzene L3 6

† F_,Cl_,Br_,I_ halo (e.g., chloro) X X 2

H_ hydrido X 2 O † H3CCO_ acetato AcO X 2 L L R _ e.g.,H3C _ alkyl (e.g., methyl) X 2 3 3 ' ‡ _ C N cyano CN X 2 3 3 _ H2CA CH CH2 allyl LX 4** L 2

cyclopentadienyl Cp L2X6 _ 2 *The sum of all electrons in the bond(s) between the ligand and the metal. †Only one electron pair is involved in the ligand–metal interaction. ‡Only the electron pair on carbon is involved in the ligand–metal interaction. §Ethylene is listed as a prototype for many alkenes. §§Donation of the nitrogen unshared election pair. **Allyl can also bind to metals as an X-type ligand. In such a situation, the p bond is not involved in coordination and the electron count is 2 (as with alkyl).

consider the Cp case to understand this. The cyclopentadienyl anion was discussed in Sec. 15.7D as an example of an aromatic ion with six p electrons. Table 18.1 indicates that Cp is

an example of an L2X ligand. What this means is that one X-type bond accounts for the fact that Cp takes on one negative charge when removed with a bonding pair from the metal, and that the four remaining p electrons (that is, two double bonds) take part in two L-type bonds. In other words, we can think of a metal–Cp complex in the following way (M metal): =

(18.21)

M We know that the p electrons in Cp are completely delocalized, and they remain delocalized

in metal complexes. (See the structure of ferrocene [Cp2Fe] on p. 728, which shows this delo- 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 835

18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 835

calization.) Consequently, a more accurate picture of such a complex would show the “L” and “X” character of the bonds parceled out equally over all five carbons, with each carbon partic- ipating in 20% of an X-type bond (5 0.20 1.0 X-type bond) and 40% of an L-type bond (5 0.40 2.0 L-type bonds). But Xthis delocalization= can be ignored for the bookkeeping purposesX discussed= in this section.

PROBLEM 18.6 Noting the LX character of the allyl ligand in Table 18.1, sketch the allyl–metal interaction, showing both L-type and X-type bonds. Use M as a general metal.

B. Oxidation State The oxidation state of the metal is an important concept in organometallic chemistry. Oxida- tion state can be conceptualized as the charge the metal would have if all covalently bonded atoms—that is, all X-type ligands—were dissociated with their bonding electron pairs. For ex- ample, if the oxygens of manganese dioxide, OAMnAO, were to dissociate from the man- ganese with their bonding electron pairs, the oxygens would each take on a 2 charge, and the manganese would take on a 4 charge. Therefore, the manganese has an- oxidation state of 4. (This process is essentially+ the same as the one used for assigning oxidation numbers to carbon+ atoms in oxidation–reduction reactions [Sec. 10.5A]). Eq. 18.22 formalizes this idea:

Oxidation state of M number of X-type ligands QM (18.22) = + In this equation, QM is the actual charge that M has before the fictitious dissociation of the X- type ligands. To illustrate a situation involving QM, consider the hexachloroplatinate dianion, 2 [Pt(Cl6)] _. For this species, QM 2. If the six chlorines were to dissociate with their bond- ing electrons, the charge on Pt would=- be 4, because Pt starts out with a 2 charge before the

fictitious dissociation. To be sure that the+ oxidation state and the actual charge- QM are not con- fused, the oxidation state is sometimes indicated with Roman numerals. Hence, the name of 2 the ion [Pt(Cl6)] _ is hexachloroplatinate(IV). L-type ligands do not contribute to the oxidation state. You should verify that the oxidation

state of platinum in the neutral complex Cl2Pt(PPh3)2 is 2. +

PROBLEMS 18.7 Calculate the oxidation state of the metal in each of the following complexes.

(a) O (b) Pd(PPh3)4 (c) Cp2Fe S A tetrakis(triphenylphosphine)palladium ferrocene O Mn O_ S L O permanganate 18.8 What is the oxidation state of the metal in the starting material in the following reaction? How does it change, if at all, as a result of the reaction?

PPh3 PPh3 Ph3P PPh3 Ph P Rh" PPh H LLRh" 3 3 2 L L L + HCl "Cl L"H chlorotris(triphenylphosphine)rhodium 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 836

836 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

C. The dn Notation In understanding the reactions of main-group elements that follow the octet rule, it is impor- tant in applying acid–base concepts for us to know whether the element undergoing a transfor- mation has unshared valence electrons. Often these unshared electrons are shown explicitly. In transition-metal chemistry, it is also important to know whether the metal has unshared va- lence electrons. In many cases, the metal has so many unshared valence electrons that it would be impractical or confusing to draw them all. Instead, we use a convenient algorithm to calcu- late the number of unshared valence electrons. The number of unshared valence electrons on the metal is the number n in a notation called dn. For example, if the metal in a complex has eight unshared valence electrons, we say that the complex is a d8 complex. We calculate n by determining the number of valence electrons remaining on the metal after removing all ligands with their electron pairs. We start with the number of valence elec- trons in the neutral transition element (from Fig. 18.3). We remove an electron for each posi- tive charge, add an electron for each negative charge, and then subtract one electron for each bond to an X-type ligand. L-type ligands have no effect on d n.

n valence electrons in neutral M QM number of X-type ligands (18.23) = - - Introducing the definition of oxidation state in Eq. 18.22, Eq. 18.23 becomes n valence electrons in neutral M oxidation state of M (18.24) = -

Study Problem 18.1 n Calculate n in the d notation for ferrocene, Cp2Fe.

Solution We’ll make this calculation with both Eqs. 18.23 and 18.24. From Fig. 18.3 we see that neutral Fe has eight valence electrons. The charge of the iron is zero, and Table 18.1 shows that each Cp ligand has one X-type bond; the iron thus has two bonds to X-type ligands. Hence, n 8 2 6, and ferrocene is thus a d6 complex. =From - Eq.= 18.24, we calculate that with two X-type ligands and zero charge, the Fe in fer- rocene has a 2 oxidation state; hence, Eq. 18.24 gives the value of n as 8 2 6. + - =

PROBLEM 18.9 What is dn for each of the following complexes? 2 (a) [W(CO)5] _ (b) Pd(PPh3)4 (c) PPh3 Ph3P PPh3 L Rh" L HClL L"H

D. Electron Counting: The 16- and 18-Electron Rules In main-group chemistry, we use the octet rule as one indicator of reactivity. For example, we know that if a main-group element in a compound has fewer than an octet of electrons, it can ac- cept an electron pair from a Lewis base in a Lewis acid–base association reaction. In other words, main-group elements have a tendency to complete their octets. Recall that counting for the octet involves adding an element’s unshared valence electrons to the number of electrons in all bonds to the element. The electron count in transition-metal complexes is also important and is determined in a similar manner. 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 837

18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 837

To determine the electron count for a transition-metal complex, we start with the n elec- trons in the dn count—the unshared electrons—and add two electrons for every ligand (both L-type and X-type). Thus,

electron count n 2(number of all ligands) (18.25) = + The multiplier 2 is required because there are two electrons per bond. The rationale for this formula is that the number n is the number of unshared valence electrons on the metal; the total electron count is the unshared valence electrons plus all electrons in bonds, just as in counting for the octet rule. Using Eq. 18.24, we can rewrite this formula in terms of the oxidation state of the metal: electron count valence electrons in neutral M oxidation state of M = 2(number- of all ligands) (18.26) + By incorporating the definition of oxidation state (Eq. 18.27), we obtain yet another equiva- lent formula:

electron count valence electrons in neutral M QM = number of X-type ligands -2(number of all ligands) (18.27) - + Recognizing that all ligands X-type ligands L-type ligands, we finally obtain a very use- ful formula for electron count:= +

electron count valence electrons in neutral M QM = number of X-type ligands -2(number of L-type ligands) (18.28) + + Thus, to obtain the electron count in a complex, we start with the electron count of the neutral metal from Fig. 18.3; we subtract the charge on the metal (taking into account its algebraic sign); we add the number of X-type ligands; and we add twice the number of L-type ligands. Don’t let the mathematical derivation of Eq. 18.28 obscure its rationale. Remember that the goal is to count all unshared and bonding electrons about the metal. Because an X-type ligand by definition has one electron in its M X bond assigned to X, we have to add this electron back to obtain the total number of electronsL in the bond. Because both electrons in the dative bond to an L-type ligand are assigned to the ligand, we have to multiply each L-type ligand by 2 to count both of these bonding electrons. Let’s use Eq. 18.28 to calculate some electron counts. For example, the electron count of

Ni(CO)4 is 10 0 0 2(4) 18. This is an 18-electron complex. (This compound, tetracarbonylnickel(0),- - is a+ very stable= complex of Ni.)

The electron count of Cl2Pd(PPh3)2 is 10 0 2 2(2) 16. This is a 16-electron complex. - + + = In transition-metal chemistry, the most stable complexes in many cases have electron

counts of 18 electrons. This statement is called the 18-electron rule. Ni(CO)4, a very stable complex of Ni(0), is an example of the 18-electron rule. Just as the octet represents the num- ber of valence electrons (8) in the outermost s and p orbitals of the nearest noble gas, 18 elec- trons is also the number of total s p d valence electrons in the nearest noble gas. Exceptions to the 18-electron +rule +occur, and an important type of exception occurs fre- quently with transition metals in the 8–11 valence-electron group (Fig. 18.3), which includes Ni and Pd, two metals of prime importance in the transition-metal-catalyzed reactions dis- cussed in this section. Although a number of stable complexes of these metals have 18 elec- trons, others contain 16 electrons. The tendency of these metals to surround themselves with

16 electrons can be called the 16-electron rule. The Cl2Pd(PPh3)2 example shows the opera- tion of the 16-electron rule. 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 838

838 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

PROBLEMS 18.10 What is the electron count for the Rh complex shown in Problem 18.9c? 18.11 How many CO ligands would be accommodated by Fe(0) if we assume that the resulting complex follows the 18-electron rule?

18.12 Using the 18-electron rule, explain why V(CO)6 can be easily reduced to [V(CO)6]_.

We used hybridization arguments to understand the basis of the octet rule in main-group chemistry (Sec. 1.9). Thus, the main-group element carbon has four valence orbitals (for ex- ample, four sp3 hybrid orbitals) that can either form two-electron bonds or house unshared electron pairs. We can justify the 18-electron rule in a similar way. Consider, for example, the 3 complex ion [Co(CN)6] _.UsingEq.18.22,weseethattheoxidationstateofCois 3, and, from Eq. 18.24, that this is a d6 complex. This means that Co(III) in this complex has+ six un- 3 shared electrons. Let’s imagine building this complex from a “naked” Co | ion. Start with the electronic configuration of this ion, as shown in Fig. 18.4a. Allow all the electrons to pair, as shown in Fig. 18.4b. Because this electron pairing violates Hund’s rules, it requires energy. This electron pairing leaves two 3d,one4s,andthree4p orbitals unoccupied. These are hybridized, as shown in Fig. 18.4c, to give six equivalent d2sp3 hybrid orbitals. Six equivalent hybrid or- bitals are directed to the corners of a regular octahedron in the same sense that sp3 carbon or- bitals in methane are directed to the corners of a regular tetrahedron. Hybridization also requires energy. Each of these empty hybrid orbitals can accept an electron pair from a cyanide ion

(_CN). Because these hybrid orbitals are directed in space, they can form stronger bonds to

Unhybridized Co3 (d6): | 4p 4p LL LL LL pair the LL LL LL electrons 4s 4s

ENERGY ENERGY 1 1 1 LL 1 LL 3d6 11111 3d6 1 1 1 LL LL (a) LL LL LL LL LL (b) LL LL LL

hybridize the empty orbitals; add 6 _CN

CN d2sp3 hybrid NC CN orbitals ÅÅ ÅÅÅÅÅÅÅÅ ÅÅ LL

LL LLÅÅ = LL cyanide LL electrons LL LL Co

1 1 6 1 NC CN ENERGY 3d 1 1 1 LL LL LL LL CN (unaffected by hybridization) (d) (c)

3 Figure 18.4 Development of the hybrid orbital description of [Co(CN)6] _,an 18-electron complex ion.(a)The 3 electronic configuration of Co |.(b) The Co electrons are arranged in pairs;the empty orbitals that remain (red) are used to form hybrid orbitals.(c) The empty orbitals are hybridized into six equivalent d2sp3 hybrid orbitals.Each of

these orbitals can accept an electron pair (symbolized by XX) from a _CN ion. (d) The hybrid orbitals and, hence, the six _CN that bind to them are oriented to the corners of a regular octahedron. (The edges of the octahedron are indicated with blue dashed lines.) 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 839

18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 839

cyanide than unhybridized orbitals, and the strength of these bonds more than compensates for 3 the energy cost of electron pairing and hybridization. The result is the octahedral [Co(CN)6] _ complex shown in Fig. 18.4d. In other words, the 18-electron rule results from the rehybridiza- 3 tion and maximal occupancy of all valence orbitals of the Co | ion. The 16-electron rule is important in square planar complexes of the 10-electron elements

Ni, Pd, and Pt. For example, consider the antitumor drug cis-platin, Cl2Pt(NH3)2 (p. 832). This 8 2 is a 16-electron d complex of Pt(II) that has square planar geometry. If we start with a Pt | ion and arrange its eight electrons in pairs within four 5d orbitals, this leaves a single 5d or- bital, a 6s orbital, and three 6p orbitals empty. It turns out that hybridization of four of the five empty orbitals to give four dsp2 hybrid orbitals and one relatively high-energy 6p orbital is a particularly favorable hybridization:

6p (empty, unhybridized)LL

ÅÅ= ligand electrons dsp2 hybrid

ENERGY orbitals ÅÅ ÅÅ ÅÅ ÅÅ (square planar)

LL LL LL LL 1 1 1

5d8 1 1 1 1 1 LL LL LL LL The four hybrid orbitals are directed to the corners of a square and accept the electron pairs from the four ligands to give a square planar complex. The element platinum can also adopt 18- electron configurations, but the point is that the 16-electron configuration is reasonably stable. As in main-group chemistry, hybridization arguments are useful for visualizing electrons in bonds, but they are inferior to molecular orbital arguments for detailed understanding of mol- ecular energies. The branch of molecular orbital theory that deals with transition-metal com- plexes is called ligand field theory. We need not explore this theory here; but suffice it to say that this theory provides excellent support for the 16- and 18-electron rules.

PROBLEM 2 18.13 Use a hybridization argument to predict the geometry of (a) the [Zn(CN)4] _ ion; (b) the neutral compound Pd(PPh3)4.

E. Fundamental Reactions of Transition-Metal Complexes We have now been introduced to the preliminaries that we need to understand the mechanistic basis of some transition-metal-catalyzed reactions, and we’re ready to look at these reactions in detail. It turns out that transition-metal complexes undergo a relatively small number of fun- damental reaction types, and many reactions are readily understood simply as combinations of these fundamental processes. The goal of this section is to introduce a few of these.

Ligand Dissociation–Association; Ligand Substitution One of the most common re- actions of transition-metal complexes is ligand dissociation and its reverse, ligand associa- tion. In ligand dissociation, a ligand simply departs from the metal with its pair of electrons, leaving a vacant site (orbital) on the metal.

Pd(PPh3)4 Pd(PPh3)3 PPh3 (18.29) Pd(0) Pd(0) + 3 an 18e complex a 16e complex _ _

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840 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

This process does not change the oxidation state of the metal, but it does change the electron count. A ligand substitution can occur by the dissociation of one ligand and the association of

another, which is somewhat analogous to an SN1 reaction in alkyl halides, or by a direct sub- stitution, in which one ligand displaces another, somewhat analogous to the SN2 reaction.

Ph3P I Ph3P CH3 LL LL Pd H3C Li Pd L Li| I_ (18.30) + L + Ph3P LLPh Ph3P L Ph Pd(II) Pd(II) a 16e complex a 16e complex _ _ In the most common ligand substitution reactions, ligands of the same type are exchanged: X- type ligands for X-type ligands and L-type ligands for L-type ligands.

Oxidative Addition In oxidative addition, a metal M reacts with a compound X Y to form a compound X M Y; the metal “inserts” into the X Y bond. We have alreadyL stud- ied an important reactionL Lof this type: the formation of a GrignardL reagent from Mg metal and an alkyl halide (Sec. 8.8A).

R Br Mg RMg Br (18.31a) L +Mg(0)2 L Mg(II)L

As the term “oxidative addition” implies, the Mg is oxidized. An electron count for Mg in this reaction is 2 for a metal and 4 in the Grignard reagent. (Remember, though, that Mg is a main- group metal and is not subject to the 16- or 18-electron rules.) An example of oxidative addition from transition-metal chemistry is the insertion of Pd into the carbon–halogen bond of iodobenzene:

Ph3P Ph3P Ph L LL Pd Ph I Pd (18.31b) + L Ph3P L Ph3P LLI Pd(0); Pd(II): a 14e complex a 16e complex _ _

Both of new bonds are X-type bonds. As a result of this reaction, both the electron count and the oxidation number of the metal increase by two units. Oxidative addition is a remarkable reaction that lies at the heart of transition-metal catalysis with aryl and vinylic halides. Why is it that a metal can break a sigma bond in this way? Mole- cular orbital theory provides a simple way to understand this process, as shown in Fig. 18.5. Think of the carbon–halogen bond as a localized bond for simplicity, and imagine a molecular orbital treatment of this bond much like the molecular orbital treatment of the H Hbondin

H2 (Sec. 1.8A). The carbon–halogen bond has an associated bonding molecular orbital,L which is occupied by the two bonding electrons, and an antibonding molecular orbital, which is un- occupied. The bonding molecular orbital can serve as a ligand, donating its electrons to one of the empty hybrid orbitals on the metal. At the same time, one of the filled d orbitals of the metal overlaps with the antibonding molecular orbital of the carbon–halogen bond. This additional overlap strengthens the metal–ligand interaction, but weakens the carbon–halogen bond, be- cause addition of electrons to an antibonding molecular orbital removes the energetic advan- tage of bonding. (See Fig. 1.14. p. 35.) The carbon–halogen bond is weakened sufficiently that it actually breaks. Hence, electrons flow from the aryl halide to the metal and, at the same time, 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 841

18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 841

filled metal d orbital

C % (

bonding antibonding M MO of MO of C X C X L empty metal L hybrid orbital X

Figure 18.5 An orbital description of concerted oxidative addition.The bonding MO of the C X bond donates electrons to an empty hybrid orbital on the metal.(The shape of this hybrid orbital is simplified.)L These orbitals are shown in purple,the thin purple lines indicate the electronic overlap,and the purple arrow shows the direction of electron flow.At the same time,a filled d orbital on the metal donates electrons to an antibonding MO of the C X bond.(A 3d orbital is used for simplicity.) Because the peaks and troughs of the d orbital (shown in blue and green,L respectively) match the peaks and troughs of the antibonding MO of the C X bond,this is a bonding interaction. The electronic overlap in this interaction is indicated with thin blue and greenL lines, and the blue and green ar- rows show the direction of electron flow. Because this interaction populates the antibonding MO of C X, this bond is weakened, and it breaks. L

from the metal to the aryl halide. We can approximate the process as follows with the curved- arrow notation (L other ligands): = L Ar L Ar L L M "X M L (18.32) 3 L L L L L X electron pair in a d orbital

Oxidative addition can occur by a variety of mechanisms, but a concerted (one-step) process is fairly common.

Reductive Elimination Reductive elimination is conceptually the reverse of oxidative addition, and the orbital interactions involved are the same, only in reverse. In reductive elim- ination, then, two ligands bond to each other and their bonds to the metal are broken; X M Y X Y M. An example of this process is the formation of a carbon–carbon bondL betweenL LT two ligandsL + within a Ni complex:

H CH2CH2CH2CH3 L Ph3P CCA L Ph3P H CH2CH2CH2CH3 LL L L L Ni L H Ni CCA L (18.33) + L Ph3P L CH3 Ph3P L H3C L H Ni(II) Ni(0) a 16e complex a 14e complex _ _

18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 842

842 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

Because two X-type ligands are lost, the metal is reduced, and its electron count is decreased. Notice in this particular example that the alkene stereochemistry is retained. Reductive elimi- nation in general occurs with retention of stereochemistry. Because this process is the reverse of oxidative addition, it follows that concerted oxidative addition also occurs with retention of stereochemistry. The electron flow in Fig. 18.5 or Eq. 18.32 is consistent with this observation.

Ligand Insertion In this process, a ligand inserts into a metal–ligand bond; that is, L M R M L R. Notice the difference between oxidative addition and ligand in- sertion.L L BothLT are insertionL L processes. In an oxidative addition, the metal inserts into a chemi- cal bond within a compound not initially associated with the metal. In a ligand insertion, a lig- and L inserts into the bond between the metal and a different ligand R, and the inserting ligand gains a bond. Two types of ligand insertion are most frequently observed in transition-metal chemistry. In a 1,1-insertion, the new bond is formed at the same atom that was bound to the metal. In- sertions of CO ligands are frequently observed examples of this type. The first reaction below is an example.

CH3 CH3 CO CH3 CO (CO) Mn" CO (CO) Mn "CO (CO) Mn" "CO(18.34) 4 1,1-ligand 4 ligand 4 3 insertion L association L Mn(I) Mn(I) Mn(I) an 18e complex a 16e complex an 18e complex _ _ _

In this reaction, the methyl group migrates, with its bonding electrons, to the carbon of the car- bonyl ligand, which in turn forms an X-type bond to the metal. This migration is possible be- cause the carbon of the carbon monoxide ligand is electron-deficient. Hence, the carbonyl car-

bon inserts into the Mn CH3 bond. Notice that this insertion leaves a vacant site (that is, an empty orbital) on the metal,L as we can see from the reduction in the electron count. In the sec- ond reaction, this empty metal orbital is filled by another molecule of the ligand from solution. Another type of ligand insertion is 1,2-insertion. In this process, the migrating group moves to an atom adjacent to the one bound to the metal. A common example of this process is the following, in which an ethylene ligand inserts into a Pd–aryl bond.

Ar PPh3 CH2 PPh3 Br Pd" Br Pd CH CH Ar Br Pd" CH CH Ar (18.35) 1,2-ligand 2 2 ligand 2 2 L insertion L L L L CH2 association "PPh3 "PPh3 "PPh3 Pd(II) Pd(II) Pd(II) a 16e complex a 14e complex a 16e complex _ _ _

Again, notice that the electron count is reduced by two; that is, the process results in an empty orbital on the metal. This orbital can then gain another electron pair by ligand association, as shown by the second step in Eq. 18.35, thus fulfilling the 16-electron rule. We can approximate the ligand insertion process in the curved-arrow notation as follows:

Ar CH2

Br Pd" Br Pd CH2CH2 Ar (18.36a) L L L L CH2 "PPh3 "PPh3 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 843

18.5 INTRODUCTION TO TRANSITION-METAL CATALYZED REACTIONS 843

As Eq. 18.36a illustrates, 1,2-ligand insertion is essentially a concerted addition of the metal (Pd in this case) and the migrating ligand (Ar in this case) to the alkene p bond. Because 1,2- ligand insertion is a concerted intramolecular addition reaction, the two new bonds must be formed at the same face of the p bond. Hence, ligand insertion is a syn-addition. This be- comes evident when the carbons of the alkene double bond are stereocenters, as they are in cy- clohexene. In this case, syn-addition requires that the Pd and the aryl group have a cis relation- ship in the insertion product.

Br

Ar Pd" Ph P Br Pd" 3 L (18.36b) syn 1,2-ligand

L e 0 insertion Ar "PPh3 Pd and aryl are cis

b-Elimination In b-elimination, a group b to the metal migrates with its bonding electron pair to the metal. This process is conceptually the reverse of ligand insertion. (Run Eq. 18.36a backward mentally and you will see the b-elimination of ethylene by migration of aryl.) It often happens that b-elimination involves a hydride migration. For example, the product of Eq. 18.36a (with Ar phenyl) can undergo b-elimination with hydride migration as follows: = H Ph Ph H %CH % "CHC Br Pd "CH2 Br Pd" (18.37) LL L CH2 "PPh3 "PPh3 Pd(II) Pd(II) a 14e complex a 16e complex _ _

Notice that b-elimination requires an empty orbital on the metal, because, as a result of this process, the electron count is increased by two units. We studied another type of b-elimination, the E2 reaction, in Sec. 9.5. The elimination re- action in Eq. 18.37 looks superficially similar, but it is quite different. In the E2 reaction, a proton is eliminated. In the b-elimination of Eq. 18.37, a hydride—a hydrogen with its bond- ing electrons—is eliminated by migration to the metal. We can stress this point with the curved-arrow notation:

H Ph Ph Ph CH H H % % "CH "CH Br Pd "CH2 b-elimination Br Pd" ligand Br Pd" (18.38) L L L CH2 association L CH2 "PPh3 "PPh3 "PPh3 Pd(II) Pd(II) Pd(II) a 14e complex a 14e complex a 16e complex _ _ _

Because this b-elimination is intramolecular—within the same molecule—it must occur as a syn-elimination. This makes sense because this reaction is conceptually the reverse of 1,2- 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 844

844 CHAPTER 18 • THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES, AND PHENOLS. TRANSITION-METAL CATALYSIS

ligand insertion, which is a syn-addition (Eq. 18.36b). Contrast the stereochemistry of this b- elimination with that of the E2 elimination, which is a bimolecular reaction and occurs with anti stereochemistry (Sec. 9.5E).

Study Problem 18.2 Consider the following mechanism for Eq. 18.18 on p. 831. Identify the process associated with each step. Counting electrons at each stage may help you.

Pd(PPh3)4 Pd(PPh3)3 Pd(PPh3)2 (18.39a)

PPh+ 3 PPh+ 3

H H

H H C

Ph P " % 3 S $

$ C

L L

L Pd(PPh3)2 $C A C $ L Pd (18.39b) + "Ph Br Ph Ph3P Br

H H H H C C Ph3P " % Ph3P " %

S C S C

L L L L L L

Pd Li| CH3CH2S_ L Pd Li| Br_ (18.39c) "Ph + "Ph + Ph P Ph P L 3 Br 3 SCH2CH3

H H

C H H

Ph P " % 3 S $ C $

L L L Pd $C A C) Pd(PPh3)2 (18.39d) "Ph + Ph P L 3 SCH2CH3 CH3CH2S Ph

Solution Step 18.39a consists of two successive ligand dissociations that reduce the electron

count around the Pd from 18e_ to 14e_. This “makes room” for the vinylic halide, which under- goes an oxidative addition with retention of configuration in Step 18.39b. This step takes the

electron count to 16e_, and oxidizes the Pd(0) to Pd(II). Step 18.39c is a ligand substitution of a bromo ligand with an ethylthio ligand. It might occur by a prior dissociation of the Br ligand, by

association of the CH3CH2S_ with the Pd to give an 18e_ complex followed by dissociation of

Br_, or by a concerted mechanism reminiscent of the SN2 reaction. Finally, Step 18.39d is a re- ductive elimination, which forms the product with retention of stereochemistry and regenerates

the catalytic Pd(0) species Pd(PPh3)2.

Let’s use the example in Study Problem 18.2 to take stock of what the Pd is actually doing—why it makes a vinylic or aryl substitution possible. Ligation to the Pd brings two groups—the vinylic group and the nucleophile—into proximity. The oxidative addition step is the key step that makes this possible, and, as we have seen (Fig. 18.5), it is driven by the si- multaneous presence of filled and empty metal orbitals that can interact with the vinylic halide

so that the carbon–halogen bond is broken. The nucleophile CH3CH2S_ and the vinylic group are then connected by reductive elimination. The orbital interactions are essentially the same as in oxidative addition. The role of the metal finds analogy in the slider of a zipper: it brings two groups together, causes them to join and lock, and then moves on to do the same thing over again. 18_BRCLoudon_pgs4-3.qxd 11/26/08 9:09 AM Page 845

18.6 EXAMPLES OF TRANSITION-METAL-CATALYZED REACTIONS 845

PROBLEMS 18.14 A student has written the following ligand substitution reaction, claiming that it changes the oxidation state of the metal by one unit. What is wrong with this reasoning?

Cl_ Pd(PPh3)4 ClPd(PPh3)3 PPh3 + LT + 3 18.15 The Wilkinson catalyst chlorotris(triphenylphosphine)rhodium(I), ClRh(PPh3)3, brings about the catalytic hydrogenation of an alkene in homogeneous solution: R R

ClRh(PPh3)3 $CHA C) 2 RCH2CH2R (18.40) + HH) $ (a) Using the following mechanistic steps as your guide, draw structures of the transition- metal complexes involved in each step. Give the electron count and the metal oxidation state at each step.

1. oxidative addition of H2 to the catalyst 2. ligand substitution of one PPh3 by the alkene 3. 1,2-insertion of the alkene into a Rh H bond and readdition of the previously ex-

pelled PPh3 ligand L 4. reductive elimination of the alkane product to regenerate the catalyst (b) According to the known stereochemistry of the 1,2-ligand insertion and reductive elim-

ination steps, what would be the stereochemistry of the product if D2 were substituted for H2 in the reaction?

18.6 EXAMPLES OF TRANSITION-METAL-CATALYZED REACTIONS

A. The Heck Reaction In the Heck reaction, an alkene is coupled to an aryl bromide or aryl iodide under the influence of a Pd(0) catalyst.

H3C $ Pd P L Lc 34 CH3 (catalyst) CH3 (CH3CH2)3N % H2C A CH2 % HBr (18.41) CH C' N (solvent) i ++3 i neutralized by Br 18 h, 125 °C CH A CH % % 2 the (CH CH ) N (86% yield) 3 2 3

(The aryl substituents of the phosphine ligands used in the catalyst in this case are o-tolyl (that is, o-methylphenyl) groups rather than phenyl groups, but phenyl groups are also sometimes used.) The reaction is named for Richard F. Heck (b. 1931), who discovered the reaction in the early 1970s while a professor of chemistry at the University of Delaware. (A Japanese chemist, T. Mizoroki, simultaneously discovered the reaction, but it is generally known as the Heck reaction.) The Heck reaction has proven to be one of the most useful processes for form- ing carbon–carbon bonds to aromatic rings and even, occasionally, to vinylic groups. The mechanism of the Heck reaction is outlined in the following equations. You should identify the process or processes involved in each step (L tri-o-tolylphosphine ligands; the steps in Eq. 18.42b are numbered for reference). =