《有机金属化学》课件Chapter 2

1、Chapter 2Complexes with metal-carbon s bondsMetal alkyls, aryl, hydride and related s-bonded ligandsOutline1. The stability of transition metal-carbon s bonds2. Preparation of complexes with metal-carbon s bonds3. Chemical properties of complexes with metal-carbon s bonds4. Related s-bonded ligand c

2、omplexes (other metal s-bonded complexes) M-X (X = -BR2, -SiR3, -NR2, -OR, -F, -Cl, etc.) 5. Metal hydridesSilyl amino References and suggested readings: The Organometallic Chemistry of the Transition Metals, Robert H. Crabtree, 3rd Edition, 2001, Chapter 3 Organometallic Chemistry, G. O. Spessard,

3、G. L. G. L. Miessler, Prentice-Hall: New Jersey, 1997, Chapter 6 Organotransition Metal Chemistry, Akio Yamamoto, 1986. Chapters 3.1, 4.2.1. The stability of transition metal-carbon s bonds 1). Thermal stability of transition metal-carbon bonds2). Factors affecting M-C bond strength3). Origin of ear

4、lier failure in obtaining TM alkyls: Kinetic consideration. Prior to 1960s: Non-transition metal alkyls are well known, e.g.AlMe3ZnEt2Transition metal alkyls are rare. Only a few transition metal alkyls were prepared. e.g.Cubic 立方的Prior to 1960s，transition metal alkyls are rare.Are transition metal-

5、carbon bonds weak and unstable?But today: Transition metal-carbon bonds are known for virtually all transition metals. Questions: Are transition metal-carbon s bonds really unstable? What affects M-C bond strength? What causes the earlier failure to obtain transition metal alkyls? 1). Thermal stabil

6、ity of transition metal-carbon bondsThe thermal stability can be related to bond dissociation energy.M-X (g) M(g) + X(g) DH = BDE (M-X) If BDE (M-X) is large, M-X is thermally more stable.Range of M-C bond dissociation energy (BDE).(See next two pages)BDE (M-C) values in MRn complexes (KJ/mol)Non-tr

7、ansition metals Transition metals M-C bond BDE Compound M-C bond BDELi-Et209Ti(CH2C(CH3)4Ti-CH2R 170Li-Bu248Ti(CH2Ph)4 Ti-CH2R 240Zn-Me176Ti(CH2Si(CH3)4 Ti-CH2R 250Zn-Et145Cp2Ti(CH3)2 Ti-CH3 250Cd-Me139Cp2TiPh2 Ti-Ph 350Hg-Me122(CO)5MnCH3 Mn-CH3 150Hg-Et101(CO)5ReCH3 Re-CH3 220Hg-i-Pr 89Zr(CH2C(CH3)

8、4 Zr-CH2R 220Hg-Ph136Zr(CH2Ph)4 Zr-CH2R 380B-Me363Zr(CH2Si(CH3)4 Zr-CH2R 225B-Et342CpPt(CH3)3Pt-CH3 165Al-Me276(Et3P)2PtPh2 Pt-Ph 250Al-Et242Ta(CH3)5 Ta-CH3 260Ga-Et237W(CH3)6 W-CH3 160Non-transition metals:BDE(M-C): 89-280 kJ/mol Except B-CTransition metals:BDE (M-alkyl) :150 - 260 kJ/molBDE (M-ary

9、l): 250 - 350 kJ/molNormally, BDE (M-alkyl) M-Br M-H = or M-I)Reason: Electron density: Cl Br IM-H M-CF3 M-Ar M-CH3 M-CO-CH3M-C-R: stability increases with more electron-withdrawing R.* M-H is stronger than M-CR3 s orbital is indirectional and more dense.* M-CF3 is stronger than M-CR3 -CF3 is electr

10、on withdrawing.* M-Ar is stronger than M-CR3. More effective back-donation for M-Ar.Would you expect M-CH3 bond be stronger or weaker than (answer: M-CH3 M-CH=CH2 M-CCH)Why?* M-CH3: no dp-pp interaction; M-CHCH2: one dp-pp interactions; M-CCH: two dp-pp interactions* Evidence: Stability of the follo

11、wing M-C bonds:M-CH3 D(M-CF3) D(M-Ph) D(M-CH3) D(M-CH2CH3) D(M-CH2Ph).The earlier failure in obtaining TM alkyls is not due to thermodynamic reasons !Exercises Which one has the strongest stability of the following complexes ?Answer: a) last one b) last one c) first one3). Origin of earlier failure

12、in obtaining TM alkyls: (Kinetic consideration)General questions.Q1. Why could many TM alkyl complexes not be isolated? The reason for the instability of TM alkyl complexes is that there are many pathways that TM alkyl complexes can be decomposed. Q2. What are the common decomposing pathways for tra

13、nsition metal alkyls?*. b-elimination (The most common route!) *. Reductive elimination*. Others, e.g. a, g-eliminations (intramolecular reactions)Q3. How can we obtain stable M-alkyl complexes? Stable M-alkyl complexes could be obtained: if the above decomposing pathways can be prevented. a) b-elim

14、ination.The most common decomposition pathway for alkyls is b-H elimination, which converts a metal-alkyl into a hydrido-olefin complex.hydrido-olefin complexWhat is the condition for b-H elimination to occur ?A. There must be a b-H.B. The complex should be able to form four-membered co-planar trans

15、ition state. (see (A) C. There is a vacant site cis to the alkyl (e.g. (B). but not (C). (need an empty orbital to accept C-H electron.)D. The olefin formed is stable.E. The metal has e- to go to s*(C-H).Exercises. (a). It is difficult for the following compounds to undergo b-hydrogen eliminations.

16、Suggest main reasons. answer: (1) can not form a four-membered TS(2) no vacant site cis to the alkyl group, and CO is usually tightly bound to metal.(3) the vacant site is not cis to the alkyl group.Further notes on b-elimination. (i) b-H elimination on d0 (electron deficient) metal complexes may no

17、t occur. e.g. The first complexno d electron, so cant undergo -elimination.The second complexhas d electrons, so can undergo -elimination.s- donation-backdonation(ii) Other b-elimination is also possible, e.g.because: M-H M-CH3 (by 15-25 kcal/mol); M-F M-Hb) Reductive elimination* Coordination numbe

18、r decreased by 2. * Oxidation number decreased by 2. The process is the second most common decomposition pathway for metal alkyls. e.g. More detailed discussion will be given later.c) Other pathwaysa-H eliminationExample,?g-H elimination. e.g. Summary: Common pathways of decomposing M-alkyls: b-H el

19、imination Reductive elimination others.Exercises. 1) For each pair of complexes listed below, which one is less stable?answer: (1) a reductive elimination. (2) b b-H elimination. (3) a b-H elimination (easier) 2) Which one is least stable ? answer: (b)3) which one is most stable ?A. Synthesis by alk

20、yl transfer reactions.M-R- + M-X M-R + X- (M = Li, Na, Al, Zn, BrMg, etc; M = TM)B. Synthesis from anionic transition metal complexes. M- + R-X M-R + X-C. Synthesis by oxidative addition reactions.M + R-X X-M-RD. Synthesis involving insertion reactions. M-X + A M-A-XE. Synthesis involving eliminatio

21、n reactions. M-A-R M-R + AF. Synthesis by attack on coordinated ligands2. Preparation of complexes with metal-carbon s bonds.Synthesis by alkyl transfer reactions (nucleophilic attack on the metal)R- + M-X M-R + X-Typical R- reagents: RLi, RMgX, AlR3, AlR2X, ZnR2, etc. Examples:B. Synthesis from ani

22、onic transition metal complexes. M- + R-X M-R + X-Examples:Acyl: 酰基C. Synthesis by oxidative addition reactions. Typical oxidative addition reactions:Examples of preparation of complexes. So bulkyD. Synthesis involving insertion Reactions. M-X + A M-A-XExamples:PEt3: triethylphosphine PPh3: tripheny

23、lphosphineE. Synthesis involving elimination reactions. M-A-R M-R + A (A should be very stable)A = CO, CO2, SO2, N2, e.g. CO: carbon monoxideCO2: carbon dioxideSO2: sulfur dioxideF. Synthesis by attack on coordinated ligands We will go to more details later.3. Chemical properties of complexes with m

24、etal-carbon s bonds(1) b-hydrogen eliminatione.g. (2) Reductive eliminatione.g. (3) a, g - elimination, e.g. (4) Migratory insertion(toluene)( bond metathesis)(5) Electrophilic attack on alkylsExamples. (Electrophilic attack on alkyls)(6) Special properties of some M-R complexes Small metallocycles

25、show interesting rearrangement.4. Related s-bonded ligand complexes A. Comparison between M-CR3 and M-SiR3M-CR3M-SiR3 Which bond is stronger ? * BDE(M-SiR3) BDE(M-CR3) Reason: a) some -interaction (Si has empty d orbitals)SinHm: silicon hydride (silane)-SiR3: silicon alkyl (silyl)b) M-SiR3 is less s

26、terically congested.c) M-SiR3 is difficult to undergo b-elimination. (Si=C is less stable than C=C bond.) So, M-SiR3 complexes are much more mumerous than M-CR3.M-SnR3 are also known.SnCl3: high trans effect catalytically active.(can help labilize other ligands, so create site for substrate binding.

27、)Can LnM-SiCH2R undergo b-H elimination?Yes.In general, M-SiR3 has properties similar to M-CR3.Comparison between M-CR3 and M-OR, M-NR2, M-F, M-Cl, etc. (1) Difference between M-CR3 and M-X M-CR3M-XM-X no dp-pp interactionpossible dp-pp interaction -OR: alkoxy or alkoxyl-NH2: amidoM-X bond is destab

28、ilized for 18e speciesM-X bond is stabilized for 16e or less speciesM-NR2 species (16e or less):M-NR2 species (18e):M-X can stabilize complexes with less than 18e. e.g. Consequences:Why?Bu: butanep bonding can make these complexes to formally have 18e. (2) Similarity between M-CR3 and M-OR: Both can

29、 undergo b-H elimination.Olefin: only with one C=C bondAlkene: with one or more C=C bondsCH2=CH2: ethyleneCH3CH2=CH2:propeneRHC=O: aldehydeH2C=O:formaldehydeR2C=O: ketoneUse of the reaction.C. M-BR2 complexes5. Metal hydrides, M-H (1955-1964)A. Structural types of metal hydrides.(1) Terminal hydride

30、 or classical hydride, e.g.(2) Non-classical hydride or molecular hydrogen complexes (1984)Q1. Would you expect the H2 molecule in the following complex can rotate freely? why? (see next page.)Consider the bonding between M and H2. Rotation will break the p bond between M and H2Q2. LnMH2 can have tw

31、o forms, depending on M and L. Rationalize the structural difference in the following pairs.Electron-donating; electron-withdrawingIn terms of bonding:For (a), Os has stronger backdonation ability than (b).For (b), Et is more electron-releasing than Ph, leading to stronger back donation.(3) Hydrogen

32、-bonding type interaction: H-X s complexesOften the interaction can be represented as:Note: Compared with the cases below:For example, agosticagostic(4) Bridging hydrides Examples:B. Synthesis of metal hydride complexes. (1) By protonationBF4-: boron tetrafluoride anion (requir a basic metal complex)(2) From hydride donorsM-X + H- donor M-H + X-Typical H- donors: NaH, LiAlH4, NaBH4, LiBHEt3 .Examples. WCl6 + LiBHEt3 + PR3 WH6(PR3)3 (3) From H2 Examples.(4) From ligands Answer:C. Chemical properties of