Organomettalic chemistry

Organomettalic chemistry

• 1.
  18 electron rule : How to count electrons The rule states that thermodynamically stable transition metal organometallic compounds  are formed when the sum of the metal  d electrons and the electrons  conventionally considered as being supplied by the surrounding ligands  equals 18.  In general, the conditions favoring adherence to the 18 electron rule are, an electron rich  metal  (one that is in a low oxidation state) and ligands that are good π‐acceptors  The hapto symbol, η,with a numerical superscript, provides a topological description  by indicating the connectivity between the ligand and the central atom. For example,  if all the five carbon atoms of a cyclopentadienyl moiety are equidistant from a metal  atom, we term it as η5‐cyclopentadienyl The symbol μ indicates bridging;  normally we have μ2 and rarely  μ3 bridging Examples:  μ2‐CO, μ3‐CO, μ2‐CH3, μ2‐H, μ2‐Cl, , μ3‐Cl,  μ2‐OR, μ2‐PR2, μ2‐NR2 Examples: η1‐R, η1‐Ar  η2‐C2R4 η1‐allyl,   η3‐allyl, η4‐ Cb, η5‐Cp, η6‐C6H6 η8‐C8H8 η2‐C60, η5‐ R5C60.
• 2.
  Ligand Neutral atom Oxidation stateLigand Neutral atom Oxidation state Electron contributi on Formal charge Electron contribu tion Formal charge Carbonyl (M–CO) 2 2 0 Halogen ( M–X) 1 2 –1 Phosphine (M–PR3) 2 2 0 Alkyl (M–R) 1 2 –1 Amine (M–NR3 ) 2 2 0 Aryl (M–Ar) 1 2 –1 Amide (M–NR2 ) 1 2 –1 acyl (M–C(O)–R 1 2 –1 Hydrogen (M–H) 1 2 –1 η1-cyclopentadienyl 1 2 –1 Alkene (sidewise) η2- 2 2 0 η1-allyl 1 2 –1 Alkyne (sidewise) η2- 2 2 0 η3-allyl 3 4 –1 η2-C60 2 2 0 η5-cyclopentadienyl 5 6 –1 Nitrosyl bent 1 2 –1 η6-benzene 6 6 0 Nitrosyl linear 3 2 +1 η7-cycloheptatrienyl 7 6 +1 Carbene (M=CR2) 2 4 –2 Carbyne (M≡CR) 3 6 –3 Alkoxide (M–OR) 1 2 –1 Thiolate (M–SR) 1 2 –1 μ-CO (M–(CO)–M) 2 2 0 μ-H 1 2 –1 μ-alkyne 4 4 0 μ-X (M–X–M) X = halogen 3 4 –1 μ-alkyl 1 2 –1 μ-amido (M–(NR2)–M 3 4 –1 μ-phosphido (M–(PR2)–M 3 4 –1 μ-alkoxide (M–(OR)–M 3 4 –1 Methods of counting: Neutral atom method  & Oxidation  state method 
• 3.
  Ru CO PPh3 PPh3 neutral atom method oxidation state method Ru8 6 (Ru +2) η3 - allyl 3 4 2 PPh3 4 4 CO 2 2 charge -1 16 16 not required Fe N Me Me Fe 8 6 (Fe +2) 2 η5 -Cp 10 12 18 18 Neutral atom method: Metal is taken as in zero oxidation state for counting purpose Oxidation state method: We first arrive at the oxidation state of the metal by considering the  number of anionic ligands present and overall charge of the complex Suggestion: Focus on one counting method till you are confident
• 4.
  Easy way to remember ligand electron contribution for neutral atom counting method Electron contribution Neutral terminal :   CO, PR3, NR32 electrons Anionic terminal :  X‐, H‐, R‐, Ar‐, R2N‐, R2P‐, RO‐ 1 electron Hapto ligands  : η2‐C2R4 η2‐C2R2, η4‐C2R2 ,η1‐allyl,  η3‐allyl, η4‐ Cb, η5‐Cp, η6‐C6H6 η7‐C7H7 η8‐C8H8 η2‐C60, η5‐R5C60 same as hapticity bridging neutral μ2‐CO, μ3‐CO 2 electrons Bridging anionic μ2‐CH3, μ2‐H ( no lone pairs) 1 electron Bridging anionic μ2‐Cl, , μ2 ‐OR, μ2‐PR2, μ2‐NR2 3 electrons (with 1 lone pair)  μ3‐Cl( 2 l.p) 5 electrons  Bridging alkyne 4 electrons NO linear 3 electrons NO bent ( l. p on nitrogen) 1 electron Carbene M=C 2 electron Carbyne M≡C 3 electron 
• 5.
  Determine the total valence electrons (TVE) in the entire molecule (that is, the number of valence  electrons of the metal plus the number of electrons from each ligand and the charge); say, it is A. Subtract this number from n ×18 where n is the number of metals in the complex, that is, (n × 18) – A;  say, it is B. (a) B divided by 2 gives the total number of M–M bonds in the complex. (b) A divided by n gives the number of electrons per metal. If the number of electrons is 18, it indicates  that there is no M–M bond; if it is 17 electrons, it indicates that there is 1 M–M bond; if it is 16  electrons, it indicates that there are 2 M–M bonds and so on.  How to determine the total number of metal ‐ metal bonds  Fe Fe Fe Co Co Co Co Molecule TVE (A) (18 × n) – A (B) Total M–M bonds (B/2) Bonds per metal Basic geometry of metal atoms Fe3(CO)12 48 54 – 48 = 6 6/2 = 3 48/3 = 16; 2 Co4(CO)12 60 72 – 60 = 12 12/2 = 6 60/4 = 15; 3 [η5-CpMo(CO)2]2 30 36 – 30 = 6 6/2 = 3 30/2 = 15; 3 Mo≡Mo (η4-C4H4)2Fe2(CO)3 30 36 – 30 = 6 6/2 = 3 30/2 = 15; 3 Fe≡Fe Fe2(CO)9 34 36 – 34 = 2 2/2 = 1 34/2 = 16; 1 Fe–Fe
• 6.
  The following organometallic compounds are stable and  has a  second  row  transition  metal at its centre. Find out the  metal and  its oxidation state Problem solving
• 7.
  A few worked out examples Understanding  metal –metal bond electron count become easier if you compare  and  see how octet is attained by each Cl atom of Cl2
• 8.
  • Square planar organometallic complexes of the late transition  metals (16e).  •Some organometallic complexes of the early transition metals (e.g. Cp2TiCl2, WMe6, Me2NbCl3, CpWOCl3) [ A possible reason for the same is that some of the orbitals of these complexes are too high in energy for effective utilization in bonding or the ligands are mostly σ donors.] • Some high valent d0 complexes have a lower electron count  than 18. • Sterically demanding bulky ligands force complexes to have  less than 18 electrons.  • The 18 electron rule fails when bonding of organometallic clusters of  moderate to big sizes (6 Metal atoms and above) are considered.  • The rule is not applicable to organometallic compounds of  main  group metals as well as to those of lanthanide and actinide metals.  Exceptions to the 18 electron rule
• 9.
  CO CO Ni OC CO CO OC Fe CO CO CO CO Cr OC OC CO CO CO CO Mn OC OC CO CO OC Mn OC CO CO CO OC Co OC OC OC Co O C CO CO CO C O Os OsOs OC OC CO CO OC CO CO CO CO CO CO Ir Ir Ir Ir OC OC OC OC CO CO CO CO CO CO OC OC Coordination number around the metal normally remains six or lesser.  17 electron  species such as  Mn(CO)5, Co(CO)4 dimerize to gain 18 electrons V(CO)6 does not dimerize.  Metal carbonyls OM  L2‐S1
• 10.
  Why study metal carbonyls ? Simplest of organometallic compounds where M‐C σ bonding is well understood.  CO is one  of the strongest  πacceptor ligands.  Back bonding       (π bonding) and variation in electronic  properties of CO can be monitored very efficiently by Infrared spectroscopy A range of metal carbonyls are used as catalysts in Chemical Industry Hydroformylation Alkene to Aldehyde Methanol to Acetic acid  Process MeOH + HI MeI + H2O MeI CO [Rh(CO)2I2] C O I H3C C O I H3C H2O C O OH H3C C CH2 R H CO, H2 CH CH2 HC R O H HCo(CO)4 AJELIAS L2‐S2
• 11.
  2s 2s 2p 2p CO O C HOMO LUMO σ σ σ π π * * σ * 32.4ev 19.5 ev 15.9 ev 10.7 ev The highest occupied molecular orbital (HOMO) of CO is weakly antibonding (compared with the O atomic orbitals) and is an MO which is carbon based. Secondly, the π* antibonding orbital which is the lowest unoccupied molecular orbital (LUMO) is also of comparatively lower energy which makes it possible to interact with metal t2g orbitals for π bonding. There exists a strong back bonding of metal electrons to the π * antibonding orbitals of CO Molecular Orbital diagram of CO Why does  CO bind a metal through its less  electronegative carbon atom  than its more  electronegative oxygen ? What makes it a  good π acceptor ? AJELIAS L2‐S3
• 12.
  Counting the electrons helps  to predict stability of metal carbonyls. But it will not  tell you whether a CO is bridging or terminal  AJELIAS L2‐S4
• 13.
  Infrared (IR) spectroscopyis one of the most common spectroscopic techniques used by organic and inorganic chemists. Simply, it is the absorption measurement of different IR frequencies by a compound positioned in the path of an IR beam. The main goal of IR spectroscopic analysis is to determine the chemical functional groups in the sample. Functional groups are identified based on vibrational modes of the groups such a stretching, bending etc. Different vibrational modes absorb characteristic frequencies of IR radiation. An infrared spectrophotometer is an instrument that passes infrared light through a molecule and produces a spectrum that contains a plot of the amount of light transmitted on the vertical axis against the wavelength of infrared radiation on the horizontal axis. Absorption of radiation lowers the percentage transmittance value. Infrared Spectroscopy‐ A spectro‐analytical tool in chemistry  AJELIAS L2‐S5
• 14.
  Infrared Spectroscopy‐ Spectra of Metal Carbonyls   Mn OC OC CO CO OC Mn OC CO CO CO OC Fe OC OC OC Fe O C CO CO CO C O O C AJELIAS L2‐S6 The range in which  the band appears  decides bridging or  terminal . The number of  bands is only  related to the  symmetry of the  molecule bridging terminal terminal