Carbon monoxide, as a two electron donor ligand, coordinates to transition metals to form metal carb — Organic Chemistry Chemistry Question
Metal Carbonyl Compounds
Carbon monoxide, as a two electron donor ligand, coordinates to transition metals to form metal carbonyl compounds. For example, iron forms the pentacarbonyl metal complex, Fe(CO)5. Nickel tetracarbonyl, Ni(CO)4, has been used for the purification of Ni metal in the Mond process. Electron counts of these metal carbonyl complexes show that they obey the 18-electron rule. Cobalt and manganese react with CO to form dinuclear complexes Co2(CO)8 and Mn2(CO)10, respectively. (Electronic configuration of Mn is [Ar](3d)5(4s)2) A metal–metal bond between the metal centres is considered essential in order for the compounds to obey the 18 electron rule. The cyclopentadienyl anion C5H5– has also been widely used as a η5–ligand. For example, ferrocene (C5H5)2Fe, a classical compound, obeys the 18 electron rule.
The reaction of W(CO)6 with sodium cyclopentadienide NaC5H5 yields an air sensitive compound A. Oxidation of A with FeSO4 yields compound B. Compound A can also be prepared from the reaction of B with Na/Hg, a strong reducing agent. In the 1600 – 2300 cm–1 region of the IR spectrum, A shows absorption bands at 1744 and 1894 cm–1 and B absorption bands at 1904, and 2010 cm–1. Compound A is a strong nucleophile and a good starting material for the synthesis of organometallic compounds containing metal–carbon bonds. The reaction of A with propargyl bromide (HC≡CCH2Br) gives compound C containing a metal–carbon σ–bond. At room temperature compound C undergoes a transformation to yield compound D. The same chemical composition was found for compounds C and D. The chemical shifts (δ) of the CH2 and CH resonances and coupling constants JH–H of propargyl bromide, C and D in the respective 1H NMR spectra are listed in the following table.
1H NMR | HC≡CCH2Br | C | D
δ (CH2) | 3.86 | 1.90 | 4.16
δ (CH) | 2.51 | 1.99 | 5.49
JH–H (Hz) | 2.7 | 2.8 | 6.7
Explain the differences in the IR spectra of A and B.
Model Answer
Compound A is anionic, the absorption bands attributed to CO stretching appear at lower frequency because of stronger back donation of the anionic charge to the anti bonding orbital of CO thus weakening the CO bond. For the neutral species B, absorption bands appear at the higher frequency.
Draw chemical structures for A, B, C and D.
The transformation of C to D involves a migration of the metal on the propargyl ligand. If DC≡CCH2Br is used for the synthesis of C, draw the structures of C and D.