Reactions and Mechanism Flashcards
The reaction of cyclooctatetraene with iron pentacarbonyl is shown below. All products A, B and C are 18-electron complexes; all of the complexes have IR bands in the region 1900 – 1980 cm-1. Clearly illustrate the structures of A, B and C (Hint: A and B belong to
the Cs point group while C belongs to C2h).
A= 3 CO’s in 3-legged piano stool. C8H8 has hapticity of 4, shape like a pringle chip
B= Remaining 2 CO and C8H* with hapticity of 6
C= two Fe with 3 CO (3-legged piano stool) bonding to same C8H8 with hapticity of 4 each.
- When Mn2(CO)10 is reacted with excess sodium-amalgam (Na/Hg) in THF, a mononuclear salt of the formula C5MnNaO5 (A) is produced; the IR spectrum of A shows a strong CO band at 1870 cm-1. In THF solution, this salt reacts with 3-chloropropene to generate a neutral monomeric complex of the formula C8H5MnO5 (B) with a CO band at 1930 cm-1. When B is heated at 80C or photolyzed at 25C, a new complex C of formula C7H5MnO4 can be isolated. Addition of PMe3 to complex C results in the formation of D having the composition C10H14MnO4P. Clearly illustrate the structures of compounds A – D and show that they all obey the 18-electron rule.
A metal M forms a carbonyl complex Mx(CO)y that contains 18.40% of carbon and in its mass spectrum exhibits a molecular ion, the isotopic pattern of which shows its most intense peak at m/z = 652. The carbonyl complex reacts with triphenylphosphine to produce two compounds; A, which has a metal-phosphorus ratio of 1:1, and B, for which the corresponding ratio is 2:1. A reacts with bromine to form a single product C, whereas B with bromine produces both C and D. D, which is identical with the sole product from the reaction of Mx(CO)y with bromine, reacts with triphenylphosphine to produce E, which is isomeric with C. C and D, but not E, have a C4 axis of symmetry. All the compounds are diamagnetic. Draw conclusions from each piece of evidence, deduce the identity of M, the values of x and y and the structures of A, B, C, D and E.
Heating [(η5-C5H5)Fe(CO)3]+ with NaH in solution affords organometallic compound A which has a room-temperature 1H NMR spectrum consisting of two singlets of relative intensity 1:5 at approximately δ = -12 ppm and δ = 5 ppm, respectively.
Compound A is thermally unstable under ambient conditions and rapidly evolves a colourless gas B and forms a purple-brown solid C having the empirical formula C7H5O2Fe. The Nujol-mull IR spectrum of solid C exhibits strong absorptions at 1961, 1942, and 1790 cm-1.
Treatment of C with an equimolar amount of iodine generates a brown solid D which is stable in air and is soluble in polar organic solvents. Its Nujol-mull IR spectrum exhibits strong absorptions at 2042 and 1992 cm-1.
Reaction of D with an equimolar amount of CH3MgI in diethyl ether affords caramel-coloured crystals of E. Compound E is unstable in air, but is soluble in common organic solvents. Its Nujol-mull IR spectrum exhibits strong absorptions at 2010 and 1965 cm-1, and its 1H NMR spectrum consists of two singlets of relative intensity 3:5 at approximately δ = 0 ppm and δ = 4 ppm.
Reaction of D with an equimolar amount of NaC5H5 produces an orange solid F with the formula C12H10O2Fe. Compound F is fairly stable in air, and its 1H NMR spectrum at 30 ºC consists of two equal intensity singlets at approximately δ = 4.5 ppm and δ = 5.7 ppm.
On heating, F evolves a colourless gas and converts to an orange solid G that is stable in air and is soluble in most organic solvents. Its 1H NMR spectrum at all temperatures consists of a single sharp resonance at approximately δ = 4 ppm.
Identify compounds A through G and sketch their molecular structures.
Identify and sketch in the frames provided the molecular structures of compounds A,
B, and C which are three different types of organometallic complexes that could
result from the reaction of (η2-H2C=CH2)Re(CO)4Cl with one equivalent of NaBH4
Name the reaction and Draw the structure of the bolded compound
Ni + 4 CO → Ni(CO)4
Adduct formation (addition reaction)
Name the reaction and Draw the structure of the bolded compound
Mn2(CO)10 + 2 Na → 2 Na[Mn(CO)5]
Reduction
Name the reaction and Draw the structure of the bolded compound
W(CO)6 refluxed in CH3CN → (CH3CN)3W(CO)3 + 3 CO
ligand substitution
Name the reaction and Draw the structure of the bolded compound
H2 + Ir(CO)(PPh3)2Cl → Ir(H)2(CO)(PPh3)2Cl
oxidative addition
Draw the molecular structures for the expected organometallic product(s) in each of the following reactions. The equations all indicate the stoichiometry
trans-Ir(PPh3)2(CO)Cl + O2 →
→
Draw the molecular structures for the expected organometallic product(s) in each of the following reactions. The equations all indicate the stoichiometry
W(CO)5[C(C6H5)(OC2H5)] + BBr3 →
Draw the molecular structures for the expected organometallic product(s) in each of the following reactions. The equations all indicate the stoichiometry
Na2[Fe(CO)4] + MeC(O)Cl →
Draw the molecular structures for the expected organometallic product(s) in each of the following reactions. The equations all indicate the stoichiometry
3 CrCl3 + 3 Al + 6 C6H6 →
Draw the molecular structures for the expected organometallic product(s) in each of the following reactions. The equations all indicate the stoichiometry
Cp*W(NO)(η3-CH2CHCHMe)(CH2CMe3) + n-hexane →
Draw the product for A-J. All obey the 18 electron rule
Draw all structures for A-I
Draw all unknown structure. Note, [] means solvent/reagent
Fe(CO)5 =[2Na in THF]=> A
A =[HCl in Et2O]=> B.
B =[Ph3SnCl in THF]=> C
A= Na2[Fe(CO)4] with the dianion having Td symmetry
B= Na[HFe(CO)4] with the monoanion having a “trigonal
bipyramidal” structure.
Neutral
C= Neutral (OC)4Fe(H)(SnPh3) having an “octahedral”
structure.
Fe(CO)5 =[reflux in C5H5]=> D
D=[2 Na in THF]=> E
E=[2CH2=CHCH2Br in THF]=> F
F =[hv, -CO in pentane]=> G
D= [CpFe(CO)2]2 having the molecular structures
E= 2 Na[CpFe(CO)2] with the anion having a “twolegged
piano-stool” molecular structure.
F= 2 CpFe(CO)2(η1-CH2CH=CH2) having the “threelegged
piano-stool” molecular structure
G = 2 CpFe(CO)(η3-C3H5) having the “three-legged
piano-stool” molecular structure:
Fe(CO)5 =[Cl2 in CCl4]=> H
H=[2 MeMgBr in Et2O]=> I
I =[CO in pentane]=>J
J=[warm in pentane]=> K (trimetallic compound) and L (organic compound)
H= Fe(CO)4Cl2 with an
“octahedral” structure.
I= Fe(CO)4Me2 with an
“octahedral” structure.
J= Fe(CO)4[C(O)Me]Me with an
“octahedral” structure.
K is Fe3(CO)12 with the
structure as shown.
L is acetone, Me2C=O.
For each of the following sequences of reactions, identify and sketch the molecular structures of each of
the organometallic cobalt compounds A to K which all satisfy the 18-valence-electron rule. Note that
concomitant gas evolution is indicated for two of the conversions
A= TBP Co(CO)4Br
B= two-legged piano stool (η5-C5H5)Co(CO)2
C=two-legged piano stool (η5-C5H5)Co(CO)[C(=OLi)Et]
D= two-legged piano stool (η5-C5H5)Co(CO)[C(=OEt)Et], Fischer carbene complex
E = two-legged piano stool (η5-C5H5)Co(Br)[C≡Et], Fischer carbyne complex
F= two-legged piano stool (η5-C5H5)Co(CO)[=C(NHR)Et], Fischer carbene complex
G = Na[Co(CO)4] in which the anion has a tetrahedral structure
H is TBP Co(CO)4H
I is TBP Co(CO)4Et
J is TBP (OC)3Co(PPh3)[C(=O)Et]
K is bimetallic (OC)4Co-Mn(CO)5 with the geometry being TBP at Co and Oh at Mn
Complete each of the following chemical transformations and identify the reaction type
involved. Note specifically the initial and final oxidation states of any element
undergoing a redox reaction during a particular transformation. Sketch the molecular
structures of the product complexes.
Re2(CO)10 + 2 Na →
2 Na[Re(CO)5]
Reaction type: Reduction
Molecular structure: The [Re(CO)5]- anion has TBP structure