Freie Universität Berlin, Institut für Chemie und Biochemie
Coordination Chemistry
Andrey Petrov (AG Müller, F208) 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. a) For acetylacetonate as the ligand draw the two enantiomers of a complex M(acac)3. 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. a) For acetylacetonate as the ligand draw the two enantiomers of a complex M(acac)3. 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2]
3- [Cr(ox)3]
2+ trans-[PtCl2(en)2]
2+ [Ni(phen)3]
- [RuBr4(phen)]
+ cis-[RuCl(py)(phen)2] 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2]
3- [Cr(ox)3]
2+ trans-[PtCl2(en)2] No plane symmetry chiral
2+ [Ni(phen)3]
- [RuBr4(phen)]
+ cis-[RuCl(py)(phen)2] 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2] No plane symmetry chiral
3- [Cr(ox)3]
2+ trans-[PtCl2(en)2]
2+ [Ni(phen)3]
- [RuBr4(phen)]
+ cis-[RuCl(py)(phen)2] 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2]
3- has a plane symmetry [Cr(ox)3]
2+ trans-[PtCl2(en)2]
2+ [Ni(phen)3]
- [RuBr4(phen)]
+ cis-[RuCl(py)(phen)2] 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2]
3- [Cr(ox)3]
2+ trans-[PtCl2(en)2]
2+ [Ni(phen)3]
- [RuBr4(phen)]
+ cis-[RuCl(py)(phen)2] 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2]
3- [Cr(ox)3]
2+ trans-[PtCl2(en)2]
2+ [Ni(phen)3]
- [RuBr4(phen)] has a plane symmetry
+ cis-[RuCl(py)(phen)2] 1) Octahedral transition metal complexes can be chiral, although the corresponding ligands are achiral. b) Which of the following octahedral complexes are chiral:
+ cis-[CoCl2(en)2]
3- [Cr(ox)3]
2+ trans-[PtCl2(en)2]
2+ [Ni(phen)3]
- [RuBr4(phen)]
+ cis-[RuCl(py)(phen)2] no plane symmetry 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6]
2- (b) [Mn(CN)6]
2+ (c) [Cr(en)3]
3- (d) [Fe(ox)3]
2- (e) [Pd(CN)4]
2- (f) [CoCl4]
2- (g) [NiBr4] 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
3- (d) [Fe(ox)3] (5)
2- (e) [Pd(CN)4] (0)
2- (f) [CoCl4] (3)
2- (g) [NiBr4] (2).
Octahedral complex, CN- is a strong field ligand 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
3- (d) [Fe(ox)3] (5)
2- (e) [Pd(CN)4] (0)
2- (f) [CoCl4] (3)
2- (g) [NiBr4] (2).
Octahedral complex, CN- is a strong field ligand 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
3- (d) [Fe(ox)3] (5)
2- (e) [Pd(CN)4] (0)
2- (f) [CoCl4] (3)
2- (g) [NiBr4] (2).
Octahedral complex, en not so strong, low charge on Cr, 1st row TM 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
(d) [Fe(ox)3]3- (5)
2- (e) [Pd(CN)4] (0)
2- (f) [CoCl4] (3)
2- (g) [NiBr4] (2).
Octahedral complex, oxalate is weak, 1st row TM 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
(d) [Fe(ox)3]3- (5)
(e) [Pd(CN)4]2- (0)
2- (f) [CoCl4] (3)
2- (g) [NiBr4] (2).
Square planar complex 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
(d) [Fe(ox)3]3- (5)
(e) [Pd(CN)4]2- (0)
(f) [CoCl4]2- (3)
2- (g) [NiBr4] (2).
Tetrahedral complex 2) In each of the following complexes, rationalize the number of observed unpaired electrons (stated after the formula):
4- (a) [Mn(CN)6] (1)
2- (b) [Mn(CN)6] (3)
2+ (c) [Cr(en)3] (4)
(d) [Fe(ox)3]3- (5)
(e) [Pd(CN)4]2- (0)
2- (f) [CoCl4] (3)
(g) [NiBr4]2- (2).
Tetrahedral complex 3) Discuss the factors that contribute to the preference for forming either a high- or a low-spin d4 complex. How would you distinguish experimentally between the two configurations? 3) Discuss the factors that contribute to the preference for forming either a high- or a low-spin d4 complex. How would you distinguish experimentally between the two configurations?
e eg g
t2g t2g
In general: if the energy required to pair two electrons is greater than the energy
cost of placing an electron in an eg, Δ, high spin splitting occurs.
P> Δ or P< Δ
• Period where the metal is • Charge on the metal (Fe2+ and Co3+) • Ligand field
Distinguish experimentally: with a magnetic susceptibility balance, EPR 4) What is the expected ordering of values of Doct for
2+ [Fe(H2O)6]
3- [Fe(CN)6]
4- [Fe(CN)6] ? 4) What is the expected ordering of values of Doct for
2+ rd [Fe(H2O)6] 3
3- [Fe(CN)6]
4- [Fe(CN)6] ? 4) What is the expected ordering of values of Doct for
2+ rd [Fe(H2O)6] 3
3- st [Fe(CN)6] 1
4- nd [Fe(CN)6] ? 2
Higher oxidation state leads to higher Δ 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6]
2- (b) [FeCl4]
(c) [CoCl3(py)3]
2+ (d) [Ni(en)3]
3+ (e) [Ti(H2O)6]
3- (f) [VCl6]
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4]
(c) [CoCl3(py)3]
2+ (d) [Ni(en)3]
3+ (e) [Ti(H2O)6]
3- (f) [VCl6]
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4] 2+, d6
(c) [CoCl3(py)3]
2+ (d) [Ni(en)3]
3+ (e) [Ti(H2O)6]
3- (f) [VCl6]
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4] 2+, d6
(c) [CoCl3(py)3] 3+, d6
2+ (d) [Ni(en)3]
3+ (e) [Ti(H2O)6]
3- (f) [VCl6]
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4] 2+, d6
(c) [CoCl3(py)3] 3+, d6
2+ (d) [Ni(en)3] 2+, d8
3+ (e) [Ti(H2O)6]
3- (f) [VCl6]
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4] 2+, d6
(c) [CoCl3(py)3] 3+, d6
2+ (d) [Ni(en)3] 2+, d8
3+ (e) [Ti(H2O)6] 3+, d1
3- (f) [VCl6]
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4] 2+, d6
(c) [CoCl3(py)3] 3+, d6
2+ (d) [Ni(en)3] 2+, d8
3+ (e) [Ti(H2O)6] 3+, d1
3- (f) [VCl6] 3+, d2
(g) [Cr(acac)3] 5) For each of the following complexes, give the oxidation state of the metal and its dn configuration:
4- (a) [Mn(CN)6] 2+, d5
2- (b) [FeCl4] 2+, d6
(c) [CoCl3(py)3] 3+, d6
2+ (d) [Ni(en)3] 2+, d8
3+ (e) [Ti(H2O)6] 3+, d1
3- (f) [VCl6] 3+, d2
(g) [Cr(acac)3] 3+, d3 6) Fill in the gaps
Ligand Structure (Metal = M) e-count of the ligand
h4-cyclobutadiene
h5-cyclopentadienyl
h6-arene
Hydride
Carbonyl
Alkyl
Alkene
Cyanide 6) Fill in the gaps:
η4-cyclobutadiene: 4 electron donor
. . - η5-cyclopentadienyl: 6 electron donor, anionic 6) Fill in the gaps:
η6-arene: 6 electron donor
carbonyl: 2 electron donor :C≡O
cyanide: 2 electron donor, anionic -C≡N 6) Fill in the gaps:
hydride: 2 electron donor, anionic H-
alkyl: 2 electron donor, anionic R-
alkene: 2 electron donor 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom. 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Fe(II) 6 + 2x6 = 18 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Fe(II) Ni(0) 6 + 2x6 = 18 10 + 4x2 = 18 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Fe(II) Ni(0) Rh(I) 6 + 2x6 = 18 10 + 4x2 = 18 8 + 4x2 = 16 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Fe(II) Ni(0) Rh(I) 6 + 2x6 = 18 10 + 4x2 = 18 8 + 4x2 = 16
Co(0) 9 + 3x2 + 3x1 = 18 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Fe(II) Ni(0) Rh(I) 6 + 2x6 = 18 10 + 4x2 = 18 8 + 4x2 = 16
Co(0) Ni(0) 9 + 3x2 + 3x1 = 18 10 + 3x2 = 16 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom. 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Ti(IV) 0 + 2x6 + 2 = 14 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Ti(IV) Co(I) 0 + 2x6 + 2 = 14 8 + 6 + 2x2 = 18 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Ti(IV) Co(I) Pt(II) 0 + 2x6 + 2 = 14 8 + 6 + 2x2 = 18 8 + 2x4 = 16 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Ti(IV) Co(I) Pt(II) 0 + 2x6 + 2 = 14 8 + 6 + 2x2 = 18 8 + 2x4 = 16
Ir(I) 8 + 2 + 2x2 + 2 = 16 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Ti(IV) Co(I) Pt(II) 0 + 2x6 + 2 = 14 8 + 6 + 2x2 = 18 8 + 2x4 = 16
Ir(I) Fe(0) 8 + 2 + 2x2 + 2 = 16 8 + 3x2 + 2x2 = 18 9a) Consider the following equilibrium:
CoBr2(CO)L2 CoBr2L2 + CO
The equilibrium constants (relative values) for the reaction with various phosphine ligands L are given in the table below. Explain the trends in Kd.
-1 L Kd νCO (cm )
PEt3 1 1985
PEt2Ph 2.5 1990
PPh3 750 1995 8a) Consider the following equilibrium:
CoBr2(CO)L2 CoBr2L2 + CO
The equilibrium constants (relative values) for the reaction with various phosphine ligands L are given in the table below. Explain the trends in Kd.
-1 L Kd νCO (cm )
PEt3 1 1985
PEt2Ph 2.5 1990
PPh3 750 1995
PPh3 is a weaker electron-donating ligand than PEt3 (you can see it from the IR as well). Hence there is less electron density on the metal, so less back-donation to C-O and therefore a weaker M-C bond, so the dissociation is favoured. 8a) Consider the following equilibrium:
CoBr2(CO)L2 CoBr2L2 + CO
The equilibrium constants (relative values) for the reaction with various phosphine ligands L are given in the table below. Explain the trends in Kd.
-1 L Kd νCO (cm )
PEt3 1 1985
PEt2Ph 2.5 1990
PPh3 750 1995
PPh3 is a weaker electron-donating ligand than PEt3 (you can see it from the IR as well). Hence there is less electron density on the metal, so less back-donation to C-O and therefore a weaker M-C bond, so the dissociation is favoured.
Moreover PPh3 is more bulky than PEt3 ( cone angle is 145° vs 132°), so the more bulky the ligands the easier it is to stabilize a complex with a lower coordination number. 8b) The CO molecule can be regarded as a harmonic oscillator. Below is given the vibration equation for a harmonic oscillator:
1 k m m = = 1 2 2 m1 + m2
13 How does the frequency νCO of the C-O vibration change in the IR spectrum, if CO instead of unlabeld CO is used? 8b) The CO molecule can be regarded as a harmonic oscillator. Below is given the vibration equation for a harmonic oscillator:
1 k m m = = 1 2 2 m1 + m2
13 How does the frequency νCO of the C-O vibration change in the IR spectrum, if CO instead of unlabeld CO is used?
For 12C, μ = 6.85
For 13C, μ = 7.17
13 → The frequency νCO of the C-O vibration in case of CO is lower. 8c) Compare NO+ (nitrosyl cation) with CO as a ligand in a transition metal complex. Is NO+ a stronger, similar or weaker π-acceptor ligand than CO? 8c) Compare NO+ (nitrosyl cation) with CO as a ligand in a transition metal complex. Is NO+ a stronger, similar or weaker π-acceptor ligand than CO?
NO+ and CO are isoelectronic, but one is charged. 8c) Compare NO+ (nitrosyl cation) with CO as a ligand in a transition metal complex. Is NO+ a stronger, similar or weaker π-acceptor ligand than CO? NO+ and CO are isoelectronic. N has a higher electronegativity than C: (χ(N) > χ(C))→ the atomic orbitals of N are lower in energy: E{2p(N)} < E{2p(C)}. O+ compared to O is positively charged → O+ has a higher electronegativity → atomic orbitals of O+ are lower in energy Linear combination of atomic orbitals yields the molecular orbitals. In this case the anti- bonding 2p π* molecular orbital is responsible for the π-accepting properties. Since the energy of the 2p(N) and 2p(O+) orbitals are lower than the 2p(C) and 2p(O), the corresponding 2p π* molecular orbital of NO+ is also lower in energy and therefor NO+ is a stronger π-acceptor ligand 9) Which effect on νCO has a) a negative charge at the metal center, b) a positive charge at the metal center? 9) Which effect on νCO has a) a negative charge at the metal center, b) a positive charge at the metal center? 9) Which effect on νCO has a) a negative charge at the metal center, b) a positive charge at the metal center?
A negative charge makes the metal center electron rich → increased back-donation to pi* of CO → weaker C-O bond → lower frequency
A positive charge makes the metal center electron poor → decreased back-donation to pi* of CO → stronger C-O bond → higher frequency 10) a) Electron withdrawing ligands L, b) electron donating ligands L have generally which effect on CO dissociation from the metal center? 10) a) Electron withdrawing ligands L, b) electron donating ligands L have generally which effect on CO dissociation from the metal center?
Me P M CO (MeO)3P M CO 3
I II 10) a) Electron withdrawing ligands L, b) electron donating ligands L have generally which effect on CO dissociation from the metal center?
Me P M CO (MeO)3P M CO 3
I II
Electron withdrawing ligands decrease back-donation to CO, so M-C is weaker.
Electron donating ligands increase back-donation to CO, so M-C is stronger. 11) Compare the two transition metal carbonyl complexes. Which one shows the higher
C-O stretching frequency νCO?
Me P M CO (MeO)3P M CO 3
I II 11) Compare the two transition metal carbonyl complexes. Which one shows the higher
C-O stretching frequency νCO?
Me P M CO (MeO)3P M CO 3
I II
I, because the phosphite is a pi-acceptor, while the phosphine is a sigma-donor.
The phosphine donates more to the metal, which then can do more pi-back-donation since is more electron-rich. 11) Compare the two transition metal carbonyl complexes. Which one shows the higher
C-O stretching frequency νCO?
Me P M CO (MeO)3P M CO 3
I II
Compared to phosphites (e.g. P(OMe)3), phosphine ligands, such as PMe3, dramatically slow down CO dissociation from the metal center. Why? 11) Compare the two transition metal carbonyl complexes. Which one shows the higher
C-O stretching frequency νCO?
Me P M CO (MeO)3P M CO 3
I II
Compared to phosphites (e.g. P(OMe)3), phosphine ligands, such as PMe3, dramatically slow down CO dissociation from the metal center. Why?
Same as before:
Electron withdrawing ligands decrease back-donation to CO, so M-C is weaker.
Electron donating ligands increase back-donation to CO, so M-C is stronger. 8) How does the stretching frequency νCO change in going from situation 1→ 3? 8) How does the stretching frequency νCO change in going from situation 1→ 3?
The stretching frequency is proportional to the strenght of the C-O bond.
The higher the frequency, the stronger the bond.
Roughly 2100 - 1700 -1200 for triple, double and single C-O bonds respectively (in cm-1). 8) How does the stretching frequency νCO change in going from situation 1→ 3?
Back-donation means filling pi* orbitals, making the C-O bond weaker, so lower in IR. Freie Universität Berlin, Institut für Chemie und Biochemie
Coordination Chemistry №2
Andrey Petrov (AG Müller, F208) 7) Count the number of valence electrons at each metal center. Select the oxidation state of the transition metal atom.
Fe(II) Ni(0) Rh(I) 6 + 2x6 = 18 10 + 4x2 = 18 8 + 4x2 = 16
Co(0) 9 + 3x2 + 3x1 = 18 1) Why is the photochemical CO-substitution reaction a common synthetic method for the introduction of a ligand into a metal complex? Give a reason by means of a simplified molecular orbital scheme. 1) Why is the photochemical CO-substitution reaction a common synthetic method for the introduction of a ligand into a metal complex? Give a reason by means of a simplified molecular orbital scheme.
• Easy (UV light)
• Effective (also weak ligands)
• Clean (CO leaves the solution)
• no other chemicals involved 1) Why is the photochemical CO-substitution reaction a common synthetic method for the introduction of a ligand into a metal complex? Give a reason by means of a simplified molecular orbital scheme.
• Easy (UV light)
• Effective (also weak ligands)
• Clean (CO leaves the solution)
• no other chemicals involved 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest.
Define cone angle. How is it defined for equal substituents, and for unequal substituents?
Chem. Rev. 1977, 77, 313. 3) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest.
Define cone angle θ. How is it defined for equal substituents, and for unequal substituents? is a measure of the steric bulk of a ligand in a transition metal complex. It is defined as the solid angle formed with the metal at the vertex and the outermost edge of the van der Waals spheres of the ligand atoms at the θ perimeter of the cone
Chem. Rev. 1977, 77, 313. 3) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest.
How is it defined for unequal substituents?
θ1 θ2 θ3
R3 R1 R2
Chem. Rev. 1977, 77, 313. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. a) Which phosphine has the largest Tolman cone angle?
PF3 PPh3 P(C6H11)3
Chem. Rev. 1977, 77, 313. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. a) Which phosphine has the largest Tolman cone angle?
PF3 PPh3 P(C6H11)3
104 143 170
Chem. Rev. 1977, 77, 313. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. a) Which phosphine has the largest Tolman cone angle?
PF3 PPh3 P(C6H11)3
104 143 170
b) Give an example of a phosphine having a larger cone angle than P(CH2CH3)3.
Chem. Rev. 1977, 77, 313. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest.
a) Which phosphine has the largest Tolman cone angle?
PF3 PPh3 P(C6H11)3
104 143 170
b) Give an example of a phosphine having a larger cone angle than P(CH2CH3)3. t P(CH2CH3)3 P( Bu)3 132 182
Chem. Rev. 1977, 77, 313. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest.
The χ parameter is the difference between the trans-CO stretching frequency of t-
Bu3PNi(CO)3 (t-Bu3P is assigned a χ value of 0) and that of the nickel carbonyl complex of the phosphine of interest.
A1 = Value of the stretching of the A1 band
χ = A1 - 2056.1 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest.
The χ parameter is the difference between the trans-CO stretching frequency of t-
Bu3PNi(CO)3 (t-Bu3P is assigned a χ value of 0) and that of the nickel carbonyl complex of the phosphine of interest.
ν(CO) L χ cm−1
P(t-Bu)3 2056.1 0
PMe3 2064.1 8
PPh3 2068.9 13 A1 = Value of the stretching of the A1 band P(OEt)3 2076.3 20
PCl3 2097.0 41
χ = A1 - 2056.1 PF3 2110.8 55 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. c) Provide a detailed description of how the IR stretching frequency of the CO ligand can be used to determine the electron donating ability of a phosphine. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. c) Provide a detailed description of how the IR stretching frequency of the CO ligand can be used to determine the electron donating ability of a phosphine.
see last exercise… 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. d) Describe whether increasing electron donation by the phosphine increases or decreases the CO stretching frequency, and why this change occurs. Your answer should discuss specific orbital interactions that are responsible for changes in the CO bond strength due to the trans-phosphine ligand. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. d) Describe whether increasing electron donation by the phosphine increases or decreases the CO stretching frequency, and why this change occurs. Your answer should discuss specific orbital interactions that are responsible for changes in the CO bond strength due to the trans-phosphine ligand.
*
Stronger donor, more back-donation, lower frequency 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. e) If a phosphine is relatively electron donating, then what can you say about its electronic parameter? 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. e) If a phosphine is relatively electron donating, then what can you say about its electronic parameter?
Strong donor, high back-donation, weaker C-O bond, low-frequency, low χ. 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. e) If a phosphine is relatively electron donating, then what can you say about its electronic parameter?
Strong donor, high back-donation, weaker C-O bond, low-frequency, low χ. f) Which phosphine is the best σ-donor?
PMe3 PF3 PPh3 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. e) If a phosphine is relatively electron donating, then what can you say about its electronic parameter?
Strong donor, high back-donation, weaker C-O bond, low-frequency, low χ. f) Which phosphine is the best σ-donor?
PMe3 PF3 PPh3 8 55 13
see lecture… 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. e) If a phosphine is relatively electron donating, then what can you say about its electronic parameter?
Strong donor, high back-donation, weaker C-O bond, low-frequency, low χ. f) Which phosphine is the best σ-donor?
PMe3 PF3 PPh3 8 55 13 g) Explain how the cone angle influences an equilibrium of the following formula: 2) Phosphines are ubiquitous ligands in organometallic chemistry. In describing the steric and electronic features of a phosphine, there are two key parameters of interest. e) If a phosphine is relatively electron donating, then what can you say about its electronic parameter?
Strong donor, high back-donation, weaker C-O bond, low-frequency, low χ. f) Which phosphine is the best σ-donor?
PMe3 PF3 PPh3 8 55 13 g) Explain how the cone angle influences an equilibrium of the following formula:
Steric-demanding ligands (big cone angle) shift the equilibrium to the right 4) The Hard and Soft Acid and Base Principle is a simple concept introduced by Ralph Pearson in 1965. Briefly describe the HSAB idea and describe the main characteristics of the four general types of acids and bases: Hard Acid, Soft Acid, Hard Base, Soft Base. 4) The Hard and Soft Acid and Base Principle is a simple concept introduced by Ralph Pearson in 1965. Briefly describe the HSAB idea and describe the main characteristics of the four general types of acids and bases: Hard Acid, Soft Acid, Hard Base, Soft Base.
A Lewis acid is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct.
A Lewis base, then, is any species that has a filled orbital containing an electron pair which is not involved in bonding but may form a dative bond with a Lewis acid to form a Lewis adduct. 3) The Hard and Soft Acid and Base Principle is a simple concept introduced by Ralph Pearson in 1965. Briefly describe the HSAB idea and describe the main characteristics of the four general types of acids and bases: Hard Acid, Soft Acid, Hard Base, Soft Base.
Acid and base are following the Lewis definitions.
Hard Soft • small atomic/ionic radius • large atomic/ionic radius • high oxidation states (mainly to acids) • low oxidation states • high electronegativity (for bases). • low electronegativity • weakly polarizable • strongly polarizable
hard bases have HOMO of low energy, Soft bases have HOMO of higher and hard acids have LUMO of high energy. energy than hard bases, and soft acids The affinity of hard acids and hard bases have LUMO of lower energy than hard for each other is mainly ionic. acids.
Generally soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and form stronger bonds with hard bases,
4a) What is the Wilkinson Catalyst? Show that Wilkinson's catalyst obeys the 16-electron rule and explain why it does not require 18 electrons to be stable. 4a) What is the Wilkinson Catalyst? Show that Wilkinson's catalyst obeys the 16-electron rule and explain why it does not require 18 electrons to be stable.
Rh(I) 4d8 4a) What is the Wilkinson Catalyst? Show that Wilkinson's catalyst obeys the 16-electron rule and explain why it does not require 18 electrons to be stable.
Total e- = 8 + 2 + 2 + 2 + 2 = 16 e- d8, 4-coordinate metal = square planar
Energy gap between d and s,p orbitals increasing 4a) What is the Wilkinson Catalyst? Show that Wilkinson's catalyst obeys the 16-electron rule and explain why it does not require 18 electrons to be stable.
Total e- = 8 + 2 + 2 + 2 + 2 = 16 e- d8, 4-coordinate metal = square planar
4b) What species serves as the reducing agent in the synthesis of Wilkinson's catalyst?
RhCl3(H2O)3 + 4 PPh3 → RhCl(PPh3)3 + OPPh3 + 2 HCl + 2 H2O 8. Wilkinson’s catalyst is a complex compound containing the transition metal rhodium, bonded to one chloride and three triphenylphosphine ligands. It can be represented by the formula [RhCl(PPh3)3]. In solution, this compound is a homogeneous catalyst for the hydrogenation of alkenes. a) State three characteristic features of a catalyst.
• Activity • Selectivity • Stability d) In the mechanism of the reaction of cyclohexene with hydrogen, catalysed by Wilkinson’s catalyst, the intermediate [RhCl(H)2(PPh3)3] is formed. Write an overall equation for the reaction of cyclohexene with hydrogen. d) In the mechanism of the reaction of cyclohexene with hydrogen, catalysed by Wilkinson’s catalyst, the intermediate [RhCl(H)2(PPh3)3] is formed. Write an overall equation for the reaction of cyclohexene with hydrogen. d) In the mechanism of the reaction of cyclohexene with hydrogen, catalysed by Wilkinson’s catalyst, the intermediate [RhCl(H)2(PPh3)3] is formed. Write an overall equation for the reaction of cyclohexene with hydrogen. d) In the mechanism of the reaction of cyclohexene with hydrogen, catalysed by Wilkinson’s catalyst, the intermediate [RhCl(H)2(PPh3)3] is formed. Write an overall equation for the reaction of cyclohexene with hydrogen. d) In the mechanism of the reaction of cyclohexene with hydrogen, catalysed by Wilkinson’s catalyst, the intermediate [RhCl(H)2(PPh3)3] is formed. Write an overall equation for the reaction of cyclohexene with hydrogen. d) In the mechanism of the reaction of cyclohexene with hydrogen, catalysed by Wilkinson’s catalyst, the intermediate [RhCl(H)2(PPh3)3] is formed. Write an overall equation for the reaction of cyclohexene with hydrogen. Freie Universität Berlin, Institut für Chemie und Biochemie
Coordination Chemistry 3
Andrey Petrov (AG Müller, F208) 1. Identify the following reactions by their type (oxidative addition, reductive elimination, b-hydride elimination, ligand addition, ligand dissociation, migratory insertion).
CpFe(CO)2(Me) + PPh3 CpFe(CO)(acyl)(PPh3)
PPh3 Fe ligand addition OC Me OC
Migratory insertion
Migratory insertion + ligand addition 1. Identify the following reactions by their type (oxidative addition, reductive elimination, b-hydride elimination, ligand addition, ligand dissociation, migratory insertion).
CpFe(CO)2(Me) + PPh3 CpFe(CO)(acyl)(PPh3)
Fe OC Me OC
Migratory insertion + ligand addition
h Cp*Ir(CO)2 + CH4 Cp*Ir(CO)(H)(CH3) + CO
Ir(I) Ir(III) Ir OC OC
Ligand dissociation + oxidative addition 1. Identify the following reactions by their type (oxidative addition, reductive elimination, b-hydride elimination, ligand addition, ligand dissociation, migratory insertion).
H2RhCl(PPh3)2(CH2CH3) HRhCl(PPh3)2 + CH3CH3
H Ph P PPh 3 Rh 3 H CH CH Cl 2 3
Reductive elimination 1. Identify the following reactions by their type (oxidative addition, reductive elimination, b-hydride elimination, ligand addition, ligand dissociation, migratory insertion).
H2RhCl(PPh3)2(CH2CH3) HRhCl(PPh3)2 + CH3CH3
H Ph P PPh 3 Rh 3 H CH CH Cl 2 3
Reductive elimination
HRu(CH2CH3)(CO)3 + CO H2Ru(CO)4 + CH2=CH2
OC CO Ru β- hydride
OC CH2CH3 H
β-hydride elimination + ligand exchange 1. Identify the following reactions by their type (oxidative addition, reductive elimination, b-hydride elimination, ligand addition, ligand dissociation, migratory insertion).
+ + [Cp2TiCH3] + CH2=CH2 [Cp2Ti(CH2CH2CH3)]
+
Ti CH3
Ligand addition + migratory insertion 2. Consider the reaction shown below: How could the conversion to the product species proceed? Describe the reaction in 4 steps.
CH CH 2 3 H C CH OC PR OC PR 2 2 Rh 3 Rh 3 + I CH I CO 3 + CH4 CO 2. Consider the reaction shown below: How could the conversion to the product species proceed? Describe the reaction in 4 steps.
Ligand Reductive dissociation elimination
- CH4
Ligand exchange 2. Consider the reaction shown below: How could the conversion to the product species proceed? Describe the reaction in 4 steps.
Ligand Reductive dissociation elimination
- CH4
Ligand exchange 2. Consider the reaction shown below: How could the conversion to the product species proceed? Describe the reaction in 4 steps.
Ligand Reductive dissociation elimination
- CH4
Ligand Beta-hydride elimination exchange 2. Consider the reaction shown below: How could the conversion to the product species proceed? Describe the reaction in 4 steps.
Ligand Reductive dissociation elimination
- CH4
Ligand Beta-hydride elimination exchange 2. Consider the reaction shown below: How could the conversion to the product species proceed? Describe the reaction in 4 steps.
Ligand Reductive dissociation elimination
- CH4
Ligand Beta-hydride elimination exchange 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory insertion 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory insertion
Ligand association 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory insertion
Ligand association
Migratory insertion 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory insertion
Ligand association
Migratory insertion
Ligand association 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory insertion
Ligand association
Migratory Reductive insertion elimination
Ligand association 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory insertion
Ligand Oxidative association addition
Migratory Reductive insertion elimination
Ligand association 3. Label the reaction steps in the following catalytic cycle for the hydrothiocarbonylation of an alkyne. Migratory Ligand insertion association
Ligand Oxidative association addition
Migratory Reductive insertion elimination
Ligand association 4. Sketch out 4 steps in the isomerization of 1-butene to 2-butene on
HRh(CO)(PPh3)2.
[Rh] 4. Sketch out 4 steps in the isomerization of 1-butene to 2-butene on
HRh(CO)(PPh3)2.
Ligand addition Ligand dissociation
Beta-hydride elimination Migratory insertion 4. Sketch out 4 steps in the isomerization of 1-butene to 2-butene on
HRh(CO)(PPh3)2.
Ligand addition Ligand dissociation
Beta-hydride elimination Migratory insertion 4. Sketch out 4 steps in the isomerization of 1-butene to 2-butene on
HRh(CO)(PPh3)2.
Ligand addition Ligand dissociation
Beta-hydride elimination Migratory insertion 4. Sketch out 4 steps in the isomerization of 1-butene to 2-butene on
HRh(CO)(PPh3)2.
Ligand addition Ligand dissociation
Beta-hydride elimination Migratory insertion 4. Sketch out 4 steps in the isomerization of 1-butene to 2-butene on
HRh(CO)(PPh3)2. [Rh]
Ligand addition Ligand dissociation
Beta-hydride elimination Migratory insertion 5) a) In hydroformylation reaction using [Rh(PPh3)3(CO)(H)] as the catalyst precursor, addition of excess PPh3 would : 1-increase rate 2.decrease rate 3.not influence 4.stop the reaction 5) a) In hydroformylation reaction using [Rh(PPh3)3(CO)(H)] as the catalyst precursor, addition of excess PPh3 would : 1-increase rate 2.decrease rate 3.not influence 4.stop the reaction b) The hydroformylation of a-olefins (CH2=CHR) using carbon monoxide CO and hydrogen H2 leads to commercially valuable chemical intermediates. In the mechanism illustrated below (PH3 as ligand), ethene is converted to 1-propanal by catalytic hydroformylation using RhH(CO)(PPh3)3state and number as a catalyst of valence precursor. electrons Calculate for complex the oxidation A, complex state B andand complexnumber E.of valence electrons for complex A, complex B and complex E.
A
B E C
D
6) n-BuLi reacts violently with water. Explain shortly why. What are the products of the b) The hydroformylation of a-olefins (CH2=CHR) using carbon monoxide CO and hydrogen H2 leads to commercially valuable chemical intermediates. In the mechanism illustrated below (PH3 as ligand), ethene is converted to 1-propanal by catalytic hydroformylation using RhH(CO)(PPh3)3state and number as a catalyst of valence precursor. electrons Calculate for complex the oxidation A, complex state B andand complexnumber E.of valence electrons for complex A, complex B and complex E.
Rh(0) 5s1 4d8 A Rh(I) 18e
B E C
D
6) n-BuLi reacts violently with water. Explain shortly why. What are the products of the b) The hydroformylation of a-olefins (CH2=CHR) using carbon monoxide CO and hydrogen H2 leads to commercially valuable chemical intermediates. In the mechanism illustrated below (PH3 as ligand), ethene is converted to 1-propanal by catalytic hydroformylation using RhH(CO)(PPh3)3state and number as a catalyst of valence precursor. electrons Calculate for complex the oxidation A, complex state B andand complexnumber E.of valence electrons for complex A, complex B and complex E.
Rh(0) 5s1 4d8 A Rh(I) 18e
B E C Rh(I) 16e
D
6) n-BuLi reacts violently with water. Explain shortly why. What are the products of the b) The hydroformylation of a-olefins (CH2=CHR) using carbon monoxide CO and hydrogen H2 leads to commercially valuable chemical intermediates. In the mechanism illustrated below (PH3 as ligand), ethene is converted to 1-propanal by catalytic hydroformylation using RhH(CO)(PPh3)3state and number as a catalyst of valence precursor. electrons Calculate for complex the oxidation A, complex state B andand complexnumber E.of valence electrons for complex A, complex B and complex E.
Rh(0) 5s1 4d8 A Rh(I) 18e
B E C Rh(I) 16e Rh(III) 18e
D
6) n-BuLi reacts violently with water. Explain shortly why. What are the products of the 9. n-BuLi reacts violently with water. Explain shortly why. What are the products of the reaction? 9. n-BuLi reacts violently with water. Explain shortly why. What are the products of the reaction?
Due to the large difference between the electronegativities of carbon (2.55) and lithium (0.98), the C-Li bond is highly polarized. n-BuLi can often be considered to react as the butylanion, n-Bu−, and a lithium cation, Li+.
pKa ≈ 50 9. n-BuLi reacts violently with water. Explain shortly why. What are the products of the reaction?
Due to the large difference between the electronegativities of carbon (2.55) and lithium (0.98), the C-Li bond is highly polarized. n-BuLi can often be considered to react as the butylanion, n-Bu−, and a lithium cation, Li+.
(n-BuLi) pKa ≈ 50 (H2O) pKa ≈ 14
C4H9Li + H2O → C4H10 + LiOH 10. If TMEDA (tetramethylethylenediamine) is added to a solution of n-BuLi, the reactivity of the lithiation-reagent is enhanced drastically. Explain shortly why. 10. If TMEDA (tetramethylethylenediamine) is added to a solution of n-BuLi, the reactivity of the lithiation-reagent is enhanced drastically. Explain shortly why. 10. If TMEDA (tetramethylethylenediamine) is added to a solution of n-BuLi, the reactivity of the lithiation-reagent is enhanced drastically. Explain shortly why.
TMEDA has a high affinity for Li ions. See Prof. Müller’s group webpage for answers.