Download 1. Acetone can bind to transition metals via oxygen only, or via both

1. Acetone can bind to transition metals via oxygen only, or via
both the carbonyl carbon and the oxygen. Sketch and indicate
the η and κ nomenclatures for these two possibilities. What
types of metals would you expect to favor each bonding mode?
Would you expect either bonding mode to activate the carbonyl
carbon toward nucleophilic attack?
Low-valent, late transition metals that favor soft π-ligands should favor η2coordination. High-valent, early transition metals that favor hard ligands (such as
oxygen ligands) should favor η1-coordination. Large ancillary ligands should
favor η1-coordination. Both forms should promote nucleophilic attack at the
carbonyl carbon. In the η2-form, the ketone is a net electron donor toward the
metal, so it should be more electrophilic than a free ketone. The η1-form is
activated essentially the same way it is activated by a Lewis acidic main
group reagent like AlR3 or Li+, making the carbonyl carbon more electrophilic.
2. Me2CHMgBr reacts with [IrClL3] (L = PMe2Ph) to give [IrHL3].
Explain the reaction and predict the other products.
3. Provide the structures of the products of these reactions.
Reaction stoichiometries are not necessarily balanced.
a. [IrHCl2(PPh3)3] + LiAlH4 --- THF ---> [IrH3(PPh3)3]
mer-isomer is the main product. The fac- isomer is also a known
compound.
b. [IrHCl2(PBu2Me)3] + NaBH4 --- EtOH --->
Why are the outcomes of a and b so different? BH4- is a more
strongly coordinating ligand than AlH4-. In general, the more polar Al-H
bonds are more inclined to cleave than the less polar B-H bonds. In
addition, PBu2Me is a better electron-donor ligand, making the metal more
basic and therefore more inclined to hold on to the Lewis acid, BH3.
c. [RuCl2(PPh3)3] + NaOMe --- MeOH --->
This is a "deceptively simple" problem.
Under mild conditions, [RuHCl(PPh3)3] would be expected. This is a
sixteen electron complex. CO abstraction from the solvent can occur
under harsher conditions to give [RuHCl(CO)(PPh3)3]. There are yet other
possibilities. Conversion of the second chloride to hydride can occur to
give very reactive [RuH2(PPh3)3], which can scavenge CO to give
[RuH2(CO)(PPh3)3] or a phosphine to give [RuH2(PPh3)4]. This is a
surprisingly complicated system, as I hope you discovered by looking in
the literature.
What products would you expect from reactions of
[RuCl2(PPh3)3] with these reagents?
i. NaOCD2CH3/ HOCD2CH3
ii. NaOCH2CD3/ HOCH2CD3
iii. NaOCH2CH3/ DOCH2CH3
Using the simplest possible product, [RuHCl(PPh3)3], reagent i will give
[RuDCl(PPh3)3] but ii and iii will give [RuHCl(PPh3)3]. The hydride
originates from the β-position.
d.
e.
4. The η5-C5H5 ligand is susceptible to both electrophilic and
nucleophilic attack. Predict the relative reactivity of ferrocene
and cobalticenium complexes in these types of reactions.
5. Predict the product of the following reactions (explain your
reasoning):
6. [(η5-C5Me5)Ir(acetone)3]2+ reacts with CH2=CH2 in acetone at
room temperature to produce [(η5-C5Me5)Ir(C5H9)]+, which has
the structure depicted below. Explain the NMR data.
Cp*
+
Ir
1
H NMR: δ(splitting, relative intensity): 4.11(tt, 1H), 3.79 (dd, 2H),
3.11(s, 2H), 2.60 (s, 2H), 2.04(dd, 2H), 1.91(s, 15H).
13
C NMR: δ(splitting, relative intensity): 100.6 (q), 86.3 (d), 47.6 (t), 47.0
(br t), 8.5 (q)
7. Propose a mechanism for the stoichiometric decarbonylation
reaction below:
O
Rh(PPh3 )3Cl
PhCH2
C
Cl
- CO
PhCH2
Cl
8. Predict the kinetically-controlled product of the following
reactions:
(a)
2
+ CN-
Ru+
η6-bound
(b)
Fe(CO)3
!4 -bound
+ [Ph3 C]BF4
4
9. Give the product (A) and propose a mechanism for the following
reaction:
Fe
OC
1
SiMe3
CO
- 78 ˚C
1. BuLi
2. MeI
A
H NMR (CS2): δ 4.67-4.90 (m , 4H), 0.31 (s, 9H), 0.18 (s, 3H)
IR (CHCl3, cm-1): 1991, 1937
Explain the characterization data.