For each of the following pairs of complexes, which is least likely to undergo b-hydride elimination? Choose only one from each pair.

Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter14: Applications Of Ultraviolet-visible Molecular Absorption Spectrometry
Section: Chapter Questions
Problem 14.16QAP
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For each of the following pairs of complexes, which is least likely to undergo b-hydride elimination?
Choose only one from each pair.
ML
A
LMT
A
C
Select one:
a.
Me
Me
O b.
O C.
O d.
or
Ph3P
ос
ML
Me
or
Ph3P
OC
Identify complexes A and B and processes 1 and 2:
H3C-CH3
1
A
H
CI
B
A
Ph3PCI
OC
A
LM
A HIKIC
CO
PPh3
CI
Ph3P Ir-
PPh3
OC
PPh3
D
CO
PPh3
Ph3P
OC
H
.. Et
Ph3P
OC
PhaPr
OC
Pha
B
OC
PPh 3
BH
CI
PhaPr
OC
B
Pi...
B
H
LM
Pr
CI
Pri
CI
PPh3
H
PPh3
MCI
PPh3
E
PPh3
-M-
G
Pr
Pr
Pr
2
or
or
LM
1
Ligand
Substitution
Oxidative
Addition
F
Oxidative
Addition
B
1
Ligand
Substitution
H
Time left 103:15:10
Reductive
Elimination
B-Hydride
Elimination
2
Reductive
Elimination
2
B-Hydride
Elimination
Transcribed Image Text:For each of the following pairs of complexes, which is least likely to undergo b-hydride elimination? Choose only one from each pair. ML A LMT A C Select one: a. Me Me O b. O C. O d. or Ph3P ос ML Me or Ph3P OC Identify complexes A and B and processes 1 and 2: H3C-CH3 1 A H CI B A Ph3PCI OC A LM A HIKIC CO PPh3 CI Ph3P Ir- PPh3 OC PPh3 D CO PPh3 Ph3P OC H .. Et Ph3P OC PhaPr OC Pha B OC PPh 3 BH CI PhaPr OC B Pi... B H LM Pr CI Pri CI PPh3 H PPh3 MCI PPh3 E PPh3 -M- G Pr Pr Pr 2 or or LM 1 Ligand Substitution Oxidative Addition F Oxidative Addition B 1 Ligand Substitution H Time left 103:15:10 Reductive Elimination B-Hydride Elimination 2 Reductive Elimination 2 B-Hydride Elimination
For the reaction of the sodium enolate with chlorotrimethylsilane shown below, what is the
thermodynamic driving force for this reaction?
+ CISiMe3
Me Si-O
NaO
+ NaCl
Select one:
A. The formation of strong covalent bonds between Sodium and Chloride ions.
B. The breaking of weak Na-O and Si-Cl bonds.
C. Silicon-oxygen silicon bonds are stronger than silicon-carbon single bonds.
D. The formation of the NaCl lattice and release of the lattice enthalpy, coupled with the
formation of a strong Si-O bond.
For organolithium clusters of the general formula (RLi)6, reaction with additional Lewis basic
donor ligands leads to deaggregation and formation of smaller rings than the 6 membered
rings seen in the original cluster. For internally solvated clusters, however, 6 membered rings
are possible. Why?
Select one:
O A. Internal solvation does not significantly increase the effective steric demand of the
organic fragment, whilst maximising R-Li contacts.
B. Internal solvation is sterically demanding, which forces larger rings to be adopted.
OC. The organolithium deprotonates the Lewis base, leading to loss of reagent and so
smaller rings.
OD. Internal solvation disrupts the normal solvation sphere, leading to lower coordination
numbers at Li.
Transcribed Image Text:For the reaction of the sodium enolate with chlorotrimethylsilane shown below, what is the thermodynamic driving force for this reaction? + CISiMe3 Me Si-O NaO + NaCl Select one: A. The formation of strong covalent bonds between Sodium and Chloride ions. B. The breaking of weak Na-O and Si-Cl bonds. C. Silicon-oxygen silicon bonds are stronger than silicon-carbon single bonds. D. The formation of the NaCl lattice and release of the lattice enthalpy, coupled with the formation of a strong Si-O bond. For organolithium clusters of the general formula (RLi)6, reaction with additional Lewis basic donor ligands leads to deaggregation and formation of smaller rings than the 6 membered rings seen in the original cluster. For internally solvated clusters, however, 6 membered rings are possible. Why? Select one: O A. Internal solvation does not significantly increase the effective steric demand of the organic fragment, whilst maximising R-Li contacts. B. Internal solvation is sterically demanding, which forces larger rings to be adopted. OC. The organolithium deprotonates the Lewis base, leading to loss of reagent and so smaller rings. OD. Internal solvation disrupts the normal solvation sphere, leading to lower coordination numbers at Li.
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