Metal-Carbonyls/Metal-Carbon π-bonds Flashcards

(27 cards)

1
Q

How does the bonding occur between a metal-carbonyl complex?

A
  • The HOMO is CO is the sσ’ orbital containing a pair of E-, which are donated to the metal (sigma donation)
  • Empty π’ orbitals can reciece E- from the metal through π-back donation - this affect dominates
  • (known as synergeric bonding which results in a strengthening of the M-C bond)
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2
Q

What affect does the π-back donation have on the strength of the C-O bond in a metal-carbonyl complex

A

Weakens the C-O
This will result in a longer C-O and a decreased vCO on the IR spectra

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3
Q

the more electron rich the metal is in the metal carbonyl complex, the ……. stable the metal-carbonyl complex is

A

more stable
The more π-back donation can occur, which stabilises the CO bound to the metal

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4
Q

What happens to the vibrating frequency of the C-O when bonded to a metal?

A
  • The vibrating frequency decreases with the more M-CO bonds added
  • Does depend on the metal due to more electro rich metals π-back donating, resulting in weaker M-CO bonds
  • The H₃B-CO shows the effect of just sigma donation = higher frequency vibrations
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5
Q

What happens to the IR vibration frequency of the C-O, when the bonded metal-carbonyl moves to a lower oxidation state

A

The lower the oxidation state, will mean they are electron rich and hence will π-back donate more
Resulting in weaker C-O and hence lower frequency stretches

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6
Q

What happens to the IR vibration frequency of the C-O when electron donating ligands are also bonded to the metal?

A

The more electron donating the other ligands are, the lower frequency vibration of the C-O there is

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7
Q

What is one of the simplest ways to make a metal-carbonyl

A

Through reacting the metal with carbon monoxide gas

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8
Q

What is another way you can make a metal-carbonyl

A
  • Through using a metal halide and reduce (Na) it to a lower oxidation state, which will make it favourable for CO to bind
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9
Q

What is two ways to remove a CO ligands from a metal?

A
  • CO is easily eliminated by heat or photolysis
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10
Q

CO can be replaced in a metal complex by

A

Other neutral, two electron donors

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11
Q

What type of reaction is this?

A

oxidation
Goes to a Mn +1 when we formed the manganese iodide species
(both species are 18 electrons)

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12
Q

Why is reduction a more common process for metal-carbonyl species

A

The carbonyl species are good at withdrawing electron density through the pi-system
So they stabilise metals in low oxidation states i.e. stabilise metals with a lot of electron density

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13
Q

How would you reduce a metal-carbonly species?

A
  • Either: by reduction using sodium
  • OR: Nu attack using sodium hydroxides which will attack the CO ligand
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14
Q

What is the end product of this reaction?

A
  • R group will form a new bond to one of the carbonyls
    (1,1-migratory insertion)
  • The negative charge on the metal makes it nucleophilic for this reaction to occur
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15
Q

How does the bonding occur in a Metal-Alkene complex?

A
  • The Alkene donates to the metal through the HOMO π-bonding orbital
  • The alkene accepts π-back-donation into the LUMO π’ antibonding orbital
  • (synergic again- the two components (back and forward donation) of the bonding are better balanced than the metal-carbonyl complexes)
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16
Q

What affect does the metal-alkene bonding have on the bonding in the alkene

A
  • Lengthening of the C=C bond - as it is weaker due to the back donation
  • Reduction of angles at C from 120° (sp²) to 109° (sp³) if there is enough back donation due to the breaking of the C=C
17
Q

What does the extent of backbonding depend on?

A
  • The energy of the frontier orbital of the M fragment
  • Steric effects
  • The alkene acceptor ability
18
Q

Electron withdrawing group affect π-back donation how?

A

Electron withdrawing groups increase π-backdonation and decrease σ-donation

19
Q

Alkene ligands can be switched in with which type of ligands?

A
  • Switched for other neutral two electron donors
20
Q

How do metal-alkene complexes react with nucleophiles

A
  • Through a 1,2-migratory insertion
  • This is due to so much electron density can be donated from the alkene to the metal which results in it being positively charged - hence more susceptible to Nu attack
21
Q

True or false?
Alkynes are weaker π-acceptors than alkenes

A

False
Alkynes are stronger π-acceptors than alkenes

22
Q

Alkynes can donate how many electrons as a ligand?

A

Alkynes have two orthogonal π-bonds and can act as 2 or 4 electron donor ligands

23
Q

Significant amounts of back donation from the metal, can result in an alkynes being

A

sp² hybridised rather than sp hybridised
notes the longer bond too

24
Q

Alkyne ligands can be swithced in/out by which type of ligand?

A
  • Neutral 2 or 4 electron ligands
25
Usually triple bonds with benzene are unstable What allows this compound to exist?
Through binding to Zirconium it is closer to being sp² hybridised (back donation) hence is more stable
26
What it the product of this reaction
Formation of an aromatic ring through combining the alkene with CO groups
27
What is the product of this reaction?
through a series of oxidative couplings