Assignment 2 answer sheet

Total: 32 marks

1. Neutral ligand method first, then oxidation state method (5 marks)

a) Fe (8e-) + 5 CO (5 x 2e-) = yes (18e-)

or Fe(0) (8e-) + 5 CO (5 x 2e-) = yes (18e-)

b) Rh (9e-) + 2 bipy (2 x 4e-) + Cl (1e-) – (1e-) = no (17e-)

or Rh(II) (7e-) + 2 bipy (2 x 4e-) + Cl (2e-) = no (17e-)

c) Re (7e-) + η5-Cp* (5e-) + 3 O (3 x 2e-) = yes (18e-)

or Re(VII) (0e-) + η5-Cp* (6e-) + 3 O (3 x 4e-) = yes (18e-)

d) Re (7e-) + 2 PPh3 (2 x 2e-) + 2 Cl (2 x 1e-) + N (3e-) = no (16e-)

or Re (V) (2e-) + 2 PPh3 (2 x 2e-) + 2 Cl (2 x 2e-) + N (6e-) = no (16e-)

e) Os (8e-) + CO (2e-) + phenylcarbyne (3e-) + 2 PPh3 (2 x 2e-) + Cl (1e-) = yes (18e-)

or Os (IV) (4e-) + CO (2e-) + phenylcarbyne (6e-) + 2 PPh3 (2 x 2e-) + Cl (2e-) = yes (18e-)

2. (4 marks)

a) each titanium atom is oxidation state (III)

b) Ti (4e-)

+ η5-Cp (5e-)

+ ½ fulvalene dianion (this behaves exactly like a Cp ligand to each titanium atom) (5e-)

+ ½ of 2 bridging hydride ligands (0.5*2e-= 1e-)

= 15e-

Or..

Ti(III) (1e-)

+ η5-Cp (6e-)

+ ½ fulvalene dianion (this behaves exactly like a Cp ligand to each titanium atom) (6e-)

+ ½ of 2 bridging hydride ligands (0.5*4e-= 2e-)

= 15e-

c) The dotted line is drawn to imply the existence of a Ti-Ti bond due to each Ti atom being short only

1e- of 16e- which is a more stable electron configuration than 15e-.

Marks were given to anyone who adequately defended their position. From this reference:

https://doi.org/10.1016/0022-328X(92)83204-U

The crystal structure of the compound shows that the distance between the two Ti atoms is shorter than

the sum of their Van der Waal’s radii, suggesting that some kind of bonding interaction is taking place.

The 15th electron in a Ti(III) complex should cause the complex to be paramagnetic, however this

particular compound is diamagnetic at room temperature, which further suggests that the 15th electron

is spin-paired with another one in a (weak) covalent bond.

3. (6 marks)

a) 2, 1, 16, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 14, 11, 15

b)

c) The π electron accepting properties of an Arduengo carbenes can be estimated by the amount of

electron density present in the empty p-orbital of the carbene carbon.

In the first compound, the p orbital of the carbene carbon (as in the above example shown in this

answer sheet) is part of a 6-electron ring where 2e- are contributed from the double bond, and 2e- each

from the nitrogen lone pairs. These nitrogen atoms are forced into a planar orientation by the structure

of the 5-membered ring. That means the p-orbital is part of an aromatic system (aromaticity = 4n + 2). It

already contains a considerable amount of electron density from this aromaticity and so it can’t accept

much from a π-electron donor.

The carbene p-orbital on the second compound is also part of a 6e- aromatic ring, but the nitrogen

atoms are bound to phenyl groups which help to delocalize the electron density of the inner-most ring.

Compared to the alkyl groups of the first, the p orbital contains less electron density due to this

delocalization effect.

The third compound is more π-acidic because now there is no aromaticity in the inner-most ring, and

therefore the carbene p orbital is electron poor and able to accept more electron density compared to

the previous NHC.

The fourth compound no longer has a nitrogen lone pair adjacent to the carbene, so the p orbital

contains significantly less electron density than previously. This makes the carbene p orbital less

electron rich, and it can accept more electron density.

The fifth compound is the only acyclic carbene in the series, meaning that its nitrogen atom is free to

rotate about the carbene-nitrogen axis. That means the nitrogen atom can only donate electron density

to the empty p orbital of the carbene carbon when it is in the correct orientation to do so. That

rotational freedom makes acyclic Arduengo carbenes in general much more π-acidic than cyclic

Arduengo carbenes.

Due to the overwhelming number of “steric hindrance” type answers, I would like to clearly state:

Steric hinderance of the carbene carbon does not influence its π-acidity.

Steric hindrance does play a small role in kinetically protecting a free carbene from dimerization

according to the Wanzlick equilibrium, but even that scenario is more heavily influenced by electronic

factors.

4. (5 marks)

a) U(IV),

Electron count: Cyclooctatetreenyl dianions are aromatic (aromaticity = 4n + 2) and donate electrons in

exactly the same way as Cp ligands do, except that these are -2 charged.

U(6e-) + 2 x cyclooctatetraenyl dianions (2 x 8e-) = 22e-

Or U(IV) (2e-) + 2 x cyclooctatetraenyl dianions (2 x 10e-) = 22e-

b) Here is the structure of Fe(COT)2 which has the correct structure. Any metal atom is worth full marks.

c) As the metal center changes from a first-row to second-row to third-row metal, the large COT2- ligands

should form increasingly better overlaps with the metal. This should cause the η value of the COT2-

ligands to increase and donate more electrons to the metal center.

5. (12 marks)

a) b)

Top row:

Co (9e-) + 2 NHC (2 x 2e-) + 2 Cl (2 x 1e-) = 15e- oxidation state (II)

Or Co(II) (7e-) + 2 NHC (2 x 2e-) + 2 Cl (2 x 2e-) = 15e- oxidation state (II)

Bridging Cl- groups donate 3 e- between 2 metal centers using the neutral ligand method, or 4 e-

between 2 metal centers using the oxidation state method.

Co (9e-) + NHC (2e-) + σ-Cl (1e-) + ½ of 2 µ-Cl (0.5 * 2 * 3e-) = 15e- oxidation state (II)

Or Co(II) (7e-) + NHC (2e-) + σ-Cl (2e-) + ½ of 2 µ-Cl (0.5 * 2 * 4e-) = 15e- oxidation state (II)

Co (9e-) + NHC (2e-) + 2 σ-Cl (2 x 1e-) + µ-Cl (2e-) = 15e- oxidation state (II)

Or Co(II) (7e-) + NHC (2e-) + σ-Cl (2e-) + 2x µ-Cl (2x 2e-) = 15e- oxidation state (II)

Bottom row:

Co (9e-) + 2 CH3 (2 x 1e-) + 2 NHC (2 x 2e-) = 15 e- oxidation state (II)

Or Co(II) (7e-) + 2 CH3 (2 x 2e-) + 2 NHC (2 x 2e-) = 15 e- oxidation state (II)

Co (9e-) + CH3 (1e-) + CH3CO (1e-) + 2 NHC (2 x 2e-) + CO (2e-) = 17e- oxidation state (II)

Or Co(II) (7e-) + CH3 (2e-) + CH3CO (2e-) + 2 NHC (2 x 2e-) + CO (2e-) = 17e- oxidation state (II)

Co (9e-) + 2 NHC (2 x 2e-) + 2 CO (2 x 2e-) = 17e- oxidation state (0)

Or Co(0) (9e-) + 2 NHC (2 x 2e-) + 2 CO (2 x 2e-) = 17e- oxidation state (0)

Co (9e-) + NHC (2e-) + 3 CO (3 x 2e-) + Co-Co bond (1e-) = 18e- oxidation state (0)

Or Co(0) (9e-) + NHC (2e-) + 3 CO (3 x 2e-) + Co-Co bond (1e-) = 18e- oxidation state (0)

Co (9e-) + NHC (2e-) + 3 CO (3 x 2e-) + Co-Co bond = 18e- oxidation state (0)

Or Co(0) (9e-) + NHC (2e-) + 3 CO (3 x 2e-) + Co-Co bond (1e-) = 18e- oxidation state (0)

c) A bond is drawn because it satisfies the 18e- rule by a covalent sharing of electrons between the two

17e- cobalt complexes. As mentioned for question 2, further analysis is required to prove the existence

of a Co-Co bond.

d) All compounds are expected to be paramagnetic except the last one:

This compound should be diamagnetic because the bond between cobalt centers would cause the 17th

electron from one Co to be spin-paired with the 17th electron from the other Co, resulting in an 18e-

diamagnetic complex.