CHEM4680 - Organometallic Chemistry

Rough answers

1. Counting.

You have to decide where the "2+" charge belongs. First (arbitrarily) assign one "+" charge to each metal atom. This will give FeIII, 17-e and PdI, 17-e. The alternatives are: both charges on Fe (FeIV, 16-e; Pd(0), 18-e) or both on Pd (FeII, 18-e; PdII, 16-e). A 17-e electron count on both metals is unlikely if there is any way to avoid it. PdII would be happy with a 16-e count (square-planar!), while FeIV would not; moreover, FeIV is a rather high oxidation state. Also, the water ligands, being "hard" ligands, would not bind well to a zero-valent metal. Therefore, the FeII/PdII formulation is the most likely one.

2. Main group metals and more

(A)

·  The F atom, like a OMe or NMe2 group, promotes ortho-lithiation. Therefore:

(apparently, F is a stronger directing factor for o-lithiation than PPh2, else the final product would have had two neighbouring PPh2 groups...)

·  The o-fluorophenyllithium intermediate is 4-e, hence would not be a monomer. It could form a dimer via bridging phenyl or (less likely) fluorine. That would bring it up to 6-e, so it would probably go on to a tetramer with e.g. 4-c-2-e bridging phenyl groups.

·  The o-fluorophenyllithium is likely to eliminate LiF (very stable!), generating the extremely reactive benzyne, which might go on to many different products.

·  Note that the central PHPh group of the ligand has been deprotonated! Rewrite the structure to distinguish between covalent and dative bonds, e.g. (there are more ways of drawing the arrows in the central 6-membered ring):

If you assign the + charge to the Pd having one covalent and three dative bonds, all Pd atoms come out PdII, 16-e. If you do it differently, you will get different counts and oxidation states for each Pd, obviously less favourable.

(B)

·  Three times the same reaction: Br/Li exchange, followed by nucleophilic substitution at P.

·  Since on the outer rings only the central Br has reacted, this seems to be the most reactive Br atom. A possible sequence would be (but ordinary Wurtz coupling could lead to the same product):

3. C-C coupling

·  X:

· 

·  The iPr groups are more electron donating that Ph groups, and slightly larger (at least close to the metal). The larger size might hinder the chloride/alkyl exchange. The stronger electron-donating capability will accelerate the oxidative addition and hinder the reductive elimination. Since the oxidative-addition product is not even observed for the original system, the reductive elimination is the slow step, so the net effect should be to accelerate the second reaction.

4. p-Complexation

·  In both complexes, the ring acts as a 6e-donor (1 e from each carbon, 2 e from N, 0 e from B is one way of counting this). So, both have Cr(0), 18-e.

·  On heating, enough energy is provided to pass over a barrier and form the most stable isomer, which apparently is 2a.

·  Deprotonation at N makes the BN ring more electron-rich, so it becomes a better donor and the Cr(CO)3 fragment (which is electron-poor because of the CO ligands) prefers to be bound to that ring. On protonation, this situation is apparently "trapped" so the less stable isomer of the protonated form is isolated. Y must be:

5. C-H activation

(note: the original question had an error in the structure of the product)

· 

·  The intermediate alkyl would have b-hydrogens, would b-eliminate, and the next olefin would insert. Eventually this would lead e.g. to olefin dimerization rather than methane addition.

·  Olefin and CH4 must compete for coordination to Sc, and CH4 is a very poor donor. Increasing the olefin concentration would hinder the CH4 coming in and hence block the s-bond metathesis step.

6. b-Aryl elimination (with the Hartwig paper)

·  Rh = 9, 2 e each for 2 phosphines, "C=C" of arene, 1 for Rh-O: RhI, 16-e

·  Insertion of a C=C bond converts a p-bond into a s-bond, which liberates about 23 kcal/mol of energy (a C=C bond is weaker than two single C-C bonds). But a C=O bond is twice as strong as two single C-O bonds, so there is no net gain in energy.

·  Conjugation usually means extra stabilization, so you would expect elimination of Ph2C=O to be favoured. However, in 2a appears to be assisted by incipient coordination of the phenyl ring to Rh. This is not possible (or at least not as favourable) for a methyl group, so elimination of a phenyl group is kinetically preferred.

·  The elimination breaks a C-C bond, forms an M-phenyl bond, converts a M-O s-bond into a dative bond. Ir tends to have stronger bonds, so if an existing bond has to be broken appreciably before a new one can be formed one would expect Ir to react slower. In this case, there is already a bonding interaction between the metal and the de-inserting phenyl group, so it looks like bond breaking and making can proceed simultaneously -> no reason to expect a large effect on the rate. The oxidation state of the metal does not change.

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