CHEM4680 - Organometallic Chemistry
Final exam, April 13, 2009: rough answers
1. Counting (16 marks)
a) The Pd-N bonds are dative, as are half of the Pd-Cl bonds.
Pd 10
C 1
Cl 1
N→ 2
Cl→ 2
16e
PdII, expected to be stable.
b) You have figure out whether the Ga-Fe bond is dative or covalent. Since Ga is in group 3 and Cp is 1, assuming a single covalent bond would leave the Ga (and the Fe) with an unpaired electron. A Ga→Fe dative bond seems more reasonable: CpGa, when taken alone, is 8e and has a lone pair at the metal that can be used for donation.
Ga 3
Cp 5
8e
GaI, count OK, low oxidation state, probably reactive and a strong reductor (the real complex has a Cp* ligand that provides more steric protection).
Fe 8
4*CO→ 8
Ga→ 2
18e
Fe(0), count OK, low oxidation state, but the strong p-acceptor CO ligands stabilize the low oxidation state.
c)
Ti 4
6*H2O→ 12
16e
Ti(0), count not too bad, but the low oxidation state in combination with pure s-donor ligands will make this highly unstable.
d) The key is noticing that there must be a + charge on the Ph2P group (this is a phosphonium ion). Therefore, there must be a charge associated with the metal.
Rh 9
1
2*Cl 2
2*P→ 4
16e
RhI, count OK, oxidation state a bit on the low side, stable but would easily undergo oxidative addition.
2. (12 marks)
a) 1 and 3 have identical counts and oxidation states. Counting of NO is sometimes ambiguous: it can be a 1-e or 3-e donor, and for oxidation state it can be counted as NO+ or NO. Counting it as 3-e NO:
W 6
2*C 2
NO 3
Cp* 5
16e
WIV, count on the low side (middle TM want to be 18-e!) but this is quite crowded so there might not be room for more, oxidation state reasonable.
b) The most reasonable intermediate would be an alkylidene complex:
c) Formation of X via "a-elimination" or s-bond metathesis, reaction with olefin via the standard 2+2 mechanism that is also the first step of olefin metathesis (or you can call it insertion of the olefin in one of the two WC bonds).
3. (12 marks)
The "H transfer" steps can be seen as s-bond metathesis or intramolecular electrophilic attack (H+ on the N/C of an M-N/M-C bond) or its reverse. Insertion of the olefin could in principle go 1,2 or 2,1; 1,2 (as shown here) is generally preferred because the most stable carbanion is the terminal one. Insertion in the M-N instead of the M-C bond is less likely because early transition metals form strong bonds to electronegative elements like N.
b) Deprotonation of the ethyl CH2N group by the neighbouring amide functionality would be more difficult than deprotonation of CH3N, again because having the charge on a terminal C is best. So, the NEt should react quite a bit slower.
4. (20 marks)
a) The most logical intermediate would be a Fischer carbene complex:
This could give the ketone via oxidation, the thioketone in a similar manner.
b)
Attack of the external organoLi nucleophile on a coordinated CO, then intramolecular attack of the alkoxide on the C-Br.
c) There is no coordinating group available for directed lithiation, so the most plausible reaction is metal-halogen exchange. This could work with either the Br2 or BrI compound, but the former might also give dilithiation since the two Br atoms would have similar reactivity. So, the most likely organic compound would be:
d) OrganoLi compound 4 would definitely add to the C=O double bond acetone.
What happens after that is less clear. One possibility is intramolecular cyclization, similar to the Cr reaction:
But it might also be that the alkoxide is stable by itself. In that case, work-up would give either the alcohol or its water elimination product (depending on how gentle the work-up is done):
Of the side reactions sometimes associated with RLi addition to C=O, single-electron transfer is unlikely (no large conjugated system on the ketone), reduction via b-hydrogen transfer is impossible (no b-H in the RLi). Reaction as a base (deprotonation) is a realistic possibility: this would give the enolate of acetone (plus RH) and on work-up give back acetone.
5. (24 marks)
a) It is a long series of steps, but it only needs 3 coordination sites around Ru (ru = L2Ru(H2) in the figure) and works systematically towards the CO-containing product. The rest of the ethanol gets eliminated at some stage as CH4. In the sequence below, the oxidation state varies between +2 and 0, and the electron count between 18 and 16. Several permutations of steps are possible.
b) Weighted average of the original shifts: (2*-6 + -9.5 + -10.5)/4 = -8 ppm
c) The H atoms of the two hydrides and the H2 ligand somehow switch places. The most logical mechanism is:
Repeated often enough, this will exchange all four hydrogens over all three positions.
d) Determination of H atom positions by X-ray is not very reliable, since there is so little electron density associated with them. If you are lucky you see them, but it is hard to be sure. However, the positions of the CO ligand and the two phosphines can be established easily. The NMR data indicate one H2 and two hydride ligands, and since the "visible" atoms of the complex are in a T-shape arrangement the less visible H/H2 ligands must also be in such an arrangement (6-coordination is virtually always octahedral). That leaves only 6 and 7. But 7 would have the two hydrides equivalent, which does not agree with the NMR observation of 2:1:1 signals. Therefore only 6 is left as a possibility, but you require both NMR and Xray data to be sure of this.
e) Hydrides have a very large trans effect and so would not want to be opposite each other (as they are in 7).
6. (16 marks)
a) Metal-catalyzed C-C coupling is a well-established technique in organic synthesis, but it usually does not work for CF3 groups. And CF3 groups are important in many new pharmaceuticals and agrochemicals.
b) Some decomposition of the Cu-CF3 complex occurs, producing F, which then attacks Si species.
c) Standard oxidative addition/reductive elimination is the most obvious possibility:
d) They state that KOtBu also reacts directly with CF3SiMe3 to produce CF3H, and this side reaction competes with the desired reaction that would regenerate the CuCF3 species (CuI → CuOtBu → CuCF3) and so complete the cycle.
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