# 31.E: Transition Metal Organic Compounds (Exercises)

Exercise 31-1 If the ferrocene rings in $$3$$ were not free to rotate, how many different dichloroferrocene isomers would be expected (including chiral forms)? How could the substitution method (Section 1-1F) be used to determine which of the isomers was which?

Exercise 31-2 The cyclobutadiene iron complex, $$10$$, has been prepared optically active, and when oxidized with $$\ce{Ce}$$(IV) in the presence of tetracyanoethene gives a mixture of cyclobutadiene cycloadducts, all of which are optically inactive.

a. Draw the other chiral form of $$10$$.

b. Write structures for the cycloadducts that would be expected to be formed if $$10$$ were oxidized with $$\ce{Ce}$$(IV) in the presence of tetracyanoethene

c. How does formation of optically inactive products indicate that the cycloadducts are formed from the cyclobutadiene corresponding to $$10$$?

d. Is cycloaddition of an alkene to cyclobutadiene best regarded as a [2 + 2] or a [4 + 2] reaction?

Exercise 31-3* Assuming the molecular formula of $$9$$ is established as $$\ce{C_{40}H_{60}N_6Zr_2}$$, explain how the proposed structure is consistent with $$\ce{^{15}N}$$ NMR spectra as follows. Made with $$\ce{^{15}N \equiv ^{14}N}$$, $$9$$ shows three widely separated resonance lines of equal intensity. However, when $$9$$ is made with $$\ce{^{15}N \equiv ^{15}N}$$, two of the peaks become doublets with a spacing of $$6 \: \text{Hz}$$.

Exercise 31-4 The diphenylethyne complex with $$\ce{Pt}$$(0), analogous to $$12$$, has been shown by x-ray diffraction analysis to have $$\ce{C-C \equiv C}$$ bond angles of about $$140^\text{o}$$ and a central $$\ce{C-C}$$ bond distance of $$1.32 \: \text{Å}$$. Explain which of the formulations, $$11a$$, $$11b$$, or $$11c$$, seems most reasonable to account for the x-ray data for this complex.

Exercise 31-5 Write the sequence of steps whereby $$\ce{(Cp)_2ZrClH}$$ reacts with 2-methyl-2-pentene to form $$\ce{(Cp)_2Zr(Cl)CH_2CH_2CH_2CH(CH_3)_2}$$. Why is there no appreciable amount of $$\ce{(Cp)_2Zr(Cl)CH_2CH(CH_3)CH_2CH_2CH_3}$$ in the product?

Exercise 31-6 Show how $$\ce{(Cp)_2ZrClH}$$ could be used to achieve the following conversions:

a.

b. $$\ce{CH_3CH_2CH_2CH_2C \equiv CH} \rightarrow \ce{CH_3CH_2CH_2CH2CH=CHCHO}$$

c.* $$\ce{(CH_3)_3CC \equiv CH} \rightarrow \ce{(CH_3)_3CCH_2CH_2COCH_3}$$

Exercise 31-7* The stereochemistry of reactions in which $$\ce{Zr-C}$$ bonds are formed and cleaved can be deduced from the results of the following reactions, where $$\ce{D}$$ is hydrogen of mass 2.

The $$\ce{CH-CH}$$ coupling constants in the proton NMR spectra of $$14$$ and $$15$$ are about $$13 \: \text{Hz}$$. Work out the favorable conformations and the likely configurations of $$14$$ and $$15$$ and the stereochemistry of the addition and cleavage reactions. (Review Section 9-10H.)

Exercise 31-8 Explain how 2-methylpropanal could be formed in substantial amount in the cycle of Figure 31-3 with propene as the starting alkene.

Exercise 31-9 Explain how an alkene-metathesis catalyst might convert a cycloalkene into (a) a long-chain unsaturated polymer, (b) a mixture of large-ring polymers, and (c) a catenane (interlocking carbon rings like two links in a chain).

Exercise 31-10 The NMR spectrum of 2-propenylmagnesium bromide in ether is shown in Figure 31-4. With the aid of the discussion in Sections 9-10C and 9-10E and the knowledge that the $$\ce{CH_2}$$ resonance of ethylmagnesium bromide comes at $$38 \: \text{Hz}$$ upfield from tetramethylsilane, sketch the NMR spectrum you would expect for $$\ce{CH_2=CHCH_2MgBr}$$. Consider possible ways of reconciling your expected spectrum with the actual spectrum shown in Figure 31-4. (Review Section 27-2.)

Figure 31-4: NMR spectrum of 2-propenylmagnesium bromide in diethyl ether solution at $$60 \: \text{MHz}$$ with reference to tetramethylsilane at $$0 \: \text{Hz}$$. The off-scale bands are due to the diethyl ether, and the signals designated $$\ce{C_6H_{10}}$$ are due to 1,5-hexadiene (coupling product resulting during formation of the Grignard reagent).

Exercise 31-11 When one mole of azabenzene (pyridine), which is a good ligand, is added to a solution of one mole of $$20$$ in diethyl ether, a complex of composition $$\ce{(C_3H_5)_2NiNC_5H_5}$$ is formed in which the very complex proton spectrum of the $$\ce{C_3H_5}$$ groups of $$20$$ becomes greatly simplified and essentially like that of Figure 31-4. Explain how complexation of one mole of azabenzene with nickel in $$20$$ could so greatly simplify the proton NMR spectrum.

Exercise 31-12* $$\pi$$-Propenyl(ethyl)nickel decomposes at $$-70^\text{o}$$ to give propene and ethene. If the ethyl group is labeled with deuterium as $$\ce{-CH_2CD_3}$$, the products are $$\ce{C_3H_5D}$$ and $$\ce{CD_2=CH_2}$$. If it is labeled as $$\ce{-CD_2-CH_3}$$, the products are $$\ce{C_3H_6} + \ce{CD_2=CH_2}$$. Are these the products expected of a radical decomposition, or of a reversible hydride-shift followed by decomposition as in the mechanism of Section 31-2B? Suppose the hydride-shift step were not reversible, what products would you expect then?

Exercise 31-13 Palladium has many interesting uses in organic syntheses. The following sequence of reactions also could be achieved by forming and carbonating a Grignard reagent, but would not be stereospecific as it is with palladium. Devise mechanistic steps for the reaction that account for the stereochemical result [$$\ce{L}$$ is $$\ce{(C_6H_5)_3P}$$]. Review Sections 31-2, 31-3, and 31-4.

Exercise 31-14*

a. When a metal is complexed with an alkene, there are two possible ways for nucleophiles to become attached to carbon, as illustrated here with palladium:

Show how these mechanisms in combination with others described in this chapter can explain how $$\ce{PdCl_2}$$ can convert $$\ce{CH_2=CH_2}$$ to $$\ce{CH_3CHO}$$ (Wacker process). Your mechanism must be in accord with the fact that, when the reaction is carried out in $$\ce{D_2O}$$, there is no deuterium in the ethanal formed.

$\ce{CH_2=CH_2} + \ce{PdCl_2} + \ce{H_2O} \rightarrow \ce{CH_3CHO} + 2 \ce{HCl} + \ce{Pd} \left( 0 \right)$

[This reaction is used for large-scale production by oxidizing the $$\ce{Pd}$$(0) back to $$\ce{Pd}$$(II) with $$\ce{Cu}$$(II). Thus $$\ce{Pd} \left( 0 \right) + 2 \ce{Cu} \left( II \right) \rightarrow \ce{Pd} \left( II \right) + 2 \ce{Cu} \left( I \right)$$, and then the $$\ce{Cu}$$(I) is converted back to $$\ce{Cu}$$(II) with $$\ce{O_2}$$. The overall result is $$\ce{CH_2=CH_2} + \frac{1}{2} \ce{O_2} \rightarrow \ce{CH_3CHO}$$.]

b. The balance between the competitive nucleophilic reactions described in Part a is a delicate one as judged from the following results:

Write mechanistic steps that will account for the difference in stereochemical results of these reactions, noting that in one case there is a single carbonylation reaction and in the other a dicarbonylation reaction.

## Contributors

• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."