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Ferrocene has two structures (1 and 2). It is bonded to the aromatic cyclopentadienyl anion. Aromatic species such as benzene, cyclopentadienyl anion undergo electrophilic aromatic substitutions.

Then, will ferrocene undergo electrophilic aromtic substitution?

Which conformer between 1 and 2, would preferentially undergo Friedel Crafts acylation reaction?

enter image description here

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  • $\begingroup$ The question about which conformation is more reactive is irrelevant (and drawing the ferrocene molecule as an ionic compound is incorrect: the bonding is covalent, though complicated). The barrier to rotation is very small so there will be no significant preference between them at room temperature. $\endgroup$
    – matt_black
    Dec 19, 2022 at 11:17

2 Answers 2

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Due to the aromatic character of the cyclopentadienyl ligands, ferrocene can undergo electrophilic aromatic substitution reactions typical of aromatic compounds such as benzene. Ferrocene shows very high reactivity towards electrophiles, making it more closely comparable to phenol than to benzene, references and examples here and here.

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  • $\begingroup$ This does not answer weather conformer 2 or conformer 1 is more reactive towards acylation..In my question i have already mentioned that cyclopentadienyl anion is aromatic.So cyclopentadienyl does undergo electrophillic aromatic substitution. $\endgroup$ May 26, 2019 at 8:55
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    $\begingroup$ Ferrocene is formed from two planar cyclopentadienyl rings, not anions. The two rings are staggered with respect to each other, but are otherwise equivalent so there is no answer to your question. $\endgroup$
    – Waylander
    May 26, 2019 at 9:02
  • $\begingroup$ @Waylander true, but what matters is not that the most stable conformation is staggered but that the barrier to rotation is very small and all conformations will readily convert under normal reaction conditions. $\endgroup$
    – matt_black
    Dec 19, 2022 at 11:22
  • $\begingroup$ @matt_black Thank you for the additional information $\endgroup$
    – Waylander
    Dec 19, 2022 at 11:39
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Refer Inorganic chemistry by Huheey (4th edition, Page No. 646) They have proceeded with conformer 1 and successive acylation on both rings will result in acyl group attached on the same side of both CP rings enter image description here

However, in my notes, i have proceeded with conformer 2 and the 2 equivalents of acyl are attached on opposite sides of the two CP ring. enter image description here

The question about stability vs reactivity here. Stability - The doubly acylated product of conformer 2 will be more stable (Less strained transition state as well as trans looking product and thus better packing in crystal).

Reactivity - I don't have the answer, neither did i find its explanation anywhere. So sorry i can't help you there.

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  • $\begingroup$ Just because that is how the molecule is drawn in a textbook means nothing. The barrier to rotation is very small and, essentially, irrelevant to the ring reactivity. So, the "other side" is meaningless after double acylation. $\endgroup$
    – matt_black
    Dec 19, 2022 at 11:20
  • $\begingroup$ In the Solid state, Staggered conformation will be present predominantly (since symmetry will lead to more efficient packing) while in the Gaseous state, an equilibrium between staggered and eclipsed form exists because the energy gap between them is very less (4.2 kJ/mol). So, essentially, the question comes down to which conformer will be more reactive in the Gaseous state. I haven't found the answer to that, but whatever i did find, i updated here. $\endgroup$
    – VaibhavRaj
    Dec 19, 2022 at 12:00
  • $\begingroup$ Given the low barrier to rotation, it is unlikely that–even if the reactivity were different–you would notice it in a real reaction. And most reactions of this type are done in solution not solid or gas states, so those are irrelevant. $\endgroup$
    – matt_black
    Dec 19, 2022 at 14:42
  • $\begingroup$ The dominance of the eclipsed ferrocene conformer in solutions at room temperature has been reported by "Differentiation of ferrocene D5d and D5h conformers using IR spectroscopy" by N. Mohammadi et al. $\endgroup$
    – VaibhavRaj
    Dec 20, 2022 at 2:38

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