Expt 3 - Friedel Crafts Acylation of Anisole

Relevant textbook reading - Klein, Chapter 19.1-19.10, especially 19.6

Reference for Procedure - Miles, W. H.; Nutaitis, C. F.; Anderton, C. A.  J. Chem. Ed., 1996, 73, 272.

Overview - You will react anisole with propionyl chloride in the presence of iron(III) chloride as a catalyst. The overall procedure is described in the experimental part of the above Journal of Chemical Education reference.

Procedure Notes

1. Run the reaction in the apparatus as described in the paper and diagrammed in Fig 7.4a (p 63) in Mohrig. Set the apparatus toward the back of the fume hood and it will not be necessary to use an inverted funnel as a gas trap.
   
2. Evaporation of the solvent will be carried out using the rotary evaporator.
3. Obtain the ¹H  NMR and IR spectra of the final product. One group will also be asked to obtain a C-13 NMR.

Assigned Questions

1. Why is iron(III) chloride attractive as an alternative to the traditional AlCl₃ as a catalyst for Friedel Crafts reactions? Do the results of this experiment prove that it should be
   
2. The methoxy group is generally considered an o,p-director in electrophilic aromatic substitution reactions. Discuss the theory behind this, writing the mechanism, and discussing factors that stabilize the intermediates of ortho and para attack of the electrophile.
3. o,p-Directing substituents often give mixtures of both ortho and para products, with the ortho isomer being statistically favored but sterically hindered. Thus, bulkier substituents tend to give lower yields of ortho product. (a) Did the reaction carried out here give detectable amounts of ortho product? (Look carefully at the proton NMR for impurity peaks that could be due to the ortho isomer.) (b) The methoxy group is not, by any stretch, a "bulky substituent", so why is so little ortho product formed? (c) Does the "accepted mechanism" given in Klein adequately explain these results? If warranted, propose an alternative mechanism that better explains the very large role that steric hindrance plays in this reaction.