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Propiophenone + Br2/AlCl3 in Et2O

I thought that the answer would be a simple electrophillic aromatic substitution and as the carbonyl is a meta director, the bromine would be present meta to the group.

But the given answer was the α-brominated compound and is also found out an article which says that this happens in case a catalytic amount of $\ce{AlCl3}$ is used while the bromine is substituted in the ring if a larger amount is used. How does this happen? Why does the amount of $\ce{AlCl3}$ matter?

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    $\begingroup$ Hmmm... Seems to me that something is wrong. The alpha-halogenation of the ketone is a reaction that takes place in aqueous medium with acid/base catalysis. This is because the reaction depends on generation of the enol/enolate form of the ketone. It is unlikely that alpha-halogenation would take place in an ether solvent with only $\ce {AlCl3}$ as a catalyst. $\endgroup$ May 20 '20 at 9:41
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    $\begingroup$ Catalytic AlCl3 enolizes the ketone. Bromination ensues liberating HBr which now acts as a catalyst while its concentration increases. Mix excess ketone with bromine in ether. Nothing happens and then the solution suddenly turns colorless. The enol content of the ketone is small but slowly bromination occurs forming more and more HBr. $\endgroup$
    – user55119
    May 20 '20 at 22:08
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The key in this reaction is ketone in propiophenone is acting as Lewis base to $\ce{AlCl3}$ Lewis acid. If $\ce{AlCl3}$ is in catalytic amount, it activate $\alpha$-$\ce{H}$ in side chain sufficiently enough (similar to acid would do to a carbonyl compound) and progress to produce $\alpha$-substituted product. If $\ce{AlCl3}$ is in excess, it seemingly deactivates aliphatic side chain completely (Ref.1), and ring substitution predominates to give $meta$-substituted product exclusively as a result (Ref.2 and 3):

AlCl3 in acetophenone

The reaction with excess $\ce{AlCl3}$ should necessarily be done in solvent free conditions, and excess $\ce{AlCl3}$ is needed such that one equivalent is used to complex with carbonyl oxygen and another equivalent is used to activate $\ce{Br2}$ molecule. As a result, benzene nucleus would be substituted and 3-bromoacetophenone would be obtained exclusively (Ref.1). The crude liquid product is purified as an oil after fractional distillation in 70-75% yield (Ref.2). The reaction with trace amount of $\ce{AlCl3}$ has done in dry ether and solid crude yield has been 88-96%. When recrystallized by methanol, the yield has been reduced to 64-66% (Ref.3). Another stark different between these two reactions are addition order. In the excess reaction, acetophenone has been added to $\ce{AlCl3}$ while in the catalytic reaction, $\ce{AlCl3}$ has been added to etherial acetophenone solution.

References:

  1. D. E. Pearson, H. W. Pope, W. W. Hargrove, W. E. Stamper, "The Swamping Catalyst Effect. II. Nuclear Halogenation of Aromatic Aldehydes and Ketones," J. Org. Chem. 1958, 23(10), 1412–1419 (https://doi.org/10.1021/jo01104a003).
  2. D. E. Pearson, H. W. Pope, W. W. Hargrove (Checked by B. C. McKusick, D. W. Wiley), "3-Bromoacetophenone," Org. Synth. 1960, 40, 7 (DOI: 10.15227/orgsyn.040.0007) (Organic Syntheses, Coll. Vol. 5, p.117 (1973)).
  3. R. M. Cowper, L. H. Davidson (Checked by Lee Irvin Smith, E. W. Kaiser), "Phenacyl Bromide ($\alpha$-Bromoacetophenone)," Org. Synth. 1939, 19, 24 (DOI: 10.15227/orgsyn.019.0024)(Organic Syntheses, Coll. Vol. 2, p.480 (1944)).
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The carbonyl group attached to the benzene ring is an electron-withdrawing group and deactivates the ring towards electrophilic substitution. Thus, $\ce{AlCl3}$ in excess is required to brominate the ring in this case. If a catalytic amount is used, then the $\alpha-\ce{H}$ atom is substituted by a standard Ketone halogenation mechanism.

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