S.K. Jana et al. / Journal of Catalysis 247 (2007) 214–222
221
Scheme 2.
Although the mechanism of oxygen activation and trans-
fer by the Mn-containing hydrotalcite-based basic catalyst for
alkylaromatic oxidation remains incompletely understood, the
finding of traces of ethylbenzene hydroperoxide in the reac-
tion mixture suggests that the reaction proceeds through the
formation of ethylbenzene hydroperoxide as an active interme-
diate. The formation of ethylbenzene hydroperoxide through
redox processes was confirmed by analyzing the presence of
Mn species in two different oxidation states, +2 and +3 (as ob-
served from XPS analysis), in the catalyst obtained after ethyl-
benzene oxidation. The transformation of ethylbenzene to eth-
ylbenzene hydroperoxide by free-radical pathway is confirmed
by the drastic decrease in ethylbenzene oxidation activity of
Mn-MgAl hydrotalcite on addition of hydroquinone as a free-
radical scavenger (Table 2, entry 7). Based on the four identified
products in the ethylbenzene oxidation reaction we speculate
that the ethylbenzene oxidation involves the following reactions
over the Mn-based hydrotalcite catalysts:
alytic activity in the oxidation of various alkylaromatics to their
corresponding benzylic ketones with atmospheric pressure of
molecular oxygen as the sole oxidant under solvent-free and
mild reaction conditions. The catalytic performance of Mn-
MgAl hydrotalcite postsynthesized by adopting the “memory
effect” was found to be much higher than that of MnAl hydro-
talcite prepared by a conventional coprecipitation technique or
MnO−4 -exchanged MgAl hydrotalcite synthesized by adopting
the “memory effect.” We have confirmed that the Mn-MgAl hy-
drotalcite is stable and reusable in the reaction.
Acknowledgment
S.K.J. acknowledges the Japan Society for the Promotion of
Science (JSPS) for a postdoctoral research fellowship as a for-
eign researcher.
References
C6H5–CH2–CH3 + O2 → C6H5–CH(OOH)–CH3,
C6H5–CH(OOH)–CH3 → C6H5–CO–CH3 + H2O,
C6H5–CH(OOH)–CH3 → C6H5–CHO + CH3OH,
C6H5–CH(OOH)–CH3 → C6H5–CH(OH)–CH3 + 0.5O2.
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