Oxidation of alkanols
Russ.Chem.Bull., Int.Ed., Vol. 58, No. 11, November, 2009 2403
Scheme 2
bromide anions with water molecules, and therefore
deceleration of their oxidation by CeIV. The excess of
LiBr compensates this effect. In the oxidation of secondary
alkanols to alkanones, a catalytic amount of LiBr suffices;
stoichiometric amount of LiBr leads to αꢀbromo ketones.
At a molar ratio 3 : CeIII : LiBr : H2O2 equal to 1 : 2 : 1 : 10,
2ꢀbromopentanꢀ3ꢀone is formed in a yield of 77 %.
Oxidation of alkanols with a Ce(NO3)3•6H2O—LiBr—H2O2
system (general procedure). To an aqueous solution of an alkanol
(1 mmol), Ce(NO3)3, and LiBr (10 mL), portions (0.5–0.6 mL)
of 35% H2O2 (overall. 10 mL, the reactant ratio is given above)
were added at 65–70 °C and vigorous stirring. After addition of
the first portion, yellow coloration typical of CeIV salts appeared,
only after complete discoloration of the reaction mixture the
next portion of H2O2 was added. Next, the reaction mixture was
cooled, extracted with ether (3×15 mL), washed with aqueous
NaHCO3 solution and water, and dried with MgSO4. The yields
of the reaction products and conversion were determined by
GLC with authentic samples of the internal standards. For the
preparative isolation of the products, column chromatography
on silica gel was used.
R = C5H11 (a), C6H13 (b), C8H17 (c)
i.
; ii. CeIV, ; iii. CeIV—LiBr
By the example of the oxidation of heptanꢀ1ꢀol (1b) in
water, it was established that no reaction occurs in the
absence of one of mediators. At slow uniform addition of
H2O2 to the mixture in 4 h at 65–70 °C at a molar ratio
of reactants 1b : CeIII : LiBr : H2O2 equal to 1 : 0.5 : 5 : 10
the conversion of alkanol 1b was 81 %, the selectivity of
its transformation to heptyl heptanoate (2b) was 98%.
With doubled amounts of CeIII and LiBr, these characterꢀ
istics were almost unchanged, while with halved amounts
the conversion of alkanol 1b noticeably decreased. Subꢀ
stitution of NaBr and KBr for LiBr decreases the converꢀ
sion as well, but the selectivity remained unchanged.
Like compound 1b, hexanꢀ1ꢀol (1a) and nonanꢀ1ꢀol
(1c) oxidize to form esters C5H11COOC6H13 (2a) and
C8H17COOC9H19 (2c), respectively, with the same yield
and selectivity.
This work was financially supported by the Russian
Foundation for Basic Research (Project No. 09ꢀ03ꢀ00292a)
and the Council on Grants at the president of the Russian
Federation (Program for State Support of Leading Scienꢀ
tific Schools of the Russian Federation. Grant 2942.2008.3).
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Secondary alkanols oxidize more readily than primary
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In the case of primary alkanols, an excess of LiBr is
required, which probably is caused by solvation of the
Received September 21, 2009