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ALCOHOLS FROM CARBONYL COMPOUNDS 229
Canceling common terms, we get:
3RCH2OH + 2Cr2O7
2− + 16H+ −→ 3RCO2H + 4Cr3+ + 11H2O
This shows that the oxidation of 3 mol of a primary alcohol to a carboxylic acid requires
2 mol of dichromate.
EXAMPLE 2
Write a balanced equation for the oxidation of styrene to benzoate ion and carbonate ion
by MnO4
− in alkaline solution.
Reduction half-reaction:
MnO4
− + 4H+ + 3e− = MnO2 + 2H2O (in acid)
Since this reaction is carried out in basic solution, we must add 4 HO− to neutralize the
4H+ on the left side, and, of course, 4 HO− to the right side to maintain a balanced equa-
tion.
MnO4
− + 4H+ + 4 HO−+ 3e− = MnO2 + 2H2O + 4 HO−
or, MnO4
−+ 2H2O + 3e− = MnO2 + 4 HO−
Oxidation half-reaction:
ArCH CH2 + 5H2O = ArCO2
−+ CO3
2− + 13H+ + 10e−
We add 13 HO− to each side to neutralize the H+ on the right side,
ArCH CH2+ 5H2O + 13HO− = ArCO2
− + CO3
2−+ 13H2O + 10e−
The least common multiple is 30, so we multiply the reduction half-reaction by 10 and the
oxidation half-reaction by 3 and add:
3ArCH CH2 + 39HO− + 10MnO4
− + 20H2O = 3ArCO2
− + 3CO3
2− +
24H2O + 10MnO2 + 40HO−
Canceling:
3ArCH CH2 + 10MnO4
− −→ 3ArCO2
− + 3CO3
2− + 4H2O + 10MnO2 + HO−
SAMPLE PROBLEMS
Using the ion-electron half-reaction method, write balanced equations for the following
oxidation reactions.
(a) Cyclohexene + MnO4
− + H+ (hot)−→ HO2C(CH2)4CO2H + Mn2+ + H2O
(b) Cyclopentene + MnO4
− + H2O
(cold)−→ cis-1,2-cyclopentanediol + MnO2+ HO−
(c) Cyclopentanol + HNO3
(hot)−→ HO2C(CH2)3CO2H + NO2 + H2O
(d) 1,2,3-Cyclohexanetriol + HIO4
(cold)−→ OCH(CH2)3CHO + HCO2H + HIO3