BenzaIdehyde (C_6H_5CHO) is oxidized to benzoic acid by potassium permanganate i

Question

BenzaIdehyde (C_6H_5CHO) is oxidized to benzoic acid by potassium permanganate in an acid medium. The rate of the reaction is found to be directly proportional to the concentrations of H^+, benzaldehyde, and KMnO_4. When the oxidation is carried out with C_6H_5CDO, the rate is much slower than with C_6H_5 CHO. The oxidation of C_6H_5CHO in H_2O^18 and MnO_4^- gives the product C_6H_5COH, whereas the same reaction in the presence of Mn^18 O_4^- gave benzoic acid containing^18 O. What important mechanistic conclusion can you draw from the above experimental observation and results?

Explanation / Answer

When the oxidation is carried out using C6H5CDO instead of C6H5CHO , the rate of reaction was much slower this implies that in the rate determining step there involve breaking of C-H bond of aldedyde group (-CHO).

When the H is replaced with heavier isotope D the C-D bond becomes stronger. Thus reaction rate becomes slower as it will take more time to break a stronger bond.

When O18 labelled water was used product obtained was C6H5COOH, but when Mn18O4- (labelled with O18) was used product obtained was C6H5CO18OH (oxygen labelled with O18) was obtained.

That means that oxygen inserted in -CHO group to form -COOH comes fom MnO4- , not from H2O.

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