The increasing order of the rate of oxidation with $H I O_{4}$ of the following is:

IV < III < II < I

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I < II < III < IV

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IV < III = II < I

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I < II = III < IV

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Solution

The correct option is A IV < III < II < I
1,2- or Vicinal diols are cleaved by periodic acid, $H I O_{4}$ , into two carbonyl compounds. The periodate ester undergoes a rearrangement of the electrons, cleaving the $C - C$ bond, and forming two $C = O$ . The order of rate of oxidation can be explained on the basis of steric effect. The heavier the $d i o l$ less the chance to undergo oxidation by $H I O_{4}$ therefore the order is $I V < I I I < I I < I$ .

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The increasing order of the rate of oxidation with HIO4 of the following is:

The increasing order of the rate of oxidation with $H I O_{4}$ of the following is: