Chromic acid oxidation of alcohol (Section 11-2A) occurs in two steps: formation of the chromate ester, followed by an elimination of chromium. Which step do you expect to be rate-limiting? Careful kinetic studies have shown that Compound A undergoes chromic acid oxidation over 10 times as fast as Compound B. Explain this large difference in rates.

The product formed by the oxidation of alcohol by chromic acid is ketone. Chromic acid has one acidic proton that plays a major role in converting an alcohol to a ketone.

The reaction of compound A is faster with chromic acid and the reaction of compound B is slower with chromic acid. The reaction happens in two steps:

• Formation of ester
• Loss of hydrogen and chromate to form C=O.

The chemical reactions taking place are given below.

The reactions take place in axial and equatorial positions. The substituents present in the equatorial systems are more stable when compared to the axial position. This is because the axial positions have 1, 3-diaxial interactions.

We can expect the equatorial chromate ester formed is faster than the axial chromate ester. It is a contrary assumption. So, step 1 is not rate-determining.

The step 2, if the rate-limiting is the elimination, we can expect the base approach to the hydrogen present in the equatorial position is faster than the base approach of axial hydrogen.

But axial hydrogen tends to leave due to steric congestion. This leads to the conclusion that the second step of the mechanism is rate-determining.