The oxidation of a secondary alkanol with Cr(VI) leads to the formation of :

Options

A. An alkanone and Cr(II)
B. A ketone and Cr (III)
C. An alkanal and Cr (III)
D. A ketone and Cr (II)

Answer

To solve the question regarding the oxidation of a secondary alkanol with Cr(VI), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**:
- A secondary alkanol is a secondary alcohol, which means that the hydroxyl group (-OH) is attached to a carbon that is connected to two other carbon atoms.

2. **Understand the Oxidation Process**:
- When secondary alcohols are oxidized, they typically convert to ketones. This is a key characteristic of secondary alcohols during oxidation.

3. **Recognize the Oxidizing Agent**:
- Cr(VI) refers to chromium in the +6 oxidation state, which is a strong oxidizing agent. During the oxidation process, Cr(VI) is reduced to Cr(III) (chromium in the +3 oxidation state).

4. **Determine the Product**:
- Since we have established that secondary alcohols oxidize to ketones, we can conclude that the product of the oxidation of a secondary alkanol with Cr(VI) will be a ketone.

5. **Eliminate Incorrect Options**:
- If options are provided, we can eliminate those that do not correspond to a ketone or those that suggest the formation of an aldehyde or other incorrect products.

6. **Final Conclusion**:
- The correct answer to the question is that the oxidation of a secondary alkanol with Cr(VI) leads to the formation of a ketone, and Cr(VI) is reduced to Cr(III).