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 determine the product of the oxidation of a secondary alkanol with Cr(VI), we can follow these steps:

### Step 1: Identify the Type of Alcohol
- A secondary alkanol (or secondary alcohol) is characterized by the presence of the hydroxyl group (-OH) attached to a carbon atom that is connected to two other carbon atoms.

### Step 2: Understand the Oxidation Process
- When secondary alcohols undergo oxidation, they are converted into ketones. This is a key distinction, as primary alcohols oxidize to aldehydes, and tertiary alcohols do not oxidize easily.

### Step 3: Identify the Oxidizing Agent
- In this case, Cr(VI) (chromium in the +6 oxidation state) is the oxidizing agent. Commonly, Cr(VI) is used in the form of chromium trioxide (CrO3) or potassium dichromate (K2Cr2O7).

### Step 4: Write the Reaction
- The oxidation of a secondary alcohol (R2CHOH) with Cr(VI) can be represented as:
\[ R_2CHOH + [O] \rightarrow R_2C=O + H_2O \]
Here, R2C=O represents the ketone formed.

### Step 5: Determine the Change in Oxidation State
- During the reaction, Cr(VI) is reduced to Cr(III) (chromium in the +3 oxidation state). This reduction is crucial as it indicates that Cr(VI) has accepted electrons during the oxidation of the alcohol.

### Conclusion
- Therefore, the oxidation of a secondary alkanol with Cr(VI) leads to the formation of a ketone and the reduction of chromium from +6 to +3.

### Final Answer
- The oxidation of a secondary alkanol with Cr(VI) leads to the formation of a **ketone** and the reduction of chromium to **Cr(III)**.

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