Ethyl alcohol on oxidation with acidified `K_(2)Cr_(2)O_(7)` gives

Options

A. Acetic acid
B. Acetaldehyde
C. Formaldehyde
D. Formic acid

Answer

To solve the question regarding the oxidation of ethyl alcohol (ethanol) with acidified potassium dichromate (K₂Cr₂O₇), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
Ethyl alcohol, also known as ethanol, has the chemical formula C₂H₅OH.

2. **Understand the Oxidation Process**:
When ethanol is oxidized, it can first be converted into an aldehyde. In this case, ethanol is oxidized to acetaldehyde (ethanal).

3. **Determine the Reaction Conditions**:
The oxidation is carried out using acidified potassium dichromate (K₂Cr₂O₇), which is a strong oxidizing agent.

4. **First Oxidation Step**:
The oxidation of ethanol (C₂H₅OH) leads to the formation of acetaldehyde (CH₃CHO). This involves the removal of hydrogen from the hydroxyl group (OH) and the formation of a carbonyl group (C=O).

5. **Further Oxidation**:
If the oxidation continues, acetaldehyde can be further oxidized to form ethanoic acid (acetic acid, CH₃COOH). This step also involves the addition of an oxygen atom and the removal of hydrogen.

6. **Final Product**:
The final product of the complete oxidation of ethanol with K₂Cr₂O₇ is ethanoic acid (acetic acid).

7. **Conclusion**:
Therefore, the answer to the question is that ethyl alcohol on oxidation with acidified K₂Cr₂O₇ gives ethanoic acid (acetic acid).

### Final Answer:
**Ethanoic acid (acetic acid)**