`A underset(H_(2)SO_(4))overset(K_(2)Cr_(2)O_(7))to (CH_(3))_(2)C=Ooverset(O)toCH_(3) COOH` ,
Here [A] is :

Options

A. Isopropyl alcohol
B. Ethyl alcohol
C. Tert butyl alcohol
D. Propyl alcohol

Answer

To determine the compound A that reacts with acidified K2Cr2O7 to form a ketone, which is then oxidized to a carboxylic acid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- K2Cr2O7 in acidic medium acts as an oxidizing agent. It can oxidize primary alcohols to aldehydes and secondary alcohols to ketones. Tertiary alcohols do not undergo oxidation.

2. **Identifying the Product**:
- The product of the first reaction is a ketone, which is given as (CH3)2C=O (acetone). This indicates that the starting compound A must be a secondary alcohol since secondary alcohols are oxidized to ketones.

3. **Structure of Secondary Alcohol**:
- The general structure of a secondary alcohol is R1R2C(OH)R3, where R1 and R2 are alkyl groups, and R3 can be hydrogen or another alkyl group.

4. **Finding the Correct Secondary Alcohol**:
- The ketone produced is (CH3)2C=O, which corresponds to acetone. The secondary alcohol that would yield acetone upon oxidation is isopropyl alcohol (2-propanol), which has the structure CH3CHOHCH3.

5. **Verification**:
- When isopropyl alcohol (2-propanol) is oxidized by K2Cr2O7, it forms acetone (the ketone).
- If acetone is further oxidized, it can be converted to acetic acid (CH3COOH), which is a carboxylic acid.

6. **Conclusion**:
- Therefore, the compound A is isopropyl alcohol (2-propanol).

### Final Answer:
**A = Isopropyl alcohol (2-propanol)**