$\mathrm{Isopropylalcohol} \stackrel{{\mathrm{H}}^{+}/{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}{\to } \mathrm{final} \mathrm{product}$ 

Isopropyl alcohol (CH₃CH(OH)CH₃) undergoes oxidation when treated with an oxidizing agent like K₂Cr₂O₇ (Potassium Dichromate) in acidic conditions. Potassium dichromate is a powerful oxidizing agent that can oxidize alcohols in a two-step process. Let’s break down the oxidation process in detail:

Step 1: Oxidation to Aldehyde

In the presence of K₂Cr₂O₇ and an acidic medium, isopropyl alcohol undergoes the first stage of oxidation. The hydroxyl group (-OH) in isopropyl alcohol is oxidized to form an aldehyde. The reaction proceeds as follows:

CH₃CH(OH)CH₃  →  CH₃COH (Acetaldehyde)    

Step 2: Further Oxidation to Carboxylic Acid

Upon continued oxidation, the aldehyde (Acetaldehyde) undergoes further oxidation to form a carboxylic acid. The second stage of oxidation converts the aldehyde into acetic acid (ethanoic acid). This transformation is facilitated by the presence of K₂Cr₂O₇ in acidic conditions:

CH₃COH (Acetaldehyde)  →  CH₃COOH (Acetic acid)    

Overall Reaction

The complete reaction, where isopropyl alcohol is oxidized to acetic acid (ethanoic acid) by K₂Cr₂O₇, is shown below:

CH₃CH(OH)CH₃ + H+ / K₂Cr₂O₇  →  CH₃COOH + CH₄    

Conclusion

Hence, when isopropyl alcohol is treated with K₂Cr₂O₇ in the presence of an acidic medium, it undergoes oxidation to form acetic acid (ethanoic acid). Therefore, the correct final product is ethanoic acid.

The correct answer is option (D) Ethanoic Acid.