1. Identify the transformation

We need to convert pent-3-en-2-ol into pent-3-en-2-one.

The substrate is: $\text{CH}_3-\text{CH(OH)}-\text{CH}=\text{CH}-\text{CH}_3$

The product is: $\text{CH}_3-\text{CO}-\text{CH}=\text{CH}-\text{CH}_3$

So this is simply oxidation of a secondary alcohol to a ketone.

2. Important selectivity issue

The molecule also contains a C=C double bond. Therefore, the oxidizing reagent must:

• oxidize the secondary alcohol to a ketone
• not attack the double bond

So we need a mild, selective oxidizing agent.

3. Check each option

Option A: Acidic permanganate

Acidic $\mathrm{KMnO_4}$ is a strong oxidizing agent. It can oxidize alcohols, but it also tends to react with the double bond, causing further oxidation/cleavage under acidic conditions.

So this is not suitable.

Option B: Acidic dichromate

Acidic dichromate $\left(\mathrm{K_2Cr_2O_7/H^+}\right)$ also oxidizes secondary alcohols to ketones, but being strongly acidic and relatively harsh, it is less suitable in the presence of an alkene, where side reactions may occur.

So this is not the best reagent.

Option C: Chromic anhydride in glacial acetic acid

Chromic anhydride ($\mathrm{CrO_3}$) in glacial acetic acid is also an oxidizing system, but it is not as selective as PCC for oxidizing an alcohol in the presence of an alkene.

So this is not the best choice.

Option D: Pyridinium chlorochromate (PCC)

PCC is a mild and selective oxidizing agent. It converts:

• primary alcohols $\to$ aldehydes
• secondary alcohols $\to$ ketones

while generally leaving the C=C double bond untouched.

Thus, $\text{pent-3-en-2-ol} \xrightarrow{\mathrm{PCC}} \text{pent-3-en-2-one}$

So this is the best reagent.

4. Final answer

The correct option is: $\boxed{\text{D: Pyridinium chlorochromate}}$