Which of the following reagent is used for the following reaction?
$$CH_3CH_2CH_3 \to CH_3CH_2CHO$$

Step 1: Identify the type of reaction.
The conversion of propane to propanal involves removal of hydrogen and introduction of one oxygen atom. This is called oxidative dehydrogenation (ODH).

Step 2: Write the balanced chemical equation for ODH of propane.
Using one‐half mole of oxygen per mole of propane yields propanal and water:
$$CH_3CH_2CH_3 + \frac{1}{2}O_2 \rightarrow CH_3CH_2CHO + H_2O$$
$$-(1)$$

Step 3: State the requirement for selective partial oxidation.
We need a catalyst that activates O-H and C-H bonds without over-oxidizing the aldehyde to an acid. Many strong oxidants or high-temperature metals give complete oxidation or cracking.

Step 4: Examine each option.
Option A: Copper at high temperature and pressure tends to crack alkanes or give CO/CO_2 rather than stop at an aldehyde.
Option B: Manganese acetate is used in radical allylic oxidations (Kharasch-Sosnovsky), not for propane.
Option D: Potassium permanganate is a very strong oxidant and would oxidize propanal further to propionic acid.

Step 5: Select the correct catalyst.
Molybdenum trioxide ($$MoO_3$$) is known to promote selective partial oxidation (ODH) of light alkanes to aldehydes with minimal over-oxidation. It activates $$O_2$$ gently and removes hydrogen without decomposing the aldehyde.

Therefore, the reagent used is Option C: Molybdenum oxide ($$MoO_3$$).