Step 1: Identify the transformation needed.
Allyl alcohol ($CH_2{=}CHCH_2OH$) must be oxidized selectively at the $-OH$ group to give propenal ($CH_2{=}CHCHO$), without breaking the $C{=}C$ double bond or over-oxidizing to the acid.
Step 2: Why activated $MnO_2$ is the correct choice.
Activated $MnO_2$ selectively oxidizes allylic and benzylic alcohols to aldehydes while leaving the carbon-carbon double bond intact. Stronger oxidants such as $KMnO_4$ would cleave the double bond, and $K_2Cr_2O_7$ would over-oxidize to the carboxylic acid.
Step 3: Reaction.
\[ CH_2{=}CHCH_2OH \xrightarrow{MnO_2} CH_2{=}CHCHO \] \[ \boxed{MnO_2 \text{ (Activated Manganese Dioxide)}} \]