Propane yields propene, which cleaves to formic and acetic acids

Steps:

• Alkane C4H8 likely erroneous for C3H8 (propane); it has two nonequivalent hydrogens, yielding 1-chloropropane and 2-chloropropane via free radical chlorination.
• Ethanolic KOH dehydrohalogenates both to the same alkene T: propene (CH3CH=CH2).
• Hot concentrated dilute KMnO4 oxidatively cleaves propene's double bond.
• The =CH2 group forms HCOOH; the =CHCH3 group forms CH3COOH.

Why B is correct:

• Oxidative cleavage of terminal alkene CH3CH=CH2 yields CH3COOH from the substituted carbon and HCOOH from =CH2 under dilute conditions preventing further oxidation to CO2.

Why the others are wrong:

• A: =CH2 oxidizes to HCOOH (dilute KMnO4), not CO2 (vigorous conditions).
• C: Ignores HCOOH from =CH2; only CH3COOH would imply symmetrical internal alkene like but-2-ene.
• D: Same as C; duplicate option, incomplete products.

Final answer: B