When OXIDISED with strong oxidising agents like acidified $\mathrm{KMnO_4}$, the product of a PRIMARY alcohol is

1. NCERT §7.4 (Oxidation of alcohols): the oxidation product depends on the alcohol's class AND the reagent strength. 2. PRIMARY alcohol ($\mathrm{R{-}CH_2OH}$): with MILD oxidants (e.g. PCC) → aldehyde $\mathrm{R{-}CHO}$, stops there. With STRONG oxidants (acidified $\mathrm{KMnO_4}$, $\mathrm{K_2Cr_2O_7}$): aldehyde is further oxidised to CARBOXYLIC ACID $\mathrm{R{-}COOH}$. 3. SECONDARY alcohol ($\mathrm{R_2CHOH}$): gives a KETONE $\mathrm{R_2C{=}O}$. Cannot be oxidised further without breaking C-C bonds. 4. TERTIARY alcohol ($\mathrm{R_3COH}$): RESISTS oxidation — no H on the OH carbon for removal. 5. Option A correctly describes the primary alcohol path. Option B is the secondary alcohol result. Options C and D describe unrelated reactions. _Source: NCERT Class 12 Chemistry Part 2, Ch 7, §7.4 (Oxidation of alcohols), p. 13–14._