When 1-bromopropane is heated with concentrated alcoholic KOH, the major product is

1. NCERT §6.5 (Reactions): aqueous KOH gives SUBSTITUTION (alcohol product); ALCOHOLIC KOH gives ELIMINATION (alkene product). 2. The reason: in aqueous solution, KOH provides $\mathrm{OH^-}$ as a NUCLEOPHILE attacking the carbon (SN reaction). In alcoholic solution, the alkoxide $\mathrm{RO^-}$ from KOH dissolves and is bulkier — it prefers to act as a BASE abstracting a $\beta$-hydrogen (E2 elimination). 3. For $\mathrm{CH_3CH_2CH_2Br}$ (1-bromopropane), E2 abstracts an H from C2, eliminates Br from C1, gives $\mathrm{CH_3CH{=}CH_2}$ (PROPENE). 4. Option A and B describe substitution products from AQUEOUS KOH. Option C requires a second OH addition that doesn't happen in one step. 5. Industrially: alcoholic KOH is the standard reagent for converting alkyl halides to alkenes. _Source: NCERT Class 12 Chemistry Part 2, Ch 6, §6.5 (Elimination reaction — alcoholic KOH conditions), p. 13._