MHTCET 20 Apr 2026 Shift 1 Paper

Concept:Moist $\mathrm{Ag_2O}$ converts a quaternary ammonium halide into the corresponding quaternary ammonium hydroxide, which on heating undergoes Hofmann elimination to give the least substituted alkene.Explanation:Step (i) uses moist $\mathrm{Ag_2O}$, which exchanges the halide ion with hydroxide.Thus, substrate A must first be a quaternary ammonium halide.Step (ii) is heating, which causes Hofmann elimination of the quaternary ammonium hydroxide.In Hofmann elimination, one alkyl group is removed as an alkene and a $\beta$-hydrogen is lost.The alkene formed here is $\mathrm{CH_2=CH_2}$, so the eliminated group must be an ethyl group.The remaining amine is $\mathrm{CH_3CH_2N(CH_3)_2}$, containing one ethyl and two methyl groups on nitrogen.So the original quaternary ammonium ion contained two ethyl and two methyl groups:$$\left[\left(\mathrm{C_2H_5}\right)_2\mathrm{N}\left(\mathrm{CH_3}\right)_2\right]^+$$Its halide salt is diethyldimethylammonium halide.Therefore, substrate A is the halide, not the hydroxide, because the hydroxide forms only after step (i).Answer:Option A — Diethyldimethylammonium halide.