AP EAPCET 21 May 2025 Shift 2 Paper

Statement-I:$\mathrm{CH}_3\mathrm{NH}_2$ is more basic than $\mathrm{NH}_3$ but $\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$ is less basic than $\mathrm{NH}_3$.Explanation:In methylamine $(\mathrm{CH}_3\mathrm{NH}_2)$, the $-\mathrm{CH}_3$ group is electron donating ($+\mathrm{I}$ effect), which increases the electron density on the nitrogen atom and makes it more basic than ammonia.In aniline $(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2)$, the lone pair on nitrogen is delocalised into the benzene ring through resonance, so it is much less available to accept a proton, making aniline less basic than ammonia.Hence, Statement-I is correct.Statement-II:The order $(\mathrm{C}_2\mathrm{H}_5)_3\mathrm{N}>(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}>\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2$ is the order of basic strength in the gas phase, i.e. the order expected from the $+\mathrm{I}$ effect alone.In the aqueous phase, the basicity is governed by two opposing factors:(i) $+\mathrm{I}$ effect of alkyl groups, which increases as tertiary $>$ secondary $>$ primary, and(ii) solvation (hydration) of the conjugate ammonium ion, which is strongest for the primary amine (maximum number of $\mathrm{N}-\mathrm{H}$ bonds available for hydrogen bonding) and decreases as primary $>$ secondary $>$ tertiary.Because of this competing solvation effect, the observed order in water is not the gas-phase order.The actual order in aqueous solution, from the $\mathrm{p}K_b$ values, is:$(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}\ (\mathrm{p}K_b=3.00)>(\mathrm{C}_2\mathrm{H}_5)_3\mathrm{N}\ (\mathrm{p}K_b=3.25)>\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2\ (\mathrm{p}K_b=3.29)>\mathrm{NH}_3\ (\mathrm{p}K_b=4.75)$A lower $\mathrm{p}K_b$ value means a stronger base.So the order written in Statement-II is incorrect for the aqueous phase; it is the gas-phase order.Hence, Statement-II is not correct.Therefore, Statement-I is correct, but Statement-II is not correct.The correct option is C.