JEE Chemistry 2025 Question with Solution

Given: Five amines A, B, C, D, E are to be arranged in descending order of basicity.

Find: The correct order of basicity and the matching option.

Aromatic amines are generally less basic than aliphatic amines because the lone pair on nitrogen in aromatic amines gets delocalized into the benzene ring, so it is less available for protonation.

Compound B has a methoxy substituent, $-OCH_3$, which is an electron-donating group. It increases electron density and makes B more basic than unsubstituted aniline A.

Compound C has a nitro substituent, $-NO_2$, which is a strong electron-withdrawing group. It decreases electron density strongly, so C is the least basic.

Compound E, $(CH_3)_2NH$, is an aliphatic amine and is more basic because the lone pair on nitrogen is directly available for protonation.

Compound D, $CH_3NH_2$, is also an aliphatic amine and is more basic than the aromatic amines.

Therefore, the descending order of basicity is $E > D > B > A > C$ This matches option B.

Given:

• A = aniline
• B = $p$-methoxy aniline
• C = $p$-nitro aniline
• D = $CH_3NH_2$
• E = $(CH_3)_2NH$

Find: Arrange them in decreasing basic strength.

Basicity depends on the availability of the lone pair on nitrogen for protonation. Alkyl groups show +I effect and increase basicity. In aromatic amines, the lone pair is delocalized into the ring, decreasing basicity. Electron-donating groups on the ring increase basicity, while electron-withdrawing groups decrease it.

So,

• E is most basic because two methyl groups increase electron density on nitrogen.
• D is next because it is an aliphatic amine with one methyl group.
• B is more basic than A because the methoxy group donates electron density by resonance.
• A is less basic than aliphatic amines because of resonance delocalization.
• C is least basic because the nitro group strongly withdraws electron density.

Hence, $E > D > B > A > C$ So the correct option is B.