Arrange the following in the decreasing order of basic strength in aqueous solution:
A. \(CH_3NH_2\)
B. \((CH_3)_3N\)
C. \((CH_3)_2NH\)
D. \(NH_3\)
E. \(C_6H_5NH_2\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C > A > B > D > E
In water, the basic strength of the methylamines is set by three effects together: the +I effect of the alkyl groups (more electron density on N), solvation of the protonated amine by H-bonding (more N–H bonds give better solvation and a more stable cation), and steric hindrance to that H-bonding.
C. \( \ce{(CH3)2NH} \) (first, strongest) — two +I groups, and its cation still has two N–H bonds for solvation.
A. \( \ce{CH3NH2} \) (second) — one +I group; its cation has three N–H bonds and is solvated best of all.
B. \( \ce{(CH3)3N} \) (third) — three +I groups, but its cation has only one N–H for H-bonding and is crowded, so in water it is weaker than A.
D. \( \ce{NH3} \) (fourth) — no electron-releasing group, so it is weaker than all the alkylamines.
E. \( \ce{C6H5NH2} \) (last, weakest) — the lone pair on N is delocalised into the benzene ring by resonance, so it is least available for protonation.
NCERT (2026-27 edition):
Alkanamines versus Ammonia – “… The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows:
(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
(CH3)2NH > CH3NH2 > (CH3)3N > NH3”