For the same alkyl group, all the three primary, secondary and tertiary amines are more basic than ammonia in aqueous solutions. Their relative basicity, however, depends upon a combination of three factors: +I-effect of alkyl groups, the stability of the ammonium cation (formed after accepting a proton) due to H-bonding with water molecules and the steric effect of the alkyl groups which tend to reduce the extent of H-bonding. All these three factors are favourable for secondary amines thereby making them the strongest bases. If, however, the alkyl group is bigger than the methyl group, +I-effect of the alkyl group outweighs stability of the ammonium cation due to H-bonding thereby making tertiary amines more basic than the primary amines. Due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength.
An aqueous solution of methanamine is added to an aqueous solution of iron(III) chloride, and a precipitate is formed. Which of the following correctly describes the precipitate and how it is formed?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Hydrated ferric oxide, precipitated by the OH− ions that methanamine produces in water
Step 1 – Methanamine gives OH− ions in water: methanamine is a base. Its nitrogen takes a proton from a water molecule, and hydroxide ions are formed, just as with ammonia:
\( \displaystyle \ce{CH3NH2 + H2O <=> CH3NH3^+ + OH^-} \)
Step 2 – Precipitation of iron(III): these hydroxide ions precipitate the iron(III) ions:
\( \displaystyle \ce{Fe^3+ + 3OH^- -> Fe(OH)3 v} \)
Final product: the precipitate is written as Fe(OH)3, or as hydrated ferric oxide, Fe2O3·xH2O (Option 1). Adding the two steps gives the full reaction:
\( \displaystyle \ce{FeCl3 + 3CH3NH2 + 3H2O -> Fe(OH)3 v + 3CH3NH3^+Cl^-} \)
the precipitate is formed by the OH− ions that methanamine produces as a base;
iron stays in the +3 state; methanamine acts as a base, not as a reducing agent.
Why the other options are incorrect
Option (2) Methylammonium chloride, formed from methanamine and the chloride ions — Incorrect. Methylammonium chloride does stay in the solution, but it is an amine salt and amine salts are soluble in water, so it is not the precipitate.
Option (3) Iron(II) hydroxide, formed after methanamine reduces Fe3+ ions to Fe2+ ions — Incorrect. Methanamine acts here as a base, not as a reducing agent; iron stays in the +3 state.
Option (4) Metallic iron, formed when methanamine reduces Fe3+ ions — Incorrect. No reduction takes place, so no iron metal is formed; the precipitate is an iron(III) compound.
NCERT (2026-27 edition):
9.6 Chemical Reactions – “Basic character of amines can be better understood in terms of their Kb and pKb values as explained below:”
\( \displaystyle \ce{R-NH2 + H2O <=> R-NH3^+ + OH^-} \)
Exercises – “9.3 Account for the following: … (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.”