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Organic Compounds Containing Nitrogen - Basicity of amines

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Amines act as Lewis bases due to the presence of a lone pair of electrons on the nitrogen atom, which can be donated to an electron-deficient species.

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Basicity is quantitatively expressed by the basic dissociation constant KbK_b. A larger KbK_b or a smaller pKbpK_b value indicates a stronger base.

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In the gaseous phase, the basicity of amines follows the order of the +I+I (inductive) effect: Tertiary amine (R3NR_3N) > Secondary amine (R2NHR_2NH) > Primary amine (RNH2RNH_2) > NH3NH_3.

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In aqueous solution, the basicity is determined by a combination of the +I+I effect, solvation effect (hydrogen bonding with water), and steric hindrance of alkyl groups.

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For methyl-substituted amines in aqueous solution, the order of basicity is: (CH3)2NH>CH3NH2>(CH3)3N>NH3(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3.

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For ethyl-substituted amines in aqueous solution, the order of basicity is: (C2H5)2NH>(C2H5)3N>C2H5NH2>NH3(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3.

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Aromatic amines like aniline (C6H5NH2C_6H_5NH_2) are significantly less basic than ammonia because the lone pair on nitrogen is delocalized into the benzene ring through resonance, making it less available for protonation.

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Substituents on the benzene ring affect basicity: Electron-donating groups (e.g., −OCH3-OCH_3, −CH3-CH_3) increase basicity, while electron-withdrawing groups (e.g., −NO2-NO_2, −CN-CN, −X-X) decrease basicity.

📐Formulae

R−NH2+H2O⇌R−NH3++OH−R-NH_2 + H_2O \rightleftharpoons R-NH_3^+ + OH^-

Kb=[RNH3+][OH−][RNH2]K_b = \frac{[RNH_3^+][OH^-]}{[RNH_2]}

pKb=−log⁡10KbpK_b = -\log_{10} K_b

Basicity (Gaseous): R3N>R2NH>RNH2>NH3\text{Basicity (Gaseous): } R_3N > R_2NH > RNH_2 > NH_3

Basicity (Aqueous Methyl): (CH3)2NH>CH3NH2>(CH3)3N>NH3\text{Basicity (Aqueous Methyl): } (CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3

💡Examples

Problem 1:

Arrange the following in increasing order of their basic strength in aqueous solution: CH3NH2CH_3NH_2, (CH3)2NH(CH_3)_2NH, (CH3)3N(CH_3)_3N, and NH3NH_3.

Solution:

NH3<(CH3)3N<CH3NH2<(CH3)2NHNH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH

Explanation:

In aqueous solution, the basicity is a result of +I+I effect, solvation effect, and steric hindrance. For methyl groups, the steric hindrance in (CH3)3N(CH_3)_3N and the reduced solvation of its cation make it less basic than the primary and secondary amines, despite having more alkyl groups.

Problem 2:

Why is aniline (C6H5NH2C_6H_5NH_2) a weaker base than methylamine (CH3NH2CH_3NH_2)?

Solution:

In aniline, the lone pair of electrons on the NN atom is in conjugation with the π\pi-electron system of the benzene ring. This delocalization (resonance) makes the lone pair less available for donation to a proton. In CH3NH2CH_3NH_2, the +I+I effect of the methyl group increases the electron density on NN.

Explanation:

Resonance structures of aniline show that the lone pair is distributed over the ortho and para positions of the ring, decreasing its availability. CH3NH2CH_3NH_2 has no such delocalization.

Problem 3:

Between pp-nitroaniline and pp-toluidine, which is more basic and why?

Solution:

pp-toluidine (CH3−C6H4−NH2CH_3-C_6H_4-NH_2) is more basic than pp-nitroaniline (NO2−C6H4−NH2NO_2-C_6H_4-NH_2).

Explanation:

The methyl group (−CH3-CH_3) in pp-toluidine is an electron-donating group (+I+I and hyperconjugation), which increases electron density on the NH2NH_2 group. The nitro group (−NO2-NO_2) in pp-nitroaniline is a strong electron-withdrawing group (−I-I and −R-R effects), which significantly reduces the availability of the lone pair on nitrogen.