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Amines - Physical and Chemical Properties

Grade 12CBSEChemistry

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

πŸ”‘Concepts

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Amines are derivatives of NH3NH_3 obtained by replacing one or more HH atoms with alkyl or aryl groups. Their physical state ranges from gases (lower aliphatic amines like CH3NH2CH_3NH_2) to liquids and solids.

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Boiling point order: Primary (1∘1^\circ) > Secondary (2∘2^\circ) > Tertiary (3∘3^\circ). This is because primary and secondary amines can form intermolecular hydrogen bonds due to the presence of HH atoms on NN, whereas tertiary amines lack such HH atoms.

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Solubility: Lower aliphatic amines are soluble in H2OH_2O because they can form hydrogen bonds with water molecules. Solubility decreases as the size of the hydrophobic alkyl part increases.

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Basicity: Amines are Lewis bases because of the lone pair of electrons on the NN atom. In the gaseous phase, basicity follows the order: 3∘>2∘>1∘>NH33^\circ > 2^\circ > 1^\circ > NH_3 due to the +I+I effect of alkyl groups.

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Basicity in Aqueous Solution: The order is influenced by the +I+I effect, solvation effect, and steric hindrance. For methyl-substituted amines, the order is (CH3)2NH>CH3NH2>(CH3)3N>NH3(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3. For ethyl-substituted amines, the order 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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Aniline is less basic than NH3NH_3 because the lone pair on NN is involved in resonance with the benzene ring, making it less available for protonation.

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Carbylamine Reaction: Only aliphatic and aromatic primary amines react with CHCl3CHCl_3 and ethanolic KOHKOH to form foul-smelling isocyanides (Rβˆ’NCR-NC). This is a test for 1∘1^\circ amines.

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Hinsberg's Reagent: Benzenesulphonyl chloride (C6H5SO2ClC_6H_5SO_2Cl) is used to distinguish between 1∘1^\circ, 2∘2^\circ, and 3∘3^\circ amines. 1∘1^\circ amines form a precipitate soluble in alkali, 2∘2^\circ amines form a precipitate insoluble in alkali, and 3∘3^\circ amines do not react.

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Electrophilic Substitution: In Aniline, the βˆ’NH2-NH_2 group is highly activating and ortho/para directing. Bromination with bromine water yields 2,4,6βˆ’tribromoaniline2,4,6-tribromoaniline.

πŸ“Formulae

Rβˆ’NH2+CHCl3+3KOHβ†’Ξ”Rβˆ’NC+3KCl+3H2OR-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O

C6H5NH2+NaNO2+2HClβ†’273βˆ’278K[C6H5N2]+Clβˆ’+NaCl+2H2OC_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} [C_6H_5N_2]^+Cl^- + NaCl + 2H_2O

pKb=βˆ’log⁑10KbpK_b = -\log_{10} K_b

C6H5SO2Cl+Rβˆ’NH2β†’C6H5SO2NHR+HClC_6H_5SO_2Cl + R-NH_2 \rightarrow C_6H_5SO_2NHR + HCl

Arβˆ’NH2+(CH3CO)2Oβ†’PyridineArβˆ’NHCOCH3+CH3COOHAr-NH_2 + (CH_3CO)_2O \xrightarrow{Pyridine} Ar-NHCOCH_3 + CH_3COOH

πŸ’‘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, NH3NH_3.

Solution:

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

Explanation:

In aqueous solution, basicity is determined by a combination of the inductive effect, solvation effect (hydration), and steric hindrance. For the methyl group, the steric hindrance is low, but the solvation of the 1∘1^\circ cation is higher than the 3∘3^\circ cation, making the 3∘3^\circ amine ((CH3)3N(CH_3)_3N) less basic than expected from the +I+I effect alone.

Problem 2:

Why does aniline not undergo Friedel-Crafts reaction?

Solution:

Aniline is a Lewis base and reacts with the Lewis acid catalyst AlCl3AlCl_3 used in Friedel-Crafts reactions.

Explanation:

The reaction forms a salt: C6H5NH2+AlCl3β†’C6H5NH2+AlCl3βˆ’C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2^+AlCl_3^-. The positive charge on NN in the salt makes the benzene ring highly deactivated for electrophilic substitution.

Problem 3:

Convert Aniline to Fluorobenzene.

Solution:

C6H5NH2β†’NaNO2/HCl,273KC6H5N2+Clβˆ’β†’HBF4C6H5N2+BF4βˆ’β†’Ξ”C6H5FC_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 273K} C_6H_5N_2^+Cl^- \xrightarrow{HBF_4} C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F

Explanation:

This is known as the Balz-Schiemann reaction. Aniline is first converted to a diazonium salt, then reacted with fluoroboric acid (HBF4HBF_4) to form benzene diazonium fluoroborate, which decomposes upon heating to yield fluorobenzene.

Physical and Chemical Properties - Revision Notes & Key Formulas | CBSE Class 12 Chemistry