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Organic Compounds Containing Nitrogen - Methods of preparation and properties

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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Classification of Amines: Nitrogen compounds are classified as primary (1∘1^\circ), secondary (2circ2^circ), and tertiary (3∘3^\circ) based on the number of alkyl or aryl groups attached to the nitrogen atom: RNH2RNH_2, R2NHR_2NH, and R3NR_3N.

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Preparation from Nitro Compounds: Nitro compounds are reduced to amines using hydrogen gas in the presence of finely divided nickel, palladium, or platinum, or by using metals in acidic medium like Sn/HClSn/HCl or Fe/HClFe/HCl.

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Gabriel Phthalimide Synthesis: This method is used for the preparation of pure primary aliphatic amines. Phthalimide reacts with ethanolic KOHKOH to form potassium phthalimide, which then reacts with an alkyl halide followed by alkaline hydrolysis.

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Hofmann Bromamide Degradation Reaction: An amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, resulting in a primary amine with one carbon atom less than the parent amide: Rβˆ’CONH2β†’Rβˆ’NH2R-CONH_2 \rightarrow R-NH_2.

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Basic Character of Amines: Amines are basic due to the presence of a lone pair of electrons on the nitrogen atom. In the aqueous phase, the basic strength order for methyl-substituted amines is (CH3)2NH>CH3NH2>(CH3)3N>NH3(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3.

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Carbylamine Reaction: Aliphatic and aromatic primary amines, when heated with chloroform and ethanolic potassium hydroxide, form isocyanides (carbylamines) which have a foul smell. This serves as a test for primary amines.

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Hinsberg's Test: Used to distinguish between 1∘1^\circ, 2∘2^\circ, and 3∘3^\circ amines using benzenesulfonyl chloride (C6H5SO2ClC_6H_5SO_2Cl). 1∘1^\circ amines form a precipitate soluble in alkali; 2∘2^\circ amines form a precipitate insoluble in alkali; 3∘3^\circ amines do not react.

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Diazonium Salts: Formed by the reaction of aniline with nitrous acid (NaNO2+HClNaNO_2 + HCl) at 273βˆ’278K273-278 K. They are vital intermediates for synthesizing substituted benzene derivatives via Sandmeyer or Gattermann reactions.

πŸ“Formulae

Rβˆ’NO2+3H2β†’Pd/C,ethanolRβˆ’NH2+2H2OR-NO_2 + 3H_2 \xrightarrow{Pd/C, \text{ethanol}} R-NH_2 + 2H_2O

Rβˆ’CONH2+Br2+4NaOHβ†’Rβˆ’NH2+Na2CO3+2NaBr+2H2OR-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O

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’278KC6H5N2+Clβˆ’+NaCl+2H2OC_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

ArN2+Clβˆ’+CuCl/HClβ†’ArCl+N2ArN_2^+Cl^- + CuCl/HCl \rightarrow ArCl + N_2

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

pKb=βˆ’log⁑10KbpK_b = -\log_{10} K_b

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, the basic strength is determined by a combination of the inductive effect (+I+I), solvation effect (hydrogen bonding with water), and steric hindrance. For the methyl group, the (CH3)2NH(CH_3)_2NH (secondary) is the strongest base, followed by the primary, then the tertiary (where steric hindrance reduces stability), and finally ammonia.

Problem 2:

How will you convert Benzenediazonium chloride to Fluorobenzene?

Solution:

C6H5N2+Clβˆ’+HBF4β†’C6H5N2+BF4βˆ’β†’Ξ”C6H5F+BF3+N2C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + BF_3 + N_2

Explanation:

This is known as the Balz-Schiemann reaction. Benzenediazonium chloride is treated with fluoroboric acid (HBF4HBF_4) to precipitate benzenediazonium fluoroborate, which on heating decomposes to yield fluorobenzene.

Problem 3:

Identify the product 'A' in the reaction: CH3CH2CONH2β†’Br2/KOHACH_3CH_2CONH_2 \xrightarrow{Br_2/KOH} A.

Solution:

CH3CH2NH2CH_3CH_2NH_2 (Ethylamine)

Explanation:

The reaction is a Hofmann Bromamide Degradation. The amide (CH3CH2CONH2CH_3CH_2CONH_2) reacts with Br2Br_2 and KOHKOH to lose the carbonyl carbon atom as K2CO3K_2CO_3, resulting in an amine (CH3CH2NH2CH_3CH_2NH_2) with one fewer carbon atom.

Methods of preparation and properties Revision - Class 12 Chemistry ICSE