Why is aniline less basic than methylamine or ammonia in aqueous solution?

In aniline ($C_6H_5NH_2$), the lone pair of electrons on nitrogen is delocalized into the benzene ring through resonance across 5 canonical structures. This reduces electron density on nitrogen for protonation. In contrast, in methylamine ($CH_3NH_2$), the $+I$ inductive effect of methyl group pushes electron density onto nitrogen without resonance delocalization, making it significantly more basic than aniline and ammonia ($NH_3$).

Why cannot primary aromatic amines ($Ar\text{-}NH_2$) be prepared by Gabriel Phthalimide Synthesis?

Gabriel Phthalimide synthesis relies on an $S_N2$ nucleophilic substitution where potassium phthalimide anion attacks an organic halide ($R\text{-}X$). Aryl halides ($Ar\text{-}X$) cannot undergo $S_N2$ displacement because the $C\text{-}X$ bond possesses partial double-bond character due to resonance with the aromatic ring and steric hindrance from the $\pi$-electron cloud.

How do I rapidly distinguish $1^\circ$, $2^\circ$, and $3^\circ$ amines in exam questions?

Use a 3-step test elimination: 1) Carbylamine Test ($CHCl_3 + KOH$): Positive (foul smell of $R\text{-}NC$) ONLY for $1^\circ$ amines. 2) Hinsberg Test ($C_6H_5SO_2Cl$): $1^\circ$ gives product soluble in $NaOH$; $2^\circ$ gives product insoluble in $NaOH$; $3^\circ$ does not react. 3) Nitrous acid test ($HNO_2$): $1^\circ$ aliphatic releases $N_2$ gas; $2^\circ$ forms yellow oily nitrosoamine; $3^\circ$ forms water-soluble quaternary salt.

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Key Formulas & High-Yield Equations

Hoffmann Bromamide Degradation

$R\text{-}CONH_2 + Br_2 + 4NaOH \rightarrow R\text{-}NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$

Valid strictly for

1^\circ

amides yielding primary amines with one less carbon atom. Proceeds via an intermediate alkyl isocyanate (

R\text{-}N=C=O

) with retention of configuration at migrating alkyl carbon.

Basic Strength Relationship

$pK_b = -\log_{10} K_b \quad \text{and} \quad pK_a + pK_b = 14 \quad (\text{at } 298\text{ K})$

Applies strictly to conjugate acid-base pairs in aqueous media at

25^\circ\text{C}

. Higher

K_b

or lower

pK_b

signifies greater basic strength.

Gabriel Phthalimide Reaction

$\text{Phthalimide} \xrightarrow{KOH} \text{Potassium Phthalimide} \xrightarrow{R\text{-}X} N\text{-alkylphthalimide} \xrightarrow{NaOH(aq)} R\text{-}NH_2 + \text{Phthalate ion}$

Valid exclusively for pure

1^\circ

aliphatic amines. Aryl halides (

Ar\text{-}X

) cannot be used as

Ar\text{-}X

fails to undergo

S_N2

nucleophilic substitution.

Diazotization Reaction

$Ar\text{-}NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\text{ K}} Ar\text{-}N_2^+ Cl^- + NaCl + 2H_2O$

Valid strictly at low temperatures (

0-5^\circ\text{C}

). Elevated temperatures lead to decomposition into phenol (

Ar\text{-}OH

) and release of

N_2

gas.

Hinsberg Separation Test

$R\text{-}NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR \xrightarrow{NaOH} [C_6H_5SO_2NR]^- Na^+ \text{ (Soluble)}$

Valid for distinguishing classes of amines:

1^\circ

amine product dissolves in aqueous

NaOH

2^\circ

amine product remains insoluble in

NaOH

, and

3^\circ

amine does not react.

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2024

In the Hoffmann bromamide degradation reaction, $R\text{-}CONH_2 + Br_2 + 4KOH \rightarrow R\text{-}NH_2 + K_2CO_3 + 2KBr + 2H_2O$ , the intermediate species formed during the structural rearrangement step is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Base abstracts a proton from $R\text{-}CONH_2$ followed by bromination to yield $N$ -bromamide ( $R\text{-}CONHBr$ ).

Step 2

Abstraction of second acidic proton yields an anion which undergoes concerted intramolecular migration of alkyl group $R$ to nitrogen along with loss of $Br^-$ .

Step 3

This rearrangement directly forms alkyl isocyanate ( $R\text{-}N=C=O$ ), which upon subsequent alkaline hydrolysis generates $1^\circ$ amine $R\text{-}NH_2$ .

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Question #2JEE Main 2023

An organic compound $A$ ( $C_3H_9N$ ) reacts with $C_6H_5SO_2Cl$ to yield a clear solution which on acidification gives an insoluble precipitate. Compound $A$ on treatment with $HNO_2$ releases $N_2$ gas. Compound $A$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Reaction with $HNO_2$ accompanied by evolution of $N_2$ gas confirms that compound $A$ is a primary aliphatic amine ( $1^\circ$ amine).

Step 2

Reaction of $1^\circ$ amine with Hinsberg reagent ( $C_6H_5SO_2Cl$ ) produces $N$ -alkylbenzenesulphonamide containing an acidic $N\text{-}H$ hydrogen, which dissolves in $NaOH$ forming a soluble sodium salt solution.

Step 3

Acidification reprecipitates the neutral insoluble sulphonamide; among the given primary amine options with molecular formula $C_3H_9N$ , compound $A$ is Propan-1-amine ( $CH_3CH_2CH_2NH_2$ ).

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Hoffmann Bromamide Degradation

Basic Strength Relationship

Gabriel Phthalimide Reaction

Diazotization Reaction

Hinsberg Separation Test

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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