(a) (I) Give reasons : (3 + 2) (i) Aniline on nitration gives good amount of m-nitroaniline, though -NH₂ group

Question

Q. (a) (I) Give reasons : (3 + 2)

(i) Aniline on nitration gives good amount of m-nitroaniline, though

-NH_2

group is o/p directing in electrophilic substitution reactions.

(ii)

(CH_3)_2NH

is more basic than

(CH_3)_3N

in an aqueous solution.

(iii) Ammonolysis of alkyl halides is not a good method to prepare pure primary amines. (II) Write the reaction involved in the following :

(i) Carbyl amine test

(ii) Gabriel phthalimide synthesis

(OR)

(b) (I) Write the structures of A, B and C in the following reactions : (3 + 1 + 1)

(i)

C_6H_5N_2^+Cl^-

(benzenediazonium chloride — a benzene ring bearing

-N_2^+Cl^-

, drawn as a structure)

\xrightarrow{CuCN}

\xrightarrow{H_2O/H^+}

\xrightarrow[\Delta]{NH_3}

(ii) Nitrobenzene (

C_6H_5NO_2

— a benzene ring bearing

-NO_2

, drawn as a structure)

\xrightarrow{Fe/HCl}

\xrightarrow[273\ K]{NaNO_2 + HCl}

\xrightarrow{C_2H_5OH}

C (II) Why aniline does not undergo Friedel-Crafts reaction ? (III) Arrange the following in increasing order of their boiling point :

C_2H_5OH

C_2H_5NH_2

(C_2H_5)_3N

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Concept understanding — Amine Basicity Order

Amine Basicity Order – From Intuition to Precision

You already know that a base is something that accepts a proton ( $H^+$ ). An amine does this through the lone pair on its nitrogen. The more readily that lone pair grabs a proton, the stronger the base. So the question becomes: what makes that lone pair more or less willing to accept a proton?

The answer depends on three competing effects: inductive effect (electron pushing/pulling from nearby groups), steric hindrance (bulky groups blocking the proton), and solvation (how water molecules stabilise the protonated amine). And critically, the order flips depending on whether you are in the gas phase or in water.

The Intuition – What Would You Expect?

Imagine a nitrogen with three hydrogens: ammonia ( $NH_3$ ). Now replace one hydrogen with a methyl group ( $CH_3$ ). Methyl is an electron-donating group — it pushes electron density toward the nitrogen. That makes the lone pair richer, so it should grab a proton more easily. So methylamine ( $CH_3NH_2$ ) should be a stronger base than ammonia.

Replace another hydrogen with a second methyl group. Now you have dimethylamine ( $(CH_3)_2NH$ ). Even more electron density on nitrogen — stronger base still. Replace the third hydrogen: trimethylamine ( $(CH_3)_3N$ ). Maximum electron donation — so you would expect it to be the strongest base of all.

That is the inductive effect prediction: more alkyl groups → stronger base. In the gas phase, this is exactly what happens. The order is:

Gas phase: $NH_3 < 1^\circ < 2^\circ < 3^\circ$

But in water, the observed order is different. Why?

The Complication – Solvation and Steric Hindrance

When an amine accepts a proton, it becomes a positively charged ammonium ion ( $RNH_3^+$ ). In water, that positive charge is stabilised by hydrogen bonding with water molecules. The more hydrogens on the nitrogen, the more hydrogen bonds it can form — and the more stable the protonated form becomes.

A primary amine ( $1^\circ$ ) has three N–H hydrogens after protonation. A secondary ( $2^\circ$ ) has two. A tertiary ( $3^\circ$ ) has only one. So solvation stabilises the protonated form in the order: $1^\circ > 2^\circ > 3^\circ$ .

At the same time, bulky alkyl groups physically block the approach of a proton to the nitrogen — this is steric hindrance. A tertiary amine has three bulky groups crowding the nitrogen, making it harder for $H^+$ to reach the lone pair.

So in water, two effects oppose the inductive effect: solvation (which favours more N–H bonds) and steric hindrance (which favours less crowding). The result is a compromise.

The Precise Statement – Aqueous Phase Order

In water, the typical order of basic strength for aliphatic amines is:

Aqueous phase: $NH_3 < 3^\circ < 1^\circ < 2^\circ$

That is, secondary amines are the strongest bases in water, followed by primary, then tertiary, then ammonia.

Important

The order in water is not simply "more alkyl groups = stronger base". The secondary amine wins because it has a good balance: enough electron donation from two alkyl groups, but still two N–H hydrogens for solvation, and not too much steric hindrance.

The Full Picture – A Table

Amine	Gas-phase order (inductive only)	Aqueous order (all effects)	Reason
$NH_3$	Weakest	Weakest	No alkyl donation; only 3 H-bonds
$1^\circ$ ( $RNH_2$ )	Second	Second	One alkyl group donates; 3 H-bonds after protonation
$2^\circ$ ( $R_2NH$ )	Third	Strongest	Two alkyl groups donate; 2 H-bonds; steric hindrance still low
$3^\circ$ ( $R_3N$ )	Strongest	Third	Three alkyl groups donate most, but only 1 H-bond and high steric hindrance

(a) Acidic nitration protonates aniline to the meta-director $-NH_3^+$ (m-nitroaniline); $(CH_3)_2NH$ more basic in water (better solvated cation); ammonolysis over-alkylates; carbylamine ( $RNC$ ) and Gabriel give amines.

(b) (i) A=benzonitrile, B=benzoic acid, C=benzamide; (ii) A=aniline, B=benzenediazonium chloride, C=benzene; aniline+ $AlCl_3$ prevents F–C; b.p. $(C_2H_5)_3N < C_2H_5NH_2 < C_2H_5OH$ .

(I)(i) Appreciable m-nitroaniline

Nitration uses conc. $HNO_3$ / $H_2SO_4$ . The basic aniline is largely protonated:

C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+

The $-NH_3^+$ group is positively charged, strongly deactivating and meta-directing, so a substantial fraction of the product is m-nitroaniline (about half), the o/p isomers coming from the small unprotonated fraction.

(I)(ii) $(CH_3)_2NH$ more basic than $(CH_3)_3N$ in water

Aqueous basicity is governed by stability (solvation) of the protonated cation. $(CH_3)_2NH_2^+$ retains two N–H bonds and is well hydrogen-bonded to water; $(CH_3)_3NH^+$ has only one N–H and three bulky methyls hindering solvation. The better-solvated dimethylammonium ion makes $(CH_3)_2NH$ the stronger base in water, even though the pure +I effect would favour the tertiary amine.

(I)(iii) Ammonolysis gives impure $1°$ amine

RX \xrightarrow{NH_3} RNH_2 \xrightarrow{RX} R_2NH \xrightarrow{RX} R_3N \xrightarrow{RX} R_4N^+X^-

Because the first-formed amine is itself nucleophilic, a mixture of primary, secondary, tertiary amines and quaternary salt results; separating pure $1°$ amine is difficult.

(II) Named reactions

Carbylamine (isocyanide) test — a test for $1°$ amines:

RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O

The foul-smelling $RNC$ confirms a primary amine; $2°$ / $3°$ amines give no reaction.

Gabriel phthalimide synthesis — pure $1°$ amines:

\text{Phthalimide} \xrightarrow{KOH} \text{K-phthalimide} \xrightarrow{RX} N\text{-alkylphthalimide} \xrightarrow{H_3O^+} RNH_2 + \text{phthalic acid}

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Concept understanding — Diazonium Salt Reactions

Diazonium Salt Reactions – A First Look

Imagine you have a benzene ring, and you want to attach a new group — say a chlorine, a bromine, a cyano group, or even a hydroxyl — directly onto the ring. The benzene ring is stubborn; it doesn't easily let go of its hydrogen atoms for simple substitution. But there is a clever trick: first convert the ring into a diazonium salt, a highly reactive intermediate that will let you swap in almost any group you want.

That is the core idea. A diazonium salt is a temporary, energetic handle on the benzene ring that you can then replace with a wide variety of substituents. It is one of the most powerful tools in aromatic synthesis.

What is a Diazonium Salt?

A diazonium salt has the general formula Ar–N₂⁺ X⁻, where Ar is an aryl group (like phenyl, C₆H₅–), N₂⁺ is a diazonium cation (two nitrogen atoms triple-bonded, with a positive charge on the terminal nitrogen), and X⁻ is a counterion like chloride, bromide, or hydrogensulfate.

The key structural feature: the –N₂⁺ group is attached directly to the benzene ring. This group is unstable — it wants to leave as N₂ gas. That instability is exactly what makes it useful: when the N₂ leaves, the ring is left with a highly reactive carbocation-like intermediate that can be attacked by a nucleophile.

Watch out

Diazonium salts are thermally unstable and can explode if dried. They are almost always prepared and used in cold solution (0–5 °C) without isolation.

How Do You Make One? (Diazotization)

You start with a primary aromatic amine (Ar–NH₂). Treat it with nitrous acid (HNO₂) at low temperature (0–5 °C). The reaction is:

\text{Ar–NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} \text{Ar–N}_2^+ \text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

The nitrous acid is generated in situ from sodium nitrite and a mineral acid. The amine gets converted into the diazonium salt almost instantly. You must keep the solution cold; if it warms up, the diazonium salt decomposes and you get phenol and nitrogen gas.

Two Major Classes of Reactions

Once you have the diazonium salt in solution, you can do two fundamentally different things with it:

1. Substitution Reactions (N₂ leaves)

Here the –N₂⁺ group is replaced by another group. The nitrogen gas bubbles away, and the ring gets a new substituent. This is called dediazoniation. The leaving group is N₂, which is extremely stable, so the reaction is thermodynamically driven.

The most common substitutions:

Reagent/Condition	Product	Name
CuCl / HCl, heat	Ar–Cl	Sandmeyer reaction
CuBr / HBr, heat	Ar–Br	Sandmeyer reaction
CuCN / KCN, heat	Ar–CN	Sandmeyer reaction
KI, heat	Ar–I	Direct substitution
H₂O, heat	Ar–OH	Hydrolysis
H₃PO₂ (hypophosphorous acid)	Ar–H	Reduction (replaces N₂ with H)
Cu₂O, Cu(NO₃)₂, H₂O	Ar–NO₂	Replacement with nitro group

Tip

The Sandmeyer reaction uses copper(I) halide or cyanide as a catalyst. The copper helps transfer the halide or cyanide to the ring. Without copper, the reaction is much slower or gives different products.

The mechanism for Sandmeyer: the diazonium salt accepts an electron from Cu⁺, forming an aryl radical, which then abstracts a halogen from CuX₂. The N₂ leaves as a gas.

2. Coupling Reactions (N₂ stays)

Here the diazonium salt keeps its N₂ group and attacks another aromatic ring (usually an activated one like phenol or aniline). The result is an azo compound with the general structure Ar–N=N–Ar'. These compounds are intensely coloured — many are used as dyes.

The reaction is an electrophilic aromatic substitution. The diazonium cation is a weak electrophile, so it only attacks rings that are strongly activated (with –OH, –NH₂, –NHR, –NR₂ groups). The coupling occurs at the para position if available; otherwise ortho.

Example: coupling with phenol in alkaline medium:

\text{C}_6\text{H}_5\text{–N}_2^+ \text{Cl}^- + \text{C}_6\text{H}_5\text{–OH} \xrightarrow{\text{NaOH, 0–5}^\circ\text{C}} \text{C}_6\text{H}_5\text{–N=N–C}_6\text{H}_4\text{–OH} \text{ (p-hydroxyazobenzene, orange dye)}

Important

Coupling requires the coupling component (phenol or aniline) to be in its reactive form: phenol is used in alkaline solution (phenoxide ion is more activating), aniline is used in slightly acidic or neutral solution (to avoid protonation of the amino group).

Why Are Diazonium Salts So Versatile? …

(b) (i) A=benzonitrile, B=benzoic acid, C=benzamide; (ii) A=aniline, B=benzenediazonium chloride, C=benzene; aniline+ $AlCl_3$ prevents F–C; b.p. $(C_2H_5)_3N < C_2H_5NH_2 < C_2H_5OH$ .

(I)(i) Structures

C_6H_5N_2^+Cl^- \xrightarrow{CuCN} \underset{A}{C_6H_5CN} \xrightarrow{H_2O/H^+} \underset{B}{C_6H_5COOH} \xrightarrow[\Delta]{NH_3} \underset{C}{C_6H_5CONH_2}

A = benzonitrile (Sandmeyer), B = benzoic acid (nitrile hydrolysis), C = benzamide.

(I)(ii) Structures

$C_6H_5NO_2 \xrightarrow{Fe/HCl} \underset{A}{C_6H_5NH_2} \xrightarrow[273K]{NaNO_2+HCl} \underset{B}{C_6H_5N_2^+Cl^-} \xrightarrow{C_2H_5OH} \underset{C}{C_6H_6}$ …