(a) Give reasons: (i) Although -NH₂ group is o/p directing in electrophilic substitution reactions, yet anilin

Question

Q. (a) Give reasons:

(i) Although

-NH_2

group is o/p directing in electrophilic substitution reactions, yet aniline, on nitration gives good yield of m-nitroaniline.

(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.

(b) Distinguish between the following:

(i)

CH_3CH_2NH_2

and

(CH_3CH_2)_2NH

(ii) Aniline and

CH_3NH_2

(OR)

(a) Write the structures of A and B in the following reactions:

(i)

C_6H_5\overset{+}{N_2}Cl^- \xrightarrow{CuCN} A \xrightarrow{H^+/H_2O} B

(ii)

CH_3COOH \xrightarrow{NH_3} A \xrightarrow{NaOBr} B

(b) Write the chemical reaction of methyl amine with benzoyl chloride and write the IUPAC name of the product obtained.

pK_b

values:

C_6H_5NH_2

NH_3

C_2H_5NH_2

(C_2H_5)_2NH

CBSECBSE Class XII Board 2020Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Basicity Order Amines

Basicity of Amines: From Intuition to Precision

Imagine you have a nitrogen atom with a lone pair of electrons — that pair is like a key that can grab a proton ( $H^+$ ). The more willing the nitrogen is to share that pair, the stronger the base. That's the core idea.

The Intuition: What Makes a Nitrogen "Want" a Proton?

Amines are organic derivatives of ammonia ( $NH_3$ ), where one or more hydrogen atoms are replaced by alkyl groups ( $R$ ). Alkyl groups are electron-donating — they push electron density toward the nitrogen. More alkyl groups = more electron density on nitrogen = stronger base.

So you'd expect: tertiary > secondary > primary > ammonia. That's the inductive effect argument.

But reality is more interesting. In water (the usual solvent for basicity measurements), the order is:

Secondary > Primary > Tertiary > Ammonia

Why the reversal for tertiary amines? Because basicity isn't just about the free amine — it's about the stability of the conjugate acid (the ammonium ion $RNH_3^+$ ) once the proton is grabbed.

The Precise Explanation: Three Factors at Play

1. Inductive Effect (pushes electron density)

Alkyl groups donate electrons through sigma bonds. More alkyl groups = more electron density on N = stronger base. This favours: tertiary > secondary > primary > ammonia.

2. Solvation Effect (stabilises the conjugate acid)

Once the amine grabs a proton, the resulting ammonium ion is positively charged. Water molecules surround it, stabilising the charge through hydrogen bonding. The more hydrogens on the nitrogen, the more H-bonds can form.

Primary ammonium ( $RNH_3^+$ ) has 3 H's → best solvation.

Secondary ( $R_2NH_2^+$ ) has 2 H's → good solvation.

Tertiary ( $R_3NH^+$ ) has 1 H → poor solvation.

This favours: primary > secondary > tertiary.

3. Steric Hindrance (blocks solvation)

Bulkier alkyl groups physically block water molecules from approaching the charged nitrogen. This further reduces solvation in tertiary amines.

The Net Result: The Actual Order in Water

Amine Type	Inductive Effect	Solvation	Steric Hindrance	Net Basicity (pK $_b$ )
Ammonia ( $NH_3$ )	Weakest	Best (3 H's)	None	4.75
Primary ( $RNH_2$ )	Moderate	Good (2 H's)	Minimal	~3.4
Secondary ( $R_2NH$ )	Strong	Moderate (1 H)	Some	~3.2
Tertiary ( $R_3N$ )	Strongest	Poor (0 H's)	Significant	~4.2

Note

pK $_b$ is the negative log of the base dissociation constant. Lower pK $_b$ = stronger base. So secondary (pK $_b$ ≈ 3.2) is the strongest, then primary (≈3.4), then tertiary (≈4.2), then ammonia (4.75).

The Final Answer

The basicity order of amines in aqueous solution is:

Secondary > Primary > Tertiary > Ammonia

This is the standard order for simple alkyl amines (methyl, ethyl, etc.). The inductive effect dominates from ammonia to secondary, but the loss of solvation and steric hindrance in tertiary amines drops them below primary.

A Quick Check: Why Not Tertiary? …

Why this formula?

Basicity Order of Amines: Why It Holds

Let's build this from first principles — understanding why amines have different basicities is essential for exams and for deeper chemistry.

1. What Does "Basicity" Mean Here?

Basicity of an amine is its ability to accept a proton ( $H^+$ ).

The stronger the base, the more readily it grabs $H^+$ .

The equilibrium is:

R_3N + H_2O \rightleftharpoons R_3NH^+ + OH^-

The base dissociation constant $K_b$ measures this:

K_b = \frac{[R_3NH^+][OH^-]}{[R_3N]}

A larger $K_b$ means a stronger base.

2. The Key Factor: Electron Density on Nitrogen

The lone pair on nitrogen is what accepts $H^+$ .

More electron density on nitrogen → stronger base.

Why? Because a proton ( $H^+$ ) is attracted to negative charge. If the nitrogen's lone pair is more "available" (less tightly held), it binds $H^+$ more easily.

3. The Two Competing Effects

✓ Inductive Effect (Electron-Donating)

Alkyl groups ( $-CH_3$ , $-C_2H_5$ , etc.) are electron-donating via the inductive effect.

They push electron density toward nitrogen.

More alkyl groups → more electron density on N → stronger base.

This suggests:

$3^\circ > 2^\circ > 1^\circ > NH_3$

✗ Solvation Effect (Hydration of the Conjugate Acid)

When the amine accepts $H^+$ , it forms $R_3NH^+$ (a positively charged ion).

This ion is stabilized by water molecules (hydration).

More $H$ atoms on the $NH^+$ group → more hydrogen bonding with water → better stabilization of the conjugate acid.
Better stabilization of $R_3NH^+$ → stronger base (because the equilibrium shifts toward protonation).

This suggests:

$1^\circ > 2^\circ > 3^\circ$ (since $1^\circ$ has 3 $H$ 's, $2^\circ$ has 2, $3^\circ$ has 1)

4. The Actual Order (Aqueous Solution)

In water, the observed order for aliphatic amines is:

\textbf{2° > 1° > 3° > NH}_3

Wait — that's not simply "more alkyl = stronger". Why?

The Reasoning (Step by Step)

From $NH_3$ to $1^\circ$ : Adding one alkyl group increases electron density (inductive effect) → stronger base.

So $1^\circ > NH_3$ .
From $1^\circ$ to $2^\circ$ : Adding a second alkyl group further increases electron density → even stronger base.

So $2^\circ > 1^\circ$ .
From $2^\circ$ to $3^\circ$ : Here, the solvation effect becomes dominant.

The $3^\circ$ amine has only one $H$ on the $NH^+$ group → poor hydration of the conjugate acid.

The inductive effect is still present, but the loss of solvation outweighs the gain in electron density.

So $3^\circ$ is weaker than $2^\circ$ .

Thus, the net order is:

\textbf{2° > 1° > 3° > NH}_3

5. The Key Formula(e) to Remember

For gas phase (no solvent):

Only inductive effect matters:

\textbf{3° > 2° > 1° > NH}_3

For aqueous solution (common in exams): …

Part (b)Concept understanding — Sandmeyer Reaction

The Core Intuition

Imagine you have a benzene ring with an amino group ( $-NH_2$ ) attached. You want to replace that amino group with a chlorine, bromine, or cyano group. Direct substitution is nearly impossible — the amino group is stubbornly attached. But there's a clever detour.

The trick is to first convert the amino group into a diazonium salt ( $Ar-N_2^+$ ). This diazonium group is a fantastic leaving group — it's practically begging to leave. The problem is that if you just heat it, it leaves as nitrogen gas and you get a mess. You need to control what replaces it.

Sandmeyer found that if you use a cuprous halide ( $CuCl$ , $CuBr$ ) or cuprous cyanide ( $CuCN$ ), the copper helps transfer the halogen or cyano group onto the ring in a clean, high-yielding reaction. The copper acts like a molecular taxi — it picks up the halogen and drops it off exactly where the diazonium group left.

The Precise Statement

Ar-N_2^+ X^- \xrightarrow{CuX \text{ or } CuCN} Ar-X + N_2

The Sandmeyer Reaction is the replacement of the diazonium group ( $-N_2^+$ ) on an aromatic ring by chlorine, bromine, or a cyano group, using the corresponding cuprous halide or cuprous cyanide as the reagent.

The Mechanism (What Actually Happens)

The reaction proceeds through a radical mechanism, not a simple substitution. Here's the sequence:

The cuprous ion ( $Cu^+$ ) transfers one electron to the diazonium ion. This breaks the $N \equiv N$ bond, releasing nitrogen gas and generating an aryl radical ( $Ar^\bullet$ ).
The cupric ion ( $Cu^{2+}$ ) that forms now has a halogen attached. The aryl radical abstracts this halogen from copper, giving you the final product $Ar-X$ and regenerating $Cu^+$ to continue the cycle.

Note

The key insight: copper changes oxidation state ( $Cu^+ \rightarrow Cu^{2+} \rightarrow Cu^+$ ), shuttling electrons to break the diazonium bond and then delivering the halogen. This is why the reaction works — the copper mediates a radical pathway that simple heating cannot achieve.

Why This Matters for Exams

Scope: The reaction works only for $Cl$ , $Br$ , and $CN$ . You cannot get $F$ or $I$ this way — fluorine is too reactive, and iodine requires different conditions (the Gattermann reaction using copper powder instead of cuprous salts).

Yields: Sandmeyer gives excellent yields (70-90%), much better than the older Gattermann method. This is why it's the preferred industrial route for these substituted aromatics. …

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Part (a): aniline nitrates to $m$ -nitroaniline because acid protonates $-NH_2$ to meta-directing $-\overset{+}{N}H_3$ ; $(CH_3)_2NH>(CH_3)_3N$ in water (solvation); ammonolysis over-alkylates. Distinguish by carbylamine and azo-dye tests. Part (b): A/B are benzonitrile→benzoic acid and acetamide→methylamine; methylamine + benzoyl chloride → $N$ -methylbenzamide; increasing $pK_b$ is $(C_2H_5)_2NH<C_2H_5NH_2<NH_3<C_6H_5NH_2$ .

Part (a)

(i) Why aniline gives $m$ -nitroaniline

The free $-NH_2$ group is $o/p$ -directing, but nitration requires the strongly acidic mixture $\text{conc. HNO}_3+\text{H}_2\text{SO}_4$ . Here the nitrogen lone pair is protonated:

C_6H_5NH_2 + H^+ \longrightarrow C_6H_5\overset{+}{N}H_3.

The anilinium ion bears a positive nitrogen that withdraws electrons ( $-I$ ), deactivating the ring and directing the electrophile meta. Hence appreciable $m$ -nitroaniline is obtained (alongside the $o/p$ product from unprotonated aniline).

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

Two opposing effects operate:

Inductive: more alkyl groups push electron density onto N (gas-phase order $3^\circ>2^\circ>1^\circ>NH_3$ ).
Solvation: the conjugate acid is stabilised by H-bonding to water at its N–H bonds. $(CH_3)_2\overset{+}{N}H_2$ has two N–H bonds; $(CH_3)_3\overset{+}{N}H$ has only one, and its bulky methyls block water.

In water the poor solvation of $(CH_3)_3\overset{+}{N}H$ outweighs its extra $+I$ , so $(CH_3)_2NH$ is the stronger base.

(iii) Ammonolysis is not good for pure 1° amines

R\text{–}X + NH_3 \to R\text{–}NH_2 + HX.

The product $RNH_2$ is more nucleophilic than $NH_3$ , so it reacts with more $R\text{–}X$ to give $R_2NH$ , then $R_3N$ , then the quaternary salt $R_4N^+X^-$ . The result is a mixture that is difficult to separate; Gabriel synthesis is preferred for pure 1° amines.

(b) Distinguishing tests

(i) $CH_3CH_2NH_2$ (1°) vs $(CH_3CH_2)_2NH$ (2°) — carbylamine test. Only a primary amine gives the offensive-smelling isocyanide:

CH_3CH_2NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3CH_2NC + 3KCl + 3H_2O.

The secondary amine gives no such reaction. (The Hinsberg test also works: 1° → alkali-soluble sulphonamide, 2° → alkali-insoluble.) …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 36 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$O_2N-C_6H_4-NH_2$
›Reveal solutionSolution
Basicity of amines depends on electron density on nitrogen. Benzylamine has an alkyl group (electron-donating) attached to the amino group, making it the strongest base among the given aromatic amines. The correct option is (B).

Why Basicity Order Matters Here

The question asks you to compare the basic strength of four amines. In organic chemistry, basicity is directly linked to how readily the nitrogen atom can donate its lone pair. The more electron-rich the nitrogen, the stronger the base. For aromatic amines, the key factor is resonance and substituent effects — electron-donating groups increase basicity, while electron-withdrawing groups decrease it.

Let’s break down each compound.

Aniline (D) — $C_6H_5NH_2$

The lone pair on nitrogen is delocalised into the benzene ring through resonance. This makes the nitrogen less available to accept a proton, so aniline is a weaker base than aliphatic amines.

4-nitroaniline (A) — $O_2N-C_6H_4-NH_2$

The nitro group ( $-NO_2$ ) is a strong electron-withdrawing group. It pulls electron density away from the nitrogen via both inductive and resonance effects. This drastically reduces the electron density on nitrogen, making it the weakest base among the four.

4-methylaniline (C) — $CH_3-C_6H_4-NH_2$

The methyl group is electron-donating (hyperconjugation + inductive effect). It pushes electron density toward the ring, which slightly increases electron density on nitrogen compared to aniline. So 4-methylaniline is a stronger base than aniline, but still weaker than benzylamine.

Benzylamine (B) — $C_6H_5CH_2NH_2$ …
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CBSE 2026Set ANNUAL1 mark
$Q. Why is methanamine a stronger base than ammonia?$
›Reveal solutionSolution

Basicity of an amine depends on how available its nitrogen lone pair is to accept a proton; an alkyl group's electron-donating ( $+I$ ) inductive effect increases that availability compared with plain ammonia.

In ammonia, $NH_3$ , the nitrogen is bonded only to three hydrogens, which contribute no electron-donating effect of their own. In methanamine, $CH_3-NH_2$ , the methyl group is electron-releasing (+I effect): it pushes electron density through the $C-N$ sigma bond onto the nitrogen atom, increasing the electron density available in its lone pair.

A more electron-rich lone pair is a better proton acceptor (Lewis base), so protonation is favoured more strongly for methanamine than for ammonia:

$CH_3NH_2 + H^+ \rightleftharpoons CH_3NH_3^+ \quad(\text{favoured more than})\quad NH_3 + H^+ \rightleftharpoons NH_4^+$ …

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CBSE 2026Set ANNUAL1 mark
$C_2H_5NH_2$
›Reveal solutionSolution
Aqueous basicity of amines balances +I electron release (favours more alkyl groups), steric hindrance to solvation of the protonated ion (disfavours bulky/3° amines), and resonance delocalisation of the lone pair (drastically weakens aniline).

Factors at play

+I effect: each ethyl group pushes electron density onto N, making the lone pair more available and increasing basicity — this alone would predict $3^\circ > 2^\circ > 1^\circ$ .

Steric hindrance to solvation: basicity in water is effectively measured by how well the protonated ( $R_3NH^+$ ) ion is stabilised by H-bonding with water. A bulky, highly alkyl-substituted ammonium ion like $(C_2H_5)_3NH^+$ is harder to solvate, which lowers its effective basicity in water — this pulls $3^\circ$ amines down.

Aromatic ring delocalisation: in aniline, $C_6H_5NH_2$ , the lone pair on N is delocalised into the benzene ring by resonance, making it far less available for protonation — anilines are always much weaker bases than aliphatic amines. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one of the following is not Sandmeyer's reaction?$
(a) C6H5N2Cl + Cu2Br2 --HCl-->
(b) C6H5N2Cl + Cu (powder) --HCl, heat-->
(c) C6H5N2Cl + Cu2Cl2 --HCl-->
(d) C6H5N2Cl + CuCN --KCN-->
›Reveal solutionSolution
Sandmeyer's reaction specifically uses cuprous salts (Cu2Cl2, Cu2Br2, CuCN) to convert a diazonium salt into the corresponding aryl halide/nitrile; using copper powder itself instead of a cuprous salt is the Gattermann reaction, a related but distinct method.

Benzenediazonium chloride (C6H5N2Cl), formed by diazotisation of aniline, can be converted to various aryl derivatives:

C6H5N2Cl + Cu2Cl2/HCl → C6H5Cl (chlorobenzene) — Sandmeyer's reaction

C6H5N2Cl + Cu2Br2/HCl → C6H5Br (bromobenzene) — Sandmeyer's reaction

C6H5N2Cl + CuCN/KCN → C6H5CN (benzonitrile) — Sandmeyer's reaction …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is most basic?$
(a) C6H5NH2
(b) NH3
(c) C2H5NH2
(d) (C2H5)2NH
›Reveal solutionSolution

In aqueous solution, basicity of these amines follows secondary > primary > NH3 > aniline — aniline is weakest because its nitrogen lone pair is delocalised into the benzene ring, while a secondary alkylamine's two electron-donating ethyl groups make it the strongest base here.

Electron-donating alkyl (+I) groups on nitrogen increase electron density on N, making the lone pair more available to accept a proton, so alkyl-substituted amines are more basic than NH3. Between primary and secondary alkylamines in water, the extra +I contribution from the second ethyl group in (C2H5)2NH outweighs its slightly greater steric hindrance/solvation penalty, making it the strongest base of this set: (C2H5)2NH > …

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CBSE 2025Set ANNUAL1 mark
$Q. The reaction (benzenediazonium chloride, C6H5-N2+Cl-, treated with Cu/HCl) giving (chlorobenzene, C6H5-Cl, + N2 + CuCl) is named as ........... .$
›Reveal solutionSolution

Replacing a benzenediazonium salt's -N2+Cl- group with -Cl using elemental COPPER POWDER and the corresponding halogen acid (HCl) is the Gattermann reaction — a variant of the better-known Sandmeyer reaction, which instead uses a cuprous salt (Cu2Cl2) rather than copper metal.

C6H5-N2+Cl- --Cu/HCl--> C6H5-Cl (chlorobenzene) + N2 + CuCl

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Benzenediazonium chloride (structure shown, a benzene ring with a -N2+Cl- substituent) + Cu2Cl2 --HCl--> ? The product of this reaction is (structures shown as options):$
(a) Benzene ring with an –OH substituent (phenol)
(b) Benzene ring with an –NCl substituent
(c) Benzene ring with a –Cu substituent
(d) Benzene ring with a –Cl substituent (chlorobenzene)
›Reveal solutionSolution

Treating benzenediazonium chloride with $Cu_2Cl_2$ (cuprous chloride) and HCl replaces the $-N_2^+Cl^-$ group with $-Cl$ , giving chlorobenzene — a classic Sandmeyer reaction.

The diagram shows benzenediazonium chloride reacting with $Cu_2Cl_2$ /HCl. This is the Sandmeyer reaction, in which the diazonium group of an aryldiazonium salt is replaced by a halogen (Cl or Br) using the corresponding cuprous halide as catalyst:

$C_6H_5N_2^+Cl^- \xrightarrow[HCl]{Cu_2Cl_2} C_6H_5Cl + N_2\uparrow$

…

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CBSE 2025Set ANNUAL1 mark
$C_6H_5NH_2$
›Reveal solutionSolution
Diethylamine is the strongest base (two electron-donating ethyl groups, still well solvated), followed by ethylamine, then unsubstituted ammonia, with aniline the weakest because the ring delocalises the nitrogen lone pair by resonance.

Basicity of an amine depends on how available the nitrogen lone pair is to accept a proton, which in aqueous solution is governed by three competing effects: (i) the +I (electron-donating) effect of alkyl groups, which pushes electron density onto N and increases basicity; (ii) steric hindrance and the extent of solvation (H-bonding) of the resulting ammonium cation, which is reduced by bulky/more numerous alkyl groups and lowers basicity; and (iii) resonance delocalisation of the lone pair, which sharply lowers basicity when N is attached to an aromatic ring.

$(C_2H_5)_2NH$ (diethylamine, 2°): two ethyl groups give a strong +I effect, and being only disubstituted it is still reasonably well solvated in water — the strongest base of the set.

$C_2H_5NH_2$ (ethylamine, 1°): one +I-donating ethyl group, well solvated — a stronger base than plain ammonia but weaker than the disubstituted amine above. …
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CBSE 2024Set 56/3/11 markMCQ
$CH_3NH_2$
›Reveal solutionSolution
In aqueous solution, basicity of amines depends on a balance between the inductive effect (which increases electron density on nitrogen) and solvation of the conjugate acid (which stabilises it). For methyl-substituted amines, the order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > C_6H_5NH_2$ , so the correct option is (A).

The question asks for the increasing order of basicity in aqueous media — that’s the key. In water, basicity is not just about how much the nitrogen “wants” to donate its lone pair; it’s also about how stable the resulting ammonium ion is once it forms. Two effects compete here: the inductive effect of alkyl groups (which push electrons toward nitrogen, making it more basic) and the solvation effect (water molecules stabilise the charged ammonium ion by hydrogen bonding — more hydrogens on the nitrogen mean better solvation).

For aniline ( $C_6H_5NH_2$ ), the lone pair on nitrogen is delocalised into the aromatic ring, making it far less available for protonation. That’s why it’s always the weakest base among these four — no contest.

Now, among the methylamines, the trend in the gas phase (no solvent) is clear: more methyl groups → more electron donation → stronger base. So gas-phase order would be $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$ . But in water, the story changes because the conjugate acid of trimethylamine, $(CH_3)_3NH^+$ , has only one N–H bond — it can form only one strong hydrogen bond with water. The conjugate acid of dimethylamine, $(CH_3)_2NH_2^+$ , has two N–H bonds, so it’s better solvated and more stabilised. This extra stabilisation outweighs the extra inductive effect of the third methyl group, making dimethylamine the strongest base in water.

Let’s walk through the reasoning step by step.

Identify the weakest base first.

Aniline (IV) has its lone pair conjugated with the benzene ring — resonance delocalisation reduces electron density on nitrogen drastically. It is by far the least basic. So IV must come first in the increasing order. That eliminates options (B) and (C), which place aniline later.

Compare the three methylamines in water.

The inductive effect of methyl groups increases electron density on nitrogen, favouring basicity: more methyl groups → stronger base, all else equal. But “all else” is not equal in water. The conjugate acid’s ability to be stabilised by solvation depends on the number of N–H bonds: each N–H can hydrogen-bond with water.

$(CH_3)_3NH^+$ has one N–H.

$(CH_3)_2NH_2^+$ has two N–Hs.

$CH_3NH_3^+$ has three N–Hs.

More N–H bonds mean better solvation, which lowers the energy of the conjugate acid and thus makes the base stronger. So solvation favours the opposite order: more hydrogens → stronger base.

The net effect in water is a compromise. …
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CBSE 2024Set D1 markMCQ
$Q. Which of the following is the most basic?$
(a) C6H5NH2
(b) (C6H5)2NH
(c) C2H5NH2
(d) (C2H5)2NH
›Reveal solutionSolution
Aliphatic amines > aromatic; secondary diethylamine is most basic here.

Basicity depends on availability of the nitrogen lone pair:

Aromatic amines C6H5NH2 (aniline) and (C6H5)2NH (diphenylamine) are weak bases because the lone pair is delocalised into the benzene ring(s); diphenylamine is the weakest.

Aliphatic amines are stronger bases due to the electron-releasing (+I) alkyl groups. …
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CBSE 2024Set B1 mark
$Q. Fill in the blank: Methyl amine is ______ acidic than ethyl amine.$
›Reveal solutionSolution

Ethyl amine is a stronger base than methyl amine because the larger ethyl group has a greater +I (electron-releasing) effect than methyl, so relative to ethylamine, methylamine is more acidic/less basic.

Basicity of simple aliphatic amines increases as the alkyl group attached to nitrogen becomes a better electron donor (+I effect), because this raises electron density on nitrogen and better stabilises the positive charge on the protonated ammonium ion formed.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. One of the reagents for Sandmeyer reaction is$
(a) Cu/HBr
(b) CuBr/HBr
(c) Cu at 573K
(d) None of these
›Reveal solutionSolution

The Sandmeyer reaction converts an aryl diazonium salt to an aryl halide using a cuprous halide (Cu(I) salt) together with the corresponding hydrohalic acid.

ArN2+ Cl- --(CuBr / HBr)--> ArBr + N2 + ...

CuBr/HBr installs -Br (Sandmeyer); analogously CuCl/HCl installs -Cl, and CuCN/KCN installs -CN.

…

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Q.(a) Give reasons:

Part (a)

(i) Why aniline gives mmm-nitroaniline

(ii) (CH3)2NH(CH_3)_2NH(CH3​)2​NH more basic than (CH3)3N(CH_3)_3N(CH3​)3​N in water

(iii) Ammonolysis is not good for pure 1° amines

(b) Distinguishing tests

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(i) Why aniline gives $m$ -nitroaniline

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

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