How will you obtain the following from aniline ? Give chemical equations only. (a) Sulphanilic acid (b) Phenyl

Question

Q. How will you obtain the following from aniline ? Give chemical equations only.

(a) Sulphanilic acid

(b) Phenylisocyanide

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Aromatic Synthesis Route

Aromatic Synthesis Route – First Principles

Imagine you are a chef who has been given a plain wooden board and told to carve a specific shape out of it. You can cut away wood, but you cannot add wood back. That is exactly the problem in aromatic synthesis: you start with a simple, cheap aromatic ring (like benzene) and you need to attach specific groups at specific positions. The ring itself is already there — you cannot rearrange its carbon skeleton. So the entire challenge is where to put the next group, and how to get it there.

The "route" is the sequence of reactions you choose. The order matters enormously because the groups already on the ring control where the next group will go. A wrong order can give you the wrong isomer, or force you into a dead end.

The Core Idea: The Ring Directs

Every substituent already on a benzene ring has a directing effect — it tells the next incoming group to go to certain positions. There are two families:

Ortho/para directors (e.g., –OH, –NH₂, –CH₃, –Cl) — they push the next group to positions 2 and 4 (ortho and para).
Meta directors (e.g., –NO₂, –CN, –CHO, –SO₃H) — they push the next group to position 3 (meta).

Watch out

A common mistake is to think you can just "add any group in any order." The ring is not passive — it has a memory of what is already attached. If you ignore directing effects, you will get a mixture of products, often with the wrong isomer as the major one.

The Precise Statement

An aromatic synthesis route is a planned sequence of electrophilic aromatic substitution (EAS) reactions, chosen so that each new substituent is introduced at the correct position relative to the existing ones. The route must account for:

Directing effects of all current substituents.
Activation/deactivation — some groups make the ring more reactive (activators), some make it less reactive (deactivators). You cannot do a reaction on a strongly deactivated ring without special conditions.
Order of introduction — sometimes you must introduce a meta director first, then an ortho/para director, or vice versa, to get the desired final pattern.

A Concrete Example: Making 4-Nitrobenzoic Acid

You want a benzene ring with –COOH at position 1 and –NO₂ at position 4 (para to each other).

–COOH is a meta director. –NO₂ is also a meta director. If you put –COOH first and then nitrate, the –COOH will send the –NO₂ to the meta position (3), not para (4). That gives the wrong isomer.

Correct route:

Nitrate benzene first → nitrobenzene ( –NO₂ is meta directing).
Then oxidise the methyl group (if you started with toluene) or use a different method to introduce –COOH. But wait — –NO₂ deactivates the ring strongly. So you cannot easily do Friedel-Crafts acylation on nitrobenzene.

So the actual correct route is different:

Start with toluene (methylbenzene). The –CH₃ is an ortho/para director and an activator.
Nitrate toluene → you get a mixture of ortho and para nitrotoluene. Separate the para isomer.
Oxidise the –CH₃ to –COOH using KMnO₄. The –NO₂ survives this oxidation.

Note

The order here is: introduce the ortho/para director first ( –CH₃ ), then nitrate to get para, then convert the –CH₃ to –COOH. If you had tried to put –COOH first, you would have a meta director that would send –NO₂ to the wrong place.

The General Strategy

When planning a route, ask yourself in order:

What is the final substitution pattern? (1,2- ; 1,3- ; 1,4- ; etc.)
Which groups are ortho/para directors and which are meta directors?
Can I introduce the meta director first, then the ortho/para director? (Often yes, because meta directors deactivate the ring, making further substitution harder — so you want to do the deactivating step last if possible.)
If I need a 1,3 pattern, I usually put a meta director first, then an ortho/para director. If I need a 1,4 pattern, I usually put an ortho/para director first, then a meta director (because the ortho/para director will send the next group to para, and the meta director will then be at the correct position). …

Why this formula?

Aromatic Synthesis Route: Understanding the Why Behind the Key Principles

In organic chemistry, an aromatic synthesis route refers to a sequence of reactions designed to construct or modify an aromatic ring (typically benzene or its derivatives). The "key formulas" here are not single equations but rather rules and principles that govern reactivity and orientation. Let's break down the reasoning behind the most critical ones.

1. The 4n+2 Hückel Rule — Why Aromaticity Exists

Formula: A planar, cyclic, conjugated molecule is aromatic if it has $(4n+2)$ π electrons, where $n = 0,1,2,\dots$

Why this holds (the derivation):

In a cyclic conjugated system, the π electrons occupy molecular orbitals (MOs) that form a ring.
The energy levels of these MOs are given by the Frost circle (or polygon rule):
- For a regular polygon with $N$ vertices (atoms), inscribe it in a circle with one vertex at the bottom.
- The energy of each MO corresponds to the vertical coordinate of each vertex.
For benzene ( $N=6$ $N = 6$ ), the MOs split into:
- 1 low-energy bonding orbital
- 2 degenerate bonding orbitals
- 2 degenerate antibonding orbitals
- 1 high-energy antibonding orbital
Key insight: The 6 π electrons fill the 3 bonding MOs completely. This gives a closed-shell, highly stable configuration — the aromatic stabilization energy (~150 kJ/mol for benzene).
For $N=4$ (cyclobutadiene), the MO pattern gives 2 degenerate non-bonding orbitals — filling with 4 electrons creates an open-shell, antiaromatic (unstable) system.

Takeaway: The $(4n+2)$ rule is not arbitrary — it emerges from the symmetry of cyclic π systems and the filling of bonding MOs.

2. Electrophilic Aromatic Substitution (EAS) — The Reactivity Formula

General reaction:

\text{Ar-H} + \text{E}^+ \xrightarrow{\text{catalyst}} \text{Ar-E} + \text{H}^+

Why this is the only viable route for aromatic rings:

Aromatic rings are electron-rich (due to the π cloud) but resistant to addition — addition would break aromaticity.
Mechanism reasoning:
1. The electrophile $\text{E}^+$ attacks the ring, forming a σ-complex (arenium ion) — this step is slow (rate-determining).
2. The σ-complex is non-aromatic (4 π electrons in the ring) — it is high-energy and unstable.
3. To regain aromaticity, the complex loses a proton ( $\text{H}^+$ ) — this step is fast and thermodynamically driven.
Why substitution, not addition: Addition would permanently destroy aromaticity; substitution restores it.

Key formula: The rate law is $\text{Rate} = k[\text{Ar-H}][\text{E}^+]$ — first order in both, because the slow step involves both reactants.

3. Orientation Rules — Why Substituents Direct Where the Next Group Goes

Rule:

Activating groups (e.g., $-\text{OH}, -\text{NH}_2, -\text{OCH}_3$ ) direct to ortho/para positions.
Deactivating groups (e.g., $-\text{NO}_2, -\text{CN}, -\text{CHO}$ ) direct to meta positions.

Why this happens (resonance + inductive reasoning):

For ortho/para directors:

The substituent has a lone pair or π bond that can donate electrons into the ring via resonance.
Draw the resonance structures of the σ-complex for attack at ortho, meta, and para:
- Ortho attack: The positive charge can be delocalized onto the substituent (e.g., $-\text{OH}$ becomes $=\text{OH}^+$ ). This stabilizes the intermediate.
- Para attack: Similar stabilization — charge delocalized to the substituent.
- Meta attack: The positive charge cannot reach the substituent — less stable.
Result: Ortho/para intermediates are lower in energy → faster reaction.

For meta directors:

The substituent is electron-withdrawing (by induction or resonance, e.g., $-\text{NO}_2$ ).
Draw resonance for ortho attack: The positive charge is placed directly on the carbon bearing the withdrawing group — this is highly destabilizing (like putting a + charge next to a + pole).
For meta attack: The positive charge is never on the carbon with the withdrawing group — relatively more stable.
Result: Meta attack is the least destabilized → preferred. …

Start your 14-day free trial to unlock the full solution →

Aniline is a versatile starting material for aromatic synthesis. Sulphanilic acid is made by sulphonation, phenylisocyanide by the carbylamine reaction, and acetanilide by acetylation. The key is understanding how the amino group directs substitution and reacts with specific reagents.

Let’s walk through each transformation, focusing on the chemistry behind the reactions.

1. Sulphanilic acid from aniline

Aniline has a strongly activating and ortho/para-directing amino group ( $-NH_2$ ). When we treat it with concentrated sulphuric acid, the amino group gets protonated first to form anilinium hydrogen sulphate. This is crucial — the $-NH_3^+$ group is meta-directing and deactivating. So why does sulphonation still happen at the para position?

The trick is that at high temperatures (around 180–200°C), the reaction proceeds via a different mechanism. The anilinium salt loses a proton to form aniline again, and the actual electrophile ( $SO_3$ from $H_2SO_4$ ) attacks the free aniline. The $-NH_2$ group then directs the sulphonic acid group to the para position (and some ortho, but para is the major product under controlled conditions).

The reaction is:

\ce{C6H5NH2 + H2SO4 ->[180-200°C] H2N-C6H4-SO3H + H2O}

The product is sulphanilic acid (4-aminobenzenesulphonic acid). It exists as a zwitterion in solid state.

Watch out

A common mistake is to think the $-NH_3^+$ group directs the incoming $-SO_3H$ group. It doesn't — the reaction happens via the free aniline at high temperature. At low temperature, you'd get mostly meta product from the anilinium ion, but that's not how sulphanilic acid is made.

2. Phenylisocyanide from aniline

This is the carbylamine reaction (also called the isocyanide test). Aniline, when heated with chloroform ( $CHCl_3$ ) and alcoholic potassium hydroxide ( $KOH$ ), gives a foul-smelling product — phenylisocyanide.

The mechanism involves the formation of dichlorocarbene ( $:CCl_2$ ) from chloroform under basic conditions. This highly reactive carbene attacks the lone pair on the nitrogen of aniline. After elimination of HCl and rearrangement, the isocyanide ( $-NC$ ) group is formed.

The balanced equation is:

\ce{C6H5NH2 + CHCl3 + 3KOH ->[alc., \Delta] C6H5NC + 3KCl + 3H2O}

Tip

This reaction is specific to primary amines. Secondary and tertiary amines do not give this test. It's a classic way to distinguish primary amines from others in the lab.

3. Acetanilide from aniline …

Start your 14-day free trial to unlock the full solution →

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

CBSE 2025Set 56/4/11 markMCQ
$ArNO_2$
›Reveal solutionSolution
The Hoffmann bromamide degradation converts a primary amide ( $\ce{RCONH2}$ ) into a primary amine with one fewer carbon atom using bromine and base. The starting material is $\ce{ArCONH2}$ , option (C).

The Hoffmann bromamide degradation is one of the most elegant reactions in organic chemistry for shortening a carbon chain while introducing an amino group. Understanding what it does—and what it requires—makes the answer obvious.

The Core Concept

The reaction takes a primary amide ( $\ce{RCONH2}$ ) and, through treatment with bromine in aqueous or alcoholic alkali (typically $\ce{NaOH}$ or $\ce{KOH}$ ), removes the carbonyl carbon entirely. The product is a primary amine ( $\ce{RNH2}$ ) with one carbon atom less than the starting amide.

The key transformation:
$\ce{RCONH2 ->[Br2/NaOH][-CO2] RNH2}$
The carbonyl carbon is lost as carbonate, and the nitrogen that was part of the amide becomes the amino group of the amine. This is why it's called a "degradation"—you're stepping down the carbon skeleton.

Why This Mechanism Matters

The reaction proceeds through several intermediates:

N-bromoamide formation: Bromine in base first converts the amide to an N-bromoamide ( $\ce{RCONHBr}$ ).

Isocyanate intermediate: Base abstracts the remaining N–H proton, and the resulting anion loses bromide to form an isocyanate ( $\ce{R-N=C=O}$ ). This is the crucial rearrangement step—the alkyl/aryl group migrates from carbon to nitrogen.

Hydrolysis: The isocyanate is rapidly hydrolyzed by the aqueous base to give a carbamic acid ( $\ce{RNHCOOH}$ ), which immediately decarboxylates to yield the primary amine ( $\ce{RNH2}$ ).

Important

The starting material must be a primary amide ( $\ce{RCONH2}$ ). Secondary or tertiary amides do not undergo this reaction because the mechanism requires an N–H hydrogen for the rearrangement.

Analyzing the Options

Now let's see which option fits:

(A) $\ce{ArNO2}$ — This is a nitro compound (nitrobenzene and derivatives). It has no amide group and cannot undergo Hoffmann degradation. Nitro groups are reduced to amines by other methods (catalytic hydrogenation, metal/acid reduction). …
Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/3/11 markMCQ
$Q. Aspirin is obtained by the acetylation of which of the following compounds? (A) Salicylaldehyde (B) Salicylic acid (C) Acetyl salicylic acid (D) Phenol$
›Reveal solutionSolution
Aspirin is synthesized by acetylating the hydroxyl group of salicylic acid with acetic anhydride or acetyl chloride, producing acetylsalicylic acid. The answer is (B).

Understanding Aspirin Synthesis

Aspirin, chemically known as acetylsalicylic acid, is one of the most widely used analgesic and anti-inflammatory drugs. The key to answering this question lies in understanding what acetylation means and recognizing the structural relationship between aspirin and its precursor.

Acetylation is the process of introducing an acetyl group ( $\ce{-COCH3}$ ) into a molecule, typically by replacing an active hydrogen (usually on a hydroxyl or amino group) with $\ce{-COCH3}$ .

When we say aspirin is "obtained by acetylation," we're looking for the compound that, when treated with an acetylating agent, gives us aspirin.

Step-by-Step Analysis

Identify aspirin's structure

Aspirin is acetylsalicylic acid, which has:

A benzene ring

A carboxylic acid group ( $\ce{-COOH}$ ) at position 1

An acetyl ester group ( $\ce{-OCOCH3}$ ) at position 2 (ortho to the carboxyl)

Work backwards from the product

If aspirin has an ester linkage ( $\ce{-OCOCH3}$ ), then before acetylation, that position must have had a free hydroxyl group ( $\ce{-OH}$ ). The acetylation reaction converts $\ce{-OH}$ to $\ce{-OCOCH3}$ .

Examine each option

(A) Salicylaldehyde has a benzene ring with $\ce{-CHO}$ (aldehyde) and $\ce{-OH}$ groups. Acetylating this would give an acetylated aldehyde derivative, not aspirin (which has a carboxylic acid group).

(B) Salicylic acid has:

A carboxylic acid group ( $\ce{-COOH}$ )

A hydroxyl group ( $\ce{-OH}$ ) at the ortho position

When the $\ce{-OH}$ is acetylated, we get exactly aspirin's structure.

(C) Acetyl salicylic acid is aspirin itself. You can't obtain aspirin by acetylating aspirin. …
Start your 14-day free trial to see the full solution →

Q.How will you obtain the following from aniline ? Give chemical equations only.

1. Sulphanilic acid from aniline

2. Phenylisocyanide from aniline

3. Acetanilide from aniline …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days