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Organic Compounds Containing Nitrogen – JEE Main Formula Sheet & Class 12 Notes
This is the complete JEE Main Formula Sheet and Class 12 Formula Sheet for Organic Compounds Containing Nitrogen — Chapter 15 from the Aakash Rapid Revision & Formula Bank. The chapter is built around two central topics: the chemical reactions of aniline — covering salt formation, acylation, alkylation, ring reactions, and diazotisation — and the chemistry of diazonium salts, including their substitution reactions and the coupling reactions that produce azo dyes like methyl orange. These topics consistently contribute 3–5 questions in JEE Main every year and carry equal weight in CBSE Class 12 board exams. Download the Free PDF below to keep all key reactions, conditions, and product patterns within reach during last-minute revision.
Topics Covered in This Formula Sheet
JEE Main Formula Sheet PDF Preview
Scroll to explore all Organic Compounds Containing Nitrogen formulas — JEE Main & Class 12 Formula Sheet
Introduction: Why This Chapter Is Essential for JEE Main and Class 12
Nitrogen-containing organic compounds represent some of the most reactive and synthetically useful molecules in all of organic chemistry. Among them, amines — and particularly aniline — occupy a special place. Aniline is not just another functional group to memorise; it is a chemical hub. A single aniline molecule can be converted into an enormous range of products — salts, amides, isocyanates, azo dyes, aryl halides, phenols, and more — depending on the reagent used. For a JEE Main student, this means that mastering aniline's reactions is like unlocking a large portion of aromatic organic chemistry in one place.
The second major theme — diazonium salt chemistry — takes the importance of this chapter up another level. Benzene diazonium chloride (BDC) is prepared from aniline and serves as a gateway to an extraordinary range of aromatic substitution products that cannot be made any other way. The substitution reactions of BDC connect directly back to Chapter 13 (aryl halide preparation via Sandmeyer's and Balzschiemann reactions), making this chapter a true synthesis node in the Class 12 organic chemistry map.
The coupling reactions that close the chapter — where diazonium salts react with phenols or tertiary amines to form highly coloured azo dyes — bring in a real-world application that is tested in both JEE Main MCQs and CBSE board short-answer questions. Methyl orange, one of the most famous acid-base indicators in chemistry, is formed by exactly these reactions. Download the Free PDF to access all reactions from this chapter in one structured, exam-ready reference.
Overview of Organic Compounds Containing Nitrogen
The chapter focuses primarily on two compound families. The first is aromatic amines — specifically aniline (C₆H₅NH₂) — and the entire reaction landscape that opens up from this single starting material. The second is diazonium salts — specifically benzene diazonium chloride — and how it functions as one of the most versatile reactive intermediates in aromatic chemistry.
What makes aniline particularly interesting is the interaction between the –NH₂ group and the benzene ring. The lone pair on nitrogen donates into the ring through resonance, making the ring electron-rich. This means aniline undergoes electrophilic substitution at ortho and para positions much more readily than benzene — but it also means the nitrogen's basicity is reduced compared to aliphatic amines. This dual character of aniline — activated ring, weakened base — runs through virtually every reaction in the chapter and is a JEE Main concept that appears in comparison and reasoning-based questions every year.
Diazonium salts are inherently unstable and must be used immediately after preparation. They are prepared at 0–5°C from aniline and NaNO₂ in acidic medium. At higher temperatures, they decompose. This temperature sensitivity is itself a JEE Main question — it explains why all diazotisation reactions are carried out in an ice bath. Download the Free PDF for a complete visual summary of all reactions in this chapter.
Key Concepts Covered in This Chapter
Chemical Reactions of Aniline — Salt Formation and Basicity
Why Aniline's Salt-Forming Reactions Are Important for JEE Main
Aniline is a weak base because the lone pair on nitrogen is involved in resonance with the benzene ring, reducing its availability for protonation. Despite this, aniline reacts with strong acids to form crystalline salts. With HCl it gives anilinium chloride [C₆H₅NH₃⁺]Cl⁻; with H₂SO₄ it gives anilinium sulphate [C₆H₅NH₃⁺]₂SO₄²⁻; with chlorauric acid (HAuCl₄) it gives the corresponding gold salt; and with chloroplatinic acid (H₂PtCl₆) it gives the platinum salt. These salt formations are tested in JEE Main as product identification questions and also form the basis of understanding why aniline needs protection (acetylation) before nitration — the acidic nitrating mixture protonates the –NH₂ group, deactivating the ring and directing substitution to the meta position instead of ortho/para.
Alkylation and Acylation of Aniline
Why N-Substitution Reactions Are Important for JEE Main
Aniline undergoes alkylation at the nitrogen atom. Reaction with CH₃I (methyl iodide) first gives N-methylaniline (C₆H₅NHCH₃), and a second equivalent of CH₃I gives N,N-dimethylaniline — a tertiary amine (C₆H₅N(CH₃)₂). A further equivalent of CH₃I converts this to the quaternary ammonium salt C₆H₅N⁺(CH₃)₃I⁻. This progressive alkylation sequence is a common JEE Main multi-step product prediction question.
Acylation of aniline is equally important. Reaction with acetyl chloride (CH₃COCl) gives acetanilide (C₆H₅NHCOCH₃). Reaction with benzoyl chloride (C₆H₅COCl) gives benzanilide (C₆H₅NHCOC₆H₅). Acetylation is strategically important in synthesis — it temporarily protects the –NH₂ group by converting it to a less activating –NHCOCH₃ group, allowing controlled nitration at ortho and para positions. After nitration, the acetyl group is removed by hydrolysis to regenerate the –NH₂ group. This protection-deprotection strategy is a classic JEE Main synthesis planning question. Download the Free PDF for all N-substitution reactions in one reference.
Carbylamine Reaction, Hinsberg's Test, and Reactions with CS₂ and COCl₂
Why These Reactions Are Important for JEE Main and Board Exams
The carbylamine reaction is a qualitative test for primary amines — both aliphatic and aromatic. When a primary amine is heated with chloroform (CHCl₃) and alcoholic KOH, a foul-smelling isocyanide (carbylamine) is formed. Aniline gives phenyl isocyanide (C₆H₅NC). Secondary and tertiary amines do not give this reaction — making it a clean, selective test used in both JEE Main and Class 12 practical chemistry questions.
Hinsberg's test uses benzenesulphonyl chloride (C₆H₅SO₂Cl) to distinguish primary, secondary, and tertiary amines. Aniline (primary) gives a sulphonamide soluble in NaOH; secondary amines give a sulphonamide insoluble in NaOH; tertiary amines do not react. With carbon disulphide (CS₂), aniline forms a dithiocarbamate, and further heating gives phenyl isothiocyanate (C₆H₅–N=C=S). With phosgene (COCl₂), aniline gives diphenyl urea (C₆H₅–NH–CO–NH–C₆H₅). With CS₂ and HCl, mustard oil (phenyl isothiocyanate) is produced. These are factual, one-mark questions that reward thorough preparation.
Ring Reactions of Aniline — Bromination, Nitration, and Oxidation
Why Electrophilic Substitution of Aniline Is a JEE Main Staple
Because the –NH₂ group is a powerful ortho/para director, aniline undergoes electrophilic aromatic substitution with exceptional ease. With bromine water (Br₂/H₂O), aniline immediately gives 2,4,6-tribromoaniline as a white precipitate — all three available ortho and para positions are brominated simultaneously without any catalyst. This is a direct contrast to benzene, which requires a Lewis acid catalyst for bromination. The white precipitate test for aniline using bromine water is both a qualitative test and a JEE Main favourite.
With Br₂/CS₂ (a non-polar medium), mono-bromination is possible and gives a mixture of 2-bromoaniline and 4-bromoaniline. Nitration directly with HNO₃/H₂SO₄ gives predominantly meta-nitroaniline, because under acidic conditions the –NH₂ is protonated to –NH₃⁺, which is meta-directing. To obtain ortho and para nitroaniline, aniline must first be acetylated to acetanilide, then nitrated (giving o- and p-nitroacetanilide), and finally hydrolysed to regenerate the –NH₂ group. Oxidation of aniline with Na₂Cr₂O₇/H₂SO₄ first gives azobenzene (C₆H₅–N=N–C₆H₅) with mild oxidation and then gives quinone (benzoquinone, a cyclic diketone) with more vigorous oxidation. Download the Free PDF for all ring reaction products of aniline in one visual reference.
Diazotisation of Aniline — Preparation of Benzene Diazonium Chloride
Why Diazotisation Is One of the Most Important Reactions in This Chapter
Diazotisation is the reaction that converts aniline into benzene diazonium chloride (BDC) — the most synthetically useful reactive intermediate in aromatic chemistry. Aniline is treated with sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at 0–5°C to give benzene diazonium chloride (C₆H₅N₂⁺Cl⁻). The temperature must be kept below 5°C throughout the reaction, because diazonium salts decompose rapidly at higher temperatures.
The diazonium group (–N₂⁺) is an exceptionally good leaving group — far better than any halide — which is why BDC can be converted into such a wide range of aromatic compounds. Reaction with Na metal gives sodium phenoxide anilide (C₆H₅⁻NHNa). Reaction with C₆H₅CHO (benzaldehyde) gives a Schiff's base (C₆H₅CH=N–C₆H₅, benzylideneaniline). Complete reduction with Ni/H₂ gives cyclohexylamine (C₆H₁₁NH₂). These reactions of aniline collectively form a reaction map that is one of the most tested formats in JEE Main organic chemistry — a central molecule with radiating arrows to multiple products, and the student must identify reagent or product from a given transformation. Download the Free PDF to study this complete reaction map.
Substitution Reactions of Benzene Diazonium Chloride
Why BDC Substitution Reactions Are Direct JEE Main and Board Exam Questions
Benzene diazonium chloride (BDC) reacts with a remarkably wide range of reagents to replace the –N₂⁺ group with different functional groups. Each transformation is a named reaction or a specific product that JEE Main and CBSE Class 12 test directly. With C₂H₅OH (ethanol), benzene is formed along with N₂, HCl, and acetaldehyde. With H₃PO₂ and H₂O in the presence of Cu⁺, benzene is again formed — this is the standard method for removing a diazonium group (reductive deamination).
With HCl/CuCl → chlorobenzene + N₂ (Sandmeyer's reaction). With HBr/CuBr → bromobenzene + N₂ + HCl (Sandmeyer's). With NaCN + CuCN → cyanobenzene (benzonitrile) + N₂ + NaCl. With H₂O/H₂SO₄ (heating) → phenol + N₂. With KI (heating) → iodobenzene + N₂ + KCl. With HBF₄ → benzenediazonium fluoroborate, then heating gives fluorobenzene + N₂ + BF₃ (Balzschiemann reaction). With C₆H₆/NaOH → biphenyl (Gomberg-Bachmann reaction).
This complete list of substitution reactions of BDC is one of the most fact-dense and exam-heavy sections in the entire Class 12 organic chemistry syllabus. Every single transformation here has appeared in JEE Main or board exams. Download the Free PDF to have this full reaction map in one place for rapid revision.
Coupling Reactions of Diazonium Salts — Azo Dyes and Methyl Orange
Why Coupling Reactions and Azo Dye Formation Are Important for JEE Main
Coupling reactions are electrophilic substitution reactions where the diazonium cation (a weak electrophile) attacks highly activated aromatic compounds — phenols or tertiary aromatic amines — to form brightly coloured azo compounds (–N=N–). These reactions are the industrial basis of the azo dye industry and are tested directly in JEE Main and board exams.
Coupling of BDC with phenol occurs in slightly basic medium (pH ≈ 8), giving p-hydroxy azobenzene — an orange dye. Coupling with N,N-dimethylaniline in slightly acidic medium (pH ≈ 5) gives p-dimethylaminoazobenzene (butter yellow). Coupling with β-naphthol gives phenylazo-β-naphthol (a red dye). The pH dependence of coupling reactions — basic for phenols, acidic for amines — is a specific JEE Main question point. At the wrong pH, phenol is not sufficiently activated (in acidic medium) or aniline is protonated and deactivated (in basic medium).
Methyl orange synthesis is the most complete coupling reaction sequence in this chapter. p-Sulphanilic acid (HO₃S–C₆H₄–NH₂) is first diazotised with NaNO₂/HCl at 0°C to give the diazonium salt. This then couples with N,N-dimethylaniline to give methyl orange — a well-known pH indicator that is red in acidic and yellow in alkaline solution. The full synthesis — diazotisation followed by coupling — is a two-step sequence that board exams ask students to write out completely. Download the Free PDF for a clean step-by-step synthesis of methyl orange alongside all other coupling reactions.
Download Free PDF — Organic Compounds Containing Nitrogen Formula Sheet
All the reactions of aniline, substitution reactions of benzene diazonium chloride, coupling reactions, and azo dye syntheses from this chapter are compiled in the Aakash Rapid Revision & Formula Bank PDF. It is structured specifically for JEE Main, CBSE boards, and NEET — covering every reaction with reagents, conditions, and products in a layout designed for exam-day recall.
Why This Chapter Is Important for Students and Exams
There are four specific reasons why Chapter 15 deserves focused preparation time from every JEE Main aspirant and Class 12 student.
Aniline is the most reaction-rich single molecule in Class 12. No other compound in the syllabus has as many direct reactions — salt formation with five different acids, alkylation, acylation, carbylamine reaction, Hinsberg's test, Schiff's base formation, diazotisation, bromination, nitration, and oxidation — all testable individually. Students who master aniline's reaction map can answer a large portion of JEE Main nitrogen chemistry questions from a single compound.
Diazonium salt substitutions are a guaranteed exam section. The ten-plus substitution reactions of benzene diazonium chloride are among the most consistently tested content in JEE Main and Class 12 boards. These are finite, learnable reactions with clear products and conditions — pure marks waiting for the prepared student.
Coupling reactions connect to real-world applications. Azo dyes — formed by coupling reactions — are the largest class of synthetic dyes used in the textile industry. Questions about azo dye formation, the pH conditions required, and methyl orange synthesis appear in both conceptual MCQs and board descriptive questions. The real-world grounding makes these facts easier to remember and harder to mix up.
Cross-chapter connections reward integrated preparation. This chapter directly connects back to Chapter 13 (Sandmeyer's and Balzschiemann reactions used BDC as starting material) and forward to dye chemistry. Students who see these connections answer multi-step synthesis questions in JEE Main more confidently. Download the Free PDF to reinforce all these connections in one revision session.
Who Should Use This Formula Sheet?
Learning Outcomes After Completing This Chapter
After working through this chapter using the formula sheet and the notes above, a student should be able to accomplish the following with confidence.
For aniline reactions: write the product of aniline with any of the reagents on its reaction map — HCl, H₂SO₄, CH₃I, CH₃COCl, C₆H₅COCl, CHCl₃/KOH, CS₂, COCl₂, Na, C₆H₅CHO, NaNO₂/HCl, Br₂/H₂O, Br₂/CS₂, HNO₃/H₂SO₄, Na₂Cr₂O₇/H₂SO₄, and Ni/H₂. Explain why direct nitration gives meta-nitroaniline and how acetylation circumvents this. Apply Hinsberg's test to distinguish primary, secondary, and tertiary amines.
For diazonium salt chemistry: write the preparation of BDC including temperature conditions and explain why low temperature is essential. Identify the product of BDC with each substitution reagent — ethanol, H₃PO₂, CuCl/HCl, CuBr/HBr, NaCN/CuCN, H₂O/H₂SO₄, KI, HBF₄, and C₆H₆/NaOH.
For coupling reactions: explain why phenol requires basic medium and amines require acidic medium for coupling, write the structures of p-hydroxy azobenzene, p-dimethylaminoazobenzene, phenylazo-β-naphthol, and methyl orange, and reproduce the two-step synthesis of methyl orange from p-sulphanilic acid. Download the Free PDF to test yourself against these outcomes before your exam.
Get the Free PDF for Quick Revision
Whether you are preparing for JEE Main, CBSE Class 12 boards, NEET, or BITSAT, having a focused formula sheet for this chapter ensures that every reaction, every product, and every condition is accessible during revision. The Aakash Rapid Revision & Formula Bank PDF for Organic Compounds Containing Nitrogen is structured precisely for exam-day recall.
Conclusion — Learn the Logic Behind the Reactions
Organic Compounds Containing Nitrogen is one of those chapters where the number of reactions can feel overwhelming at first glance. The key is to organise them into a logical structure rather than a flat list. Aniline's reactions divide cleanly into three groups: reactions at nitrogen (salt formation, alkylation, acylation, diazotisation), reactions that use aniline as a test molecule (carbylamine, Hinsberg's), and reactions at the ring (bromination, nitration, oxidation). Once you see these groupings, the reactions stop feeling random.
Diazonium chemistry is even more systematic. Every substitution reaction of BDC follows the same pattern: the –N₂⁺ leaving group is replaced by a nucleophile or by a radical-based pathway, releasing N₂ gas as the driving force. Knowing that N₂ is released in almost every BDC reaction helps you verify your answers. Coupling reactions follow a different logic — electrophilic attack on an activated ring — and the pH conditions make sense once you understand that you need the phenol to be ionised (basic medium) or the amine to be free (acidic medium) for the reaction to work.
Use this page, the subtopic breakdowns, and the Free PDF Download as your revision anchor. The reactions here are finite, learnable, and highly predictable in exams — invest the time, and this chapter will reward you with consistent marks in every JEE Main paper and Class 12 board exam.
Frequently Asked Questions
Why is aniline less basic than aliphatic amines like methylamine?
In aliphatic amines like methylamine (CH₃NH₂), the lone pair on nitrogen is fully available to accept a proton, making them relatively strong bases. In aniline (C₆H₅NH₂), the lone pair on nitrogen is in conjugation with the π-system of the benzene ring — it participates in resonance, delocalising into the ring. This reduces the availability of the lone pair for protonation. As a result, aniline accepts protons less readily and is a weaker base than aliphatic amines. The basicity order is: aliphatic amines > ammonia > aniline. This resonance-reduced basicity also explains why aniline's –NH₂ group is a less powerful activator in EAS than an alkyl amine's –NH₂ attached to an aliphatic chain.
Why does direct nitration of aniline give meta-nitroaniline instead of the expected ortho/para products?
The –NH₂ group in aniline is normally an ortho/para director. However, the nitrating mixture (conc. HNO₃ + conc. H₂SO₄) is strongly acidic. Under these acidic conditions, the –NH₂ group of aniline is completely protonated to –NH₃⁺ before the nitration step even begins. The –NH₃⁺ group is an electron-withdrawing, meta-directing group. So when electrophilic substitution occurs, the NO₂⁺ electrophile attacks the meta position, giving predominantly meta-nitroaniline. To obtain ortho and para nitro products, aniline must first be acetylated to acetanilide, which resists protonation. Nitration of acetanilide gives ortho and para nitroacetanilide, and subsequent hydrolysis regenerates the –NH₂ group.
What is the carbylamine reaction and which class of amines gives a positive test?
The carbylamine reaction is a qualitative test for primary amines. When a primary amine — either aliphatic or aromatic — is heated with chloroform (CHCl₃) and alcoholic potassium hydroxide (KOH), it gives an isocyanide (also called carbylamine), which is characterised by an extremely unpleasant, foul smell. For aniline, the product is phenyl isocyanide (C₆H₅–N≡C). Secondary and tertiary amines do not react under these conditions, so the carbylamine reaction is exclusively positive for primary amines. This selectivity makes it one of the most useful distinguishing tests in the chapter and a regular one-mark question in both JEE Main and Class 12 board exams.
Why must diazotisation of aniline be carried out at 0–5°C?
Benzene diazonium chloride (BDC) is inherently unstable. The diazonium group (–N₂⁺) is a good leaving group, and at temperatures above 5°C, the diazonium salt begins to decompose spontaneously. At room temperature or above, BDC hydrolyses rapidly to give phenol and nitrogen gas (C₆H₅N₂Cl + H₂O → C₆H₅OH + N₂ + HCl). Keeping the reaction mixture in an ice bath (0–5°C) stabilises the diazonium salt long enough for it to be used in subsequent coupling or substitution reactions. If the temperature rises even slightly during diazotisation, the yield drops significantly and the product decomposes. This temperature requirement is a direct JEE Main question about the conditions for diazotisation.
What are the Sandmeyer's reaction products from benzene diazonium chloride?
Sandmeyer's reaction is used to replace the diazonium group (–N₂⁺) of benzene diazonium chloride with a halide using a copper(I) halide catalyst. Treating BDC with CuCl in HCl gives chlorobenzene and nitrogen gas. Treating BDC with CuBr in HBr gives bromobenzene and nitrogen gas. For cyanobenzene (benzonitrile), BDC is treated with NaCN and CuCN in pyridine at 470 K. The iodide reaction (giving iodobenzene) does not require copper — BDC simply reacts with KI on heating to give iodobenzene directly. Sandmeyer's reaction is valuable because it allows the introduction of chloro, bromo, or cyano groups at a specific position on a benzene ring that was originally occupied by an amino group, offering positional control that direct halogenation cannot.
Why does coupling of diazonium salt with phenol require basic medium while coupling with aniline requires acidic medium?
The diazonium cation is a weak electrophile, so it can only attack very electron-rich aromatic rings. For coupling with phenol, the ring needs to be maximally activated. In basic medium (pH ≈ 8), phenol is partially converted to the phenoxide ion (C₆H₅O⁻), which donates electrons much more strongly into the ring than the –OH group alone. This makes the ring reactive enough for coupling at the para position. In acidic medium, phenol remains as –OH (not –O⁻) and is less activated, so coupling is slower or does not occur. For coupling with N,N-dimethylaniline, acidic medium (pH ≈ 5) is needed. In basic medium, the dimethylamine group would not be protonated, but more importantly, the medium needs to be controlled to prevent decomposition of the diazonium salt at higher pH. The acidic medium keeps the diazonium salt stable while keeping the tertiary amine unprotonated and free to couple.
How is methyl orange synthesised, and why is it used as a pH indicator?
Methyl orange synthesis proceeds in two steps. In the first step, p-sulphanilic acid (HO₃S–C₆H₄–NH₂) is diazotised with NaNO₂ and HCl at 0°C to give the corresponding diazonium salt. In the second step, this diazonium salt undergoes a coupling reaction with N,N-dimethylaniline in slightly acidic medium to give methyl orange — a sodium salt of an azo compound containing both a sulphonate group and a dimethylamino group. Methyl orange is used as a pH indicator because the azo linkage (–N=N–) is sensitive to protonation. In acidic solution (pH below 3.1), the dimethylamino nitrogen is protonated, which shifts the conjugation pattern and changes the colour to red. In alkaline solution (pH above 4.4), the free amine form is present and the dye appears yellow. Between these pH values, a range of orange colour is observed.
What happens when aniline is treated with bromine water, and how does this differ from bromination of benzene?
When aniline is treated with bromine water (Br₂/H₂O), an immediate white precipitate of 2,4,6-tribromoaniline is formed. All three available ortho and para positions (two ortho and one para) are brominated simultaneously, and no catalyst is needed. This is because the –NH₂ group donates electrons very powerfully into the ring through resonance, making the ring so electron-rich that even the weak electrophile Br₂ in water can attack all three activated positions without any Lewis acid help. Benzene, by contrast, requires a Lewis acid catalyst (FeBr₃ or AlCl₃) to polarise Br₂ and generate a sufficiently reactive electrophile for EAS. This dramatic difference illustrates the activating power of the –NH₂ group and is a direct JEE Main comparison question.
What is Hinsberg's test and how does it distinguish primary, secondary, and tertiary amines?
Hinsberg's test uses benzenesulphonyl chloride (C₆H₅SO₂Cl) to classify amines. A primary amine reacts with benzenesulphonyl chloride to give a sulphonamide that has one N–H bond remaining. This sulphonamide is acidic enough to dissolve in aqueous NaOH, giving a clear solution. A secondary amine also reacts with benzenesulphonyl chloride but gives a sulphonamide with no N–H bond. This product is neutral and insoluble in NaOH, giving a precipitate or turbid solution. A tertiary amine does not react with benzenesulphonyl chloride at all because it has no N–H bond available for reaction. So: primary amine → soluble in NaOH; secondary amine → insoluble in NaOH; tertiary amine → no reaction. This three-way distinction is tested both as a concept and as a practical chemistry question in Class 12 board exams.
What product is formed when benzene diazonium chloride reacts with ethanol, and why?
When benzene diazonium chloride reacts with ethanol (C₂H₅OH), benzene is formed along with nitrogen gas, hydrochloric acid, and acetaldehyde (CH₃CHO). The diazonium group is replaced by hydrogen — the ethanol acts as a reducing agent, donating a hydrogen atom to the aryl radical or intermediate and being oxidised to acetaldehyde in the process. This reaction is the standard method for converting an aryl diazonium salt back to an unsubstituted arene — effectively removing the amino group from aniline through diazotisation followed by treatment with ethanol. This overall transformation (aniline → BDC → benzene) is used synthetically when an amino group was used as a temporary ortho/para director and needs to be removed after the substitution is complete.
JEE Main Formula Hub — Chapter-Wise Formulas (Chemistry | Physics | Maths)
Access chapter-wise formula sheets for all three JEE Main subjects. Click on any chapter below to get complete formulas, key concepts, and free PDF downloads prepared from the Aakash Rapid Revision Formula Bank.
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