JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Classification Preparation
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Classification Preparation
🎓 Class 12
Chemistry
CBSE
Theory
Ch 6 – Haloalkanes and Haloarenes
⏱ ~14 min
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Classification Preparation
Introduction: A World Held Together by C–X Bonds
Substitute a hydrogen of a hydrocarbon with a halogen atom and you create a brand-new family of compounds — the haloalkanes and haloarenes. Nature makes only a handful of these (thyroxine carries iodine; marine algae produce brominated terpenes), yet human chemistry has engineered thousands: the anaesthetic halothane, the topical antiseptic iodoform, the refrigerant Freon-12, the polymer Teflon, and once-ubiquitous pesticides such as DDT and BHC.
The same strength of the carbon–halogen bond that makes these molecules useful also makes many of them environmentally persistent. Chlorofluorocarbons survived in the stratosphere long enough to carve the ozone hole; DDT accumulated up food chains to thin raptors' eggshells. Understanding how the C–X bond is made and broken is therefore both a central organic-chemistry topic and a problem of planetary scale.
What lies ahead: We will classify these compounds, learn their IUPAC names, examine the electronic nature of the C–X bond, and master six preparative routes for haloalkanes and four for haloarenes.
6.1 Classification
6.1.1 On the Basis of the C–X Bond
- Alkyl halides (haloalkanes, R–X): halogen attached to an sp³-hybridised carbon. Example: CH₃Cl, CH₃CH₂Br.
- Aryl halides (haloarenes, Ar–X): halogen attached to an sp²-hybridised carbon of an aromatic ring. Example: chlorobenzene C₆H₅Cl.
- Allylic halides: X on an sp³ carbon adjacent to a C=C double bond. Example: CH₂=CH–CH₂–Cl (allyl chloride).
- Benzylic halides: X on an sp³ carbon attached to a benzene ring. Example: C₆H₅–CH₂Cl (benzyl chloride).
- Vinylic halides: X attached directly to an sp² carbon of C=C. Example: CH₂=CH–Cl (vinyl chloride).
6.1.2 On the Basis of Number of Halogen Atoms
A molecule may carry one, two, three or more halogens: mono-, di-, tri- or poly-halides. Dihalides are further classified as gem-dihalides (both X on the same carbon, e.g., CH₃CHCl₂ ethylidene chloride) or vic-dihalides (on adjacent carbons, e.g., CH₂Cl–CH₂Cl ethylene dichloride).
6.1.3 On the Basis of the Carbon Bearing the Halogen
An alkyl halide is primary (1°), secondary (2°) or tertiary (3°) depending on whether the carbon bonded to X is attached to one, two or three other carbons.
6.2 Nomenclature
6.2.1 IUPAC Rules (Haloalkanes)
- Pick the longest carbon chain that contains the C–X carbon — it is the parent alkane.
- Number the chain so that the halogen (or the first substituent encountered) gets the lowest locant.
- Name the halogen as a prefix: fluoro-, chloro-, bromo-, iodo-. Multiple halogens receive di-, tri- prefixes and are listed alphabetically.
| Structure | Common name | IUPAC name |
|---|---|---|
| CH₃Cl | Methyl chloride | Chloromethane |
| CH₃CH₂Br | Ethyl bromide | Bromoethane |
| (CH₃)₂CHI | Isopropyl iodide | 2-Iodopropane |
| (CH₃)₃CBr | tert-Butyl bromide | 2-Bromo-2-methylpropane |
| CH₂=CHCl | Vinyl chloride | Chloroethene |
| C₆H₅CH₂Cl | Benzyl chloride | (Chloromethyl)benzene |
| CHCl₃ | Chloroform | Trichloromethane |
| CHI₃ | Iodoform | Triiodomethane |
6.2.2 Haloarenes
For a disubstituted benzene, the relative positions can be indicated either by numerical locants (1,2-; 1,3-; 1,4-) or by the historical prefixes ortho (o-), meta (m-) and para (p-). Thus 1,4-dichlorobenzene is the same compound as p-dichlorobenzene (used in mothballs).
6.3 Nature of the C–X Bond
Halogens are more electronegative than carbon, so the C–X bond is polar covalent with partial positive charge on carbon and partial negative on the halogen:
| Bond | Bond length (pm) | Bond enthalpy (kJ mol⁻¹) | Dipole moment (D) |
|---|---|---|---|
| C–F | 139 | 452 | 1.85 |
| C–Cl | 178 | 351 | 1.90 |
| C–Br | 193 | 293 | 1.79 |
| C–I | 214 | 234 | 1.64 |
Reactivity trend: As we descend the group, the C–X bond becomes longer and weaker. Consequently, the ease of homolytic or heterolytic cleavage — and hence the rate of nucleophilic substitution — runs R–I > R–Br > R–Cl > R–F, i.e., opposite to bond strength.
6.4 Methods of Preparation of Haloalkanes
6.4.1 From Alcohols (most general)
The –OH group of an alcohol is a poor leaving group, but after protonation or by conversion with a halogenating agent, it is readily displaced by halide ion.
R–OH + HX → R–X + H₂O (HCl needs ZnCl₂; HI is the most reactive)
3 R–OH + PCl₃ → 3 R–Cl + H₃PO₃
R–OH + PCl₅ → R–Cl + POCl₃ + HCl
R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑ (Darzens method — preferred)
Thionyl chloride (SOCl₂) is the cleanest of these, because both by-products are gases that escape, leaving a pure alkyl chloride. Reactivity of alcohols toward HX follows carbocation-stability order: 3° > 2° > 1°. The mixture of concentrated HCl and anhydrous ZnCl₂ is the Lucas reagent — a classic qualitative test for 1°, 2° and 3° alcohols.
6.4.2 From Alkanes — Free-Radical Halogenation
R–H + X₂ → R–X + H–X (hν or heat, X = Cl, Br)
The reaction goes by a chain mechanism (initiation → propagation → termination) and yields a mixture of mono-, di-, tri- and polyhaloalkanes. Useful at industrial scale but rarely for clean laboratory synthesis.
6.4.3 From Alkenes
(a) Addition of HX follows Markovnikov's rule: the H goes to the carbon that already bears more hydrogens; X ends up on the more substituted carbon (the one giving the more stable carbocation).
CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane, major)
(b) Addition of X₂ across C=C gives a vicinal dihalide — and the dropping of red-brown bromine colour is a classical test for unsaturation.
CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br (1,2-dibromoethane)
6.4.4 Halogen Exchange — Finkelstein Reaction
R–Cl (or R–Br) + NaI dry acetone→ R–I + NaCl↓ (or NaBr↓)
NaI is soluble in acetone but NaCl and NaBr are not — the precipitation of the sodium halide pulls the equilibrium forward. This is the cleanest way to make alkyl iodides.
An aromatic analogue uses heavy-metal fluorides: Ar–Cl + AgF / Hg₂F₂ → Ar–F (the Swarts reaction).
6.4.5 Hunsdiecker Reaction
R–COOAg + Br₂ CCl₄, Δ→ R–Br + CO₂↑ + AgBr↓
The silver salt of a carboxylic acid loses one carbon (as CO₂) to give an alkyl bromide with one fewer C atom. It is effective for 1° silver carboxylates only.
6.5 Methods of Preparation of Haloarenes
6.5.1 Direct Electrophilic Halogenation
C₆H₆ + X₂ FeX₃ (dark)→ C₆H₅–X + HX (X = Cl, Br)
The Lewis acid (FeCl₃, FeBr₃ or AlCl₃) polarises X₂ into X⁺ ... FeX₄⁻. The electrophile X⁺ attacks the π-cloud of benzene. Iodination is reversible and needs an oxidising agent (HNO₃) to drive it; fluorination is too violent.
6.5.2 From Arenediazonium Salts
Primary aromatic amines diazotised with NaNO₂/HCl at 0–5 °C give the reactive salt Ar–N₂⁺Cl⁻. Replacing N₂⁺ by a halogen opens the following routes:
Worked Examples
Example 6.1 — IUPAC naming of a branched haloalkane
Give the IUPAC name of (CH₃)₂CH–CH(Br)–CH₃.
Step 1 — longest chain containing the C–Br carbon = 4 carbons (butane).
Step 2 — numbering. Numbering from right: Br at C-2, methyl at C-3. Numbering from left: Br at C-3, methyl at C-2. The first-point-of-difference (2,3) vs (2,3) — tie; use the lower locant for the principal characteristic group (halogen in a haloalkane is treated as substituent, but we choose lower locant for the first cited substituent in alphabetical order — "bromo"). Thus Br gets C-2.
Answer: 2-bromo-3-methylbutane.
Example 6.2 — Common ↔ IUPAC interconversion
Translate: (i) isopropyl iodide, (ii) sec-butyl bromide, (iii) neopentyl chloride.
(i) (CH₃)₂CHI → 2-iodopropane
(ii) CH₃CH₂CH(Br)CH₃ → 2-bromobutane
(iii) (CH₃)₃C–CH₂Cl → 1-chloro-2,2-dimethylpropane
Example 6.3 — Markovnikov addition
Predict the product(s) when propene reacts with HBr in the absence of peroxides.
Protonation of the double bond can give two carbocations: CH₃–⁺CH–CH₃ (2°, stable) or ⁺CH₂–CH₂–CH₃ (1°, unstable). The more stable 2° carbocation captures Br⁻.
CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane)
[In the presence of peroxides, radical "anti-Markovnikov" addition reverses the selectivity and gives 1-bromopropane — the Kharasch effect.]
Example 6.4 — Finkelstein synthesis
How would you convert 1-bromobutane into 1-iodobutane?
CH₃CH₂CH₂CH₂Br + NaI dry acetone→ CH₃CH₂CH₂CH₂I + NaBr↓
Precipitation of NaBr removes product from equilibrium.
Example 6.5 — Hunsdiecker
Show how pentanoic acid (CH₃CH₂CH₂CH₂COOH) is converted into 1-bromobutane.
Step 1 — make the silver salt with AgOH / Ag₂O.
Step 2 — heat with Br₂ in CCl₄.
CH₃(CH₂)₃COOAg + Br₂ → CH₃(CH₂)₃Br + CO₂ + AgBr
Example 6.6 — Preparing chlorobenzene from aniline
Two steps: diazotisation then Sandmeyer.
C₆H₅NH₂ NaNO₂/HCl, 0-5°C→ C₆H₅N₂⁺Cl⁻ CuCl/HCl→ C₆H₅Cl + N₂↑
Example 6.7 — Arranging by reactivity
Arrange CH₃F, CH₃Cl, CH₃Br, CH₃I in order of decreasing rate of reaction with aqueous NaOH.
Reactivity tracks C–X bond strength inversely. Weakest bond (C–I) breaks fastest:
Order: CH₃I > CH₃Br > CH₃Cl > CH₃F.
Activity 6.1 — Why does thionyl chloride give the purest product?L4 Analyse
Take three 5 mL samples of ethanol. Treat one with conc. HCl + ZnCl₂, one with PCl₅, and one with SOCl₂. Write out the by-products of each reaction.
Predict: Which reaction will leave the lab flask cleanest and why?
- List phase/volatility of each by-product.
- Identify which by-products escape as gases vs. remain in the liquid.
- Decide which method needs the least purification afterwards.
HCl/ZnCl₂: by-product H₂O stays in mixture; ZnCl₂ residue also contaminates.
PCl₅: by-products POCl₃ (liquid) and HCl (gas); POCl₃ must be separated.
SOCl₂: by-products SO₂ (gas) and HCl (gas) — both escape; the alkyl chloride remains almost pure. Hence SOCl₂ is the preferred laboratory reagent.
Interactive — Preparation-Route Picker
Choose a starting material and the target halogen; the tool will recommend the best preparation route.
Result will appear here …
Competency-Based Questions — Classification & Preparation
A student is designing a two-step sequence to prepare 2-bromobutane in high yield starting from 1-butene. She has available HBr, Br₂ in CCl₄, NaI/acetone, peroxides and aqueous NaOH.
1. The most direct one-step product of CH₃CH₂CH=CH₂ + HBr (no peroxide) is:
(b) 2-bromobutane. Markovnikov addition places Br on the internal (more substituted) carbon giving the more stable 2° carbocation.
2. Classify CH₂=CH–CH₂–Br.
An allylic halide (Br on sp³ C adjacent to C=C). Also a 1° alkyl halide.
3. State whether True or False: SOCl₂ is preferred over PCl₅ because both by-products of its reaction with an alcohol are gases.
True. SO₂ and HCl are both gaseous; the alkyl chloride remains pure.
4. (Fill in the blank) The IUPAC name of (CH₃)₃CBr is ________.
2-bromo-2-methylpropane.
5. Why is Finkelstein done in acetone rather than water?
NaI is soluble in acetone while NaCl / NaBr are not. The sodium halide precipitates out, shifting the equilibrium to the right. Water would dissolve all three salts, killing the driving force.
Assertion–Reason Questions
Assertion (A): Among the four methyl halides, CH₃I is the most reactive in nucleophilic substitution.
Reason (R): The C–I bond is the weakest of the four C–X bonds.
Answer: A. Weakest bond breaks fastest.
Assertion (A): Direct iodination of benzene with I₂/FeI₃ is not a practical synthesis of iodobenzene.
Reason (R): The reaction is reversible; HI formed reduces iodobenzene back to benzene unless an oxidising agent is added.
Answer: A. HNO₃ or HIO₃ is added to oxidise HI and pull the reaction forward.
Assertion (A): The Hunsdiecker reaction gives an alkyl bromide with one fewer carbon atom than the starting acid.
Reason (R): The mechanism releases CO₂ after homolysis of the Ag–O bond generates a carboxyl radical.
Answer: A. Loss of CO₂ is precisely why the product chain is shortened by one carbon.
Frequently Asked Questions - Classification Preparation
What is the main concept covered in Classification Preparation?
In NCERT Class 12 Chemistry Chapter 6 (Haloalkanes and Haloarenes), "Classification Preparation" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Classification Preparation useful in real-life or applied chemistry?
Real-life applications of "Classification Preparation" from NCERT Class 12 Chemistry Chapter 6 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Classification Preparation?
Key reactions in "Classification Preparation" (NCERT Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 6?
NCERT Class 12 Chemistry Chapter 6 (Haloalkanes and Haloarenes) is structured so each part builds chemical understanding sequentially. "Classification Preparation" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Classification Preparation?
CBSE board questions from "Classification Preparation" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Classification Preparation" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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