CLASS 12 CHEMISTRY · CHAPTER 6 · STUDY NOTES

Haloalkanes and Haloarenes

A comprehensive study of organohalogen compounds, covering their classification, nomenclature, methods of preparation, physical and chemical properties, and environmental impact.

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Introduction and Classification

Locate the C–X bond before classifying the molecule.

Find the carbon holding the halogen

The starred sp³ carbon bears Br and has 1 directly attached carbon neighbours: primary. Count these neighbours, not every carbon in the molecule. The sketch is a connectivity view, not bond angles.

Haloalkane: An organic compound with halogen bonded to an sp³ carbon of an alkyl group.

Count halogens first: mono-, di- and polyhalogen compounds contain one, two and more than two halogens. For an alkyl halide, primary/secondary/tertiary counts carbon neighbours directly attached to the carbon bearing X. A branched molecule can still have a primary C–X carbon. Methyl halides are a separate zero-carbon-neighbour case.

Allylic means the sp³ carbon bearing X is next to C=C; benzylic means it is next to an aromatic ring. Vinylic and aryl halides instead have X directly attached to an sp² alkene or aromatic carbon. Geminal dihalides have both X on one carbon; vicinal ones have X on adjacent carbons.

Let’s look a little closer.

Go deeper: Classify the bond, not the whole molecule

The simple CₙH₂ₙ₊₁X formula describes saturated acyclic monohaloalkanes; it is not a universal formula for rings or unsaturated halides. Several halogen classes can occur in one larger molecule, so identify each attachment separately.

Organohalogen structures occur in biological molecules, medicines and synthetic materials. Such examples motivate the chemistry but do not imply every halogen-containing compound has similar reactivity, usefulness or safety. A ring next to CH₂Cl does not make its C–Cl bond an aryl bond.

Take a moment. What do you think?

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Question

In C₆H₅CH₂CH₂Cl, is the carbon holding Cl benzylic? Explain by tracing its neighbours.

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NCERT reference: chapter PDF pages 1, 2, 3.

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Nomenclature and Nature of C-X Bond

Name the carbon skeleton, then interpret partial charge and measured bond data.

Read the structure and the bond

Number the four-carbon chain from the end giving Br the lower locant: 2-bromobutane. Numbers label chain positions, not atom charges. A halogen is a prefix substituent; a double bond in an unsaturated parent affects the numbering rules.

Polar covalent C–X bond: A shared-electron bond with electron density biased toward the more electronegative halogen.

Property	Meaning
Bond length	Nuclear separation, measured in pm
Bond enthalpy	Energy per mole to break the specified bond
Dipole moment	Measure of charge distribution and separation

In IUPAC names, fluoro-, chloro-, bromo- and iodo- are substituent prefixes on the selected parent. Give appropriate low locants while respecting parent unsaturation. Alphabetical order does not count di-/tri- prefixes. Aromatic 1,2-, 1,3- and 1,4- arrangements correspond to ortho, meta and para in suitable disubstituted benzene examples.

Carbon is δ⁺ and halogen δ⁻ in a C–X bond; these are partial charges, not a dissociated ion pair. In the chapter’s methyl-halide data, bond length rises C–F 139 pm to C–I 214 pm, while bond enthalpy falls 452 to 234 kJ mol⁻¹. The dipole moments do not form that same monotonic sequence.

Go deeper: Different properties answer different questions

Bond enthalpy, bond length, dipole moment and leaving-group ability are distinct. A long weak C–I bond contributes to reactivity, but substrate, solvent, nucleophile and mechanism also matter. A measured dipole combines charge distribution and distance; it cannot be inferred from size alone.

Common names such as tert-butyl bromide describe an alkyl group followed by the halide. A systematic name should reconstruct connectivity: 2-bromobutane fixes the attachment on a four-carbon chain. Use the structure to check whether a proposed name counts the correct carbon neighbours.

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Question

For the chapter data, C–Cl has dipole 1.860 D and C–F 1.847 D. Does this contradict the greater electronegativity of F?

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NCERT reference: chapter PDF pages 3, 4, 5, 6.

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Preparation of Haloalkanes

Choose the starting functional group and account for coproducts.

Choose a route to the C–X bond

Alcohol plus HCl replaces OH with Cl; primary and secondary cases use ZnCl₂ in the chapter account. The visual is balanced reaction accounting, not a molecular mechanism or a student preparation procedure.

Halogen exchange: Replacement of one halogen in an organic substrate by another under suitable conditions.

Alcohols give haloalkanes using hydrogen halides, phosphorus halides or thionyl chloride. In the chapter account, primary/secondary alcohols with HCl use ZnCl₂; tertiary alcohols react more readily. SOCl₂ gives alkyl chloride plus gaseous SO₂ and HCl. PBr₃ converts three ROH to three RBr with H₃PO₃. These routes do not directly convert phenols to aryl halides in the same way.

Free-radical alkane halogenation can give mixtures of positional and further-substituted products. Addition of HX across C=C usually follows Markovnikov orientation under ordinary conditions; Br₂ addition gives vicinal dibromides. Finkelstein exchange uses NaI in dry acetone with NaCl/NaBr precipitation. Swarts exchange uses suitable metallic fluorides to form alkyl fluorides.

Go deeper: Conditions belong to the reaction

Ordinary HBr addition to propene gives the Markovnikov major product. The peroxide effect is a separate radical case for HBr, not a universal reversal for HCl or HI. A reagent label without solvent, light, catalyst or other relevant conditions may not identify one unique product.

H₂SO₄ is unsuitable with KI in the chapter’s alcohol-to-iodide route because it can oxidise HI/iodide to iodine; orthophosphoric acid avoids this competing oxidation in that account. Precipitation can favour exchange but does not erase steric limitations or make all aryl and alkyl halides interchangeable.

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Question

A route uses one mole of PBr₃ with excess ROH. From the balanced net equation, what is the maximum stoichiometric amount of RBr and what phosphorus product accompanies it?

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NCERT reference: chapter PDF pages 6, 7, 8.

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Preparation of Haloarenes

Keep direct ring substitution separate from side-chain halogenation.

Attach the halogen directly to a ring

Reaction stage: 0%

Benzene + Cl₂ gives chlorobenzene + HCl under a Lewis-acid catalyst such as FeCl₃. One ring H is replaced while aromaticity is restored in the final product. The slider shows net before/after composition, not an elementary concerted displacement; F₂ is not a selective equivalent here.

Aryl diazonium salt: An aromatic structure containing an Ar–N₂⁺ group, with a counterion balancing its charge.

Chlorination or bromination of an arene with an appropriate Lewis-acid catalyst replaces ring hydrogen while restoring aromaticity. Iodination requires conditions that remove HI, such as an oxidising agent in the chapter account. Highly reactive F₂ is not used as an equivalent selective route here.

A primary aromatic amine forms a diazonium salt with NaNO₂ and mineral acid at 273–278 K. CuCl or CuBr replaces the diazonium group by Cl or Br in Sandmeyer chemistry, with N₂ released. KI enables iodide formation without a cuprous halide. The diazonium salt is a separate intermediate rather than an alkyl carbocation.

Go deeper: Locate the new bond in the product

Ring halogenation and benzylic radical halogenation create C–X bonds at different carbons. Their conditions and products must be distinguished even if both begin with an aromatic hydrocarbon. Draw the attachment before naming the halide.

Phenol’s C–O bond has conjugative and sp² contributions, so alcohol replacement methods do not simply extend unchanged to it. The chapter’s diazonium conversion offers a different route to direct aryl attachment. These are educational reaction accounts, not practical preparation instructions.

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Question

A proposed Sandmeyer drawing replaces Ar–N₂⁺ by Ar–CH₂Cl. What carbon-accounting error has been made?

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NCERT reference: chapter PDF pages 8, 9.

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Physical Properties

Use molecular size, shape and packing to explain bulk behaviour.

Molecular shape changes bulk properties

For a fixed alkyl group, boiling-point ordering generally rises RF < RCl < RBr < RI as polarizability/dispersion grows. This ladder is qualitative: its spacing is not measured degrees. The C–X dipole alone does not predict the order.

Intermolecular attraction: Attraction between separate molecules that influences boiling, melting and solubility.

For a fixed alkyl group, the usual boiling-point order is RF < RCl < RBr < RI as size and polarizability enhance dispersion. Among comparable haloalkane isomers, branching commonly lowers boiling point by reducing effective intermolecular contact. Para-dihalobenzene symmetry can favour crystal packing and a higher melting point than ortho/meta isomers.

Haloalkanes have low water solubility despite polar C–X bonds. Dissolving replaces strong water–water attractions with insufficiently favourable new ones. Solubility in suitable organic solvents can be greater. The chapter’s 1-chloropropane density is 0.89 g mL⁻¹, while 1-bromopropane is 1.335 and 1-iodopropane 1.747; heavier derivatives can lie below water.

Go deeper: Comparisons need matching structures and conditions

A boiling-point trend based on one fixed R group should not be stretched into a ranking of all unrelated compounds. Density and water solubility also vary with molecular structure and temperature; some chlorinated compounds are denser than water while others are not.

“Like dissolves like” is a useful summary, not a complete energy calculation. Very low solubility is not exactly zero, and a layer diagram approximates the bulk phases. Melting depends on crystal packing as well as attractions, so it need not track a boiling-point sequence.

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Question

A haloalkane has density 1.40 g mL⁻¹ and only slight water solubility. Where is its bulk layer relative to water, and does the diagram mean no molecules dissolve?

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NCERT reference: chapter PDF pages 9, 10, 11.

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Nucleophilic Substitution: SN2 Mechanism

Follow the electron-pair approach and simultaneous changes in two bonds.

One concerted substitution

Reaction stage: 0%

Follow OH⁻ approaching opposite Br. The three unchanged groups flatten at the transition state and invert as C–O formation accompanies C–Br cleavage. Dashed bonds are partial, not five complete stable bonds. H, CH₃ and C₂H₅ make this centre chiral; the drawing illustrates geometric inversion without assigning R/S or optical-rotation signs.

SN2 substitution: A concerted nucleophilic substitution whose simple rate depends on both substrate and nucleophile concentrations.

A nucleophile approaches the C–X carbon from the side opposite X. C–Nu bond formation accompanies C–X cleavage through one transition state, without a carbocation intermediate. Three unchanged substituents pass through a planar arrangement; a chiral substrate undergoes inversion of configuration. The transition state has partial incoming/outgoing bonds, not five complete stable bonds.

The simple rate is k[RX][Nu]. For comparable simple alkyl substrates, backside crowding gives methyl > primary > secondary, while tertiary substrates resist ordinary SN2. For fixed R, leaving-group trends commonly favour I over Br over Cl, with F poor. Solvent and nucleophile identity remain important.

Go deeper: Choose the incoming atom, not just the reagent formula

With KCN, carbon attachment gives a nitrile R–C≡N; AgCN commonly favours nitrogen-linked isonitrile R–N≡C. Nitrite similarly permits O-linked alkyl nitrite with KNO₂ and N-linked nitroalkane with AgNO₂. These are alternative donor connections of ambident nucleophiles, not two bonds made to one carbon at once.

Substitution with hydroxide gives alcohol, alkoxide gives ether, ammonia can give amine, carboxylate gives ester and hydride reduction gives hydrocarbon. Further alkylation and competing pathways may complicate a chosen case. Geometric inversion is not a universal claim of changing + rotation to −; R/S labels can also depend on priorities after replacement.

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Question

At fixed temperature, multiply substrate concentration by 1.5 and nucleophile concentration by 2. What rate factor follows the simple SN2 law, and what remains assumed unchanged?

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Question

Original exam-style task · suggested 3 marks. Explain why a chiral haloalkane undergoing SN2 shows inversion. In your answer link approach direction, bond changes and spatial arrangement.

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NCERT reference: chapter PDF pages 11, 12, 13, 14, 15, 16.

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Nucleophilic Substitution: SN1 Mechanism

Separate slow ionisation from subsequent nucleophile attack.

An intermediate opens two faces

Reaction stage: 0%

The leaving group departs first, forming a planar sp² carbocation. A nucleophile then approaches either face to make a tetrahedral product; the face selector shows opposite configurations. Ionisation controls the simple rate=k[RX]. This idealised separated-ion model explains racemisation; ion-pair/solvent effects can give unequal product proportions.

Carbocation intermediate: A positively charged carbon species occupying an intermediate energy minimum between elementary steps.

In the simple SN1 account, C–X cleavage forms a carbocation and halide; the nucleophile then attacks, with proton transfer when needed to form a neutral product. Rate-controlling ionisation gives rate=k[RX], independent of nucleophile concentration in that simple regime. Polar protic media can stabilise the ions. Tertiary carbocations are more stable than comparable secondary/primary ones; allylic and benzylic stabilisation adds resonance considerations.

A planar sp² carbocation exposes two faces, so a chiral substrate can yield both product configurations. An ideal separated-ion picture gives racemisation, but ion-pair and solvent effects can produce unequal proportions. A two-maximum energy graph has a real intermediate minimum; a transition state is a maximum and is not isolable as a stable species.

Go deeper: Chirality and optical activity are related but different

Enantiomers are non-superimposable mirror images. A tetrahedral carbon with four different substituents is a useful criterion in these simple molecules; two identical groups remove that single-centre chirality. A racemic mixture contains chiral molecules but has zero net optical rotation because equal opposite contributions cancel under matched conditions.

Retention preserves spatial arrangement; inversion changes it at the reacting centre. Optical +/− signs must be measured and are not assigned from R/S or a wedge drawing alone. Changes elsewhere in a molecule may preserve a stereocentre yet alter the sign of the product’s rotation.

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Question

An enantiomer mixture contains 70% of a form with rotation +a and 30% with −a under identical measurement conditions. Find the net rotation in units of a. Is it racemic?

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Question

Original exam-style task · suggested 3 marks. A substrate gives both product configurations through an SN1 pathway. Explain the intermediate geometry and two possible approaches, then qualify a claim that every such product mixture must be exactly 50:50.

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NCERT reference: chapter PDF pages 15, 16, 17, 18, 19, 20, 21.

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Elimination and Reactions with Metals

Trace the atoms removed and the new carbon connection.

Changing connectivity in different ways

Reaction stage: 0%

In β-elimination, the base removes H from a carbon adjacent to the C–Br carbon; the C–H pair supplies a C=C π bond as Br departs. The slider shows linked changes for an E2-like schematic. Alcoholic KOH/heat can favour elimination, while aqueous conditions often favour substitution; substrate/base/solvent still matter.

β-elimination: Loss of hydrogen from a carbon adjacent to the leaving-group carbon, forming a carbon–carbon double bond.

In dehydrohalogenation, α carbon bears X and β carbon supplies H. Alcoholic KOH with heat commonly favours alkene formation. When alternatives exist, the usual small-base Saytzeff trend favours a more alkyl-substituted alkene. Base size, substrate, solvent and geometry can alter this outcome; substitution competes rather than disappearing by a solvent label alone.

Magnesium in dry ether converts suitable RX to Grignard RMgX, with electron-rich carbon in the polar C–Mg bond. Water or another proton source consumes it to give RH. Wurtz couples alkyl halides using Na/dry ether; Wurtz–Fittig couples alkyl and aryl groups; Fittig couples aryl groups. Mixed starting partners can yield competing products.

Go deeper: Check conservation and competition

A useful elimination drawing keeps carbon atoms and removes H/X from adjacent positions. The C–H electron pair can contribute to C=C formation while the leaving group takes its bonding pair. A more detailed E2 model also requires a suitable anti-periplanar geometry; the flat schematic shows connectivity, not every conformational requirement.

Dry ether is a reaction medium, not a source of an extra carbon in a Grignard formula. Proton transfer from water gives RH + Mg(OH)X in net accounting. Symmetric Wurtz products double the carbon count of an identical alkyl partner; mixing different partners can reduce selectivity, so “always one doubled chain” is not general.

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Question

One molecule of C₃H₇MgBr encounters water. Which organic formula results, and did the carbon skeleton become longer?

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NCERT reference: chapter PDF pages 21, 22, 23.

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Chemical Reactions of Haloarenes

Distinguish a difficult C–X displacement from substitution elsewhere on the ring.

A ring changes both reactivity and orientation

The aryl sp² C–Cl bond is shorter than the chapter’s alkyl comparison (169 versus 177 pm) and has partial double-bond character through conjugation. An ordinary phenyl cation is not a benzyl cation. Chlorobenzene resists ordinary SN1/SN2 conditions; hydroxide substitution requires strong conditions or activating substituents.

Haloarene: A compound with a halogen directly bonded to an aromatic ring carbon.

Conjugation gives aryl C–Cl partial double-bond character; sp² carbon also holds the bond more tightly than the alkyl sp³ comparison. The chapter gives 169 pm for aryl C–Cl versus 177 pm for its alkyl comparison. A phenyl cation lacks the benzylic carbocation’s resonance stabilisation, so ordinary SN1/SN2 reactivity cannot simply be transferred to chlorobenzene.

Chlorobenzene conversion to phenol needs strong conditions in the chapter account (aqueous NaOH, 623 K, 300 atm, followed by appropriate acid work-up). Ortho/para NO₂ can stabilise an anionic addition intermediate by resonance and assist nucleophilic displacement. For electrophilic ring substitution, halogen deactivates overall through its inductive contribution while directing predominantly ortho/para through lone-pair donation.

Go deeper: Different intermediates explain the apparently conflicting rules

Nucleophilic aromatic substitution with activating NO₂ involves an electron-rich addition intermediate. Resonance can place negative charge at ortho/para positions relative to the attacked site, giving a route to delocalisation into NO₂ there. Meta lacks this particular resonance pathway; that does not mean it has no inductive effect at all.

Electrophilic ring reactions such as nitration, halogenation and sulphonation use a different charge pattern. Ortho/para direction describes relative orientation, not overall activation compared with benzene. Metal coupling is another pathway: aryl/alkyl partner choice distinguishes Fittig and Wurtz–Fittig connectivity.

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Question

A learner argues that an ortho/para director must react faster than benzene. Use chlorobenzene to distinguish two independent comparisons.

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Question

Original exam-style task · suggested 3 marks. Explain why para-NO₂ can assist nucleophilic displacement in a chloroarene more strongly through resonance than meta-NO₂. Focus on the addition intermediate rather than electrophilic direction.

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NCERT reference: chapter PDF pages 23, 24, 25, 26, 27, 28.

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Polyhalogen Compounds

Connect molecular properties with uses and unintended consequences.

Useful properties can create lasting impacts

Compare how many H atoms have been replaced around tetrahedral carbon: CH₂Cl₂, CHCl₃, CHI₃, CCl₄ and CCl₂F₂. These structures have different uses and hazards; “stable” does not imply environmentally harmless. This is a connectivity projection, not an exact three-dimensional bond-angle model.

Persistent organohalogen compound: A halogen-containing organic compound whose resistance to breakdown can support lasting environmental exposure.

CH₂Cl₂ has solvent applications; CHCl₃ is chloroform, CHI₃ iodoform, CCl₄ tetrachloromethane and CCl₂F₂ a CFC historically used as Freon-12. Iodoform’s historical antiseptic effect is associated with iodine release. Chloroform’s historical anaesthetic use is not current medical advice; oxidation in air/light can form hazardous phosgene: 2CHCl₃ + O₂ → 2COCl₂ + 2HCl.

Stable CFCs can reach the stratosphere, where UV releases chlorine radicals. A catalytic cycle consumes ozone and atomic oxygen while regenerating Cl·. DDT’s persistence and fat solubility support storage during repeated exposure and transfer through food webs. Useful solvent or insecticidal behaviour does not establish that a compound is harmless or currently approved.

Go deeper: Use history must not become present-day permission

The textbook’s industrial examples and historical statements describe chemistry and consequences. Regulations and contemporary usage vary; these notes do not assert current bans, approvals or universal safety categories. Stable in one chemical setting does not mean inert under high-energy UV or metabolically harmless.

Bioaccumulation describes concentration building within an organism from ongoing exposure. Biomagnification describes increasing concentrations through trophic transfer in suitable food webs. Neither is demonstrated quantitatively by decorative dot counts; actual measurements require exposure, uptake and elimination data.

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Question

Add Cl· + O₃ → ClO· + O₂ and ClO· + O → Cl· + O₂. What cancels and what is the net equation?

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NCERT reference: chapter PDF pages 29, 30, 31.

Haloalkanes and Haloarenes Class 12 Notes

Chapter notes

Haloalkanes and Haloarenes

Introduction and Classification

Find the carbon holding the halogen

Nomenclature and Nature of C-X Bond

Read the structure and the bond

Preparation of Haloalkanes

Choose a route to the C–X bond

Preparation of Haloarenes

Attach the halogen directly to a ring

Physical Properties

Molecular shape changes bulk properties

Nucleophilic Substitution: SN2 Mechanism

One concerted substitution

Nucleophilic Substitution: SN1 Mechanism

An intermediate opens two faces

Elimination and Reactions with Metals

Changing connectivity in different ways

Chemical Reactions of Haloarenes

A ring changes both reactivity and orientation

Polyhalogen Compounds

Useful properties can create lasting impacts

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