Class 12 · Chemistry · Chapter 6
Haloalkanes and Haloarenes
The first organic chapter of Class 12: how halogen compounds are classified and named, how they are made from alcohols, hydrocarbons and diazonium salts, and how they react by nucleophilic substitution (SN1 and SN2), elimination and with metals. It is the chapter where the board tests mechanisms, optical activity, reasoning questions on chlorobenzene and the uses of polyhalogen compounds, and it feeds directly into Alcohols, Amines and the conversion questions of the whole organic unit.
Before you startWhat the board asks from this chapter
Haloalkanes and Haloarenes carries 6 marks in the CBSE unit-wise weightage for the 70-mark theory paper. In practice that is one or two 1-mark items (an MCQ or an assertion-reason), a 2- or 3-mark reasoning or mechanism question, and the chapter usually contributes one or two steps to the 5-mark conversion or name-reaction question drawn from the organic unit as a whole.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Which halide reacts fastest by SN1 or SN2, which molecule is chiral, the reagent for Finkelstein or Swarts reaction, the major product of alcoholic KOH, or the classification of a halide as vinylic, allylic, benzylic or vic-dihalide. |
| Assertion–Reason | 1 | Pairs on chlorobenzene being less reactive than chloroethane, tertiary halides going by SN1, KCN giving nitriles while AgCN gives isocyanides, and the peroxide effect being limited to HBr. |
| Very short answer | 2 | IUPAC names and structures; give reasons (why alkyl halides are immiscible with water, why Grignard reagents need anhydrous conditions, why p-dichlorobenzene has the higher melting point, why H2SO4 is not used with KI); write the product of a given reaction. |
| Short answer | 3 | Mechanism of SN1 or SN2 with stereochemistry; arrange halides in order of reactivity or boiling point; three what-happens-when equations; explain why haloarenes are less reactive towards nucleophiles; identify A, B, C in a reaction sequence. |
| Long answer | 5 | Usually shared with Alcohols or Amines: a set of conversions (benzene to biphenyl, propene to 1-nitropropane, ethanol to but-1-yne), name reactions with equations (Sandmeyer, Finkelstein, Swarts, Wurtz-Fittig, Fittig), plus a reasoning part. |
| Case-based | 4 | A passage on SN1 versus SN2 (rates, stereochemistry, solvent), on the chemistry of chlorobenzene (resonance, Dow's process, nitro activation) or on polyhalogen compounds and the environment, followed by four sub-questions. |
Hot topics, in order
- SN1 and SN2 mechanisms: order, steps, stereochemistry (inversion versus racemisation) and reactivity order of halides
- Why haloarenes are less reactive than haloalkanes towards nucleophilic substitution, and the effect of nitro groups
- Give-reason questions on physical properties: boiling points, p-dichlorobenzene, immiscibility with water, dipole moment of chlorobenzene
- Name reactions with equations: Sandmeyer, Finkelstein, Swarts, Wurtz, Wurtz-Fittig, Fittig, Grignard formation
- Conversions through the organic unit that pass through a haloalkane (peroxide effect, alcoholic versus aqueous KOH, KCN versus AgCN)
- Optical activity: chiral carbon, enantiomers, racemic mixture, which molecule is optically active
- IUPAC nomenclature and classification (primary, secondary, tertiary; allylic, benzylic, vinylic, aryl; gem and vic dihalides)
- Uses and environmental effects of CH2Cl2, CHCl3, CHI3, CCl4, freons and DDT
The whole chapter, in orderNotes
01 6.1 Classification PYQ 2020
A haloalkane is an alkane in which one or more hydrogens are replaced by halogen; the halogen sits on an sp3 carbon of an aliphatic chain. A haloarene has the halogen bonded directly to an sp2 carbon of an aromatic ring. Keep this distinction in mind from the first page, because everything about reactivity in this chapter follows from whether the C–X carbon is sp3 or sp2.
NCERT classifies the compounds in two ways. First, by the number of halogen atoms: mono-, di-, tri- and polyhalogen compounds. Dihalides get two more names that the board likes. A geminal dihalide (gem-dihalide) has both halogens on the same carbon, for example ethylidene chloride, CH3CHCl2, which is 1,1-dichloroethane. A vicinal dihalide (vic-dihalide) has them on adjacent carbons, for example ethylene dichloride, ClCH2CH2Cl, which is 1,2-dichloroethane.
Second, by the hybridisation of the carbon carrying the halogen. Compounds with an sp3 C–X bond are of three types. Alkyl halides (R–X) are primary (1°), secondary (2°) or tertiary (3°) according to whether the halogen-bearing carbon is joined to one, two or three other carbons. Allylic halides have the halogen on an sp3 carbon that is next to a C=C double bond, as in CH2=CH–CH2Cl (3-chloropropene). Benzylic halides have the halogen on an sp3 carbon attached to a benzene ring, as in C6H5CH2Cl. Compounds with an sp2 C–X bond are of two types: vinylic halides, where the halogen is on a carbon of a C=C double bond (CH2=CHCl), and aryl halides, where the halogen is on a benzene ring carbon (C6H5Cl).
The trap: benzyl chloride is not an aryl halide. The chlorine is on the CH2, an sp3 carbon, so it behaves like a very reactive alkyl halide. Chlorobenzene, with chlorine on the ring, is the unreactive one.
02 6.2 Nomenclature PYQ 2023 · 2019 · 2015
In the common system an alkyl halide is named as alkyl + halide (methyl chloride, isopropyl bromide, tert-butyl chloride). In the IUPAC system the halogen is a substituent, so the name is halo + alkane with the position number: CH3CH2CH2Cl is 1-chloropropane, (CH3)2CHBr is 2-bromopropane and (CH3)3CCl is 2-chloro-2-methylpropane. Two rules earn most of the marks: number the chain so the halogen and other substituents get the lowest set of locants, and when there is a tie, list substituents alphabetically and give the lower number to the one that comes first in the alphabet (bromo before chloro before iodo before methyl).
Dihalides: the common names ethylene dichloride (1,2-dichloroethane) and ethylidene chloride (1,1-dichloroethane) are asked directly. Haloarenes: C6H5Cl is chlorobenzene in both systems; for disubstituted rings the common prefixes o-, m-, p- correspond to 1,2-, 1,3- and 1,4- in IUPAC, so p-dichlorobenzene is 1,4-dichlorobenzene. For haloalkenes the double bond gets priority in numbering, so CH2=CH–CH2Br is 3-bromoprop-1-ene and CH2=CHCl is chloroethene (vinyl chloride).
Some NCERT names the board reuses: CH3CH(Cl)CH(Br)CH3 is 2-bromo-3-chlorobutane (bromo is alphabetically first so the chain is numbered from its side); CHF2CBrClF is 1-bromo-1-chloro-1,2,2-trifluoroethane; ClCH2C≡CCH2Br is 1-bromo-4-chlorobut-2-yne; (CH3)2CHCH(Cl)CH3 is 2-chloro-3-methylbutane; CH3CH=CH–C(Br)(CH3)2 is 4-bromo-4-methylpent-2-ene. Practise the reverse direction too: 4-tert-butyl-3-iodoheptane, 1-chloro-4-ethylcyclohexane and 1-bromo-4-sec-butyl-2-methylbenzene have all been set as structure-drawing questions.
03 6.3 Nature of C–X Bond PYQ 2019 · 2016
Halogens are more electronegative than carbon, so the C–X bond is polar: carbon carries a partial positive charge and the halogen a partial negative charge. This is why haloalkanes are attacked by nucleophiles at carbon and why the halide ion is a good leaving group.
Down the group the halogen atom gets bigger, so the C–X bond becomes longer and weaker. NCERT gives the numbers and the examiner accepts them as they are: bond lengths C–F 139 pm, C–Cl 178 pm, C–Br 193 pm, C–I 214 pm; bond enthalpies C–F 452, C–Cl 351, C–Br 293, C–I 234 kJ mol−1. So the C–I bond breaks most easily, which makes iodides the most reactive in substitution and elimination, and the C–F bond hardly breaks at all.
Dipole moments do not follow a simple trend. The values for the methyl halides are CH3F 1.847 D, CH3Cl 1.860 D, CH3Br 1.830 D and CH3I 1.636 D. Fluorine is the most electronegative, but the C–F bond is so short that the product of charge and distance is slightly smaller than for C–Cl. This is why CH3Cl has the highest dipole moment among the methyl halides, a favourite one-mark question.
Two comparisons from this section recur in give-reason form. Among CH2Cl2, CHCl3 and CCl4, dichloromethane has the highest dipole moment (about 1.60 D) because its two C–Cl dipoles reinforce each other, in CHCl3 the three C–Cl dipoles partly cancel (about 1.08 D), and in the tetrahedral CCl4 they cancel completely to give zero. And the C–Cl bond in chlorobenzene is shorter (about 169 pm) and stronger than in an alkyl chloride because the carbon is sp2 and the bond has partial double-bond character from resonance; that point is developed in section 6.7.2.
04 6.4 Methods of Preparation of Haloalkanes PYQ 2023 · 2020 · 2018
From alcohols. The –OH group is replaced by halogen. With halogen acids: primary and secondary alcohols need HCl with anhydrous ZnCl2 (Lucas reagent), tertiary alcohols react with concentrated HCl alone by shaking. Bromides are made with constant boiling HBr (48%) or with NaBr and concentrated H2SO4. Iodides are made with KI and 95% orthophosphoric acid, never with H2SO4, because sulphuric acid oxidises the HI formed to I2. With phosphorus halides: 3ROH + PX3 → 3RX + H3PO3 (X = Cl, Br), ROH + PCl5 → RCl + POCl3 + HCl, and for bromides and iodides red phosphorus with Br2 or I2 generates PBr3 or PI3 in situ. With thionyl chloride: ROH + SOCl2 → RCl + SO2 + HCl. The SOCl2 method is preferred for chlorides because both by-products are gases and escape, leaving pure alkyl chloride. Reactivity of alcohols towards these reagents is 3° > 2° > 1°.
From hydrocarbons. Free radical chlorination or bromination of an alkane gives a mixture of isomeric mono- and polyhaloalkanes that is hard to separate, so it is a poor route to a single product; it is still asked as 'how many monochloro products' from a given alkane. Electrophilic substitution of arenes with Cl2 or Br2 in the presence of a Lewis acid (FeCl3, FeBr3) gives ortho and para haloarenes; the isomers boil close together but the para isomer melts much higher, so they are separated by fractional crystallisation. Iodination is reversible because HI reduces the iodoarene back, so an oxidising agent such as HNO3 or HIO4 is added to destroy HI. Fluoro compounds are not made this way because fluorine is too reactive.
Sandmeyer's reaction. A primary aromatic amine is diazotised with NaNO2 and HCl at 273–278 K to give the benzenediazonium salt; treatment with Cu2Cl2/HCl or Cu2Br2/HBr replaces –N2+ by Cl or Br. Iodine needs no copper salt: shaking the diazonium solution with KI gives iodobenzene.
From alkenes. HX adds across C=C by Markovnikov's rule (propene + HBr → 2-bromopropane). In the presence of peroxide, HBr alone adds anti-Markovnikov (1-bromopropane); HCl and HI do not show the peroxide effect. X2 in CCl4 adds to give vic-dihalides, and the discharge of the reddish-brown colour of bromine is the test for unsaturation.
Halogen exchange. Finkelstein reaction: R–Cl or R–Br with NaI in dry acetone gives R–I; NaCl and NaBr are insoluble in acetone and precipitate, pulling the equilibrium forward. Swarts reaction: an alkyl chloride or bromide heated with a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3 gives the alkyl fluoride, e.g. CH3Br + AgF → CH3F + AgBr.
05 6.5 Preparation of Haloarenes PYQ 2022 · 2017
NCERT gives this as a short separate section, and it is really two of the routes above applied to the aromatic ring. Aryl halides cannot be made from phenols by replacing –OH with halogen, because the C–O bond in phenol has partial double-bond character and is not broken by HX or PX3; this is an assertion-reason favourite, so hold on to the reason.
Electrophilic substitution. Benzene or a substituted benzene reacts with Cl2 or Br2 in the presence of the Lewis acid catalyst FeCl3 or FeBr3 (in the dark, at ordinary temperature). The halogen is polarised by the catalyst, the ring attacks the positive end, and the product is the haloarene. With toluene, the methyl group directs to ortho and para, giving 2-chlorotoluene and 4-chlorotoluene, which are separated by exploiting the large difference in melting point of the para isomer. The examiner expects the note that the reaction must be done in the absence of sunlight; with sunlight and no catalyst chlorine attacks the side chain instead and gives benzyl chloride.
From diazonium salts. Aniline is diazotised with NaNO2 and HCl at 273–278 K to give benzenediazonium chloride. Sandmeyer's reaction with Cu2Cl2/HCl or Cu2Br2/HBr replaces the diazonium group by Cl or Br with loss of N2. For iodobenzene, the diazonium salt is simply warmed with potassium iodide. This is the route to use whenever a conversion asks you to place a halogen at a specific position that direct halogenation would not give, for example aniline to chlorobenzene, or benzene to 4-bromonitrobenzene (nitrate first, then brominate; the nitro group is meta-directing, so instead brominate first and then nitrate, which puts the nitro para to bromine).
The Fittig and Wurtz-Fittig reactions, which use haloarenes to build biphenyl and alkylbenzenes, are covered under the reactions of haloarenes in section 6.7.2.
06 6.6 Physical Properties PYQ 2024 · 2023 · 2020 · 2016
Pure haloalkanes are colourless, but bromides and iodides develop colour on exposure to light because a little halogen is liberated. Many volatile halogen compounds have a sweet smell.
Boiling points. The molecules are polar, so dipole-dipole attractions add to van der Waals forces and the boiling points are well above those of the parent hydrocarbons. For the same alkyl group the order is RI > RBr > RCl > RF, because the size and mass of the halogen, and hence the strength of the van der Waals forces, increase down the group. Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Boiling point rises with the number of carbon atoms, and with the number of halogen atoms (CH3Cl < CH2Cl2 < CHCl3 < CCl4). For isomers, boiling point falls with branching because the molecule becomes more compact and its surface area drops: 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane. Hence an NCERT order: chloromethane < bromomethane < dibromomethane < bromoform, and isopropyl chloride < 1-chloropropane < 1-chlorobutane.
Dihalobenzenes. The ortho, meta and para isomers have nearly the same boiling points, but the para isomer has a much higher melting point and a lower solubility. The reason the examiner wants: the para isomer is symmetrical, fits better into the crystal lattice, and so has stronger intermolecular forces in the solid; more energy is needed to melt it or to pull it apart to dissolve it.
Density. Bromo, iodo and polychloro derivatives are heavier than water. Density increases with the number of carbon atoms, the number of halogen atoms and the atomic mass of the halogen.
Solubility. Haloalkanes are only very slightly soluble in water although they are polar. To dissolve, the attractions between haloalkane molecules and the hydrogen bonds between water molecules must both be broken, and the new attractions between haloalkane and water are not strong enough to make up for this. They dissolve readily in organic solvents because the new forces are about as strong as those broken.
07 6.7.1 Reactions of Haloalkanes: Nucleophilic Substitution (S<sub>N</sub>2 and S<sub>N</sub>1) PYQ 2024 · 2023 · 2022 · 2020 · 2019 · 2017
The carbon of the C–X bond is electron deficient, so a nucleophile attacks it and the halide leaves: R–X + Nu− → R–Nu + X−. Products depend on the nucleophile: OH− (aqueous KOH) gives an alcohol, R'O− gives an ether (Williamson synthesis), KCN gives a nitrile, AgCN gives an isocyanide, KNO2 gives an alkyl nitrite R–O–N=O, AgNO2 gives a nitroalkane R–NO2, NH3 gives an amine, RCOOAg gives an ester, and LiAlH4 reduces R–X to the alkane. CN− and NO2− are ambident nucleophiles: they can attack through either of two atoms. KCN is ionic and the cyanide ion attacks through carbon (the C–C bond is stronger than C–N), while in covalent AgCN the carbon is bonded to silver, so nitrogen attacks and the isocyanide forms.
SN2 (substitution nucleophilic bimolecular). Studied by Hughes and Ingold in 1937 with CH3Cl + OH− → CH3OH + Cl−. The rate depends on both reactants: rate = k[CH3Cl][OH−], second order. It is a one-step process. The nucleophile attacks the carbon from the side opposite to the halogen, and in the transition state the carbon is bonded partially to both the incoming nucleophile and the leaving halide, with the three other groups in a plane. As the new bond forms the old one breaks and the three groups flip like an umbrella in the wind: the configuration is inverted (Walden inversion). Bulky groups around the carbon obstruct the attack, so reactivity is CH3X > 1° > 2° > 3°.
SN1 (substitution nucleophilic unimolecular). The model reaction is (CH3)3C–Br + OH− → (CH3)3C–OH + Br− in a polar protic solvent such as water, ethanol or acetic acid. Rate = k[(CH3)3CBr], first order; it depends only on the halide. The reaction is in two steps. Step 1, slow and rate determining: the C–Br bond breaks heterolytically to give a planar tert-butyl carbocation and Br−, helped by solvation. Step 2, fast: the nucleophile attacks the carbocation. Since the rate depends on how easily the carbocation forms, reactivity is 3° > 2° > 1° > CH3X, the reverse of SN2. Allylic and benzylic halides react fast by SN1 because their carbocations are stabilised by resonance.
For both mechanisms the leaving group matters: R–I > R–Br > R–Cl > R–F, following the weakening of the C–X bond. Between C6H5CH2Cl and C6H5CH(Cl)C6H5, the second is hydrolysed faster because its carbocation is stabilised by two phenyl rings.
08 6.7.1 Stereochemical Aspects of Nucleophilic Substitution PYQ 2024 · 2022 · 2019 · 2016
Plane polarised light and optical activity. Ordinary light vibrates in all planes perpendicular to its direction; passing it through a Nicol prism gives plane polarised light. A compound that rotates the plane of polarised light is optically active. If it rotates the plane clockwise (as seen by an observer looking towards the light) it is dextrorotatory, shown as d or (+); anticlockwise is laevorotatory, l or (−). The instrument is a polarimeter.
Chirality. Objects that are not superimposable on their mirror images are chiral (like your hands); those that are superimposable are achiral. A molecule with a carbon bonded to four different groups (an asymmetric or stereocentre) is chiral: 2-chlorobutane, 2-bromobutane, butan-2-ol and bromochloroiodomethane are chiral; 1-chlorobutane, propan-2-ol and 2-bromopropane are achiral. The two non-superimposable mirror-image forms are enantiomers. They have identical physical properties except that they rotate polarised light by equal amounts in opposite directions. An equimolar mixture of the two enantiomers is a racemic mixture, shown as (±) or dl; it is optically inactive because the rotations cancel (external compensation). Converting an enantiomer into a racemic mixture is called racemisation.
Retention, inversion and racemisation. If the spatial arrangement of the groups around the stereocentre is kept, the reaction proceeds with retention of configuration. If the product's configuration is the mirror image of the starting compound's, there is inversion. If equal amounts of both are formed, the product is racemic.
What the mechanisms do. In SN2 the nucleophile attacks from the side opposite the leaving group, so an optically active halide gives a product of inverted configuration: (−)-2-bromooctane with OH− gives (+)-octan-2-ol. In SN1 the intermediate carbocation is planar and sp2, so the nucleophile can attack from either face with equal probability; 50% inversion and 50% retention give a racemic mixture. So an optically active tertiary or secondary halide hydrolysed by the SN1 path gives a racemic (optically inactive) alcohol. The board asks this both ways: 'which mechanism gives inversion' and 'why does the product show no optical rotation'.
09 6.7.1 Elimination Reactions and Reaction with Metals PYQ 2024 · 2023 · 2020 · 2018
β-Elimination (dehydrohalogenation). When a haloalkane with a β-hydrogen is heated with alcoholic KOH, the halogen is removed from the α-carbon and a hydrogen from the β-carbon, and an alkene forms. If more than one β-hydrogen is available, more than one alkene can form. Saytzeff's rule decides the major product: the more highly substituted alkene (the one with more alkyl groups on the double-bond carbons) is preferred. So 2-bromopentane with alcoholic KOH gives pent-2-ene (major, 81%) and pent-1-ene (minor, 19%). 2-Bromobutane gives but-2-ene as the major product; 2-chloro-2-methylbutane gives 2-methylbut-2-ene rather than 2-methylbut-1-ene.
Aqueous versus alcoholic KOH. In water KOH ionises to give hydroxide ions, a strong nucleophile that attacks the carbon and gives an alcohol by substitution. In alcohol the ethoxide ion (from KOH + C2H5OH) is a stronger base than a nucleophile, so it removes the β-hydrogen and gives the alkene by elimination. This is a two-mark give-reason that appears almost every second year. Elimination is also favoured by higher temperature, a bulky base and a tertiary halide; substitution by a lower temperature, a small strong nucleophile and a primary halide.
Reaction with magnesium: Grignard reagents. A haloalkane reacts with magnesium metal in dry ether to give an alkyl magnesium halide, RMgX, discovered by Victor Grignard in 1900. The C–Mg bond is covalent but highly polar, with carbon towards the negative end, so the alkyl group behaves as a carbanion and adds to carbonyl compounds to build C–C bonds (used in the Alcohols chapter). Grignard reagents are extremely reactive towards any compound with an acidic hydrogen, even water: RMgX + H2O → R–H + Mg(OH)X. That is why they must be prepared under anhydrous conditions, and why the reaction with D2O is used to put a deuterium atom onto a carbon.
Wurtz reaction. Two molecules of alkyl halide react with sodium in dry ether to give an alkane with twice the number of carbon atoms: 2R–X + 2Na → R–R + 2NaX. It gives symmetrical alkanes; 1-chlorobutane with sodium gives n-octane, and isobutyl bromide gives 2,5-dimethylhexane.
10 6.7.2 Reactions of Haloarenes PYQ 2024 · 2023 · 2022 · 2020 · 2019 · 2015
Nucleophilic substitution. Aryl halides are far less reactive than haloalkanes towards nucleophiles. NCERT gives four reasons and the examiner wants at least two, ideally with the resonance structures. (1) Resonance: the lone pairs on chlorine delocalise into the ring, so the C–Cl bond acquires partial double-bond character and is harder to break. (2) Hybridisation: the ring carbon is sp2, more electronegative than an sp3 carbon, so it holds the shared pair more tightly and the bond is shorter (169 pm against 177 pm in haloalkanes) and stronger. (3) Instability of the phenyl cation: an SN1 path would need a phenyl cation, which cannot be stabilised by resonance, so it does not form. (4) Repulsion: the electron-rich ring repels an approaching electron-rich nucleophile. Chlorobenzene is converted to phenol only by heating with aqueous NaOH at 623 K and 300 atm (the Dow's process), followed by acidification.
Effect of substituents. Electron-withdrawing groups at the ortho and para positions make nucleophilic substitution easier, because the negative charge of the intermediate carbanion is delocalised onto the nitro group. The conditions for hydrolysis to phenol fall steadily: chlorobenzene needs 623 K and 300 atm; 4-nitrochlorobenzene needs 15% NaOH at 443 K; 2,4-dinitrochlorobenzene needs 15% NaOH at 368 K; 2,4,6-trinitrochlorobenzene is hydrolysed by warm water alone. A nitro group at the meta position has no such effect because none of the resonance structures puts the negative charge on the carbon bearing it.
Electrophilic substitution. Haloarenes undergo the usual reactions of benzene: halogenation (Cl2/FeCl3 gives 1,2- and 1,4-dichlorobenzene), nitration (conc. HNO3 + conc. H2SO4 gives mainly 1-chloro-4-nitrobenzene), sulphonation (conc. H2SO4 gives 4-chlorobenzenesulphonic acid), Friedel-Crafts alkylation (CH3Cl/anhyd. AlCl3 gives 4-chlorotoluene) and acylation (CH3COCl/AlCl3 gives 4-chloroacetophenone). The halogen is deactivating but ortho- and para-directing. Its −I effect withdraws electrons and deactivates the ring; but resonance places the electron density at ortho and para positions, so the attack happens there. The para product predominates because of steric hindrance at the ortho positions.
Reaction with metals. Wurtz-Fittig reaction: an aryl halide and an alkyl halide with sodium in dry ether give an alkylarene (chlorobenzene + CH3Cl + 2Na → toluene + 2NaCl). Fittig reaction: two aryl halides with sodium give a biaryl (2C6H5Cl + 2Na → C6H5–C6H5 + 2NaCl). Bromobenzene with Mg in dry ether gives phenylmagnesium bromide.
11 6.8 Polyhalogen Compounds PYQ 2023 · 2020 · 2018
Carbon compounds with more than one halogen atom are useful in industry and agriculture, and the board asks their uses and their harmful effects in one- or two-mark bits.
Dichloromethane (methylene chloride, CH2Cl2) is a solvent for paint removers, a propellant in aerosols, a process solvent in drug manufacture and a metal-cleaning and finishing solvent. It harms the human central nervous system; low exposure causes slightly impaired hearing and vision, higher doses dizziness, nausea and tingling of the fingers and toes. Skin contact causes burning and redness; direct contact with the eye damages the cornea.
Trichloromethane (chloroform, CHCl3) is a solvent for fats, alkaloids and iodine, and is used mainly to make the refrigerant R-22 (chlorodifluoromethane). It was once an anaesthetic in surgery but has been replaced by safer ones such as ether. Inhaling it depresses the central nervous system; about 900 parts per million for a short time causes dizziness, fatigue and headache, and long exposure damages the liver (where it is metabolised to phosgene) and kidneys and causes skin sores. In air and light it is slowly oxidised to the poisonous gas phosgene, CCl2O, so it is stored in closed, dark-coloured bottles filled to the brim to keep air out.
Triiodomethane (iodoform, CHI3) was used as an antiseptic; the action is due to the free iodine it liberates, not to the compound itself. It has been replaced by other iodine formulations because of its objectionable smell.
Tetrachloromethane (carbon tetrachloride, CCl4) is used to make refrigerants and aerosol propellants, as a feedstock for chlorofluorocarbons, and as a cleaning fluid, degreasing agent, spot remover and fire extinguisher. It causes liver cancer, and exposure brings dizziness, nausea and vomiting, nerve damage and in severe cases coma or death. Released into the air it reaches the stratosphere and depletes the ozone layer, increasing UV exposure, skin cancer and eye disease.
Freons are chlorofluorocarbons of methane and ethane: stable, unreactive, non-toxic, non-corrosive and easily liquefied. Freon 12 (CCl2F2), the commonest, is made from CCl4 by the Swarts reaction. They are used as aerosol propellants and in refrigeration and air conditioning, but they diffuse unchanged into the stratosphere and start radical chain reactions that break down ozone.
DDT (p,p'-dichlorodiphenyltrichloroethane) was the first chlorinated organic insecticide, synthesised in 1873; Paul Müller discovered its insecticidal action in 1939 and won the Nobel Prize in 1948. It was used against the mosquito that spreads malaria and the lice that spread typhus. Insects developed resistance, it is highly toxic to fish, and because it is chemically stable and not metabolised rapidly it accumulates in fatty tissues. Its use was banned in the United States in 1973.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Alcohol to chloride with thionyl chloride | ROH + SOCl2 → RCl + SO2 + HCl | Preferred method: both by-products are gases, so the product is pure. |
| Alcohol with phosphorus halides | 3ROH + PX3 → 3RX + H3PO3; ROH + PCl5 → RCl + POCl3 + HCl | Red P + Br2 or I2 gives PBr3/PI3 in situ for bromides and iodides. |
| Alcohol with halogen acids | ROH + HX → RX + H2O | HCl needs anhydrous ZnCl2 for 1° and 2° alcohols; HBr from NaBr + H2SO4; HI from KI + H3PO4 (not H2SO4). |
| Sandmeyer's reaction | ArN2+Cl− + Cu2Cl2/HCl → ArCl + N2; with Cu2Br2/HBr → ArBr; with KI → ArI | Diazotisation with NaNO2 + HCl at 273–278 K; no copper salt needed for iodide. |
| Finkelstein reaction | R–X + NaI → R–I + NaX↓ (X = Cl, Br; dry acetone) | NaCl/NaBr insoluble in acetone, so the reaction goes to completion. |
| Swarts reaction | R–Br + AgF → R–F + AgBr | Also Hg2F2, CoF2, SbF3. Freon 12 from CCl4 this way. |
| Addition of HBr to alkene | CH3CH=CH2 + HBr → CH3CHBrCH3; with peroxide → CH3CH2CH2Br | Markovnikov without peroxide; anti-Markovnikov with peroxide (HBr only). |
| Nucleophilic substitution products | R–X + KOH(aq) → ROH; + KCN → RCN; + AgCN → RNC; + KNO2 → RONO; + AgNO2 → RNO2; + NaOR' → ROR'; + NH3 → RNH2 | KCN/AgCN and KNO2/AgNO2 contrast is the ambident nucleophile question. |
| Rate laws | SN2: rate = k[RX][Nu−]; SN1: rate = k[RX] | SN2 order CH3X > 1° > 2° > 3°, inversion; SN1 order 3° > 2° > 1° > CH3X, racemisation. Leaving group RI > RBr > RCl > RF for both. |
| Dehydrohalogenation | R–CH2–CH(X)–R' + KOH (alc.) → R–CH=CH–R' + KX + H2O | Saytzeff rule: more substituted alkene is major. Aqueous KOH gives the alcohol instead. |
| Grignard reagent | R–X + Mg → RMgX (dry ether); RMgX + H2O → RH + Mg(OH)X | Must be prepared under anhydrous conditions. |
| Wurtz, Wurtz-Fittig and Fittig | 2RX + 2Na → R–R + 2NaX; ArX + RX + 2Na → Ar–R + 2NaX; 2ArX + 2Na → Ar–Ar + 2NaX | All in dry ether. Benzene → biphenyl: chlorinate, then Fittig. |
| Dow's process | C6H5Cl + NaOH (623 K, 300 atm) → C6H5ONa → C6H5OH (with H+) | Conditions ease with o/p nitro groups: 443 K (one NO2), 368 K (two), warm water (three). |
| C–X bond data (NCERT) | Length: C–F 139, C–Cl 178, C–Br 193, C–I 214 pm; Enthalpy: 452, 351, 293, 234 kJ mol−1 | Dipole moments: CH3Cl 1.860 > CH3F 1.847 > CH3Br 1.830 > CH3I 1.636 D. |
Practise with a pencilDiagrams that earn marks
S<sub>N</sub>2 mechanism of CH<sub>3</sub>Cl with OH<sup>−</sup>
Draw three stages left to right. Stage 1: the carbon of CH3Cl with Cl on the right and the three hydrogens fanned to the left, and HO− approaching from the left (opposite side to Cl) with a curved arrow from the oxygen lone pair to carbon and another from the C–Cl bond onto Cl. Stage 2, in square brackets with a double-dagger sign: the transition state with the three hydrogens in one plane and dotted bonds HO···C···Cl carrying δ− on both O and Cl. Stage 3: CH3OH with the hydrogens now fanned to the right, plus Cl−. Label 'backside attack', 'transition state' and 'inversion of configuration'.
S<sub>N</sub>1 mechanism of (CH<sub>3</sub>)<sub>3</sub>CBr with OH<sup>−</sup>
Two steps one under the other. Step 1 (label 'slow, rate determining'): (CH3)3C–Br with a curved arrow from the C–Br bond to Br, an equilibrium arrow, and the planar (CH3)3C+ carbocation drawn as a trigonal planar carbon plus Br−. Step 2 (label 'fast'): OH− attacking the carbocation to give (CH3)3C–OH. Add a note that the nucleophile can attack the planar carbocation from either face, so an optically active halide gives a racemic mixture.
Resonance structures of chlorobenzene
Draw five structures joined by double-headed arrows. The first is ordinary chlorobenzene with two lone pairs on Cl. In the second, a lone pair from Cl forms C=Cl+ and the ring carries a negative charge on the ortho carbon. Move the negative charge to the para carbon in the third and to the other ortho carbon in the fourth, then back to neutral chlorobenzene in the fifth. Write under it: 'partial double-bond character of C–Cl; electron density at ortho and para positions'. The same drawing is used for low reactivity to nucleophiles and for o/p-direction in electrophilic substitution.
Enantiomers of 2-chlorobutane (chirality)
Draw a vertical dashed line for the mirror. On the left draw C-2 as a tetrahedral carbon with H (wedge), Cl (dash), CH3 and C2H5 as the four groups. On the right draw its mirror image with the wedge and dash positions swapped. Label the two forms as enantiomers, mark the stereocentre with an asterisk, and note that they are non-superimposable and rotate plane-polarised light in opposite directions. For contrast, a small drawing of 2-chloropropane with two identical CH3 groups shows why it is achiral.
Effect of nitro groups on hydrolysis of chlorobenzene
Draw four benzene rings with Cl at the top: plain chlorobenzene, 4-nitro, 2,4-dinitro and 2,4,6-trinitro. Under each write the condition needed to replace Cl by OH: NaOH 623 K 300 atm; 15% NaOH 443 K; 15% NaOH 368 K; warm water. Add an arrow showing 'ease of nucleophilic substitution increases' left to right, and a footnote that a meta-nitro group has no effect.
TextbookNCERT questions the board keeps asking
6.2 Give the IUPAC names of the following compounds: (i) CH3CH(Cl)CH(Br)CH3 (ii) CHF2CBrClF (iii) ClCH2C≡CCH2Br (iv) (CCl3)3CCl (v) CH3C(p-ClC6H4)2CH(Br)CH3 (vi) (CH3)3CCH=C(Cl)C6H4I-p3 m
(i) 2-Bromo-3-chlorobutane (bromo comes first alphabetically, so it gets the lower locant).
(ii) 1-Bromo-1-chloro-1,2,2-trifluoroethane.
(iii) 1-Bromo-4-chlorobut-2-yne.
(iv) 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane.
(v) 2-Bromo-3,3-bis(4-chlorophenyl)butane.
(vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene.
Show the chain numbering on the structure if you draw it; alphabetical order decides ties.
6.4 Which one of the following has the highest dipole moment? (i) CH2Cl2 (ii) CHCl3 (iii) CCl42 m
CH2Cl2 (dichloromethane) has the highest dipole moment. In CCl4 the four C–Cl dipoles are arranged tetrahedrally and cancel completely, so μ = 0. In CHCl3 the resultant of two C–Cl dipoles opposes the third C–Cl dipole, so the net moment is small (about 1.08 D). In CH2Cl2 the resultant of the two C–Cl dipoles is reinforced by the resultant of the two C–H dipoles, giving the largest net moment (about 1.60 D). Order: CH2Cl2 > CHCl3 > CCl4.
6.5 A hydrocarbon C5H10 does not react with chlorine in the dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.2 m
C5H10 fits either an alkene or a cycloalkane. An alkene would add chlorine even in the dark, so the compound is saturated: it is a cycloalkane. Of the cycloalkanes with this formula, only cyclopentane has all ten hydrogens equivalent, so free-radical substitution in sunlight gives a single product, chlorocyclopentane. The hydrocarbon is cyclopentane.
6.6 Write the isomers of the compound having formula C4H9Br.2 m
Four isomers: (i) CH3CH2CH2CH2Br, 1-bromobutane (1°); (ii) CH3CH2CH(Br)CH3, 2-bromobutane (2°, chiral); (iii) (CH3)2CHCH2Br, 1-bromo-2-methylpropane (1°); (iv) (CH3)3CBr, 2-bromo-2-methylpropane (3°). The first two are chain isomers of the last two, and each pair are position isomers of each other.
6.7 Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.3 m
(i) CH3CH2CH2CH2OH + KI + H3PO4 → CH3CH2CH2CH2I + KH2PO4 + H2O (HI generated in situ; alternatively red P + I2).
(ii) Finkelstein reaction: CH3CH2CH2CH2Cl + NaI → CH3CH2CH2CH2I + NaCl↓ (dry acetone, heat).
(iii) Markovnikov addition of HI: CH3CH2CH=CH2 + HI → CH3CH2CH(I)CH3 gives 2-iodobutane, not the target. So first add HBr in the presence of peroxide to get 1-bromobutane (anti-Markovnikov), then do the Finkelstein reaction with NaI in acetone to get 1-iodobutane. (HI itself shows no peroxide effect.)
6.8 What are ambident nucleophiles? Explain with an example.2 m
Ambident nucleophiles are nucleophiles that have two nucleophilic centres and can attack through either of them. Cyanide ion is a resonance hybrid of −:C≡N and :C=N−, so it can bond through carbon or nitrogen. With ionic KCN the attack is through carbon and an alkyl cyanide (nitrile) forms, because the C–C bond is stronger than the C–N bond; with covalent AgCN the carbon is bonded to silver, so nitrogen attacks and the isocyanide forms: CH3CH2Br + KCN → CH3CH2CN, but CH3CH2Br + AgCN → CH3CH2NC. Nitrite ion behaves similarly: KNO2 gives the alkyl nitrite R–O–N=O, AgNO2 gives the nitroalkane R–NO2.
6.9 Which compound in each of the following pairs will react faster in SN2 reaction with OH−? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl2 m
(i) CH3I reacts faster. I− is a better leaving group than Br− because the C–I bond is longer and weaker (bond enthalpy 234 against 293 kJ mol−1) and iodide is a larger, more stable ion.
(ii) CH3Cl reacts faster. SN2 needs backside attack on the carbon; in (CH3)3CCl the three methyl groups crowd the carbon and hinder the nucleophile, so the tertiary halide is the least reactive by SN2 (it prefers SN1).
6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane.3 m
(i) β-Hydrogens are on the ring CH2 groups and on the CH3. Products: 1-methylcyclohexene (major; trisubstituted double bond inside the ring) and methylenecyclohexane (minor; disubstituted, exocyclic).
(ii) (CH3)2C(Cl)CH2CH3 loses H from C-1 or C-3. Products: 2-methylbut-2-ene, (CH3)2C=CHCH3 (major, trisubstituted), and 2-methylbut-1-ene, CH2=C(CH3)CH2CH3 (minor, disubstituted).
(iii) (CH3)3C–C(Br)(CH3)–CH2CH3 has β-hydrogens on the CH3 and on the CH2 (the tert-butyl carbon has none). Products: 3,4,4-trimethylpent-2-ene, CH3CH=C(CH3)C(CH3)3 (major, trisubstituted), and 2-ethyl-3,3-dimethylbut-1-ene, CH2=C(C2H5)C(CH3)3 (minor). In each case Saytzeff's rule picks the more substituted alkene as major.
6.12 Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride; (ii) alkyl halides, though polar, are immiscible with water; (iii) Grignard reagents should be prepared under anhydrous conditions.3 m
(i) In chlorobenzene the carbon bonded to Cl is sp2 hybridised and so more electronegative than the sp3 carbon of cyclohexyl chloride; it pulls electron density back from Cl and the C–Cl bond is less polar. Also, resonance delocalises the lone pairs of Cl into the ring, giving the C–Cl bond partial double-bond character and reducing the charge separation. Both lower the dipole moment.
(ii) For an alkyl halide to dissolve, the hydrogen bonds between water molecules and the dipole-dipole attractions between alkyl halide molecules must be broken. The new attractions between alkyl halide and water are weaker than the hydrogen bonds broken, so the energy is not recovered and the two liquids do not mix.
(iii) Grignard reagents are very reactive towards any compound with an acidic hydrogen; even the traces of water in the solvent decompose them: RMgX + H2O → RH + Mg(OH)X. Hence they are prepared in dry ether under anhydrous conditions.
6.15 Write the mechanism of the following reaction: n-BuBr + KCN → n-BuCN (in ethanol–water).3 m
1-Bromobutane is a primary halide, so the reaction goes by the SN2 mechanism in one step. KCN ionises to give CN−, an ambident nucleophile that attacks through carbon because the C–C bond that forms is stronger than a C–N bond.
Mechanism: the cyanide carbon attacks C-1 of CH3CH2CH2CH2Br from the side opposite to bromine. A transition state forms in which carbon is partially bonded to both CN and Br, with the two hydrogens and the propyl group in one plane: [NC···CH2(C3H7)···Br]δ−. The C–Br bond breaks as the C–C bond forms, bromide leaves, and the configuration at carbon inverts: NC− + CH3CH2CH2CH2Br → CH3CH2CH2CH2CN + Br−. Rate = k[n-BuBr][CN−], second order.
6.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.3 m
SN2 is slowed by crowding around the carbon carrying the halogen, so 1° > 2° > 3°, and among primary halides the one with branching nearer the C–Br carbon is slower.
(i) 1-Bromopentane (1°) > 2-bromopentane (2°) > 2-bromo-2-methylbutane (3°).
(ii) 1-Bromo-3-methylbutane (1°) > 3-bromo-2-methylbutane (2°) > 2-bromo-2-methylbutane (3°).
(iii) All are primary, so compare the branching: 1-bromobutane (no branch) > 1-bromo-3-methylbutane (branch at C-3) > 1-bromo-2-methylbutane (branch at C-2) > 1-bromo-2,2-dimethylpropane (two branches at C-2, most hindered).
6.18 p-Dichlorobenzene has higher melting point and lower solubility than those of o- and m-isomers. Discuss.2 m
The para isomer is symmetrical, so its molecules pack closely and fit well into the crystal lattice. The intermolecular forces in the solid are therefore stronger than in the less symmetrical ortho and meta isomers. More energy is needed to break the lattice, so the melting point is higher; and the solvent must also supply more energy to pull the molecules out of the lattice, so the solubility is lower.
6.20 The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.2 m
In aqueous solution KOH is fully ionised and gives OH− ions, which are strong nucleophiles. They attack the electron-deficient carbon of C–Cl and displace Cl− (substitution), so an alcohol forms: RCH2CH2Cl + KOH(aq) → RCH2CH2OH + KCl.
In alcoholic solution KOH reacts with ethanol to give ethoxide ions, C2H5O−, which are stronger bases than nucleophiles and also bulkier. They prefer to abstract a β-hydrogen, and the halide leaves from the α-carbon (β-elimination), giving the alkene: RCH2CH2Cl + KOH(alc.) → RCH=CH2 + KCl + H2O. Also, water being a better solvating solvent stabilises the transition state of substitution, while alcohol does not.
6.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18, which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.3 m
There are two primary isomers of C4H9Br: 1-bromobutane and 1-bromo-2-methylpropane. Since (d) differs from n-octane (the Wurtz product of n-butyl bromide), (a) must be 1-bromo-2-methylpropane (isobutyl bromide), (CH3)2CHCH2Br.
(a) → (b): (CH3)2CHCH2Br + KOH (alc.) → (CH3)2C=CH2 (2-methylpropene) + KBr + H2O.
(b) → (c): (CH3)2C=CH2 + HBr → (CH3)3CBr, 2-bromo-2-methylpropane (Markovnikov addition), a positional isomer of (a).
(a) → (d): 2(CH3)2CHCH2Br + 2Na → (CH3)2CHCH2CH2CH(CH3)2 (2,5-dimethylhexane) + 2NaBr, in dry ether (Wurtz reaction). n-Butyl bromide would give CH3(CH2)6CH3, n-octane, a different compound.
6.22 What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN?3 m
(i) Dehydrohalogenation gives but-1-ene: CH3CH2CH2CH2Cl + KOH (alc.) → CH3CH2CH=CH2 + KCl + H2O.
(ii) Phenylmagnesium bromide, a Grignard reagent, forms: C6H5Br + Mg → C6H5MgBr.
(iii) Chlorobenzene is not hydrolysed under ordinary conditions; with NaOH at 623 K and 300 atm it gives sodium phenoxide, which on acidification gives phenol: C6H5Cl + NaOH → C6H5ONa → C6H5OH.
(iv) Substitution gives ethanol: C2H5Cl + KOH (aq) → C2H5OH + KCl.
(v) Wurtz reaction gives ethane: 2CH3Br + 2Na → CH3CH3 + 2NaBr.
(vi) Ethanenitrile (methyl cyanide) forms: CH3Cl + KCN → CH3CN + KCl.
6.4 Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields (i) a single monochloride (ii) three isomeric monochlorides (iii) four isomeric monochlorides.3 m
C5H12 has three isomers: n-pentane, 2-methylbutane and 2,2-dimethylpropane.
(i) A single monochloride: 2,2-dimethylpropane (neopentane), C(CH3)4. All twelve hydrogens are equivalent, so only 1-chloro-2,2-dimethylpropane forms.
(ii) Three isomeric monochlorides: n-pentane, CH3CH2CH2CH2CH3, which has three sets of equivalent hydrogens (C-1/C-5, C-2/C-4, C-3), giving 1-chloro-, 2-chloro- and 3-chloropentane.
(iii) Four isomeric monochlorides: 2-methylbutane, (CH3)2CHCH2CH3, which has four sets of hydrogens, giving 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane and 1-chloro-3-methylbutane.
6.6 Arrange each set of compounds in order of increasing boiling points: (i) Bromomethane, bromoform, chloromethane, dibromomethane (ii) 1-Chloropropane, isopropyl chloride, 1-chlorobutane.2 m
(i) Boiling point rises with the mass and number of halogen atoms, because van der Waals forces get stronger: chloromethane < bromomethane < dibromomethane < bromoform.
(ii) Boiling point rises with the number of carbons and falls with branching: isopropyl chloride (branched, C3) < 1-chloropropane (straight, C3) < 1-chlorobutane (C4).
Previous years and frequently askedBoard questions with model answers
MCQ Which of the following is an example of a vic-dihalide?- Dichloromethane
- 1,2-Dichloroethane
- Ethylidene chloride
- Allyl chloride
Frequently asked1 m
MCQ Which of the following alkyl halides will undergo SN1 reaction most readily?- (CH3)3C–F
- (CH3)3C–Cl
- (CH3)3C–Br
- (CH3)3C–I
Frequently asked1 m
MCQ Which of the following molecules is chiral?- 2-Bromopropane
- 1-Bromobutane
- 2-Bromobutane
- 2-Bromo-2-methylpropane
Frequently asked1 m
MCQ The reaction of ethyl bromide with NaI in dry acetone to give ethyl iodide is called- Swarts reaction
- Finkelstein reaction
- Wurtz reaction
- Sandmeyer's reaction
Frequently asked1 m
MCQ Which of the following has the highest boiling point?- CH3CH2CH2Cl
- CH3CH2CH2Br
- CH3CH2CH2I
- (CH3)2CHBr
Frequently asked1 m
MCQ Ethyl bromide on treatment with AgCN gives mainly- Propanenitrile
- Ethyl isocyanide
- Ethyl cyanide
- Ethanamine
Frequently asked1 m
MCQ Chlorobenzene on heating with aqueous NaOH at 623 K and 300 atm followed by acidification gives- Benzene
- Phenol
- Benzoic acid
- Chlorophenol
Frequently asked1 m
Assertion–Reason Assertion (A): Chlorobenzene is less reactive than chloroethane towards nucleophilic substitution. Reason (R): The C–Cl bond in chlorobenzene acquires partial double-bond character due to resonance.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Tert-butyl bromide undergoes hydrolysis by the SN1 mechanism. Reason (R): The tert-butyl carbocation is stabilised by three electron-releasing methyl groups.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Addition of HBr to propene in the presence of peroxide gives 1-bromopropane. Reason (R): HCl and HI also show the peroxide effect.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Aryl halides cannot be prepared by replacing the hydroxyl group of phenol by halogen using HX or PX3. Reason (R): The C–O bond in phenol has partial double-bond character and is not easily broken.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer Why is sulphuric acid not used during the reaction of alcohols with KI?Frequently asked2 m
Concentrated H2SO4 is a strong oxidising agent. The HI produced in situ from KI is a reducing agent and is oxidised by H2SO4 to I2 (2HI + H2SO4 → I2 + SO2 + 2H2O), so it is no longer available to convert the alcohol into the alkyl iodide. Non-oxidising 95% orthophosphoric acid is used instead. (1 mark for the oxidising nature, 1 for HI → I2 and the alternative acid.)
Very short answer Give reasons: (i) Alkyl halides, though polar, are immiscible with water. (ii) Grignard reagents should be prepared under anhydrous conditions.Frequently asked2 m
(i) To dissolve, the hydrogen bonds between water molecules and the dipole-dipole forces between alkyl halide molecules must be broken. The new attractions between alkyl halide and water are weaker than the hydrogen bonds that would be lost, so the energy is not recovered and the liquids stay separate. (1 mark)
(ii) Grignard reagents react with any compound that has an acidic hydrogen, and water is enough: RMgX + H2O → R–H + Mg(OH)X. Any moisture would destroy the reagent, so it is made in dry ether under anhydrous conditions. (1 mark)
Very short answer Out of SN1 and SN2, which reaction occurs with (i) inversion of configuration (ii) racemisation? Give reason in each case.Frequently asked2 m
(i) SN2. The nucleophile attacks from the side opposite to the leaving group in a single step, so the three other groups flip over like an umbrella and the product has the inverted configuration. (1 mark)
(ii) SN1. The intermediate carbocation is planar (sp2), so the nucleophile can attack from either face with equal probability, giving equal amounts of the two enantiomers, a racemic mixture. (1 mark)
Very short answer Write the structure of the major product formed when (i) 2-bromopentane is heated with alcoholic KOH; (ii) chlorobenzene is treated with CH3Cl in the presence of anhydrous AlCl3.Frequently asked2 m
(i) Dehydrohalogenation; by Saytzeff's rule the more substituted alkene is major: pent-2-ene, CH3CH=CHCH2CH3 (pent-1-ene is minor). (1 mark)
(ii) Friedel-Crafts alkylation; chlorine is ortho/para-directing and the para product predominates: 4-chlorotoluene (1-chloro-4-methylbenzene), with a little 2-chlorotoluene. (1 mark)
Very short answer Which one in each pair will react faster and why? (i) SN1 reaction: C6H5CH2Cl or C6H5CH(Cl)C6H5 (ii) SN2 reaction: 1-bromobutane or 2-bromobutaneFrequently asked2 m
(i) C6H5CH(Cl)C6H5 reacts faster by SN1. Its carbocation, (C6H5)2CH+, is stabilised by resonance with two phenyl rings, whereas the benzyl cation has only one. (1 mark)
(ii) 1-Bromobutane reacts faster by SN2. It is a primary halide, so the carbon is less hindered and the nucleophile can attack from the back side more easily than at the secondary carbon of 2-bromobutane. (1 mark)
Short answer Explain why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions.Frequently asked3 m
Any three of the following, with the resonance structures of chlorobenzene drawn for full credit:
- Resonance effect: the lone pairs on chlorine are delocalised into the benzene ring, so the C–Cl bond acquires partial double-bond character. It is shorter (169 pm against 177 pm) and stronger than in haloalkanes and is difficult to break.
- Hybridisation: in haloarenes the halogen is attached to an sp2 carbon, which has more s-character and is more electronegative than the sp3 carbon of haloalkanes; it holds the bonding pair more tightly, shortening and strengthening the bond.
- Instability of phenyl cation: an SN1 route would require a phenyl cation, which is not stabilised by resonance (the empty orbital is perpendicular to the π system), so it does not form.
- Repulsion: the electron-rich aromatic ring repels the approaching electron-rich nucleophile.
Hence chlorobenzene needs 623 K and 300 atm with NaOH to give phenol, while chloroethane is hydrolysed by warm aqueous KOH.
Short answer Write the mechanism of the hydrolysis of tert-butyl bromide by aqueous KOH. What happens to an optically active alkyl halide in such a reaction?Frequently asked3 m
Tertiary halides react by SN1 in two steps.
Step 1 (slow, rate determining): the C–Br bond breaks heterolytically, helped by the polar protic solvent, to give the planar tert-butyl carbocation and Br−: (CH3)3C–Br → (CH3)3C+ + Br−. The carbocation is stabilised by the +I effect and hyperconjugation of three methyl groups.
Step 2 (fast): the hydroxide ion attacks the carbocation: (CH3)3C+ + OH− → (CH3)3C–OH. Rate = k[(CH3)3CBr], first order, independent of [OH−]. (2 marks for the two steps with the rate law)
Stereochemistry: the carbocation is sp2 and planar, so OH− can bond from either face with equal probability. An optically active halide therefore gives equal amounts of the two enantiomeric alcohols, a racemic mixture, and the product is optically inactive (racemisation). (1 mark)
Short answer What happens when (i) ethyl chloride is treated with AgNO2, (ii) chlorobenzene is treated with Cl2 in the presence of FeCl3, (iii) methyl bromide is treated with KCN followed by treatment with H2/Ni? Write the chemical equations.Frequently asked3 m
(i) AgNO2 is covalent, so nitrogen attacks and nitroethane forms: C2H5Cl + AgNO2 → C2H5NO2 + AgCl. (With KNO2 the product would be ethyl nitrite, C2H5ONO.)
(ii) Electrophilic substitution; Cl is ortho/para-directing, so 1,4-dichlorobenzene (major) and 1,2-dichlorobenzene (minor) form: C6H5Cl + Cl2 → C6H4Cl2 + HCl (anhydrous FeCl3).
(iii) KCN gives ethanenitrile, which is then reduced to ethanamine: CH3Br + KCN → CH3CN + KBr; CH3CN + 2H2 → CH3CH2NH2 (Ni). Note that the chain grows by one carbon, a step used in conversions.
Short answer Arrange the following in the order indicated and give one reason for each: (i) Increasing boiling point: 1-chlorobutane, 1-bromobutane, 2-bromo-2-methylpropane, 1-chloropropane. (ii) Increasing reactivity towards SN2: CH3Br, (CH3)2CHBr, CH3CH2Br. (iii) Increasing ease of nucleophilic substitution: chlorobenzene, 4-nitrochlorobenzene, 2,4-dinitrochlorobenzene.Frequently asked3 m
(i) 1-Chloropropane < 2-bromo-2-methylpropane < 1-chlorobutane < 1-bromobutane. Boiling point rises with the number of carbons and with the mass of the halogen, and falls with branching; the tertiary bromide is compact, so despite bromine it boils below 1-chlorobutane (about 346 K against 351 K), while 1-bromobutane (about 375 K) is highest. (1 mark)
(ii) (CH3)2CHBr < CH3CH2Br < CH3Br. SN2 needs backside attack, and steric hindrance around the carbon increases from methyl to primary to secondary. (1 mark)
(iii) Chlorobenzene < 4-nitrochlorobenzene < 2,4-dinitrochlorobenzene. Electron-withdrawing nitro groups at ortho/para positions delocalise the negative charge of the intermediate carbanion and make substitution easier; the conditions fall from 623 K/300 atm to 443 K to 368 K. (1 mark)
Short answer An alkyl halide (A), C4H9Br, is optically active. On treatment with alcoholic KOH it gives (B), which on ozonolysis gives two aldehydes. On treatment with aqueous KOH, (A) gives (C), an optically active alcohol. Identify (A), (B) and (C) and write the reactions.Frequently asked3 m
(A) is optically active, so it must contain a chiral carbon: of the four isomers of C4H9Br only 2-bromobutane, CH3CH(Br)CH2CH3, has one. (1 mark)
(A) + KOH (alc.) → (B): by Saytzeff's rule the major alkene is but-2-ene, CH3CH=CHCH3. Ozonolysis of but-2-ene gives two molecules of ethanal, CH3CHO; a symmetrical internal alkene gives 'two aldehydes' in the sense of two aldehyde molecules, which fits. (1 mark)
(A) + KOH (aq) → (C): substitution gives butan-2-ol, CH3CH(OH)CH2CH3, which is optically active because C-2 still carries four different groups. Since a secondary halide reacts largely by SN2, the alcohol has the inverted configuration. (1 mark)
Long answer (a) Describe the SN2 mechanism with the example of chloromethane and hydroxide ion, including the rate law and the stereochemical outcome. (b) Compare SN1 and SN2 mechanisms on any three points. (c) Which of CH3Cl, (CH3)2CHCl and (CH3)3CCl would react fastest by SN1, and why?Frequently asked5 m
(a) CH3Cl + OH− → CH3OH + Cl−. The reaction is second order: rate = k[CH3Cl][OH−], so both species take part in the single rate-determining step. The hydroxide ion attacks the carbon from the side opposite to chlorine. In the transition state carbon is partially bonded to both OH and Cl, with the three hydrogens in a plane perpendicular to the O···C···Cl line; the negative charge is shared between O and Cl. As the C–O bond forms the C–Cl bond breaks and chloride leaves. The three hydrogens swing over like an umbrella turned inside out, so if the carbon is a stereocentre the product has the inverted configuration (Walden inversion). Reactivity: CH3X > 1° > 2° > 3°, because bulky groups hinder the backside attack. (2 marks: mechanism with transition state, rate law and inversion)
(b) SN1: two steps via a carbocation; rate = k[RX], first order; reactivity 3° > 2° > 1° > CH3X; racemisation; favoured by polar protic solvents and weak nucleophiles. SN2: one step via a transition state; rate = k[RX][Nu−], second order; reactivity CH3X > 1° > 2° > 3°; inversion; favoured by strong nucleophiles and polar aprotic solvents. (2 marks for any three contrasts)
(c) (CH3)3CCl reacts fastest by SN1. The rate depends on the stability of the carbocation formed in the slow step, and the tertiary carbocation is the most stable because of the +I effect and hyperconjugation of three methyl groups. Order: (CH3)3CCl > (CH3)2CHCl > CH3Cl. (1 mark)
Long answer How will you bring about the following conversions? Write the reactions. (i) Benzene to biphenyl (ii) Propene to 1-nitropropane (iii) Ethanol to but-1-yne (iv) Aniline to chlorobenzene (v) Toluene to benzyl alcoholFrequently asked5 m
(i) Chlorinate, then Fittig: C6H6 + Cl2 → C6H5Cl (anhydrous FeCl3); 2C6H5Cl + 2Na → C6H5–C6H5 + 2NaCl (dry ether). (1 mark)
(ii) Anti-Markovnikov HBr, then AgNO2: CH3CH=CH2 + HBr → CH3CH2CH2Br (peroxide); CH3CH2CH2Br + AgNO2 → CH3CH2CH2NO2 + AgBr. Without peroxide you would get 2-bromopropane and hence 2-nitropropane. (1 mark)
(iii) Make ethyl bromide and couple it with sodium acetylide: C2H5OH + HBr → C2H5Br + H2O; separately, ethanol → ethyl bromide → (alc. KOH) ethene → (Br2) 1,2-dibromoethane → (NaNH2) ethyne → (NaNH2) sodium acetylide, HC≡C−Na+; then HC≡CNa + C2H5Br → HC≡C–CH2CH3 (but-1-yne) + NaBr. (1 mark)
(iv) Diazotise, then Sandmeyer: C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl− + NaCl + 2H2O (273–278 K); C6H5N2+Cl− + Cu2Cl2/HCl → C6H5Cl + N2. (1 mark)
(v) Side-chain chlorination in sunlight, then aqueous hydrolysis: C6H5CH3 + Cl2 → C6H5CH2Cl + HCl (hν); C6H5CH2Cl + KOH (aq) → C6H5CH2OH + KCl. (1 mark)
Long answer (a) Explain why the halogen atom in a haloarene is deactivating yet ortho- and para-directing in electrophilic substitution. (b) Write the equations for the nitration and the Friedel-Crafts acylation of chlorobenzene. (c) Give one use each of CH2Cl2, CHCl3, CHI3, CCl4 and freon 12, and one harmful effect of freons.Frequently asked5 m
(a) Two opposing effects operate. The −I (inductive) effect of the electronegative halogen withdraws electron density from the ring, making it less reactive than benzene towards electrophiles: the halogen is deactivating. The +R (resonance) effect of the halogen lone pairs pushes electron density into the ring, and the resonance structures place the negative charge on the ortho and para carbons only. Attack at those positions is therefore favoured relative to the meta position, so the halogen is o/p-directing; the para product predominates because of steric hindrance at the ortho positions. (2 marks; draw the resonance structures)
(b) Nitration: C6H5Cl + HNO3 (conc. H2SO4) → 1-chloro-4-nitrobenzene (major) + 1-chloro-2-nitrobenzene (minor) + H2O. Acylation: C6H5Cl + CH3COCl (anhydrous AlCl3) → 4-chloroacetophenone (major) + 2-chloroacetophenone (minor) + HCl. (1 mark)
(c) CH2Cl2: paint remover or aerosol propellant. CHCl3: solvent for fats, alkaloids and iodine; used to make the refrigerant R-22. CHI3: antiseptic (due to the iodine it liberates). CCl4: manufacture of refrigerants and propellants; cleaning fluid and fire extinguisher. Freon 12 (CCl2F2): refrigerant in refrigeration and air conditioning, aerosol propellant. Harmful effect: freons diffuse unchanged into the stratosphere and initiate radical chain reactions that deplete the ozone layer. (2 marks)
Case-based Read the passage and answer the questions that follow. Nucleophilic substitution in haloalkanes proceeds by two mechanisms. In the SN2 mechanism, studied by Hughes and Ingold, the nucleophile attacks the carbon from the side opposite to the leaving group in a single step; the rate depends on the concentration of both the halide and the nucleophile, and the configuration at the carbon is inverted. In the SN1 mechanism the halide first ionises slowly to a planar carbocation, which is then attacked rapidly by the nucleophile; the rate depends only on the halide, and an optically active substrate gives a racemic product. Polar protic solvents favour SN1 while polar aprotic solvents favour SN2.
(i) Write the rate law for the reaction of bromomethane with OH−.
(ii) Arrange CH3CH2Br, (CH3)2CHBr and (CH3)3CBr in increasing order of reactivity towards SN1, with reason.
(iii) (−)-2-Bromooctane reacts with aqueous NaOH to give (+)-octan-2-ol. What does this tell you about the mechanism?
(iv) Why do allyl chloride and benzyl chloride react readily by SN1 although they are primary halides?Frequently asked4 m
(i) Bromomethane is a methyl halide and reacts by SN2: rate = k[CH3Br][OH−], second order. (1 mark)
(ii) CH3CH2Br < (CH3)2CHBr < (CH3)3CBr. SN1 goes through a carbocation, and carbocation stability rises from primary to secondary to tertiary because of the +I effect and hyperconjugation of alkyl groups. (1 mark)
(iii) The product has the opposite sign of rotation and the inverted configuration, so the reaction went with inversion, characteristic of SN2 backside attack; an SN1 path would have given a racemic, optically inactive mixture. (1 mark)
(iv) The carbocations they form (allyl and benzyl cations) are stabilised by resonance: the positive charge is delocalised over the C=C double bond or the benzene ring, so the slow ionisation step is easy despite the halides being primary. (1 mark)
Case-based Read the passage and answer the questions that follow. Chlorobenzene is much less reactive than chloroethane towards nucleophiles. The lone pairs on chlorine are delocalised into the ring, so the C–Cl bond has partial double-bond character and is shorter (169 pm) than in chloroalkanes (177 pm). Chlorobenzene is converted into phenol only by heating with aqueous NaOH at 623 K and 300 atm (Dow's process). Electron-withdrawing groups at the ortho and para positions make the substitution easier: 4-nitrochlorobenzene reacts with 15% NaOH at 443 K, 2,4-dinitrochlorobenzene at 368 K, and 2,4,6-trinitrochlorobenzene is hydrolysed by warm water. In contrast, chlorobenzene readily undergoes electrophilic substitution, the halogen directing the new group to the ortho and para positions.
(i) Draw or describe the resonance structures that explain the partial double-bond character of the C–Cl bond.
(ii) Why does a nitro group at the meta position not increase the ease of nucleophilic substitution?
(iii) Why is the phenyl cation not formed, ruling out an SN1 path?
(iv) Name the major product when chlorobenzene is treated with concentrated H2SO4.Frequently asked4 m
(i) Five structures: neutral chlorobenzene, and four in which a chlorine lone pair forms a C=Cl+ double bond while a negative charge sits on an ortho carbon, the para carbon or the other ortho carbon, returning to the neutral form. The C–Cl bond is thus a hybrid of single and double bonds. (1 mark)
(ii) In the intermediate carbanion formed by nucleophilic attack, the negative charge is delocalised only over the ortho and para carbons relative to the point of attack. A nitro group at those positions can take the charge onto its oxygens and stabilise the intermediate; a meta nitro group never carries the charge in any resonance structure, so it gives no stabilisation. (1 mark)
(iii) In the phenyl cation the empty orbital lies in the plane of the ring, perpendicular to the π system, so the positive charge cannot be delocalised by resonance. The cation is highly unstable and the C–Cl bond does not ionise. (1 mark)
(iv) Sulphonation gives 4-chlorobenzenesulphonic acid as the major product (with a little of the 2-isomer), because chlorine is ortho/para-directing and the para position is less hindered. (1 mark)
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Calling benzyl chloride an aryl halide. The chlorine is on the sp3 CH2, so it is a benzylic (alkyl-type) halide and is highly reactive; only chlorobenzene, with Cl on the ring carbon, is an aryl halide.
- Writing that CH3F has the highest dipole moment among methyl halides because fluorine is the most electronegative. NCERT's values give CH3Cl the highest (1.860 D) because the very short C–F bond reduces the charge separation.
- Giving the SN1 and SN2 reactivity orders the wrong way round, or forgetting that both increase from fluoride to iodide. SN2: CH3X > 1° > 2° > 3°; SN1: 3° > 2° > 1° > CH3X; leaving group RI > RBr > RCl > RF for both.
- Reporting only the major alkene in a dehydrohalogenation question that asks for all products, or picking the less substituted alkene as major. Saytzeff's rule gives the more substituted alkene as the major product.
- Applying the peroxide effect to HCl or HI. Only HBr adds anti-Markovnikov in the presence of peroxide; conversions such as propene to 1-bromopropane depend on this.
- Mixing up the ambident nucleophile pairs: KCN gives nitriles (R–CN) and AgCN gives isocyanides (R–NC); KNO2 gives alkyl nitrites (R–O–N=O) and AgNO2 gives nitroalkanes (R–NO2).
- Explaining the low reactivity of chlorobenzene with only one reason and no resonance structures. The scheme rewards resonance, sp2 hybridisation, instability of the phenyl cation and repulsion, each as a separate point.
- Nitrating benzene before brominating when asked for 4-bromonitrobenzene. Nitro is meta-directing, so bromination must come first and nitration second.
- Forgetting the conditions: dry ether for Grignard, Wurtz and Fittig reactions; dry acetone for Finkelstein; 273–278 K for diazotisation; 623 K and 300 atm for the Dow's process. Missing conditions cost half a mark each.
How to prepareStrategy for this chapter
Treat this chapter as the foundation of the organic unit rather than as six marks on its own: half of what you learn here (alcoholic versus aqueous KOH, KCN versus AgCN, Sandmeyer, Grignard, the peroxide effect) reappears in Alcohols, Aldehydes and Amines conversions. Start with classification and nomenclature in one sitting, doing NCERT 6.1, 6.2 and 6.3 completely because the board sets those exact structures.
Then spend your longest sitting on section 6.7.1. Write the SN2 and SN1 mechanisms from memory, with the transition state, the carbocation and the stereochemical result, until each takes under four minutes. Attach to them the two reactivity orders, the leaving-group order, and the three chirality examples (2-bromobutane, butan-2-ol, bromochloroiodomethane). Next learn the chlorobenzene story as one drawing: the five resonance structures answer the low reactivity question, the o/p-direction question and the shorter C–Cl bond question.
Memorise: the name reactions with reagents and conditions (Sandmeyer, Finkelstein, Swarts, Wurtz, Wurtz-Fittig, Fittig, Dow's process), the give-reason answers in sections 6.3, 6.4 and 6.6, the four conditions for hydrolysis of nitrochlorobenzenes, and one use plus one harmful effect for each polyhalogen compound. Derive rather than memorise: dehydrohalogenation products by Saytzeff's rule, and the products of any nucleophile from the table in 6.7.1.
In the last two days do NCERT 6.10, 6.11, 6.14, 6.16, 6.19, 6.21 and 6.22 on paper. In the exam, give conditions over every arrow, name the mechanism when you write one, and for give-reason questions write the two marking phrases the examiner is looking for rather than a paragraph.