NCERT Solutions for Class 12 Chemistry Chapter 6: Haloalkanes and Haloarenes

Concept and approach. An IUPAC name is a complete recipe for the structure: the parent chain or ring (pentane, cyclohexane, heptane, butene, benzene) is drawn first and numbered; each locant then tells the carbon on which a substituent (halo, alkyl) is placed; the suffix –ene shows a C=C and its number shows where the double bond begins. Halogens are treated as substituent prefixes (chloro-, bromo-, iodo-), exactly like alkyl groups, and are arranged alphabetically in the name. In the answer below, every structure is first written as a condensed formula and then checked by counting carbons against the parent name.

Why the details matter. The name 2-chloro-3-methylpentane gives six carbons in all (five from the chain plus one from the methyl), and a molecular formula C₆H₁₃Cl — if the structure you draw does not give this count, the structure is wrong. In (iii) “tert-butyl” is a branched four-carbon group, (CH₃)₃C–, written with prefix alphabetisation under “b” (not “t”), whereas in (v) “sec-butyl” is –CH(CH₃)CH₂CH₃ and also sorts under “b”; in the benzene name the substituents are numbered so that the locants are as low as possible (1, 2, 4) and the first-cited prefix in alphabetical order (bromo) receives locant 1.

Classification of the halides drawn. (i) is a secondary alkyl halide (Cl on a carbon bonded to two other carbons); (iii) is a secondary alkyl halide with a very bulky neighbouring tert-butyl group, which hinders S_N2 attack; (iv) is an allylic halide at both ends, hence very reactive towards substitution; (v) is an aryl halide (halogen on an sp² carbon of the ring) and therefore very unreactive towards nucleophiles.

Conclusion: The five compounds are CH₃CH(Cl)CH(CH₃)CH₂CH₃, 1-chloro-4-ethylcyclohexane, CH₃CH₂CH(I)CH[C(CH₃)₃]CH₂CH₂CH₃, BrCH₂CH=CHCH₂Br and the 1,2,4-trisubstituted benzene Br/CH₃/sec-butyl, each drawn directly from the locants in its IUPAC name.

Background. Alkyl iodides are prepared from alcohols by heating them with an alkali-metal iodide in the presence of a non-oxidising acid (95 % orthophosphoric acid). The acid protonates the alcohol’s –OH to make a better leaving group (R–OH₂⁺), and the generated hydrogen iodide (from KI + acid) supplies the iodide nucleophile: ROH + HI → RI + H₂O. The success of the method depends on HI remaining intact as a reagent.

Why H₂SO₄ fails. Concentrated sulphuric acid is a strong oxidising agent. It first converts KI to HI (KI + H₂SO₄ → KHSO₄ + HI), but HI is an excellent reducing agent, so it is immediately oxidised by the H₂SO₄ to iodine, while the acid is itself reduced to SO₂ (and, with more HI, even to S or H₂S):

2KI + 2H₂SO₄ → 2KHSO₄ + 2HI; 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Because the hydrogen iodide is consumed as it forms, the concentration of I⁻ available to displace the protonated –OH is destroyed, the violet vapour of iodine is seen, and the yield of alkyl iodide is negligible. By contrast, orthophosphoric acid (H₃PO₄) is a non-oxidising acid of adequate strength, so HI survives and reacts with the alcohol.

Conclusion: Sulphuric acid cannot be used because it oxidises the HI formed from KI to iodine, so the iodide needed for substitution is lost; non-oxidising H₃PO₄ is used instead.

Concept. Dihalogen derivatives of propane (C₃H₈) are obtained by replacing two of its eight hydrogen atoms by halogen atoms. Propane has two kinds of carbon — the two terminal CH₃ carbons (C1, C3) and the central CH₂ carbon (C2) — so the two halogens can be placed on the same carbon (geminal or gem-dihalides) or on different carbons (vicinal if adjacent, otherwise 1,3-). For the same halogen X, a systematic count of the possible locant pairs gives exactly four constitutional isomers.

Why there are only four. Placing the two halogens on C1 and C3 and on C3 and C1 describes the same molecule, because propane is symmetrical; likewise 2,3- is the same as 1,2- after renumbering from the other end. With two different halogens (for example Cl and Br) the number of isomers rises (1-bromo-1-chloro, 1-bromo-2-chloro, 2-bromo-1-chloro, 1-bromo-3-chloro, 2-bromo-2-chloro … ), but for the commonly asked case the four above are the answer. Of these four, 1,2-dichloropropane (CH₃CHClCH₂Cl) has a chiral carbon (C2 bears H, CH₃, Cl and CH₂Cl) and therefore exists as a pair of enantiomers.

Conclusion: Four dihalogen derivatives of propane exist: 1,1-, 1,2-, 1,3- and 2,2-dihalopropane (Cl₂CHCH₂CH₃, ClCH₂CHClCH₃, ClCH₂CH₂CH₂Cl and CH₃CCl₂CH₃).

Background. Photochemical (free-radical) chlorination replaces a hydrogen atom of an alkane by chlorine, and the number of different monochlorides equals the number of chemically distinct sets of hydrogen atoms in the alkane (equivalent hydrogens give the same product). Pentane (C₅H₁₂) has three structural isomers — n-pentane, isopentane (2-methylbutane) and neopentane (2,2-dimethylpropane) — and the symmetry of each decides how many sets of H there are.

Reasoning for each. In neopentane the four methyl groups are identical and so are all twelve hydrogens; replacing any one gives the same compound, 1-chloro-2,2-dimethylpropane. In n-pentane the two CH₃ ends are equivalent (set a), the C2 and C4 methylene groups are equivalent (set b), and the central CH₂ is unique (set c): three monochlorides. In isopentane the two CH₃ groups on C2 are equivalent (a), the tertiary H on C2 is unique (b), the CH₂ at C3 is unique (c), and the terminal CH₃ at C4 is unique (d): four monochlorides (one of them, 2-chloro-3-methylbutane, is chiral).

Extra depth. Free-radical chlorination is not selective (Cl· attacks 1°, 2° and 3° hydrogens with relative rates of roughly 1 : 3.8 : 5), so a mixture of all possible isomers is actually obtained — which is why this method is poor for preparing a single haloalkane, except in a case such as neopentane (or cyclopentane) in which only one product is possible.

Conclusion: (i) 2,2-dimethylpropane (neopentane) gives one monochloride; (ii) n-pentane gives three; (iii) 2-methylbutane (isopentane) gives four.

Concept. The question tests six different routes to monohalo compounds: conversion of an alcohol (–OH → –Cl), side-chain free-radical halogenation, selective substitution at a benzylic –OH, Markovnikov addition of HX to an alkene, Finkelstein halogen exchange, and allylic bromination. For each reaction the first thing to decide is which bond of which carbon reacts: that decides the product.

Extra depth. In (iv) the intermediate carbocation formed by protonation of the C=C is tertiary, so iodine ends on the ring carbon that carries the methyl group; with HI, a tertiary alkyl iodide forms. In (vi) the substitution at the allylic position is selective because the allylic radical is resonance-stabilised; at low temperature in the dark Br₂ would instead add across the double bond to give trans-1,2-dibromocyclohexane (the bromine-water/CCl₄ test for unsaturation).

Conclusion: (i) chlorocyclohexane; (ii) 1-(1-bromoethyl)-4-nitrobenzene; (iii) 4-(chloromethyl)phenol; (iv) 1-iodo-1-methylcyclohexane; (v) iodoethane; (vi) 3-bromocyclohexene.

Concept. The boiling point of a liquid depends on the strength of the intermolecular forces that must be overcome to vaporise it. In haloalkanes these are dipole–dipole attractions (the C–X bond is polar) and van der Waals (London dispersion) forces, which grow with the size, the molecular mass and the number of electrons of the molecule and with the surface area available for contact. Two practical rules therefore apply: (1) for a given alkyl group, b.p. rises with the size/mass of the halogen (RF < RCl < RBr < RI) and with the number of halogen atoms; (2) for isomers, branching lowers the boiling point because the molecule becomes more spherical, the surface area falls and the van der Waals contact weakens.

(i) Bromomethane, bromoform, chloromethane, dibromomethane. Molar masses: CH₃Cl 50.5, CH₃Br 95, CH₂Br₂ 174, CHBr₃ 253 g mol⁻¹. Each step adds more electrons and more halogen atoms, so the dispersion forces grow strongly; the actual boiling points are about 249 K, 277 K, 370 K and 422 K respectively. Chloromethane and bromomethane are gases at room temperature, whereas dibromomethane and bromoform are liquids.

CH₃Cl < CH₃Br < CH₂Br₂ < CHBr₃

(ii) 1-Chloropropane, isopropyl chloride, 1-chlorobutane. 1-Chlorobutane (C₄H₉Cl) is the heaviest, with one more CH₂ group. The two C₃H₇Cl isomers have identical mass, but isopropyl chloride (2-chloropropane) is branched and more compact, whereas 1-chloropropane is a straight chain with larger surface area; thus the b.p. of the isomers is isopropyl chloride (≈ 308 K) < 1-chloropropane (≈ 319 K); 1-chlorobutane boils at about 351 K.

Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane

Conclusion: Increasing boiling points: (i) chloromethane < bromomethane < dibromomethane < bromoform; (ii) isopropyl chloride < 1-chloropropane < 1-chlorobutane.

Concept. An S_N2 (substitution nucleophilic bimolecular) reaction is a one-step process in which the nucleophile attacks the carbon bearing the halogen from the side opposite to the leaving group, while the C–X bond breaks at the same time. The rate = k[RX][Nu⁻]. Because the nucleophile must reach the back side of the carbon, steric hindrance by bulky groups on or near the reacting carbon slows the reaction dramatically. Hence the order for simple alkyl halides is CH₃X > primary > secondary > tertiary (Fig. 6.3 of NCERT gives relative rates).

Extra depth. Notice how branching can slow S_N2 even in a primary halide: neopentyl bromide, (CH₃)₃CCH₂Br, is a primary halide but almost inert in S_N2 because of the three β-methyl groups. Since the stability of the carbocation does not matter in S_N2, the order is exactly the reverse of the S_N1 order.

Conclusion: (i) 1-bromobutane; (ii) 2-bromobutane; (iii) 1-bromo-3-methylbutane — in each case the less sterically hindered halide reacts faster by S_N2.

Concept. An S_N1 (substitution nucleophilic unimolecular) reaction proceeds in two steps: a slow ionisation of R–X to a carbocation and a halide ion (the rate-determining step), followed by rapid attack of the nucleophile on the carbocation. Because the carbocation is formed in the slow step, the more stable the carbocation, the faster the S_N1 reaction. The stability order of carbocations is 3° > 2° > 1° > CH₃⁺ because alkyl groups release electrons by the +I (inductive) effect and by hyperconjugation (σ C–H electrons overlapping the empty p orbital), dispersing the positive charge. Polar protic solvents (water, alcohols, acetic acid) favour S_N1 by solvating both ions.

Additional points. In the NCERT figure for (ii) the secondary chloride has the chlorine on an interior carbon, and the primary chloride has it at the end of the chain. The same reasoning is applied for allylic and benzylic halides, which react very fast by S_N1 because their carbocations are stabilised by resonance (CH₂=CH–CH₂⁺ ↔ ⁺CH₂–CH=CH₂). A final check: the leaving-group ability follows I > Br > Cl > F in both mechanisms, so tertiary iodides are the most reactive of all.

Conclusion: (i) tert-Butyl chloride and (ii) the secondary halide undergo the faster S_N1 reaction because they give the more stable (tertiary and secondary) carbocations.

Concept. Haloalkanes react with magnesium metal in dry ether to give Grignard reagents, R–Mg–X, in which the C–Mg bond is covalent but strongly polar (Cᵟ⁻–Mgᵟ⁺), so the carbon behaves like a carbanion (a strong base and nucleophile). Any proton source (H₂O, D₂O, alcohols, amines) protonates the carbon to give the hydrocarbon: R–MgX + H–OH → R–H + Mg(OH)X. Using D₂O places a deuterium on the carbon that had Mg, which allows the structure of R to be deduced. The Wurtz reaction (R–X + Na, dry ether) joins two R groups: 2RX + 2Na → R–R + 2NaX.

(a) Cyclohexyl bromide + Mg (dry ether) → A = cyclohexylmagnesium bromide (C₆H₁₁MgBr); A + H₂O → B = cyclohexane (C₆H₁₂) + Mg(OH)Br.

(b) R–Br + Mg → C = R–MgBr; C + D₂O → R–D. Since the product is CH₃CHDCH₃ (propane with D on the middle carbon), R must be the isopropyl group, (CH₃)₂CH–; therefore R–Br is 2-bromopropane and C is isopropylmagnesium bromide, (CH₃)₂CHMgBr.

(c) The Wurtz product (CH₃)₃C–C(CH₃)₃ (2,2,3,3-tetramethylbutane) is made by joining two tert-butyl groups, so R¹ = (CH₃)₃C– (tert-butyl) and R¹–X is tert-butyl halide, (CH₃)₃C–X. The same halide with Mg gives D = (CH₃)₃C–MgX, which on hydrolysis gives E = (CH₃)₃CH, 2-methylpropane (isobutane).

Conclusion: A = cyclohexyl-MgBr, B = cyclohexane, C = (CH₃)₂CHMgBr with R = isopropyl, R¹ = tert-butyl, D = (CH₃)₃CMgX, E = isobutane (2-methylpropane).

Concept and method. To name a halide by the IUPAC system: (1) select the longest continuous carbon chain (containing the C=C if there is one) as the parent; (2) number it from the end that gives the lowest locants to the double bond first, then to the substituents (halo and alkyl groups together); (3) list the substituents alphabetically as prefixes. To classify it, look at the carbon that carries the halogen: if it is an sp³ carbon bonded to one, two or three other carbons the halide is primary, secondary or tertiary alkyl; if that sp³ carbon is next to a C=C it is allylic; if it is attached to a benzene ring it is benzylic; if the halogen is on a C=C carbon it is vinylic; and if on a ring carbon it is an aryl halide.

Explanation of the less obvious cases. In (ii) the chain is numbered from the Cl end so that Cl gets locant 3 and the methyl 4 (the other direction would give 4-chloro-3-methyl, which has a higher locant for the first-cited prefix at the point of difference). In (viii) the double bond is given the lowest possible locant (2), and the chlorine sits on a doubly bonded carbon so the compound is a vinylic halide — the C–Cl bond is shortened by resonance and sp² character and is very unreactive to nucleophiles. In (ix) the C–Br carbon is sp³ and adjacent to the C=C, i.e. allylic; because it also bears two methyl groups it is a tertiary allylic halide (extremely reactive in S_N1 because its carbocation is resonance-stabilised). In (iv) and (xi) the halogen-bearing carbon is directly attached to the benzene ring, so these are benzylic halides; (xi) is primary and (iv) is secondary.

Conclusion: The twelve halides are named and classified in the table: four secondary alkyl [(i), (ii), (v)] and tertiary/primary alkyl [(iii), (vi), (vii)], two benzylic [(iv), (xi)], one vinylic (viii), one allylic (ix) and two aryl halides [(x), (xii)].

Concept. Polyhalogen and multiply substituted halides are named like any substituted hydrocarbon: find the longest chain containing the multiple bond (if any), give the lowest locants to the multiple bond, then to all the substituents together, and write the prefixes in alphabetical order (bromo, chloro, fluoro, iodo …). When two numberings give the same set of locants, the substituent that comes first alphabetically receives the lower number.

Why these names. In (iv), because the central carbon bears one Cl and two CCl₃ groups, the longest chain is propane (C1 and C3 are the carbons of the two CCl₃ groups), and the third CCl₃ group is a trichloromethyl substituent on C2; the locant set 1,1,1,2,3,3,3 accounts for seven chlorine atoms on the chain. In (iii) the alkyne takes priority for the lowest locant (2), then the substituents are numbered {1,4} either way, and alphabetical order (bromo before chloro) decides which end is C1. In (v) the prefix “bis” is used for the two identical, complex substituent groups (4-chlorophenyl), each enclosed in parentheses.

Conclusion: (i) 2-bromo-3-chlorobutane; (ii) 2-bromo-2-chloro-1,1,2-trifluoroethane; (iii) 1-bromo-4-chlorobut-2-yne; (iv) 2-(trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane; (v) 3-bromo-2,2-bis(4-chlorophenyl)butane; (vi) 1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene.

Concept. Converting a name to a structure is the reverse of naming: draw the parent chain or ring, number it, then attach every prefix at the stated locant. For rings, the carbon atoms are corners of the polygon and a circle inside a hexagon represents the benzene π-system; “p-, o-, m-” correspond to positions 1,4-; 1,2-; 1,3- on the ring.

Notes on drawing. In (iv) the chain numbering starts at the carbon bearing iodine (C1) and the aryl group is on C2, giving ICH₂–CH(Ar)–CH₂CH₂CH₂CH₂CH₂CH₃ with Ar = 2-chlorophenyl (Cl on the ring carbon next to the point of attachment). In (vi), tert-butyl is –C(CH₃)₃ attached at C4, so C4 is bonded to C3, C5 and the quaternary carbon of the tert-butyl group. In (viii) the C=C is between C2 and C3, with a Br on each end carbon; the compound can exist as cis and trans isomers because each alkene carbon has two different groups.

Conclusion: The eight structures are given in the table (and the ring compounds are drawn above), each following directly from the parent skeleton and locants in its IUPAC name.

Concept. The dipole moment of a molecule is the vector sum of the individual bond moments; it depends on both the polarity of the bonds and on the molecular geometry. In the chloromethanes every C–Cl bond is polar (Clᵟ⁻), and the C–H bond is only very weakly polar, but the geometry about carbon is tetrahedral (bond angle about 109.5°), so the C–Cl vectors may partly or completely cancel.

Extra depth. Carbon tetrachloride, although it has four polar bonds, is non-polar and is therefore immiscible with water and a good solvent for non-polar substances (this is one reason why it was used as a degreasing agent). Chloroform is polar enough to dissolve fats and alkaloids. The fact that CH₂Cl₂ has the highest dipole moment shows that the dipole moment does not simply increase with the number of polar bonds: it depends on their relative orientation.

Conclusion: Dichloromethane (CH₂Cl₂) has the highest dipole moment; CCl₄ has zero because its four bond moments cancel by symmetry.

Concept and approach. A formula CₙH₂ₙ for a hydrocarbon (here C₅H₁₀) corresponds to either an alkene or a cycloalkane (one degree of unsaturation). Two observations decide which: (1) no reaction with chlorine in the dark — alkenes add Cl₂ readily across the C=C bond even in the dark (as with Br₂ in CCl₄), so the hydrocarbon cannot be an alkene; cycloalkanes, like alkanes, are inert to Cl₂ without light. (2) Bright sunlight gives a single monochloro compound, C₅H₉Cl — a free-radical substitution that replaces one H by Cl; a single monochloride means all ten hydrogen atoms are equivalent.

Identification. Of the possible C₅H₁₀ cycloalkanes — cyclopentane, methylcyclobutane, 1,1-dimethylcyclopropane, ethylcyclopropane and the 1,2-dimethylcyclopropanes — only cyclopentane has all hydrogens equivalent (five identical CH₂ groups with a plane/axis of symmetry), so replacement of any one hydrogen gives the same product, chlorocyclopentane. The other isomers contain different kinds of CH₂, CH or CH₃ hydrogens and would give two or more monochloro derivatives.

C₅H₁₀ (cyclopentane) + Cl₂ —(hν, sunlight)→ C₅H₉Cl (chlorocyclopentane) + HCl

Conclusion: The hydrocarbon is cyclopentane; it is saturated (no reaction in the dark) and all ten hydrogens are equivalent, giving only chlorocyclopentane on free-radical chlorination.

Concept. C₄H₉Br is a saturated monobromoalkane (an alkyl bromide, CₙH₂ₙ₊₁Br with n = 4). Its isomers arise from (a) the carbon skeleton of butane (straight chain, n-butyl) and of 2-methylpropane (branched chain, isobutyl) and (b) the position of the bromine atom on that skeleton. There are two skeletons — n-butane (CH₃CH₂CH₂CH₃) and isobutane ((CH₃)₃CH) — and each has two kinds of carbon to which bromine can be attached.

Extra depth. 2-Bromobutane contains a stereocentre (C2 bears H, Br, CH₃ and C₂H₅), so it exists as a pair of enantiomers, (R)- and (S)-2-bromobutane — including these, there are five stereoisomers of C₄H₉Br in all. The four constitutional isomers differ in their reactivity: tert-butyl bromide undergoes S_N1 and E1 readily; n-butyl and isobutyl bromides undergo S_N2 (isobutyl more slowly because of β-branching); sec-butyl bromide can follow either route. Their boiling points also differ — n-butyl bromide (375 K) > sec-butyl (364 K) > isobutyl (364 K) > tert-butyl (346 K) — showing the effect of branching.

Conclusion: Four structural isomers of C₄H₉Br exist: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane (2-bromobutane also has two enantiomers).

Concept. 1-Iodobutane, CH₃CH₂CH₂CH₂I, is a primary alkyl iodide. Alkyl iodides are best made either by replacing the –OH of an alcohol with iodine (using NaI/KI with a non-oxidising acid, or red phosphorus and iodine, which generate PI₃ in situ) or by halogen exchange from the chloride/bromide (Finkelstein reaction). From an alkene one must add HX with the correct orientation: the iodine must end up on the terminal carbon, which means anti-Markovnikov addition of HBr (peroxide effect) followed by halogen exchange.

(i) From 1-butanol — heating the alcohol with sodium (or potassium) iodide in 95 % orthophosphoric acid; HI is generated in situ and the –OH is replaced by iodine (H₂SO₄ must not be used because it oxidises HI to I₂). Alternatively, red phosphorus + I₂ (PI₃ in situ):

CH₃CH₂CH₂CH₂OH + NaI + H₃PO₄ —(heat)→ CH₃CH₂CH₂CH₂I + NaH₂PO₄ + H₂O

3CH₃CH₂CH₂CH₂OH + PI₃ → 3CH₃CH₂CH₂CH₂I + H₃PO₃ [PI₃ from 2P + 3I₂]

(ii) From 1-chlorobutane — Finkelstein reaction: treatment with NaI in dry acetone. Sodium chloride is insoluble in acetone and precipitates, which drives the equilibrium to the right (Le Chatelier’s principle):

CH₃CH₂CH₂CH₂Cl + NaI —(dry acetone)→ CH₃CH₂CH₂CH₂I + NaCl↓

(iii) From but-1-ene — a two-step route. Direct addition of HI would follow Markovnikov’s rule and give 2-iodobutane, so first add HBr in the presence of a peroxide (anti-Markovnikov, free-radical addition) to obtain 1-bromobutane, and then carry out a Finkelstein exchange with NaI in acetone:

CH₃CH₂CH=CH₂ + HBr —(peroxide)→ CH₃CH₂CH₂CH₂Br

CH₃CH₂CH₂CH₂Br + NaI —(dry acetone)→ CH₃CH₂CH₂CH₂I + NaBr↓

Conclusion: 1-Iodobutane is obtained (i) from 1-butanol with NaI/H₃PO₄ (or red P + I₂), (ii) from 1-chlorobutane with NaI in dry acetone, and (iii) from but-1-ene via HBr/peroxide to 1-bromobutane and then NaI/acetone.

Definition and background. A nucleophile is an electron-rich species that attacks an electron-deficient centre (here the carbon of R–X) using a lone pair or an anionic charge. An ambident nucleophile (from the Latin ambi, “both”, and dens, “tooth”) is a nucleophile that has two different atoms (two nucleophilic centres) through either of which it can donate an electron pair to the substrate, so that it can attack in two different ways and lead to two different products. Typical examples are the cyanide ion (CN⁻) and the nitrite ion (NO₂⁻).

Example 1 — the cyanide ion. CN⁻ is a resonance hybrid of two contributing structures, ⁻:C≡N ↔ :C=N⁻ (a lone pair on carbon and a lone pair on nitrogen). Attack through carbon gives an alkyl cyanide (nitrile) R–C≡N; attack through nitrogen gives an alkyl isocyanide R–N⁺≡C⁻. The ionic reagent KCN gives mainly the nitrile, because the free ion attacks through the carbon (the C–C bond formed is more stable than the C–N bond), whereas the largely covalent AgCN has no free cyanide ions and nitrogen donates its lone pair, giving mainly the isocyanide (Example 6.5 of NCERT).

R–X + KCN → R–C≡N (alkyl cyanide) + KX R–X + AgCN → R–N⁺≡C⁻ (alkyl isocyanide) + AgX

Example 2 — the nitrite ion. O=N–O⁻ can bond through oxygen or nitrogen: KNO₂ (ionic) attacks through O to give an alkyl nitrite R–O–N=O, whereas AgNO₂ (covalent) attacks through N to give a nitroalkane R–NO₂.

Conclusion: Ambident nucleophiles (e.g. CN⁻, NO₂⁻) have two nucleophilic centres and can give two different products depending on which atom attacks — KCN gives alkyl cyanides and AgCN gives isocyanides.

Concept. In an S_N2 reaction with hydroxide ion (a strong nucleophile) the rate = k[R–X][OH⁻] depends on two factors: (a) the leaving-group ability of the halide — the weaker the C–X bond and the more stable the halide ion X⁻ that leaves, the faster the reaction (I⁻ > Br⁻ > Cl⁻ ≫ F⁻); and (b) steric hindrance at the carbon being attacked (CH₃ > 1° > 2° > 3°).

(i) CH₃Br or CH₃I — CH₃I reacts faster. The alkyl group is the same, so only the leaving group matters. The C–I bond (bond enthalpy 234 kJ mol⁻¹, length 214 pm) is the longest and weakest of the C–X bonds (C–Br: 293 kJ mol⁻¹, 193 pm), and iodide, being larger and more polarisable, is a better leaving group and a more stable anion (HI is the strongest of the halogen acids). The transition state is therefore reached more easily.

(ii) (CH₃)₃CCl or CH₃Cl — CH₃Cl reacts faster. Methyl chloride has three small hydrogen atoms on the carbon, so the hydroxide ion can approach the back of the carbon easily. In tert-butyl chloride the three methyl groups shield the carbon bearing chlorine (steric hindrance), the transition state with five groups crowded around the carbon is of very high energy, and the compound reacts instead by S_N1 or by elimination to give 2-methylpropene.

Conclusion: (i) CH₃I (better leaving group) and (ii) CH₃Cl (unhindered primary carbon) react faster with –OH by the S_N2 mechanism.

Concept. Sodium ethoxide in ethanol is a strong base, so alkyl halides containing β-hydrogens undergo β-elimination (dehydrohalogenation): a base removes a proton from a carbon β to the C–X carbon while the halide leaves, forming a C=C. When more than one kind of β-hydrogen is available, a mixture of alkenes can result, and the major product is given by Saytzeff’s (Zaitsev’s) rule: the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbons (the more substituted, more stable alkene).

(i) 1-Bromo-1-methylcyclohexane. The C–Br carbon (a ring carbon bearing a CH₃ group) has β-hydrogens on the CH₃ group and on the two ring CH₂ groups (which are equivalent). Elimination towards the ring gives 1-methylcyclohexene (trisubstituted alkene, major); elimination from the CH₃ gives methylenecyclohexane (disubstituted exocyclic alkene, minor).

(ii) 2-Chloro-2-methylbutane, (CH₃)₂C(Cl)CH₂CH₃. β-hydrogens are present on the two equivalent CH₃ groups and on the CH₂ of the ethyl group. Loss from the CH₂ gives 2-methylbut-2-ene, (CH₃)₂C=CHCH₃ (trisubstituted, major); loss from a CH₃ gives 2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃ (disubstituted, minor).

(iii) 2,2,3-Trimethyl-3-bromopentane, (CH₃)₃C–C(Br)(CH₃)–CH₂CH₃. The carbon next to C–Br on one side is the quaternary carbon C2 (no hydrogens), so β-hydrogens exist only on the C3-CH₃ and on the CH₂ of the ethyl group. Elimination through the CH₂ gives 2,2,3-trimethylpent-3-ene, (CH₃)₃C–C(CH₃)=CHCH₃ (trisubstituted, major); elimination through the methyl gives 2-ethyl-3,3-dimethylbut-1-ene [(CH₃)₃C–C(=CH₂)CH₂CH₃, disubstituted, minor].

Conclusion: Major alkenes by Saytzeff’s rule: (i) 1-methylcyclohexene, (ii) 2-methylbut-2-ene, (iii) 2,2,3-trimethylpent-3-ene; the minor products are the corresponding less substituted alkenes.

Concept. A conversion is solved by identifying the type of change (carbon chain lengthening, change of functional group, change of position of a double bond, introduction of a halogen) and choosing standard reactions in the right sequence: alcohol → halide (PBr₃, SOCl₂); halide → alkene (alcoholic KOH); alkene → vic-dihalide (Br₂) → alkyne (alc. KOH/NaNH₂); halide → alcohol (aq. KOH); halide → nitrile (KCN); halide + metal → Wurtz/Fittig/Grignard; alkane or side-chain → halide (X₂, hν).

Reasoning behind the choice. In (i) a terminal alkyne is deprotonated (sodium acetylide) and alkylated by the primary bromoethane in an S_N2 reaction, which lengthens the chain by two carbons; in (ii) the second dehydrohalogenation is stopped at the vinyl bromide by using one equivalent of base; in (iii) the peroxide effect puts bromine on the terminal carbon and the covalent AgNO₂ gives the nitro (not the nitrite) compound; in (iv) the reaction in light takes place on the side-chain methyl and the benzylic halide is hydrolysed to the alcohol; in (vii) the nitrile and Grignard reagent supply the two methyl groups on each side of the carbonyl; in (viii) the position of the double bond is moved by adding HBr (Markovnikov) and eliminating HBr again (Saytzeff).

Conclusion: All ten conversions are accomplished with the sequences in the table, using only the reactions of haloalkanes and haloarenes studied in this unit.

(i) Dipole moment of chlorobenzene < that of cyclohexyl chloride. The dipole moment depends on the magnitude of the partial charges and on the C–Cl bond length. In chlorobenzene the carbon bonded to chlorine is sp² hybridised, which has more s-character (33 %) than the sp³ carbon (25 %) of cyclohexyl chloride, so it is more electronegative and holds the shared pair more tightly; the electronegativity difference between C and Cl is smaller and so is the polarity. In addition, the lone pair on chlorine is delocalised into the ring (resonance, +R effect), which pushes electron density towards the ring and partly opposes the inductive (–I) shift towards chlorine. The C–Cl bond is shorter in chlorobenzene (169 pm, with partial double-bond character) than in cyclohexyl chloride (about 177 pm), reducing the charge separation. All these effects lower the dipole moment of chlorobenzene.

(ii) Alkyl halides are immiscible with water. To dissolve an alkyl halide in water, energy is needed to break the attractions between the halide molecules and the strong hydrogen bonds between water molecules. The new interactions formed between haloalkane and water molecules (dipole–dipole) are weaker than the O–H···O hydrogen bonds being broken, so less energy is released than absorbed and dissolution is unfavourable. Alkyl halides cannot form hydrogen bonds with water; their large hydrophobic alkyl part is also non-polar. They dissolve readily in organic solvents because the new interactions are of similar strength to those broken.

(iii) Grignard reagents must be prepared under anhydrous conditions. In RMgX the C–Mg bond is covalent but highly polar (Cᵟ⁻–Mgᵟ⁺), so R behaves as a very strong base (like a carbanion). It reacts with any source of protons — even water, alcohols and amines — to give the hydrocarbon: RMgX + H₂O → R–H + Mg(OH)X. Traces of moisture would therefore destroy the reagent, so the reaction is done in dry ether with perfectly dry apparatus (ether also solvates and stabilises the reagent by co-ordinating to Mg).

R–MgX + H–OH → R–H + Mg(OH)X ; R–MgX + R′OH → R–H + Mg(OR′)X

Conclusion: (i) sp² carbon and resonance reduce the polarity of C–Cl in chlorobenzene; (ii) haloalkanes cannot form hydrogen bonds with water, so the energy balance is unfavourable; (iii) Grignard reagents are destroyed by water (protonation), so dry ether and dry apparatus are required.

Background. Polyhalogen compounds contain more than one halogen atom and are valuable in industry, agriculture and medicine, but several of them accumulate in the environment or damage the ozone layer, so their uses have been restricted.

Environmental perspective. Freon, DDT and CCl₄ are stable and resist breakdown by soil bacteria; they persist and travel (to the stratosphere or up the food chain), so their use is now limited and safer substitutes (hydrofluorocarbons, biodegradable pesticides) are preferred.

Conclusion: Freon 12: refrigerant and aerosol propellant; DDT: insecticide; CCl₄: solvent, CFC feedstock and former fire extinguisher; iodoform: antiseptic — each with environmental or health drawbacks.

Concept. Each reaction illustrates one of the characteristic reactions of haloalkanes or of the methods of making them: halogen exchange, elimination versus substitution, nucleophilic substitution by cyanide, Williamson ether synthesis, conversion of alcohols with thionyl chloride, and Markovnikov addition of HBr to alkenes. For every reaction the product is decided by the substrate class (1°, 2°, 3°), the strength/nature of the reagent and the conditions.

Additional points. In (ii) the substrate is tertiary, so an S_N2 attack is impossible (steric hindrance); a strong base like KOH removes a β-hydrogen (from any of the three equivalent CH₃ groups) to give only one alkene. In (iii) a secondary halide can undergo both substitution and elimination; aqueous NaOH favours substitution, whereas alcoholic KOH favours elimination (see Ex. 6.20). In (vii) and (viii) the formation of the more stable carbocation intermediate explains Markovnikov’s rule; in the presence of a peroxide, (vii) would give 1-bromobutane (anti-Markovnikov) instead.

Conclusion: (i) CH₃CH₂CH₂I; (ii) (CH₃)₂C=CH₂; (iii) CH₃CH(OH)CH₂CH₃; (iv) CH₃CH₂CN; (v) C₆H₅OC₂H₅; (vi) CH₃CH₂CH₂Cl; (vii) CH₃CH₂CHBrCH₃; (viii) CH₃CH₂C(Br)(CH₃)₂.

Concept. n-Butyl bromide is a primary alkyl halide and cyanide (from ionic KCN) is a good nucleophile. A primary substrate, unhindered at the α-carbon, with a good nucleophile reacts by the S_N2 (substitution nucleophilic bimolecular) mechanism: the reaction is a one-step process, the rate = k[n-BuBr][CN⁻] (second order), no intermediate is formed and the carbon undergoes inversion of configuration (not visible here because the carbon is not a stereocentre).

Mechanism. (1) The cyanide ion, with the lone pair on carbon, approaches the carbon atom bonded to bromine from the side opposite to the leaving group (backside attack). (2) In the transition state the carbon is partly bonded to both CN and Br ([NC···CH₂(C₃H₇)···Br]⁻); the three other substituents (two H and C₃H₇) lie in a plane perpendicular to the Nu–C–Br axis. (3) The C–Br bond breaks as the new C–C bond is completed, and bromide leaves as Br⁻.

N≡C⁻ + CH₃CH₂CH₂CH₂–Br → [N≡C···CH₂(CH₂CH₂CH₃)···Br]⁻ (transition state) → N≡C–CH₂CH₂CH₂CH₃ + Br⁻

Why not S_N1 or elimination? A primary carbocation is very unstable, so ionisation (S_N1) is unfavourable; cyanide is only weakly basic, so elimination is minor. The product, an alkyl cyanide, is useful because the –CN group can be hydrolysed to –COOH or reduced to –CH₂NH₂, giving a route for increasing the carbon chain by one carbon.

Conclusion: n-BuBr reacts with KCN by a single-step S_N2 mechanism involving backside attack by CN⁻ and simultaneous departure of Br⁻, giving n-butyl cyanide (pentanenitrile).

Concept. In S_N2 reactions the nucleophile must approach the carbon bearing the leaving group from the back; therefore the less crowded the carbon (and its neighbouring carbon), the faster the reaction. The usual order CH₃X > 1° > 2° > 3° is modified by branching at the β-carbon (next to the α-carbon), which also hinders the approach: a primary halide with β-branching is slower than an unbranched or γ-branched primary halide.

(i) 1-Bromopentane is primary (CH₃CH₂CH₂CH₂CH₂Br), 2-bromopentane is secondary, and 2-bromo-2-methylbutane is tertiary. Reactivity decreases from 1° to 3°:

1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane

(ii) 1-Bromo-3-methylbutane [(CH₃)₂CHCH₂CH₂Br] is primary; 2-bromo-3-methylbutane [(CH₃)₂CHCHBrCH₃] is secondary (with an adjacent isopropyl branch); 2-bromo-2-methylbutane is tertiary:

1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane

(iii) All four are primary bromides and differ only in the position of the methyl branches: 1-bromobutane has no branch; 1-bromo-3-methylbutane has a branch at γ (far from C–Br); 1-bromo-2-methylbutane has a branch at β (adjacent to the α-carbon), which hinders attack; and 1-bromo-2,2-dimethylpropane (neopentyl bromide) has two β-methyl groups (three groups on the β-carbon) and is almost inert in S_N2:

1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane

Conclusion: S_N2 reactivity: (i) 1-bromopentane > 2-bromopentane > 2-bromo-2-methylbutane; (ii) 1-bromo-3-methylbutane > 2-bromo-3-methylbutane > 2-bromo-2-methylbutane; (iii) 1-bromobutane > 1-bromo-3-methylbutane > 1-bromo-2-methylbutane > 1-bromo-2,2-dimethylpropane.

Concept. Hydrolysis of a benzylic halide by aqueous KOH (a weak-base, nucleophilic, polar protic medium) generally proceeds through the S_N1 mechanism in which the rate-determining step is the formation of a carbocation: R–Cl → R⁺ + Cl⁻, followed by attack of water/OH⁻. Therefore the more stable the carbocation, the faster the hydrolysis. Both compounds are benzylic halides, so the carbocations are stabilised by resonance with the benzene ring (the positive charge is delocalised onto the ortho and para carbons); the extent of stabilisation differs.

Extra depth. The carbocation (C₆H₅)₂CH⁺ can be drawn with the positive charge on six different ring carbons (three on each ring), whereas the benzyl cation has only three such resonance forms. This extra delocalisation lowers the energy of the carbocation and of the transition state leading to it. In an S_N2 sense the opposite would be expected (benzyl chloride is less hindered), but in aqueous KOH benzylic secondary halides ionise so readily that S_N1 dominates; the NCERT answer is therefore diphenylmethyl chloride.

Conclusion: C₆H₅CHClC₆H₅ (diphenylmethyl chloride) is hydrolysed more easily by aqueous KOH because it forms the more stable, doubly resonance-stabilised carbocation.

Concept. The melting point of a molecular solid depends mainly on how well the molecules pack in the crystal lattice and the strength of intermolecular forces; the boiling point depends mainly on the polarity (dipole–dipole forces) and mass of molecules in the liquid. Isomeric dihalobenzenes have the same molar mass and very similar polarity, so their boiling points are nearly the same; but their melting points differ greatly because of differences in symmetry.

Explanation. p-Dichlorobenzene (1,4-) is a highly symmetrical molecule (centrosymmetric with the Cl atoms at opposite ends). Because of its symmetry, the molecules fit very well into the crystal lattice, packing closely and uniformly, so the intermolecular forces in the crystal are stronger and more energy (higher temperature) is needed to break the lattice. The ortho- and meta-isomers are less symmetrical, do not pack as efficiently, have weaker lattice forces and therefore melt at lower temperatures.

Practical consequence. The large difference in melting points of the ortho and para isomers allows them to be separated: when a mixture is cooled, the para isomer crystallises out first (as in the electrophilic halogenation of arenes, where an o/p mixture is obtained and the p-isomer is separated by crystallisation).

Conclusion: p-Dichlorobenzene melts higher than the o- and m-isomers because its greater symmetry allows better packing in the crystal lattice and stronger lattice forces.

Concept and strategy. Multi-step conversions are solved by working backwards from the target and forwards from the substrate, choosing reagents from a small “toolbox”: (a) Markovnikov addition of HX to alkenes, and anti-Markovnikov (peroxide) addition of HBr; (b) alcoholic KOH (elimination) versus aqueous KOH (substitution); (c) KCN (adds one carbon, gives nitrile → acid by hydrolysis); (d) Wurtz and Fittig reactions (C–C bond formation, doubling the carbon number); (e) side-chain free-radical halogenation (X₂, hν) versus ring halogenation (X₂, Fe); (f) Sandmeyer reaction of diazonium salts; (g) carbylamine reaction. Every step below is a reaction from this unit.

Reasoning behind key steps. To move a halogen from the 1- to the 2-position (iii) or the reverse (xvi, xix) the halide is first dehydrohalogenated to the alkene (alcoholic KOH) and HBr is then re-added with the required orientation — Markovnikov (no peroxide) for the 2-isomer, anti-Markovnikov (peroxide) for the 1-isomer. To convert a secondary halide into a primary alcohol (xiii) the same “alkene detour” is followed, ending with aqueous KOH. In (v) and (xv) the order of steps matters: halogen first (o,p-directing), then nitration; in (xv) the nitro group at the para position activates the C–Cl bond so that aqueous NaOH can displace Cl. In (xx) the isocyanide arises from the carbylamine reaction of a primary aromatic amine with chloroform and alcoholic KOH, with dichlorocarbene as the reactive species.

Conclusion: All twenty conversions follow from the reactions in the table, using substitution, elimination, addition, Wurtz/Fittig coupling, Sandmeyer and carbylamine reactions in suitable sequences.

Concept. An alkyl halide that has β-hydrogens can react with a base/nucleophile in two competing ways: nucleophilic substitution (the group –OH replaces –X, giving an alcohol) or elimination (loss of H from the β-carbon and X from the α-carbon, giving an alkene). Which route wins depends on the nature of the reagent in solution, which is determined by the solvent: this is the principle “a chemical reaction is a race won by the fastest runner”.

Aqueous KOH. In water, KOH is almost completely ionised to K⁺ and OH⁻, and the hydroxide ion is strongly hydrated (surrounded by H₂O molecules through hydrogen bonding). The hydration shell reduces its basicity so that OH⁻ is a good nucleophile but only a poor proton acceptor; it therefore attacks the electrophilic carbon (substitution) rather than abstracting a β-proton:

R–CH₂–CH₂–Cl + KOH(aq) → R–CH₂–CH₂–OH + KCl (substitution, alcohol)

Alcoholic KOH. Dissolving KOH in ethanol gives an equilibrium in which some ethoxide ion, C₂H₅O⁻, is formed (KOH + C₂H₅OH ⇌ C₂H₅OK + H₂O). Ethoxide is a much stronger and bulkier base than hydroxide, and there is hardly any water to hydrate (stabilise) it. It readily abstracts a proton from the β-carbon (and the solvent also favours elimination, particularly when heated), so the alkene forms (β-elimination or dehydrohalogenation):

R–CH₂–CH₂–Cl + KOH(alc.) —(heat)→ R–CH=CH₂ + KCl + H₂O (elimination, alkene)

Extra depth. Substitution versus elimination also depends on the substrate: primary halides prefer S_N2, secondary give both, tertiary prefer elimination or S_N1; a bulkier base favours elimination because it cannot approach the α-carbon easily and instead abstracts the more accessible β-hydrogen. Higher temperatures also favour elimination (a higher entropy product).

Conclusion: In aqueous KOH, hydrated OH⁻ acts as a nucleophile and gives alcohols, whereas in alcoholic KOH the stronger base ethoxide abstracts a β-hydrogen and gives alkenes.

Concept and strategy. This is a deduction problem. The four isomers of C₄H₉Br are n-butyl, sec-butyl, isobutyl and tert-butyl bromide; (a) is stated to be primary, so it is either n-butyl bromide or isobutyl bromide. The last clue — its Wurtz product (d) is different from that of n-butyl bromide (which gives n-octane) — rules out n-butyl bromide and leaves isobutyl bromide, 1-bromo-2-methylpropane.

Step-by-step deduction.

1. (a) is a primary C₄H₉Br with a different Wurtz product from n-BuBr ⇒ (a) = (CH₃)₂CHCH₂Br, isobutyl bromide (1-bromo-2-methylpropane).
2. (a) + alcoholic KOH → dehydrohalogenation ⇒ (b) = (CH₃)₂C=CH₂ (2-methylpropene).
3. (b) + HBr (Markovnikov addition, no peroxide) ⇒ (c) = (CH₃)₃CBr, tert-butyl bromide — an isomer of (a) as required.
4. (a) + Na (dry ether, Wurtz) ⇒ (d) = (CH₃)₂CHCH₂–CH₂CH(CH₃)₂ = 2,5-dimethylhexane (C₈H₁₈), which differs from n-octane obtained from n-butyl bromide.

(a) (CH₃)₂CHCH₂Br + KOH(alc.) → (b) (CH₃)₂C=CH₂ + KBr + H₂O

(b) (CH₃)₂C=CH₂ + HBr → (c) (CH₃)₃C–Br [Markovnikov: H adds to CH₂, Br to the tertiary carbon]

(a) 2(CH₃)₂CHCH₂Br + 2Na —(dry ether)→ (d) (CH₃)₂CHCH₂CH₂CH(CH₃)₂ + 2NaBr

Why not n-butyl bromide? If (a) were n-butyl bromide, alcoholic KOH would give but-1-ene, which with HBr would give 2-bromobutane (sec-butyl bromide, also an isomer of (a)); but its Wurtz product would be n-octane, which is the compound that the question says (d) is not. Hence only isobutyl bromide fits all the data.

Conclusion: (a) is isobutyl bromide (CH₃)₂CHCH₂Br; (b) 2-methylpropene; (c) tert-butyl bromide; (d) 2,5-dimethylhexane.

Concept. The six reagents illustrate the major reaction types of haloalkanes and haloarenes: elimination (alcoholic KOH), organometallic formation (Mg, dry ether), difficult nucleophilic substitution of aryl halides (hydrolysis), nucleophilic substitution (aqueous KOH, KCN), and Wurtz coupling (Na, dry ether).

Explanation and extra points. In (i) the alcoholic medium contains the strong base ethoxide, which removes a β-hydrogen; in (iv) aqueous KOH gives substitution. In (ii) the C–Mg bond is covalent but polar (Cᵟ⁻–Mgᵟ⁺), which makes C₆H₅MgBr a source of the nucleophilic phenyl carbanion (used for making alcohols, acids etc.). In (iii) the contrast with (iv) shows the low reactivity of aryl halides to nucleophiles; the presence of an electron-withdrawing nitro group at o/p positions would make the reaction easy. In (v) the Wurtz reaction is limited to symmetrical alkanes (using two different halides gives a mixture). In (vi) the product CH₃CN is the nitrile that can be hydrolysed to acetic acid.

Conclusion: (i) but-1-ene; (ii) phenylmagnesium bromide; (iii) phenol (after drastic hydrolysis and acidification); (iv) ethanol; (v) ethane; (vi) methyl cyanide (acetonitrile).