Haloalkanes & Haloarenes Class 12 NCERT Solutions PDF Download

Structural observation. Treat naming as a two-pass problem: first pick the parent (longest chain through the halogen for aliphatics, the ring for aromatics with short branches), then assign locants. Classification is independent of the IUPAC name: just look at the hybridisation of the C attached to X and what else is on it.

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• Alkyl/Benzyl/Allyl: X is on an sp³ carbon.
• Vinyl/Aryl: X is on an sp² carbon.
• Further split sp³ into 1^∘/2^∘/3^∘ by counting carbons on the C–X carbon.

Alternative approach: classification first. A useful shortcut for board questions is to do classification before naming. Drawing the carbon skeleton already tells you the hybridisation of the C–X carbon, and one glance gives 1^∘/2^∘/3^∘, allyl, benzyl, vinyl or aryl. Once classified, the IUPAC name becomes almost mechanical: aryl/vinyl need the unsaturation/ring in the parent, allyl/benzyl give either choice of parent, and alkyl is always named as an alkane.

1. Group the questions: (i), (ii), (v), (vii) are open-chain sp³ halides, so they are named as substituted alkanes. (iii), (vi) have a quaternary carbon adjacent to CH2X; the longest chain runs through one of the long arms, the other becomes a substituent.
2. (iv), (xi) have the halogen on the benzylic CH2 or CHR: benzyl class. Their names use benzene as the parent because the side chain is short.
3. (viii) has Cl on sp² C=C carbon, so vinyl. (ix) has Br on an sp³ carbon directly attached to C=C, so allyl. Watch the distinction: it is not the geometry of the bond, it is which carbon (allylic = next-door to C=C; vinylic = part of C=C).
4. (x), (xii) have X on the ring carbon itself, so aryl. The chains hanging off the ring are substituents, named as isobutyl (= 2-methylpropyl) and sec-butyl (= butan-2-yl).
5. Re-read every assigned name to check alphabetical order of prefixes, lowest locants, correct -ene/-yne for unsaturated chains.
6. Reactivity preview (concept linkage). The class label tells you how the molecule will behave later in the chapter: 3^∘ alkyl and benzyl/allyl tertiary go through S_N1; 1^∘ alkyl go through S_N2; vinyl and aryl halides are nearly unreactive toward nucleophilic substitution because the C–X carbon is sp² and the bond has partial double-bond character.

Why this matters. The two questions ``what is the IUPAC name?'' and ``what type of halide is it?'' are independent and need two passes over the structure. JEE/NEET routinely test the allyl vs. vinyl and benzyl vs. aryl distinction because one carbon shift changes everything (reactivity, stability, mechanism). A typical 2-mark CBSE question asks ``classify the following halide and write its IUPAC name'' - both parts get full marks only if the class label matches what you draw.

(i) 2-chloro-3-methylbutane, 2^∘ alkyl;  (ii) 3-chloro-4-methylhexane, 2^∘ alkyl;  (iii) 1-iodo-2,2-dimethylbutane, 1^∘ alkyl;  (iv) (1-bromo-3,3-dimethylbutyl)benzene, 2^∘ benzyl;  (v) 2-bromo-3-methylbutane, 2^∘ alkyl;  (vi) 1-bromo-2-ethyl-2-methylbutane, 1^∘ alkyl;  (vii) 3-chloro-3-methylpentane, 3^∘ alkyl;  (viii) 3-chloro-5-methylhex-2-ene, vinyl;  (ix) 4-bromo-4-methylpent-2-ene, 3^∘ allyl;  (x) 1-chloro-4-(2-methylpropyl)benzene, aryl;  (xi) 1-(chloromethyl)-3-(2,2-dimethylpropyl)benzene, 1^∘ benzyl;  (xii) 1-bromo-2-(butan-2-yl)benzene, aryl.

Strategic angle. Polyhalogen names are a discipline problem, not a chemistry problem. Step through three checks: (a) identify the parent chain, (b) number so the substituents take the lowest set, (c) write substituents alphabetically.

Alphabetisation rule, expanded. When two substituents tie for the locant, the first letter of the substituent root (not the multiplying prefix) decides which gets the smaller number. So dimethyl alphabetises as ``m''; trichloromethyl alphabetises as ``t''; bis(4-chlorophenyl) alphabetises as ``c'' (chlorophenyl). For (iv), the substituent trichloromethyl comes after chloro alphabetically, so chloro locants are written first.

1. For (i), the chain is short and the two halogens differ: bromo alphabetises before chloro, so it must take the smaller locant. Number from the right.
2. For (ii), the parent is ethane. Pick the end with the larger substituent cluster. 1-Br + 1-Cl + 1-F + 2,2-(F,F) gives three F atoms, so the trifluoro prefix is 1,2,2.
3. For (iii), the locant of the triple bond and the locant of the lowest-set substituents are both 2: tie. Use alphabet: bromo < chloro, so bromo gets the 1 position.
4. For (iv), the chain is just two carbons. Three CCl3 groups become three trichloromethyl substituents. Locants: 1,2,2,2-tetrachloro plus 1,1-bis(trichloromethyl). Note ``bis'' (not ``di'') because the substituent name trichloromethyl is itself a compound name.
5. For (v) and (vi), the aryl groups go in parentheses with their substitution locant: 4-chlorophenyl, 4-iodophenyl. In (vi) the C=C is between C1 and C2 of a but-1-ene chain.
6. Cross-check. Re-read each name and rebuild the structure on paper; if you can recover the input formula unambiguously, the name is correct. This back-translation is the surest way to catch wrong locants.

Why this matters. The 2026–27 syllabus emphasises modern IUPAC formatting (locants attached to the suffix, e.g. ``but-2-ene'' not ``2-butene''), and numbering questions appear every year in board papers as 1- and 2-mark questions.

Same six names as in the main solution.

Picture-first. Name → structure is the inverse of naming. Read the parent first (chain length and saturation), then walk through the locants and attach substituents at the named positions. Treat aryl substituents like 2-chlorophenyl as sub-named groups: a phenyl ring with a chloro at the ortho (2-) position.

Decoding nested substituent names. When a name has a substituent in parentheses, both the name outside and inside need parsing. ``2-(2-chlorophenyl)-1-iodooctane'' decodes as: parent = octane (8 C); at C1 attach iodine; at C2 attach ``2-chlorophenyl'' (a phenyl ring where its own C2 carries Cl). The parenthesised number always refers to the substituent's own numbering, not the parent chain's.

1. For (i), (v), (vi), (viii) the parent is an open-chain alkane/alkene. Draw the chain, number left to right by default, drop substituents at the listed locants.
2. For (ii), (vii) the parent is benzene. Reduce the substituent names to fragments (Br, Cl, CH3, sec-butyl). Place at the listed ring positions. Remember para = 1,4 and ortho = 1,2.
3. For (iii) the parent is a 6-membered ring (cyclohexane); place substituents diametrically.
4. For (iv) the parent is an 8-C chain (octane). The phrase ``2-(2-chlorophenyl)'' attaches a benzene ring at C2 of the octane chain, and the benzene ring itself carries a Cl at its own C2 (the ring carbon adjacent to the link).
5. For (vi) the common-name fragment tert-butyl = -C(CH3)3 attaches like any other substituent; sec-butyl in (vii) is -CH(CH3)CH2CH3. Memorise the four C4 substituent fragments (n-, iso-, sec-, tert-) since they recur in CBSE questions.
6. Sanity-check: re-derive the IUPAC name from each drawn structure to confirm it matches. If the recovered name differs in any locant, your placement is wrong.

Why this matters. Reversibility (name ↔ structure) is a key NCERT exam ask. Practising both directions on the same compound trains the eye. In the 2-mark NCERT format, half marks are for the structure and half for showing the locant assignment; neat structural drawings get full marks even without a re-derivation step.

Structures as drawn in the main solution.

Picture-first. Draw the tetrahedron and add bond-dipole vectors from C toward the more electronegative end of each bond. The vector sum is the molecular dipole.

Vector intuition. Treat each bond dipole as a vector of length _bond pointing from the less- to the more-electronegative atom. For symmetric arrangements (tetrahedral, trigonal-planar, linear) where all four bonds are equal, the resultant is zero. The moment a single atom is swapped (say Cl for H), one of the cancelling vectors is replaced by a different vector, and the resultant becomes non-zero. The size of the resultant depends on the angle between the surviving cluster vectors.

1. CCl4: four equal C-Cl vectors arranged tetrahedrally. The sum of vectors from the centre of a regular tetrahedron to its four vertices is the zero vector. Hence μ = 0 D.
2. CHCl3: replace one Cl of CCl4 with H. The C-H bond dipole points toward C (since _C > _H), opposite to the C-Cl dipole that was removed. The single C-H dipole partly opposes the net C-Cl dipole, so the molecule has a small net moment, ∼ 1.04 D.
3. CH2Cl2: two C-Cl dipoles and two C-H dipoles, with the two halves on opposite faces of the tetrahedron. The vector sums of the two C-Cl dipoles and of the two C-H dipoles point in the same direction, so they add. Net μ ≈ 1.60 D, the largest of the three.
4. Experimental dipole moments (NCERT Table 6.2): CH3Cl = 1.86 D, CH2Cl2 = 1.60 D, CHCl3 = 1.04 D, CCl4 = 0 D. The trend CH3Cl > CH2Cl2 > CHCl3 > CCl4 is governed by symmetry, not by the count of polar bonds.
5. Common pitfall to avoid. ``More C-Cl bonds means higher dipole'' is wrong. The dipole of CH3Cl is even larger than CH2Cl2 although it has fewer Cl atoms, because in CH3Cl all three C-H dipoles add constructively with the single C-Cl along the C-Cl axis.
6. Numerical cross-check. A back-of-envelope estimate for CH2Cl2: each C-Cl bond moment is about 1.5 D; with a Cl–C–Cl angle of ∼ 109.5^∘ the resultant of the two C-Cl dipoles is 2 × 1.5 cos(54.7^∘) ≈ 1.73 D, then add the two C-H contributions (∼ 0.4 D each, in the same direction) and you reach the observed ∼ 1.6 D.

Why this matters. Counting polar bonds is a beginner's trap. Always sketch the geometry and add vectors. The same vector logic explains why p-dichlorobenzene has μ = 0 while o- and m-isomers do not (Q 6.18), why NH3 and NF3 differ in dipole sign, and why CO2 (μ = 0) and H2O (μ ≠ 0) behave so differently.

CH2Cl2 has the highest dipole moment.

Strategic angle. Two clues, two filters: ``no dark reaction with Cl2'' eliminates alkenes; ``single monochloride'' demands all-equivalent hydrogens.

Symmetry counting in detail. To check whether all H's in a cycloalkane are equivalent, draw the ring and mark each ring carbon. Use the ring's rotational and mirror symmetry to test if any two H's can be carried into each other by a symmetry operation. For cyclopentane, the D_5h symmetry has C₅ rotational axis: any ring CH2 can be carried onto any other by a 72^∘ rotation, and the two H's on each CH2 are related by the in-plane mirror. All 10 H's are equivalent.

1. Filter 1: ``no reaction in dark with Cl2'' means no C=C. Since the formula C5H10 has one degree of unsaturation, that unsaturation must be a ring.
2. Filter 2: ``single monochloride'' means there is only one kind of C-H. For five-carbon cycloalkanes, the candidates are cyclopentane (all 10 H's equivalent), methylcyclobutane (4 kinds of H), and ethylcyclopropane (5 kinds of H). Only cyclopentane survives.
3. Reaction is photochemical (Cl2 hν 2 Cl^.); the Cl^. radical abstracts an H to give a 5-membered radical, which reacts with Cl2 to give the monochloride. The overall stoichiometry is C5H10 + Cl2 -> C5H9Cl + HCl.
4. Confirm by counting hydrogens: cyclopentane has 10 equivalent C-H bonds, so abstracting any of them gives the same radical and the same product.
5. Common pitfall to avoid. Don't forget that methylcyclobutane and ethylcyclopropane also satisfy C5H10 and have one ring (so Ω = 1, no C=C). It is the ``single monochloride'' clue that rules them out, not the dark-test clue.
6. Concept linkage. The same reasoning identifies neopentane (2,2-dimethylpropane, C5H12) from radical bromination giving a single monobromide; both compounds have T_d-related symmetry that puts all C–H's in one equivalence class.

Why this matters. Substitution-selectivity problems on hydrocarbons reduce to symmetry counting. Practice on C4H10 (only n-butane and isobutane), C5H12 (the three pentanes), and C5H10. In JEE-Mains the question is often phrased ``which hydrocarbon gives only one monobromide?'' - the answer is always the most symmetric one.

Cyclopentane.

Structural observation. C4H9Br is saturated with no rings: enumerate by carbon skeleton, then by H-replacement position.

Two-step enumeration recipe. For any C_n H_2n+1 X: (1) list all carbon skeletons of C_n H_2n+2; for n = 4 that is 2 (n-butane, isobutane); for n = 5 that is 3 (n-pentane, isopentane, neopentane). (2) On each skeleton, identify each set of equivalent H's and substitute one at a time. The number of constitutional isomers equals the sum of inequivalent H sets across skeletons.

1. For C4 the two carbon skeletons are n-butane (linear) and isobutane (branched).
2. In n-butane, two non-equivalent H positions exist (terminal CH₃ and internal CH₂): 1-bromobutane and 2-bromobutane.
3. In isobutane, two non-equivalent H positions exist (terminal CH₃ on any of the three equivalent arms, and the central CH): 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
4. 2-Bromobutane has a stereocentre at C2 (four different groups: H, Br, methyl, ethyl): it is chiral and exists as a pair of enantiomers.
5. Identity check. Confirm each isomer by counting H's: 1-Br-butane = C4H9Br (3+2+2+2 H + 0 on Br C = 9 H, ); 2-Br-butane = 3+2+1+3 = 9 H ; isobutyl bromide = 6+1+2 = 9 H ; t-butyl bromide = 9 + 0 = 9 H . All match the molecular formula.
6. Class of each. 1-Br-butane (1^∘), 2-Br-butane (2^∘), isobutyl bromide (1^∘), t-butyl bromide (3^∘). The two primaries give slow S_N1; the secondary mixes S_N1/S_N2; the tertiary gives only S_N1 (and E1).

Why this matters. Counting isomers is a common JEE/CBSE Q. The recipe (skeletons × non-equivalent H sites) generalises to any C_n H_2n+1 X. The same enumeration template also produces all isomers of C5H12 (3), C5H11Br (8), and other simple haloalkane problems.

Four constitutional isomers; 2-bromobutane adds an R/S pair for a total of five stereoisomers.

Strategic angle. Aim at the same end product CH3CH2CH2CH2I from three different starting points: a hydroxyl, a chloride, and an alkene.

Alternative reagents (board-friendly substitutes). For (i), SOCl2 then NaI/acetone is a longer but equally valid two-step route to the chloride first, then Finkelstein. PCl5 or PCl3 would give the chloride from the alcohol if you prefer working through 1-chlorobutane. NCERT explicitly mentions P + I2/red P for direct conversion to the iodide.

1. For 1-butanol, the practical lab reagent is P + I2, which yields HI in situ. The overall stoichiometry is 3 ROH + P + 3/2 I2 -> 3 RI + H3PO3. Each ROH gets attacked by HI in two steps: protonate OH, then S_N2 by I-.
2. For 1-chlorobutane, Finkelstein. The driving force is differential solubility: NaI is soluble in dry acetone, but the product NaCl is not. Le Chatelier pulls the equilibrium fully to the right.
3. For but-1-ene, do not try direct HI addition (it gives 2-iodobutane, not the desired 1-iodobutane). Instead, Kharasch-add HBr/peroxide to get 1-bromobutane (anti-Markovnikov), then Finkelstein to swap Br for I.
4. Confirm regioselectivity in step (iii): Br^. adds to the terminal CH₂ (less substituted carbon) because that puts the radical on the more substituted (more stable) secondary carbon.
5. Why HI peroxide doesn't work. The H–I bond (∼ 297 kJ/mol) is much weaker than C–I (∼ 234 kJ/mol), so the chain-transfer step R^. + HI -> R-H + I^. is exothermic (favourable) but the alkene-addition step I^. + C=C -> I-C-C^. is too endothermic to sustain a chain. So peroxide effect runs cleanly only for HBr, not HI or HCl.
6. Concept linkage. This question previews retrosynthesis: every chain-modifying disconnection in organic chemistry comes back to one of these named transformations (alcohol → halide → Finkelstein, alkene → Markovnikov/anti-Markovnikov HX).

Why this matters. These three routes summarise the three canonical disconnections to a primary alkyl iodide. In a CBSE board question, examiners explicitly state ``write three different conversions'' to test whether students can pick the correct reagent for each starting material.

(i) P/I₂ on 1-butanol;  (ii) NaI/acetone on 1-chlorobutane;  (iii) HBr/peroxide on but-1-ene, then NaI/acetone.

Structural observation. Whenever you see a nucleophile written with delocalised charge over two atoms (O/N, C/N, S/N), suspect ambident behaviour.

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• CN-: C or N.
• NO2-: N or O.
• SCN-: S or N.

HSAB applied carefully. Hard nucleophiles (small, electronegative, weakly polarisable) prefer hard electrophiles (small, positive). Soft nucleophiles (large, polarisable, often neutral) prefer soft electrophiles (large, partial positive). In CN-, the N end is harder (more electronegative); the C end is softer (more polarisable). For an alkyl halide with δ⁺ on a soft carbon, C attack dominates. For early (S_N1-like) transition states with a more localised charge, the N attack route opens up.

1. For CN- with KCN, the ionic K–C bond is fully ionised; the C lone pair is free and attacks R-X in S_N2, producing the nitrile.
2. For CN- with AgCN, Ag forms a covalent Ag-C bond, locking the C lone pair and freeing the N lone pair to attack, producing the isocyanide.
3. HSAB explains it: hard N- prefers harder electrophiles (early transition states, S_N1-like), soft C prefers softer electrophiles (later transition states, S_N2).
4. Selectivity also depends on solvent: protic solvents hydrogen-bond the more electronegative end, slowing it down and shifting attack to the other end.
5. Exam-tip framing. In NCERT exercise answers, always cite the example as CN- with KCN vs. AgCN - this is the textbook canonical pair and is the one CBSE markers expect. Mention NO2-, SCN- as additional examples for the 3-mark variant.
6. Concept linkage. Ambident nucleophiles appear again in chapter 7 (Alcohols, Phenols, Ethers): the phenoxide C6H5O^- is ambident between O and the ortho/para ring carbons, leading to Kolbe–Schmidt and Reimer–Tiemann reactions where C- rather than O-attack dominates.

Why this matters. Same starting material plus the same nucleophile, two different products. NCERT exam writes this as a one-mark or short-answer question. The deeper idea - that delocalised charge enables more than one nucleophilic site - recurs through enolates, phenoxides, carboxylates and is foundational for much of the rest of class 12 organic chemistry.

Same conclusion: CN- (and NO2-, SCN-) is ambident; R-X + KCN -> R-CN but R-X + AgCN -> R-NC.

Strategic angle. Two single-question filters: leaving-group ability for (i), steric accessibility for (ii).

Quantitative perspective. Relative S_N2 rates with OH- (taking CH3I = 1): CH3I ≈ 1.0, CH3Br ≈ 0.02 - a ∼ 50× rate difference. For the steric pair: CH3Cl/(CH3)3CCl ≈ 10⁶ - six orders of magnitude. These ratios underline how different the two controlling factors are in magnitude.

1. Leaving-group ability for halides follows I^- > Br^- > Cl^- > F^-. The rule of thumb: the weaker the conjugate base, the better the leaving group, and HI (pK_a ≈ -10) is much more acidic than HBr (pK_a ≈ -9).
2. For (i), same carbon, different X: CH3I wins because the C–I bond is weaker and I^- is a better leaving group.
3. Steric hindrance to backside attack scales as 3^∘ ≫ 2^∘ > 1^∘ ≈ methyl. The S_N2 transition state has a crowded 5-coordinate C; more substituents means worse crowding.
4. For (ii), same X, very different steric situation: methyl chloride is essentially fully exposed; tert-butyl chloride has three methyls blocking the backside. Methyl chloride wins by orders of magnitude.
5. Alternative perspective: bond strength. The C–X bond strengths (kJ/mol) are roughly C–F (485), C–Cl (327), C–Br (285), C–I (213). Weaker bond = easier breaking = better leaving group. This trend reinforces (does not contradict) the basicity argument.
6. Concept linkage. The same orderings carry into E2 eliminations (I- fastest, F- slowest leaving) and into haloarene reactivity (where F on benzene is actually fastest in the addition-elimination mechanism of nucleophilic aromatic substitution - the opposite of the S_N2 trend! See chapter 6 sections on aryl halides).

Why this matters. A two-line answer per pair is enough for NCERT marking; the underlying reasoning (LG ability + steric access) is universal across S_N2 questions. CBSE markers expect both the order and one reason (LG ability OR steric) for full marks.

(i) CH3I;  (ii) CH3Cl.

Strategic angle. Identify every β-hydrogen on a distinct β-carbon, draw the alkene formed by removing each in turn, then rank by alkene substitution (Saytzeff).

Mechanism perspective. E2 is concerted: the C–H bond, the C–C bond and the C–X bond all move together in one transition state. The base (EtO-) abstracts H as X- leaves; the two carbons rehybridise from sp³ to sp². A key geometric requirement is that the H and X sit antiperiplanar (dihedral ≈ 180^∘). In open-chain systems this is achieved by rotation; in cyclic systems it constrains which H can be removed (this matters for ring systems with conformational locks like methylcyclohexane derivatives).

1. For (i), the two ring CH2 are equivalent: they give one product (1-methylcyclohex-1-ene). The methyl β-H gives methylenecyclohexane. Trisubstituted vs disubstituted; major is the trisubstituted ring alkene.
2. For (ii), the C1 methyl and the C2-methyl substituent give the same terminal alkene 2-methylbut-1-ene; the C3 CH2 gives 2-methylbut-2-ene. Major: 2-methylbut-2-ene (trisubstituted).
3. For (iii), the quaternary C2 has no β-H. The C4 CH2 gives an internal trisubstituted alkene; the methyl on C3 gives a terminal disubstituted alkene. Major: 3,4,4-trimethylpent-2-ene.
4. Count the substitution: trisubstituted has 3 alkyl groups on C=C; disubstituted has 2. Saytzeff says ``more''.
5. Numerical check (alkene stability). Heats of hydrogenation (kJ/mol, smaller = more stable alkene): monosubstituted ≈ -126, disubstituted ≈ -116, trisubstituted ≈ -113, tetrasubstituted ≈ -111. The ∼ 5 kJ/mol difference per added substituent is the energetic basis of Saytzeff's rule.
6. Common pitfall in (i). Students sometimes count only the ring CH2 groups and forget the methyl. The methyl is itself a β-position (carbons one bond from C-Br), and removing one of its H's gives the terminal alkene methylenecyclohexane. Both products are valid; only the relative amount differs.
7. Concept linkage. Saytzeff governs E1 as well (carbocation → most substituted alkene). When the base is bulky (like t-BuO-, hindered amines), Hofmann elimination kicks in and the least substituted alkene becomes major because the bulky base can only reach the least crowded β-H.

Why this matters. Saytzeff vs Hofmann is a one-mark selection problem in NCERT/JEE. The recipe: list β-H sites, count alkene substitution, pick the most substituted (with a small base). For 3-mark questions, also draw the alkenes' structural formulae and label the substitution pattern.

Same Saytzeff majors as in main solution.

Strategic angle. Each conversion is a retrosynthesis. Walk backward from the target to the starting material; identify the disconnection(s); pick standard reagents.

1. For (i) but-1-yne, the disconnection C-C between an ethyl and the terminal alkyne gives ethyne + CH3CH2Br. Ethyne is built from ethanol via ethylene, dibromide, double elimination.
2. For (ii) bromoethene, the disconnection at C=C-Br suggests CHBr=CH2 from CH2Br-CH2Br by single elimination. Build the dibromide from ethylene; ethylene from bromoethane (radical bromination of ethane).
3. For (iii), R-NO2 from R-Br + AgNO2. The R-Br is from propene by Kharasch addition of HBr with peroxide.
4. For (iv), Ph-CH2OH from Ph-CH2Cl via hydrolysis. Get Ph-CH2Cl from toluene by photochemical side-chain chlorination.
5. For (v), double dehydrohalogenation of a vicinal dibromide. Build the dibromide from propene and Br2.
6. For (vi), since HF cannot substitute -OH directly, route through bromide and swap to fluoride with AgF.
7. For (vii), nitrile-to-ketone sequence: CH3Br -> CH3CN -> CH3-CO-CH3 via Grignard addition and hydrolysis.
8. For (viii), Markovnikov HBr gives 2-bromobutane; alc. KOH gives Saytzeff but-2-ene.
9. For (ix), Wurtz couples two molecules of R-Cl with Na in dry ether to give the R-R dimer.
10. For (x), Wurtz–Fittig couples two Ph-Br with Na in dry ether to give biphenyl.

Why this matters. These ten conversions span the major named reactions of this chapter (Wurtz, Wurtz–Fittig, Finkelstein, Saytzeff, Kharasch, Grignard). Multi-mark CBSE questions sometimes recombine 4–5 of them as a single ``starting from X, give Y'' chain; practising all ten in advance puts you in a position to recognise any sub-chain instantly.

Concept linkage. Notice that conversions (i), (ii), (iv) all use a haloalkane intermediate; (v) uses two elimination steps in sequence. Most multistep retrosynthesis in class 12 organic chemistry passes through one or more haloalkanes - they are the chapter's pivot point because every C–X bond breaks (substitution) or its β-H leaves (elimination) under known conditions.

Same ten sequences with same reagents.

Strategic angle. Three short ``why?'' questions, three named effects: hybridisation + resonance; H-bond mismatch; acid–base reactivity.

Diagram thinking for (i). Sketch chlorobenzene with the Cl lone pair drawn as an arrow into the ring (the +M resonance contributor); the partial negative on the ring carbons opposes the C-Cl dipole. Add the C-Cl bond-length annotation: ∼ 169 pm (chlorobenzene) vs ∼ 177 pm (cyclohexyl chloride). Both shorter bond and weaker partial charge reduce μ = q · d.

1. For (i), pair the two factors. Higher s-character at the sp² ring C makes it more electronegative; the C-Cl bond is also shorter (∼ 169 pm vs ∼ 177 pm for sp³). Combined with +M donation from Cl, the net dipole is reduced.
2. For (ii), break down miscibility into Δ H_soln and Δ S_soln. The positive Δ H from breaking water H-bonds dominates over the small ion–dipole attractions an alkyl halide can offer. Two phases is the lower-energy state.
3. For (iii), use a pK_a argument. The carbanion derived from a Grignard is the conjugate base of an alkane (pK_a ≈ 50). Water (pK_a = 15.7) is a much stronger acid. The protonation lies ∼ 10³⁴ to the right toward R-H.
4. Numerical reinforcement for (i). Even though chlorobenzene has higher s-character at C (so the C-Cl bond is more polar in some sense), the resonance donation reduces the partial negative on Cl. Both effects together drop μ from ∼ 2.05 D (cyclohexyl chloride) to ∼ 1.69 D (chlorobenzene) - an 18% decrease.
5. Practical note on (iii). The Grignard rule ``no moisture'' extends to apparatus: glassware must be flame-dried; solvents (dry ether or THF) must be distilled over Na/benzophenone; Mg turnings should be activated with a tiny crystal of I2 to remove surface oxide. A single drop of water can quench a Grignard run that took hours to set up.
6. Concept linkage. The same protic-incompatibility of Grignards extends to alcohols (ROH also kills Grignards), amines (R2NH), terminal alkynes (HC#CR), and acids - any O-H, N-H, S-H, or terminal sp-CH with pK_a < 50 will protonate R-MgX.

Why this matters. Each of these three explanations is a self-contained one-mark answer NCERT loves to ask. The reasoning templates (resonance + hybridisation; energetics of solvation; pK_a logic) recur in chapters on Alcohols, Amines and Carbonyls.

Same three explanations, with the quantitative arguments above.

Quick reading. Four common polyhalogen compounds, four common uses, four corresponding environmental footnotes.

Structure → property logic. Each compound's utility comes from a structural feature you can read off the formula: freon-12 (CCl2F2) has small molecular mass and zero net H-bonding → low b.p., volatile, good refrigerant. CCl4 is heavy and symmetric (zero dipole, no H-bond) → dense solvent for non-polar oils. CHI3 is heavy (low volatility) yet hydrolytically unstable → slow I2 release. DDT's chlorinated aromatic core is lipophilic and persistent → long-lasting insecticide.

1. Freon-12: refrigerant. Property used: low boiling (-29.8^∘C), non-flammable.
2. DDT: insecticide, especially anti-malarial vector control. Property used: lipophilic, neurotoxic to insects.
3. CCl4: solvent for fats/oils, formerly a fire extinguisher. Property used: dense, non-flammable.
4. CHI3: antiseptic (releases I2, which is the actual germicide). Property used: slow I2 release.
5. Environmental caveats (board-asked). CFCs catalytically destroy stratospheric O3 (Rowland–Molina mechanism). DDT bioaccumulates (lipid-soluble, long biological half-life). CCl4 is hepatotoxic and forms COCl2 (phosgene) in heat. CHI3 causes skin irritation and has an unpleasant odour.
6. Replacements (modern chemistry). Freon-12 has been replaced by HFCs (e.g. CH2FCF3) and HFOs that contain no Cl (so no ozone depletion). CCl4 in fire-fighting has been replaced by CO2 and dry chemical extinguishers. DDT use is heavily restricted but the WHO still permits indoor residual spraying against malaria mosquitoes in some regions.

Why this matters. NCERT pairs the use with an environmental caveat in each case; expect a one-mark question or a part of a multi-mark question asking ``write one use and one disadvantage''. The Montreal Protocol (1987) and Stockholm Convention (2001) are international agreements students should be aware of for the 2-mark ``mention one international protocol'' extension.

Same uses as the main solution.

Strategic angle. Eight one-line transformations, each matching a named reaction. Identify the reaction class, write the product directly.

Reaction-class index. Build a mental table: (i) Finkelstein (halide exchange); (ii) E2 (tertiary + alcoholic base); (iii) S_N (secondary + aqueous base); (iv) S_N with ambident nucleophile (CN- via C end); (v) Williamson ether synthesis; (vi) SOCl2 clean chlorination; (vii), (viii) Markovnikov HX addition. Reading the reagent first makes the answer ``write itself''.

1. (i) Finkelstein in dry acetone; primary chloride → iodide.
2. (ii) 3^∘ bromide + alcoholic KOH → E2; only one Saytzeff alkene possible.
3. (iii) 2^∘ bromide + aqueous NaOH → substitution to give the secondary alcohol.
4. (iv) KCN's C end attacks primary bromide → nitrile.
5. (v) Williamson: phenoxide + primary alkyl halide → ether.
6. (vi) SOCl2 converts 1^∘ alcohol to 1^∘ chloride; gaseous by-products.
7. (vii) Markovnikov HBr addition to but-1-ene; Br to the more substituted C.
8. (viii) Markovnikov HBr addition to 2-methylbut-2-ene; Br to the C with two methyl substituents.
9. Common pitfall in (ii). A tertiary halide with aqueous KOH would also give some elimination because of the substrate's steric crowding; the textbook answer focuses on alcoholic KOH where the elimination channel dominates cleanly. Always read the solvent specification.
10. Why SOCl2 is preferred over HCl/PCl5 in (vi). The by-products are gases (SO2, HCl) which escape; no separation step needed, and the product is high-purity alkyl chloride. PCl₅ leaves POCl3 as a liquid contaminant.

Why this matters. A perfect drill on S_N2 vs E2 vs addition, plus the Williamson and SOCl₂ shortcuts. Many of these reactions are the building blocks for the multistep conversions in Q 6.11 and Q 6.19, so mastering them in isolation lets you handle the chains downstream.

Same eight products as in main solution.

Picture-first. Walden's inversion is the visual signature of S_N2: think ``umbrella inversion'' at the carbon. The three non-leaving groups pass through a planar arrangement and emerge on the opposite side.

Energy-profile perspective. Plot energy vs. reaction coordinate: starts at the reactant level (R-Br + CN-), rises smoothly to a single transition state (the trigonal-bipyramidal arrangement with partial bonds to both Nu and LG), and falls to the product level (R-CN + Br-). Only one maximum, no intermediate. Contrast with S_N1 (two maxima, a carbocation valley between them).

1. Identify the substrate as primary, the nucleophile as strong (CN-); both push toward S_N2.
2. Draw the curly arrow from the C of CN- to the C bearing Br; simultaneously draw the curly arrow from the C-Br bond out onto Br.
3. In the transition state, show the three remaining substituents (C3H7, H, H) flattened in a plane through C, with partial bonds to both incoming CN and outgoing Br.
4. Complete the inversion: the CN group is now on the opposite side from where Br was. Since the C in this substrate is not a stereocentre (two H's on it), the inversion is not observable here, but in a chiral case it would invert configuration.
5. Confirm by the rate law: rate = k[RBr][CN^-] is bimolecular.
6. Alternative mechanism check. An S_N1 pathway would require n-BuBr to ionise to n-Bu+ first. Primary carbocation CH3CH2CH2CH2+ has no π-stabilisation and lies ∼ 280 kJ/mol above the ionised state of a tertiary carbocation. The ionisation step is forbiddingly slow, so S_N1 does not compete.
7. Stereochemistry test for S_N2. If the substrate were chiral (e.g. (R)-2-bromobutane), S_N2 would invert the configuration to give (S)-2-cyanobutane. This single observation (inversion, not retention, not racemisation) is what distinguishes S_N2 from S_N1 experimentally.

Why this matters. Backside attack with Walden inversion is one of the most diagnostically clean reactions in organic chemistry. It generalises: any bimolecular nucleophilic substitution at sp³ C inverts the configuration. Modern synthesis uses this for stereocontrolled construction of stereocentres (Mitsunobu reaction, S_N2 ring-opening of epoxides with full inversion).

Same: S_N2, single step, backside attack, inversion, rate = k[RBr][CN^-].

Structural observation. Reactivity in S_N2 is a steric contest. Rank by (a) class of substrate at α, (b) crowding at β for ties.

Quantitative scale. Relative S_N2 rates (typical C-Br, NaOEt in EtOH, set methyl = 1): methyl ≈ 1, ethyl (1^∘) ≈ 0.03, 1^∘ with β-methyl (isobutyl) ≈ 4 × 10^-4, neopentyl (β-tBu) ≈ 4 × 10^-6. The drop is dramatic - neopentyl is slower than the corresponding tertiary halide for S_N2!

1. Class hierarchy: methyl > 1^∘ > 2^∘ > 3^∘. Three classes in (i), three in (ii); rank as is.
2. In (iii), all are 1^∘, so move to the β-C. Count methyl substituents on β: 1-Br-butane has 0; 1-Br-3-methylbutane has 0 at β (branch at γ); 1-Br-2-methylbutane has 1 methyl at β; neopentyl has 3 at β. Order is the reverse of β-substitution.
3. Confirm by mental drawing of the TS for each: more substituents on the β-C means more atoms in the way of the incoming Nu.
4. Visualisation. Picture the S_N2 transition state as a hand reaching from behind the C, while the other three substituents lie flat. For methyl, only three small H's are flat - easy. For neopentyl, three methyls hang off the C next to the planar layer and actively block the approach path of the incoming Nu.
5. γ-branching is fine. 1-Bromo-3-methylbutane has a branch at γ (two carbons from Br), which is far enough that it does not crowd the backside; its rate is essentially the same as that of n-pentyl bromide. Only β (and α) branching matters.
6. Concept linkage. The same steric reasoning reverses sign for S_N1: tertiary halides ionise fastest because the more substituted carbocation is more stable. Reactivity scales opposite to S_N2: 3^∘ > 2^∘ > 1^∘ > methyl.

Why this matters. Steric reasoning is the only tool you need to rank S_N2 rates within a homologous series. The same ``count substituents at β'' technique solves any JEE/CBSE ranking question on S_N2 within a closed family of primary halides.

Same three orderings as the main solution.

Strategic angle. The fast-hydrolysis question on benzyl-type halides is always answered by counting cation resonance forms.

Resonance counting recipe. For each candidate carbocation, draw the structure with + on C, then push every adjacent π system or lone pair toward the +. Count distinct contributors that place + on different atoms. For PhCH2+: + on the benzylic C, then + on o (2 of these), then + on p, then back to benzylic - 4 contributors. For Ph2CH+: + on the central C, then on o/p of each of the two phenyl rings - 7 contributors. Roughly twice the delocalisation, much more stable.

1. In aqueous KOH, polar protic solvent stabilises the ionisation intermediate. Both substrates can in principle undergo S_N1.
2. PhCH2+ has one phenyl ring to delocalise into; the resonance contributors place positive charge at the ortho (2 positions) and para of the ring.
3. Ph2CH+ has two phenyl rings; positive charge delocalises across both, doubling the stabilising resonance.
4. Lowering the carbocation energy lowers Δ G for ionisation; rate goes up.
5. Hence Ph2CHCl reacts faster than PhCH2Cl with aqueous KOH.
6. Mechanism perspective. Hydrolysis proceeds via S_N1: rate-determining ionisation gives the carbocation, which is then trapped by H2O (or OH-) in a fast step. So the rate depends only on [RCl], not on [OH-]. This means the rate ratio Ph₂CHCl/PhCH₂Cl would be the same in pure water or in dilute KOH.
7. Triphenyl extension. For triphenylmethyl (trityl) chloride Ph3CCl, the cation is stabilised by three rings - exceptionally stable. Trityl chloride hydrolyses in pure water at room temperature, whereas PhCH2Cl needs warm aqueous KOH.
8. Concept linkage. The same logic explains why allyl, benzyl, and especially diphenyl/triphenyl methyl substrates react via S_N1 even when they are primary or secondary - substantial cation π-stabilisation overrides the standard 1^∘ < 2^∘ < 3^∘ progression based purely on inductive effects.

Why this matters. The trick is to spot that we are comparing carbocation stabilities, not steric accessibilities; benzhydryl wins decisively because of double resonance. CBSE short-answer questions on ``which halide hydrolyses faster?'' are almost always carbocation stability problems in disguise.

C6H5CHClC6H5 (benzhydryl chloride) hydrolyses faster.

Structural observation. M.p. is a crystal property; b.p. is a liquid–gas property. They can move in opposite directions when symmetry and dipole disagree.

Sketch of the crystal-packing argument. In a molecular crystal, the lattice energy comes from (a) dispersion (London) forces between heavy atoms, (b) dipole–dipole alignment, and (c) shape complementarity (how snugly the molecules fit). Higher shape symmetry = more efficient packing density = more neighbouring contacts per molecule = larger total lattice attraction. Even though p-dichlorobenzene has zero dipole, its shape (linear-rod) packs into a denser unit cell than the bent shapes of o- and m-isomers.

1. Sketch each isomer; mark the two C-Cl dipoles and the molecular symmetry axis.
2. Only the para-isomer is centrosymmetric; its μ = 0 and it has the highest symmetry.
3. In the solid, high symmetry means molecules stack like flat coins into close-packed columns. Lattice energy is therefore largest for the para-isomer.
4. Heating breaks the lattice; the para lattice requires the most heat to disrupt, hence the highest m.p.
5. The o- and m-isomers have higher dipole moments but poorer packing; net result is a lower m.p. but a competitive (sometimes higher) b.p.
6. Boiling-point cross-check. o-dichlorobenzene: b.p. 180^∘C (μ ≈ 2.5 D, highest); m-: b.p. 173^∘C (μ ≈ 1.7 D); p-: b.p. 174^∘C (μ = 0 D). B.p. correlates with dipole moment, not with crystal packing - exactly the opposite trend to m.p.
7. Concept linkage. The same idea generalises to n-alkanes (even-numbered alkanes pack better than adjacent odd ones, so their m.p. is locally higher), to symmetric vs. asymmetric isomers of disubstituted aromatics in general, and to fullerenes (C60 packs like a sphere, very high m.p.).
8. Exam-tip. If asked ``why is the m.p. of A higher than that of B?'', invoke crystal-packing/symmetry. If asked ``why is the b.p. of A higher than that of B?'', invoke dipole moment and/or molecular mass (dispersion).

Why this matters. A classic two-mark NCERT question. Use symmetry + packing for m.p.; dipole + vdW for b.p. The counter-intuitive result that the most symmetric (zero-dipole) isomer has the highest m.p. trains the eye to separate ``melting'' (crystal → liquid) from ``boiling'' (liquid → gas) in problems on physical properties.

Higher symmetry → tighter crystal → higher m.p. of p-dichlorobenzene compared with the o- and m-isomers.

Strategic angle. A pattern recognition drill. Group by the reaction type and write reagents from memory.

• [leftmargin=*]
• Hydroboration–oxidation: (i), (xiii).
• Multistep alkyne building: (ii).
• Wurtz / Wurtz–Fittig: (ix), (xvii), (xviii).
• Sandmeyer / diazonium: (viii).
• Kolbe nitrile: (vi), (vii), (xi).
• Carbylamine: (xx).
• Haloform: (xiv).
• Halide-shift through alkene + Kharasch: (iii), (xii), (xvi), (xix).
• Markovnikov: (x).
• EAS sequence on benzene: (v), (xv).
• Free-radical side-chain chlorination + hydrolysis: (iv).

Halide-shift pattern explained. Conversions (iii), (xvi), (xix) all involve relocating a halogen from one carbon to a neighbouring carbon. The universal recipe is: (1) eliminate to the alkene with alc. KOH, (2) re-add HX. If Markovnikov gives the desired product, use plain HX; if anti-Markovnikov is needed, use HX/peroxide. This two-step ``halide-shift'' is the only way to relocate a halogen in NCERT-level chemistry.

1. Walk through each in ∼ 1 minute: identify the functional group change, pick the standard reagent.
2. Pay attention to directing effects in aromatic substitutions (Cl, Br are o,p-directors; NO2 is a m-director).
3. For halide-shift problems (xiii, xvi, xix), the recipe is always: E2 to give alkene, then add HX with peroxide if you want anti-Markovnikov.
4. Order of operations in (v) and (xv). For C6H6 → 4-bromonitrobenzene, brominate before nitrating because Br is o,p-directing but NO2 is m-directing - the order determines which isomer dominates. For chlorobenzene → 4-nitrophenol, nitrate first to activate the ring toward nucleophilic substitution of Cl; without the para NO2 the Cl would not be hydrolysed under reasonable conditions.
5. Wurtz vs. Wurtz–Fittig in (ix), (xvii), (xviii). Wurtz: R-X + R-X + 2 Na -> R-R + 2 NaX (alkyl coupling). Wurtz–Fittig: Ar-X + R-X + 2 Na -> Ar-R + 2 NaX (mixed alkyl-aryl) or 2 Ar-X + 2 Na -> Ar-Ar + 2 NaX (aryl-aryl). For (xviii) we use the latter on PhBr to give biphenyl.
6. Spotting Sandmeyer in (viii). Aniline (ArNH2) → aryl halide requires the diazonium intermediate Ar-N2+X-, which is then displaced by CuX (Sandmeyer) or by aqueous HCl boil (Gattermann). Aniline can never be converted directly to chlorobenzene with HCl or NaCl.

Why this matters. Master the table of ``reaction class → reagent'' and you can produce these from a clean slate. NCERT Q 6.19 with its twenty parts is the chapter's flagship pattern-drill; working through it builds the foundation for organic-chemistry retrosynthesis throughout class 12. JEE-Advanced 1-mark MCQs are nearly always drawn from this list of named transformations.

Same twenty reagent sequences.

Strategic angle. Same reagent, two different solvents, two different products. The discriminator is which species is solvated vs. free.

Acid–base picture of solvent effect. In water, the equilibrium KOH(aq) <=> K⁺ + OH^- favours the right completely; OH- is the only meaningful nucleophile. In ethanol, KOH dissolves and also establishes OH^- + C2H5OH <=> C2H5O^- + H2O, which has K_eq ≈ 1 because pK_a(H₂O)≈ 15.7 and pK_a(EtOH)≈ 16. So a mixture of OH- and C2H5O- is present in alcoholic KOH. Ethoxide, being slightly bulkier and slightly stronger, dominates the elimination channel.

1. In water, OH- is the active species: small, strong nucleophile, moderate base. With a primary or secondary alkyl chloride, it does S_N to give the alcohol.
2. In ethanol, KOH dissolves with deprotonation of ethanol and C2H5O- (ethoxide) becomes the active species. Ethoxide is bulkier and a stronger base, so it abstracts a β-H (E2) in preference to attacking the C.
3. Polar protic solvents (water more so than ethanol) stabilise the build-up of charge in the substitution TS; less polar protic solvents (ethanol) tilt the balance toward elimination.
4. Hence: R-Cl + KOH(aq) gives R-OH (substitution); R-Cl + KOH(alc.) gives alkene (elimination).
5. Temperature also matters. Heating (Δ) favours elimination over substitution because E2 has a higher activation entropy (more degrees of freedom in the TS) and is favoured at higher T. So warm alcoholic KOH gives the cleanest alkene yield.
6. Substrate sensitivity. The substitution-vs- elimination balance also depends on substrate class: 1^∘ halides prefer S_N2 even with bulky bases; 3^∘ halides prefer E2 even with small bases; 2^∘ halides are in the borderline zone where solvent choice dictates the outcome.

Why this matters. The same idea generalises: choose aqueous for substitution, alcoholic for elimination, bulky alkoxide (t-BuOK) for Hofmann elimination. The trio ``aqueous KOH, alcoholic KOH, t-BuOK'' covers nearly every substitution–elimination scenario in class 12 organic chemistry.

Aqueous KOH gives R-OH; alcoholic KOH gives alkene.

Strategic angle. The two clues that nail down (a) are: (1) primary halide, (2) Wurtz product differs from n-octane. Combined, they force (a) = isobutyl bromide.

Logic-puzzle framework. This is a classic ``identify the compound from the reactions'' puzzle. Whenever an NCERT problem gives you a series of clues, write the candidates that satisfy each clue, then take the intersection. Here, ``primary C4H9Br'' has 2 candidates; ``Wurtz ≠ n-octane'' has 1 candidate (since n-BuBr Wurtz = n-octane). The intersection is unique.

1. Filter 1: primary C4H9Br: candidates are n-BuBr and isobutyl bromide.
2. Filter 2: Wurtz on (a) gives a C8H18 different from the Wurtz product of n-BuBr (= n-octane). So (a) is isobutyl bromide.
3. Alc. KOH gives the only Saytzeff alkene available: 2-methylprop-1-ene.
4. HBr Markovnikov: H to the less substituted C (=CH2), Br to the more substituted C (C(CH3)2). Product: t-BuBr, an isomer of (a). Tick.
5. Wurtz coupling: two molecules of (CH3)2CHCH2Br couple at the CH2Br centres to give (CH3)2CHCH2-CH2CH(CH3)2 = 2,5-dimethylhexane.
6. Counter-check (a) by self-consistency. If (a) were n-BuBr instead, the Wurtz product would be n-octane (CH3(CH2)6CH3) - but the problem explicitly says (d) is different from that. So n-BuBr is excluded. The alc. KOH product of n-BuBr is but-1-ene, whose Markovnikov HBr addition gives 2-bromobutane (not an isomer of (CH3)2CHCH2Br in the sense the question intends; it is an isomer of n-BuBr though). Confusing, which is why the Wurtz check is the cleaner filter.
7. Concept linkage. This problem packs four named reactions into one identification puzzle: dehydrohalogenation (E2 Saytzeff), Markovnikov HX addition, Wurtz coupling. Each clue is itself a one-mark sub-question. CBSE 5-mark questions of this format are routine.

Why this matters. A logic puzzle: each clue removes one candidate. The skill of intersecting structural clues (primary, halide, Wurtz product different from a specific isomer) is exactly the analytical thinking that JEE and AIIMS test repeatedly. The same puzzle template applies to any ``mystery compound A reacts as follows; identify A'' question.

Same (a)–(d) assignments as in main solution.

Quick reading. Six small reactions, one named transformation each. Identify the reagent class, predict the product.

Reagent → named-reaction table. Alc. KOH → E2 (dehydrohalogenation); Mg/dry ether → Grignard; NaOH at 623 K/300 atm → Dow process (phenol from chlorobenzene); Aq. KOH → S_N (alcohol); Na/dry ether → Wurtz coupling; KCN → ambident, C-attack, nitrile.

1. (i) Alc. KOH = E2 ⇒ alkene. The only Saytzeff product from n-BuCl is but-1-ene.
2. (ii) Mg in dry ether = Grignard formation; C6H5MgBr is one of the most useful carbanion equivalents.
3. (iii) Chlorobenzene is essentially inert to ordinary NaOH because the C-Cl bond has partial double-bond character (resonance) and the ring C is sp². Industrial Dow process (623 K, 300 atm NaOH) gives phenol.
4. (iv) Aq. KOH = S_N on a primary halide → alcohol. Ethyl chloride gives ethanol.
5. (v) Na in dry ether = Wurtz. Methyl bromide gives CH3CH3 (ethane).
6. (vi) CN- attacks C end; primary halide gives nitrile. Methyl chloride gives acetonitrile.
7. Why chlorobenzene resists hydrolysis (iii). Three factors: (a) sp² C is more electronegative than sp³, so the C-Cl bond is shorter and stronger; (b) lone-pair donation from Cl into the ring gives partial double-bond character to C-Cl; (c) the backside of the sp² C is occupied by the π-cloud, so S_N2 backside attack is impossible. The only way around all three is the harsh Dow conditions, which proceed via a benzyne intermediate.
8. Grignard's silent partner is dry ether. The ether oxygen lone pairs solvate Mg^2+, stabilising the reagent in solution. THF is even better than diethyl ether for unreactive aryl halides; without coordinating solvent, the Grignard would not form.
9. Concept linkage. (i) connects to Q 6.20; (ii) to Q 6.12; (iii) to the haloarenes section discussion; (iv) also to Q 6.20; (v) to Q 6.21's Wurtz argument; (vi) to Q 6.8's ambident nucleophile analysis. The six parts of Q 6.22 effectively summarise the entire chapter in capsule form.

Why this matters. Six standard reactions in six lines. Master the reagent → outcome mapping. In CBSE board exams, ``what happens when X is treated with Y?'' is the most frequent 2-mark question format; six such pairs in one question is essentially a comprehensive end-of-chapter test.

Same six products as in main solution.