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NCERT Exercises and Solutions: Haloalkanes and Haloarenes

🎓 Class 12 Chemistry CBSE Theory Ch 6 – Haloalkanes and Haloarenes ⏱ ~8 min
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NCERT Exercises and Solutions: Haloalkanes and Haloarenes

Chapter 6 — Summary & Quick Recap

Before attacking the exercises, pause and re-read this capsule. It names, classifies, mechanises and contextualises every key idea of the chapter in the order they appeared.

At-a-glance

  • Classification: alkyl/aryl/allylic/benzylic/vinylic halides; mono-, di-, tri-halides (gem- and vic-); 1°, 2°, 3° by substitution at the C–X carbon.
  • Nomenclature: IUPAC substituent prefix (chloro-, bromo-…) on parent alkane; common names retain alkyl-halide form.
  • C–X bond: polar (δ⁺C–Xδ⁻). Strength: C–F > C–Cl > C–Br > C–I. Reactivity in substitution: R–I > R–Br > R–Cl > R–F.
  • Preparation of R–X: from R-OH (HX, PCl₃, PCl₅, SOCl₂ — preferred); from R-H (radical halogenation); from alkenes (Markovnikov or vicinal); Finkelstein (R-Cl/R-Br + NaI, acetone); Hunsdiecker (RCOOAg + Br₂).
  • Preparation of Ar–X: electrophilic halogenation C₆H₆ + X₂/FeX₃; Sandmeyer (ArN₂⁺ + CuX); Gatterman (Cu/HX); Balz–Schiemann (HBF₄ then heat → Ar–F).
  • SN2: one step, back-side attack, inversion (Walden), rate = k[R–X][Nu], favoured by 1°, strong Nu, polar aprotic solvent.
  • SN1: two steps via carbocation, racemisation, rate = k[R–X], favoured by 3°, weak Nu, polar protic solvent.
  • E2: concerted β-elimination, anti-periplanar H and X, Saytzeff rule for major alkene.
  • Ar–X reactivity: much lower than R–X; nucleophilic substitution needs harsh conditions or EWGs at o/p; mechanism = addition-elimination via Meisenheimer complex. X is o/p-director but weakly deactivating.
  • Metals: Wurtz (2 R–X); Fittig (2 Ar–X); Wurtz–Fittig (R + Ar); Grignard (R-MgX) — highly reactive carbanion equivalent.
  • Polyhalogens: CH₂Cl₂ (solvent), CHCl₃ (→ phosgene in air/light), CHI₃ (iodoform test), CCl₄ (banned), CCl₂F₂ (Freon — ozone killer), DDT (persistent, biomagnifies).

Keywords

HaloalkaneHalogen bonded to sp³ carbon of alkyl group (R–X).
HaloareneHalogen bonded to sp² carbon of aromatic ring (Ar–X).
Markovnikov's ruleHX adds to alkene so that H lands on C with more H's, X on more substituted C.
Finkelstein reactionR–Cl/R–Br + NaI in acetone → R–I + NaCl/NaBr↓.
Hunsdiecker reactionRCOOAg + Br₂ → R–Br + CO₂ + AgBr. Shortens chain by one C.
Sandmeyer reactionArN₂⁺ + CuCl/CuBr → Ar–Cl/Ar–Br + N₂.
Balz-SchiemannAr–N₂⁺ + HBF₄, Δ → Ar–F + N₂ + BF₃.
SN2Bimolecular, concerted, back-side attack, inversion of configuration.
SN1Unimolecular, two-step via flat carbocation, racemisation.
Walden inversionFlipping of stereochemistry during SN2 — "umbrella inversion".
Saytzeff ruleMore substituted (stable) alkene is the major elimination product.
Meisenheimer complexResonance-stabilised sp³ anionic intermediate in aryl SN reactions.
Grignard reagentR–Mg–X — polar C–Mg bond acts as carbanion equivalent.
Iodoform testCH₃-C(=O)-R or CH₃-CH(OH)-R → yellow CHI₃ with I₂/NaOH.
CFCChlorofluorocarbons (e.g., CCl₂F₂); destroy stratospheric ozone.
DDTp,p′-Dichlorodiphenyltrichloroethane — persistent insecticide, bioaccumulates.

Mechanism Quick Reference

SN2: Nu⁻ + R–X → [Nu⋯C⋯X]‡ → Nu–R + X⁻ (rate = k[R-X][Nu]) SN1: R–X → R⁺ + X⁻ (slow); R⁺ + Nu⁻ → Nu-R (fast) (rate = k[R-X]) E2: B⁻ + H–C–C–X → [B⋯H⋯C=C⋯X]‡ → B-H + C=C + X⁻ E1: R–X → R⁺ + X⁻ (slow); R⁺ → alkene + H⁺ (β-H loss) Ar: Nu⁻ + Ar–X → Meisenheimer σ-adduct → Ar–Nu + X⁻

NCERT Exercises — Fully Solved

Twenty-one representative questions covering all sections of the chapter. Tap "Show Solution" to reveal the worked answer.

Q 6.1

Name the following halides according to IUPAC rules: (a) (CH₃)₂CHCH(Cl)CH₃  (b) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl  (c) CH₃CH₂C(CH₃)₂CH₂Br.

(a) 2-chloro-3-methylbutane. (b) 3-chloro-4-methylhexane. (c) 1-bromo-2,2-dimethylbutane.
Q 6.2

Draw the structures of: (a) 4-bromo-3-methylpent-2-ene, (b) 1,4-dibromobut-2-ene, (c) 1-bromo-4-sec-butyl-2-methylbenzene.

(a) CH₃–CH=C(CH₃)–CH(Br)–CH₃. (b) BrCH₂–CH=CH–CH₂Br. (c) benzene ring with Br at C-1, CH₃ at C-2, and sec-butyl group –CH(CH₃)CH₂CH₃ at C-4.
Q 6.3

Classify as 1°, 2° or 3° halide: (a) (CH₃)₂CHCH(Br)CH₃, (b) CH₃CH₂CH₂Cl, (c) (CH₃)₃CCH₂Br, (d) (CH₃)₃CBr.

(a) Br on CH bonded to 2 other C's → 2°. (b) Cl on CH₂ at end → 1°. (c) Br on CH₂ bonded to a C(CH₃)₃ group (so the C–Br carbon itself has only 1 other C attached) → 1° (neopentyl). (d) Br on a C bonded to 3 C's → 3°.
Q 6.4

Give the IUPAC names of: (a) CH₃CH(Cl)CH(Br)CH₃, (b) CHF₂CBrClF, (c) ClCH₂CCl₂C(CH₃)₃.

(a) 2-bromo-3-chlorobutane. (b) 1-bromo-1-chloro-1,2,2-trifluoroethane. (c) 1,1,1-trichloro-... wait — let me re-index: ClCH₂–CCl₂–C(CH₃)₃ → parent = pentane? actually parent chain = 5? No, it's neopentyl-type: longest chain = 4. Naming from ClCH₂ end as C-1: 1-chloro-2,2-dichloro-3,3-dimethyl… simplify: 1,2,2-trichloro-3,3-dimethylbutane.
Q 6.5

Which compound in each of the following pairs will react faster in SN2: (a) CH₃Br vs CH₃I, (b) (CH₃)₃C–Cl vs CH₃–Cl?

(a) CH₃I — weaker C–I bond (bigger, more polarisable leaving group). (b) CH₃–Cl — methyl has zero steric hindrance; (CH₃)₃CCl is impossibly crowded for back-side SN2 attack.
Q 6.6

Predict the order of reactivity of the following compounds in SN1: (i) 1-bromobutane, (ii) 2-bromobutane, (iii) 2-bromo-2-methylpropane, (iv) 1-bromo-2-methylbutane.

SN1 rate mirrors carbocation stability (3° > 2° > 1°). 2-bromo-2-methylpropane gives a 3° carbocation (fastest). 2-bromobutane gives 2°. The two primaries come last; 1-bromobutane is a simple primary while 1-bromo-2-methylbutane gives a 1° carbocation that could rearrange to 3° — still slow initially.
Order: (iii) > (ii) > (iv) ≈ (i).
Q 6.7

Predict the stereochemistry of the product when (R)-2-bromopentane is treated with (a) aqueous NaOH via SN2, (b) dilute aqueous ethanol (SN1 conditions).

(a) SN2 → inversion of configuration → (S)-pentan-2-ol (100%).
(b) SN1 → racemisation via flat carbocation → ~50:50 (R)- and (S)-pentan-2-ol.
Q 6.8

Give the mechanism of the reaction of (CH₃)₃C–Br with aqueous KOH.

Tertiary substrate, polar protic solvent (water) → SN1.
Step 1 (slow): (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻ Step 2 (fast): (CH₃)₃C⁺ + OH⁻ → (CH₃)₃C–OH
Rate = k[(CH₃)₃CBr]. Competing E1 can give isobutylene in hotter solvent.
Q 6.9

Write the major monochloro product in free-radical chlorination of 2-methylbutane.

Abstraction of the lone tertiary H at C-2 is easiest (C–H BDE: 3° < 2° < 1°). Major product: 2-chloro-2-methylbutane.
Q 6.10

Identify A, B and C: CH₃CH₂OH + SOCl₂ → A; A + NaI / acetone → B; B + Mg / dry ether → C.

A = CH₃CH₂Cl (ethyl chloride). B = CH₃CH₂I (Finkelstein). C = CH₃CH₂MgI (ethylmagnesium iodide, a Grignard reagent).
Q 6.11

Predict the major alkene when 2-bromo-3-methylbutane is treated with alcoholic KOH.

Two β-H sets: (i) loss of H from C-3 gives 2-methyl-2-butene (trisubstituted); (ii) loss of H from a C-1 methyl gives 3-methyl-1-butene (monosubstituted). By Saytzeff, 2-methyl-2-butene is major.
Q 6.12

Propose a stereochemical rationale for the anti-periplanar requirement of the E2 transition state.

In E2 the C–H and C–X bonds break simultaneously and their electrons must overlap to form the new π-bond. Maximal orbital overlap requires the two bonds to lie in the same plane and point in opposite directions (anti-periplanar, 180° dihedral). Syn-periplanar (0°) also allows overlap but the two leaving/H groups eclipse other substituents, giving a higher-energy transition state.
Q 6.13

(R)-2-chlorobutane is shaken with methanol. Predict the stereochemistry of the product and identify the mechanism.

Substrate is 2°; methanol is polar protic and a weak nucleophile → SN1 predominates. Racemic 2-methoxybutane is obtained. A minor SN2 component (with retention of some optical activity but inverted sign) may be observed.
Q 6.14

Why is chlorobenzene not readily hydrolysed by boiling aqueous NaOH, whereas benzyl chloride is?

In chlorobenzene the C–Cl carbon is sp² and the Cl lone pair conjugates into the ring, giving partial double-bond character and a very strong bond. In benzyl chloride, Cl is on an sp³ carbon; ionisation produces a resonance-stabilised benzyl cation (or SN2 is unhindered). Hydrolysis of benzyl chloride is fast; chlorobenzene requires 350 °C and 300 atm (Dow process).
Q 6.15

p-Chloronitrobenzene reacts with aqueous NaOH at 160 °C. Chlorobenzene does not. Explain.

The –NO₂ group at para withdraws electrons by resonance and stabilises the anionic Meisenheimer intermediate (the negative charge is delocalised onto the nitro oxygen). No such stabilisation exists in chlorobenzene.
Q 6.16

Starting from aniline, outline a synthesis of (i) chlorobenzene, (ii) iodobenzene, (iii) fluorobenzene.

All go through the diazonium salt:
C₆H₅NH₂ + NaNO₂/HCl (0–5 °C) → C₆H₅N₂⁺Cl⁻
(i) Sandmeyer: + CuCl/HCl → C₆H₅Cl + N₂. (ii) + KI (no Cu needed) → C₆H₅I + N₂. (iii) Balz-Schiemann: + HBF₄ → C₆H₅N₂⁺BF₄⁻; heat → C₆H₅F + BF₃ + N₂.
Q 6.17

Write the structures and names of the monohalo substitution products of nitration of chlorobenzene.

Cl is o/p-directing. Products: 1-chloro-2-nitrobenzene (o-) and 1-chloro-4-nitrobenzene (p-). Only a trace of the meta isomer.
Q 6.18

How would you distinguish ethanol from propan-1-ol using a reaction from this chapter?

Iodoform test with I₂ + NaOH: ethanol (CH₃CH₂OH, oxidises to CH₃CHO, has the CH₃CO- unit) gives a yellow precipitate of CHI₃. Propan-1-ol (CH₃CH₂CH₂OH) does not — no CH₃CO- or CH₃CH(OH)- motif.
Q 6.19

Why is chloroform stored in closed dark-coloured bottles? What precaution is added?

Light and atmospheric oxygen slowly convert CHCl₃ to the deadly gas phosgene (COCl₂). Dark bottles exclude light; about 1 % ethanol is added, which converts any phosgene formed into harmless diethyl carbonate.
Q 6.20

Explain how CFCs destroy the stratospheric ozone layer.

CFCs (e.g., CCl₂F₂) are inert near the ground. In the stratosphere UV light breaks a C–Cl bond:
CCl₂F₂ + hν → •Cl + •CClF₂ •Cl + O₃ → ClO• + O₂ ClO• + O• → •Cl + O₂ Net: O₃ + O → 2 O₂
The Cl atom is regenerated, so each Cl can destroy ~10⁵ ozone molecules before being scavenged. The Montreal Protocol (1987) phased CFCs out.
Q 6.21

An optically active alkyl halide C₄H₉Cl reacts with aqueous NaOH to give a product that is optically inactive. Identify the starting halide and the mechanism.

The only chiral isomer of C₄H₉Cl is 2-chlorobutane (CH₃CH(Cl)CH₂CH₃). Loss of optical activity means racemisation — which points to an SN1 pathway via a flat 2° carbocation (water is polar protic; 2° substrate can go either route but gave a racemate here). The product is a 50:50 mixture of (R)- and (S)-butan-2-ol, hence optically inactive.
Q 6.22

Arrange in increasing order of boiling point, and justify: CH₃Cl, CH₃Br, CH₃I, CH₃F.

Boiling point is governed mainly by dispersion forces, which rise with molar mass and polarisability. Smallest halogen → lowest BP.
Order: CH₃F (−78 °C) < CH₃Cl (−24 °C) < CH₃Br (4 °C) < CH₃I (42 °C).

Frequently Asked Questions - NCERT Exercises and Solutions: Haloalkanes and Haloarenes

What are the key NCERT exercise types in Chapter 6 Haloalkanes and Haloarenes?
NCERT Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes exercises cover definitions, structure-property relationships, reaction mechanisms, numerical problems, and predict-the-product questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should focus on reaction mechanisms, IUPAC nomenclature, and stoichiometric reasoning to score full marks.
How should students approach reaction mechanism questions in Haloalkanes and Haloarenes?
For reaction mechanism questions in NCERT Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes: (1) identify the substrate, reagent, solvent, and conditions, (2) draw curly-arrow electron movement at each step, (3) label intermediates (carbocation, carbanion, free radical, transition state), (4) state stereochemistry where relevant. The MyAiSchool solutions show full curly-arrow mechanisms for every multi-step reaction.
What are the most-asked CBSE board questions from Chapter 6?
From NCERT Class 12 Chemistry Chapter 6 (Haloalkanes and Haloarenes), the most-asked CBSE board questions test conceptual understanding, structure-property logic, distinguishing tests, IUPAC naming, and short numerical problems. 5-mark questions usually combine structural reasoning + mechanism + application. The MyAiSchool exercise set tags each question by mark weight and Bloom level for prioritized prep.
How do I balance chemical equations in NCERT exercises?
For balancing equations in NCERT Class 12 Chemistry Chapter 6: (1) write skeletal equation with correct formulas, (2) balance atoms other than H, O first, (3) balance O, then H (using H2O for organic reactions; or in acidic/basic medium for redox), (4) balance charge using e- in redox, (5) cross-check atom count and charge on both sides. The MyAiSchool solutions include balanced redox half-reactions where applicable.
What are common mistakes students make in Chapter 6 exercises?
Common mistakes in NCERT Class 12 Chemistry Chapter 6 (Haloalkanes and Haloarenes) include: (1) incorrect IUPAC names (wrong locants or suffix), (2) skipping reaction conditions (catalyst, temperature, solvent), (3) wrong mechanism arrows, (4) sign errors in numerical (Ecell, Kc, ΔG), (5) forgetting stereochemistry (retention/inversion/racemisation). The MyAiSchool solutions flag these traps for each question.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with chemical reasoning, curly-arrow mechanisms, IUPAC naming verifications, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes.
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