Chemistry Notes

Haloalkanes and Haloarenes — NEET Notes

Replace a hydrogen in a hydrocarbon with a halogen and you unlock the largest playground in organic chemistry. Haloalkanes (R–X) and haloarenes (Ar–X) are the gateway compounds — they undergo substitution, elimination, and metalation reactions that connect to almost every other functional group on the syllabus. NEET tests this chapter every single year. SN1 vs SN2 is the single most asked mechanism in organic chemistry; Saytzeff's rule, the C–X bond enthalpy order, Grignard reagents, and the dichotomy between haloalkane and haloarene reactivity all reappear with predictable frequency. This chapter teaches you the rules of the game.

Classification of haloalkanes & haloarenes

Halogen-containing organic compounds are classified along two axes — by the number of halogen atoms they carry, and by the nature of the carbon bearing the halogen. The first axis divides them into mono-, di-, tri-, and polyhalogen compounds. The second is where NEET focuses, because reactivity depends almost entirely on which kind of carbon holds the halogen.

The carbon-halogen carbon can be sp3 or sp2, and within sp3 it can be alkyl, allylic, or benzylic. Within sp2 it can be vinylic or aryl. Each kind behaves differently in substitution and elimination — so this opening classification is not pedantry, it is the map of the entire chapter.

Alkyl halide (R–X)

sp3 C–X

simple alkyl carbon

Halogen on a saturated alkyl chain. Sub-classified by carbon as 1°, 2°, 3° — primary if the C–X carbon is bonded to one other carbon, secondary to two, tertiary to three.

Allylic halide

sp3 next to C=C

allyl-position carbon

Halogen on an sp3 carbon directly attached to a C=C double bond. Resonance-stabilised carbocation makes these very reactive in SN1.

Benzylic halide

sp3 next to aryl

benzyl-position carbon

Halogen on an sp3 carbon attached to an aromatic ring. Carbocation stabilised by ring resonance — also very SN1-reactive.

Vinylic halide

sp2 on C=C

vinyl carbon

Halogen on an sp2 carbon of a C=C double bond. Very unreactive to nucleophilic substitution — partial double-bond character of C–X.

Aryl halide (Ar–X)

sp2 on ring

aromatic carbon

Halogen directly on the aromatic ring. Unreactive to SN1 and SN2 — resonance + sp2 hybridisation shorten and strengthen the C–X bond.

Gem- & vic-dihalides

2 halogens

same / adjacent C

Gem-dihalide: both halogens on the same carbon (alkylidene halide). Vic-dihalide: both halogens on adjacent carbons (alkylene dihalide).

NEET 2023 asked exactly this — given a structure with halogen on an sp3 carbon adjacent to a C=C bond, identify the class. The answer is allylic halide. Vinylic would require the halogen on the sp2 carbon itself. Benzylic would need an aromatic ring instead of an alkene. Memorising this matrix solves a free mark.

IUPAC nomenclature

Common names of alkyl halides simply name the alkyl group followed by halide — methyl chloride, ethyl bromide, tert-butyl bromide. IUPAC names them as halosubstituted hydrocarbons. The halogen takes a locant, the lowest-numbered position. The parent chain is chosen as the longest carbon chain containing the halogen-bearing carbon. Common names of dihaloalkanes give the bonding type explicitly — alkylidene halide for gem-dihalides, alkylene dihalide for vic-dihalides. IUPAC writes both as dihaloalkanes with numerical locants.

For benzene-ring substitution, common system uses o-, m-, p- prefixes for the disubstituted derivatives; IUPAC uses numerical locants 1,2; 1,3; 1,4. NCERT Table 6.1 is the canonical reference: tert-butyl bromide = 2-bromo-2-methylpropane; chloroform = trichloromethane; CCl4 = tetrachloromethane; methylene chloride = dichloromethane; allyl bromide = 3-bromopropene; vinyl chloride = chloroethene. Memorise the common-to-IUPAC pairs — questions on naming are basic NCERT-spotters.

Nature of the C–X bond

Halogens are more electronegative than carbon, so the C–X bond is polarised: carbon carries a δ+ partial charge and halogen carries δ. This polarity is the engine of every nucleophilic substitution reaction in this chapter — nucleophiles attack the electron-deficient carbon, the polarised bond breaks heterolytically, and the halogen leaves as a halide ion.

Down the halogen group, atomic size rises sharply. Fluorine is the smallest, iodine the largest. Three consequences follow.

Bond length

F < Cl < Br < I

C–X grows with halogen size

C–F is the shortest C–X bond; C–I the longest. Longer bonds are weaker and easier to break — explaining why R–I is the most reactive haloalkane.

Bond enthalpy

F > Cl > Br > I

opposite to bond length

CH3–F ≈ 452, CH3–Cl ≈ 351, CH3–Br ≈ 293, CH3–I ≈ 234 kJ/mol. Asked verbatim in NEET 2021.

NEET 2021 — exact order tested

Dipole moment

Cl > F ≈ Br > I

balance of charge and length

CH3Cl has the highest dipole moment among methyl halides — fluorine is more electronegative but the C–F bond is shorter, so charge separation is smaller.

Reactivity in both SN1 and SN2 follows the same order — R–I > R–Br > R–Cl >> R–F. The driving factor is the leaving-group ability of the halide ion, which improves down the group (weaker C–X bond, more stable halide ion in solution).

Methods of preparation — haloalkanes

NCERT lays out three routes to haloalkanes. Each has a signature reagent and a signature reaction name that NEET examiners love to test.

1. From alcohols (the workhorse route)

Concentrated HX, phosphorus halides (PCl5, PBr3, PI3), or thionyl chloride (SOCl2) all replace the –OH group with halogen. SOCl2 is preferred industrially because the by-products are gaseous SO2 and HCl — both escape, leaving pure alkyl chloride. Reactivity order of alcohols with HX is 3° > 2° > 1°. With 1° and 2° alcohols + HCl, anhydrous ZnCl2 catalyst is needed; with 3° alcohols, simply shaking with conc. HCl works at room temperature (basis of the Lucas test).

For alkyl iodides, mixing alcohol with NaI/KI in 95% orthophosphoric acid gives good yields. Note: H2SO4 is not used here — it oxidises HI back to I2, defeating the reaction.

2. From hydrocarbons

(a) Free-radical halogenation of alkanes — alkane + Cl2/Br2 under sunlight or heat. Gives a complex mixture of mono- and polyhalogenated products, useful in industry but poor in selectivity. (b) Addition of HX to alkenes — Markovnikov's rule applies: H adds to the carbon with more H atoms, halogen to the more substituted one. (c) Addition of X2 to alkenes — Br2 in CCl4 giving vic-dibromide is the standard laboratory test for unsaturation (the red-brown bromine is decolorised).

3. Halogen exchange — Finkelstein & Swarts

Preparation of haloarenes

Haloarenes cannot be prepared from phenols because the C–O bond in phenol has partial double-bond character and is too strong to cleave. Two routes work instead.

(i) Electrophilic substitution. Benzene + Cl2/Br2 in the presence of a Lewis acid (Fe, FeCl3, FeBr3) gives chlorobenzene or bromobenzene. The Lewis acid generates the electrophile (Cl+ or Br+), which attacks the ring. Iodination needs an oxidising agent (HNO3, HIO4) to consume HI as it forms. Fluorination cannot be done this way — fluorine is too reactive. This is the route NEET 2022 tested directly: chlorobenzene is made from benzene + Cl2 + anhydrous FeCl3.

(ii) From amines — Sandmeyer's reaction. Primary aromatic amine + NaNO2 + cold dilute HCl gives a diazonium salt; treating that diazonium salt with CuCl, CuBr replaces –N2+ with –Cl or –Br. For iodination, plain KI is enough (no copper salt needed).

Physical properties

Lower haloalkanes (methyl chloride, methyl bromide, ethyl chloride, chlorofluoromethanes) are gases at room temperature; higher members are liquids or solids. Boiling points are higher than the parent hydrocarbon because of stronger dipole-dipole and van der Waals forces. For the same alkyl group: R–I > R–Br > R–Cl > R–F (rises with halogen mass and polarisability). Branching lowers the boiling point — among isomeric C4H9Br, tert-butyl bromide has the lowest b.p. Bromo, iodo and polychloro compounds are denser than water. Haloalkanes are nearly insoluble in water (their dipoles cannot replace water's hydrogen-bond network) but dissolve freely in organic solvents.

For dihalobenzenes, the para-isomer melts at the highest temperature while o-, m-, and p- have similar boiling points. The para-isomer's symmetry packs the crystal lattice efficiently, raising m.p. — a classic NEET reasoning question (NCERT exercise 6.18).

Nucleophilic substitution — the central reaction

The δ+ carbon of a haloalkane is the electrophilic site. A nucleophile (Nu) — anything with a lone pair and an inclination to share it — attacks the carbon, displaces the halide ion as the leaving group, and a new C–Nu bond replaces the old C–X bond. This is the nucleophilic substitution reaction. The bond that breaks (C–X) is broken heterolytically: both bonding electrons leave with the halide. The same set of substrates can be converted into alcohols (with OH), ethers (with OR), amines (with NH3), nitriles (with CN), nitroalkanes (with NO2), and so on — Table 6.4 in NCERT.

Cyanide (CN) and nitrite (NO2) are ambident nucleophiles — they can attack from either of two atoms. KCN (ionic) gives mainly alkyl cyanides via C-attack; AgCN (covalent) gives mainly isocyanides via N-attack. The rule: ionic salt → C-attack → cyanide; covalent silver salt → N-attack → isocyanide.

SN1 vs SN2 — the most-tested distinction in organic chemistry

Substitution can proceed by either of two mechanisms, and they are opposite in almost every dimension. NEET tests this contrast every year — sometimes by direct comparison, sometimes hidden inside a reactivity-order or stereochemistry question.

SN2 mechanism — one step, inversion

Hughes and Ingold (1937) proposed the SN2 mechanism for the reaction of methyl chloride with hydroxide ion. The nucleophile attacks the δ+ carbon from the side opposite the leaving group. As the new C–OH bond forms, the C–Cl bond simultaneously weakens. In the transition state the carbon is partly bonded to five atoms — three to the original substituents (which start moving away like an umbrella catching wind), one to the incoming OH, and one to the departing Cl. As the new bond fully forms and the old one fully breaks, the three other substituents "flip" through the plane of the carbon. The configuration inverts — Walden inversion.

Because the transition state involves the nucleophile attaching close to the carbon, steric bulk kills SN2. Methyl halides are fastest (only three small hydrogens). Tertiary halides are dead — the three alkyl groups block the back-side approach. Hence the order: CH3X > 1° > 2° > 3°.

SN1 mechanism — two steps, carbocation, racemisation

Because the rate depends only on the C–X bond breaking, anything that stabilises the carbocation accelerates SN1. Tertiary carbocations are the most stable (three alkyl groups donating electrons inductively and through hyperconjugation), so 3° haloalkanes are fastest. Allylic and benzylic halides are also very fast because their carbocations enjoy resonance stabilisation — the positive charge delocalises onto the double bond or aromatic ring.

Stereochemistry of substitution & the Walden inversion

If the substrate carries a stereocentre — a carbon with four different groups — the difference between SN1 and SN2 becomes geometrically visible. Three outcomes are possible at a stereocentre when a bond is broken: retention (configuration preserved), inversion (configuration flipped), or racemisation (50:50 mixture). SN2 always gives inversion. SN1 always gives racemisation (because the planar carbocation is attacked equally from both faces).

The classic example: (–)-2-bromooctane + NaOH gives (+)-octan-2-ol via SN2 — the –OH ends up on the opposite face from where Br was. The optical rotation has flipped sign because the configuration has flipped. This is the Walden inversion, the stereochemical fingerprint of SN2.

The carbon is turned inside out like an umbrella caught in a strong wind.

NCERT description of Walden inversion in SN2

For SN1: (±)-butan-2-ol forms when optically active 2-bromobutane is hydrolysed under SN1 conditions. The carbocation intermediate is flat; the nucleophile attacks from either face; the product mixture rotates light through equal and opposite angles — net zero rotation. Racemisation. Inversion = SN2; racemisation = SN1. Memorise it as a two-line rule.

Elimination reactions — E1 & E2

When a haloalkane carrying a β-hydrogen is heated with alcoholic KOH, hydrogen leaves the β-carbon, halogen leaves the α-carbon, and a C=C double bond forms between them. An alkene is produced. Because the H is lost from the β-position, this is called β-elimination or dehydrohalogenation.

α-carbon: the carbon bearing the halogen. β-carbon: the carbon adjacent to it. β-hydrogen: any hydrogen on a β-carbon — only these are eliminated.

Rate-determining step + stereochemistry + substrate preference — the three NEET-relevant dimensions for SN1, SN2, E1 and E2:

SN1

Slow: ionisation

first-order kinetics

Stereo: racemisation (planar carbocation)

Best substrate: 3° > 2° > 1°

SN2

Slow = only step

second-order kinetics

Stereo: inversion (Walden)

Best substrate: CH3X > 1° > 2° > 3°

E1

Slow: ionisation

first-order kinetics

Stereo: Saytzeff product (more substituted alkene)

Best substrate: 3° > 2° > 1°

E2

Slow = only step

second-order kinetics

Stereo: anti-periplanar; Saytzeff alkene major

Best substrate: 3° > 2° > 1° (alc. KOH)

Saytzeff (Zaitsev) rule

When more than one β-hydrogen is available, more than one alkene can form. The Russian chemist Alexander Saytzeff (also spelled Zaitsev) noted in 1875 that the major product is the more substituted alkene — the one with more alkyl groups on the doubly bonded carbons. This is the more stable alkene (greater hyperconjugation, greater inductive donation). Example: 2-bromopentane gives pent-2-ene (di-substituted, major) over pent-1-ene (mono-substituted, minor). NEET 2021 asked this directly; NEET 2020 asked it as part of a statement-correctness question.

Substitution vs elimination — which wins?

A haloalkane with a β-hydrogen has a choice: substitution (SN1 or SN2) or elimination (E1 or E2). The outcome depends on substrate, base, and conditions. The default rules:

  • Primary haloalkanes prefer SN2 (back-side attack easy, no carbocation). With strong bulky base — elimination.
  • Secondary haloalkanes are the most ambiguous — SN2, E2, or E1 depending on conditions. Strong base + heat favours E2.
  • Tertiary haloalkanes avoid SN2 (steric blockade); they choose between SN1 and E1 depending on conditions. With heat and strong base, E2.
  • Aqueous KOH favours substitution → alcohol. Alcoholic KOH favours elimination → alkene. NEET 2016 tested this exact reagent contrast.

Reactions with metals — Grignard, Wurtz

Haloalkanes are the starting point for organometallic chemistry. With Mg in dry ether, R–X gives an alkyl magnesium halide R–MgX — the Grignard reagent (Victor Grignard, 1900; Nobel 1912). The C–Mg bond is highly polar — carbon carries a partial negative charge and behaves as a nucleophilic carbanion. Grignard reagents react with carbonyls, epoxides, CO2, and many other electrophiles to forge new C–C bonds. They are how organic chemists build up molecules from small fragments.

The catch: Grignard reagents are destroyed by even trace moisture. Water, alcohols, amines, terminal alkynes — anything with a proton acidic enough to be plucked — protonates the carbanion, converting R–MgX into R–H (the hydrocarbon). Hence the rule: Grignard reactions must be run under strictly anhydrous conditions, in dry ether.

Wurtz reaction. Alkyl halide + sodium in dry ether → alkane with double the number of carbons. Two R–X molecules couple: 2 R–X + 2 Na → R–R + 2 NaX. Useful for symmetric higher alkanes; gives mixtures when two different R–X are used.

Reactions of haloarenes

Haloarenes resist the SN1 and SN2 reactions that haloalkanes love. Four reasons:

  1. Resonance. The halogen's lone pair conjugates with the ring's π-electrons, putting partial double-bond character on the C–X bond — harder to break.
  2. Hybridisation. The ring carbon is sp2 (33% s-character) vs sp3 in haloalkane (25% s-character). More s-character means more electronegative — the bonding pair is held more tightly. C–Cl bond length: chlorobenzene 169 pm, methyl chloride 177 pm.
  3. Phenyl cation instability. SN1 is impossible — the phenyl cation cannot be stabilised by resonance (the empty orbital is orthogonal to the π-system).
  4. Repulsion. The electron-rich nucleophile is repelled by the electron-rich aromatic ring.

Forcing conditions (NaOH, 623 K, 300 atm) convert chlorobenzene to phenol. An electron-withdrawing group (e.g. –NO2) at ortho/para position increases reactivity by stabilising the intermediate carbanion (Meisenheimer complex). At meta — no effect.

Electrophilic substitution of haloarenes

Haloarenes still undergo the standard electrophilic substitutions — halogenation, nitration, sulphonation, Friedel-Crafts — but more slowly than benzene. The halogen is slightly deactivating (electron-withdrawing through induction) but ortho-, para-directing (electron-donating through resonance for the σ-complex stabilisation). This dichotomy is the most asked reasoning question in haloarene chemistry: reactivity controlled by the inductive effect, orientation controlled by resonance.

Wurtz-Fittig & Fittig reactions

A mixture of alkyl halide + aryl halide + Na in dry ether gives an alkylarene — Wurtz-Fittig reaction. Two aryl halides + Na in dry ether give a biaryl — Fittig reaction. Bromobenzene + Na gives biphenyl (Fittig); bromobenzene + methyl bromide + Na gives toluene (Wurtz-Fittig).

Polyhalogen compounds — uses & concerns

Compounds with two or more halogen atoms have outsized industrial importance — and outsized environmental cost. The six NCERT-named members:

  • Dichloromethane (CH2Cl2, methylene chloride): solvent, paint remover, aerosol propellant, drug-manufacture solvent. Harms CNS; high exposure causes dizziness, nausea, numbness; direct skin contact burns.
  • Trichloromethane (CHCl3, chloroform): solvent for fats, alkaloids, iodine; historically a general anaesthetic, now replaced by safer ones. Major industrial use: feedstock for freon R-22. Slowly oxidised by air + light to phosgene (COCl2) — extremely poisonous. Stored in dark bottles, completely filled, with a little ethanol as stabiliser.
  • Triiodomethane (CHI3, iodoform): formerly antiseptic, replaced now. Its antiseptic action came from liberated iodine, not the molecule itself. Used in the iodoform test (yellow precipitate identifies CH3CO– or CH3CH(OH)– groups).
  • Tetrachloromethane (CCl4): feedstock for chlorofluorocarbons, pharmaceutical manufacture, general solvent. Once used as cleaning fluid and fire extinguisher. Damages the liver and CNS; releases to atmosphere deplete the ozone layer.
  • Freons (chlorofluorocarbons, CFCs): CFCl3, CCl2F2 etc. Stable, non-toxic, non-flammable, easily liquefiable — perfect refrigerants and propellants. Freon-12 (CCl2F2) is manufactured from CCl4 by Swarts reaction. Released CFCs reach the stratosphere and initiate radical chain reactions that destroy ozone — banned globally under the Montreal Protocol.
  • DDT (p,p'-dichlorodiphenyltrichloroethane): the first chlorinated organic insecticide. Paul Müller won the 1948 Nobel Prize in Medicine for showing its effectiveness against malaria-carrying mosquitoes. But DDT is fat-soluble, persistent, and bioaccumulates — banned in the US in 1973; still used in malaria control elsewhere.

The shared concern across these compounds: persistence. The chlorinated and fluorinated C–C and C–Cl bonds resist biodegradation. CFCs and CCl4 deplete ozone. DDT bioaccumulates up the food chain. Modern green chemistry replaces them where possible.

NEET PYQ Snapshot

Real NEET previous-year questions — solve before moving on.

NEET 2023

The given compound CH2=CH–CH(X)–CH2CH3 is an example of __________.

  1. Vinylic halide
  2. Benzylic halide
  3. Aryl halide
  4. Allylic halide
Answer: (4) Allylic halide

Why: The halogen-bearing carbon is sp3 and is directly next to a C=C double bond — by definition, allylic. Vinylic would require the halogen on the sp2 carbon of the C=C itself; benzylic would require an aromatic ring; aryl would require the halogen directly on the ring.

NEET 2022

Which of the following sequence of reactions is suitable to synthesize chlorobenzene?

  1. Phenol, NaNO2, HCl, CuCl
  2. Aniline, NaNO2, HCl, CuCl
  3. Aniline, HCl, heating
  4. Benzene, Cl2, anhydrous FeCl3
Answer: (4) Benzene + Cl2 + anhydrous FeCl3

Why: Electrophilic aromatic substitution on benzene with Cl2 in the presence of FeCl3 (Lewis acid) gives chlorobenzene. Option (2) — Sandmeyer's reaction from aniline — would also work in principle, but the question's option (2) was stated incompletely. Option (4) is the direct, textbook route.

NEET 2021

The major product formed in dehydrohalogenation reaction of 2-bromopentane is pent-2-ene. This product formation is based on?

  1. Huckel's Rule
  2. Saytzeff's Rule
  3. Hund's Rule
  4. Hofmann Rule
Answer: (2) Saytzeff's Rule

Why: Saytzeff's rule states that the major elimination product is the more substituted alkene. Pent-2-ene is di-substituted (CH3–CH=CH–CH2–CH3); pent-1-ene is mono-substituted. The more substituted alkene is more stable due to hyperconjugation. Hofmann's rule is the reverse pattern, applies to quaternary ammonium hydroxides — wrong here.

NEET 2021

The correct sequence of bond enthalpy of 'C–X' bond is:

  1. CH3–Cl > CH3–F > CH3–Br > CH3–I
  2. CH3–F < CH3–Cl < CH3–Br < CH3–I
  3. CH3–F > CH3–Cl > CH3–Br > CH3–I
  4. CH3–F < CH3–Cl > CH3–Br > CH3–I
Answer: (3) C–F > C–Cl > C–Br > C–I

Why: Bond length increases down the halogen group (F < Cl < Br < I), and bond strength is inversely related — shorter bonds are stronger. Values: CH3–F 452, CH3–Cl 351, CH3–Br 293, CH3–I 234 kJ/mol.

NEET 2020

Elimination reaction of 2-bromopentane to form pent-2-ene is: (a) β-elimination, (b) follows Zaitsev rule, (c) dehydrohalogenation reaction, (d) dehydration reaction. Choose:

  1. (a), (c), (d)
  2. (b), (c), (d)
  3. (a), (b), (d)
  4. (a), (b), (c)
Answer: (4) (a), (b), (c)

Why: H is lost from the β-carbon, halogen from the α-carbon — β-elimination, yes. Major product is the more substituted alkene — Zaitsev, yes. H + halogen lost as HBr — dehydrohalogenation, yes. Not dehydration, since no water is removed. (d) is the trap.

Expert FAQs

Questions NEET has asked from this chapter, answered straight.

What is the difference between SN1 and SN2 reactions?
SN2 is a one-step bimolecular reaction whose rate depends on both substrate and nucleophile; it proceeds through a single five-coordinate transition state, gives inversion of configuration, and is fastest for primary substrates (CH3X > 1° > 2° > 3°). SN1 is a two-step unimolecular reaction whose rate depends only on the substrate; it goes through a planar carbocation intermediate, gives racemisation, and is fastest for tertiary substrates (3° > 2° > 1°).
Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?
Four reasons. (i) Resonance — the halogen's lone pair conjugates with the ring, giving the C–X bond partial double-bond character. (ii) Hybridisation — the ring carbon is sp2 (more s-character, more electronegative), holding the bonding pair more tightly; C–Cl in haloarene is 169 pm versus 177 pm in haloalkane. (iii) Phenyl cation instability — SN1 is ruled out because the planar phenyl cation lacks resonance stabilisation. (iv) Electrostatic repulsion between the electron-rich nucleophile and the electron-rich aromatic ring.
What is Saytzeff's rule?
In dehydrohalogenation reactions, the preferred (major) alkene is the one with the greater number of alkyl groups attached to the doubly bonded carbon atoms — that is, the more substituted, more stable alkene. For example, 2-bromopentane gives pent-2-ene (di-substituted) as the major product over pent-1-ene (mono-substituted). NEET 2021 and NEET 2020 both tested this.
Why are Grignard reagents prepared under anhydrous conditions?
Grignard reagents (R–MgX) carry a strongly polarised C–Mg bond with carbon as a carbanion-like nucleophile. Even traces of water — or any compound with an acidic proton such as alcohols, amines, or carboxylic acids — protonate the carbanion, converting the reagent into the corresponding hydrocarbon (R–H). This destroys the reagent before it can react. That is why dry ether is the standard solvent.
What is the correct order of C–X bond enthalpies?
C–F > C–Cl > C–Br > C–I. Bond enthalpies for methyl halides are approximately CH3–F 452, CH3–Cl 351, CH3–Br 293, CH3–I 234 kJ/mol. Bond length increases F < Cl < Br < I, and bond strength falls in the same direction. NEET 2021 tested this exact order.
What is the Finkelstein reaction?
The Finkelstein reaction prepares alkyl iodides by treating an alkyl chloride or bromide with NaI in dry acetone. The reaction is driven forward by the precipitation of NaCl or NaBr — both poorly soluble in acetone, while NaI is soluble. Le Chatelier's principle pulls the equilibrium toward the alkyl iodide product.
What is the Swarts reaction?
The Swarts reaction prepares alkyl fluorides by heating an alkyl chloride or bromide with a metallic fluoride such as AgF, Hg2F2, CoF2, or SbF3. The chloride/bromide is exchanged for fluoride. This is how freon-12 (CCl2F2) is manufactured industrially from tetrachloromethane.
Why is chloroform stored in dark-coloured bottles filled to the top?
Chloroform (CHCl3) is slowly oxidised by atmospheric oxygen in the presence of light to phosgene (COCl2), a highly poisonous gas. To prevent this, chloroform is stored in dark-coloured bottles to exclude light, and the bottle is filled completely to keep air out. A small amount of ethanol is also added as a stabiliser — it converts any phosgene formed into harmless diethyl carbonate.

Go Deeper

Drill into the subtopics that NEET asks most often.