Haloalkanes and haloarenes for NEET

Haloalkanes are where organic chemistry starts asking "how" as well as "what". The NMC syllabus for this unit names the nature of the C–X bond and the mechanisms of substitution reactions, so NEET questions go beyond products: which halide reacts fastest by SN2, why a chiral halide gives a racemic product, why chlorobenzene ignores aqueous NaOH. The syllabus also names the uses and environmental effects of chloroform, iodoform, freons and DDT, which are easy marks if you have read the last pages of the NCERT chapter.

Everything here builds on carbocation stability from general organic chemistry and on the alkene reactions in hydrocarbons.

The C–X bond

Halogen is more electronegative than carbon, so the C–X bond is polar: carbon carries a partial positive charge and is attacked by nucleophiles. In haloalkanes the carbon is sp³.

So for the same alkyl group, reactivity is R–I > R–Br > R–Cl > R–F: the weaker bond breaks more easily. Dipole moment does not follow the same order. NCERT gives CH₃Cl > CH₃F > CH₃Br > CH₃I, because in CH₃F the very short bond offsets fluorine's higher charge.

Physical properties worth remembering:

• Boiling point rises with molar mass (RI > RBr > RCl > RF) and falls with branching among isomers.
• Among the dichlorobenzenes, the para isomer has the highest melting point because its symmetrical molecules pack best in the crystal.
• Bromo, iodo and polychloro compounds are denser than water.

Preparation

From alcohols (the usual laboratory route):

• R–OH + HCl with anhydrous ZnCl₂ (Groves process); reactivity of alcohols is 3° > 2° > 1°.
• R–OH + PCl₅ → R–Cl + POCl₃ + HCl; 3R–OH + PCl₃ → 3R–Cl + H₃PO₃.
• R–OH + SOCl₂ → R–Cl + SO₂ + HCl. Thionyl chloride is preferred because both by-products are gases and escape, leaving a pure product.

From hydrocarbons: free-radical halogenation of alkanes (gives mixtures, so it is a poor method) and addition of HX or X₂ to alkenes.

Halogen exchange:

• Finkelstein reaction: R–Cl or R–Br + NaI in dry acetone → R–I. NaCl and NaBr are insoluble in dry acetone and precipitate, which pulls the reaction forward (Le Chatelier).
• Swarts reaction: heating an alkyl chloride or bromide with a metal fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃ gives the alkyl fluoride.

Haloarenes: electrophilic substitution of benzene with Cl₂ and a Lewis acid (FeCl₃), or from a diazonium salt: Sandmeyer (Cu₂Cl₂/HCl or Cu₂Br₂/HBr), Gattermann (copper powder with HCl or HBr) and KI for iodobenzene.

Nucleophilic substitution: SN2 and SN1

SN2 in words. The nucleophile approaches from the side opposite the halogen. In the transition state the carbon is bonded partly to both the incoming nucleophile and the leaving halide, with the other three groups in one plane. As the halide leaves, the three groups flip over like an umbrella in the wind. That is why an optically active halide gives a product with inverted configuration. Bulky groups block the back-side approach, so tertiary halides hardly react by SN2.

SN1 in words. The C–X bond ionises slowly to give a planar carbocation (the rate-determining step), then the nucleophile attacks quickly from either face. Attack from both faces gives both enantiomers, so an optically active halide gives a racemic (or partly racemic) product. Allylic and benzylic halides react fast by SN1 because their carbocations are resonance-stabilised.

Ambident nucleophiles

Some nucleophiles can attack through two different atoms.

Elimination and reactions with metals

β-Elimination. Heating a haloalkane with alcoholic KOH removes H from the β-carbon and X from the α-carbon to give an alkene. When more than one alkene can form, Saytzeff's (Zaitsev's) rule says the more substituted alkene is the major product.

Substitution or elimination? Aqueous KOH supplies OH⁻, a nucleophile, and gives mainly the alcohol. Alcoholic KOH supplies the alkoxide ion, a stronger base, and gives mainly the alkene. Tertiary halides and bulky bases also favour elimination.

Grignard reagent. R–X + Mg in dry ether → RMgX. The carbon bonded to Mg behaves as a carbanion, so the reagent reacts with any source of a proton: RMgX + H₂O → R–H + Mg(OH)X. That is why it must be made and used under dry conditions.

Coupling with sodium:

• Wurtz: 2R–X + 2Na → R–R + 2NaX (dry ether).
• Wurtz–Fittig: Ar–X + R–X + 2Na → Ar–R + 2NaX.
• Fittig: 2Ar–X + 2Na → Ar–Ar + 2NaX.

Why haloarenes resist nucleophilic substitution

Chlorobenzene does not react with aqueous NaOH under ordinary conditions. Four reasons, all in NCERT:

1. Resonance: a lone pair on chlorine is delocalised into the ring, giving the C–Cl bond partial double-bond character.
2. Hybridisation: the carbon is sp², with more s-character, so the C–X bond is shorter and stronger.
3. Unstable phenyl cation: an SN1 path would need one, and it is not resonance-stabilised.
4. Repulsion: the electron-rich ring repels an approaching nucleophile, and the ring blocks back-side attack.

Forcing conditions work. The Dow process converts chlorobenzene to phenol with NaOH at 623 K and 300 atm. Electron-withdrawing –NO₂ groups at the ortho and para positions make it much easier: 4-nitrochlorobenzene reacts at 443 K, 2,4-dinitrochlorobenzene at 368 K, and 2,4,6-trinitrochlorobenzene with warm water. A nitro group at the meta position has no such effect.

In electrophilic substitution, halogens are deactivating but ortho/para directing (see the directing-group table in hydrocarbons).

Polyhalogen compounds: uses and environmental effects

Worked NEET-style questions

Example 1: SN2 order

Arrange for SN2 reactivity: 2-bromobutane, bromomethane, 2-bromo-2-methylpropane, 1-bromobutane.

• SN2 depends on steric crowding: methyl < 1° < 2° < 3° in crowding.
• Bromomethane > 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane.
• For SN1 the order would reverse, following carbocation stability.

Example 2: aqueous or alcoholic KOH

What does 2-bromobutane give with (i) aqueous KOH, (ii) alcoholic KOH?

• (i) Substitution: butan-2-ol.
• (ii) Elimination: but-1-ene and but-2-ene can both form. But-2-ene has more alkyl groups on the double bond, so but-2-ene is the major product (Saytzeff).

Example 3: stereochemistry

(–)-2-Bromooctane is treated with NaOH under SN2 conditions. What is the product's configuration?

• SN2 means back-side attack and inversion. The product, octan-2-ol, has the inverted configuration.
• If the same halide reacted by SN1, you would expect a largely racemic alcohol.

Example 4: an ambident nucleophile

CH₃CH₂Br is treated separately with KCN and AgCN. Name the products.

• KCN: ionic, the carbon end attacks: propanenitrile (CH₃CH₂CN).
• AgCN: the Ag–C bond is largely covalent, so nitrogen attacks: ethyl isocyanide (CH₃CH₂NC).

Practice MCQs

1. The best reagent for converting ethanol to chloroethane with gaseous by-products only is: (a) HCl (b) PCl₅ (c) SOCl₂ (d) PCl₃
2. NaI in dry acetone converts CH₃CH₂Br to CH₃CH₂I. This is the: (a) Finkelstein reaction (b) Swarts reaction (c) Sandmeyer reaction (d) Wurtz reaction
3. Which halide reacts fastest by SN1? (a) CH₃Br (b) CH₃CH₂Br (c) (CH₃)₂CHBr (d) (CH₃)₃CBr
4. An SN2 reaction at a chiral carbon gives: (a) retention (b) racemisation (c) inversion (d) no change in optical activity
5. Chloroform is stored in dark, completely filled bottles to prevent formation of: (a) CCl₄ (b) COCl₂ (c) CH₂Cl₂ (d) Cl₂
6. Which is most reactive towards aqueous NaOH? (a) chlorobenzene (b) 3-nitrochlorobenzene (c) 4-nitrochlorobenzene (d) 2,4-dinitrochlorobenzene
7. CH₃MgBr reacts with water to give: (a) CH₄ (b) CH₃OH (c) C₂H₆ (d) CH₃Br
8. The major product of 2-bromo-2-methylbutane with alcoholic KOH is: (a) 2-methylbut-1-ene (b) 3-methylbut-1-ene (c) 2-methylbut-2-ene (d) 2-methylbutan-2-ol

Answers

1. (c) SO₂ and HCl both escape as gases.
2. (a) The precipitating NaBr drives it forward.
3. (d) The tertiary carbocation is the most stable.
4. (c) Back-side attack flips the configuration.
5. (b) Light and air oxidise CHCl₃ to phosgene.
6. (d) Two –NO₂ groups at ortho and para positions withdraw the most electron density.
7. (a) The CH₃ group takes a proton from water.
8. (c) The trisubstituted alkene is more substituted than the disubstituted one (Saytzeff).

What to do next

• Draw the SN2 transition state and the SN1 carbocation for one chiral halide, and write the stereochemical outcome under each.
• Make a one-page table of every named reaction here with its reagent and condition.
• Learn the polyhalogen table; it is pure recall and often skipped.
• Move on to alcohols, phenols and ethers, where the same SN1/SN2 ideas return in ether cleavage.