Haloalkanes and HaloarenesNEET Chemistry · Class 12 · NCERT Chapter 6

SN1 (unimolecular) is a two-step reaction: the C-X bond breaks first to form a carbocation, then the nucleophile attacks. SN2 (bimolecular) is a one-step concerted reaction: the nucleophile attacks the back side of the carbon as the leaving group leaves simultaneously. To predict which occurs, check three things: (1) Substrate: 3° always SN1 (carbocation stable), 1° and CH3 always SN2 (no steric block), 2° can be either. (2) Nucleophile: strong nucleophile (CN-, OH-, I-) drives SN2; weak nucleophile (water, alcohol) allows SN1. (3) Solvent: polar protic (water, alcohols) stabilises carbocation, favours SN1; polar aprotic (DMSO, acetone, DMF) leaves nucleophile free and reactive, favours SN2. The stereochemical outcome is key: SN1 gives racemisation; SN2 gives Walden inversion (complete inversion of configuration).

In haloarenes (like chlorobenzene), the halogen is directly attached to the aromatic ring. The lone pairs on the halogen can overlap with the pi system of the ring through resonance. This resonance delocalises electron density from the halogen into the ring, giving the C-X bond partial double-bond character. Because of this partial double bond, the C-X bond in chlorobenzene (bond length 169 pm) is shorter and stronger than in cyclohexyl chloride (179 pm). This extra strength means the C-X bond is harder to break, making haloarenes much less reactive than haloalkanes towards nucleophilic substitution.

The Finkelstein reaction converts an alkyl chloride or bromide to an alkyl iodide using sodium iodide in dry acetone: R-Cl + NaI (dry acetone) → R-I + NaCl. The driving force is solubility. Sodium iodide is soluble in acetone, but the products NaCl and NaBr are practically insoluble in acetone and precipitate out of solution. By Le Chatelier's principle, this removes products from the equilibrium and drives the reaction forward to almost complete conversion. You can't use water as solvent because all sodium halides are soluble in water, so the equilibrium doesn't shift far.

Optical isomers (enantiomers) are non-superimposable mirror images of each other. They arise when a carbon atom has four different groups attached (a chiral centre). The two enantiomers rotate plane-polarised light in opposite directions: one is dextrorotatory (+) and the other is laevorotatory (-). An equal mixture of both is a racemic mixture and shows no net optical activity. The classic NEET example is 2-bromobutane: CH3-CHBr-C2H5. The middle carbon carries four different groups (CH3, Br, H, and C2H5), making it a chiral centre. It exists as (R)-2-bromobutane and (S)-2-bromobutane. SN2 on (R)-2-bromobutane with OH- gives (S)-butan-2-ol (inversion). SN1 gives a racemic mixture of both.

This seems contradictory but makes sense when you separate two effects. The halogen withdraws electrons from the ring through the sigma bond (inductive effect, -I), which deactivates the ring overall and makes it less reactive than benzene. But the halogen also donates lone pair electrons into the pi system by resonance (+M), and this donation specifically activates the ortho and para positions. The positive charge in the resonance structures appears at ortho and para positions on the ring, stabilising the electrophilic attack intermediate at those positions. The net result: ring is deactivated (slower than benzene) but the little reaction that does occur is directed to ortho and para positions. Electrophilic attack at meta would not be stabilised by resonance from the halogen.

A Grignard reagent (R-MgX) is made by reacting a haloalkane with magnesium metal in dry diethyl ether: R-X + Mg → R-MgX. The C-Mg bond is highly polar (carbon is nucleophilic), so Grignard reagents act as powerful nucleophiles and strong bases. Key reactions: (1) Water: R-MgX + H2O → R-H + Mg(OH)X (destroyed — must keep it dry). (2) Aldehyde: R-MgX + RCHO → R-CHOH-R (secondary alcohol after hydrolysis). (3) Ketone: R-MgX + R'COR' → tertiary alcohol. (4) Formaldehyde (HCHO): gives primary alcohol. (5) CO2: R-MgX + CO2 → RCOOH (carboxylic acid, chain extended by 1C). Grignard reagents are essential for making C-C bonds and are one of the most important synthetic tools in organic chemistry.

The iodoform test detects the presence of a CH3CO- (methyl ketone) or CH3CHOH- (secondary alcohol with methyl group) group. The reagent is NaOI (made in situ from I2 + NaOH). Positive test: a yellow precipitate of CHI3 (iodoform) with its characteristic antiseptic smell forms. Compounds that give a positive iodoform test: (1) Acetaldehyde (CH3CHO) — only aldehyde that gives positive. (2) All methyl ketones (CH3COR). (3) Ethanol (CH3CH2OH). (4) All secondary alcohols of the type CH3CHOHR. (5) Acetone. (6) Isopropanol (CH3CHOHCH3). Compounds that do NOT give iodoform test: benzaldehyde, formaldehyde, other aldehydes, 1° alcohols (except ethanol), ketones without CH3CO- group.

All three reactions use sodium metal to join two organic halide molecules by eliminating NaX. The difference is in the starting halides: Wurtz reaction: two molecules of the same alkyl halide (2 R-X + 2Na → R-R + 2NaX). Used to make symmetric alkanes. Example: 2 CH3Br + 2Na → C2H6 + 2NaBr. Works only for haloalkanes. Fittig reaction: two molecules of aryl halide (2 Ar-X + 2Na → Ar-Ar + 2NaX). Used to make symmetric biaryls (diaryl compounds). Example: 2 C6H5Br + 2Na → biphenyl + 2NaBr. Wurtz-Fittig reaction: one aryl halide + one alkyl halide + 2Na → Ar-R + 2NaX. Used to attach an alkyl group to an arene. Example: C6H5Br + CH3Br + 2Na → toluene (C6H5CH3) + 2NaBr.