Quick Recap — Haloalkanes & Haloarenes

  • Haloalkanes (R−XR-X): polar C−-X bond (Cδ+−Xδ−C^{\delta+}-X^{\delta-}); classified 1∘^\circ/2∘^\circ/3∘^\circ; undergo nucleophilic substitution and elimination.
  • Reactivity toward substitution: R−I>R−Br>R−ClR-I>R-Br>R-Cl (bond strength C−I<C−Br<C−ClC-I<C-Br<C-Cl).
  • Haloarenes (e.g. chlorobenzene) are far less reactive (resonance gives partial double-bond character).
  • Reactions: aq. KOH →\to alcohol; alc. KOH →\to alkene; Na (Wurtz) →\to alkane; Mg →\to Grignard RMgXRMgX. Uses/hazards: CHCl3CHCl_3, CHI3CHI_3, DDT, freons (ozone-depleting).

Beyond-NCERT JEE Essentials

1. SN1 vs SN2 - the master comparison

SN2 (bimolecular): one concerted step. The nucleophile attacks the electrophilic carbon (delta+) from the side exactly opposite the leaving group (backside, 180 degrees); the C-Nu bond forms as C-X breaks, through a trigonal-bipyramidal transition state.

  • Kinetics: Rate = k[R-X][Nu], overall second order (first order in each).
  • Stereochemistry: inversion of configuration - the three spectator bonds flip like an umbrella in a gale (Walden inversion). A single enantiomer gives a single, inverted product.
  • Substrate order: methyl > primary > secondary > tertiary (bulky groups block the backside approach; a tertiary halide essentially does not react by SN2).

SN1 (unimolecular): two steps - slow ionisation of R-X to a carbocation + X- (rate-determining), then fast capture of the flat cation by the nucleophile.

  • Kinetics: Rate = k[R-X], first order; completely independent of the nucleophile's nature or concentration.
  • Stereochemistry: racemisation - the planar (sp2) carbocation is attacked from both faces, so an optically pure substrate gives a nearly 50:50 (+/-) racemate. (Beyond-NCERT: ion-pairing leaves a small excess of inversion, so racemisation is rarely 100 percent.)
  • Substrate order: tertiary > secondary > primary > methyl (order of carbocation stability). Allyl and benzyl rank high because resonance spreads the positive charge.

[JEE Tip] Read the substrate first: methyl/primary -> think SN2; tertiary -> think SN1/E1; secondary -> it depends on nucleophile and solvent (the genuine "could be either" case examiners love).

2. The three dials: nucleophile, solvent, leaving group

Nucleophile (matters only for SN2, since it sits in the rate law): a strong, concentrated, anionic nucleophile drives SN2; a weak, neutral one (H2O, ROH) lets SN1 win by default.

  • [JEE Tip] Nucleophilicity order FLIPS with the solvent. In polar protic solvents (H2O, ROH) the small F- is caged by hydrogen bonds, so the order is I- > Br- > Cl- > F-. In polar aprotic solvents (acetone, DMF, DMSO) the anion is naked and basicity takes over, reversing it to F- > Cl- > Br- > I-.

Solvent: polar protic solvents hydrogen-bond to and stabilise both the carbocation and the departing anion, so they favour SN1/E1 (and they cage the nucleophile, slowing SN2). Polar aprotic solvents solvate only the cation, freeing a hot, bare nucleophile - they favour SN2.

Leaving group (helps BOTH mechanisms): the weaker the base left behind, the better it leaves, so I- > Br- > Cl- >> F- (R-F is almost inert). This, together with falling C-X bond strength, sets the overall reactivity R-I > R-Br > R-Cl > R-F - the exact opposite of the electronegativity order.

3. Elimination E1/E2 and regiochemistry

E2 (bimolecular): concerted; a base removes a beta-H that is anti-periplanar to the leaving group as C-X breaks. Rate = k[R-X][base], second order; stereospecific (needs anti-periplanar geometry). E1 (unimolecular): the same slow ionisation to a carbocation as SN1, then a base removes a beta-H. Rate = k[R-X]; competes with SN1.

  • Saytzeff (Zaitsev): the MORE substituted (more stable) alkene predominates - normal with small bases (OH-, EtO-).
  • Hofmann: the LESS substituted (terminal) alkene predominates - forced by a bulky base (potassium tert-butoxide) or a bulky substrate.

Substitution vs elimination - how to call it:

  • Aqueous KOH -> substitution (alcohol); alcoholic KOH -> elimination (alkene). Same OH-, opposite roles set by the solvent - memorise this.
  • A strong nucleophile that is a weak base (I-, CN-, RS-) -> substitution; a strong, bulky base (tert-butoxide) -> elimination.
  • Raising the temperature always tips the balance toward elimination.
  • Primary -> mostly substitution; tertiary + base -> mostly elimination.

4. Which C-X reacts: the sp3 vs sp2 divide

  • Reactive (halogen on sp3 carbon): alkyl, allylic and benzylic halides undergo SN1/SN2/elimination.
  • [JEE Tip] Allylic (CH2=CH-CH2-X) and benzylic (C6H5-CH2-X) halides are the FASTEST SN1 substrates - the cation is resonance-delocalised onto the double bond/ring - and are good SN2 substrates too. Treat them as super-reactive.
  • Essentially unreactive (halogen on sp2 carbon): vinyl (CH2=CH-X) and aryl (C6H5-X) halides. Three reasons: (i) a halogen lone pair conjugates into the pi system, giving C-X partial double-bond character (a shorter, stronger bond); (ii) the sp2 carbon holds its electrons more tightly; (iii) the vinyl/aryl cation needed for SN1 is far too unstable, and the electron-rich pi cloud repels an incoming nucleophile, shutting down SN2.

5. Named reactions (know the reagent AND the product)

  • Finkelstein: R-Cl (or R-Br) + NaI -> R-I + NaCl, in dry acetone. Halide exchange driven by NaCl/NaBr precipitating out of acetone (Le Chatelier).
  • Swarts: R-X + AgF / Hg2F2 / SbF3 / CoF2 -> R-F. The standard route to alkyl fluorides.
  • Wurtz: 2 R-X + 2 Na -> R-R + 2 NaX (dry ether). A symmetrical alkane, carbon count doubles; a mixed pair gives a useless statistical mixture.
  • Wurtz-Fittig: Ar-X + R-X + 2 Na -> Ar-R (an alkylarene). e.g. C6H5Br + CH3Br -> toluene.
  • Fittig: 2 Ar-X + 2 Na -> Ar-Ar (a biaryl). e.g. 2 C6H5Br -> biphenyl.
  • Sandmeyer: Ar-N2+ X- + CuX/HX -> Ar-X (X = Cl, Br), or + CuCN -> Ar-CN; Cu(I) catalyses.
  • Gattermann: Ar-N2+ X- + Cu powder + HX -> Ar-X. The same Cl/Br products as Sandmeyer, cheaper Cu powder, lower yield.
  • Balz-Schiemann: Ar-N2+ Cl- + HBF4 -> Ar-N2+ BF4- (a solid), then heat -> Ar-F + N2 + BF3. The route to aryl FLUORIDES (Sandmeyer fails for F).
  • [JEE Tip] From a diazonium salt: F by Balz-Schiemann; Cl, Br by Sandmeyer/Gattermann; I needs NO copper - just warm with KI (Ar-N2+ + KI -> Ar-I + N2).
  • Dow process (industrial phenol): chlorobenzene + NaOH, 623 K, about 300 atm -> sodium phenoxide, then H+ -> phenol. The brutal conditions are the price of aryl-halide inertness.

6. Grignard reagents

Formation: R-X + Mg -> R-Mg-X (dry ether); reactivity R-I > R-Br > R-Cl; aryl/vinyl halides need THF. The carbon is carbanion-like (R is delta-, Mg is delta+), a strong nucleophile and base, so everything must be scrupulously anhydrous. On acidic work-up:

  • Any active-H compound (H2O, R-OH, R-COOH, N-H, terminal alkynes) -> R-H (alkane); this is the basis of the Zerewitinoff active-hydrogen estimation.
  • HCHO -> a primary alcohol; other RCHO -> a secondary alcohol; a ketone -> a tertiary alcohol.
  • CO2 (dry ice) -> R-COOH (adds one carbon).
  • Epoxide (oxirane) -> R-CH2CH2-OH (adds two carbons).

7. Optical activity, Walden inversion and activated haloarenes

  • A carbon carrying four different groups is a stereocentre, so the molecule is chiral and optically active (e.g. 2-bromobutane).
  • Walden inversion: SN2 on a chiral substrate inverts the configuration; SN1 racemises it. The SAME chiral bromide therefore gives an inverted product with a strong nucleophile in DMF (SN2) but a racemate in aqueous ethanol (SN1) - a favourite explain-the-stereochemistry question.
  • [JEE Tip] Aryl halides DO undergo nucleophilic substitution when strong electron-withdrawing groups sit ortho/para to the halogen. One o/p-NO2 (needs heat), two (2,4-dinitrochlorobenzene, mild) or three (picryl chloride -> picric acid, very easy) let NaOH or NH3 displace the halide by ADDITION-ELIMINATION (SNAr): the nucleophile adds first, the negative charge parks on the o/p-NO2 oxygen (the Meisenheimer intermediate), then X- leaves. A meta-NO2 does NOT activate - it cannot deliver the charge by resonance.
  • [JEE Tip] In this SNAr route the halide order REVERSES to F > Cl > Br > I, because C-F makes the ring carbon most electrophilic and C-X cleavage is not the slow step. (With no activating group, an unactivated aryl halide plus a very strong base such as NaNH2 instead reacts through a benzyne intermediate.)
  • [JEE Tip] Ambident-nucleophile trap: KCN gives the nitrile R-CN (attack through C) but AgCN gives the isocyanide R-NC (attack through N); likewise KNO2 gives the nitrite ester R-O-N=O while AgNO2 gives the nitroalkane R-NO2.

Worked Examples - Beyond-NCERT Essentials

Example 1 - Rank an SN2 series. Q: Arrange CH3Br, CH3CH2Br, (CH3)2CHBr and (CH3)3CBr in decreasing rate of reaction with KCN in DMF, and name the mechanism. KCN is a strong anionic nucleophile and DMF is polar aprotic - textbook SN2 conditions. The SN2 rate is decided by steric access to the backside carbon, so the least hindered reacts fastest. Bulk rises methyl < primary < secondary < tertiary, so the rate falls in the same order. Answer: CH3Br > CH3CH2Br > (CH3)2CHBr > (CH3)3CBr; mechanism SN2. The tertiary bromide is effectively inert to SN2 (with a base it would eliminate instead).

Example 2 - Same substrate, two fates (stereochemistry). Q: Predict the mechanism and stereochemistry when a single enantiomer of 2-bromobutane reacts (i) in aqueous ethanol with no added nucleophile, and (ii) with concentrated NaCN in DMSO. 2-Bromobutane is secondary, so the conditions decide. (i) A weak nucleophile (only the solvent) in a polar protic medium -> SN1: the planar secondary carbocation is attacked from both faces, giving racemic butan-2-ol (optically inactive). (ii) A strong nucleophile (CN-) in polar aprotic DMSO -> SN2: backside attack inverts the configuration (Walden), giving a single enantiomer of 2-methylbutanenitrile, CH3CH(CN)CH2CH3. Answer: (i) SN1, racemisation; (ii) SN2, inversion.

Example 3 - Most reactive toward a nucleophile (with a twist). Q: Among CH3CH2F, CH3CH2Cl, CH3CH2Br and CH3CH2I, which reacts fastest with a nucleophile? How does the order change for an activated aryl halide? For aliphatic SN1/SN2 the slow event is C-X breaking, governed by bond strength and leaving-group ability: C-I is weakest and I- is the best leaving group, so CH3CH2I is fastest and CH3CH2F almost inert - order R-I > R-Br > R-Cl > R-F. Beyond-NCERT twist: for nucleophilic AROMATIC substitution on an activated ring (e.g. 1-halo-2,4-dinitrobenzene + NH3) the nucleophile's ADDITION is rate-determining, and C-F makes the ring carbon most electrophilic, so the order REVERSES to F > Cl > Br > I. Answer: aliphatic - iodide fastest; activated aromatic - fluoride fastest.

Example 4 - Diazonium toolbox (Sandmeyer and friends). Q: Starting from aniline, give reagents for (a) chlorobenzene, (b) bromobenzene, (c) iodobenzene, (d) fluorobenzene, (e) benzonitrile. First diazotise: aniline + NaNO2/HCl at 273-278 K -> benzenediazonium chloride (C6H5N2+ Cl-). Then replace the -N2+ group: (a) CuCl/HCl (Sandmeyer) -> chlorobenzene. (b) CuBr/HBr (Sandmeyer) -> bromobenzene. (c) KI, warm - no copper needed -> iodobenzene. (d) HBF4, then heat the dry diazonium fluoroborate (Balz-Schiemann) -> fluorobenzene. (e) CuCN/KCN (Sandmeyer) -> benzonitrile (C6H5CN). Every step expels N2. (Cu powder + HX in place of CuX is the Gattermann variant for a and b.)

Example 5 - Grignard products. Q: Give the final product after acidic work-up when CH3CH2MgBr reacts with (a) HCHO, (b) CH3CHO, (c) propan-2-one, (d) CO2, (e) ethanol. The ethyl carbanion adds to the carbonyl carbon; work-up then protonates the alkoxide. (a) HCHO -> CH3CH2CH2OH (propan-1-ol, a primary alcohol). (b) CH3CHO -> CH3CH2CH(OH)CH3 (butan-2-ol, a secondary alcohol). (c) (CH3)2CO -> (CH3)2C(OH)CH2CH3 (2-methylbutan-2-ol, a tertiary alcohol). (d) CO2 then H+ -> CH3CH2COOH (propanoic acid; one carbon added). (e) ethanol - its O-H is acidic, so it simply protonates the reagent -> ethane (CH3CH3) + Mg(OEt)Br; the Grignard is wasted, no addition occurs. Answer: the aldehyde/ketone class controls the alcohol class, while an active-H solvent destroys the reagent.

Example 6 - Why chlorobenzene is unreactive. Q: Chlorobenzene is untouched by aqueous NaOH under ordinary conditions, yet benzyl chloride hydrolyses easily and 1-chloro-2,4-dinitrobenzene reacts with NaOH on gentle warming. Explain.

  • Chlorobenzene: Cl sits on an sp2 ring carbon. Its lone pair conjugates into the ring, giving C-Cl partial double-bond character (shorter, stronger); the sp2 carbon grips its electrons; an aryl cation is far too unstable for SN1; and the pi cloud blocks a backside SN2. Hence it yields only under the Dow process (623 K, 300 atm).
  • Benzyl chloride (C6H5CH2Cl): Cl is on an sp3 carbon and the benzylic cation is resonance-stabilised, so SN1 (and SN2) proceed easily to benzyl alcohol.
  • 1-Chloro-2,4-dinitrobenzene: the two o/p-NO2 groups withdraw electrons; OH- adds to give a carbanion whose charge is delocalised onto the nitro oxygens (the Meisenheimer intermediate), then Cl- leaves (addition-elimination). Product: 2,4-dinitrophenol.

Example 7 - Saytzeff vs Hofmann. Q: 2-Bromobutane undergoes E2 elimination. Which alkene dominates with (i) sodium ethoxide and (ii) potassium tert-butoxide? 2-Bromobutane, CH3-CHBr-CH2-CH3, has beta-H atoms on C1 (a terminal CH3 -> but-1-ene) and on C3 (a CH2 -> but-2-ene). (i) The small base ethoxide follows Saytzeff, giving the more substituted, more stable but-2-ene (mainly the trans isomer). (ii) The bulky base tert-butoxide cannot reach the crowded internal position easily, so it abstracts the accessible primary beta-H and follows Hofmann, giving but-1-ene as the major product. Answer: (i) but-2-ene (Saytzeff); (ii) but-1-ene (Hofmann).