Methods of Preparation of Haloalkanes

Why So Many Routes Exist

A haloalkane is, in essence, a hydrocarbon skeleton fitted with one reactive handle — the carbon–halogen bond. Because the halogen is more electronegative than carbon, that bond is polarised: the carbon carries a partial positive charge and the halogen a partial negative charge. This polarity is precisely what makes haloalkanes such versatile starting materials for substitution, elimination and organometallic chemistry, and it also explains why chemists keep several distinct routes to them in reserve.

NCERT opens its treatment with a practical note: alkyl halides are best prepared from alcohols, which are cheap and readily available. But alcohols are not the only handle nature provides. Alkanes can be halogenated directly, alkenes add hydrogen halides and halogens across their double bonds, and an existing haloalkane can have one halogen swapped for another. The four families below cover the entire NEET-relevant scope.

It helps to keep the underlying logic in view as you learn the equations. In the alcohol routes, a poor leaving group (–OH) is converted into a good one (–X); in the alkane route, a C–H bond is broken homolytically and replaced by C–X; in the alkene routes, a π bond is opened and the two new groups add across it; and in halogen exchange, a leaving-group equilibrium is pushed in the desired direction. Recognising which of these four mechanisms a question is testing usually makes the correct reagent obvious, even when the substrate looks unfamiliar.

From Alcohols

The hydroxyl group of an alcohol is replaced by a halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride. All three deliver the same transformation — $\ce{R-OH -> R-X}$ — but they differ in by-products, purity of the product and the conditions each substrate demands. NCERT §6.4.1 treats this as the headline method, and for good reason: it is the route most often probed in conversions and reagent-matching questions.

Alcohols with Halogen Acids (HX)

The simplest reagent is the halogen acid itself. Alkyl chlorides are made either by passing dry hydrogen chloride gas through a solution of the alcohol, or by heating the alcohol with concentrated aqueous HCl.

$$\ce{C2H5OH + HCl ->[\text{anhyd. } ZnCl2] C2H5Cl + H2O}$$

Primary and secondary alcohols are sluggish with HCl and need a catalyst — anhydrous zinc chloride, the active component of Lucas reagent (concentrated HCl + anhydrous ZnCl₂). The ZnCl₂ also absorbs the water formed, suppressing the reverse reaction. Tertiary alcohols, by contrast, react simply on shaking with concentrated HCl at room temperature, with no catalyst required. This difference in pace underlies the laboratory test of alcohol class.

For bromides, constant boiling with 48% HBr is used; NIOS notes that bromoethane is obtained by refluxing ethanol with HBr using a little concentrated H₂SO₄ as catalyst. Iodides are obtained in good yield by heating the alcohol with sodium or potassium iodide in 95% orthophosphoric acid.

$$\ce{C2H5OH + HBr ->[\text{conc. } H2SO4][\Delta] C2H5Br + H2O}$$

Phosphorus tribromide and triiodide are usually generated in situ by treating red phosphorus with bromine or iodine in the reaction mixture, which sidesteps the need to handle these unstable reagents directly. NCERT is explicit that the HX route does not extend to aryl halides: the carbon–oxygen bond in phenols has partial double-bond character and is too strong to break this way.

Phosphorus Halides and SOCl₂

Phosphorus halides convert alcohols to haloalkanes cleanly and are widely used in the laboratory. NIOS §25.2.1 gives the three standard equations:

$$\ce{3 C2H5OH + PCl3 -> 3 C2H5Cl + H3PO3}$$ $$\ce{C2H5OH + PCl5 -> C2H5Cl + POCl3 + HCl}$$ $$\ce{3 C2H5OH + PBr3 -> 3 C2H5Br + H3PO3}$$

The standout reagent, however, is thionyl chloride. NCERT singles it out: it is preferred because the alkyl chloride forms alongside two gaseous by-products, SO₂ and HCl, both of which escape from the flask. Because nothing contaminates the product, the reaction gives a pure alkyl halide without a separate purification step.

$$\ce{R-OH + SOCl2 -> R-Cl + SO2 ^ + HCl ^}$$

From Hydrocarbons

The second family starts from the hydrocarbon skeleton itself. NCERT §6.4.2 splits this into halogenation of alkanes and addition to alkenes — two very different beasts in terms of how clean a product they deliver.

Free-radical halogenation of alkanes

Direct chlorination or bromination of an alkane proceeds through a free-radical chain mechanism, initiated by sunlight or heat. NIOS gives chloroethane from ethane as the standard example:

$$\ce{CH3-CH3 + Cl2 ->[\text{sunlight}] CH3-CH2Cl + HCl}$$

The trouble, as NCERT stresses, is selectivity. Free-radical halogenation gives a complex mixture of isomeric mono- and polyhaloalkanes that is difficult to separate, so the yield of any single pure compound is low. The method is preparatively useful only when every replaceable hydrogen is equivalent — for instance, neopentane (2,2-dimethylpropane), which gives a single monochloride. NIOS adds two practical limits: direct iodination is reversible and so does not proceed, and direct fluorination is too violently exothermic to control.

From Alkenes — Addition

Alkenes offer a far cleaner route, and one with rich mechanistic content. There are two distinct additions.

(i) Addition of hydrogen halides. An alkene adds HCl, HBr or HI to give the corresponding alkyl halide. With a symmetrical alkene the product is unambiguous, but with an unsymmetrical alkene such as propene, two products are possible and one predominates according to Markovnikov's rule: the hydrogen attaches to the doubly bonded carbon already bearing the greater number of hydrogen atoms, and the halogen lands on the more substituted carbon.

$$\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}$$

The selectivity is governed by carbocation stability. Protonation of propene can give either a primary or a secondary carbocation; the secondary cation is more stable, so it forms preferentially, and the bromide adds to the carbon bearing the positive charge.

(ii) Addition of halogens. Adding bromine dissolved in CCl₄ across a double bond produces a colourless vic-dibromide. The disappearance of the reddish-brown bromine colour is a classic laboratory test for unsaturation.

$$\ce{CH2=CH2 + Br2 ->[CCl4] CH2Br-CH2Br}$$

The two halogen atoms add to adjacent carbons, which is why the product is described as vicinal (vic-). The reaction is fast, quantitative and easy to monitor by eye, so it doubles as both a synthesis of dihaloalkanes and a diagnostic test — a useful overlap to remember when a NEET question pairs a colour change with a structural deduction. In the 2016 paper, recognising that bromine addition across a C=C is an addition (not substitution or elimination) was the entire point of the question.

Halogen Exchange

The third family takes an existing haloalkane and swaps one halogen for another. Two named reactions cover the cases that direct methods handle badly: iodides and fluorides.

Finkelstein reaction. An alkyl chloride or bromide is heated with sodium iodide in dry acetone to give the alkyl iodide.

$$\ce{R-Cl + NaI ->[\text{dry acetone}] R-I + NaCl v}$$

The driving force is solubility. Sodium iodide dissolves in acetone, but the sodium chloride (or bromide) produced is insoluble and precipitates out. Removing a product continuously shifts the equilibrium forward by Le Chatelier's principle, so the exchange goes to completion even though the reaction is inherently reversible.

Swarts reaction. Alkyl fluorides are best made by heating an alkyl chloride or bromide with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃. This indirect route exists precisely because direct fluorination of hydrocarbons is uncontrollable.

$$\ce{CH3-Br + AgF -> CH3-F + AgBr}$$

The Swarts reaction is more than a textbook curiosity: it is the industrial route to the chlorofluorocarbon Freon-12 (CCl₂F₂), manufactured from tetrachloromethane by stepwise fluorine-for-chlorine exchange. Both Finkelstein and Swarts illustrate the same principle that runs through halogen exchange — a reaction that is thermodynamically marginal can be driven to completion by removing one product, whether by precipitation (NaCl in Finkelstein) or by forming an insoluble metal halide (AgBr in Swarts).

Selectivity and Reagent Choice

For NEET, the real skill is not reciting equations but choosing the right route for a target. The decisive factors are the halogen you want and the purity you need. The table below distils the practical logic NCERT and NIOS describe.

One bond-strength fact threads through the chapter and has appeared verbatim in a recent NEET paper. As the halogen gets larger down the group, the C–X bond lengthens and weakens, so C–X bond enthalpy falls in the order $\ce{CH3-F} > \ce{CH3-Cl} > \ce{CH3-Br} > \ce{CH3-I}$ (452, 351, 293 and 234 kJ mol⁻¹ respectively). The same trend explains why iodine is the best leaving group and why iodides are the most reactive haloalkanes in subsequent substitution chemistry.