JEE Mains Chemistry · Haloalkanes and Haloarenes

Preparation of Haloalkanes and Haloarenes

Haloalkanes come from alcohols (HX, PCl₅ or SOCl₂), from alkenes (HX, Markovnikov unless HBr meets a peroxide) and from C–H bonds by free-radical halogenation; haloarenes come from ring halogenation with a Lewis acid or from diazonium salts, never from phenol.

All Haloalkanes and Haloarenes notes JEE Mains Chemistry strategy

Why this matters

Eleven PYQs, one numerical, two from 2026. Seven follow a halogen onto an alcohol, an alkene or a C–H bond and ask where it lands, or what stops the reaction running backwards; four match named reactions to their reagents or products, or ask which routes can make an aryl halide.

Concept 1 of 2: Haloalkanes from alcohols, alkenes and hydrocarbons

There are three places a halogen can come from. It can replace the OH of an alcohol, it can add to a C=C together with a hydrogen, or it can replace a hydrogen on a C–H bond through radicals. In the first two the product is decided by the most stable carbocation; in the third it is decided by the most stable radical.

Definition

From alcohols: $\mathrm{R{-}OH + HX}$ . Tertiary alcohols react with conc. HCl at room temperature; primary and secondary need anhydrous $\mathrm{ZnCl_2}$ (Lucas reagent) and heat. Also $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ , $\mathrm{PBr_3}$ and $\mathrm{PI_3}$ (made from red P with $\mathrm{Br_2}$ or $\mathrm{I_2}$ ). $\mathrm{SOCl_2}$ is preferred because both by-products are gases.
The HX route goes through a carbocation, so an allylic alcohol can give a rearranged allylic halide.
A benzylic OH is replaced easily; a phenolic OH is not replaced at all, because the C–O bond of phenol has partial double-bond character.
From alkenes: HX adds by Markovnikov's rule, X going to the carbon that gives the more stable carbocation. HBr with a peroxide adds the other way (anti-Markovnikov, through a radical). $\mathrm{Br_2}$ in $\mathrm{CCl_4}$ gives a vicinal dibromide.
From C–H bonds: $\mathrm{Cl_2}$ or $\mathrm{Br_2}$ with light gives substitution through radicals. Bromine is selective for the 3° C–H; an allylic C–H is replaced rather than the C=C attacked.
Iodination is reversible, because HI reduces the alkyl iodide back. An oxidising agent such as $\mathrm{HIO_3}$ , $\mathrm{HNO_3}$ or $\mathrm{HIO_4}$ removes the HI.
On an arene: $\mathrm{Cl_2}$ with $\mathrm{FeCl_3}$ in the dark substitutes the ring (ortho and para); $\mathrm{Cl_2}$ with light or heat and no catalyst substitutes the side chain of toluene.

Alcohol to alkyl chloride with thionyl chloride; alcohol reactivity with HX

\mathrm{R{-}OH + SOCl_2 \longrightarrow R{-}Cl + SO_2\uparrow + HCl\uparrow} \qquad \text{reactivity with HX: } 3^\circ > 2^\circ > 1^\circ

Worked example

Give the major product when but-1-ene reacts with HBr (a) in the dark with no peroxide, (b) in the presence of benzoyl peroxide.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · 2021 Compilation Paper 16 · Q21Moderate

Example 1 · Haloalkanes and Haloarenes · Preparation of Haloalkanes and Haloarenes

CH_{4}+ I_{2}\rightleftharpoons CH_{3}- I + HI

Which reagent can stop backward reaction?

Only HBr shows the peroxide effect

A peroxide reverses the addition of HBr only. HCl and HI still add by Markovnikov's rule when a peroxide is present.

Phenol does not give an aryl halide with HX

The C–O bond of phenol has partial double-bond character and does not break. Aryl halides are made by ring halogenation or from diazonium salts. Phenol does not react violently with halogen acids either.

Light chlorinates the side chain, iron(III) chloride the ring

Toluene with

\mathrm{Cl_2}

and light gives benzyl chloride; with

\mathrm{Cl_2}

and

\mathrm{FeCl_3}

in the dark it gives 2- and 4-chlorotoluene.

Drill 6 more on haloalkanes from alcohols, alkenes and hydrocarbons

Concept 2 of 2: Named reactions that make or couple organic halides

Each named reaction is a fixed pair of reagent and change. Halogen exchange swaps one halogen for another (Finkelstein for iodides, Swarts for fluorides). Diazonium salts give aryl halides (Sandmeyer with copper(I) salts, Gattermann with copper powder). Sodium in dry ether couples aryl halides (Wurtz-Fittig and Fittig).

Definition

Finkelstein: $\mathrm{R{-}Cl}$ or $\mathrm{R{-}Br}$ + NaI in dry acetone → $\mathrm{R{-}I}$ . NaCl and NaBr do not dissolve in acetone, so they precipitate and pull the reaction forward.
Swarts: $\mathrm{R{-}Cl}$ or $\mathrm{R{-}Br}$ heated with a metal fluoride (AgF, $\mathrm{Hg_2F_2}$ , $\mathrm{CoF_2}$ or $\mathrm{SbF_3}$ ) → $\mathrm{R{-}F}$ .
Sandmeyer: $\mathrm{ArN_2^+}$ with $\mathrm{Cu_2Cl_2/HCl}$ → ArCl, with $\mathrm{Cu_2Br_2/HBr}$ → ArBr, with CuCN/KCN → ArCN.
Gattermann: $\mathrm{ArN_2^+}$ with copper powder and HCl or HBr → ArCl or ArBr.
$\mathrm{ArN_2^+}$ + KI → ArI, with no copper needed.
Wurtz-Fittig: ArX + RX + Na in dry ether → Ar–R. Fittig: 2 ArX + 2 Na in dry ether → Ar–Ar.

Reaction	Reagent	Change	Example
Finkelstein	NaI in dry acetone	R–Cl or R–Br → R–I	$\mathrm{CH_3CH_2Br \to CH_3CH_2I}$
Swarts	AgF, $\mathrm{Hg_2F_2}$ , $\mathrm{CoF_2}$ or $\mathrm{SbF_3}$	R–Cl or R–Br → R–F	$\mathrm{CH_3Br + AgF \to CH_3F}$
Sandmeyer	$\mathrm{Cu_2Cl_2/HCl}$ , $\mathrm{Cu_2Br_2/HBr}$ or CuCN/KCN	$\mathrm{ArN_2^+}$ → ArCl, ArBr or ArCN	Benzenediazonium chloride → chlorobenzene
Gattermann	Copper powder with HCl or HBr	$\mathrm{ArN_2^+}$ → ArCl or ArBr	Benzenediazonium chloride → bromobenzene
Iodide from a diazonium salt	KI (no copper)	$\mathrm{ArN_2^+}$ → ArI	Benzenediazonium chloride → iodobenzene
Wurtz-Fittig	Na in dry ether	ArX + RX → Ar–R	Chlorobenzene + methyl chloride → toluene
Fittig	Na in dry ether	2 ArX → Ar–Ar	Chlorobenzene → biphenyl

Sandmeyer uses copper(I) salts; Gattermann uses copper powder.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q40Moderate

Example 2 · Haloalkanes and Haloarenes · Preparation of Haloalkanes and Haloarenes

Match the List-I with List-II

	List-I Name of reaction		List-II Reagent or catalyst used
(A)	Finkelstein reaction	(I)	${SbF}_{3}$
(B)	Swarts reaction	(II)	Na, dry ether
(C)	Sandmeyer's reaction	(III)	NaI
(D)	Fittig reaction	(IV)	${Cu}_{2}{Cl}_{2}$

Choose the correct answer from the options given below :

Gattermann makes aryl chlorides and bromides, not cyanides

In the NCERT scheme the Gattermann reaction uses copper powder with HCl or HBr and gives ArCl or ArBr. The aryl cyanide comes from the Sandmeyer reaction with CuCN. A statement that both reactions give aryl cyanides is false by this scheme.

Finkelstein runs because the salt precipitates

NaI dissolves in dry acetone but NaCl and NaBr do not. Their precipitation removes a product and drives the exchange forward; in water the reaction would not go to completion.

Drill 3 more on named reactions that make or couple organic halides

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Haloalkanes from alcohols, alkenes and hydrocarbons
Alcohol to alkyl chloride with thionyl chloride; alcohol reactivity with HX
$\mathrm{R{-}OH + SOCl_2 \longrightarrow R{-}Cl + SO_2\uparrow + HCl\uparrow} \qquad \text{reactivity with HX: } 3^\circ > 2^\circ > 1^\circ$

Reference tables (1)

Named reactions that make or couple organic halides7 rows

Reaction	Reagent	Change	Example
Finkelstein	NaI in dry acetone	R–Cl or R–Br → R–I	$\mathrm{CH_3CH_2Br \to CH_3CH_2I}$
Swarts	AgF, $\mathrm{Hg_2F_2}$ , $\mathrm{CoF_2}$ or $\mathrm{SbF_3}$	R–Cl or R–Br → R–F	$\mathrm{CH_3Br + AgF \to CH_3F}$
Sandmeyer	$\mathrm{Cu_2Cl_2/HCl}$ , $\mathrm{Cu_2Br_2/HBr}$ or CuCN/KCN	$\mathrm{ArN_2^+}$ → ArCl, ArBr or ArCN	Benzenediazonium chloride → chlorobenzene
Gattermann	Copper powder with HCl or HBr	$\mathrm{ArN_2^+}$ → ArCl or ArBr	Benzenediazonium chloride → bromobenzene
Iodide from a diazonium salt	KI (no copper)	$\mathrm{ArN_2^+}$ → ArI	Benzenediazonium chloride → iodobenzene
Wurtz-Fittig	Na in dry ether	ArX + RX → Ar–R	Chlorobenzene + methyl chloride → toluene
Fittig	Na in dry ether	2 ArX → Ar–Ar	Chlorobenzene → biphenyl

Sandmeyer uses copper(I) salts; Gattermann uses copper powder.

Watch out for (5)

Only HBr shows the peroxide effect
→ Haloalkanes from alcohols, alkenes and hydrocarbons
Phenol does not give an aryl halide with HX
→ Haloalkanes from alcohols, alkenes and hydrocarbons
Light chlorinates the side chain, iron(III) chloride the ring
→ Haloalkanes from alcohols, alkenes and hydrocarbons
Gattermann makes aryl chlorides and bromides, not cyanides
→ Named reactions that make or couple organic halides
Finkelstein runs because the salt precipitates
→ Named reactions that make or couple organic halides

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