Preparation of Alkyl Halides (Halo Alkanes) & Aryl Halides (Halo Arenes)

Q: What is the best method to prepare alkyl chlorides from alcohols?

Heating the alcohol with thionyl chloride, SOCl2, usually with pyridine, is the best method. The by-products, sulfur dioxide and hydrogen chloride, are both gases and escape. The alkyl chloride is therefore left almost pure, with no separation step needed, unlike with PCl5, where liquid POCl3 must be removed.

Q: Which preparation reactions of haloalkanes are important for NEET?

For NEET, focus on alcohols with HCl and ZnCl2, PCl5 and SOCl2, Markovnikov and peroxide addition to alkenes, and the Finkelstein, Swarts and Sandmeyer reactions. NEET questions usually give a reagent and ask for the product, or name a reaction and ask for its reagents.

ChemistryHaloalkanes And HaloarenesFor JEE aspirants

Preparation of alkyl halides mostly starts from alcohols, alkanes or alkenes: the OH group is replaced by a halogen, a hydrogen is substituted by a halogen, or HX adds across a double bond. Halogen exchange (Finkelstein and Swarts reactions) and the Hunsdiecker reaction give particular halides. Aryl halides are made differently, by halogenating benzene with a Lewis acid or from diazonium salts. The preparation of alkyl halides and haloarenes is tested in JEE Main and NEET through named reactions, reagents and major products.

Key Formulas - Quick Reference

$R - OH + HCl anhyd. ZnC l_{2} R - Cl + H_{2} O$ (Groves process)
$R - OH + SOC l_{2} ⟶ R - Cl + S O_{2} ↑ + HCl ↑$ (best method)
$R - OH + PC l_{5} ⟶ R - Cl + POC l_{3} + HCl$ ; $3R - OH + P X_{3} ⟶ 3R - X + H_{3} P O_{3}$
$R - H + X_{2} h ν R - X + HX$ (gives a mixture)
Markovnikov: in HX addition, X goes to the double-bond carbon with fewer H atoms; with peroxide, HBr adds the opposite way
Finkelstein: $R - X + NaI dry acetone R - I + NaX$ (X = Cl, Br)
Swarts: $R - X + AgF ⟶ R - F + AgX$
Hunsdiecker: $RCOOAg + B r_{2} CC l_{4} R - Br + C O_{2} + AgBr$ (one carbon less)
Sandmeyer: $Ar N_{2}^{+} X^{-} C u_{2} X_{2} /HX Ar - X + N_{2}$ (X = Cl, Br)
Balz-Schiemann: $Ar N_{2}^{+} B F_{4}^{-} Δ Ar - F + B F_{3} + N_{2}$

1. Overview of Preparation Methods

Haloalkanes and haloarenes are made by different routes. Alkyl halides can be made by replacing the OH of an alcohol, by substituting a hydrogen of an alkane, by adding to an alkene, or by exchanging one halogen for another. Aryl halides cannot be made from phenols in the same way, so they come from benzene itself or from diazonium salts.

Figure 1: Where alkyl halides (R-X) and aryl halides (Ar-X) come from, with the key reagent for each route.

2. From Alcohols

Alcohols are the most common starting material. The OH group is replaced by a halogen atom using a hydrogen halide, a phosphorus halide or thionyl chloride.

2.1 Using hydrogen halides (HX)

R - OH + HX R - X + H_{2} O

HCl: primary and secondary alcohols need anhydrous $ZnC l_{2}$ as a catalyst (Groves process). Tertiary alcohols react with concentrated HCl at room temperature.
HBr: used as 48% HBr, or produced in the flask from NaBr and $H_{2} S O_{4}$ .
HI: produced in the flask from NaI or KI and 95% $H_{3} P O_{4}$ .
Reactivity order: of alcohols, 3° > 2° > 1°; of acids, HI > HBr > HCl.

C H_{3} C H_{2} OH + HCl anhyd. ZnC l_{2} C H_{3} C H_{2} Cl + H_{2} O

(C H_{3})_{3} C - OH + HCl conc., room temp. (C H_{3})_{3} C - Cl + H_{2} O

R - OH + NaBr + H_{2} S O_{4} Δ R - Br + NaHS O_{4} + H_{2} O

2.2 Using phosphorus halides

3R - OH + PC l_{3} 3R - Cl + H_{3} P O_{3}

R - OH + PC l_{5} R - Cl + POC l_{3} + HCl

For bromides and iodides, red phosphorus and bromine or iodine are used. They form $PB r_{3}$ or $P I_{3}$ in the flask, which then reacts with the alcohol:

2P + 3 X_{2} X = Br, I 2P X_{3}

3R - OH + P X_{3} 3R - X + H_{3} P O_{3}

2.3 Using thionyl chloride (the preferred method)

R - OH + SOC l_{2} pyridine R - Cl + S O_{2} ↑ + HCl ↑

Both by-products are gases and escape from the mixture, so the alkyl chloride is left almost pure.

Reagent	Product	By-products	Remark
HCl + anhyd. $ZnC l_{2}$	R-Cl	$H_{2} O$	Needed for 1° and 2° alcohols
$PC l_{3}$	R-Cl	$H_{3} P O_{3}$	Three alcohol molecules per $PC l_{3}$
$PC l_{5}$	R-Cl	$POC l_{3}$ , HCl	$POC l_{3}$ must be separated
Red P + $B r_{2}$ or $I_{2}$	R-Br, R-I	$H_{3} P O_{3}$	$P X_{3}$ formed in the flask
$SOC l_{2}$	R-Cl	$S O_{2}$ , HCl (gases)	Purest product, best method

Exam Trick

"Thionyl is tidy." With $SOC l_{2}$ , everything except the alkyl chloride leaves as gas, so no separation step is needed. That is why it is the answer to "best method to convert an alcohol to an alkyl chloride".

Phenols do not behave like alcohols here. The C-O bond in phenol has partial double-bond character due to resonance with the ring, so it is hard to break. Chlorobenzene cannot be made from phenol and HCl.

3. From Hydrocarbons

3.1 From alkanes: free radical halogenation

Alkanes react with halogens in the presence of light (or heat) to give alkyl halides:

R - H + C l_{2} h ν R - Cl + HCl

C H_{4} + C l_{2} h ν C H_{3} Cl + HCl

The alkyl halide formed reacts further with halogen to give di-, tri- and tetra-halogen compounds: $C H_{3} Cl$ $\to$ $C H_{2} C l_{2}$ $\to$ $CHC l_{3}$ $\to$ $CC l_{4}$ . Higher alkanes also give a mixture of isomers, so this method is poor for making one pure halide:

C H_{3} C H_{2} C H_{2} C H_{3} + C l_{2} h ν C H_{3} C H_{2} C H_{2} C H_{2} Cl + C H_{3} C H_{2} CH (Cl) C H_{3}

3.2 From alkenes: addition of hydrogen halides

Alkenes add halogen acids to give alkyl halides. For a symmetrical alkene there is only one product:

C H_{2} = C H_{2} + HCl C H_{3} - C H_{2} Cl

Markovnikov's rule: in the addition of HX to an unsymmetrical alkene, the negative part (X) attaches to the double-bond carbon that has fewer hydrogen atoms.

C H_{3} - CH = C H_{2} + HCl C H_{3} - CH (Cl) - C H_{3}

Peroxide (Kharasch) effect: when HBr is added in the presence of a peroxide, the addition goes against Markovnikov's rule. This is called anti-Markovnikov addition.

C H_{3} - CH = C H_{2} + HBr peroxide C H_{3} - C H_{2} - C H_{2} - Br

Both results follow from the same idea: the reaction goes through the more stable 2° intermediate. Without peroxide the first species to add is $H^{+}$ , giving a carbocation. With peroxide it is a bromine atom, giving a free radical.

Figure 2: HBr adds to propene through the more stable 2° intermediate in both cases. Without peroxide,

H^{+}

adds first (Markovnikov product); with peroxide,

Br \cdot

adds first (anti-Markovnikov product, Kharasch effect).

Why only HBr shows the peroxide effect: the H-Cl bond is too strong to be broken by the radical chain, and iodine atoms prefer to join into

I_{2}

rather than add to the double bond. So HCl and HI always add in the Markovnikov way. The reactivity of HX towards alkenes is HI > HBr > HCl.

3.3 From alkenes: addition of halogens

C H_{2} = C H_{2} + B r_{2} CC l_{4} BrC H_{2} - C H_{2} Br

The product is a vicinal dihalide (1,2-dibromoethane). The reddish-brown colour of bromine disappears during the reaction, which is used as a test for a C=C bond.

JEE Advanced

At high temperature, chlorine substitutes at the allylic position instead of adding to the double bond:

C H_{2} = CH - C H_{3} + C l_{2} 773 K C H_{2} = CH - C H_{2} Cl + HCl

In the lab, N-bromosuccinimide (NBS) is used to put Br on an allylic or benzylic carbon while leaving the C=C untouched.

4. Halogen Exchange Reactions

4.1 Finkelstein reaction

Alkyl chlorides and bromides react with sodium iodide in dry acetone to give alkyl iodides:

R - Cl + NaI dry acetone R - I + NaCl ↓

C H_{3} C H_{2} Br + NaI dry acetone C H_{3} C H_{2} I + NaBr ↓

The reaction works because NaI is soluble in acetone but NaCl and NaBr are not. They precipitate out, which removes a product and pushes the equilibrium forward (Le Chatelier's principle).

4.2 Swarts reaction

Alkyl fluorides are best made by heating an alkyl chloride or bromide with a metallic fluoride such as AgF, $H g_{2} F_{2}$ , $Co F_{2}$ or $Sb F_{3}$ :

C H_{3} Br + AgF C H_{3} F + AgBr

5. From Silver Salts of Carboxylic Acids (Hunsdiecker Reaction)

The silver salt of a carboxylic acid reacts with bromine in carbon tetrachloride to give an alkyl bromide. This is the Borodine-Hunsdiecker reaction.

RCOOAg + B r_{2} CC l_{4}, Δ R - Br + C O_{2} + AgBr

C H_{3} COOAg + B r_{2} CC l_{4} C H_{3} Br + C O_{2} + AgBr

Carbon dioxide is lost, so the alkyl bromide has one carbon atom less than the acid. The reaction works best for bromides.

6. Preparation of Aryl Halides (Haloarenes)

6.1 Electrophilic substitution of arenes

Benzene reacts with chlorine or bromine in the dark in the presence of a Lewis acid (a halogen carrier) such as iron, $FeC l_{3}$ or $FeB r_{3}$ :

C_{6} H_{6} + C l_{2} anhyd. FeC l_{3} dark, cold C_{6} H_{5} Cl + HCl

Toluene gives a mixture of o- and p-chlorotoluene, which are easy to separate because the para isomer has a much higher melting point.

Iodination is reversible because the HI formed reduces the product. It is carried out with an oxidising agent such as $HN O_{3}$ or $HI O_{4}$ , which destroys the HI.
Fluorination is too violent to control, so aryl fluorides are made from diazonium salts.

Exam Trick

"SSS and CCC." Sunlight, high temperature: Side-chain substitution. Cold, Catalyst: substitution in the ring (the Core). Toluene with $C l_{2}$ in sunlight gives benzyl chloride; with $C l_{2}$ and $FeC l_{3}$ in the dark it gives o- and p-chlorotoluene.

6.2 From diazonium salts

A primary aromatic amine is first converted to a diazonium salt by treating it with sodium nitrite and hydrochloric acid at 273-278 K (diazotisation):

C_{6} H_{5} N H_{2} + NaN O_{2} + 2HCl 273-278 K C_{6} H_{5} N_{2}^{+} C l^{-} + NaCl + 2 H_{2} O

The diazonium group is then replaced by a halogen, with loss of nitrogen gas:

Figure 3: One diazonium salt, five routes to aryl halides. Nitrogen gas is lost in every replacement step.

Reaction	Reagent	Product
Sandmeyer	$C u_{2} C l_{2}$ / HCl or $C u_{2} B r_{2}$ / HBr	ArCl or ArBr
Gattermann	Copper powder / HCl or HBr	ArCl or ArBr
Iodination	KI, warm (no copper needed)	ArI
Balz-Schiemann	$HB F_{4}$ , then heat the dry salt	ArF

C_{6} H_{5} N_{2}^{+} C l^{-} C u_{2} C l_{2} / HCl C_{6} H_{5} Cl + N_{2} ↑

C_{6} H_{5} N_{2}^{+} C l^{-} + KI warm C_{6} H_{5} I + KCl + N_{2} ↑

C_{6} H_{5} N_{2}^{+} C l^{-} + HB F_{4} C_{6} H_{5} N_{2}^{+} B F_{4}^{-} + HCl

C_{6} H_{5} N_{2}^{+} B F_{4}^{-} Δ C_{6} H_{5} F + B F_{3} + N_{2} ↑

Exam Trick

Sandmeyer uses Salts, Gattermann uses Grains. Sandmeyer needs copper(I) salts ( $C u_{2} X_{2}$ ); Gattermann needs copper powder with HX. Iodide needs neither.

7. Solved Examples

Solved Example 1

Dry gaseous hydrogen halides, and not their aqueous solutions, are used to prepare alkyl halides from alkenes. Explain.

Solution:

In water, HX ionises completely and the proton is present as $H_{3} O^{+}$ , which is a weaker electrophile than dry HX. So the addition is slower. Also, in aqueous solution water can act as a nucleophile and attack the carbocation, giving an alcohol instead of the alkyl halide.

Solved Example 2

What is the function of anhydrous

ZnC l_{2}

in the reaction of alcohols with concentrated HCl (Lucas reagent)?

Solution:

Anhydrous $ZnC l_{2}$ is a Lewis acid. It coordinates to the oxygen atom of R-OH, which weakens the C-O bond. The bond then breaks to give a carbocation, which combines with $C l^{-}$ . Anhydrous $ZnC l_{2}$ also acts as a dehydrating agent and helps the reaction go in the forward direction.

Solved Example 3

Iodine reacts with alcohols to form alkyl iodides only in the presence of phosphorus. Explain.

Solution:

Iodine alone cannot replace the OH group. Phosphorus reacts with iodine to form $P I_{3}$ , and it is $P I_{3}$ that replaces the OH group:

2P + 3 I_{2} 2P I_{3}

3R - OH + P I_{3} 3R - I + H_{3} P O_{3}

Solved Example 4

How would you prepare 1-iodopropane from (i) propene (ii) propan-1-ol?

Solution:

(i) Add HBr in the presence of peroxide to get 1-bromopropane, then use the Finkelstein reaction:

C H_{3} CH = C H_{2} + HBr peroxide C H_{3} C H_{2} C H_{2} Br

C H_{3} C H_{2} C H_{2} Br + NaI acetone reflux C H_{3} C H_{2} C H_{2} I + NaBr

Adding HI directly would give 2-iodopropane, because HI does not show the peroxide effect.

(ii) Dehydrate the alcohol to propene, then continue as in (i):

C H_{3} C H_{2} C H_{2} OH conc. H_{2} S O_{4} Δ C H_{3} CH = C H_{2}

C H_{3} CH = C H_{2} HBr, peroxide C H_{3} C H_{2} C H_{2} Br

C H_{3} C H_{2} C H_{2} Br NaI, acetone C H_{3} C H_{2} C H_{2} I

A shorter route is to heat propan-1-ol with red phosphorus and iodine, which gives 1-iodopropane in one step.

Solved Example 5

How is chlorobenzene obtained from (i) benzene (ii) benzenediazonium chloride?

Solution:

(i) By direct chlorination of benzene in the presence of a halogen carrier:

C_{6} H_{6} + C l_{2} FeC l_{3} C_{6} H_{5} Cl + HCl

(ii) By treating benzenediazonium chloride with cuprous chloride and HCl (Sandmeyer reaction):

C_{6} H_{5} N_{2}^{+} C l^{-} C u_{2} C l_{2} / HCl C_{6} H_{5} Cl + N_{2}

Solved Example 6

Give the major product when 2-methylpropene reacts with HBr (a) without peroxide (b) with peroxide.

Solution:

(a) $H^{+}$ adds to the $C H_{2}$ end, giving the 3° carbocation $(C H_{3})_{3} C^{+}$ . Bromide adds to it: the product is 2-bromo-2-methylpropane, $(C H_{3})_{3} CBr$ .

(b) A bromine atom adds to the $C H_{2}$ end, giving the 3° radical. It then takes H from HBr: the product is 1-bromo-2-methylpropane, $(C H_{3})_{2} CHC H_{2} Br$ .

Solved Example 7

Identify A and B:

C H_{3} C H_{2} COOAg + B r_{2}

(in

CC l_{4}

) gives A; A with NaI in dry acetone gives B.

Solution:

The Hunsdiecker reaction removes one carbon as $C O_{2}$ , so A is bromoethane, $C H_{3} C H_{2} Br$ . The Finkelstein reaction then swaps Br for I, so B is iodoethane, $C H_{3} C H_{2} I$ .

Solved Example 8

Why is thionyl chloride preferred over

PC l_{5}

for converting an alcohol into an alkyl chloride?

Solution:

With $SOC l_{2}$ , the by-products $S O_{2}$ and HCl are gases and escape, so the alkyl chloride is obtained pure. With $PC l_{5}$ , the by-product $POC l_{3}$ is a liquid (boiling point about 379 K) that must be separated by distillation.

Solved Example 9

What happens when toluene reacts with chlorine (a) in the dark with

FeC l_{3}

(b) in sunlight?

Solution:

(a) Ring substitution: a mixture of o-chlorotoluene and p-chlorotoluene.

(b) Side-chain substitution by free radicals: benzyl chloride, $C_{6} H_{5} C H_{2} Cl$ . With excess chlorine, benzal chloride ( $C_{6} H_{5} CHC l_{2}$ ) and benzotrichloride ( $C_{6} H_{5} CC l_{3}$ ) form.

Practice Questions

Write the equation for the Borodine-Hunsdiecker reaction.Answer: $RCOOAg + B r_{2}$ in $CC l_{4}$ gives $RBr + C O_{2} + AgBr$
Write the equation for the Finkelstein reaction.Answer: $R - Cl + NaI$ in dry acetone gives $R - I + NaCl$
Write the equations for the Balz-Schiemann reaction.Answer: $Ar N_{2}^{+} C l^{-} + HB F_{4}$ gives $Ar N_{2}^{+} B F_{4}^{-}$ ; on heating this gives $ArF + B F_{3} + N_{2}$

Common Mistakes to Avoid

Watch out

Applying the peroxide effect to HCl or HI. Only HBr adds anti-Markovnikov in the presence of peroxide.
Forgetting that the Hunsdiecker product has one carbon less than the acid.
Reversing the Finkelstein logic. NaI dissolves in acetone; NaCl and NaBr precipitate.
Trying to make chlorobenzene from phenol and HCl. The phenolic C-O bond is too strong.
Chlorinating toluene in sunlight when a ring-substituted product is wanted. Use $FeC l_{3}$ in the dark.
Mixing up Sandmeyer ( $C u_{2} X_{2}$ with HX) and Gattermann (copper powder with HX).
Adding a copper salt for iodobenzene. KI alone replaces the diazonium group.
Choosing free radical halogenation to make one pure alkyl halide. It gives a mixture.

Frequently Asked Questions

What is the best method to prepare alkyl chlorides from alcohols?

Heating the alcohol with thionyl chloride, $SOC l_{2}$ , usually with pyridine, is the best method. The by-products, sulfur dioxide and hydrogen chloride, are both gases and escape. The alkyl chloride is therefore left almost pure, with no separation step needed, unlike with $PC l_{5}$ , where liquid $POC l_{3}$ must be removed.

What is the Finkelstein reaction?

The Finkelstein reaction converts an alkyl chloride or bromide into an alkyl iodide by heating it with sodium iodide in dry acetone. It works because sodium iodide dissolves in acetone while sodium chloride and sodium bromide precipitate, which drives the equilibrium towards the alkyl iodide.

What is the Swarts reaction?

The Swarts reaction is used to prepare alkyl fluorides. An alkyl chloride or bromide is heated with a metallic fluoride such as AgF, $H g_{2} F_{2}$ , $Co F_{2}$ or $Sb F_{3}$ , and the halogen is exchanged for fluorine. For example, bromomethane with silver fluoride gives fluoromethane and silver bromide.

Why does only HBr show the peroxide effect?

In the presence of peroxide, HBr adds through a free radical chain that puts Br on the carbon with more hydrogens. The H-Cl bond is too strong to take part in this chain, and iodine atoms combine to form iodine instead of adding to the alkene. So HCl and HI always follow Markovnikov's rule.

Why can't aryl halides be prepared from phenols like alkyl halides from alcohols?

In phenol, the lone pair on oxygen is in resonance with the benzene ring. This gives the C-O bond partial double-bond character, making it shorter and stronger than in alcohols. Reagents such as HCl cannot break it, so aryl halides are made by halogenating benzene or from diazonium salts instead.

What is the difference between the Sandmeyer and Gattermann reactions?

Both replace the diazonium group of an arenediazonium salt with chlorine or bromine. The Sandmeyer reaction uses copper(I) chloride or bromide with the matching acid, while the Gattermann reaction uses copper powder with HCl or HBr. For iodobenzene no copper is needed, and potassium iodide alone is used.

Which preparation reactions of haloalkanes are important for NEET?

For NEET, focus on alcohols with HCl and $ZnC l_{2}$ , $PC l_{5}$ and $SOC l_{2}$ , Markovnikov and peroxide addition to alkenes, and the Finkelstein, Swarts and Sandmeyer reactions. NEET questions usually give a reagent and ask for the product, or name a reaction and ask for its reagents.

What should JEE Main aspirants focus on in the preparation of haloalkanes and haloarenes?

JEE Main often asks multi-step conversions, such as making 1-iodopropane from propene, and reagent-based product prediction. Master the peroxide effect, the loss of one carbon in the Hunsdiecker reaction, ring versus side-chain chlorination of toluene, and the diazonium routes to each aryl halide.

Previous year questions on Preparation of Alkyl Halides (Halo Alkanes) & Aryl Halides (Halo Arenes)

32 questions from past papers, each with a step-by-step solution.

Show all 32 questions

Ready to master Haloalkanes And Haloarenes?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test