Preparation of Haloarenes

Preparing haloalkanes from alcohols or hydrocarbons is relatively straightforward, but haloarenes pose a tougher challenge. The aromatic ring, with its stable delocalised electron cloud, resists change. You cannot simply swap out the hydroxyl group of a phenol the way you would with an alcohol, because resonance strengthens the $C-O$ bond in phenols far beyond what halogenating reagents can break. Two entirely different strategies are used instead: attacking the ring directly with a halogen, and routing through a reactive intermediate called a diazonium salt.

Direct Halogenation by Electrophilic Aromatic Substitution

The most straightforward approach is to react the arene with molecular chlorine ($Cl_2$) or bromine ($Br_2$) in the presence of a Lewis acid (an electron-pair acceptor that activates the halogen). Common Lewis acid catalysts for this reaction include metallic iron ($Fe$) and iron(III) chloride ($FeCl_3$).

The aromatic ring is electron-rich due to its delocalised $\pi$ system, but $Cl_2$ or $Br_2$ alone is not electrophilic enough to react with it. The Lewis acid polarises the halogen molecule, generating a species electrophilic enough to attack the ring. One hydrogen on the ring is then replaced by the halogen through electrophilic substitution (a reaction where an electron-seeking species replaces a hydrogen on the ring).

Consider toluene as an example:

The methyl group ($-CH_3$) is an ortho, para director: it steers the incoming electrophile toward the positions immediately adjacent to it (ortho) and the position directly across the ring (para). The reaction therefore yields two products:

• ortho-Halotoluene (halogen on a carbon next to the methyl group)
• para-Halotoluene (halogen on the carbon directly opposite the methyl group)

Separating the Two Products

These two isomers share the same molecular formula, but their crystal packing differs because of their different shapes. This gives them a large difference in melting points, making separation by fractional crystallisation simple and efficient.

Special Cases: Iodine and Fluorine

Chlorination and bromination of arenes work cleanly, but iodine and fluorine each bring complications:

• Iodination is reversible. The $HI$ formed as a by-product can react with the aryl iodide and regenerate the starting arene. To prevent this, an oxidising agent such as nitric acid ($HNO_3$) or periodic acid ($HIO_4$) is added to the reaction. The oxidising agent converts $HI$ into $I_2$ as fast as it forms, removing it from the equilibrium and pushing the reaction forward.

• Fluorination is impractical. Fluorine ($F_2$) is the most reactive halogen. It attacks aromatic rings so violently that the reaction cannot be controlled, producing a chaotic mixture of products rather than a clean aryl fluoride. Aryl fluorides must be prepared by other routes.

The Sandmeyer Reaction: Going Through a Diazonium Salt

When direct electrophilic substitution is unsuitable, or when a halogen needs to be placed at a specific ring position, the Sandmeyer reaction provides an elegant alternative. It starts from a primary aromatic amine (a compound with an $-NH_2$ group attached directly to the benzene ring) and proceeds through two steps.

Step 1: Forming the Diazonium Salt (Diazotisation)

The amine is dissolved or suspended in cold aqueous mineral acid, and sodium nitrite ($NaNO_2$) is added. The temperature must be kept between 273 and 278 K (0 to 5 degrees Celsius) using an ice bath. Under these conditions, the amino group is converted into a diazonium group ($-N_2^+$), producing a diazonium salt:

Keeping the temperature low is not optional. Diazonium salts (compounds containing the $-N_2^+$ group) are highly unstable and decompose rapidly if the solution warms up. The ice bath ensures the intermediate survives long enough for the next step.

Step 2: Replacing the Diazonium Group with a Halogen

The freshly prepared diazonium salt solution is mixed with cuprous chloride ($Cu_2Cl_2$) or cuprous bromide ($Cu_2Br_2$). The diazonium group departs as nitrogen gas ($N_2$), and the halogen from the cuprous salt takes its place on the ring:

The result is a clean aryl chloride or aryl bromide with nitrogen gas bubbling off as the only by-product.

Iodination Without Cuprous Halide

Introducing iodine through this route is even simpler. No cuprous halide is needed. The diazonium salt solution is shaken with potassium iodide ($KI$), and the iodide ion directly displaces the diazonium group:

This makes the Sandmeyer route particularly valuable for preparing aryl iodides, which are difficult to obtain by direct electrophilic iodination of the ring.

Why Sulfuric Acid Cannot Be Paired with Potassium Iodide

When converting alcohols to alkyl iodides (covered in the previous topic), the choice of acid matters. A natural thought is to use sulfuric acid ($H_2SO_4$), since it is readily available and strongly acidic. The problem is that $H_2SO_4$ is also a strong oxidising agent. It oxidises the hydrogen iodide ($HI$) formed during the reaction back into molecular iodine ($I_2$). This destroys the reagent you need, making the preparation unproductive.

Phosphoric acid ($H_3PO_4$) is used instead. It provides the acidic environment needed for the reaction without oxidising $HI$, allowing the alkyl iodide product to accumulate.

Counting Monochlorides: A Test of Structural Awareness

Free radical monochlorination replaces one hydrogen atom with chlorine. The number of distinct monochlorinated products depends entirely on how many structurally different types of hydrogen the starting alkane has. The three isomers of $C_5H_{12}$ illustrate this nicely:

• Neopentane (2,2-dimethylpropane, $C(CH_3)_4$): every one of its twelve hydrogen atoms sits in the same chemical environment. Replace any single hydrogen with chlorine and you get the same compound, 1-chloro-2,2-dimethylpropane. One monochloride.

• n-Pentane ($CH_3CH_2CH_2CH_2CH_3$): the molecule has a mirror plane through the central carbon. Hydrogens on $C_1$ and $C_5$ are equivalent, those on $C_2$ and $C_4$ are equivalent, and those on $C_3$ are unique. Three distinct hydrogen types give three monochlorides.

• Isopentane (2-methylbutane, $(CH_3)_2CHCH_2CH_3$): this branched isomer has four different hydrogen environments: the two equivalent methyl groups on the branch, the lone hydrogen on the branching carbon, the $-CH_2-$ group, and the terminal $-CH_3$. Four monochlorides.

The takeaway: symmetry reduces the number of unique products, while branching without symmetry increases it.

Worked Example 6.4: Predicting Addition Products

Problem: Write the products of the following reactions.

(i) Cyclohexene $+ \ HBr \longrightarrow$ ?

(ii) $CH_3CH_2CH{=}CH_2 + HCl \longrightarrow$ ?

(iii) $C_6H_5CH_2C(=CH_2)H + HBr \xrightarrow{\text{Peroxide}}$ ?

Solution:

(i) Cyclohexene + $HBr$

Cyclohexene is a symmetrical alkene: both carbons of the double bond carry identical substituents (each is bonded to a $-CH_2-$ group on either side). Because the two ends of the double bond are equivalent, Markovnikov’s rule does not come into play. $HBr$ simply adds across the double bond, placing $H$ on one carbon and $Br$ on the adjacent carbon.

Product: Bromocyclohexane.

(ii) $CH_3CH_2CH{=}CH_2 + HCl$

But-1-ene is an unsymmetrical alkene (the two carbons of the double bond carry different groups). Applying Markovnikov’s rule: the hydrogen from $HCl$ adds to the terminal $=CH_2$, which already has more hydrogen atoms. The chlorine goes to the internal carbon ($C_2$).

Product: 2-Chlorobutane ($CH_3CH_2CHClCH_3$).

(iii) With Peroxide Present

A peroxide switches the mechanism from ionic to free radical. In this pathway, the bromine radical adds first and preferentially attaches to the less substituted carbon of the double bond. This gives the anti-Markovnikov product, with bromine on the terminal carbon rather than the internal one.

Product: The terminal bromide, with $Br$ on the carbon at the end of the chain (anti-Markovnikov addition).