Haloalkanes and Haloarenes

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 6.

Same C–Cl bond, two completely different personalities

Shake ethyl chloride with aqueous sodium hydroxide and it hydrolyses without fuss. Do the same with chlorobenzene and nothing happens, unless you heat it to 623 K at 300 atm. Both molecules contain a carbon–chlorine bond. The difference is the carbon it sits on, and that single idea runs through the whole chapter. A halogen on an sp³ carbon is a good leaving group waiting to be replaced. A halogen on an sp² carbon is held tightly and resists. Learn why, and the reactions, the mechanisms and most of the exceptions follow.

Classifying halogen compounds

Haloalkanes and haloarenes are hydrocarbons in which one or more hydrogens are replaced by a halogen X (F, Cl, Br or I). By the number of halogens they are mono-, di-, tri- or polyhalogen compounds. By the carbon that carries X, they fall into very different classes.

Two halogens on the same carbon make a gem-dihalide (CH₃CHCl₂). Two on neighbouring carbons make a vic-dihalide (ClCH₂CH₂Cl).

The first three classes are reactive in nucleophilic substitution, since X is on an sp³ carbon. The last two have C(sp²)–X and are not.

Naming them

In IUPAC names, the halogen is a prefix (fluoro, chloro, bromo, iodo) on the parent hydrocarbon. Number the longest chain so the substituents get the lowest locants, and list prefixes alphabetically. In benzene derivatives, the common o-, m- and p- become 1,2-, 1,3- and 1,4-.

Worked example: naming. Name CH₃–CH(C₂H₅)–CH₂–Br and CH₃CHClCH₂CH(CH₃)₂.

The common wrong step is to choose a chain that avoids the branch, giving "1-bromo-2-ethylpropane". The ethyl group extends the chain, so butane, not propane, is the parent.

The C–X bond: polar, longer and weaker as you go down

Halogens are more electronegative than carbon, so the carbon is slightly positive and X is slightly negative. That positive carbon is the target for attack by nucleophiles. Going down the group, the bond gets longer and weaker.

The dipole moment does not simply follow electronegativity, because dipole moment is charge multiplied by distance. Fluorine has the largest partial charge, but the C–F bond is shortest. The weak C–I bond is why iodides are the most reactive and fluorides the least.

Making haloalkanes and haloarenes

Balanced equations to know:

Beyond the textbook. For the exam: Finkelstein works because NaCl or NaBr precipitates from acetone. In reality: the real reason is solubility. NaI dissolves in dry acetone, while NaCl and NaBr hardly do. By Le Chatelier's principle the equilibrium keeps moving to the right as the salt leaves the solution. Why it matters: if the solvent were water, where all three salts dissolve, the reaction would not go to completion.

Physical properties and what causes them

Haloalkanes are polar, so molecules attract each other by dipole–dipole forces on top of London dispersion forces. They cannot form hydrogen bonds with water, which is why they dissolve poorly in water but well in organic solvents.

Under normal conditions CH₃Cl, CH₃Br, C₂H₅Cl and some chlorofluoromethanes are gases.

Nucleophilic substitution: the heart of the chapter

A nucleophile (Nu⁻, an electron pair donor) attacks the δ+ carbon and pushes out the halide, the leaving group:

R–X + Nu⁻ → R–Nu + X⁻

Because X is replaced by a nucleophile, many products can be made from one haloalkane.

Ambident nucleophiles. CN⁻ and NO₂⁻ can attack through either of two atoms. The ionic salts (KCN, KNO₂) release free ions, and the more nucleophilic atom, C in CN⁻ and O in NO₂⁻, attacks. The covalent salts (AgCN, AgNO₂) have the metal bonded to the more electronegative end, so the lone pair on the other atom, N in AgCN and N in AgNO₂, attacks. That is why KCN gives R–CN but AgCN gives R–NC, and KNO₂ gives R–O–N=O but AgNO₂ gives R–NO₂.

Pause and predict: ethyl bromide is warmed with AgCN. What is the main product?

Check your answer: ethyl isocyanide, C₂H₅NC. AgCN is covalent, so nitrogen attacks. Ethyl cyanide comes from KCN.

SN2: one step, back side attack

SN2 stands for substitution, nucleophilic, bimolecular. The nucleophile attacks the carbon from the side opposite the leaving group, while the halide leaves at the same time. The rate depends on both reactants, so the reaction is second order:

$$\text{rate} = k\,[\text{R-X}]\,[\text{Nu}^-]$$

Reactivity in SN2: methyl > 1° > 2° > 3°. The back side attack is blocked by bulky groups around the carbon. This is steric hindrance, and a tertiary halide is almost unreactive by SN2.

SN1: two steps, a carbocation in the middle

SN1 stands for substitution, nucleophilic, unimolecular. The slow first step is the halide leaving by itself. The nucleophile arrives afterwards. The rate depends only on the haloalkane, so the reaction is first order:

$$\text{rate} = k\,[\text{R-X}]$$

Reactivity in SN1: 3° > 2° > 1° > methyl. It follows the stability of the carbocation. Alkyl groups push electrons towards the positive carbon and stabilise it. Allylic and benzylic halides are fast in SN1 too, because their carbocations are stabilised by resonance. Polar protic solvents (water, alcohol, acetic acid) favour SN1 since they stabilise the ions.

The leaving group matters in both: R–I > R–Br > R–Cl >> R–F. The weaker the C–X bond and the more stable X⁻, the faster the reaction.

Pause and predict: tert-butyl bromide is warmed with water. Which mechanism operates, and what is the product?

Check your answer: SN1. Water is a polar protic solvent, and the tertiary carbocation is stable. The product is tert-butyl alcohol, 2-methylpropan-2-ol.

Worked example: ranking SN2 reactivity. Arrange in increasing order of SN2 reactivity: 1-bromobutane, 2-bromobutane, 2-bromo-2-methylpropane, 1-bromo-2-methylpropane.

Answer: 2-bromo-2-methylpropane < 2-bromobutane < 1-bromo-2-methylpropane < 1-bromobutane. The common wrong step is to rank all primary halides equally. A branch on the neighbouring carbon still slows the back side attack.

Stereochemistry: inversion and racemisation

A carbon with four different groups is a chiral (asymmetric) centre. A chiral molecule cannot be superimposed on its mirror image. The two mirror images are enantiomers, and they rotate plane polarised light by equal amounts in opposite directions: one dextrorotatory (+), the other laevorotatory (−). A 50:50 mixture of the two is a racemic mixture, written (±) or dl, and it has no net rotation. Turning one enantiomer into this mixture is racemisation.

Note that inversion in SN2 refers to the spatial arrangement. The sign of rotation (+ or −) can change or not, and cannot be predicted from the mechanism alone.

Beyond the textbook. For the exam: SN1 gives a racemic mixture, 50:50. In reality: most SN1 reactions give slightly more inversion than retention. The leaving group stays near the carbocation for a moment as an ion pair and shields the face it left from, so the nucleophile finds the back face easier. Why it matters: the textbook 50:50 is the limit for a completely free carbocation, and it explains why a question may say "mainly racemic".

Elimination: the other thing a haloalkane can do

Heat a haloalkane that has a hydrogen on a β carbon with alcoholic KOH and it loses HX to give an alkene. This is dehydrohalogenation, or β-elimination. When two alkenes are possible, the Saytzeff (Zaitsev) rule says the major product is the more substituted alkene, the one with more alkyl groups on the C=C, since it is more stable.

Substitution or elimination? Aqueous KOH favours substitution. Alcoholic KOH favours elimination. A strong and bulky base such as tert-butoxide removes a proton rather than attacking carbon.

Worked example: predicting products. Give the main organic product in each case.

Reactions with metals

Grignard reagents. A haloalkane reacts with magnesium in dry ether: R–X + Mg → R–MgX. The C–Mg bond is covalent but highly polar (carbon δ−, Mg δ+), so the carbon behaves like a carbanion and is a strong nucleophile and a strong base. It reacts violently with water, alcohols and amines to give the alkane, for example C₂H₅MgBr + H₂O → C₂H₆ + Mg(OH)Br, so everything must be perfectly dry.

Wurtz reaction. Two alkyl halides join in dry ether with sodium to give an alkane with double the carbon count: 2R–X + 2Na → R–R + 2NaX. For example 2CH₃Br + 2Na → C₂H₆ + 2NaBr.

Haloarenes: why the halogen won't leave

Now back to the opening puzzle. Chlorobenzene resists nucleophilic substitution for four reasons.

Under extreme conditions the reaction goes: C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O at 623 K and 300 atm, and acidification of the sodium phenoxide gives phenol.

An electron withdrawing nitro group changes this, but only if it is ortho or para. In the intermediate, the negative charge from the incoming nucleophile can be spread onto the carbon bearing NO₂, which stabilises it. From the meta position the nitro group cannot take part in this delocalisation. The more nitro groups at the ortho and para positions, the easier the reaction becomes, and the conditions mellow from drastic to warm water with three of them.

Electrophilic substitution goes the other way. A halogen is deactivating (its −I effect pulls electron density out of the ring, so reactions need harsher conditions than for benzene) but ortho and para directing (its lone pairs push electron density, by +R, to the ortho and para positions through resonance). Reactivity is controlled by −I, orientation by +R. Nitration, halogenation, sulphonation and Friedel–Crafts reactions of chlorobenzene give mainly the para product with some ortho.

Coupling with sodium. In dry ether:

Polyhalogen compounds

Freon 12 is made from carbon tetrachloride by the Swarts reaction: CCl₄ + 2HF → CCl₂F₂ + 2HCl (SbF₃ acts as the fluorinating agent or catalyst). Chloroform, not CCl₄, is the source of freon R-22: CHCl₃ + 2HF → CHClF₂ + 2HCl.

Beyond the textbook. For the exam: freons are stable, non-toxic gases that deplete the ozone layer. In reality: their stability is exactly the problem. They survive for decades, drift into the stratosphere, and there ultraviolet light splits off chlorine atoms. Each chlorine atom can destroy many thousands of ozone molecules, in a chain reaction: Cl• + O₃ → ClO• + O₂, followed by ClO• + O → Cl• + O₂. The Montreal Protocol of 1987 phased out CFCs. Why it matters: it is the standard example of a chemical with a perfect lab profile and a serious global cost.

Exam traps

Reaction sheet

60-second revision

Where this chapter connects

• Before this: Hydrocarbons gives the free radical halogenation and the addition of HX that make haloalkanes, and Organic Chemistry: Some Basic Principles and Techniques supplies the inductive effect, resonance and carbocation stability used throughout.
• Next: Alcohols, Phenols and Ethers is the other side of the same reactions: haloalkanes are made from alcohols and turn back into alcohols and ethers.
• Forward link: Amines uses the diazonium salts behind the Sandmeyer reaction, and Chemical Kinetics explains the order of reaction behind the SN1 and SN2 rate laws.