6. Haloalkanes and Haloarenes

What you will learn

• Classify C–X environments and name halo compounds.
• Choose preparation routes with attention to positional control.
• Connect physical properties with whole-molecule interactions.
• Compare substitution and elimination mechanisms.
• Use paired 2D/3D drawings to explain chirality, inversion and symmetry.
• Explain haloarene behaviour and contextualise important polyhalogen compounds.

Big idea: the carbon around a C–X bond decides which reaction routes are accessible.

A halogen can turn a hydrocarbon into a useful starting point for further transformations. Yet replacing a group at a crowded saturated carbon is different from replacing one on an aromatic ring. A meaningful prediction therefore starts with structure, not simply the halogen symbol.

Use the model pairs to track labelled groups through space. A flat formula counts attachments; wedges, dashed bonds and a rotatable model explain their alignment. In an exam, preserve that information in a clear, labelled 2D drawing.

The tetrahedral-carbon ideas associated with van ’t Hoff and Le Bel linked molecular connectivity with spatial arrangement. They helped explain why some compounds have mirror-related forms even when ordinary formula counting suggests only one structure. That connection remains central to stereochemical substitution mechanisms.

First locate the carbon bonded to the halogen. Its bonding environment often predicts reactivity better than the total number of carbon atoms.

A monohalo compound contains one halogen substituent; dihalo and polyhalo names indicate more. A molecule may also contain different halogens. The count does not tell you whether a halogen is bonded to a saturated carbon or an aromatic ring.

In a haloalkane, the relevant C–X bond usually involves an $sp^3$ carbon. A primary, secondary or tertiary label counts carbon groups attached directly to that carbon, not hydrogens and not halogens. A carbon bearing the halogen is primary if it has one carbon neighbour, secondary if it has two, and tertiary if it has three.

A vinylic halide has halogen directly attached to a double-bond carbon. A haloarene has halogen directly attached to an aromatic carbon. A benzylic halide is different from a haloarene even though both contain a benzene ring. This distinction becomes central to substitution reactions.

Classify benzyl bromide by the carbon directly attached to Br.

Follow the Br bond to the side-chain carbon. That carbon is saturated and lies beside the aromatic ring.

Result: It is benzylic, not an aryl halide. The presence of a ring alone does not make the C–Br bond aryl.

1. What is the difference between allylic and vinylic halides?
2. Does a tertiary haloalkane mean three halogen atoms are present?

Answers

1. Allylic halogen is on a saturated carbon beside a double bond; vinylic halogen is directly on a double-bond carbon.
2. No. It describes three carbon neighbours at the halogen-bearing carbon.

A structural name is a route back to the molecule. Choose the parent, number it carefully and preserve every position that distinguishes an isomer.

For a simple substituted alkane, use fluoro-, chloro-, bromo- or iodo- as prefixes with appropriate locants. Number the parent to obtain the correct lowest set of locants under the naming rules. List different substituent names alphabetically, while ordinary multiplicative prefixes do not control that ordering.

For an aromatic parent, ring positions distinguish constitutional isomers. The familiar ortho, meta and para descriptions correspond to relative positions, but numerical locants are widely useful in systematic names. Common names may be recognised, but they do not replace an unambiguous structural formula.

Name $\mathrm{CH_3CH_2CHClCH_3}$.

The longest chain contains four carbons. Number from the end nearer the chlorine-bearing carbon.

Result: 2-Chlorobutane. The locant is $2$, not three.

Draw the condensed structure of 1-bromo-3-methylbutane.

Draw the four-carbon parent first. Place Br at the first carbon and a methyl branch at the third. Finally check that each carbon has its correct valence.

Result: $\mathrm{BrCH_2CH_2CH(CH_3)CH_3}$.

A useful self-check is to read your structure from both ends. Does the selected numbering give the intended locants? Does a ring substituent attach directly to the ring or to a side chain? Are apparent repeated names actually the same structure viewed from a different end? These checks prevent false isomer counting.

1. Which name uses the shorter locant for CH₃CHBrCH₂CH₃?
2. What must be added to a basic ring formula to distinguish positional isomers?

Answers

1. 2-Bromobutane.
2. Substituent positions, expressed by suitable locants or an appropriate established positional name.

A carbon–halogen bond is polar, but bond polarity alone does not determine which haloalkane reacts fastest.

Halogens attract bonding electron density strongly. The carbon end of a C–X bond is therefore electron-poor relative to the halogen end. A nucleophile can attack a suitable carbon centre, while the halogen may leave with the bonding pair. Whether that route is fast depends on the structure and conditions.

Across comparable carbon–halogen bonds, bond length generally increases as halogen size increases. C–F is short and strong; C–I is longer and weaker. In many ordinary nucleophilic substitutions of alkyl halides, leaving-group behaviour commonly follows iodide better than bromide better than chloride, while fluoride is poor. This is not predicted by simply choosing the most polar bond.

An aryl C–X bond differs from an ordinary alkyl C–X bond. The bonded carbon is part of an aromatic system, and conjugative effects give the bond partial multiple-bond character in the familiar resonance description. The shorter, stronger bond and the mechanistic difficulties of substitution help explain lower reactivity under ordinary alkyl-halide conditions.

Why is “fluoroalkanes should substitute fastest because C–F is most polar” an incomplete argument?

A reacting molecule must overcome a pathway barrier. Bond strength, leaving-group stability, steric access, solvent and mechanism all affect that barrier.

Result: The strong C–F bond and poor ordinary leaving-group behaviour can outweigh the simple polarity argument.

Use dipole moment carefully too. A molecular dipole is a vector sum over the whole structure, not a direct reading of one bond’s polarity. Geometry can reinforce or cancel contributions. That is another reason to connect a flat drawing with a spatial model before making a property prediction.

1. Why can molecular geometry affect dipole moment?
2. What additional features should accompany a C–X polarity explanation of substitution?

Answers

1. Individual bond contributions add as vectors and may reinforce or cancel.
2. Bond strength, leaving-group ability, steric access, solvent and the proposed pathway.

A preparation method is useful only when it gives the desired skeleton and substitution pattern without an uncontrolled mixture.

Replacing an alcohol hydroxyl group with halogen gives a haloalkane under suitable reagent conditions. Hydrogen halides and selected phosphorus or sulfur halide reagents are classical routes. The substrate can affect mechanism, rearrangement and competing elimination. Do not assume every alcohol behaves by the same path.

$$\mathrm{CH_{3}CH_{2}CH_{2}OH} + \mathrm{HBr} \longrightarrow \mathrm{CH_{3}CH_{2}CH_{2}Br} + \mathrm{H_{2}O}$$

Free-radical halogenation of an alkane replaces hydrogen through a chain process. Initiation generates radicals, propagation carries the chain and termination consumes radicals. Multiple substitution and different possible hydrogen sites often create mixtures. A balanced overall equation does not display these competing products.

$$\mathrm{CH_{4}} + \mathrm{Cl_{2}} \xrightarrow{h\nu} \mathrm{CH_{3}Cl} + \mathrm{HCl}$$

Hydrogen halide addition across an unsymmetrical alkene needs regioselectivity. Under the ordinary ionic addition pattern, the product often follows the Markovnikov orientation, consistent with the more favourable intermediate. Peroxide-promoted radical addition is a special well-known case for HBr; it is not a universal reversal for every hydrogen halide.

$$\mathrm{CH_{3}CH=CH_{2}} + \mathrm{HBr} \longrightarrow \mathrm{CH_{3}CHBrCH_{3}}$$

A halide can be exchanged for another under favourable reaction and solvent conditions. In a Finkelstein-type conversion, a suitable alkyl chloride or bromide reacts with iodide; precipitation of the accompanying sodium salt in an appropriate solvent can help drive the change. A Swarts-type fluorination uses a suitable metal fluoride to replace another halogen.

$$\mathrm{CH_{3}CH_{2}CH_{2}Br} + \mathrm{NaI} \xrightarrow{\mathrm{acetone}} \mathrm{CH_{3}CH_{2}CH_{2}I} + \mathrm{NaBr}$$

$$\mathrm{CH_{3}CH_{2}Br} + \mathrm{AgF} \longrightarrow \mathrm{CH_{3}CH_{2}F} + \mathrm{AgBr}$$

You need one specific monohalogenated product from an unsymmetrical alkane. Why might direct radical substitution be a poor first choice?

Different hydrogen sites may react and the initial product can be further halogenated. Even a strong preference is not necessarily complete selectivity.

Result: A route from an appropriately placed functional group can offer better positional control than an uncontrolled radical mixture.

1. What feature can help drive a halogen-exchange equilibrium in a suitable solvent?
2. Why must rearrangement be considered for some alcohol-to-halide conversions?

Answers

1. Removal of a product, for example by precipitation of an accompanying salt.
2. A pathway involving a carbocation can allow the carbon framework or substitution pattern to rearrange.

Attaching halogen directly to an aromatic ring uses chemistry different from substituting a normal saturated carbon.

An aromatic ring can replace a ring hydrogen with halogen under suitable electrophilic substitution conditions. A Lewis-acid catalyst helps activate chlorine or bromine in familiar examples. This reaction preserves aromaticity in the product; it is not ordinary addition across one printed ring double bond.

$$\mathrm{C_{6}H_{6}} + \mathrm{Br_{2}} \xrightarrow{\mathrm{FeBr_3}} \mathrm{C_{6}H_{5}Br} + \mathrm{HBr}$$

Ring substituents already present affect rate and position. Distinguish ring halogenation from side-chain radical halogenation of an alkylbenzene: catalyst and irradiation conditions can change which site reacts. A structural formula plus conditions is more informative than saying “benzene derivative plus chlorine”.

An aromatic amine can first be converted into a diazonium salt under the appropriate cold acidic conditions. Replacement of the diazonium group then provides several aryl-halide routes. Sandmeyer-type conversions use suitable copper(I) reagents for selected replacements. Iodide can also replace the diazonium group in a familiar route without requiring the same copper reagent.

$$\mathrm{C_{6}H_{5}N_{2}Cl} \xrightarrow{\mathrm{CuCl/HCl}} \mathrm{C_{6}H_{5}Cl} + \mathrm{N_{2}}$$

$$\mathrm{C_{6}H_{5}N_{2}}^{+} + \mathrm{I}^{-} \longrightarrow \mathrm{C_{6}H_{5}I} + \mathrm{N_{2}}$$

Why can a diazonium route give positional control that direct ring halogenation may not provide?

The amino group already identifies the ring position to be transformed. Replacing its derived diazonium group keeps the substitution at that position rather than relying only on electrophilic orientation.

Result: It converts an already located functional group into a new substituent at the same ring carbon.

The detailed preparation and reaction scope of diazonium salts is developed in the amines chapter. Here the useful planning principle is functional-group replacement at a known aromatic position.

1. What leaves an aromatic diazonium group during these replacements?
2. Does aromatic halogenation normally mean adding halogen across one isolated double bond of the ring?

Answers

1. Nitrogen gas.
2. No. The familiar route is electrophilic aromatic substitution.

Physical properties reflect the whole molecular structure, including shape and the interactions that structure supports.

Halogen substitution changes mass, polarisability and dipole contributions. For closely comparable molecules, increased polarisability and stronger dispersion forces often raise boiling point. Branching can reduce effective contact between molecules and lower boiling point relative to a less-branched isomer. Avoid comparing unrelated structures using only molecular mass.

Melting is especially sensitive to crystal packing and symmetry. A more symmetrical positional isomer may pack differently and melt higher even if its boiling behaviour is less distinctive. Boiling and melting therefore need not follow the same simple ordering.

Many haloalkanes and haloarenes dissolve poorly in water because forming their new interactions with water does not sufficiently compensate for disrupting the water network. They can be more soluble in suitable organic solvents. “Contains a polar bond” does not automatically mean “freely soluble in water”.

Density depends on both mass and molecular packing. Heavily halogenated compounds can be denser than water, but it is not safe to state that every haloalkane lies below or above a particular density without identifying the compound.

Two haloalkane isomers have the same formula but different branching. Which one often has the lower boiling point, and why?

Compare the molecular shapes rather than their identical molecular masses. Greater branching often reduces the contact area that supports dispersion interactions.

Result: The more highly branched isomer often boils lower, for an otherwise comparable pair.

An exam property explanation should name the interaction and specify what is being compared. If measured values are given, use them. If they are not, state a justified trend and avoid presenting an exception-prone trend as a law.

1. Why can a symmetrical ring isomer have an unexpectedly high melting point?
2. Why is molecular mass alone insufficient for comparing boiling points?

Answers

1. Its crystal packing can be especially favourable.
2. Shape, polarisability and other intermolecular interactions also contribute.

Substitution exchanges a group; elimination creates a multiple bond. The conditions and molecular environment decide which route competes most effectively.

A nucleophile is an electron-pair donor. In substitution it forms a bond to a carbon as the leaving group departs. Different nucleophiles give different functional groups: hydroxide can give an alcohol, cyanide can extend the carbon skeleton as a nitrile, and alkoxide can give an ether. The full reaction conditions control which alternative dominates.

$$\mathrm{CH_{3}CH_{2}Br} + \mathrm{OH}^{-} \longrightarrow \mathrm{CH_{3}CH_{2}OH} + \mathrm{Br}^{-}$$

$$\mathrm{CH_{3}CH_{2}Br} + \mathrm{CN}^{-} \longrightarrow \mathrm{CH_{3}CH_{2}CN} + \mathrm{Br}^{-}$$

An $S_\mathrm N2$ event couples bond formation and bond breaking in one elementary step. The nucleophile approaches from the side opposite the leaving group. Steric crowding at the reacting carbon inhibits this approach. For a suitable simple case, rate depends on both substrate and nucleophile concentrations.

$$v=k[\mathrm{RX}][\mathrm{Nu^-}]$$

Backside attack reverses the labelled spatial arrangement at a stereogenic reacting centre. This is inversion. A carbon is stereogenic in the simplest tetrahedral case when its four attached groups differ. The R/S descriptor also depends on priority order, so a statement about inversion is more fundamental than automatically changing one letter to the other.

An $S_\mathrm N1$ model begins with a slow ionisation step that forms a carbocation and leaving group. The simple rate law is first order in substrate. A planar carbocation permits approach from more than one face. A chiral substrate can give substantial racemisation, though ion pairing and solvent effects can make the mixture depart from an exactly equal pair.

Carbocation stability and solvent stabilisation are central. Tertiary alkyl substrates commonly support this route better than simple primary ones. Allylic and benzylic systems can gain resonance stabilisation. Rearrangement becomes possible if the carbocation can form a more stable skeleton or position.

A beta carbon is adjacent to the carbon bearing the leaving group. Removal of a beta hydrogen and the leaving group can form an alkene. Heating with a suitable alcoholic base often favours elimination relative to an aqueous substitution condition. The substitution pattern, base size, solvent and temperature all matter.

$$\mathrm{CH_{3}CHBrCH_{3}} + \mathrm{KOH} \xrightarrow{\mathrm{alcohol},\Delta} \mathrm{CH_{3}CH=CH_{2}} + \mathrm{KBr} + \mathrm{H_{2}O}$$

For many ordinary eliminations, the more substituted alkene is favoured, but bulky bases and structural constraints can change selectivity. The broad preference is not permission to ignore a stated reagent or an accessible beta hydrogen.

Coupling with sodium can join suitable alkyl groups in a Wurtz reaction. A mixed pair can produce several coupled products, so it is not a clean universal route to an unsymmetrical alkane. Magnesium insertion gives a Grignard reagent under suitable dry conditions. Its carbon behaves as a strong nucleophilic/basic site; water destroys it by proton transfer.

$$2 \mathrm{CH_{3}CH_{2}Br} + 2 \mathrm{Na} \xrightarrow{\mathrm{dry\ ether}} \mathrm{CH_{3}CH_{2}CH_{2}CH_{3}} + 2 \mathrm{NaBr}$$

$$\mathrm{CH_{3}CH_{2}Br} + \mathrm{Mg} \xrightarrow{\mathrm{dry\ ether}} \mathrm{CH_{3}CH_{2}MgBr}$$

$$\mathrm{CH_{3}CH_{2}MgBr} + \mathrm{H_{2}O} \longrightarrow \mathrm{CH_{3}CH_{3}} + \mathrm{MgOHBr}$$

Ordinary backside substitution at an aromatic carbon is difficult, and a simple aryl carbocation is unfavourable for the ordinary unimolecular route. Haloarenes therefore resist many standard alkyl-halide substitution conditions. Electron-withdrawing groups at suitable ring positions or forcing conditions can enable other substitution pathways.

For chlorobenzene-to-phenol chemistry, strong conditions can give phenoxide followed by acidification. Nitro groups ortho or para to a leaving group can activate an addition–elimination route by stabilising a relevant intermediate; a meta group does not supply the same resonance connection.

$$\mathrm{C_{6}H_{5}Cl} + 2 \mathrm{NaOH} \xrightarrow{\mathrm{high\ temperature\ and\ pressure}} \mathrm{C_{6}H_{5}ONa} + \mathrm{NaCl} + \mathrm{H_{2}O}$$

Halogens on an aromatic ring are deactivating overall in electrophilic substitution but direct many incoming electrophiles towards ortho and para positions. Inductive withdrawal reduces overall reactivity, while resonance donation influences position. These are different effects, so “deactivating but ortho/para directing” is not a contradiction.

Aryl halides can also enter sodium-mediated coupling. Fittig coupling joins two aryl groups; Wurtz–Fittig coupling joins suitable aryl and alkyl groups. Product planning must again account for mixtures and competing combinations.

$$2 \mathrm{C_{6}H_{5}Br} + 2 \mathrm{Na} \xrightarrow{\mathrm{dry\ ether}} \mathrm{C_{12}H_{10}} + 2 \mathrm{NaBr}$$

A student wants to predict substitution solely from whether the substrate is primary or tertiary. What information is missing?

Nucleophile/base identity, solvent, temperature, leaving-group ability and resonance effects can alter the competition. A substrate label is useful but incomplete.

Result: Use a mechanism-based comparison that includes the stated conditions, rather than a one-word classification alone.

1. What feature makes an SN2 attack especially sensitive to crowding?
2. Why can an SN1 route rearrange?

Answers

1. The incoming nucleophile must reach the reacting carbon from the backside during the same step that the leaving bond breaks.
2. A carbocation intermediate can undergo a favourable rearrangement before nucleophile capture.

Multiple halogens can produce useful solvent properties and also persistence or toxicity. A practical use is not evidence that a substance is harmless.

Dichloromethane is a volatile chlorinated solvent. Its ability to dissolve certain organic materials explains several historical and industrial uses. Volatility also makes exposure through vapour important. In a chemical description, separate useful solvency from safety and environmental behaviour.

Trichloromethane, commonly called chloroform, is another chlorinated solvent. Its historical anaesthetic use is not a modern instruction for use. Oxidation under unsuitable storage conditions can form highly hazardous products. The chemistry illustrates that even an apparently stable solvent can change in air and light.

Triiodomethane, or iodoform, is associated with the characteristic yellow precipitate in a functional-group test. The useful observation is linked with a specific structural pattern and reaction sequence, developed in the carbonyl chapter. A positive test does not identify every detail of the original molecule.

Tetrachloromethane is a nonpolar molecule despite its polar individual bonds because a symmetrical tetrahedral arrangement cancels the bond-dipole contributions. Its toxicity and environmental behaviour limit the interpretation of older solvent uses. Structure and practical acceptability are different questions.

Some chlorofluorocarbons are sufficiently persistent to reach the upper atmosphere. Photochemical processes can then release chlorine-containing radicals that participate in ozone-destroying cycles. Their low ordinary reactivity in one environment does not imply environmental inertness everywhere.

The highly chlorinated insecticide DDT is an important historical example of useful short-term biological action combined with long-term persistence and accumulation concerns. Study the relationship between molecular persistence, fat affinity and environmental transport without treating a historical use as a recommendation.

How can a molecule with several polar bonds have no net dipole?

In a symmetric geometry, equal vector contributions can cancel. The conclusion depends on their spatial arrangement, not just their presence.

Result: The symmetric tetrachloromethane tetrahedron is an example; a less symmetric substitution pattern need not cancel.

1. Why can CCl₄ be nonpolar even though C–Cl bonds are polar?
2. Why should historical solvent or pesticide uses be described with context?

Answers

1. The symmetric tetrahedral bond-dipole vectors cancel.
2. Their usefulness does not erase toxicity, persistence or environmental consequences.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

6.1 Classify $\mathrm{CH_3CH_2CH_2Cl}$ and $\mathrm{CH_3CHClCH_3}$.

6.2 Distinguish benzyl chloride from chlorobenzene by the reacting carbon.

6.3 Name $\mathrm{CH_3CH_2CHBrCH_2CH_3}$.

6.4 Draw 1-chloro-2-methylpropane in condensed form.

6.5 Why is C–F polarity alone a poor predictor of ordinary substitution rate?

6.6 Write the expected principal ionic-addition product of HBr and propene without peroxide.

6.7 Why can radical chlorination give several products?

6.8 What chemical feature can drive a Finkelstein-type exchange in a suitable solvent?

6.9 Give one method of converting an aromatic amino position into an iodo position.

6.10 Why may branching lower boiling point among isomeric haloalkanes?

6.11 Explain why water solubility is not guaranteed by one polar C–X bond.

6.12 State the simple SN2 rate-law dependence.

6.13 State the key stereochemical feature of backside SN2 attack.

6.14 Why can an SN1 product mixture be substantially racemised?

6.15 Why is exact equal-face attack not guaranteed in every SN1 system?

6.16 Distinguish nucleophilic substitution from beta elimination.

6.17 Give the simple elimination product from $\mathrm{CH_3CHBrCH_3}$ with a suitable heated alcoholic base.

6.18 Why does a Grignard reagent fail to survive contact with water?

6.19 Predict the Wurtz product from two ethyl-halide groups.

6.20 Explain why a halogen can deactivate an aromatic ring but direct ortho/para.

6.21 Why does ordinary alkyl SN2 reasoning fail for a haloarene?

6.22 Connect tetrahedral symmetry with a physical property of tetrachloromethane.

Answers

6.1 Primary and secondary haloalkanes, respectively.

6.2 Benzyl chloride has a saturated side-chain C–Cl bond; chlorobenzene has Cl directly attached to an aromatic carbon.

6.3 3-Bromopentane.

6.4 $\mathrm{ClCH_2CH(CH_3)_2}$.

6.5 The strong bond and poor ordinary leaving-group behaviour also affect the pathway barrier.

6.6 $\mathrm{CH_3CHBrCH_3}$, under the usual Markovnikov addition model.

6.7 Different hydrogen sites and further substitution can both contribute.

6.8 Removal of an accompanying salt by precipitation can favour the exchange.

6.9 Form the corresponding aromatic diazonium salt under suitable conditions, then replace the diazonium group with iodide.

6.10 It often reduces effective intermolecular contact and dispersion attraction.

6.11 The full mixing energy and interactions must compensate for disrupting water–water interactions.

6.12 $v=k[\mathrm{RX}][\mathrm{Nu}]$ for the applicable elementary model.

6.13 Inversion of the labelled spatial arrangement at the reacting centre.

6.14 The planar carbocation can be attacked from more than one face.

6.15 Ion pairs and solvent organisation can favour one approach.

6.16 Substitution exchanges a group at a carbon; beta elimination removes groups from adjacent atoms to form a multiple bond.

6.17 Propene, $\mathrm{CH_3CH=CH_2}$.

6.18 Its basic carbon site is protonated, producing a hydrocarbon and a magnesium-containing product.

6.19 Butane, $\mathrm{CH_3CH_2CH_2CH_3}$.

6.20 Inductive withdrawal decreases overall reactivity, while resonance effects favour those positions in the substitution pathway.

6.21 Backside access and the aromatic carbon environment differ; the aryl bond has additional stabilisation and uses different substitution pathways.

6.22 Its symmetric bond arrangement cancels the molecular dipole, although individual C–Cl bonds remain polar.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

Why does a Grignard reagent fail to survive contact with water?

Its basic carbon site is protonated, producing a hydrocarbon and a magnesium-containing product.

Predict the Wurtz product from two ethyl-halide groups.

Butane, $\mathrm{CH_3CH_2CH_2CH_3}$.

Explain why a halogen can deactivate an aromatic ring but direct ortho/para.

Inductive withdrawal decreases overall reactivity, while resonance effects favour those positions in the substitution pathway.

Why does ordinary alkyl SN2 reasoning fail for a haloarene?

Backside access and the aromatic carbon environment differ; the aryl bond has additional stabilisation and uses different substitution pathways.

Connect tetrahedral symmetry with a physical property of tetrachloromethane.

Its symmetric bond arrangement cancels the molecular dipole, although individual C–Cl bonds remain polar.