Chapter 6 · Haloalkanes and Haloarenes

6.1 Classification, Nomenclature and Preparation

Replace a hydrogen of a hydrocarbon by a halogen and you get a haloalkane or a haloarene. This section sorts them by the carbon that carries the halogen, names them by IUPAC rules, explains the polar C–X bond, and shows how they are made from alcohols, hydrocarbons, other halides and diazonium salts.

The anaesthetic halothane, $\mathrm{CF_{3}CHBrCl}$, put millions of patients safely to sleep during surgery. The antimalarial chloroquine and the antibiotic chloramphenicol carry chlorine atoms; the thyroid hormone thyroxine carries four iodine atoms, and without them it does not work. Water pipes and electrical insulation are made of PVC, a polymer of chloroethene. Teflon, the non-stick coating, is a fluorocarbon. On the darker side, the chlorofluorocarbons once used in every refrigerator thinned the ozone layer, and the insecticide DDT built up in food chains. Compounds in which a halogen atom is bonded to carbon are everywhere, and for good and bad they are among the most useful molecules chemists make.

They are also the workhorses of organic synthesis. A carbon–halogen bond is polar, so the carbon is open to attack, and the halogen leaves easily. In this chapter we will see that one alkyl halide can be turned into an alcohol, an ether, an amine, a nitrile, an alkene or a longer carbon chain. This first section lays the groundwork: how halogen compounds are classified and named, what the C–X bond is like, how they are prepared, and how their physical properties follow from their structure. (X stands for any halogen: F, Cl, Br or I.)

Classification: Which Carbon Carries the Halogen?

📘 Definition 6.1 — Haloalkanes and haloarenes

A haloalkane (alkyl halide) is a compound in which one or more halogen atoms replace hydrogen atoms of an aliphatic hydrocarbon, e.g. $\mathrm{CH_{3}CH_{2}Br}$. A haloarene (aryl halide) is a compound in which one or more halogen atoms are bonded directly to a carbon of an aromatic ring, e.g. $\mathrm{C_{6}H_{5}Cl}$.

By the number of halogen atoms they are mono-, di-, tri- or polyhalogen compounds. Dihaloalkanes with both halogens on the same carbon are geminal (gem-) dihalides, e.g. $\mathrm{CH_{3}CHCl_{2}}$; with the halogens on adjacent carbons they are vicinal (vic-) dihalides, e.g. $\mathrm{ClCH_{2}CH_{2}Cl}$.

For chemistry, the most useful classification looks at the single carbon atom that carries the halogen: is it $sp^{3}$ or $sp^{2}$ hybridised, and what is it bonded to? These details decide almost everything about how the compound reacts, as Sections 6.2 and 6.3 will show.

📘 Definition 6.2 — Classification by the carbon bonded to the halogen

(a) $\mathrm{C}(sp^{3}){-}\mathrm{X}$ compounds.

  • Alkyl halides $\mathrm{R{-}X}$: the halogen is on a saturated carbon of an alkyl group. They are primary (1°), secondary (2°) or tertiary (3°) when the carbon carrying X is bonded to one, two or three other carbon atoms. $\mathrm{CH_{3}X}$ (no other carbon) is a methyl halide.
  • Allylic halides: X is on an $sp^{3}$ carbon next to a $\mathrm{C{=}C}$ double bond (the allylic carbon), e.g. $\mathrm{CH_{2}{=}CH{-}CH_{2}{-}Cl}$.
  • Benzylic halides: X is on an $sp^{3}$ carbon attached directly to an aromatic ring, e.g. $\mathrm{C_{6}H_{5}CH_{2}Cl}$.

(b) $\mathrm{C}(sp^{2}){-}\mathrm{X}$ compounds.

  • Vinylic halides: X is on one of the carbons of a $\mathrm{C{=}C}$ double bond, e.g. $\mathrm{CH_{2}{=}CH{-}Cl}$.
  • Aryl halides (haloarenes): X is on a carbon of an aromatic ring, e.g. $\mathrm{C_{6}H_{5}Cl}$.
⚠️ Allylic is not vinylic, benzylic is not aryl
In an allylic or benzylic halide the halogen sits one carbon away from the double bond or ring, on an $sp^{3}$ carbon; these halides are very reactive. In a vinylic or aryl halide the halogen is on the $sp^{2}$ carbon itself; these are very unreactive towards substitution. Always find the carbon that carries X first, then look at its neighbours. Also note that $\mathrm{(CH_{3})_{3}CCH_{2}Br}$ is a primary halide although it contains a tertiary carbon: only the carbon bearing Br counts.
✏️ Example 1 — Classify the halide
Classify each compound by the carbon that carries the halogen: (a) $\mathrm{CH_{3}CH_{2}CHBrCH_{3}}$; (b) $\mathrm{CH_{2}{=}C(Cl)CH_{3}}$; (c) $\mathrm{C_{6}H_{5}CH_{2}CH_{2}Cl}$; (d) $\mathrm{CH_{3}CH{=}CHCH_{2}Br}$; (e) $\mathrm{(CH_{3})_{2}CICH_{2}CH_{3}}$; (f) $\mathrm{Cl{-}C_{6}H_{4}{-}CH_{2}Cl}$ (the two groups para on the ring).
(a) The $\mathrm{CHBr}$ carbon is bonded to $\mathrm{CH_{2}}$ and $\mathrm{CH_{3}}$, two carbons: secondary (2°) alkyl halide.
(b) Cl is on a doubly bonded carbon: vinylic halide.
(c) The $\mathrm{CH_{2}Cl}$ carbon is bonded to one carbon ($\mathrm{CH_{2}}$), and it is two carbons from the ring: primary alkyl halide, not benzylic.
(d) $\mathrm{CH_{2}Br}$ is $sp^{3}$ and attached to a $\mathrm{C{=}C}$ carbon: allylic halide (a primary one).
(e) The carbon carrying I is bonded to two $\mathrm{CH_{3}}$ groups and a $\mathrm{CH_{2}}$: three carbons, tertiary (3°).
(f) One Cl is on the ring (aryl) and the other on the $\mathrm{CH_{2}}$ attached to the ring (benzylic). Section 6.2 shows that only the benzylic Cl is replaced easily.

Nomenclature and the Carbon–Halogen Bond

Simple halogen compounds have common names built from the alkyl group plus "halide": methyl chloride, isopropyl bromide, tert-butyl chloride, allyl bromide, benzyl chloride. IUPAC names treat the halogen as a prefix on the parent hydrocarbon, so they are systematic and work for any structure.

📐 Rule 6.1 — IUPAC names of haloalkanes and haloarenes
  1. Halogens are named as prefixes fluoro-, chloro-, bromo-, iodo-, exactly like alkyl substituents: $\mathrm{CH_{3}CH_{2}Cl}$ is chloroethane.
  2. Choose the longest carbon chain (it must contain any $\mathrm{C{=}C}$). Number it so that the double bond, if present, gets the lowest locant; then give the lowest set of locants to all substituents, halogens and alkyl groups together.
  3. If two numberings give the same set of locants, the substituent cited first in alphabetical order gets the lower number. Write the prefixes in alphabetical order (di-, tri- are ignored when alphabetising).
  4. Haloarenes are named as substituted benzenes: chlorobenzene; for two substituents use locants $1,2$-, $1,3$-, $1,4$- (common prefixes o-, m-, p-): $1$-chloro-$4$-methylbenzene ($p$-chlorotoluene). Where the halogen is on a side chain, the side chain is a substituent: $\mathrm{C_{6}H_{5}CH_{2}Cl}$ is (chloromethyl)benzene (benzyl chloride).
  5. Gem- and vic-dihalides: $\mathrm{CH_{3}CHCl_{2}}$ is $1,1$-dichloroethane (ethylidene chloride); $\mathrm{ClCH_{2}CH_{2}Cl}$ is $1,2$-dichloroethane (ethylene dichloride). Polyhalomethanes: $\mathrm{CH_{2}Cl_{2}}$ dichloromethane (methylene chloride), $\mathrm{CHCl_{3}}$ trichloromethane (chloroform), $\mathrm{CHI_{3}}$ triiodomethane (iodoform), $\mathrm{CCl_{4}}$ tetrachloromethane (carbon tetrachloride).
✏️ Example 2 — Naming halogen compounds
Give IUPAC names: (a) $\mathrm{(CH_{3})_{2}CHCHClCH_{3}}$; (b) $\mathrm{CH_{3}CH_{2}C(CH_{3})_{2}CH_{2}I}$; (c) $\mathrm{CH_{3}CHBrCH_{2}CHClCH_{3}}$; (d) $\mathrm{CH_{2}{=}CHCH_{2}Br}$; (e) $\mathrm{(CH_{3})_{3}CCH_{2}Br}$; (f) $o$-$\mathrm{BrC_{6}H_{4}CH_{3}}$.
(a) Chain of four: $\mathrm{CH_{3}{-}CH(CH_{3}){-}CHCl{-}CH_{3}}$. Either end gives locants $\{2,3\}$, so alphabetical order decides: chloro before methyl. 2-Chloro-3-methylbutane.
(b) Chain $\mathrm{ICH_{2}{-}C(CH_{3})_{2}{-}CH_{2}{-}CH_{3}}$: numbering from $\mathrm{CH_{2}I}$ gives $\{1,2,2\}$, from the other end $\{3,3,4\}$. 1-Iodo-2,2-dimethylbutane.
(c) Pentane with halogens at 2 and 4 from either end; bromo comes first alphabetically. 2-Bromo-4-chloropentane.
(d) The double bond gets locant 1, so Br is on C3. 3-Bromoprop-1-ene (allyl bromide).
(e) Longest chain three carbons with two methyls on C2. 1-Bromo-2,2-dimethylpropane (neopentyl bromide), a 1° halide.
(f) Substituents on C1 and C2 of the ring; bromo first alphabetically. 1-Bromo-2-methylbenzene ($o$-bromotoluene).
✏️ Example 3 — All the isomers of C₄H₉Br
Draw and name every structural isomer of $\mathrm{C_{4}H_{9}Br}$ and classify each as 1°, 2° or 3°. Which one has a carbon bonded to four different groups?
Skeletons Four carbons form either a straight chain (butane) or a branched one (2-methylpropane). Put Br on each different kind of carbon.
Butane $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}Br}$ 1-bromobutane (1°); $\ \mathrm{CH_{3}CH_{2}CHBrCH_{3}}$ 2-bromobutane (2°).
2-Methylpropane $\mathrm{(CH_{3})_{2}CHCH_{2}Br}$ 1-bromo-2-methylpropane (1°); $\ \mathrm{(CH_{3})_{3}CBr}$ 2-bromo-2-methylpropane (3°).
Answer Four structural isomers. In 2-bromobutane C2 carries $\mathrm{H}$, $\mathrm{Br}$, $\mathrm{CH_{3}}$ and $\mathrm{C_{2}H_{5}}$, four different groups. Section 6.2 shows that such a molecule exists as two mirror-image forms.

The C–X bond. Halogens are more electronegative than carbon, so the C–X bond is polar: $\mathrm{\overset{\delta+}{C}{-}\overset{\delta-}{X}}$. The carbon carries a partial positive charge and is attacked by electron-rich reagents (nucleophiles, Section 6.2). Down the group the halogen atom gets larger, so the bond gets longer and weaker.

📐 Rule 6.2 — Trends in the C–X bond of CH₃–X
BondC–FC–ClC–BrC–I
bond length / pm139178193214
$\mathrm{C{-}X}$ bond enthalpy / $\mathrm{kJ\,mol^{-1}}$452351293234
dipole moment of $\mathrm{CH_{3}X}$ / D1.8471.8601.8301.636

Bond length increases and bond enthalpy decreases from F to I, so the C–I bond is the easiest to break: alkyl iodides are the most reactive alkyl halides and alkyl fluorides the least. The dipole moment, charge × distance, does not follow electronegativity alone: the C–Cl bond has a smaller charge separation than C–F but a much longer bond, so $\mathrm{CH_{3}Cl}$ has the largest dipole moment.

Interactive 1 — Halide Sorter: Find the Carbon, Then Classify

A structure is drawn in skeletal form. Find the carbon carrying the halogen, look at its hybridisation and its neighbours, and press the class you think it is. The lab then rings the C–X carbon in red, marks any neighbouring $sp^{2}$ carbon in green and counts the carbons attached to it. Next gives a new compound; Play runs through all sixteen with the answers shown.

IUPAC name—
formula · molar mass—
C bearing X is—
carbons on that C—
class—
your score0 / 0

Preparing Haloalkanes

Haloalkanes are made from three kinds of starting material: alcohols (the commonest laboratory route), hydrocarbons, and other haloalkanes by exchanging one halogen for another.

From alcohols. The $\mathrm{{-}OH}$ group of an alcohol is replaced by a halogen. Hydroxide is a poor leaving group, so every reagent first converts $\mathrm{{-}OH}$ into something that leaves more easily (protonated $\mathrm{{-}OH_{2}^{+}}$, or an $\mathrm{O{-}P}$ or $\mathrm{O{-}S}$ group).

📐 Rule 6.3 — Converting alcohols into haloalkanes
$$\mathrm{R{-}OH+HCl\xrightarrow{\text{anhyd. }ZnCl_{2}}R{-}Cl+H_{2}O}$$ $$\mathrm{R{-}OH+HBr\rightarrow R{-}Br+H_{2}O}$$ $$\mathrm{R{-}OH+HI\rightarrow R{-}I+H_{2}O}$$ $$\mathrm{3R{-}OH+PX_{3}\rightarrow 3R{-}X+H_{3}PO_{3}}\quad(\mathrm{X=Cl,\,Br,\,I})$$ $$\mathrm{R{-}OH+PCl_{5}\rightarrow R{-}Cl+POCl_{3}+HCl}$$ $$\mathrm{R{-}OH+SOCl_{2}\rightarrow R{-}Cl+SO_{2}\uparrow+HCl\uparrow}$$
  • Reactivity of alcohols towards $\mathrm{HX}$: $3^{\circ}\gt2^{\circ}\gt1^{\circ}$. Tertiary alcohols react with concentrated $\mathrm{HCl}$ at room temperature just by shaking; primary and secondary alcohols need anhydrous $\mathrm{ZnCl_{2}}$ (a Lewis acid that helps the $\mathrm{C{-}O}$ bond break).
  • $\mathrm{HBr}$ is generated from $\mathrm{NaBr}$ and concentrated $\mathrm{H_{2}SO_{4}}$; $\mathrm{HI}$ from $\mathrm{KI}$ and $95\%$ $\mathrm{H_{3}PO_{4}}$ (sulphuric acid would oxidise $\mathrm{HI}$ to $\mathrm{I_{2}}$).
  • $\mathrm{PBr_{3}}$ and $\mathrm{PI_{3}}$ are made in situ from red phosphorus with bromine or iodine.
  • Thionyl chloride is preferred: both by-products are gases that escape, leaving almost pure alkyl chloride.
  • These methods do not make aryl halides from phenols: the $\mathrm{C{-}O}$ bond of a phenol has partial double-bond character and is hard to break.
✏️ Example 4 — Choosing the reagent
(a) Write a balanced equation for making chloroethane from ethanol in the way that gives the purest product, and explain the choice. (b) How many moles of ethanol react with one mole of $\mathrm{PBr_{3}}$? (c) 2-Methylpropan-2-ol turns cloudy almost at once when shaken with concentrated $\mathrm{HCl}$, but butan-1-ol does not. Explain, and say how butan-1-ol can be converted to 1-chlorobutane with $\mathrm{HCl}$.
(a) $$\mathrm{CH_{3}CH_{2}OH+SOCl_{2}\rightarrow CH_{3}CH_{2}Cl+SO_{2}+HCl}$$ $\mathrm{SO_{2}}$ and $\mathrm{HCl}$ are gases and leave the mixture, so nothing needs to be separated from the product. Check: C 2 = 2, H 6 = 5 + 1, O 2 = 2, S 1 = 1, Cl 2 = 1 + 1 ✓.
(b) $\mathrm{3CH_{3}CH_{2}OH+PBr_{3}\rightarrow3CH_{3}CH_{2}Br+H_{3}PO_{3}}$: three moles. Each P–Br bond delivers one bromine.
(c) The tertiary alcohol gives a tertiary carbocation easily in strong acid, so the chloride forms fast and, being insoluble, makes the solution cloudy. A primary alcohol cannot form a stable carbocation; it needs anhydrous $\mathrm{ZnCl_{2}}$ with concentrated $\mathrm{HCl}$ and heating: $\mathrm{CH_{3}(CH_{2})_{3}OH+HCl}$ $\mathrm{\xrightarrow{ZnCl_{2},\ \Delta}CH_{3}(CH_{2})_{3}Cl+H_{2}O}$.

From hydrocarbons. Alkanes react with chlorine or bromine in light or on heating by a free-radical chain mechanism (Class 11 Chemistry, Hydrocarbons). Any hydrogen can be replaced, so a mixture of isomeric monohalides results, together with some di- and polyhalides; the method is useful only when all the hydrogens are equivalent or the products can be separated. Alkenes add hydrogen halides and halogens across the double bond:

$$\mathrm{CH_{3}CH{=}CH_{2}+HBr\rightarrow CH_{3}CHBrCH_{3}}\quad(\text{Markovnikov})$$ $$\mathrm{CH_{3}CH{=}CH_{2}+HBr\xrightarrow{\text{peroxide}}CH_{3}CH_{2}CH_{2}Br}$$ $$\mathrm{CH_{2}{=}CH_{2}+Br_{2}\xrightarrow{CCl_{4}}BrCH_{2}CH_{2}Br}$$

In Markovnikov addition the hydrogen goes to the double-bond carbon that already has more hydrogens, because the more stable carbocation forms; with peroxides, $\mathrm{HBr}$ (only) adds the other way. Adding bromine gives a vic-dibromide; the red-brown colour of bromine disappears, which is the usual test for a $\mathrm{C{=}C}$ bond.

✏️ Example 5 — Halides from hydrocarbons
(a) Butane is chlorinated in light. Name the monochloro products. In chlorination at room temperature, a secondary hydrogen is replaced about $3.8$ times as fast as a primary one (an approximate experimental figure). Estimate the percentage of each product. (b) Give the major product of propene with $\mathrm{HBr}$, without and with benzoyl peroxide. (c) What forms when but-2-ene is treated with bromine in $\mathrm{CCl_{4}}$? Is it a gem- or vic-dihalide?
(a) Products Butane has two kinds of H: six primary (on the two $\mathrm{CH_{3}}$) giving 1-chlorobutane, and four secondary (on the two $\mathrm{CH_{2}}$) giving 2-chlorobutane.
(a) Ratio Relative amounts $=$ (number of H) $\times$ (reactivity per H): primary $6\times1=6$; secondary $4\times3.8=15.2$. Total $21.2$, so 1-chlorobutane $\tfrac{6}{21.2}=28\%$ and 2-chlorobutane $\tfrac{15.2}{21.2}=72\%$. A mixture is unavoidable.
(b) Without peroxide: 2-bromopropane (Markovnikov, via the secondary carbocation). With peroxide: 1-bromopropane (anti-Markovnikov, free-radical addition).
(c) $\mathrm{CH_{3}CH{=}CHCH_{3}+Br_{2}\rightarrow CH_{3}CHBrCHBrCH_{3}}$, 2,3-dibromobutane: the bromines are on adjacent carbons, a vic-dihalide.
Interactive 2 — Free-Radical Chlorination: Which Product, and How Much?

Choose an alkane. Its carbons are coloured by the kind of hydrogen they carry: primary, secondary or tertiary, with the number of each. Press Play: chlorine atoms strike the molecule many times, each strike removing one hydrogen, and every hit is counted for the product it gives. The coloured bars (simulated shares) grow towards the black marks (the prediction, (number of H) × (reactivity per H)). Switch the reactivities off to see what pure counting would give. The relative reactivities $1:3.8:5.0$ (1° : 2° : 3°) are approximate room-temperature figures.

kinds of H—
monochloro isomers—
Cl atoms fired—
major product (predicted)—

Halogen exchange. Alkyl iodides and fluorides are hard to make directly (iodination of alkanes is reversible; fluorination is explosive), so they are usually made from alkyl chlorides or bromides by swapping the halogen.

📘 Definition 6.3 — Finkelstein and Swarts reactions

The Finkelstein reaction converts an alkyl chloride or bromide into an alkyl iodide by heating with sodium iodide in dry acetone:

$$\mathrm{R{-}X+NaI\xrightarrow{\text{dry acetone}}R{-}I+NaX}\quad(\mathrm{X=Cl,\,Br})$$

The reaction is reversible, but $\mathrm{NaCl}$ and $\mathrm{NaBr}$ are almost insoluble in acetone while $\mathrm{NaI}$ dissolves; the precipitate removes $\mathrm{NaX}$ and, by Le Chatelier's principle, drives the reaction forward.

The Swarts reaction makes alkyl fluorides by heating an alkyl chloride or bromide with a metallic fluoride such as $\mathrm{AgF}$, $\mathrm{Hg_{2}F_{2}}$, $\mathrm{CoF_{2}}$ or $\mathrm{SbF_{3}}$:

$$\mathrm{CH_{3}Br+AgF\rightarrow CH_{3}F+AgBr}$$ $$\mathrm{2CH_{3}CH_{2}Cl+Hg_{2}F_{2}\rightarrow 2CH_{3}CH_{2}F+Hg_{2}Cl_{2}}$$

Preparing Haloarenes; Physical Properties

By electrophilic substitution. Benzene reacts with chlorine or bromine in the dark, at room temperature, when a Lewis acid catalyst such as iron(III) chloride (or iron filings, which form it) is present:

$$\mathrm{C_{6}H_{6}+Cl_{2}\xrightarrow{\text{anhyd. }FeCl_{3}}C_{6}H_{5}Cl+HCl}$$

Methylbenzene (toluene) gives a mixture of 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene ($o$- and $p$-chlorotoluene), because the methyl group directs the incoming halogen to the ortho and para positions. The para isomer has the higher melting point and is easily separated. Iodination is reversible (the $\mathrm{HI}$ formed reduces iodobenzene back) and needs an oxidising agent such as $\mathrm{HNO_{3}}$ or $\mathrm{HIO_{4}}$ to destroy the $\mathrm{HI}$; fluorination is too violent to control.

📘 Definition 6.4 — Haloarenes from diazonium salts: the Sandmeyer reaction

A primary aromatic amine dissolved in cold mineral acid reacts with sodium nitrite at $273$–$278\ \mathrm{K}$ to give an arenediazonium salt (diazotisation, Chapter 9). In the Sandmeyer reaction the diazonium group is replaced by $\mathrm{Cl}$ or $\mathrm{Br}$ by warming with copper(I) chloride in $\mathrm{HCl}$ or copper(I) bromide in $\mathrm{HBr}$, and nitrogen gas is given off:

$$\mathrm{C_{6}H_{5}NH_{2}\xrightarrow[\text{273–278 K}]{\mathrm{NaNO_{2}+HX}}C_{6}H_{5}N_{2}^{+}X^{-}}$$ $$\mathrm{C_{6}H_{5}N_{2}^{+}X^{-}\xrightarrow{\mathrm{Cu_{2}X_{2}}}C_{6}H_{5}X+N_{2}}\quad(\mathrm{X=Cl,\,Br})$$

Iodobenzene needs no copper salt: shaking the diazonium salt with aqueous potassium iodide gives $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+KI\rightarrow C_{6}H_{5}I+KCl+N_{2}}$. The diazonium route gives a single, pure isomer, because the halogen goes exactly where the $\mathrm{{-}NH_{2}}$ group was.

📐 Rule 6.4 — Physical properties of haloalkanes and haloarenes
  • State and smell. $\mathrm{CH_{3}Cl}$, $\mathrm{CH_{3}Br}$, $\mathrm{C_{2}H_{5}Cl}$ and some chlorofluoromethanes are gases at room temperature; higher members are liquids or solids. Many volatile ones smell sweet.
  • Boiling points are higher than those of hydrocarbons of similar size: the molecules are polar and heavier, so dipole–dipole and van der Waals (dispersion) attractions are stronger. For the same alkyl group: $\mathrm{RI\gt RBr\gt RCl\gt RF}$. Boiling point rises with chain length and falls with branching (a more compact molecule has less surface for attraction): 1-bromobutane $102\ ^{\circ}\mathrm{C}$ $\gt$ 2-bromobutane $91\ ^{\circ}\mathrm{C}$ $\gt$ 2-bromo-2-methylpropane $73\ ^{\circ}\mathrm{C}$.
  • Isomeric dihalobenzenes boil at almost the same temperature ($o$ $180$, $m$ $173$, $p$ $174\ ^{\circ}\mathrm{C}$), but the symmetrical para isomer melts far higher ($53\ ^{\circ}\mathrm{C}$ against about $-17$ and $-25\ ^{\circ}\mathrm{C}$): it packs better in the crystal.
  • Density increases with the number and mass of halogen atoms; bromo, iodo and polychloro compounds are denser than water. In $\mathrm{g\,mL^{-1}}$: 1-chloropropane $0.89$, 1-bromopropane $1.34$, 1-iodopropane $1.75$; $\mathrm{CH_{2}Cl_{2}}$ $1.34$, $\mathrm{CHCl_{3}}$ $1.49$, $\mathrm{CCl_{4}}$ $1.59$.
  • Solubility. They are only very slightly soluble in water: dissolving would break hydrogen bonds between water molecules, and the weaker attractions between haloalkane and water do not pay back that energy. They dissolve readily in organic solvents.

Data are rounded literature values.

✏️ Example 6 — Routes and properties
(a) Suggest a good preparation of 1-iodopropane from 1-bromopropane, and explain why the solvent must be dry acetone. (b) How would you make fluoroethane from chloroethane? (c) Why is iodobenzene better made from aniline than from benzene and iodine? (d) Arrange in order of increasing boiling point: $\mathrm{CHBr_{3}}$, $\mathrm{CH_{3}Cl}$, $\mathrm{CH_{2}Br_{2}}$, $\mathrm{CH_{3}Br}$. (e) 1-Bromopropane is shaken with water in a separating funnel. Which layer is the organic one?
(a) Finkelstein: $\mathrm{CH_{3}CH_{2}CH_{2}Br+NaI}$ $\mathrm{\xrightarrow{\text{dry acetone}}CH_{3}CH_{2}CH_{2}I+NaBr\downarrow}$. $\mathrm{NaBr}$ does not dissolve in acetone and precipitates, pulling the equilibrium to the right. Water would dissolve $\mathrm{NaBr}$ and stop that.
(b) Swarts: heat with a metallic fluoride, e.g. $\mathrm{2CH_{3}CH_{2}Cl+Hg_{2}F_{2}\rightarrow2CH_{3}CH_{2}F+Hg_{2}Cl_{2}}$ (or $\mathrm{AgF}$, $\mathrm{SbF_{3}}$).
(c) Direct iodination is reversible and needs an oxidising agent. From aniline: diazotise ($\mathrm{NaNO_{2}+HCl}$, $273$–$278\ \mathrm{K}$), then add $\mathrm{KI}$; the reaction goes to completion because $\mathrm{N_{2}}$ escapes.
(d) Boiling point increases with molar mass here (all are polar halomethanes): $\mathrm{CH_{3}Cl}$ ($50.5$) $\lt$ $\mathrm{CH_{3}Br}$ ($95$) $\lt$ $\mathrm{CH_{2}Br_{2}}$ ($174$) $\lt$ $\mathrm{CHBr_{3}}$ ($253\ \mathrm{g\,mol^{-1}}$). The boiling points are about $-24$, $4$, $97$ and $149\ ^{\circ}\mathrm{C}$.
(e) Its density ($1.34\ \mathrm{g\,mL^{-1}}$) is greater than that of water, so the organic layer is the lower one.
⚠️ Common mistakes in naming and preparation

1. Counting the wrong carbon. 1°, 2°, 3° refer to the carbon bearing X only: $\mathrm{(CH_{3})_{3}CCH_{2}Cl}$ is primary.

2. Alphabetical order. Prefixes are listed alphabetically (bromo, chloro, fluoro, iodo, methyl…), and only break a tie in locants; they never override a lower set of locants.

3. Phenol + HCl. Phenols do not give aryl halides with $\mathrm{HX}$, $\mathrm{PCl_{5}}$ or $\mathrm{SOCl_{2}}$ in the way alcohols give alkyl halides.

4. $\mathrm{PX_{3}}$ stoichiometry. One $\mathrm{PX_{3}}$ converts three alcohol molecules; $\mathrm{PCl_{5}}$ and $\mathrm{SOCl_{2}}$ convert one each.

5. Peroxide effect. Only $\mathrm{HBr}$ adds anti-Markovnikov in the presence of peroxide; $\mathrm{HCl}$ and $\mathrm{HI}$ do not.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — bromine is choosier than chlorine

In JEE Main, NEET and JEE Advanced 2026 (mechanism of halogenation of alkanes); not required by the CBSE syllabus.

Bromine atoms are much less reactive than chlorine atoms, and less reactive reagents are more selective. In radical bromination near $400\ \mathrm{K}$ the relative reactivities per H are roughly $1:82:1600$ for 1° : 2° : 3°, compared with about $1:3.8:5$ for chlorination. For 2-methylpropane (nine 1° H, one 3° H), chlorination gives $\tfrac{5}{9+5}=36\%$ of 2-chloro-2-methylpropane, but bromination gives $\tfrac{1600}{9+1600}=99.4\%$ of 2-bromo-2-methylpropane: almost a single product. This is why bromination, not chlorination, is used when one particular position must be halogenated.

Exercises

I
Classification and nomenclature.
  • 1 Classify (1°, 2°, 3° alkyl, allylic, benzylic, vinylic or aryl): (a) $\mathrm{CH_{3}CH_{2}CH_{2}CH_{2}I}$; (b) $\mathrm{(CH_{3})_{2}CHCH_{2}Cl}$; (c) $\mathrm{CH_{3}CBr(CH_{3})CH_{2}CH_{3}}$; (d) $\mathrm{C_{6}H_{5}CHBrCH_{3}}$; (e) $\mathrm{CH_{2}{=}CHBr}$; (f) $\mathrm{CH_{2}{=}CHCHClCH_{3}}$.
  • 2 Give IUPAC names: (a) $\mathrm{(CH_{3})_{2}CHCH_{2}CH_{2}Br}$; (b) $\mathrm{CH_{3}CHClCH(CH_{3})CH_{2}CH_{3}}$; (c) $\mathrm{(CH_{3})_{3}CCl}$; (d) $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}Cl}$; (e) $\mathrm{ClCH_{2}CH_{2}CH_{2}Br}$; (f) $m$-$\mathrm{C_{6}H_{4}Br_{2}}$.
  • 3 Write the structure of each compound and classify the halogen: (a) 4-bromopent-2-ene; (b) 1,4-dibromobut-2-ene; (c) 1-chloro-4-ethylcyclohexane.
  • 4 How many structural isomers have the formula $\mathrm{C_{5}H_{11}Br}$? How many of them are primary, secondary and tertiary?
  • 5 Draw and name the dichloro isomers of $\mathrm{C_{2}H_{4}Cl_{2}}$ and of $\mathrm{C_{3}H_{6}Cl_{2}}$. Label each gem, vic or neither.
  • 6 Fluorine is more electronegative than chlorine, yet $\mathrm{CH_{3}Cl}$ has a larger dipole moment than $\mathrm{CH_{3}F}$. Explain.
  • 7 Arrange the C–X bonds of $\mathrm{CH_{3}X}$ in order of increasing bond length and of increasing bond enthalpy, and say which alkyl halide you expect to react fastest when the C–X bond must break.
1 (a) 1°; (b) 1° (the $\mathrm{CH_{2}Cl}$ carbon has one carbon neighbour); (c) 3°; (d) benzylic (secondary); (e) vinylic; (f) allylic (secondary).
2 (a) 1-bromo-3-methylbutane; (b) 2-chloro-3-methylpentane (locants $\{2,3\}$ beat $\{3,4\}$); (c) 2-chloro-2-methylpropane; (d) 3-chloro-2-methylprop-1-ene; (e) 1-bromo-3-chloropropane (tie broken alphabetically); (f) 1,3-dibromobenzene.
3 (a) $\mathrm{CH_{3}CH{=}CHCHBrCH_{3}}$, allylic (2°); (b) $\mathrm{BrCH_{2}CH{=}CHCH_{2}Br}$, both Br allylic (1°); (c) a cyclohexane ring with Cl on C1 and $\mathrm{C_{2}H_{5}}$ on C4; the $\mathrm{CHCl}$ carbon has two ring carbons as neighbours, 2° alkyl.
4 Eight. From pentane: 1-bromo (1°), 2-bromo (2°), 3-bromo (2°). From 2-methylbutane: 1-bromo-2-methylbutane (1°), 2-bromo-2-methylbutane (3°), 2-bromo-3-methylbutane (2°), 1-bromo-3-methylbutane (1°). From 2,2-dimethylpropane: 1-bromo-2,2-dimethylpropane (1°). So 4 primary, 3 secondary, 1 tertiary.
5 $\mathrm{C_{2}H_{4}Cl_{2}}$: 1,1-dichloroethane (gem), 1,2-dichloroethane (vic). $\mathrm{C_{3}H_{6}Cl_{2}}$: 1,1-dichloropropane (gem), 2,2-dichloropropane (gem), 1,2-dichloropropane (vic), 1,3-dichloropropane (neither): four isomers.
6 Dipole moment $=$ charge $\times$ distance. The C–F bond has the larger partial charges but is much shorter ($139$ against $178\ \mathrm{pm}$); the longer C–Cl bond more than makes up for its smaller charges, so $\mu(\mathrm{CH_{3}Cl})=1.860\ \mathrm{D}\gt\mu(\mathrm{CH_{3}F})=1.847\ \mathrm{D}$.
7 Length: $\mathrm{C{-}F\lt C{-}Cl\lt C{-}Br\lt C{-}I}$. Enthalpy: $\mathrm{C{-}I\lt C{-}Br\lt C{-}Cl\lt C{-}F}$. $\mathrm{CH_{3}I}$ has the weakest bond and reacts fastest.
II
Haloalkanes from alcohols and hydrocarbons.
  • 1 Complete and balance: (a) $\mathrm{CH_{3}CH_{2}CH_{2}OH+SOCl_{2}\rightarrow}$; (b) $\mathrm{(CH_{3})_{2}CHOH+PCl_{5}\rightarrow}$; (c) $\mathrm{C_{2}H_{5}OH\xrightarrow{\text{red P}/I_{2}}}$ (write the equation with $\mathrm{PI_{3}}$).
  • 2 Give two reasons why thionyl chloride is often the best reagent for converting an alcohol into an alkyl chloride.
  • 3 Why must anhydrous $\mathrm{ZnCl_{2}}$ be added when ethanol is converted to chloroethane with $\mathrm{HCl}$, but not when 2-methylpropan-2-ol is used?
  • 4 Alkyl iodides are made with $\mathrm{KI}$ and $95\%$ phosphoric acid rather than $\mathrm{KI}$ and concentrated sulphuric acid. Why?
  • 5 Name all the monochloro products of 2-methylbutane formed by free-radical chlorination. How many are there?
  • 6 Using relative reactivities per hydrogen of $1:5.0$ for 1° and 3° H, estimate the percentages of the two monochloro products of 2-methylpropane.
  • 7 Write the major product of propene with (a) $\mathrm{HCl}$; (b) $\mathrm{HBr}$ and benzoyl peroxide; (c) $\mathrm{HCl}$ and benzoyl peroxide.
  • 8 Which alkene gives 2-bromo-2-methylpropane as the only product with $\mathrm{HBr}$ (no peroxide)? Write the equation.
  • 9 Ethene is bubbled through bromine dissolved in $\mathrm{CCl_{4}}$. Write the equation, name the product, and say what is seen.
1 (a) $\mathrm{CH_{3}CH_{2}CH_{2}OH+SOCl_{2}}$ $\mathrm{\rightarrow CH_{3}CH_{2}CH_{2}Cl+SO_{2}+HCl}$; (b) $\mathrm{(CH_{3})_{2}CHOH+PCl_{5}}$ $\mathrm{\rightarrow(CH_{3})_{2}CHCl+POCl_{3}+HCl}$; (c) $\mathrm{3C_{2}H_{5}OH+PI_{3}}$ $\mathrm{\rightarrow3C_{2}H_{5}I+H_{3}PO_{3}}$.
2 The by-products $\mathrm{SO_{2}}$ and $\mathrm{HCl}$ are gases that escape, so the product needs little purification; and the reaction is clean at moderate temperature, with no $\mathrm{POCl_{3}}$ or $\mathrm{H_{3}PO_{3}}$ to separate.
3 A tertiary alcohol forms a stable tertiary carbocation readily in acid, so it reacts with concentrated $\mathrm{HCl}$ at room temperature. A primary alcohol does not; the Lewis acid $\mathrm{ZnCl_{2}}$ coordinates to oxygen and weakens the C–O bond so that $\mathrm{Cl^{-}}$ can replace it.
4 Concentrated $\mathrm{H_{2}SO_{4}}$ is an oxidising agent and converts $\mathrm{HI}$ into $\mathrm{I_{2}}$; phosphoric acid is not oxidising.
5 Four: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane and 1-chloro-3-methylbutane.
6 $9\times1=9$ (1° H) and $1\times5.0=5$ (3° H); total $14$. 1-Chloro-2-methylpropane $\tfrac{9}{14}=64\%$; 2-chloro-2-methylpropane $\tfrac{5}{14}=36\%$.
7 (a) 2-chloropropane; (b) 1-bromopropane; (c) 2-chloropropane (the peroxide effect works only with $\mathrm{HBr}$).
8 2-Methylpropene: $\mathrm{(CH_{3})_{2}C{=}CH_{2}+HBr\rightarrow(CH_{3})_{3}CBr}$. H adds to the $\mathrm{CH_{2}}$ end, giving the tertiary carbocation.
9 $\mathrm{CH_{2}{=}CH_{2}+Br_{2}\rightarrow BrCH_{2}CH_{2}Br}$, 1,2-dibromoethane. The red-brown colour of bromine fades.
III
Halogen exchange and haloarenes.
  • 1 Write the equation for the Finkelstein reaction of chloroethane, naming the product.
  • 2 The Finkelstein reaction is reversible. Explain why it nevertheless goes almost to completion in dry acetone, and why it would not in water.
  • 3 Write the Swarts reaction of bromoethane with silver fluoride. Why are alkyl fluorides not made by direct fluorination of alkanes?
  • 4 Give two ways of making bromobenzene, one starting from benzene and one from aniline, with reagents and conditions.
  • 5 Methylbenzene is chlorinated with $\mathrm{Cl_{2}}$ and $\mathrm{FeCl_{3}}$ in the dark. Name the two main products and explain how the para isomer can be separated.
  • 6 Why is nitric acid (or $\mathrm{HIO_{4}}$) added when benzene is iodinated?
  • 7 How would you convert 4-methylaniline into 1-bromo-4-methylbenzene? Why is this better than brominating methylbenzene?
  • 8 Why can chlorobenzene not be prepared by heating phenol with $\mathrm{HCl}$ and $\mathrm{ZnCl_{2}}$?
1 $\mathrm{CH_{3}CH_{2}Cl+NaI}$ $\mathrm{\xrightarrow{\text{dry acetone}}CH_{3}CH_{2}I+NaCl\downarrow}$: iodoethane.
2 $\mathrm{NaCl}$ (or $\mathrm{NaBr}$) is almost insoluble in acetone and precipitates, so it is removed from the equilibrium and the forward reaction continues (Le Chatelier). In water all the sodium salts dissolve, nothing is removed, and the equilibrium mixture contains much unchanged alkyl halide.
3 $\mathrm{CH_{3}CH_{2}Br+AgF\rightarrow CH_{3}CH_{2}F+AgBr}$. Fluorine reacts explosively with hydrocarbons and breaks C–C bonds; the reaction cannot be controlled.
4 From benzene: $\mathrm{Br_{2}}$ with anhydrous $\mathrm{FeBr_{3}}$ (or Fe), in the dark at room temperature. From aniline: $\mathrm{NaNO_{2}+HBr}$ at $273$–$278\ \mathrm{K}$ to give benzenediazonium bromide, then warm with $\mathrm{Cu_{2}Br_{2}}$ in $\mathrm{HBr}$ (Sandmeyer).
5 1-Chloro-2-methylbenzene ($o$) and 1-chloro-4-methylbenzene ($p$). The more symmetrical para isomer has a much higher melting point, so it crystallises out on cooling.
6 Iodination is reversible because the $\mathrm{HI}$ formed is a strong reducing agent. The oxidising agent converts $\mathrm{HI}$ to $\mathrm{I_{2}}$, removing it, so the forward reaction goes to completion.
7 Diazotise with $\mathrm{NaNO_{2}/HBr}$ at $273$–$278\ \mathrm{K}$, then warm with $\mathrm{Cu_{2}Br_{2}/HBr}$. Bromine goes only where the $\mathrm{NH_{2}}$ group was, giving the pure para product; direct bromination of methylbenzene gives an $o$/$p$ mixture.
8 In phenol the oxygen lone pair is delocalised into the ring, giving the C–O bond partial double-bond character; it is too strong to be broken by $\mathrm{HCl}$.
IV
Physical properties and calculations. Use $M(\mathrm{C})=12$, $M(\mathrm{H})=1$, $M(\mathrm{O})=16$, $M(\mathrm{S})=32$, $M(\mathrm{Cl})=35.5\ \mathrm{g\,mol^{-1}}$.
  • 1 Arrange in order of increasing boiling point: 1-bromobutane, 2-bromo-2-methylpropane, 2-bromobutane. Explain.
  • 2 Arrange in order of increasing boiling point: $\mathrm{C_{2}H_{5}I}$, $\mathrm{C_{2}H_{5}F}$, $\mathrm{C_{2}H_{5}Br}$, $\mathrm{C_{2}H_{5}Cl}$.
  • 3 1,4-Dichlorobenzene melts at about $53\ ^{\circ}\mathrm{C}$; its 1,2- and 1,3-isomers are liquids at room temperature. Explain.
  • 4 Chloroethane is polar, but it hardly dissolves in water. Explain in terms of the energy changes on mixing.
  • 5 Which of 1-chloropropane ($0.89\ \mathrm{g\,mL^{-1}}$) and trichloromethane ($1.49\ \mathrm{g\,mL^{-1}}$) would form the upper layer when shaken with water? Suggest why density rises with the number of chlorine atoms.
  • 6 Environment. Dichloromethane ($1.34\ \mathrm{g\,mL^{-1}}$, slightly soluble in water) leaks from a storage tank into soil above groundwater. Where in the groundwater would you expect most of it to collect, and why does that make clean-up harder?
  • 7 Lab planning. A student needs $46.25\ \mathrm{g}$ of 1-chlorobutane from butan-1-ol and thionyl chloride (1 : 1). (a) What masses of the two reactants are needed at $100\%$ yield? (b) What mass of butan-1-ol is needed if the yield is only $80\%$?
1 2-Bromo-2-methylpropane ($73\ ^{\circ}\mathrm{C}$) $\lt$ 2-bromobutane ($91\ ^{\circ}\mathrm{C}$) $\lt$ 1-bromobutane ($102\ ^{\circ}\mathrm{C}$). Same molar mass; more branching makes the molecule more compact, with less surface for dispersion forces.
2 $\mathrm{C_{2}H_{5}F\lt C_{2}H_{5}Cl\lt C_{2}H_{5}Br\lt C_{2}H_{5}I}$: molar mass and polarisability increase, so van der Waals forces increase.
3 The para isomer is symmetrical and fits closely into a crystal lattice, so more energy is needed to melt it. Boiling points (liquid to gas) depend little on packing, and the three isomers boil at similar temperatures.
4 To dissolve, the haloalkane must break hydrogen bonds between water molecules. The new haloalkane–water attractions (dipole–dipole, dispersion) are much weaker than those hydrogen bonds, so dissolving is energetically unfavourable.
5 1-Chloropropane is less dense than water and floats (upper layer); trichloromethane sinks. Each chlorine adds much mass ($35.5\ \mathrm{g\,mol^{-1}}$) but little volume compared with a $\mathrm{CH_{2}}$ group, so mass per volume rises.
6 Being denser than water and only slightly soluble, it sinks through the groundwater as a separate liquid and collects in pools at the bottom of the aquifer, on the less permeable layer below. There it slowly dissolves for years, and it is hard to reach and pump out.
7 $M(\mathrm{C_{4}H_{9}Cl})=92.5$, $M(\mathrm{C_{4}H_{9}OH})=74$, $M(\mathrm{SOCl_{2}})=119\ \mathrm{g\,mol^{-1}}$. (a) $n=\dfrac{46.25}{92.5}=0.500\ \mathrm{mol}$: butan-1-ol $0.500\times74=37.0\ \mathrm{g}$; $\mathrm{SOCl_{2}}$ $0.500\times119=59.5\ \mathrm{g}$. (b) $\dfrac{0.500}{0.80}=0.625\ \mathrm{mol}$ of alcohol: $0.625\times74=46.25\ \mathrm{g}$ (about $46\ \mathrm{g}$).
🎯 Exam corner BoardJEE MainNEET

E1. Board · 2 marks Write one chemical equation each to illustrate (a) the Finkelstein reaction and (b) the Swarts reaction.

(a) $\mathrm{CH_{3}CH_{2}Br+NaI}$ $\mathrm{\xrightarrow{\text{dry acetone}}CH_{3}CH_{2}I+NaBr}$ (1 mark)
(b) $\mathrm{CH_{3}Br+AgF\rightarrow CH_{3}F+AgBr}$ (1 mark). Any correct metallic fluoride ($\mathrm{Hg_{2}F_{2}}$, $\mathrm{CoF_{2}}$, $\mathrm{SbF_{3}}$) earns the mark.

E2. NEET Which of the following is a benzylic halide? (A) $\mathrm{C_{6}H_{5}Cl}$  (B) $\mathrm{C_{6}H_{5}CH_{2}CH_{2}Br}$  (C) $\mathrm{C_{6}H_{5}CHBrCH_{3}}$  (D) $\mathrm{CH_{2}{=}CHCH_{2}Br}$

Idea Benzylic: X on an $sp^{3}$ carbon bonded directly to the ring. (A) is aryl; in (B) the Br is two carbons from the ring (1° alkyl); (D) is allylic.
Answer (C), a secondary benzylic bromide.

E3. JEE Main Numerical value. The number of structural isomers of $\mathrm{C_{5}H_{11}Cl}$ that are primary alkyl halides is ______.

Enumerate Put Cl on a $\mathrm{CH_{3}}$ end of each $\mathrm{C_{5}}$ skeleton: pentane gives 1-chloropentane; 2-methylbutane gives 1-chloro-2-methylbutane and 1-chloro-3-methylbutane (its two kinds of methyl end are different); 2,2-dimethylpropane gives 1-chloro-2,2-dimethylpropane.
Answer $\mathbf{4}$ (of the eight isomers, 3 are secondary and 1 is tertiary).

E4. JEE Main Numerical value. Free-radical monochlorination of 2,2,4-trimethylpentane, $\mathrm{(CH_{3})_{3}CCH_{2}CH(CH_{3})_{2}}$, can give how many structurally different monochloro products? ______

Kinds of H (i) the nine H of the three equivalent methyls on C2; (ii) the two H of the $\mathrm{CH_{2}}$ (C3); (iii) the single H on C4; (iv) the six H of the two equivalent methyls on C4.
Answer Four kinds of hydrogen give $\mathbf{4}$ monochloro products.

E5. JEE Main NEET Assertion–reason. Assertion (A): Thionyl chloride is preferred for converting alcohols into alkyl chlorides. Reason (R): Both by-products of the reaction of an alcohol with $\mathrm{SOCl_{2}}$ are gases. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation. (C) A is true, R is false. (D) A is false, R is true.

Check $\mathrm{ROH+SOCl_{2}\rightarrow RCl+SO_{2}\uparrow+HCl\uparrow}$: the gases escape, so the alkyl chloride is obtained pure. That is exactly why the reagent is preferred.
Answer (A).

E6. Board · Case study Halothane, $\mathrm{CF_{3}CHBrCl}$, was introduced in the 1950s as a non-flammable inhaled anaesthetic. It is a dense, volatile liquid, almost insoluble in water but very soluble in fats and oils; it acts on the fatty membranes of nerve cells. (a) Give the IUPAC name of halothane and calculate its molar mass. (2 marks) (b) Which carbon–halogen bond in halothane is the weakest? (1 mark) (c) Explain why halothane dissolves in fats but hardly in water. (1 mark) ($M$: C 12, H 1, F 19, Cl 35.5, Br 80.)

(a) Numbering from $\mathrm{CF_{3}}$ gives locants $\{1,1,1,2,2\}$ (the other way $\{1,1,2,2,2\}$); prefixes alphabetical: 2-bromo-2-chloro-1,1,1-trifluoroethane. $M(\mathrm{C_{2}HBrClF_{3}})$ $=2(12)+1+80+35.5+3(19)$ $=197.5\ \mathrm{g\,mol^{-1}}$.
(b) The C–Br bond: it is the longest of the three kinds present (C–F, C–Cl, C–Br) and has the lowest bond enthalpy.
(c) Dissolving in water would require breaking water's hydrogen bonds, which halothane cannot replace; in fats only weak dispersion forces must be overcome and similar forces are formed ("like dissolves like").
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Which reagent should a pharmaceutical plant use to turn an alcohol into an alkyl chloride, if it wants the least waste?

Why it matters. Green chemistry asks not only "what is the yield?" but "how much of the reactants' mass ends up in the product?" This is the atom economy: $\dfrac{\text{molar mass of desired product}}{\text{total molar mass of reactants}}\times100\%$, using the balanced equation. In industry, by-products have to be captured, treated or sold, and that cost often decides the route.

What to do.

  1. Write balanced equations for converting butan-1-ol into 1-chlorobutane with (i) $\mathrm{SOCl_{2}}$, (ii) $\mathrm{PCl_{5}}$, (iii) $\mathrm{PCl_{3}}$, (iv) $\mathrm{HCl}$ with $\mathrm{ZnCl_{2}}$ (a catalyst, so it does not count).
  2. Calculate the atom economy of each route ($M$: C 12, H 1, O 16, P 31, S 32, Cl 35.5).
  3. List the by-products of each route. Which are gases, which are corrosive, which could be sold or reused?
  4. Rank the routes for a large plant and for a small teaching laboratory, and explain why the rankings may differ.

Tools: A calculator or spreadsheet; the equations in Rule 6.3.

Hand in: A table of the four routes (equation, atom economy, by-products) and a 100-word recommendation.

Going further. Find out what the "E-factor" (kilograms of waste per kilogram of product) is, and why the pharmaceutical industry has some of the highest E-factors of any chemical sector.

Instructor note: $M$(butan-1-ol) $=74$, $M$(1-chlorobutane) $=92.5\ \mathrm{g\,mol^{-1}}$. Atom economies: $\mathrm{HCl}$ $\tfrac{92.5}{110.5}=83.7\%$; $\mathrm{PCl_{3}}$ (3 alcohol + 1 $\mathrm{PCl_{3}}$ $\rightarrow$ 3 $\mathrm{RCl}$ + $\mathrm{H_{3}PO_{3}}$) $\tfrac{277.5}{359.5}=77.2\%$; $\mathrm{SOCl_{2}}$ $\tfrac{92.5}{193}=47.9\%$; $\mathrm{PCl_{5}}$ $\tfrac{92.5}{282.5}=32.7\%$. $\mathrm{SOCl_{2}}$ scores poorly on atom economy but well on ease of purification, which matters most in a small laboratory; the $\mathrm{HCl}$ route is best on paper but slow for primary alcohols.

🧪 Self-check quiz

0/5 correct

Q1. Which compound is a vinylic halide?

Q2. Which reagent converts ethanol into chloroethane with only gaseous by-products?

Q3. The Finkelstein reaction uses

Q4. Which isomer of $\mathrm{C_{4}H_{9}Br}$ has the highest boiling point?

Q5. The correct order of carbon–halogen bond enthalpy in $\mathrm{CH_{3}X}$ is

← Chapter 6 — Haloalkanes and Haloarenes6.2 Nucleophilic Substitution: SN1, SN2 and Stereochemistry →