Chapter 6 · Haloalkanes and Haloarenes

6.3 Elimination, Reactions of Haloarenes and Polyhalogen Compounds

Heat a haloalkane with a strong base and it may lose HX to give an alkene; give it magnesium and it becomes a Grignard reagent. Haloarenes resist nucleophiles but still undergo ortho/para electrophilic substitution. Finally, the polyhalogen compounds we use, from chloroform to freons and DDT, and what they do to us and the planet.

In 1912 the Nobel Prize in Chemistry went to Victor Grignard for a simple discovery: an alkyl halide stirred with magnesium turnings in dry ether gives a reagent that can join carbon to carbon almost at will. More than a century later, Grignard reagents are still used to make medicines, perfumes and plastics. In the same years, chemists learned that a strong base can strip HX out of a haloalkane to make an alkene, the starting material for polymers and fuels.

This section completes the chemistry of halogen compounds. We first meet elimination, the competitor of substitution, and the rule that predicts which alkene forms. Then come the reactions with metals, the stubborn unreactivity of haloarenes towards nucleophiles and the way their ring still reacts with electrophiles. We end with the polyhalogen compounds, very useful and sometimes very harmful: dichloromethane, chloroform, iodoform, carbon tetrachloride, the freons that thinned the ozone layer, and the insecticide DDT.

Elimination: Making Alkenes, and Saytzeff's Rule

📘 Definition 6.11 — β-Elimination (dehydrohalogenation)

When a haloalkane that has a hydrogen atom on a β-carbon (a carbon next to the one carrying X, the α-carbon) is heated with alcoholic potassium hydroxide (or sodium ethoxide in ethanol), the base removes that β-hydrogen as a proton while the halide ion leaves from the α-carbon. A double bond forms between the α- and β-carbons:

$$\mathrm{{-}\overset{\beta}{C}H{-}\overset{\alpha}{C}X{-}\xrightarrow[\text{heat}]{\text{alc. KOH}}{\gt}C{=}C{\lt}+HX}$$

Because H and X are lost from neighbouring carbons, this is called β-elimination; because HX is removed, it is also called dehydrohalogenation. Example: $\mathrm{CH_{3}CH_{2}CH_{2}Br+KOH}$ $\mathrm{\xrightarrow{\text{alc.},\ \Delta}CH_{3}CH{=}CH_{2}+KBr+H_{2}O}$.

📐 Rule 6.9 — Saytzeff's (Zaitsev's) rule

If a haloalkane has more than one kind of β-hydrogen, more than one alkene can form. The major product is the more substituted alkene, the one with the greater number of alkyl groups on the doubly bonded carbons, because it is the more stable alkene. Equivalently, the hydrogen is lost preferably from the β-carbon that has fewer hydrogens.

Order of reactivity in elimination: $\mathrm{3^{\circ}\gt2^{\circ}\gt1^{\circ}}$ haloalkanes; for the same alkyl group, $\mathrm{RI\gt RBr\gt RCl}$.

Substitution or elimination? Hydroxide and alkoxide ions are both nucleophiles (they attack carbon) and bases (they remove a proton). Which role wins depends on the haloalkane, the reagent and the conditions.

📐 Rule 6.10 — Substitution versus elimination
  • Conditions. Aqueous $\mathrm{KOH}$ (or water) and moderate temperature favour substitution (alcohol). Alcoholic $\mathrm{KOH}$ or an alkoxide, a high temperature and a bulky base favour elimination (alkene).
  • Primary haloalkanes prefer SN2 substitution; they give alkenes mainly with alcoholic $\mathrm{KOH}$ on heating, or with bulky bases.
  • Secondary haloalkanes give substitution or elimination, depending on how strong the base is: strong bases and heat favour the alkene.
  • Tertiary haloalkanes cannot undergo SN2. With a weak nucleophile in a polar solvent they undergo SN1 (with some alkene); with a strong base they give mainly the alkene.
✏️ Example 1 — Which alkene?
Give the alkene(s) formed when each compound is heated with alcoholic KOH, and identify the major product: (a) 1-bromobutane; (b) 2-bromo-2-methylbutane; (c) 2-bromo-3-methylbutane; (d) 2-bromopentane.
(a) Only one β-carbon (C2): $\mathrm{CH_{3}CH_{2}CH{=}CH_{2}}$, but-1-ene, the only product.
(b) $\mathrm{(CH_{3})_{2}CBrCH_{2}CH_{3}}$ has β-H on the two $\mathrm{CH_{3}}$ groups and on C3. Losing H from C3 gives $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$, 2-methylbut-2-ene (three alkyl groups on C=C, major); losing H from a methyl gives $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$, 2-methylbut-1-ene (two alkyl groups, minor).
(c) $\mathrm{CH_{3}CHBrCH(CH_{3})_{2}}$: losing H from C3 gives $\mathrm{CH_{3}CH{=}C(CH_{3})_{2}}$, 2-methylbut-2-ene (major, trisubstituted); losing H from C1 gives $\mathrm{CH_{2}{=}CHCH(CH_{3})_{2}}$, 3-methylbut-1-ene (minor, monosubstituted).
(d) Pent-2-ene $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$ (major, mainly trans) and pent-1-ene $\mathrm{CH_{2}{=}CHCH_{2}CH_{2}CH_{3}}$ (minor).
✏️ Example 2 — Substitution or elimination?
Predict the main organic product: (a) bromoethane warmed with aqueous $\mathrm{KOH}$; (b) 2-bromopropane boiled with alcoholic $\mathrm{KOH}$; (c) 2-bromo-2-methylpropane heated with sodium ethoxide in ethanol; (d) 1-bromobutane with sodium cyanide in ethanol; (e) 2-bromo-2-methylpropane warmed with water.
(a) Primary, aqueous, moderate temperature: SN2 substitution, ethanol.
(b) Secondary, alcoholic KOH, heat: elimination, propene.
(c) Tertiary with a strong base and heat: elimination, 2-methylpropene $\mathrm{(CH_{3})_{2}C{=}CH_{2}}$ (an SN2 ether is impossible at a tertiary carbon).
(d) Primary with a good nucleophile that is only a weak base: SN2, pentanenitrile $\mathrm{CH_{3}(CH_{2})_{3}CN}$.
(e) Tertiary, weak nucleophile, polar protic solvent: SN1, 2-methylpropan-2-ol, with a little 2-methylpropene.
Interactive 1 — Substitution or Elimination? A Product Predictor

Choose a bromoalkane and a set of conditions. The structure shows the α-carbon and every β-carbon that carries hydrogen. The cards on the right list each alkene that elimination could give, with the number of alkyl groups on its double bond: Saytzeff's rule picks the most substituted one. The verdict weighs substrate class, base strength and temperature to give the typical major product. Press Play to step through the conditions.

main pathway—
major product—
other product(s)—
β-carbons with H—

Reactions with Metals: Grignard Reagents and the Wurtz Reaction

📘 Definition 6.12 — Grignard reagents

A haloalkane reacts with magnesium metal in dry ether to give an alkylmagnesium halide, a Grignard reagent:

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

Magnesium is much less electronegative than carbon, so the C–Mg bond is highly polar with carbon negative, $\mathrm{\overset{\delta-}{R}{-}\overset{\delta+}{Mg}X}$, while the Mg–X bond is essentially ionic. The carbon of a Grignard reagent behaves like a carbanion: it is a strong base and a strong nucleophile.

Grignard reagents react with any compound that has a hydrogen attached to O, N or another electronegative atom (water, alcohols, amines), taking the proton to give a hydrocarbon: $\mathrm{RMgX+H_{2}O\rightarrow RH+Mg(OH)X}$. That is why they must be prepared and kept completely free of moisture. Their reactions with carbonyl compounds to make alcohols are met in Chapters 7 and 8.

📘 Definition 6.13 — The Wurtz reaction

Haloalkanes react with sodium in dry ether to give an alkane with twice the number of carbon atoms of the alkyl group:

$$\mathrm{2RX+2Na\xrightarrow{\text{dry ether}}R{-}R+2NaX}$$

e.g. $\mathrm{2CH_{3}Br+2Na\rightarrow CH_{3}CH_{3}+2NaBr}$. The reaction is useful for symmetrical alkanes. With two different haloalkanes, $\mathrm{RX+R'X}$, three alkanes form ($\mathrm{R{-}R}$, $\mathrm{R{-}R'}$, $\mathrm{R'{-}R'}$), which are hard to separate.

✏️ Example 3 — Grignard and Wurtz
(a) Write the equations for making ethylmagnesium bromide and for its reaction with water. Name the hydrocarbon formed. (b) Why must the ether used be dry? (c) Which alkane is formed when 2-bromopropane is treated with sodium in dry ether? Give its IUPAC name. (d) A mixture of bromomethane and bromoethane is treated with sodium. Name all the alkanes formed. (e) How could you make $\mathrm{CH_{3}CH_{2}D}$ from bromoethane?
(a) $\mathrm{C_{2}H_{5}Br+Mg\xrightarrow{\text{dry ether}}C_{2}H_{5}MgBr}$; $\ \mathrm{C_{2}H_{5}MgBr+H_{2}O}$ $\mathrm{\rightarrow C_{2}H_{6}+Mg(OH)Br}$. The hydrocarbon is ethane.
(b) Any water would destroy the Grignard reagent as it forms, converting it to the alkane.
(c) $\mathrm{2(CH_{3})_{2}CHBr+2Na}$ $\mathrm{\rightarrow(CH_{3})_{2}CH{-}CH(CH_{3})_{2}+2NaBr}$: 2,3-dimethylbutane.
(d) $\mathrm{CH_{3}{-}CH_{3}}$ ethane, $\mathrm{CH_{3}{-}C_{2}H_{5}}$ propane and $\mathrm{C_{2}H_{5}{-}C_{2}H_{5}}$ butane.
(e) Make the Grignard reagent and add heavy water: $\mathrm{C_{2}H_{5}MgBr+D_{2}O\rightarrow C_{2}H_{5}D+Mg(OD)Br}$. The carbanion-like carbon takes a deuteron exactly where the bromine was.

Reactions of Haloarenes

Chlorobenzene can be boiled with aqueous sodium hydroxide for hours and nothing happens, while chloromethane is hydrolysed readily. Aryl halides (and vinylic halides) are extremely unreactive towards nucleophilic substitution.

📐 Rule 6.11 — Why haloarenes resist nucleophilic substitution
  1. Resonance. A lone pair on chlorine is delocalised into the ring. In three of the resonance structures of chlorobenzene the C–Cl bond is a double bond ($\mathrm{C{=}Cl^{+}}$) and a negative charge sits on an ortho or para carbon. The C–Cl bond therefore has partial double-bond character and is harder to break: it is $169\ \mathrm{pm}$ long in chlorobenzene against $177\ \mathrm{pm}$ in chloromethane.
  2. Hybridisation. The carbon bonded to Cl is $sp^{2}$ (more $s$ character, more electronegative) rather than $sp^{3}$; it holds the bonding pair more tightly, making the bond shorter and stronger.
  3. No SN1. Loss of $\mathrm{Cl^{-}}$ would give a phenyl cation, which is not stabilised by resonance (its empty orbital is $sp^{2}$, in the plane of the ring) and is very unstable.
  4. No SN2. The electron-rich ring repels the incoming nucleophile, and the ring itself blocks attack from the back of the C–Cl bond.

Replacement by –OH. Under forcing conditions chlorobenzene is converted into phenol (the Dow process):

$$\mathrm{C_{6}H_{5}Cl\xrightarrow[\text{623 K, 300 atm}]{\text{NaOH}}C_{6}H_{5}O^{-}Na^{+}}$$

and acidifying the sodium phenoxide gives phenol, $\mathrm{C_{6}H_{5}OH}$.

📐 Rule 6.12 — Activation by nitro groups at ortho and para positions

An electron-withdrawing $\mathrm{{-}NO_{2}}$ group ortho or para to the halogen makes the substitution much easier, and each extra nitro group lowers the temperature needed:

HaloareneConditionsProduct (after acidifying)
chlorobenzene$\mathrm{NaOH}$, $623\ \mathrm{K}$, $300\ \mathrm{atm}$phenol
1-chloro-4-nitrobenzeneaq. $\mathrm{NaOH}$, $443\ \mathrm{K}$4-nitrophenol
1-chloro-2,4-dinitrobenzeneaq. $\mathrm{NaOH}$, $368\ \mathrm{K}$2,4-dinitrophenol
2-chloro-1,3,5-trinitrobenzenewarm water, $323\ \mathrm{K}$2,4,6-trinitrophenol (picric acid)

Why. The nucleophile adds to the carbon carrying Cl, giving a negatively charged intermediate (the ring loses its aromaticity for a while); then $\mathrm{Cl^{-}}$ is expelled. The negative charge of the intermediate is spread to the carbons ortho and para to the point of attack. A nitro group on one of those carbons takes the charge onto its oxygen atoms by resonance, stabilising the intermediate. A meta nitro group sits on a carbon that never carries the charge, so it has little effect.

Electrophilic substitution. The ring of a haloarene still reacts with electrophiles, more slowly than benzene. The halogen has two opposing effects: its $-I$ effect withdraws electron density from the whole ring (deactivating), while its $+R$ effect feeds electron density into the ortho and para positions. Since the electrophile attacks where the electron density is highest, halogens are deactivating but ortho/para-directing.

📐 Rule 6.13 — Electrophilic substitution of chlorobenzene
ReactionReagentsProducts (para isomer major)
halogenation$\mathrm{Cl_{2}}$, anhyd. $\mathrm{FeCl_{3}}$1,2- and 1,4-dichlorobenzene
nitrationconc. $\mathrm{HNO_{3}}$ + conc. $\mathrm{H_{2}SO_{4}}$1-chloro-2-nitro- and 1-chloro-4-nitrobenzene
sulphonationconc. $\mathrm{H_{2}SO_{4}}$, heat2- and 4-chlorobenzenesulphonic acid
Friedel–Crafts alkylation$\mathrm{CH_{3}Cl}$, anhyd. $\mathrm{AlCl_{3}}$1-chloro-2-methyl- and 1-chloro-4-methylbenzene
Friedel–Crafts acylation$\mathrm{CH_{3}COCl}$, anhyd. $\mathrm{AlCl_{3}}$1-(2-chlorophenyl)ethan-1-one and 1-(4-chlorophenyl)ethan-1-one

The para product is major because the ortho positions are crowded by the neighbouring halogen.

📘 Definition 6.14 — Wurtz–Fittig and Fittig reactions

A haloarene and a haloalkane react with sodium in dry ether to give an alkylarene: the Wurtz–Fittig reaction.

$$\mathrm{C_{6}H_{5}Br+CH_{3}Br+2Na\xrightarrow{\text{dry ether}}C_{6}H_{5}CH_{3}+2NaBr}$$

Two molecules of a haloarene react with sodium in dry ether to give a biaryl: the Fittig reaction.

$$\mathrm{2C_{6}H_{5}Cl+2Na\xrightarrow{\text{dry ether}}C_{6}H_{5}{-}C_{6}H_{5}+2NaCl}$$
✏️ Example 4 — Reactivity of haloarenes towards nucleophiles
(a) Arrange in order of increasing reactivity towards aqueous $\mathrm{NaOH}$: 1-chloro-2,4-dinitrobenzene, chlorobenzene, 2-chloro-1,3,5-trinitrobenzene, 1-chloro-4-nitrobenzene. (b) 1-Chloro-3-nitrobenzene is much less reactive than 1-chloro-4-nitrobenzene. Explain. (c) The C–Cl bond in chlorobenzene ($169\ \mathrm{pm}$) is shorter than in chloromethane ($177\ \mathrm{pm}$). Give two reasons.
(a) More o/p nitro groups, easier substitution (lower temperature): chlorobenzene ($623\ \mathrm{K}$) $\lt$ 1-chloro-4-nitrobenzene ($443\ \mathrm{K}$) $\lt$ 1-chloro-2,4-dinitrobenzene ($368\ \mathrm{K}$) $\lt$ 2-chloro-1,3,5-trinitrobenzene ($323\ \mathrm{K}$).
(b) In the intermediate formed when $\mathrm{OH^{-}}$ adds to C1, the negative charge is spread over C2, C4 and C6. A nitro group at C4 takes the charge onto its oxygens; at C3 it cannot, because C3 never carries the charge.
(c) (i) Resonance gives the C–Cl bond partial double-bond character; (ii) the carbon is $sp^{2}$ rather than $sp^{3}$, and the greater $s$ character makes the bond shorter.
✏️ Example 5 — Products from chlorobenzene and bromobenzene
(a) Name the products of nitrating chlorobenzene, and say which is major. (b) Chlorine withdraws electrons, yet it directs the nitro group to the ortho and para positions. Explain. (c) How would you make ethylbenzene from bromobenzene? (d) What forms when chlorobenzene is heated with sodium in dry ether?
(a) 1-Chloro-2-nitrobenzene (minor) and 1-chloro-4-nitrobenzene (major).
(b) The $-I$ effect lowers the electron density of the whole ring (slower than benzene), but the $+R$ effect of the chlorine lone pairs raises the density at C2, C4 and C6 relative to C3 and C5. The electrophile attacks where the density is greatest.
(c) Wurtz–Fittig: $\mathrm{C_{6}H_{5}Br+C_{2}H_{5}Br+2Na}$ $\mathrm{\xrightarrow{\text{dry ether}}C_{6}H_{5}C_{2}H_{5}+2NaBr}$.
(d) Fittig reaction: biphenyl, $\mathrm{C_{6}H_{5}{-}C_{6}H_{5}}$.
Interactive 2 — Inside a Haloarene: Resonance, Electrophiles and Nitro Activation

Resonance mode steps through the five resonance structures of chlorobenzene and then the hybrid: watch the C–Cl bond gain partial double-bond character and the negative charge visit the ortho and para carbons. Electrophile mode shows where an electrophile enters and the two products. Nucleophile mode adds nitro groups and shows whether they can take the negative charge of the intermediate, and the conditions needed to replace Cl by OH.

showing—
C–Cl bond—
result—

Polyhalogen Compounds: Uses and Environmental Effects

Carbon compounds containing more than one halogen atom are made on a huge scale as solvents, refrigerants and pesticides. Their very stability, which makes them useful, also lets them persist in the environment.

📘 Definition 6.15 — Freons (chlorofluorocarbons) and ozone depletion

The freons are chlorofluorocarbons (CFCs) of methane and ethane, such as $\mathrm{CCl_{2}F_{2}}$ (freon-12), made from tetrachloromethane by halogen exchange with HF (a Swarts-type reaction): $\mathrm{CCl_{4}+2HF\rightarrow CCl_{2}F_{2}+2HCl}$. They are stable, unreactive, non-toxic, non-corrosive and easily liquefied, which made them ideal refrigerants, air-conditioner fluids and aerosol propellants.

Being unreactive, they survive for decades and diffuse unchanged into the stratosphere. There ultraviolet light breaks the weakest bond, C–Cl, releasing chlorine atoms that destroy ozone in a chain reaction:

$$\mathrm{CCl_{2}F_{2}\xrightarrow{h\nu}{}^{\bullet}CClF_{2}+Cl^{\bullet}}$$ $$\mathrm{Cl^{\bullet}+O_{3}\rightarrow ClO^{\bullet}+O_{2}}$$ $$\mathrm{ClO^{\bullet}+O\rightarrow Cl^{\bullet}+O_{2}}$$

The chlorine atom is regenerated, so one atom can destroy many thousands of ozone molecules. The ozone layer filters out harmful UV radiation; its thinning raises the risk of skin cancer and cataracts. The Montreal Protocol (1987) phased out CFCs worldwide, and the ozone layer is now slowly recovering.

DDT, 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane ($p,p'$-dichlorodiphenyltrichloroethane, $\mathrm{C_{14}H_{9}Cl_{5}}$), was the first chlorinated organic insecticide. Its insect-killing power, discovered by Paul Müller (Nobel Prize in Physiology or Medicine, 1948), was used during and after the Second World War against the mosquitoes that spread malaria and the lice that carry typhus, saving millions of lives. Problems soon appeared: many insects became resistant; it is highly toxic to fish; and it is chemically stable and fat-soluble, so animals cannot break it down or excrete it quickly. It is stored in fatty tissue and its concentration rises at each step up a food chain (biomagnification), which caused thinning of eggshells in birds of prey. DDT was banned in the United States in the early 1970s and later in many other countries, but it is still used in a few places to control malaria mosquitoes.

✏️ Example 6 — Polyhalogen compounds
(a) Why is chloroform stored in dark bottles filled to the brim? Write the equation for the reaction it prevents. (b) Freons are non-toxic and unreactive near the ground. How, then, do they damage the ozone layer? (c) Calculate the molar mass of DDT, $\mathrm{C_{14}H_{9}Cl_{5}}$, and its percentage of chlorine by mass. ($M$: C 12, H 1, Cl 35.5.) (d) Why does DDT accumulate in birds that eat fish?
(a) Light and air oxidise it to poisonous phosgene: $\mathrm{2CHCl_{3}+O_{2}\rightarrow2COCl_{2}+2HCl}$. Dark glass keeps out light and a full bottle leaves no air.
(b) Their stability lets them reach the stratosphere intact, where UV light splits off chlorine atoms. Each $\mathrm{Cl^{\bullet}}$ starts a chain ($\mathrm{Cl^{\bullet}+O_{3}\rightarrow ClO^{\bullet}+O_{2}}$; $\mathrm{ClO^{\bullet}+O\rightarrow Cl^{\bullet}+O_{2}}$) that destroys ozone while regenerating the chlorine atom.
(c) $M=14(12)+9(1)+5(35.5)$ $=168+9+177.5$ $=354.5\ \mathrm{g\,mol^{-1}}$. $\%\,\mathrm{Cl}=\dfrac{177.5}{354.5}\times100=50.1\%$.
(d) DDT is fat-soluble and is broken down and excreted only very slowly, so each animal keeps the DDT from everything it eats. Fish concentrate it from smaller organisms, and birds eating many fish concentrate it further: biomagnification.
⚠️ Common mistakes in this section

1. Aqueous vs alcoholic KOH. Aqueous KOH gives alcohols (substitution); alcoholic KOH on heating gives alkenes (elimination).

2. Saytzeff picks the more substituted alkene, not the one from the β-carbon with more hydrogens.

3. Grignard reagents and water. $\mathrm{RMgX}$ is destroyed by water, alcohols or acids, giving RH; never write it in aqueous solution.

4. Deactivating ≠ meta-directing. Halogens are the exception: deactivating but ortho/para-directing.

5. Meta nitro groups. Only o/p nitro groups activate a haloarene towards nucleophiles.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the halogen order turns upside down

In JEE Main, NEET and JEE Advanced 2026 (substitution mechanisms; JEE Advanced names nucleophilic aromatic substitution); not required by the CBSE syllabus.

For haloalkanes the leaving-group order is $\mathrm{I\gt Br\gt Cl\gt F}$, because the C–X bond breaks in the slow step. In activated haloarenes the slow step is the addition of the nucleophile to the ring, and C–X breaks only afterwards, in a fast step. The most electronegative halogen makes the attacked carbon most positive and best stabilises the negative intermediate by its $-I$ effect, so 1-fluoro-2,4-dinitrobenzene reacts fastest: $\mathrm{F\gg Cl\approx Br\gt I}$. This is why the fluoro compound (Sanger's reagent) was used to label the amino ends of proteins.

Exercises

I
Elimination and Saytzeff's rule.
  • 1 What is meant by β-elimination? Mark the α- and β-carbons in 2-chlorobutane.
  • 2 Give all the alkenes (ignore cis/trans) formed on heating each with alcoholic KOH, and mark the major one: (a) 2-chloro-2-methylbutane; (b) 3-bromopentane; (c) 1-bromo-2-methylpropane; (d) 2-bromo-2,3-dimethylbutane.
  • 3 Which isomers of $\mathrm{C_{4}H_{9}Br}$ give only one alkene (ignoring cis/trans) on dehydrohalogenation?
  • 4 Arrange 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane in order of increasing ease of dehydrohalogenation.
  • 5 Predict the main organic product: (a) 1-bromopropane + aqueous KOH, warm; (b) 1-bromopropane + alcoholic KOH, heat; (c) 2-bromo-2-methylbutane + sodium ethoxide, heat; (d) 2-bromobutane + NaI in acetone.
  • 6 Which is more stable, but-1-ene or but-2-ene? Which isomer of but-2-ene is more stable? Relate your answers to Saytzeff's rule.
  • 7 Industry. A plant needs but-1-ene that is free of but-2-ene. Should it dehydrobrominate 1-bromobutane or 2-bromobutane? Explain.
1 Loss of H from a β-carbon and X from the α-carbon to form a C=C bond. In $\mathrm{CH_{3}{-}CHCl{-}CH_{2}{-}CH_{3}}$, C2 is α; C1 and C3 are β.
2 (a) 2-methylbut-2-ene (major) and 2-methylbut-1-ene; (b) pent-2-ene only (both β-carbons are equivalent); (c) 2-methylpropene only; (d) 2,3-dimethylbut-2-ene (four alkyl groups, major) and 2,3-dimethylbut-1-ene.
3 Three: 1-bromobutane (but-1-ene), 1-bromo-2-methylpropane and 2-bromo-2-methylpropane (both 2-methylpropene). 2-Bromobutane gives two (but-1-ene and but-2-ene).
4 1-Bromobutane $\lt$ 2-bromobutane $\lt$ 2-bromo-2-methylpropane (1° $\lt$ 2° $\lt$ 3°).
5 (a) propan-1-ol; (b) propene; (c) 2-methylbut-2-ene (major alkene); (d) 2-iodobutane (SN2).
6 But-2-ene (two alkyl groups on the double bond, against one). trans-But-2-ene, in which the two methyl groups are far apart. Saytzeff's rule simply says the most stable alkene is formed fastest.
7 1-Bromobutane: it has only one β-carbon, so it can give only but-1-ene. 2-Bromobutane gives mainly but-2-ene.
II
Reactions with metals. Use $M$: C 12, H 1, Br 80.
  • 1 Write equations for making methylmagnesium iodide and for its reaction with water.
  • 2 What are the products of ethylmagnesium bromide with methanol?
  • 3 Give two precautions needed when preparing a Grignard reagent, with reasons.
  • 4 Name the alkane formed in the Wurtz reaction of (a) bromoethane; (b) 1-bromopropane; (c) 2-bromobutane.
  • 5 Why is the Wurtz reaction a poor way of making propane from bromomethane and bromoethane?
  • 6 What mass of butane can be made from $2.18\ \mathrm{g}$ of bromoethane by the Wurtz reaction, assuming $100\%$ yield?
  • 7 Can methane be made by the Wurtz reaction? Suggest a way of making methane from bromomethane.
1 $\mathrm{CH_{3}I+Mg\xrightarrow{\text{dry ether}}CH_{3}MgI}$; $\ \mathrm{CH_{3}MgI+H_{2}O\rightarrow CH_{4}+Mg(OH)I}$.
2 $\mathrm{C_{2}H_{5}MgBr+CH_{3}OH}$ $\mathrm{\rightarrow C_{2}H_{6}+Mg(OCH_{3})Br}$: ethane (the alcohol supplies the proton).
3 Use perfectly dry ether and apparatus, and protect from moist air: water destroys RMgX. Use an ether solvent: it is unreactive towards RMgX and its oxygen lone pairs coordinate to magnesium, stabilising the reagent.
4 (a) butane; (b) hexane; (c) 3,4-dimethylhexane, $\mathrm{CH_{3}CH_{2}CH(CH_{3})CH(CH_{3})CH_{2}CH_{3}}$.
5 Three alkanes form together, ethane, propane and butane, which must then be separated, and the yield of propane is low.
6 $n(\mathrm{C_{2}H_{5}Br})=\dfrac{2.18}{109}=0.0200\ \mathrm{mol}$; 2 mol give 1 mol butane, so $0.0100\ \mathrm{mol}\times58\ \mathrm{g\,mol^{-1}}=0.580\ \mathrm{g}$.
7 No: the Wurtz product always has at least two carbons (2 CH₃Br gives ethane). Make $\mathrm{CH_{3}MgBr}$ and add water, or reduce $\mathrm{CH_{3}Br}$ with $\mathrm{LiAlH_{4}}$.
III
Haloarenes.
  • 1 Give three reasons why chlorobenzene is much less reactive than chloromethane towards nucleophiles.
  • 2 Draw the resonance structures of chlorobenzene. On which carbons does a negative charge appear?
  • 3 Give the conditions for converting (a) chlorobenzene into phenol; (b) 1-chloro-4-nitrobenzene into 4-nitrophenol.
  • 4 Arrange in increasing order of reactivity towards aqueous NaOH: 1-chloro-2,4-dinitrobenzene, 1-chloro-3-nitrobenzene, 1-chloro-4-nitrobenzene.
  • 5 Name the main products when chlorobenzene reacts with (a) $\mathrm{Cl_{2}}$/anhydrous $\mathrm{FeCl_{3}}$; (b) concentrated $\mathrm{H_{2}SO_{4}}$; (c) $\mathrm{CH_{3}COCl}$/anhydrous $\mathrm{AlCl_{3}}$.
  • 6 Why is the para isomer the major product in the electrophilic substitution of chlorobenzene?
  • 7 How would you make propylbenzene by a Wurtz–Fittig reaction? Write the equation.
  • 8 What is formed when 4-bromotoluene is heated with sodium in dry ether?
1 Any three of: partial double-bond character of C–Cl from resonance; $sp^{2}$ carbon holds the bond more tightly; the phenyl cation is too unstable for SN1; the electron-rich ring repels the nucleophile and blocks backside attack (no SN2).
2 Five structures: two Kekulé structures with C–Cl single, and three with $\mathrm{C{=}Cl^{+}}$ and a negative charge on C2, C4 or C6 (the two ortho carbons and the para carbon).
3 (a) $\mathrm{NaOH}$ at $623\ \mathrm{K}$ and $300\ \mathrm{atm}$, then acidify; (b) aqueous $\mathrm{NaOH}$ at $443\ \mathrm{K}$, then acidify.
4 1-Chloro-3-nitrobenzene $\lt$ 1-chloro-4-nitrobenzene $\lt$ 1-chloro-2,4-dinitrobenzene.
5 (a) 1,4-dichlorobenzene (major) and 1,2-dichlorobenzene; (b) 4-chlorobenzenesulphonic acid (major) and 2-chlorobenzenesulphonic acid; (c) 1-(4-chlorophenyl)ethan-1-one (major) and 1-(2-chlorophenyl)ethan-1-one.
6 Both ortho and para positions are activated by the $+R$ effect, but the ortho positions are next to the chlorine atom, which hinders the approach of the electrophile.
7 $\mathrm{C_{6}H_{5}Cl+CH_{3}CH_{2}CH_{2}Cl+2Na}$ $\mathrm{\xrightarrow{\text{dry ether}}C_{6}H_{5}CH_{2}CH_{2}CH_{3}+2NaCl}$.
8 Fittig reaction: 4,4′-dimethylbiphenyl, $\mathrm{CH_{3}C_{6}H_{4}{-}C_{6}H_{4}CH_{3}}$.
IV
Polyhalogen compounds. Use $M$: C 12, F 19, Cl 35.5.
  • 1 Give one use and one health hazard of each: dichloromethane, trichloromethane, triiodomethane, tetrachloromethane.
  • 2 Write the equation for the oxidation of chloroform by air in light. Why is the product dangerous?
  • 3 Give the formula of freon-12. Why does a non-toxic, unreactive compound threaten the ozone layer?
  • 4 Write the two chain-propagation steps of ozone destruction by chlorine atoms and add them to find the overall change.
  • 5 Calculate the percentage of chlorine by mass in freon-12.
  • 6 What does the name 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane tell you about the structure of DDT?
  • 7 Ecology. In an illustrative food chain the DDT concentrations are: lake water $3\times10^{-6}$ ppm, zooplankton $0.04$ ppm, small fish $0.5$ ppm, large fish $2$ ppm, fish-eating birds $25$ ppm. By what factor is DDT concentrated from water to birds? Which single step gives the largest factor?
1 $\mathrm{CH_{2}Cl_{2}}$: paint remover / solvent; harms the central nervous system. $\mathrm{CHCl_{3}}$: solvent for fats and alkaloids; liver and kidney damage. $\mathrm{CHI_{3}}$: former antiseptic; objectionable smell (iodine releasing). $\mathrm{CCl_{4}}$: making refrigerants and propellants; liver damage and cancer, ozone depletion.
2 $\mathrm{2CHCl_{3}+O_{2}\rightarrow2COCl_{2}+2HCl}$; phosgene ($\mathrm{COCl_{2}}$) is extremely poisonous to the lungs.
3 $\mathrm{CCl_{2}F_{2}}$. Because it is unreactive it is not removed in the lower atmosphere, so it reaches the stratosphere, where UV light releases chlorine atoms that destroy ozone catalytically.
4 $\mathrm{Cl^{\bullet}+O_{3}\rightarrow ClO^{\bullet}+O_{2}}$ and $\mathrm{ClO^{\bullet}+O\rightarrow Cl^{\bullet}+O_{2}}$. Sum: $\mathrm{O_{3}+O\rightarrow2O_{2}}$; the chlorine atom is not used up.
5 $M(\mathrm{CCl_{2}F_{2}})=12+71+38=121\ \mathrm{g\,mol^{-1}}$; $\%\,\mathrm{Cl}=\dfrac{71}{121}\times100=58.7\%$.
6 A two-carbon chain: C1 carries three Cl atoms; C2 carries two 4-chlorophenyl groups (benzene rings with Cl para to the point of attachment) and one H: $\mathrm{Cl_{3}C{-}CH(C_{6}H_{4}Cl)_{2}}$.
7 $\dfrac{25}{3\times10^{-6}}=8.3\times10^{6}$, about eight million times. Step factors: water → zooplankton $1.3\times10^{4}$; then $12.5$, $4$ and $12.5$. The largest jump is from water into the first living organisms.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks State Saytzeff's rule and illustrate it with the dehydrobromination of 2-bromopentane.

Rule In dehydrohalogenation, when more than one alkene can form, the preferred product is the alkene with the greater number of alkyl groups on the doubly bonded carbons. (1 mark)
Example $\mathrm{CH_{3}CHBrCH_{2}CH_{2}CH_{3}\xrightarrow{\text{alc. KOH},\ \Delta}}$ $\mathrm{CH_{3}CH{=}CHCH_{2}CH_{3}}$ (pent-2-ene, major) $+$ $\mathrm{CH_{2}{=}CHCH_{2}CH_{2}CH_{3}}$ (pent-1-ene, minor). (1 mark)

E2. NEET The major product when 2-bromo-2-methylbutane is heated with alcoholic KOH is (A) 2-methylbut-1-ene  (B) 2-methylbut-2-ene  (C) 3-methylbut-1-ene  (D) 2-methylbutan-2-ol

Idea Alcoholic KOH and heat: elimination. Saytzeff: the trisubstituted alkene $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$ beats the disubstituted $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$. (C) cannot form from this substrate; (D) is the substitution product.
Answer (B).

E3. JEE Main Which compound reacts fastest with aqueous NaOH to replace Cl by OH? (A) chlorobenzene  (B) 1-chloro-3-nitrobenzene  (C) 1-chloro-4-nitrobenzene  (D) 1-chloro-2,4-dinitrobenzene

Idea Nitro groups ortho and para to Cl stabilise the anionic intermediate; two of them (positions 2 and 4) do so more than one. A meta nitro group gives no such help.
Answer (D).

E4. JEE Main Numerical value. The total number of alkenes, counting geometrical isomers separately, formed when 2-bromobutane is heated with alcoholic KOH is ______.

Structural But-1-ene (H lost from C1) and but-2-ene (H lost from C3).
Geometrical But-2-ene, $\mathrm{CH_{3}CH{=}CHCH_{3}}$, has two different groups on each doubly bonded carbon, so it exists as cis and trans isomers; but-1-ene has two H on C1 and has none.
Answer $\mathbf{3}$ (but-1-ene, cis-but-2-ene, trans-but-2-ene).

E5. JEE Advanced Matrix match. Match each named reaction in List I with the example in List II. List I: (P) Wurtz, (Q) Fittig, (R) Wurtz–Fittig, (S) Finkelstein. List II: (1) $\mathrm{C_{6}H_{5}Cl+Na}$ gives biphenyl; (2) $\mathrm{C_{2}H_{5}Br+Na}$ gives butane; (3) $\mathrm{C_{6}H_{5}Br+CH_{3}Br+Na}$ gives toluene; (4) $\mathrm{C_{2}H_{5}Br+NaI}$ in acetone gives iodoethane. Options: (A) P-2, Q-1, R-3, S-4  (B) P-1, Q-2, R-3, S-4  (C) P-2, Q-3, R-1, S-4  (D) P-3, Q-1, R-2, S-4

Match Wurtz couples two alkyl halides (2); Fittig couples two aryl halides (1); Wurtz–Fittig couples an aryl with an alkyl halide (3); Finkelstein exchanges halogen with NaI in acetone (4).
Answer (A).

E6. Board · Case study Chlorofluorocarbons were once used in refrigerators and spray cans. Being inert, they spread through the lower atmosphere and slowly rise into the stratosphere. There, ultraviolet radiation releases chlorine atoms, which take part in a chain reaction that converts ozone into oxygen. Their replacements, hydrofluorocarbons such as $\mathrm{CH_{2}FCF_{3}}$, contain no chlorine. (a) Write the equations for the formation of chlorine atoms from $\mathrm{CCl_{2}F_{2}}$ and for the two chain steps that destroy ozone. (2 marks) (b) Why can a few CFC molecules destroy a large amount of ozone? (1 mark) (c) Why do hydrofluorocarbons not deplete ozone in this way? (1 mark)

(a) $\mathrm{CCl_{2}F_{2}\xrightarrow{h\nu}{}^{\bullet}CClF_{2}+Cl^{\bullet}}$; $\ \mathrm{Cl^{\bullet}+O_{3}\rightarrow ClO^{\bullet}+O_{2}}$; $\ \mathrm{ClO^{\bullet}+O\rightarrow Cl^{\bullet}+O_{2}}$.
(b) The chlorine atom is regenerated in the second step, so it acts as a catalyst and can go round the chain thousands of times.
(c) They have no C–Cl bond. The C–F bond is much stronger, so stratospheric UV does not release halogen atoms from them (and fluorine atoms would not sustain the same chain).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. What should go inside the next generation of refrigerators and air conditioners?

Why it matters. CFCs were banned because they destroy ozone. Their first replacements, HCFCs and then HFCs, protect the ozone layer far better, but many are powerful greenhouse gases. The Kigali Amendment (2016) to the Montreal Protocol now phases down HFCs as well, and countries such as India are planning the switch for millions of new air conditioners.

What to do.

  1. Four refrigerants are CFC-12 ($\mathrm{CCl_{2}F_{2}}$), HCFC-22 ($\mathrm{CHClF_{2}}$), HFC-134a ($\mathrm{CH_{2}FCF_{3}}$) and propane ($\mathrm{C_{3}H_{8}}$). Calculate the molar mass and the percentage of chlorine of each ($M$: C 12, H 1, F 19, Cl 35.5).
  2. Published approximate values of the ozone-depletion potential (ODP, relative to CFC-11) and the 100-year global-warming potential (GWP, relative to $\mathrm{CO_{2}}$) are: CFC-12 ODP about 1, GWP about 10 900; HCFC-22 ODP about 0.055, GWP about 1 800; HFC-134a ODP 0, GWP about 1 400; propane ODP 0, GWP about 3. Plot ODP against % Cl. What do you notice?
  3. Propane has almost no environmental impact. What property of propane is the main drawback, and how do designers deal with it?
  4. Recommend a refrigerant for a household air conditioner sold in a hot, densely populated city, and justify it in terms of ozone, climate and safety.

Tools: Calculator or spreadsheet; the ozone chain reaction in Definition 6.15.

Hand in: A table of the four refrigerants (formula, $M$, % Cl, ODP, GWP) and a 150-word recommendation.

Going further. Find out how the size of the Antarctic ozone hole has changed since the Montreal Protocol, and when the ozone layer is expected to recover.

Instructor note: $M$ and % Cl — $\mathrm{CCl_{2}F_{2}}$ $121$, $58.7\%$; $\mathrm{CHClF_{2}}$ $86.5$, $41.0\%$; $\mathrm{CH_{2}FCF_{3}}$ $102$, $0\%$; $\mathrm{C_{3}H_{8}}$ $44$, $0\%$. ODP is zero without chlorine (or bromine), but it is not simply proportional to % Cl: HCFC-22 has an H atom and is largely destroyed in the lower atmosphere before reaching the stratosphere. Propane is flammable, so charge sizes are limited and equipment is designed to avoid ignition sources. The ODP and GWP figures are rounded published values.

🧪 Self-check quiz

0/5 correct

Q1. 2-Bromobutane is heated with alcoholic KOH. The major product is

Q2. Ethylmagnesium bromide reacts with water to give

Q3. In electrophilic substitution, the chlorine atom of chlorobenzene is

Q4. Bromobenzene and bromomethane heated with sodium in dry ether give mainly

Q5. Which compound releases chlorine atoms in the stratosphere and so depletes ozone?

← 6.2 Nucleophilic Substitution: SN1, SN2 and StereochemistryChapter 7 — Alcohols, Phenols and Ethers →