Haloalkanes and Haloarenes: Class 12 Chemistry · Chapter 6

Chemistry Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

110 min
1 simulation
32 practice questions
39 MCQs

What you will learn in Haloalkanes and Haloarenes

Classify organohalogen compounds by the number of halogen atoms and by the kind of carbon that carries the halogen, and name them by common and IUPAC rules
Relate the polarity, length and strength of the carbon–halogen bond to boiling point, density, solubility and reactivity
Choose reagents to make haloalkanes from alcohols, alkanes, alkenes and other halides, and haloarenes from arenes and diazonium salts
Describe the SN2 and SN1 mechanisms, predict which one a halide follows and predict the stereochemical result
Recognise chiral molecules and stereocentres, and explain optical activity, enantiomers and racemic mixtures
Predict the products of dehydrohalogenation (Zaitsev rule) and of reactions with magnesium and sodium
Explain why haloarenes resist nucleophilic substitution yet undergo ortho/para electrophilic substitution
Describe the uses and the health and environmental effects of common polyhalogen compounds

Original companion text aligned to NCERT Chemistry Class 12, Chapter 6, pp. 159–192. Not affiliated with or endorsed by NCERT.

Look around a kitchen and you will find carbon–halogen bonds in unexpected places. The grey PVC pipe under the sink is a polymer of chloroethene. The non-stick coating on a tawa is a long chain of $C F_{2}$ units. The low-calorie sweetener sucralose is table sugar in which three $- O H$ groups have been exchanged for chlorine atoms, and the gas circulating in many split air conditioners, difluoromethane, is a halogen compound too.

Each of these molecules has carbon bonded to fluorine, chlorine, bromine or iodine. Chemists split such organohalogen compounds into two families by looking at the carbon that holds the halogen. If it is an $s p^{3}$ carbon, part of an alkyl group, the compound is a haloalkane; if it is an $s p^{2}$ carbon of an aromatic ring, the compound is a haloarene. This one difference in hybridisation gives the two families strikingly different chemistry.

Definition

Haloalkane (alkyl halide)

A compound $R - X$ in which a halogen atom X is bonded to an $s p^{3}$ -hybridised carbon of an alkyl group.

ExampleBromoethane, $C H_{3} C H_{2} B r$ .

Definition

Haloarene (aryl halide)

A compound $A r - X$ in which X is bonded directly to an $s p^{2}$ -hybridised carbon of an aromatic ring.

ExampleChlorobenzene, $C_{6} H_{5} C l$ .

Why study them? In the laboratory a C–X bond is a handle: it can be exchanged for an $- O H$ , $- O R$ , $- N H_{2}$ or $- C N$ group, or used to join two carbon chains, so many syntheses start from a halide. Several halides also dissolve oils and waxes well, and many widely used medicines carry a halogen atom. The flip side is that tough carbon–halogen bonds let some of these compounds linger in soil, water and the upper air, a story we return to at the end of the chapter.

6.1 Classification

We can sort organohalogen compounds in two ways: by how many halogen atoms a molecule has, and by what kind of carbon each halogen is bonded to. The second way is the useful one for predicting reactions.

6.1.1 On the Basis of Number of Halogen Atoms

A molecule with one halogen atom is a monohalogen compound; with two, a dihalogen compound; with three or more, a polyhalogen (tri-, tetra-, …) compound. The table gives one aliphatic and one aromatic example of each kind.

Table 6.1Counting the halogen atoms

Number	On a chain	On a ring
one (mono)	$C H_{3} C H_{2} B r$	$C_{6} H_{5} I$
two (di)	$C l C H_{2} C H_{2} C l$	$C_{6} H_{4} C l_{2}$
three (tri)	$C H I_{3}$	$C_{6} H_{3} B r_{3}$
many (poly)	$C C l_{4}$	$C_{6} C l_{6}$

6.1.2 Compounds Containing sp³ C—X Bond

Alkyl halides, R–X. Here the halogen sits on a saturated ( $s p^{3}$ ) carbon of an alkyl chain. To decide whether such a halide is primary (1°), secondary (2°) or tertiary (3°), count the carbon atoms joined to the carbon that holds X: one, two or three (Fig. 6.1). Halomethanes, $C H_{3} X$ , have no carbon neighbour at all and are treated as a class of their own. Open-chain monohaloalkanes fit the general formula $C_{n} H_{2 n + 1} X$ .

Allylic and benzylic halides. Two kinds of neighbour earn special names. If the $s p^{3}$ carbon carrying X is joined to one carbon of a C=C double bond, the halide is allylic, as in 3-chloroprop-1-ene, $C H_{2} = C H - C H_{2} C l$ . If it is joined to an aromatic ring, the halide is benzylic, as in $C_{6} H_{5} C H_{2} C l$ . Each can itself be 1°, 2° or 3°: $C_{6} H_{5} C (C H_{3})_{2} B r$ is a tertiary benzylic bromide. The neighbouring π system makes these halides unusually reactive (The SN1 mechanism).

1/4Count the carbons bonded to the ringed carbon: one, two or three.

Fig. 6.1Look at the carbon that carries X: its hybridisation and its neighbours decide the class.NCERT 6.1.2 · p. 160

Five category chips run across the top: alkyl, allylic and benzylic under the heading X on an sp3 carbon, and vinylic and aryl under X on an sp2 carbon. Step 1 highlights alkyl and shows chloroethane, 2-bromopropane and 2-iodo-2-methylpropane with the carbon bearing the halogen ringed, labelled 1°, 2° and 3° by the number of carbons on it. Step 2 highlights allylic and benzylic and shows 3-bromopropene and benzyl chloride with the CH2 ringed. Step 3 highlights vinylic and shows chloroethene and a cyclohexene with bromine on a double-bond carbon. Step 4 highlights aryl and shows chlorobenzene and 4-bromotoluene with the ring carbon carrying the halogen ringed.

6.1.3 Compounds Containing sp² C—X Bond

If the carbon that carries X is $s p^{2}$ -hybridised, the halogen behaves very differently. In a vinylic halide X is on one of the two carbons of a C=C double bond, as in chloroethene, $C H_{2} = C H C l$ (vinyl chloride, the monomer of PVC). In an aryl halide X is bonded straight to a carbon of an aromatic ring, as in bromobenzene. Vinylic and aryl halides share a stubborn C–X bond, for reasons we explore in Section 6.7.2.

Quick check

1 of 3

How would you classify 2-bromo-2-methylbutane,

(C H_{3})_{2} C (B r) C H_{2} C H_{3}

primary alkyl halidesecondary alkyl halidetertiary alkyl halidevinylic halide

Show answer

Answer (C) tertiary alkyl halide

The carbon carrying Br is bonded to three other carbons (two

C H_{3}

and one

C H_{2}

), so the bromide is tertiary. It is not vinylic: there is no double bond.

6.2 Nomenclature

Every halide can be named in two ways. The older common name reads like the name of a salt: the alkyl group first, then the halide, as two words (isopropyl chloride, allyl bromide). The IUPAC name instead treats the halogen as a prefix on the parent hydrocarbon (fluoro, chloro, bromo, iodo): 2-chloropropane, 3-bromoprop-1-ene.

The usual IUPAC rules apply. Choose the longest chain (containing any C=C or C≡C), number it to give the lowest locants, and list prefixes alphabetically. In an alkane a halogen and an alkyl branch have equal standing, so if numbering from either end gives the same locant set, the prefix that comes first in the alphabet gets the lower number. In an alkene or alkyne the multiple bond takes priority for the low locant.

Rings and dihalides add two more conventions (Fig. 6.2). With one halogen on a benzene ring, both systems give the same name (iodobenzene). With two groups on the ring, a common name says where the second group sits relative to the first: o- (ortho, next door), m- (meta, one ring carbon between) or p- (para, opposite). IUPAC numbers the ring instead: 1,2-, 1,3- or 1,4-. On a chain, two identical halogens on one carbon make a geminal (gem) dihalide, called an alkylidene halide in the common system; two on adjacent carbons make a vicinal (vic) dihalide, an alkylene dihalide. IUPAC needs no special words: both are dihaloalkanes with the right locants.

Fig. 6.2Common names use o-, m-, p- and gem/vic; IUPAC names use numbers.

On the left, three benzene rings each carry two chlorine atoms: on neighbouring ring carbons (labelled ortho, o-, and 1,2-), with one ring carbon between them (meta, m-, 1,3-) and on opposite carbons (para, p-, 1,4-). On the right, separated by a line, two condensed formulas: CH3–CHBr2 with the CHBr2 group boxed, labelled gem, one carbon, 1,1-dibromoethane; and BrCH2–CH2Br with both carbons boxed, labelled vic, next-door carbons, 1,2-dibromoethane.

Table 6.2Common and IUPAC names of some halides

Structure	Common; IUPAC
$C H_{3} C H_{2} C H_{2} I$	n-propyl iodide; 1-iodopropane
$(C H_{3})_{2} C H C l$	isopropyl chloride; 2-chloropropane
$(C H_{3})_{2} C H C H_{2} B r$	isobutyl bromide; 1-bromo-2-methylpropane
$(C H_{3})_{3} C I$	tert-butyl iodide; 2-iodo-2-methylpropane
$C_{6} H_{11} B r$ (a ring)	cyclohexyl bromide; bromocyclohexane
$C H_{2} = C H C H_{2} I$	allyl iodide; 3-iodoprop-1-ene
$C F_{2} = C F_{2}$	tetrafluoroethylene; tetrafluoroethene
$C_{6} H_{5} C H_{2} B r$	benzyl bromide; (bromomethyl)benzene
$C H_{3} C H B r_{2}$	ethylidene bromide; 1,1-dibromoethane
$B r C H_{2} C H_{2} B r$	ethylene dibromide; 1,2-dibromoethane
$m$ - $B r C_{6} H_{4} C H_{3}$	m-bromotoluene; 1-bromo-3-methylbenzene

Example 6.1NCERT Example 6.1 · p. 162

Ring isomers of C₄H₇Br

A compound $C_{4} H_{7} B r$ contains either one ring or one C=C double bond. Find every structural isomer that has a ring and no double bond, name each one, and say whether it is a primary, secondary or tertiary bromide.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.2NCERT Example 6.2 · p. 163

Naming unsaturated halides

Give the IUPAC name of each compound and say whether the halogen is allylic, vinylic or on an ordinary alkyl carbon.

$C H_{2} = C H C H_{2} C H (C l) C H_{3}$
$C H_{3} C H = C (C l) C H_{3}$
$(C H_{3})_{2} C = C H C H_{2} I$
$C H_{2} = C H C H (I) C H_{2} C H_{3}$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 1 question

Practice: naming halides: questions with answers

1
Short answerEasyNCERT Intext Question 6.1
Write the condensed structural formula of each compound.
1. 2-Bromo-4-methylhexane
  Show solution
  $C H_{3} C H (B r) C H_{2}$ – $C H (C H_{3}) C H_{2} C H_{3}$
2. 1-Iodo-2,2-dimethylbutane
  Show solution
  $I C H_{2} C (C H_{3})_{2}$ – $C H_{2} C H_{3}$
3. 4-Bromo-3-ethylhept-1-ene
  Show solution
  $C H_{2} = C H$ – $C H (C_{2} H_{5})$ – $C H (B r) C H_{2}$ – $C H_{2} C H_{3}$
4. 1,3-Dichlorobut-1-ene
  Show solution
  $C l C H = C H$ – $C H (C l) C H_{3}$ : one vinylic and one allylic chlorine.
5. 2-Bromo-1-chloro-4-methylbenzene
  Show solution
  A benzene ring with Cl on C1, Br on the next carbon (C2) and $C H_{3}$ on C4, para to Cl.

Quick check

1 of 3

What is the IUPAC name of

(C H_{3})_{2} C H C H_{2} C H_{2} B r

4-bromo-2-methylbutane1-bromo-3-methylbutane2-methyl-4-bromobutane1-bromo-2-methylbutane

Show answer

Answer (B) 1-bromo-3-methylbutane

Number from the end nearer a substituent: Br on C1 and the methyl on C3 give {1, 3}, lower than {2, 4}. Prefixes go in alphabetical order, so 1-bromo-3-methylbutane. Option D puts the methyl on the wrong carbon.

6.3 Nature of C—X Bond

Start with the electronegativity row of the table below: every halogen pulls on electrons harder than carbon (2.5). The bonding pair of a C–X bond therefore sits nearer the halogen, which carries a partial negative charge (δ−), and leaves the carbon partially positive (δ+). That δ+ carbon is the target that electron-rich reagents aim for in almost every reaction of this chapter; the simulation further down lets you feel the pull.

Now read along the bond-length row. Each step from F to I adds a shell of electrons to the halogen, and the bond stretches from 138 pm to 213 pm. The longer the bond, the poorer the overlap of the bonding orbitals, so the bond enthalpy falls by almost half, from 460 to 239 kJ mol⁻¹. C–I bonds are by far the easiest of the four to break.

Table 6.3Carbon–halogen bonds in the halomethanes CH₃X (gas-phase values, rounded; EN = electronegativity of X, μ = dipole moment)NCERT Table 6.2 · p. 164

Property	F	Cl	Br	I
EN	4.0	3.2	3.0	2.7
length / pm	138	178	193	213
enthalpy / kJ mol⁻¹	460	350	294	239
μ / D	1.86	1.89	1.82	1.62

The dipole moments hold a surprise: chloromethane, not fluoromethane, has the largest. A dipole moment is charge separation multiplied by distance. Fluorine pulls hardest, but the C–F bond is so short that the product ends up slightly smaller than for the longer C–Cl bond. When several C–X bonds are present, their bond dipoles add as vectors; in tetrachloromethane the four point to the corners of a tetrahedron and cancel exactly, so $C C l_{4}$ has no dipole moment at all.

SimulationA polar C–X bond as two partial charges

Two fixed point charges sit side by side in the middle of the stage: a positive one on the left standing for the carbon atom and a negative one on the right standing for the halogen. Field lines run from the positive to the negative charge. A small negative test charge, representing an electron-rich nucleophile, can be dragged anywhere; an arrow and a readout show the size and direction of the force on it.

Fig. 6.3The + charge stands for the δ+ carbon and the − charge for the δ− halogen. The small negative test charge plays an approaching nucleophile.

Try this

Drag the negative test charge slowly around the pair. Where is it pulled hardest, and towards which atom?
Place the test charge just beyond the halogen end, on the line through both charges. Is it attracted or pushed away?
Move the test charge behind the carbon, on the side away from the halogen. Notice that this is the approach a nucleophile uses in the SN2 reaction.

Quick check

1 of 3

Arrange the C–X bonds from the strongest to the weakest.

C–Br
C–Cl
C–I
C–F

Show answer

Answer

C–F
C–Cl
C–Br
C–I

Bond enthalpy falls as the halogen gets larger and the bond longer: about 460, 350, 294 and 239 kJ per mole.

6.4 Methods of Preparation of Haloalkanes

There are three good starting points. Alcohols are cheap and easy to get; hydrocarbons, which can take up halogens by substitution or addition; and other halides, whose halogen can be swapped for a different one.

6.4.1 From Alcohols

The task is to replace the $- O H$ group by a halogen. Hydroxide is a poor leaving group, so every reagent below first turns the OH into something that departs easily. Acids protonate it to $- O H_{2}^{+}$ , which leaves as water, while phosphorus and sulfur halides convert it into a phosphite or chlorosulfite ester with an excellent leaving group.

Table 6.4Reagents that turn R–OH into R–XNCERT 6.4.1 · p. 164

Reagent	By-products	Notes
conc. $H C l$ + anhydrous $Z n C l_{2}$	$H_{2} O$	1° and 2° alcohols need the $Z n C l_{2}$ catalyst; a 3° alcohol reacts with conc. HCl alone, on shaking at room temperature
$H B r$ (48 % aqueous), or $N a B r$ + conc. $H_{2} S O_{4}$	$H_{2} O$ ( $N a H S O_{4}$ )	bromides; heat under reflux
$K I$ or $N a I$ + 95 % $H_{3} P O_{4}$	$H_{2} O$	iodides; phosphoric acid is used because $H_{2} S O_{4}$ would oxidise the HI to $I_{2}$
$P C l_{3}$ (1 mol per 3 mol ROH)	$H_{3} P O_{3}$	chlorides
$P C l_{5}$	$P O C l_{3}$ + $H C l$	chlorides
red P + $B r_{2}$ or $I_{2}$	$H_{3} P O_{3}$	$P B r_{3}$ or $P I_{3}$ is made in the flask itself (in situ)
$S O C l_{2}$ (thionyl chloride)	$S O_{2}$ (g) + $H C l$ (g)	chlorides; both by-products escape as gases, so the product needs little purifying

Thionyl chloride route

\begin{aligned} R - O H + S O C l_{2} \\ \overset{}{\to} R - C l + S O_{2} ↑ \\ + H C l ↑ \end{aligned}

With hydrogen halides, tertiary alcohols react fastest and primary ones slowest (3° > 2° > 1°), because a more substituted carbon lets go of the protonated $- O H_{2}^{+}$ group more readily. In practice a chloride is made either by passing a stream of dry HCl gas into the alcohol or by refluxing the alcohol with concentrated hydrochloric acid. Phenols are the exception to this whole section: overlap with the ring strengthens their C–O bond so much that it will not break, and aryl halides need the routes in Section 6.5.

6.4.2 From Hydrocarbons

(I) From alkanes, by free-radical halogenation. Ultraviolet light (or heat) splits $C l_{2}$ or $B r_{2}$ into atoms that pull hydrogen atoms off an alkane, one at a time, in a chain reaction. The trouble is lack of choice. A chlorine atom attacks almost any C–H bond it meets, and the monochloride it forms can be attacked again, so the flask ends up holding several isomers plus di- and trichlorides that boil close together. That is why chemists seldom use this route to make one pure haloalkane. (Bromine atoms are pickier, preferring tertiary hydrogens, but bromination still gives some mixture.)

4/4Replacing c gives a tertiary chloride; replacing d gives another primary chloride.

Fig. 6.4Hydrogens related by the molecule's symmetry give the same product, so we count kinds of hydrogen, not hydrogen atoms.

The condensed structure of 3-methylpentane is shown: CH3–CH2–CH–CH2–CH3 with a CH3 branch hanging from the central CH. Step 1 shows the molecule alone. Step 2 tags the groups: both end CH3 groups are a, both CH2 groups are b, the central CH is c and the branch CH3 is d, with the note that the same letter means the same kind of hydrogen, so four monochlorides are possible. Step 3 lists the products from replacing a and b: 1-chloro-3-methylpentane and 2-chloro-3-methylpentane. Step 4 adds the products from c and d: 3-chloro-3-methylpentane and 3-(chloromethyl)pentane.

Example 6.3NCERT Example 6.3 · p. 165

Counting monochlorination products

How many structurally different monochlorides $C_{6} H_{13} C l$ can free-radical chlorination of 3-methylpentane, $C H_{3} C H_{2} C H (C H_{3}) C H_{2} C H_{3}$ , give? Name them.

Try it yourself first. Then reveal the solution one step at a time.

(II) From alkenes. (i) Adding a hydrogen halide. HCl, HBr or HI adds across the C=C bond to give an alkyl halide. With an unsymmetrical alkene two products are possible, and the major one follows Markovnikov's rule (Class 11). The hydrogen goes to the double-bond carbon that already holds more hydrogens, because that route passes through the more stable carbocation. In the presence of a peroxide, HBr (only HBr) adds the other way round.

Markovnikov addition

\begin{aligned} R C H = C H_{2} + H B r \\ \overset{}{\to} R C H (B r) C H_{3} \end{aligned}

(ii) Adding a halogen. Shake an alkene with a few drops of red-brown bromine solution and the colour disappears within seconds: each $B r_{2}$ molecule has added across a C=C bond, giving a colourless vic-dibromide. Alkanes do not react this way, so the vanishing colour is a quick laboratory test for unsaturation.

1,2-dibromoethane

\begin{aligned} C H_{2} = C H_{2} + B r_{2} \\ \overset{}{\to} B r C H_{2} C H_{2} B r \end{aligned}

Try this

Which cooking oils have C=C bonds?

You’ll need

liquid paraffin
sunflower oil
coconut oil
dilute bromine water (handled by the teacher)
three test tubes with stoppers

Predict first

Which liquid do you expect to remove the colour of bromine water most completely?

liquid paraffinsunflower oilcoconut oil

Pour 2 mL of each liquid into separate, labelled test tubes.
Your teacher adds 1 mL of dilute bromine water to each tube.
Stopper, shake for 30 seconds and let the layers settle.
Compare the colour of each tube with a tube of bromine water alone.

Table 6.5Observations

Liquid	Colour after shaking
Liquid paraffin
Sunflower oil
Coconut oil

Your entries are saved automatically.

6.4.3 Halogen Exchange

To turn an alkyl chloride or bromide into an iodide, warm it with sodium iodide in dry acetone: the Finkelstein reaction. The exchange of halogens is reversible, but acetone tips the balance. Sodium iodide dissolves in it well, whereas sodium chloride and sodium bromide hardly dissolve and settle out as a white solid. Every ion pair that leaves the solution this way is removed from the equilibrium, so, as Le Chatelier's principle predicts, the reaction keeps running towards the alkyl iodide.

Finkelstein reaction

\begin{aligned} R C l + N a I \overset{a c e t o n e}{\to} \\ R I + N a C l ↓ \end{aligned}

Alkyl fluorides come from the Swarts reaction: an alkyl chloride or bromide is heated with an inorganic fluoride, for example antimony(III) fluoride, $S b F_{3}$ , cobalt(II) fluoride, $C o F_{2}$ , silver fluoride, $A g F$ , or mercury(I) fluoride, $H g_{2} F_{2}$ . Fluorine gas itself is useless for this job, because it attacks organic compounds explosively.

Swarts reaction

\begin{aligned} C_{2} H_{5} B r + A g F \\ \overset{}{\to} C_{2} H_{5} F + A g B r \end{aligned}

6.5 Preparation of Haloarenes

(i) Halogen straight onto the ring. On their own, chlorine and bromine are too weakly electrophilic to attack benzene. A Lewis acid catalyst, iron filings or iron(III) chloride or bromide, polarises the halogen molecule, and its positive end then replaces a ring hydrogen. When the arene has a side chain such as a methyl group, the reaction is run in the dark, because light would start a radical reaction on the side chain instead. Toluene gives a mixture of 2- and 4-halotoluene; the para isomer melts much higher and crystallises out, so the two are easy to separate.

Ring chlorination (in the dark)

\begin{aligned} C_{6} H_{6} + C l_{2} \\ \overset{F e C l_{3}}{\to} C_{6} H_{5} C l \\ + H C l \end{aligned}

The other two halogens misbehave. Iodination is reversible, because the HI it produces reduces iodobenzene back to benzene; an oxidising agent such as $H N O_{3}$ or $H I O_{4}$ is added to destroy the HI as fast as it forms. Fluorine has the opposite problem: it reacts with arenes so violently that ring fluorination cannot be controlled.

(ii) From an aromatic amine: the Sandmeyer reaction. This route replaces an $- N H_{2}$ group rather than a hydrogen, so the halogen ends up exactly where the amino group was. First the amine is diazotised: it is taken up in cold hydrochloric acid, kept at 273–278 K, and sodium nitrite is added, which gives an arenediazonium salt, $A r N_{2}^{+} X^{-}$ . These salts decompose on warming, so the solution goes straight into the next step. Copper(I) chloride or copper(I) bromide then exchanges the $- N_{2}^{+}$ group for Cl or Br, and nitrogen gas fizzes out. Iodides are even simpler to make: potassium iodide alone does the job, with no copper salt at all (Fig. 6.5).

Sandmeyer route (cold diazotisation, then CuCl)

\begin{aligned} A r N H_{2} \\ \overset{N a N O_{2}, H C l}{\to} A r N_{2}^{+} \\ \overset{C u C l}{\to} A r C l + N_{2} ↑ \end{aligned}

Fig. 6.5Ring halogenation adds X in place of H; the Sandmeyer route puts X where the amino group was.

Two rows of boxes joined by arrows. Top row, from an arene: benzene, C6H6, goes to chlorobenzene, C6H5Cl, along an arrow labelled Cl2, FeCl3, dark. Bottom row, from an amine: 4-methylaniline, CH3C6H4NH2, goes along an arrow labelled NaNO2 and HCl, cold, 273 to 278 K, to the diazonium salt CH3C6H4N2+ Cl−, drawn in a warm-coloured box to show it must be used at once. From the diazonium salt one arrow labelled CuCl leads up to 4-chlorotoluene, CH3C6H4Cl, and another labelled KI leads down to 4-iodotoluene, CH3C6H4I.

Example 6.4NCERT Example 6.4 · p. 167

Predicting addition products

Write the major organic product in each case.

$(C H_{3})_{2} C = C H_{2} + H B r$
$C H_{3} C H_{2} C H_{2} C H = C H_{2} + H B r$ , with a peroxide present
$C_{6} H_{5} C H = C H C H_{3} + H B r$
cyclopentene + $B r_{2}$

Try it yourself first. Then reveal the solution one step at a time.

6.6 Physical Properties

Freshly distilled haloalkanes are colourless. A bottle of an iodoalkane left on a sunny shelf, however, slowly turns brown: light breaks a few C–I bonds, and the iodine released colours the liquid (bromides do the same, more slowly). Only the lightest members, such as chloromethane (b.p. 249 K) and the refrigerant chlorodifluoromethane, are gases at room temperature; from bromoethane (b.p. 311 K) upwards they are liquids, and many polyhalides are solids. Low-boiling halides such as chloroform have a sweetish smell.

Boiling points. A halide boils far higher than the alkane it comes from: butane boils at 273 K, 1-chlorobutane at 352 K. Two kinds of attraction grow when a hydrogen becomes a halogen. The polar C–X bond adds dipole–dipole attraction, and the halogen's large, easily distorted electron cloud strengthens the dispersion (van der Waals) forces. (Comparing a halide with an alkane of equal molar mass can mislead. Bromine adds a lot of mass but little surface, so bromoethane, M = 109, boils at 311 K while heptane, M = 100, boils at 371 K.)

For one alkyl group the boiling point rises in the order R–F < R–Cl < R–Br < R–I. The bond dipole actually falls from Cl to I, yet the boiling point climbs, because the bigger halogen atoms carry many more electrons and the dispersion forces win. Among isomers, branching lowers the boiling point: a branched molecule is more compact, nearly spherical, and touches its neighbours over a smaller area. Fig. 6.6 shows both trends.

2/3Each larger halogen raises the boiling point of the butyl halide.

Fig. 6.6Heavier halogens raise the boiling point; branching lowers it.NCERT Fig. 6.1 · p. 168

A chart with boiling point in kelvin on the vertical axis from about 250 to 420. Step 1 shows only butane, the parent alkane, as a square at 273 K. Step 2 adds four joined points for 1-fluoro-, 1-chloro-, 1-bromo- and 1-iodobutane at 306, 352, 375 and 404 K, rising steadily. Step 3 replaces them with a dashed line through the four isomeric butyl chlorides: n-butyl 352 K, sec-butyl 341 K, isobutyl 342 K and tert-butyl 324 K, the most branched being lowest.

Melting points and symmetry. The three dichlorobenzenes boil within a few kelvin of one another (453, 446 and 447 K for the 1,2-, 1,3- and 1,4-isomers), but their melting points are very different: 256 K, 248 K and 327 K. Melting depends on how well molecules pack in a crystal, and the symmetrical para isomer stacks far more neatly. That is why 1,4-dichlorobenzene is a solid (it was once sold as moth crystals) while the other two are liquids.

Solubility. A haloalkane is polar, yet it barely mixes with water: even dichloromethane dissolves only to about 1.3 g per 100 g. Think of the energy bookkeeping. Liquid water is a tightly hydrogen-bonded network, and a haloalkane molecule can enter it only if some of those hydrogen bonds are broken. The dipole and dispersion attractions it offers water in return are much weaker, so the exchange does not pay. Mixed with another organic liquid, a haloalkane loses and gains attractions of similar size, and the two blend completely.

Density. A halogen atom is much heavier than the hydrogen it replaces, so density rises with the mass and the number of halogen atoms. A bigger alkyl part dilutes this effect, so within a series the density falls as the chain grows. Bromides, iodides and most polychloro compounds are denser than water and sink below it.

Table 6.6Densities of some halogen compounds at about 293 K (g mL⁻¹)NCERT Table 6.3 · p. 169

Trend	Compound	Density
heavier halogen	1-chlorobutane	0.89
heavier halogen	1-bromobutane	1.28
heavier halogen	1-iodobutane	1.62
longer chain	iodomethane	2.28
longer chain	iodoethane	1.95
more halogen	1,2-dichloroethane	1.25
more halogen	1,1,2-trichloroethane	1.44
more halogen	1,1,2,2-tetrachloroethane	1.59

Quick check

1 of 4

Arrange in order of increasing boiling point (lowest first).

$C H_{3} I$
$C H_{3} F$
$C H_{3} B r$
$C H_{3} C l$

Show answer

Answer

$C H_{3} F$
$C H_{3} C l$
$C H_{3} B r$
$C H_{3} I$

Same methyl group, bigger halogen: dispersion forces grow from F to I. The boiling points are about 195, 249, 277 and 316 K.

6.7 Chemical Reactions

Haloalkanes react in three main ways. A nucleophile can take the place of the halogen (nucleophilic substitution); a base can pull off HX to leave a double bond (elimination); and a metal can slip in between carbon and halogen (reaction with metals). Haloarenes follow different rules, which we take up in Section 6.7.2.

6.7.1 Reactions of Haloalkanes

Nucleophilic substitution

A nucleophile (Nu) is an electron-rich species: an anion such as $O H^{-}$ or $C N^{-}$ , or a neutral molecule with a lone pair such as $N H_{3}$ or $H_{2} O$ . Drawn to the δ+ carbon of a haloalkane, it donates an electron pair and forms a new bond there. At the same time the C–X bond breaks and the halogen departs with both bonding electrons as a halide ion, the leaving group. This exchange of one group for another is called nucleophilic substitution, and with it a single haloalkane can be turned into alcohols, ethers, amines, nitriles and more.

Nucleophilic substitution

\begin{aligned} N u^{-} + R - X \\ \overset{}{\to} R - N u + X^{-} \end{aligned}

Definition

Nucleophile

An electron-pair donor that attacks an electron-poor carbon atom and forms a new bond to it.

Example $O H^{-}$ , $C N^{-}$ , $I^{-}$ , $N H_{3}$ , $H_{2} O$ .

By choosing the nucleophile, you choose the product. The table sorts some common reagents by the atom that forms the new bond to carbon, using bromoethane as the haloalkane every time.

Table 6.7What bromoethane, CH₃CH₂Br, becomes, sorted by the attacking atomNCERT Table 6.4 · p. 170

Reagent (atom)	Product (family)
$N a O H$ or $K O H$ , aq (O)	$C H_{3} C H_{2} O H$ , ethanol (alcohol)
$C H_{3} O N a$ (O)	$C H_{3} C H_{2} O C H_{3}$ , methoxyethane (ether)
$C H_{3} C O O A g$ (O)	$C H_{3} C O O C H_{2} C H_{3}$ , ethyl ethanoate (ester)
$K N O_{2}$ (O)	$C H_{3} C H_{2} - O - N = O$ , ethyl nitrite (alkyl nitrite)
$N H_{3}$ , excess (N)	$C H_{3} C H_{2} N H_{2}$ , ethanamine (1° amine)
$C H_{3} N H_{2}$ (N)	$C H_{3} C H_{2} N H C H_{3}$ , N-methylethanamine (2° amine)
$A g N O_{2}$ (N)	$C H_{3} C H_{2} N O_{2}$ , nitroethane (nitroalkane)
$A g C N$ (N)	$C H_{3} C H_{2} N C$ , ethyl isocyanide (isocyanide)
$K C N$ (C)	$C H_{3} C H_{2} C N$ , propanenitrile (nitrile)
$H C \equiv C N a$ (C)	$C H_{3} C H_{2} C \equiv C H$ , but-1-yne (alkyne)
$N a I$ in acetone (I)	$C H_{3} C H_{2} I$ , iodoethane (alkyl iodide)
$L i A l H_{4}$ (H)	$C H_{3} C H_{3}$ , ethane (alkane)

Look again at the cyanide and nitrite rows: each ion appears twice, attacking through a different atom. Cyanide has lone pairs on both carbon and nitrogen; nitrite can bond through an oxygen or through nitrogen. Species with two such attacking sites are called ambident nucleophiles. Which site is used depends on the salt. With potassium cyanide the ion is free in solution and bonds through carbon, giving a nitrile. In silver cyanide the carbon end is already held in a largely covalent Ag–C bond, so only the nitrogen lone pair is available and an isocyanide results. Nitrite behaves in a parallel way, as the next example shows.

Definition

Ambident nucleophile

A nucleophile with two different atoms that can each donate an electron pair, so it can give two different products.

Example $C N^{-}$ (C or N), $N O_{2}^{-}$ (O or N).

Example 6.5NCERT Example 6.5 · p. 171

One salt, two products

Bromoethane gives mainly ethyl nitrite with potassium nitrite, but mainly nitroethane with silver nitrite. Explain.

Try it yourself first. Then reveal the solution one step at a time.

The SN2 mechanism

Nucleophilic substitution happens by two different routes. To tell them apart, chemists measure how the rate depends on concentrations. For bromoethane and hydroxide ion in a solvent such as acetone with a little water, doubling $[C H_{3} C H_{2} B r]$ doubles the rate, and so does doubling $[O H^{-}]$ . The rate is proportional to both, so the reaction is second order overall.

SN2 rate law

rate = k [R - X] [N u^{-}]

(6.1)

Both reactants appear in the rate law because both are present in the one and only step. Picture the nucleophile arriving from the side directly opposite the halogen, where it is shielded from the halogen's electron cloud. Bond making and bond breaking are not separate events but one smooth change. Halfway along, at the transition state, the carbon is joined by partial bonds to both Nu and X, which lie on a straight line through it. Its three other groups are pressed flat into the plane between them: five groups around one carbon. Such an arrangement exists only for an instant and can never be isolated. As X drifts away, the three groups keep moving and finish up pointing the other way, so the arrangement around the carbon has been turned inside out (Fig. 6.7).

2/3Five groups around carbon: I and Cl partly bonded, three groups in one plane.

Fig. 6.7Back-side attack turns the carbon's other three bonds inside out.NCERT Fig. 6.2 · p. 171

A carbon atom carries a hydrogen on a wedge, a methyl group in the plane and an ethyl group on a dash, all leaning left, with chlorine bonded on the right. Step 1 shows an iodide ion on the far left and a curved arrow from it to the back of the carbon, with a second curved arrow showing the C–Cl bond electrons moving to chlorine. Step 2 is the transition state in square brackets marked with a double dagger: iodine and chlorine are joined to carbon by dashed partial bonds, each labelled δ−, and the three other groups lie in a vertical plane through carbon. Step 3 shows the product: iodine bonded on the left, the three groups now leaning right, and a chloride ion drifting away on the right.

Definition

SN2 reaction

Substitution, nucleophilic, bimolecular: a one-step reaction in which bond making and bond breaking happen together, with rate $= k [R X] [N u^{-}]$ and inversion of configuration.

Structure matters because the nucleophile needs a clear path to the back of the carbon. Seen from the nucleophile's side (Fig. 6.8), a methyl halide shows an almost bare carbon; every alkyl group that replaces a hydrogen hides more of it. SN2 rates therefore fall steeply along the series $C H_{3} X$ > 1° > 2° >> 3°, and tertiary halides practically do not react this way. Crowding one carbon further out counts too: 1-bromo-2,2-dimethylpropane is a primary halide, yet its bulky neighbour slows back-side attack enormously.

3/4Two alkyl groups leave a narrow gap.

Fig. 6.8Seen from the nucleophile's side, each methyl group hides more of the target carbon.NCERT Fig. 6.3 · p. 173

A view straight along the C–Cl bond from the side a nucleophile would come from: the target carbon sits in the centre of a dashed circle, with chlorine hidden behind it. On the right a formula, its class and a horizontal bar labelled speed of attack. Step 1, chloromethane: three small hydrogens around the carbon, the carbon fully exposed, the bar long and marked fast. Step 2, chloroethane: one hydrogen is replaced by a large methyl disc that covers the top of the carbon; the bar shrinks to half, slower. Step 3, 2-chloropropane: two methyl discs, the bar much shorter. Step 4, 2-chloro-2-methylpropane: three methyl discs nearly hide the carbon, the bar almost vanishes, almost none.

The SN1 mechanism

Tertiary halides do substitute, but by another route. Warm 2-chloro-2-methylbutane, $C H_{3} C H_{2} C (C H_{3})_{2} C l$ , in aqueous ethanol and the rate depends on the concentration of the halide alone: adding more hydroxide ions does not speed it up. The reaction is first order. Reactions of this type go best in polar protic solvents, such as water and the alcohols, whose O–H groups can hydrogen-bond to the ions that form.

SN1 rate law

rate = k [R - X]

(6.2)

The mechanism has two steps. In the first, slow step the C–Cl bond breaks on its own, with both electrons going to chlorine, to give a carbocation and a chloride ion. Breaking the bond costs energy; much of it is repaid by solvation, as protic solvent molecules hydrogen-bond to the chloride ion and crowd round the cation. In the second, fast step a nucleophile, often a solvent molecule, bonds to the carbocation. Only the halide takes part in the slow step, so only its concentration appears in the rate law.

The steps (water attacks, then loses H⁺)

\begin{aligned} R_{3} C - C l \overset{s l o w}{\to} \\ R_{3} C^{+} + C l^{-} \\ R_{3} C^{+} + H_{2} O \overset{f a s t}{\to} \\ R_{3} C - O H + H^{+} \end{aligned}

where

$R_{3} C$: the 2-methylbutan-2-yl group, $(C H_{3})_{2} (C_{2} H_{5}) C$

2/4C–Br breaks: a planar carbocation forms, with an empty p orbital above and below the plane of its three groups.

Fig. 6.9The flat carbocation can be attacked from either face.

The top half is an energy profile: the curve climbs to a tall first peak labelled slow step, drops into a shallow valley labelled carbocation, rises over a small second peak labelled fast, and falls to the products. A dot marks the current point. The bottom half shows the structures for each step. Step 1: a tertiary carbon with methyl, ethyl and propyl groups leaning left and bromine bonded on the right; the dot is at the start. Step 2: bromide has moved away and the carbon is a flat cation marked + with two dashed lobes left and right; the dot sits in the valley. Step 3: hydroxide ions approach the cation from both sides; the dot is on the second peak. Step 4: two products, one with OH on the left and the groups leaning right, the other with OH on the right and the groups leaning left, labelled 50 : 50; the dot is at the end.

Definition

SN1 reaction

Substitution, nucleophilic, unimolecular: slow ionisation of R–X to a carbocation, then fast capture by the nucleophile; rate $= k [R X]$ .

Definition

Carbocation

An ion with a positively charged carbon that has only three bonds and six valence electrons; it is $s p^{2}$ -hybridised and planar.

Because ionisation is the slow step, anything that steadies the carbocation speeds up the whole reaction. Alkyl groups push electron density towards the positive carbon (by their inductive effect and by hyperconjugation), so cation stability runs 3° > 2° > 1° > $C H_{3}^{+}$ . SN1 reactivity follows the same order, 3° > 2° > 1° > methyl: exactly the reverse of the SN2 order.

Allylic and benzylic halides ionise especially easily, even when they are primary, because the positive charge of the cation is shared by resonance. In the allyl cation it is spread over both end carbons; in the benzyl cation it reaches three ring carbons as well as the $C H_{2}$ group.

Allyl cation

\begin{aligned} C H_{2} = C H C H_{2}^{+} \overset{}{\leftrightarrow} \\ _{}^{}_{}^{+} C H_{2} C H = C H_{2} \end{aligned}

Changing only the halogen moves both mechanisms in the same direction: iodides react fastest and fluorides hardly at all, $R - I$ > $R - B r$ > $R - C l$ >> $R - F$ . Two facts combine here. The C–I bond is the weakest of the four, and the iodide ion, large enough to spread its charge thinly, is the most stable halide ion to leave as.

Compare

SN2 and SN1 side by side

SN2

one step, no intermediate
rate $= k [R X] [N u^{-}]$ (second order)
back-side attack; inversion of configuration
fastest for $C H_{3} X$ and 1°
favoured by a strong nucleophile at good concentration

SN1

two steps through a carbocation
rate $= k [R X]$ (first order)
attack on either face; racemisation
fastest for 3°, allylic and benzylic
favoured by polar protic solvents and weak nucleophiles

Example 6.6NCERT Example 6.6 · p. 174

Which reacts faster by SN2?

In each pair, which compound reacts faster with a nucleophile by the SN2 route? Give the reason.

$C H_{3} C H_{2} C H_{2} C l$ or $(C H_{3})_{2} C H C l$
$C H_{3} C H_{2} C H_{2} C H_{2} C l$ or $C H_{3} C H_{2} C H_{2} C H_{2} I$

Try it yourself first. Then reveal the solution one step at a time.

Example 6.7NCERT Example 6.7 · p. 174

Ranking by both mechanisms

Arrange each set in order of reactivity towards SN1 and towards SN2.

$C H_{3} C l$ , $C H_{3} C H_{2} C l$ , $(C H_{3})_{2} C H C l$ , $(C H_{3})_{3} C C l$
$C_{6} H_{5} C H_{2} C l$ , $C_{6} H_{5} C H (C H_{3}) C l$ , $C_{6} H_{5} C (C H_{3})_{2} C l$

Try it yourself first. Then reveal the solution one step at a time.

InteractiveSubstitution lab: SN1 or SN2?

Buttons choose the substrate (methyl, primary, secondary, tertiary or benzylic bromide), the nucleophile (strong, such as hydroxide or cyanide, or weak, such as water) and the solvent (polar protic, such as water or ethanol, or acetone). Two horizontal bars compare how fast the SN2 and SN1 routes would be on an illustrative logarithmic scale, and a line names the route that wins. An energy-profile graph draws the one-hump SN2 curve and the two-hump SN1 curve with heights that follow the same scale. A mechanism panel shows the reacting carbon with its three groups and lets the learner step through, or play, the winning mechanism: back-side attack and umbrella inversion for SN2, or ionisation to a flat carbocation and attack from both faces for SN1. A final line states the stereochemical result for a chiral starting halide and warns when a strong base would cause elimination instead.

Fig. 6.10Pick a halide, a nucleophile and a solvent, then step or play through the mechanism the conditions favour.

Try this

Racing halides with silver nitrate

You’ll need

ethanolic silver nitrate solution (about 2 %)
1-chlorobutane, 1-bromobutane, 1-iodobutane and 2-chloro-2-methylpropane
four test tubes, a dropper and a beaker of warm water
a stopwatch

Predict first

Among the three straight-chain butyl halides, which will give a precipitate first?

1-chlorobutane1-bromobutane1-iodobutane

Which chloride will turn cloudy sooner?

1-chlorobutane2-chloro-2-methylpropaneboth at the same time

Stand four test tubes, each holding 1 mL of ethanolic silver nitrate, in a beaker of warm water (about 320 K).
Add 3 drops of a different halide to each tube at the same moment and start a timer.
Note the time at which each tube turns cloudy, and the colour of the precipitate.

Table 6.8Observations

Halide	Time (s), colour
1-Chlorobutane
1-Bromobutane
1-Iodobutane
2-Chloro-2-methylpropane

Your entries are saved automatically.

Optical activity

To follow what substitution does in three dimensions we need some ideas from stereochemistry. Ordinary light is a wave whose electric field vibrates in every direction at right angles to the beam. A polariser, such as a Nicol prism cut from calcite or a Polaroid sheet, lets through only the vibrations in one plane: the light that emerges is plane-polarised.

Put a solution of a suitable compound in the tube of a polarimeter (Fig. 6.11) and the light leaving the tube no longer vibrates in its original plane. The plane has been turned through an angle, which you find by rotating the analyser until the light is brightest again. Substances that do this show optical activity. If, looking towards the lamp, you have to turn the analyser clockwise, the substance is dextrorotatory, labelled (+) or d; anticlockwise means laevorotatory, (−) or l. The sign goes in front of the angle, as in +12° or −12°.

3/4A chiral solution turns that plane through an angle α.

Fig. 6.11The analyser is turned until the light is brightest again; the angle turned is the rotation.NCERT 6.7.1 (c)(i) · p. 175

A lamp on the left sends a beam through a polariser, a sample tube and an analyser to an eye on the right; small circles below the beam show the direction of vibration seen end-on. Step 1: the lamp's light vibrates in all planes, shown by three crossed lines. Step 2: after the polariser only a vertical vibration remains. Step 3: the tube fills with a chiral solution and the vibration leaving it is tilted by an angle alpha; the beam beyond the analyser is faint because the analyser is still vertical. Step 4: the analyser is turned clockwise by alpha until the light comes through brightly, and the reading +α is shown.

Definition

Optical activity

The rotation of the plane of plane-polarised light by a substance; (+) for clockwise and (−) for anticlockwise, viewed towards the source.

Chirality and enantiomers

Timeline

From twisted light to 3D molecules

1815
Liquids that turn light
Jean-Baptiste Biot finds that sugar solutions and turpentine rotate plane-polarised light.
1848
Mirror-image crystals
Louis Pasteur sorts tartrate crystals by hand into two mirror-image shapes; their solutions rotate light equally but in opposite directions.
1874
The tetrahedral carbon
J. H. van 't Hoff and J. A. Le Bel, independently, explain this with four bonds pointing to the corners of a tetrahedron.
1901
A first Nobel Prize
Van 't Hoff receives the first Nobel Prize in Chemistry, for his work on solutions and reaction rates.

A right chappal will not fit a left foot, however you turn it, although each chappal is the mirror image of the other. An object that cannot be superimposed on its own mirror image is chiral, from the Greek word for hand, the most familiar example. A cricket ball, a steel tumbler or an unsharpened pencil coincides with its reflection and is achiral. Molecules obey the same test: chiral molecules are optically active, achiral ones are not.

The commonest cause of chirality is a carbon atom bonded to four different groups, called a stereocentre (or asymmetric carbon). Its four bonds point to the corners of a tetrahedron, so the groups can be arranged in two mirror-image ways. To test a molecule, draw (or build) it beside its mirror image, turn the image as a whole and see whether it can be made to coincide with the original (Fig. 6.12). In 2-chloropropane the central carbon carries two identical methyl groups, and a half-turn makes the image fit exactly. In 2-chlorobutane, C2 carries H, Cl, $C H_{3}$ and $C_{2} H_{5}$ , and no turn can make the image fit.

4/4After the half-turn, ethyl and methyl are on the wrong sides. Chiral.

Fig. 6.12Build the mirror image, turn it, and try to lay it over the original.NCERT Figs. 6.5–6.6 · p. 176

A dashed vertical mirror divides the figure. Step 1: on the left, 2-chloropropane drawn with H up, Cl to the lower left in the plane, one methyl on a wedge and one on a dash; on the right, its mirror image. Step 2: the mirror image is turned half a turn about a vertical axis, shown by a curved arrow, and now matches the original exactly; a green panel reads identical: achiral. Step 3: the same layout for 2-chlorobutane, with a methyl on the wedge and an ethyl on the dash, and its mirror image. Step 4: the turned image has the ethyl on the wedge and the methyl on the dash; dashed rings mark the mismatched wedge groups and a panel reads not identical: chiral.

Definition

Chiral

Describes an object or molecule that cannot be superimposed on its mirror image; its opposite is achiral.

ExampleYour hand; 2-chlorobutane.

Definition

Stereocentre (asymmetric carbon)

A carbon atom bonded to four different atoms or groups.

ExampleC2 of 2-chlorobutane: H, Cl, $C H_{3}$ , $C_{2} H_{5}$ .

InteractiveChirality explorer

Four drop-down menus set the groups on a tetrahedral carbon, chosen from H, CH3, C2H5, OH, F, Cl, Br, I and COOH, with presets for 2-chlorobutane, 2-chloropropane, bromochlorofluoromethane, lactic acid and dichloromethane. The molecule is drawn on the left with two bonds in the plane, one wedge and one dash, and its mirror image on the right of a dashed mirror. A Turn button smoothly rotates the mirror image, choosing a turn that lays it exactly on the molecule whenever one exists, and otherwise a half-turn that keeps the top and left groups in place; a Reset button brings the image back. The result line says whether the turned image matches the original, rings and names the groups that are out of place, and says whether the carbon is a stereocentre and whether the molecule is chiral. A Swap button exchanges the wedge and dash groups to show that swapping two groups on a stereocentre gives the mirror-image arrangement.

Fig. 6.13Choose four groups for a carbon, make its mirror image, turn it and see whether the two match.

Let's explore

Build a stereocentre

You’ll need

clay or plasticine in four colours
toothpicks
a mirror

Predict first

With four different balls, will the model and its mirror-image model ever match?

no, neveryes, after the right turnonly if you view them from above

Roll a large ball of clay for carbon and push four toothpicks into it so they point to the corners of a tetrahedron (like a camera tripod with one leg pointing straight up).
Stick four small balls of different colours on the toothpicks.
Stand the model in front of a mirror and build a second model that matches the reflection.
Try every way you can to place one model exactly over the other.
Now make two of the balls on each model the same colour and try again.

Two stereoisomers that are non-superimposable mirror images form a pair of enantiomers. Measure the melting points, boiling points, densities or refractive indices of the two members of a pair and you get identical numbers; with ordinary achiral reagents they even react at the same rate. A polarimeter is what tells them apart: the two turn plane-polarised light by exactly the same angle, one clockwise and the other anticlockwise. Be careful with the labels, though. Whether a compound is (+) or (−) is found by experiment; it cannot be read off a wedge-and-dash drawing of its arrangement in space, its configuration.

Definition

Enantiomers

A pair of stereoisomers that are non-superimposable mirror images; they have equal and opposite optical rotations.

Example(+)- and (−)-2-chlorobutane.

What if a sample contains both enantiomers in equal amounts? For every molecule turning the light one way there is a partner turning it back, so the net rotation is zero. Such a 50 : 50 sample is a racemic mixture (or racemic modification), shown by the prefix (±) or dl, as in (±)-2-chlorobutane. Any process that turns a single enantiomer into this mixture is called racemisation.

Definition

Racemic mixture

An equimolar (50 : 50) mixture of two enantiomers; it is optically inactive and is labelled (±).

Example 6.8NCERT Example 6.8 · p. 177

Spotting chiral molecules

In each pair, which molecule is chiral? Justify your choice by looking for a stereocentre.

$C H_{3} C H_{2} C H (C l) C H_{2} C H_{3}$ and $C H_{3} C H (C l) C H_{2} C H_{2} C H_{3}$
$C H_{2} B r C l$ and $C H_{3} C H B r C l$
$(C H_{3})_{2} C H C H_{2} C H_{2} O H$ and $C H_{3} C H_{2} C H (C H_{3}) C H_{2} O H$

Try it yourself first. Then reveal the solution one step at a time.

Retention, inversion and racemisation

What happens at a stereocentre depends on whether any of its four bonds breaks. If none does, the groups keep their arrangement: retention of configuration. Take optically active 1-bromo-3-methylpentane, $B r C H_{2} C H_{2} C H (C H_{3}) C H_{2} C H_{3}$ , whose stereocentre is C3. Warmed with sodium iodide in acetone, it exchanges Br for I at C1, two carbons away, so the stereocentre is never touched and the 1-iodo-3-methylpentane formed has the same arrangement at C3. Its sign of rotation need not match that of the bromide, however: it is a different compound with its own rotation, which may be of either sign.

If a bond to the stereocentre is broken and a new group Y takes the place of X, there are three possible outcomes, set out below.

Compare

Three outcomes when Y replaces X at a stereocentre

Retention

Y takes the position X had
the product has the same configuration as the reactant

Inversion

Y enters on the side opposite to X
the other three groups flip over, like an umbrella
the product has the opposite configuration

Racemisation

both arrangements form, 50 : 50
the product shows no optical rotation

SN2 gives inversion. The nucleophile always enters from the side opposite the leaving group, so the configuration flips. When an optically active sample of 2-chlorobutane reacts with sodium iodide in acetone, the 2-iodobutane formed has the inverted configuration. In each molecule the iodine ends up on the face opposite to the one its chlorine left from.

SN1 gives (near) racemisation. Once the leaving group has gone, the flat carbocation keeps no memory of the original arrangement: it is achiral, and a nucleophile can bond to its upper face or its lower face with equal ease. Optically active 3-bromo-3-methylhexane in aqueous ethanol therefore gives 3-methylhexan-3-ol that is very nearly racemic. In practice there is often a small excess of the inverted product, because for a moment the departing bromide ion still screens the face it left from.

Elimination reactions

In a haloalkane, the carbon holding the halogen is the α-carbon and any carbon bonded to it is a β-carbon. Heat such a halide with a strong base, typically potassium hydroxide dissolved in ethanol (alcoholic KOH), and two things happen together: the base plucks a proton from a β-carbon and the halide ion departs from the α-carbon. The electrons left behind form a C=C double bond between the two carbons. Because H and X are lost as a pair the reaction is called dehydrohalogenation, and because the hydrogen comes from the β position it is also called β-elimination.

Dehydrohalogenation (alcoholic KOH, heat; KX and H₂O also form)

\begin{aligned} R C H_{2} C H_{2} X \overset{K O H}{\to} \\ R C H = C H_{2} \end{aligned}

Definition

β-Elimination

A reaction in which H is removed from a β-carbon and X from the neighbouring α-carbon, forming a C=C bond.

Example1-Bromopropane and alcoholic KOH give propene.

Many halides have hydrogens on two or more different β-carbons, so more than one alkene is possible, yet one product usually dominates. In 1875 Alexander Zaitsev, working at Kazan, spotted the pattern: the main product is the alkene whose double-bond carbons carry the most alkyl groups. Such highly substituted alkenes are more stable, and the route leading to them has the lower energy barrier (Fig. 6.14).

4/4The more substituted alkene is the major product.

Fig. 6.14The more substituted alkene is the major product (Zaitsev rule).NCERT 6.7.1 (2) · p. 180

The condensed structure CH3–CH(Br)–CH(CH3)–CH3 is shown with the bromine-bearing carbon tagged alpha and the two neighbouring carbons tagged beta. Step 1 shows only the tags. Step 2 rings the terminal CH3 and the Br and draws an arrow down-left to CH2=CH–CH(CH3)2, 3-methylbut-1-ene, with one alkyl group on C=C. Step 3 rings the branched CH and the Br and draws an arrow down-right to CH3–CH=C(CH3)2, 2-methylbut-2-ene, with three alkyl groups on C=C. Step 4 shows both products labelled minor and major respectively.

Example

Applying the Zaitsev rule

3-Bromo-3-methylpentane, $C H_{3} C H_{2} C (B r) (C H_{3}) C H_{2} C H_{3}$ , is heated with alcoholic KOH. Name the possible alkenes and the major one.

Try it yourself first. Then reveal the solution one step at a time.

Key ideaNCERT Elimination versus substitution · p. 180

Elimination or substitution?

Any base is also a potential nucleophile, so substitution and elimination compete whenever a haloalkane meets one, and whichever is faster decides the main product. Size matters: a bulky base finds it far easier to pluck an exposed hydrogen from the edge of the molecule than to squeeze in at the carbon, so it tends to eliminate. Heat and an alcoholic rather than aqueous solvent push the same way. The structure of the halide counts as well:

Table 6.9Typical outcomes

Halide	Conditions	Result
$C H_{3} X$ or 1°	good nucleophile ( $I^{-}$ , $C N^{-}$ , aqueous $O H^{-}$ )	SN2
1°	bulky strong base, e.g. $(C H_{3})_{3} C O - K^{+}$ , heat	elimination
2°	strong base in ethanol, heat	mainly elimination
2°	good nucleophile that is a weak base	SN2
3°	water or ethanol, no strong base	SN1 (some elimination)
3°	strong base (alcoholic KOH)	elimination

This explains a classic pair of results. With aqueous KOH a haloalkane mainly gives an alcohol: in water the hydroxide ion is wrapped in hydrogen-bonded water molecules, which weakens it as a base but leaves it a good nucleophile. With alcoholic KOH on heating, the medium contains ethoxide ions, $C_{2} H_{5} O^{-}$ , a stronger and less solvated base, so elimination to the alkene wins.

Reaction with metals

Chlorides, bromides and iodides of carbon react with several metals to give organometallic compounds, in which carbon is bonded directly to a metal. The most useful are the alkyl- and arylmagnesium halides, RMgX, first made by the French chemist Victor Grignard in 1900 and named Grignard reagents after him. To make one, magnesium turnings are stirred with the halide in dry diethyl ether; ether molecules bond to the magnesium and keep the reagent in solution.

Making a Grignard reagent

\begin{aligned} C_{2} H_{5} B r + M g \\ \overset{d r y e t h e r}{\to} C_{2} H_{5} M g B r \end{aligned}

Making RMgX turns the polarity of the carbon upside down. In R–X the carbon is δ+, but magnesium is far less electronegative than carbon, so in R–MgX the shared electrons lean towards carbon and it becomes δ−; the Mg–X bond is essentially ionic. A δ− carbon is a powerful base. Offer it an O–H or N–H proton, from water, an alcohol or an amine, and it takes it at once, leaving the hydrocarbon R–H. That is why Grignard reagents are prepared in dry ether in carefully dried glassware. Chemists also use this reactivity on purpose: adding water converts a halide cleanly into an alkane, and heavy water ( $D_{2} O$ ) places a deuterium atom exactly where the halogen was.

Grignard reagents are destroyed by water

\begin{aligned} R M g X + H_{2} O \overset{}{\to} \\ R H + M g (O H) X \end{aligned}

Definition

Grignard reagent

An organomagnesium halide $R M g X$ , made from R–X and Mg in dry ether; its δ− carbon acts as a strong base and nucleophile.

Wurtz reaction. Sodium in dry ether can also join two alkyl groups: two molecules of a haloalkane give up their halogens to the metal, and their alkyl groups bond together as R–R. The method suits a single halide only, because a mixture of two different haloalkanes gives three different alkanes.

Wurtz reaction

\begin{aligned} 2 R X + 2 N a \\ \overset{d r y e t h e r}{\to} R - R \\ + 2 N a X \end{aligned}

Practice · 2 questions

Practice: reactions of haloalkanes: questions with answers

1
Short answerMediumNCERT Intext Questions 6.7–6.8
In each pair, choose the compound that reacts faster by the stated mechanism, and give one reason.
1. SN2: $C H_{3} C H_{2} C H_{2}$ – $C H_{2} C H_{2} C l$ or $C H_{3} C H_{2}$ – $C H (C l) C H_{2} C H_{3}$
  Show solution
  1-Chloropentane: primary, so its back side is less crowded than that of the secondary 3-chloropentane.
2. SN2: $C H_{3} (C H_{2})_{5} I$ or $(C H_{3} C H_{2})_{2} C H C H_{2} I$
  Show solution
  1-Iodohexane. Both are primary, but in 1-iodo-2-ethylbutane two ethyl groups hang from the β-carbon, right next to the reacting carbon, and crowd the approach.
3. SN1: $(C H_{3})_{2} C (C l)$ – $C H_{2} C H_{2} C H_{3}$ or $C H_{3} C H (C l)$ – $C H_{2} C H_{2} C H_{3}$
  Show solution
  2-Chloro-2-methylpentane: it ionises to a tertiary carbocation, more stable than the secondary one from 2-chloropentane.
4. SN1: $C H_{3} C H_{2} C H_{2} C H_{2} C l$ or $C_{6} H_{5} C H_{2} C l$
  Show solution
  Benzyl chloride: its cation is stabilised by resonance with the ring.
2
Short answerHardNCERT Intext Question 6.9
Identify A, B, C and D.
1. Bromocyclopentane + Mg (dry ether) → A; A + $H_{2} O$ → B
  Show solution
  A = cyclopentylmagnesium bromide, $C_{5} H_{9} M g B r$ ; B = cyclopentane, $C_{5} H_{10}$ .
2. A halide $R - X$ gives 2,5-dimethylhexane with Na in dry ether. With Mg in dry ether it gives C, and C with water gives D. Identify R–X, C and D.
  Show solution
  2,5-Dimethylhexane is $(C H_{3})_{2} C H C H_{2}$ – $C H_{2} C H (C H_{3})_{2}$ , two isobutyl groups joined. So R–X is 1-bromo-2-methylpropane, $(C H_{3})_{2} C H C H_{2} B r$ (or the chloride/iodide); C = $(C H_{3})_{2} C H$ – $C H_{2} M g B r$ ; D = 2-methylpropane, $(C H_{3})_{3} C H$ .

6.7.2 Reactions of Haloarenes

Nucleophilic substitution

Reflux chloroethane with aqueous sodium hydroxide and you soon have ethanol; treat chlorobenzene the same way and you get your chlorobenzene back. Halogens on a ring (and on a C=C carbon) shrug off nucleophiles, and four features of the C–X bond explain why:

The bond is partly double. One of chlorine's lone pairs overlaps with the π electrons of the ring, giving resonance structures that contain $C = C l^{+}$ . The real C–Cl bond lies somewhere between single and double, and is harder to break.
An $s p^{2}$ carbon. An $s p^{2}$ orbital has 33 % s character against 25 % for $s p^{3}$ , so the ring carbon holds the bonding pair closer. The C–Cl bond is shorter and stronger: about 174 pm in chlorobenzene against about 178 pm in chloromethane.
No stable cation. If the C–X bond did ionise, the phenyl cation would have its empty orbital in the plane of the ring, where the π electrons cannot reach it. Without resonance help it is very unstable, so the SN1 route is closed.
No way in from behind. The back side of the C–X bond points into the ring, and the electron-rich π cloud repels an approaching nucleophile, so the SN2 route is closed too.

2/3Cl's lone pair delocalises into the ring: a shorter, partly double C–Cl bond.

Fig. 6.15Partial double-bond character, an sp² carbon and a ring in the way all protect the C–Cl bond.NCERT 6.7.2 (1) · p. 181

Chlorobenzene is drawn on the left and chloromethane on the right, each with three lone pairs on chlorine. The ring carbon bearing Cl is labelled sp2 and the carbon of chloromethane sp3. Step 1 shows just the two molecules. Step 2 adds a curved arrow from a chlorine lone pair towards the ring bond in chlorobenzene, a dashed second line between C and Cl showing partial double-bond character, and the bond lengths 174 pm for chlorobenzene and 178 pm for chloromethane. Step 3 shows a nucleophile arrow coming from below each molecule: under chlorobenzene it stops at the ring with a bar marked blocked, while under chloromethane it reaches the carbon, marked open.

Replacement by hydroxyl. Forced hard enough, the substitution does go: chlorobenzene heated with aqueous NaOH above 600 K under some 300 atm gives sodium phenoxide, which acidifies to phenol. A far gentler way in is to put a nitro group (or another strong electron-withdrawing group) ortho or para to the halogen. Each such group makes the reaction much easier, as the table shows, whereas a nitro group meta to the halogen barely helps.

Table 6.10How hard is it to replace Cl by OH?

Chloroarene	What it takes
chlorobenzene	NaOH(aq) above 600 K under high pressure
1-chloro-3-nitrobenzene	almost as hard: a meta NO₂ hardly helps
1-chloro-4-nitrobenzene	hot NaOH(aq), around 430 K (an industrial route to 4-nitrophenol)
1-chloro-2,4-dinitrobenzene	boiling aqueous sodium carbonate, near 373 K
2-chloro-1,3,5-trinitrobenzene	warm water alone

The reason lies in the mechanism, which is an addition followed by an elimination. In the slow first step, hydroxide adds to the carbon carrying the chlorine. That carbon becomes $s p^{3}$ , the ring loses its aromatic sextet, and it carries a negative charge that resonance spreads over the carbons ortho and para to the site of attack. In the fast second step, chloride leaves and the aromatic ring is restored. A nitro group sitting at one of those ortho or para positions can pull the negative charge onto its own oxygen atoms, stabilising the intermediate and lowering the barrier. From a meta position it never meets the charge (Fig. 6.16).

3/5A para (or ortho) NO₂ pulls part of the charge onto its oxygens.

Fig. 6.16A nitro group ortho or para to the halogen can carry part of the negative charge; a meta nitro group cannot.NCERT 6.7.2 (1) · p. 183

A benzene ring with Cl at the top and NO2 at the bottom, labelled 4-nitro isomer. Step 1: an OH⁻ ion approaches the carbon bearing Cl. Step 2: OH and Cl are both on that carbon, the inner ring circle is replaced by a dashed arc, and minus-sign bubbles appear on the two carbons next to it and on the bottom carbon. Step 3: an extra minus bubble appears on the NO2 group, showing it shares the charge. Step 4: Cl⁻ has left, only OH remains on the ring and the aromatic circle returns. Step 5: the left ring keeps the step 3 picture, and a second ring, the 3-nitro isomer, appears on the right with NO2 on a lower side carbon; its minus bubbles fall on the other carbons, never on the one carrying NO2.

Electrophilic substitution

The ring of a haloarene still takes part in the familiar electrophilic substitutions of benzene (nitration, halogenation, sulphonation and both kinds of Friedel–Crafts reaction), but the halogen changes the details. Its electronegativity drains electron density from the ring through the σ bond (the −I effect), so electrophiles attack more slowly than they attack benzene and stronger conditions are needed. Its lone pairs, though, can be shared with the ring (the +R effect), and that help is available only when the electrophile has attacked an ortho or para carbon (Fig. 6.17). The outcome is unusual: a group that deactivates the ring yet sends newcomers ortho and para, mostly para, where there is more room.

3/3Like ortho, but with more room: the main product.

Fig. 6.17Only ortho and para attack put positive charge next to the chlorine, where its lone pair can share it.NCERT 6.7.2 (2) · p. 184

Chlorobenzene with Cl and its lone pairs at the top. In each step the electrophile E and a hydrogen are drawn on the attacked ring carbon, and plus-sign bubbles mark the three ring carbons that share the positive charge of the intermediate. Step 1, ortho attack: one bubble lands on the carbon carrying Cl and an extra bubble appears beside the chlorine; the side panel reads + reaches the C–Cl carbon, so Cl can share it, favoured. Step 2, meta attack: the bubbles fall on the other three carbons and never on the C–Cl carbon; the panel reads Cl cannot help, disfavoured. Step 3, para attack: as for ortho, a bubble on the C–Cl carbon and one beside Cl; favoured (major).

Table 6.11Electrophilic substitution of bromobenzene (some ortho isomer also forms)

Reagents	Main product
nitration: conc. $H N O_{3}$ + conc. $H_{2} S O_{4}$	1-bromo-4-nitrobenzene
bromination: $B r_{2}$ , anhydrous $F e B r_{3}$	1,4-dibromobenzene
Friedel–Crafts acylation: $C H_{3} C O C l$ , anhydrous $A l C l_{3}$	1-(4-bromophenyl)ethan-1-one
Friedel–Crafts alkylation: $C H_{3} C H_{2} C l$ , anhydrous $A l C l_{3}$	1-bromo-4-ethylbenzene
sulphonation: conc. $H_{2} S O_{4}$ , heat	$p$ - $B r C_{6} H_{4} S O_{3} H$ , a sulfonic acid

Example 6.9NCERT Example 6.9 · p. 185

Reading a product ratio

Nitrating chlorobenzene gives about 30 % 1-chloro-2-nitrobenzene, 1 % 1-chloro-3-nitrobenzene and 69 % 1-chloro-4-nitrobenzene, and the reaction is roughly thirty times slower than for benzene. Explain both facts.

Try it yourself first. Then reveal the solution one step at a time.

Reaction with metals

Wurtz–Fittig reaction. Sodium in dry ether can link an aryl group to an alkyl group: heat a mixture of an aryl halide and an alkyl halide with the metal and an alkylbenzene forms. Fittig reaction. With only an aryl halide present, sodium links two aryl groups to give a biaryl.

Wurtz–Fittig: ethylbenzene

\begin{aligned} C_{6} H_{5} B r + C_{2} H_{5} B r \\ + 2 N a \overset{e t h e r}{\to} \\ C_{6} H_{5} C_{2} H_{5} + 2 N a B r \end{aligned}

Fittig: biphenyl

\begin{aligned} 2 C_{6} H_{5} B r + 2 N a \\ \overset{e t h e r}{\to} C_{6} H_{5} - C_{6} H_{5} \\ + 2 N a B r \end{aligned}

6.8 Polyhalogen Compounds

Put several halogen atoms on one or two carbons and you have a polyhalogen compound. A handful of them have been made by the million tonnes for factories, farms and homes. The very qualities that made them attractive, chemical stability and a talent for dissolving fats and oils, are also what make several of them harmful to people and slow to disappear from the environment.

6.8.1 Dichloromethane (Methylene chloride)

Dichloromethane, $C H_{2} C l_{2}$ , dissolves a huge range of organic substances, does not catch fire easily and boils at only about 313 K, so it evaporates quickly and leaves nothing behind. That combination makes it the active ingredient of many paint and varnish strippers and a standard extraction solvent in pharmaceutical factories (it was once used to decaffeinate coffee). It also degreases metal parts and used to propel aerosol sprays.

Its hazards follow from the same volatility. The dense vapour collects in poorly ventilated rooms, and breathing it acts like a slow anaesthetic. People become dizzy and sick, their fingers and toes may tingle or go numb, and even modest levels can dull hearing and sight a little. Splashes burn the skin and can injure the cornea. In the body, the liver turns part of it into carbon monoxide, which cuts the blood's oxygen supply. It is used only with good ventilation, gloves and eye protection.

6.8.2 Trichloromethane (Chloroform)

Chloroform, $C H C l_{3}$ , is sold in brown bottles filled right to the neck, and the reason is chemical. In air and light it is slowly oxidised to phosgene (carbonyl chloride), an extremely poisonous gas. A dark, brim-full bottle keeps out both light and oxygen, and about 1 % of ethanol is usually added to destroy any phosgene that still forms.

Chloroform to phosgene

\begin{aligned} 2 C H C l_{3} + O_{2} \\ \overset{l i g h t}{\to} 2 C O C l_{2} \\ + 2 H C l \end{aligned}

Today most chloroform is not used as such but converted into chlorodifluoromethane, $C H C l F_{2}$ , the refrigerant R-22, which is also the starting material for PTFE. It remains a useful laboratory solvent for waxes, rubber and many plant extracts. Its first fame was medical: in 1847 the Edinburgh obstetrician James Young Simpson introduced it as a general anaesthetic. Doctors abandoned it because the gap between a sleeping dose and a dangerous one was small. Chloroform dulls the central nervous system (drowsiness and headache after a short exposure), can disturb the heart's rhythm and, with repeated exposure, damages the liver and kidneys; the liver itself converts some of it into phosgene.

6.8.3 Triiodomethane (Iodoform)

Triiodomethane, $C H I_{3}$ , is a yellow crystalline solid with a penetrating smell. You will meet it in later chapters as the yellow precipitate of the iodoform test. Nineteenth-century surgeons sprinkled it on wounds, but it never killed microbes itself: it slowly released iodine, which did. Its smell was so persistent that it gave way to odourless iodine preparations such as povidone–iodine, the brown antiseptic used in hospitals today.

6.8.4 Tetrachloromethane (Carbon tetrachloride)

Tetrachloromethane, $C C l_{4}$ , is dense, does not burn and dissolves grease superbly. For much of the twentieth century it was therefore everywhere: in fire extinguishers, dry cleaning, household spot removers and metal degreasing, and as the raw material for CFC refrigerants and aerosol propellants. Two discoveries ended most of these uses.

First, it is a serious poison. Its vapour depresses the nervous system: mild exposure brings headache, giddiness and vomiting, while heavy exposure can cause unconsciousness, a disturbed or stopped heartbeat and death, sometimes with lasting nerve damage. It is also a notorious liver poison, linked to liver cancer, and it irritates the eyes. Second, vapour that escapes rises unchanged to the stratosphere and destroys ozone (Section 6.8.5). International agreements have since phased out almost all its production.

6.8.5 Freons

In the 1920s home refrigerators ran on ammonia, sulfur dioxide or chloromethane, all toxic, and leaks caused deaths. In 1928 Thomas Midgley and his colleagues found a safer answer in the chlorofluorocarbons (CFCs) of methane and ethane, sold as freons. These gases are non-toxic, do not burn or corrode metals, react with almost nothing and liquefy under modest pressure. For decades they filled refrigerators, air conditioners, aerosol cans and plastic foams. Freon-12, $C C l_{2} F_{2}$ , is made from tetrachloromethane by Swarts-type exchange of chlorine for fluorine.

Their stability is the problem. A freon molecule released at ground level does not break down; over a few years it drifts up into the stratosphere. There, hard ultraviolet light splits off a chlorine atom, which starts a chain of reactions that destroys ozone, the layer that screens the Earth from harmful UV (Fig. 6.18). A single chlorine atom can remove many thousands of ozone molecules before it is locked away.

4/4ClO meets an O atom, gives $O_{2}$ and frees $C l \cdot$ again: a catalytic cycle.

Fig. 6.18The chlorine atom comes out of each cycle unchanged, ready to destroy more ozone.

On the left, a column from the ground up to a shaded band labelled stratosphere. Step 1: a CCl2F2 molecule sits at ground level. Step 2: it has risen into the stratosphere, a wavy UV arrow strikes it and a chlorine atom, Cl·, is released; a Cl· node also appears at the top of a cycle on the right. Step 3: an arrow labelled + O3 leads round the right side of the cycle to a ClO· node, with the note O2 released below it. Step 4: an arrow labelled + O leads back up the left side to Cl·, closing the loop, and a second note reads Cl· comes back. The two equations are written out below the figure.

The chlorine cycle

\begin{aligned} C l \cdot + O_{3} \\ \overset{}{\to} C l O \cdot + O_{2} \\ C l O \cdot + O \\ \overset{}{\to} C l \cdot + O_{2} \end{aligned}

6.8.6 p,p'-Dichlorodiphenyltrichloroethane (DDT)

DDT, $(C l C_{6} H_{4})_{2} C H - C C l_{3}$ , was the first chlorinated organic insecticide to be used on a massive scale. From the 1940s it was sprayed in homes and on fields and marshes against the mosquitoes that carry malaria and the lice that carry typhus. India's National Malaria Control Programme, launched in 1953, relied on indoor DDT spraying, and malaria cases fell from tens of millions a year to around a hundred thousand by the mid-1960s.

Timeline

The rise and fall of DDT

1874
First made
Othmar Zeidler prepares DDT in a laboratory but finds no use for it.
1939
A powerful insecticide
Paul Müller, at Geigy in Switzerland, discovers that it kills insects on contact.
1948
Nobel Prize
Müller receives the Nobel Prize in Physiology or Medicine.
1962
Silent Spring
Rachel Carson's book describes harm to birds and wildlife.
1972
Banned in the USA
Most uses are stopped; many countries follow.
2001
Stockholm Convention
DDT is restricted worldwide to controlling disease-carrying insects.

DDT's weakness is that it does not go away. Its molecule resists chemical and biological attack, and it dissolves in fats and oils far better than in water. An animal that takes in DDT can neither break it down quickly nor flush it out in urine, so the DDT is stored in its fatty tissue and the store grows with every meal. A predator inherits the stores of all the prey it eats, so the concentration rises at each level of a food chain (Fig. 6.19). At the top, fish-eating birds laid eggs with shells too thin to survive. Two further problems showed up within a few years of mass spraying: DDT proved highly poisonous to fish, and many insect populations became resistant to it.

4/4Top predators carry about a million times the level in the water.

Fig. 6.19Illustrative numbers: each feeding level concentrates fat-soluble DDT further.

A row of five icons from left to right: water, plankton, a small fish, a big fish and a heron, joined by arrows, with dots on each icon standing for stored DDT. Under each is the concentration relative to the water on an illustrative scale. Step 1 shows water, with one dot and times 1, and plankton, with three dots and times 10 cubed. Step 2 adds the small fish with six dots and times 10 to the fourth. Step 3 adds the big fish with eleven dots and times 10 to the fifth. Step 4 adds a heron, drawn standing on two legs with a long neck, with sixteen dots and times 10 to the sixth.

Quick check

1 of 3

Match each compound with a fact about it.

Tap an item on the left, then its match on the right.

Show answer

Answer

$C H C l_{3}$ : turns into phosgene in air and light
$C H I_{3}$ : works as an antiseptic by releasing iodine
$C C l_{2} F_{2}$ : refrigerant that harms the ozone layer
DDT : is stored in body fat
$C H_{2} C l_{2}$ : paint-stripping solvent

These are the main uses and hazards described for each compound in this section.

Recap

Haloalkanes and Haloarenes: key points and summary

Haloalkanes have X on an $s p^{3}$ carbon and haloarenes on an aromatic $s p^{2}$ carbon; halides are sorted by the number of X atoms and as alkyl (1°, 2°, 3°), allylic, benzylic, vinylic or aryl (Section 6.1).
IUPAC names treat halogens as prefixes; common names use alkyl + halide, o/m/p, and gem/vic for dihalides (Section 6.2).
The C–X bond is polar (δ+ carbon) and grows longer and weaker from C–F to C–I (Section 6.3).
Haloalkanes come from alcohols (HX, PX₃, PCl₅, SOCl₂), from alkanes (free radicals, mixtures), from alkenes (HX, X₂) and by halogen exchange (Finkelstein, Swarts) (Section 6.4).
Haloarenes come from ring halogenation with a Lewis acid, or from diazonium salts (Sandmeyer; KI for iodides) (Section 6.5).
Halides boil higher than their parent alkanes; b.p. rises from F to I and falls with branching; they barely dissolve in water, and bromides, iodides and polychlorides are denser than water (Section 6.6).
SN2: one step, rate $= k [R X] [N u^{-}]$ , back-side attack and inversion, fastest for methyl and 1° halides (The SN2 mechanism).
SN1: slow ionisation to a planar carbocation, rate $= k [R X]$ , racemisation, fastest for 3°, allylic and benzylic halides (The SN1 mechanism).
A carbon with four different groups is a stereocentre; chiral molecules are optically active, enantiomers rotate light equally in opposite directions, and racemic mixtures show no rotation (Chirality and enantiomers).
Alcoholic KOH gives β-elimination, with the more substituted alkene as the major product (Zaitsev) (Elimination reactions).
Grignard reagents RMgX (dry ether) have a δ− carbon and are destroyed by water; sodium couples halides (Wurtz, Wurtz–Fittig, Fittig) (Reaction with metals).
Haloarenes resist nucleophiles unless o/p nitro groups stabilise the anionic intermediate; halogens are deactivating but o/p-directing in electrophilic substitution (Section 6.7.2).
Dichloromethane, chloroform, iodoform, carbon tetrachloride, freons and DDT are useful but can harm health, deplete ozone or build up in food chains (Section 6.8).

Flash cards · 9

Named reactions to remember

1 / 9

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Intext Question 6.1	Practice: naming halides · Q1
NCERT Intext Question 6.2	Practice: preparing halides · Q1
NCERT Intext Question 6.3	Practice: preparing halides · Q2
NCERT Intext Question 6.4	Practice: preparing halides · Q3
NCERT Intext Question 6.5	Practice: preparing halides · Q4
NCERT Intext Questions 6.7–6.8	Practice: reactions of haloalkanes · Q1
NCERT Intext Question 6.9	Practice: reactions of haloalkanes · Q2
NCERT Exercises Q6.1	Exercises · Q1
NCERT Exercises Q6.2	Exercises · Q2
NCERT Exercises Q6.3	Exercises · Q3
NCERT Exercises Q6.4	Exercises · Q4
NCERT Exercises Q6.5	Exercises · Q5
NCERT Exercises Q6.6	Exercises · Q6
NCERT Exercises Q6.7	Exercises · Q7
NCERT Exercises Q6.8	Exercises · Q8
NCERT Exercises Q6.9	Exercises · Q9
NCERT Exercises Q6.10	Exercises · Q10
NCERT Exercises Q6.11	Exercises · Q11
NCERT Exercises Q6.12	Exercises · Q12
NCERT Exercises Q6.13	Exercises · Q13
NCERT Exercises Q6.14	Exercises · Q14
NCERT Exercises Q6.15	Exercises · Q15
NCERT Exercises Q6.16	Exercises · Q16
NCERT Exercises Q6.17	Exercises · Q17
NCERT Exercises Q6.18	Exercises · Q18
NCERT Exercises Q6.19	Exercises · Q19
NCERT Exercises Q6.20	Exercises · Q20
NCERT Exercises Q6.21	Exercises · Q21
NCERT Exercises Q6.22	Exercises · Q22

Quick check: MCQs on Haloalkanes and Haloarenes

6.1.3 Compounds Containing sp² C—X Bond3 questions
6.2 Nomenclature3 questions
6.3 Nature of C—X Bond3 questions
6.4.3 Halogen Exchange4 questions
6.5 Preparation of Haloarenes3 questions
6.6 Physical Properties4 questions
The SN1 mechanism4 questions
Retention, inversion and racemisation4 questions
Reaction with metals4 questions
Reaction with metals4 questions
6.8.6 p,p'-Dichlorodiphenyltrichloroethane (DDT)3 questions

Glossary · 15

Key terms (English–Hindi)

ambident nucleophile: A nucleophile with two different atoms able to attack, such as cyanide (C or N) and nitrite (O or N).First seen in the chapter →

carbocationकार्बधनायनkaarbdhanaayan: A planar ion with a positively charged, $s p^{2}$ carbon that has only six valence electrons.First seen in the chapter →
chiralकिरलkiral: Not superimposable on its own mirror image, like a left hand; an object that is superimposable is achiral.First seen in the chapter →

enantiomer: One of a pair of stereoisomers that are non-superimposable mirror images of each other.First seen in the chapter →

Grignard reagent: An alkyl- or arylmagnesium halide, $R M g X$ , made from a halide and magnesium in dry ether; its carbon is strongly basic.First seen in the chapter →

haloalkaneहैलोऐल्केनhailoalkane: A compound in which a halogen atom is bonded to an $s p^{3}$ carbon of an alkyl group, $R - X$ .First seen in the chapter →
haloareneहैलोऐरीनhailoareen: A compound in which a halogen atom is bonded directly to an $s p^{2}$ carbon of an aromatic ring, $A r - X$ .First seen in the chapter →

leaving group: The atom or group that departs with the bonding pair during substitution or elimination; for haloalkanes it is the halide ion.First seen in the chapter →

nucleophileनाभिकस्नेहीnaabhikasnehi: An electron-rich species (an anion or a molecule with a lone pair) that attacks an electron-poor carbon and forms a new bond to it.First seen in the chapter →

optical activityध्रुवण घूर्णकताdhruvan ghoornakta: The ability of a substance to turn the plane of plane-polarised light; clockwise is (+), anticlockwise is (−).First seen in the chapter →

racemic mixtureरेसिमिक मिश्रणresimik mishran: A 50 : 50 mixture of two enantiomers; it shows no optical rotation and is written with a (±) prefix.First seen in the chapter →

SN1 reaction: Unimolecular nucleophilic substitution: slow ionisation to a carbocation, then fast capture; rate $= k [R X]$ .
SN2 reaction: Bimolecular nucleophilic substitution: one step, rate $= k [R X] [N u^{-}]$ , back-side attack and inversion of configuration.
stereocentre: A carbon atom bonded to four different groups; swapping any two of them gives the mirror-image arrangement.First seen in the chapter →

β-eliminationविलोपनvilopan: Loss of a hydrogen from the β-carbon and the halogen from the α-carbon to form a C=C double bond.First seen in the chapter →

6.1 Classification

6.1.1 On the Basis of Number of Halogen Atoms

6.1.2 Compounds Containing sp³ C—X Bond

6.1.3 Compounds Containing sp² C—X Bond

6.2 Nomenclature

6.3 Nature of C—X Bond

6.4 Methods of Preparation of Haloalkanes

6.4.1 From Alcohols

6.4.2 From Hydrocarbons

6.4.3 Halogen Exchange

6.5 Preparation of Haloarenes

6.6 Physical Properties

6.7 Chemical Reactions

6.7.1 Reactions of Haloalkanes

Nucleophilic substitution

The SN2 mechanism

The SN1 mechanism

Optical activity

Chirality and enantiomers

Retention, inversion and racemisation

Elimination reactions

Reaction with metals

6.7.2 Reactions of Haloarenes

Nucleophilic substitution

Electrophilic substitution

Reaction with metals

6.8 Polyhalogen Compounds

6.8.1 Dichloromethane (Methylene chloride)

6.8.2 Trichloromethane (Chloroform)

6.8.3 Triiodomethane (Iodoform)

6.8.4 Tetrachloromethane (Carbon tetrachloride)

6.8.5 Freons

6.8.6 p,p'-Dichlorodiphenyltrichloroethane (DDT)

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