Hydrocarbons: Class 11 Chemistry · Chapter 9

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

Published 22 Oct 2026 · Updated 22 Oct 2026

118 min
33 practice questions
41 MCQs

What you will learn in Hydrocarbons

Classify hydrocarbons and name alkanes, alkenes, alkynes and arenes by IUPAC rules, including their structural and geometrical isomers
Describe how alkanes, alkenes, alkynes and benzene are prepared, and choose a route to a given compound
Explain the free-radical mechanism of alkane halogenation and the other reactions of alkanes
Draw Sawhorse and Newman projections of ethane and compare the stability of its conformations
Predict the products of additions to alkenes and alkynes, including Markovnikov and anti-Markovnikov orientation and ozonolysis
Explain the acidity of terminal alkynes in terms of hybridisation
Explain the structure and stability of benzene, apply the Hückel rule, and describe electrophilic substitution and the directive influence of substituents
Recognise hydrocarbons as fuels and raw materials, and the toxicity of benzene and polynuclear hydrocarbons

Original companion text aligned to NCERT Chemistry Class 11, Chapter 9, pp. 295–327. Not affiliated with or endorsed by NCERT.

Light the gas stove at home and you are burning a hydrocarbon. The cylinder in most Indian kitchens holds LPG (liquefied petroleum gas), mostly propane and butane kept liquid under pressure. CNG autos and buses run on compressed natural gas, which is mainly methane, and LNG is that same gas chilled to about −162 °C so that tankers can ship it across oceans.

A hydrocarbon is a compound of carbon and hydrogen only. Almost all of them come from three underground stores, crude oil, natural gas and coal, with a little from living sources such as the biogas of a village gobar-gas plant. Besides being fuels, they are the chemical industry's raw material: the plastic of a water bottle, the synthetic fibre of a sports jersey, a painter's thinner and many medicines can all be traced back to a hydrocarbon.

This chapter works through four families — alkanes, alkenes, alkynes and aromatic hydrocarbons. For each you will learn to name the compounds, to make them, and to predict how they react, using the reaction mechanism as the main tool.

9.1 Classification

Chemists sort hydrocarbons by two features: the kind of carbon–carbon bonds present, and whether the carbon skeleton is an open chain or a ring (Fig. 9.1). The saturated family, with single bonds only, includes the open-chain alkanes and the ring-shaped cycloalkanes; unsaturated compounds have at least one multiple bond; aromatic compounds are a special group of rings typified by benzene.

3/3Benzene-like rings with evenly shared ring electrons.

Fig. 9.1Three families, sorted by bond type and by chain or ring.NCERT p. 296

A tree diagram with Hydrocarbons at the top and three branches. Step 1 shows the saturated branch: open-chain alkanes such as propane, C3H8, and ring cycloalkanes such as cyclohexane, C6H12. Step 2 adds the unsaturated branch: alkenes with a C=C bond, such as ethene C2H4, and alkynes with a C≡C bond, such as ethyne C2H2. Step 3 adds the aromatic branch: arenes such as benzene, C6H6, drawn as a hexagon with a circle.

Model it

Build hydrocarbon models

You’ll need

kneaded atta dough or clay, in two sizes of ball
used matchsticks or short pieces of broom stick
two pipe cleaners

Predict first

How many different skeletons can you build from four carbons joined only by single bonds, with no rings?

1234

Roll larger balls for carbon atoms and smaller ones for hydrogen atoms.
Give each carbon four sticks pointing as far apart as possible, and each hydrogen one stick. Build methane.
Pull off one H and plug in a new C with its own hydrogens. Repeat to build every skeleton you can with four carbons and no rings.
Join two carbons with two bendy strips instead of one stick. Try to twist one carbon while holding the other.

Quick check

1 of 2

Which of these is a saturated hydrocarbon?

Ethene,

C_{2} H_{4}

Propyne,

C_{3} H_{4}

Benzene,

C_{6} H_{6}

Cyclohexane,

C_{6} H_{12}

Show answer

Answer (D) Cyclohexane, $C_{6} H_{12}$

Cyclohexane has only single bonds, so it is saturated even though its formula fits

C_{n} H_{2 n}

like an alkene. A ring uses up two hydrogens just as a double bond does.

9.2 Alkanes

Methane, $C H_{4}$ , is the simplest alkane: it bubbles up from marshes and paddy fields, collects in coal mines, and makes up most of biogas and CNG. Add a $C H_{2}$ unit and you get ethane, $C_{2} H_{6}$ ; add another for propane, $C_{3} H_{8}$ , then butane, $C_{4} H_{10}$ , and so on. Members that differ by $C H_{2}$ form a homologous series with one general formula:

Alkanes

C_{n} H_{2 n + 2}

where

$n$: the number of carbon atoms (1, 2, 3, …)

Definition

Alkane

A saturated open-chain hydrocarbon, $C_{n} H_{2 n + 2}$ , with only C–C and C–H single bonds.

ExampleButane, $C_{4} H_{10}$ , in LPG.

Every carbon in an alkane is $s p^{3}$ hybridised. Its four bonds point to the corners of a regular tetrahedron, 109.5° apart, so methane is a tetrahedron with carbon at the centre. Head-on overlap of two $s p^{3}$ orbitals makes a C–C σ bond (154 pm), and overlap of an $s p^{3}$ orbital with a hydrogen 1s orbital makes a C–H σ bond (112 pm). Chains of such tetrahedra zigzag in space.

At room temperature alkanes ignore acids, bases and most oxidising agents. That is why their old name was paraffins, from Latin words meaning “little affinity”.

9.2.1 Nomenclature and Isomerism

Methane, ethane and propane can each be built in only one way. From four carbons on, one molecular formula can describe several skeletons. $C_{4} H_{10}$ is either a straight four-carbon chain, butane (n-butane), or a three-carbon chain with a methyl branch on the middle carbon, 2-methylpropane (isobutane). $C_{5} H_{12}$ has three skeletons: pentane, 2-methylbutane and 2,2-dimethylpropane. Compounds with the same formula but different carbon chains are chain isomers, one kind of structural isomer. Because their shapes differ, they boil and melt at different temperatures (Section 9.2.3).

Table 9.1Chain isomers of the alkanes grow rapidly in number

Formula	Name of straight-chain isomer	Number of chain isomers
$C_{4} H_{10}$	butane	2
$C_{5} H_{12}$	pentane	3
$C_{6} H_{14}$	hexane	5
$C_{7} H_{16}$	heptane	9
$C_{8} H_{18}$	octane	18
$C_{9} H_{20}$	nonane	35
$C_{10} H_{22}$	decane	75

To describe a skeleton, count the carbon neighbours of each carbon. One neighbour makes it primary (1°): every chain end is primary, and so is the lone carbon of methane. Two neighbours make it secondary (2°), three tertiary (3°) and four quaternary (4°). In 2,2,4-trimethylpentane, the “iso-octane” that sets the octane number of petrol, you can find five 1°, one 2°, one 3° and one 4° carbon.

Example 9.1NCERT Problem 9.1 · p. 297

Heptane isomers with a five-carbon chain

$C_{7} H_{16}$ has nine chain isomers. Draw and name those whose longest chain has exactly five carbons.

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

Remove one hydrogen from an alkane and you get an alkyl group, $C_{n} H_{2 n + 1}^{-}$ , often written R–. Methyl ( $C H_{3}^{-}$ ) and ethyl ( $C_{2} H_{5}^{-}$ ) have one form each. Propyl has two, because the hydrogen can be taken from an end carbon (propyl) or from the middle carbon (isopropyl, propan-2-yl).

Example 9.2NCERT Problem 9.2 · p. 298

Isomeric butyl groups

How many different butyl groups, $C_{4} H_{9}^{-}$ , are there? Draw each one and name the chloroalkane you get by attaching Cl at its free bond.

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

The IUPAC rules from Unit 8 reduce to four moves: find the longest chain; number it so that the substituents get the lowest locants; name the substituents and list them in alphabetical order; then join the parts with commas between numbers and hyphens between numbers and words. Table 9.2 shows the trickier rules at work.

Table 9.2Rules that decide an alkane name

Condensed structure	IUPAC name	Rule shown
$(C H_{3})_{2} C H C H_{2} C H (C H_{3}) C H_{2} C H_{3}$	2,4-dimethylhexane	locants {2,4} beat {3,5}
$C H_{3} C H_{2} C H (C_{2} H_{5}) C H (C H_{3}) C H_{2} C H_{3}$	3-ethyl-4-methylhexane	equal locants: first in the alphabet gets the lower number
$(C H_{3})_{3} C C H_{2} C H (C_{2} H_{5}) C H_{2} C H_{3}$	4-ethyl-2,2-dimethylhexane	di-, tri- are ignored when alphabetising
$C H_{3} C H_{2} C H (C_{2} H_{5}) C H [C (C H_{3})_{3}] C H_{2} C H_{2} C H_{3}$	4-tert-butyl-3-ethylheptane	tert- and sec- are ignored (tert-butyl files under b); iso- is not
$C H_{3} (C H_{2})_{3} C H [C H (C H_{3}) C H (C H_{3})_{2}] (C H_{2})_{3} C H_{3}$	5-(1,2-dimethylpropyl)nonane	a branched substituent is numbered from its own point of attachment and written in brackets

Example 9.3NCERT Problem 9.3 · p. 299

Naming branched alkanes

Give the IUPAC name of each: (i) $(C H_{3})_{3} C C H (C H_{3}) C H_{2} C H_{3}$ , (ii) $C_{2} H_{5} C (C H_{3})_{2} C H (C H_{3}) C_{2} H_{5}$ , (iii) $(C_{2} H_{5})_{4} C$ .

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

Going the other way, from a name to a structure, uses the same rules in reverse: draw the parent chain, number it, hang the substituents on the numbered carbons, and finally give every carbon enough hydrogens to make four bonds (Fig. 9.2).

4/4Each carbon gets hydrogens up to four bonds: $C_{10} H_{22}$ .

Fig. 9.2From name to structure: chain, numbers, substituents, hydrogens.NCERT p. 300

The name 4-ethyl-2,3-dimethylhexane is printed at the top. Step 1 draws a chain of six carbon symbols joined by bonds. Step 2 numbers them 1 to 6 from the left. Step 3 hangs a CH3 group on carbons 2 and 3 and a C2H5 group on carbon 4. Step 4 adds hydrogens so that each carbon has four bonds, giving CH3, CH, CH, CH, CH2, CH3 along the chain, and notes the formula C10H22.

Example 9.4NCERT Problem 9.4 · p. 300

Structures from names

Draw condensed formulas for (i) 3-ethyl-2,4-dimethylpentane and (ii) 5-ethyl-2,2-dimethyloctane.

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

Example 9.5NCERT Problem 9.5 · p. 300

Fixing wrong names

Each name below breaks a rule. Draw the structure it describes, find the mistake and correct the name: (i) 5-ethyl-2-methylhexane; (ii) 3,3,5-trimethylhexane.

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

Quick check

1 of 3

What is the IUPAC name of

(C H_{3})_{2} C H C H (C_{2} H_{5}) C H_{3}

2,3-dimethylpentane2-ethyl-3-methylbutane3,4-dimethylpentane2-isopropylbutane

Show answer

Answer (A) 2,3-dimethylpentane

The longest chain runs through the ethyl group (five carbons), with methyls on C2 and C3. A name containing 2-ethyl always means a longer chain was missed.

9.2.2 Preparation

Most alkanes are obtained directly from petroleum and natural gas. In the laboratory, three families of reactions give them.

From unsaturated hydrocarbons

Vanaspati ghee is made by bubbling hydrogen through vegetable oil over finely divided nickel, so that hydrogen adds to the C=C bonds of the oil. The same hydrogenation turns any alkene or alkyne into an alkane. A catalyst is needed because H–H is a strong bond: metal surfaces adsorb hydrogen and loosen it. Palladium and platinum are active enough at room temperature, whereas nickel, the cheap industrial choice, needs higher temperature and pressure.

But-2-ene and propyne give alkanes

\begin{aligned} C H_{3} C H = C H C H_{3} \\ + H_{2} \overset{P t, P d o r N i}{\to} \\ C H_{3} C H_{2} C H_{2} C H_{3} \\ C H_{3} C \equiv C H + 2 H_{2} \\ \overset{P t, P d o r N i}{\to} \\ C H_{3} C H_{2} C H_{3} \end{aligned}

From alkyl halides

(i) Zinc and dilute hydrochloric acid replace the halogen of an alkyl chloride, bromide or iodide (not a fluoride) by hydrogen:

2-Chlorobutane to butane

\begin{aligned} C H_{3} C H_{2} C H C l C H_{3} \\ + H_{2} \overset{Z n, H^{+}}{\to} \\ C H_{3} C H_{2} C H_{2} C H_{3} \\ + H C l \end{aligned}

(ii) In the Wurtz reaction, sodium metal in dry ether joins two alkyl groups: $2 R X + 2 N a \overset{}{\to} R - R + 2 N a X$ . The ether must be free of water, because sodium reacts with water. The product has twice the carbon count of the alkyl halide, so the method suits alkanes with an even number of carbons.

1-Bromopropane to hexane

\begin{aligned} 2 C H_{3} C H_{2} C H_{2} B r \\ + 2 N a \overset{d r y e t h e r}{\to} \\ C H_{3} (C H_{2})_{4} C H_{3} \\ + 2 N a B r \end{aligned}

From carboxylic acids

(i) Decarboxylation. Soda lime is sodium hydroxide mixed with calcium oxide. Heated with a sodium carboxylate, RCOONa, it removes the –COONa group as sodium carbonate and puts H in its place, so the chain comes out one carbon shorter:

Sodium propanoate to ethane

\begin{aligned} C H_{3} C H_{2} C O O N a \\ + N a O H \to_{h e a t}^{C a O} \\ C H_{3} C H_{3} + N a_{2} C O_{3} \end{aligned}

Example 9.6NCERT Problem 9.6 · p. 301

Choosing the salt

You need 2-methylpropane and have soda lime. Which sodium carboxylate should you heat with it? Give the equation.

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

(ii) Kolbe's electrolytic method. A concentrated aqueous solution of a sodium or potassium carboxylate is electrolysed. At the anode each carboxylate ion loses an electron; the radical formed throws off $C O_{2}$ , and two alkyl radicals pair up into an alkane. Hydrogen gas and alkali form at the cathode (Fig. 9.3).

Sodium propanoate to butane

\begin{aligned} 2 C H_{3} C H_{2} C O O N a \\ + 2 H_{2} O \\ \overset{e l e c t r o l y s i s}{\to} \\ C H_{3} C H_{2} C H_{2} C H_{3} \\ + 2 C O_{2} + H_{2} \\ + 2 N a O H \end{aligned}

3/4The radical loses $C O_{2}$ ; two ethyl radicals join to give butane.

Fig. 9.3Butane forms at the anode; hydrogen at the cathode.NCERT p. 301

An electrolysis cell with a positive anode on the left and a negative cathode on the right in a solution of propanoate and sodium ions. Step 1 shows the cell and the ions. Step 2 shows a propanoate ion at the anode losing an electron to become a propanoyloxy radical. Step 3 shows that radical splitting into carbon dioxide and an ethyl radical, and two ethyl radicals joining to form butane. Step 4 shows water gaining electrons at the cathode to give hydrogen gas and hydroxide ions.

Since every product molecule is built from two radicals, Kolbe's method cannot make methane: the smallest possible product, from an ethanoate, is ethane.

9.2.3 Properties

Physical properties

C–C bonds are non-polar and C–H bonds nearly so, because carbon and hydrogen have similar electronegativities. Alkane molecules therefore attract one another only through weak van der Waals (dispersion) forces. At 298 K the alkanes up to butane ( $C_{1}$ – $C_{4}$ ) are gases, those from $C_{5}$ to $C_{17}$ are liquids, and those with 18 or more carbons are waxy solids. Pure alkanes are colourless and odourless; the smell of LPG comes from a sulfur compound added deliberately so that leaks are noticed.

Being non-polar, alkanes do not dissolve in water but mix freely with other non-polar liquids: like dissolves like. That is why petroleum jelly, a soft mixture of large alkanes, will not rinse off your hands with water alone, yet a few drops of coconut oil remove it at once.

Boiling points climb steadily along the series, because larger molecules have more surface over which dispersion forces act (Fig. 9.4). Among isomers, branching lowers the boiling point. A branched molecule is more compact, almost ball-shaped, so neighbouring molecules touch over a smaller area and are easier to separate. Hexane boils at 342 K, while its most branched isomer, 2,2-dimethylbutane, boils at about 323 K.

2/2Same formula, $C_{6} H_{14}$ : the more branched, the lower the boiling point.

Fig. 9.4Longer chains boil higher; more branching boils lower.NCERT Table 9.2 · p. 302

Step 1 is a vertical bar chart of the boiling points of the straight-chain alkanes from methane to octane, rising from 112 K to 399 K, with the 298 K room-temperature line marked so that the first four are gases. Step 2 is a horizontal bar chart of the five C6H14 isomers on an axis starting at 300 K: hexane 342 K, 3-methylpentane 336 K, 2-methylpentane 333 K, 2,3-dimethylbutane 331 K and 2,2-dimethylbutane 323 K, the most branched lowest.

Chemical properties

Alkanes resist acids, bases, oxidising agents and reducing agents, but under suitable conditions they take part in the following reactions.

1. Substitution (halogenation). In ultraviolet light or diffused sunlight, or on strong heating, a halogen replaces the hydrogens of an alkane one after another. With methane and enough chlorine all four can be replaced:

Stepwise chlorination (each step also gives HCl)

\begin{aligned} C H_{4} \overset{C l_{2}, h ν}{\to} C H_{3} C l \\ \overset{C l_{2}, h ν}{\to} C H_{2} C l_{2} \\ \overset{C l_{2}, h ν}{\to} C H C l_{3} \\ \overset{C l_{2}, h ν}{\to} C C l_{4} \end{aligned}

Fluorine is so reactive that the reaction runs out of control, while iodine barely reacts and the HI it forms reverses the change, unless an oxidant such as $H I O_{3}$ or $H N O_{3}$ is present to destroy the HI. The useful halogens are therefore chlorine and bromine, and the overall order of reactivity is $F_{2} > C l_{2} > B r_{2} > I_{2}$ . Which hydrogen is replaced also matters: tertiary goes fastest, then secondary, then primary (3° > 2° > 1°). Nitration and sulphonation do not work on the lower alkanes.

Iodination needs an oxidant to remove HI

\begin{aligned} C_{2} H_{6} + I_{2} \overset{}{⇌} \\ C_{2} H_{5} I + H I \\ 5 H I + H I O_{3} \overset{}{\to} \\ 3 I_{2} + 3 H_{2} O \end{aligned}

Example

How selective is chlorination?

At 298 K a chlorine atom removes a secondary hydrogen about 3.8 times as fast as a primary one. Estimate the percentage of 2-chloropropane in the monochlorination of propane, $C H_{3} C H_{2} C H_{3}$ .

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

The mechanism. Halogenation is a free-radical chain reaction with three stages (Fig. 9.5). A free radical is an atom or group with an unpaired electron, shown by a dot.

Initiation. Light or heat splits the weakest bond present, Cl–Cl, so that each atom keeps one electron (homolysis): $C l_{2} \overset{}{\to} 2 C l^{∙}$ .
Propagation. (a) $C l^{∙} + C H_{4} \overset{}{\to} C H_{3}^{∙} + H C l$ . (b) $C H_{3}^{∙} + C l_{2} \overset{}{\to} C H_{3} C l + C l^{∙}$ . Step (b) regenerates the chlorine atom that step (a) needs, so one photon can lead to thousands of product molecules. Similar steps on $C H_{3} C l$ give $C H_{2} C l_{2}$ , and so on.
Termination. Radicals meet and pair up: $C l^{∙} + C l^{∙} \overset{}{\to} C l_{2}$ , $C H_{3}^{∙} + C H_{3}^{∙} \overset{}{\to} C_{2} H_{6}$ and $C H_{3}^{∙} + C l^{∙} \overset{}{\to} C H_{3} C l$ . The chain stops when radicals or reactants run out.

The second termination step explains why a little ethane is always found among the products of chlorinating methane.

4/4Two radicals join; ethane is a tell-tale by-product.

Fig. 9.5One photon starts a chain; radical pairing ends it.NCERT p. 303

Step 1, initiation: a Cl–Cl molecule absorbs light and splits into two chlorine atoms, each with a dot for its unpaired electron. Step 2, propagation (a): a chlorine atom takes a hydrogen from methane, giving HCl and a methyl radical. Step 3, propagation (b): the methyl radical takes a chlorine atom from Cl2, giving chloromethane and a new chlorine atom, and a curved arrow loops back to step (a) to show the chain. Step 4, termination: three ways two radicals pair up, giving Cl2, ethane or chloromethane.

2. Combustion. With plenty of air, alkanes burn completely to carbon dioxide and water and release a large amount of heat; one mole of methane gives out 890 kJ. For any alkane:

Complete combustion of an alkane

\begin{aligned} C_{n} H_{2 n + 2} + \frac{3 n + 1}{2} O_{2} \\ ⟶ n C O_{2} \\ + (n + 1) H_{2} O \end{aligned}

(9.19)

With too little air, combustion is incomplete: poisonous carbon monoxide and soot (carbon black) form instead, for example $C H_{4} + O_{2} \overset{}{\to} C + 2 H_{2} O$ . Carbon black is used in printer ink, black pigments and as a filler in rubber tyres.

Example

Energy in a kitchen cylinder

A domestic LPG cylinder holds 14.2 kg of gas. Treat it as pure butane ( $M = 58.1 g m o l^{- 1}$ , $Δ_{c} H = - 2878 k J m o l^{- 1}$ ). How much heat does the full cylinder release, and how many moles of $O_{2}$ does it use?

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

3. Controlled oxidation. With a regulated supply of oxygen, a catalyst and the right temperature and pressure, oxidation stops part-way and gives useful products:

methane and oxygen over copper at 523 K and 100 atm give methanol, $C H_{3} O H$ ;
methane and oxygen over heated $M o_{2} O_{3}$ give methanal, $H C H O$ ;
ethane and oxygen with manganese(II) ethanoate give ethanoic acid, $C H_{3} C O O H$ ;
an alkane with a tertiary hydrogen, such as 2-methylbutane, is oxidised by $K M n O_{4}$ to a tertiary alcohol, here 2-methylbutan-2-ol. Other alkanes resist $K M n O_{4}$ .

4. Isomerisation. On heating with anhydrous $A l C l_{3}$ and HCl gas, a straight chain reshuffles into branched isomers. Refineries use this because branched alkanes burn more smoothly in petrol engines.

Butane to 2-methylpropane

\begin{aligned} C H_{3} C H_{2} C H_{2} C H_{3} \\ \to_{H C l, h e a t}^{a n h y d \cdot A l C l_{3}} \\ (C H_{3})_{2} C H C H_{3} \end{aligned}

5. Aromatization (reforming). Refineries pass the vapour of straight-chain alkanes with six or more carbons over oxides of vanadium, molybdenum or chromium on alumina, at about 773 K and 10–20 atm. The chains lose hydrogen and close into rings: hexane becomes benzene and heptane becomes toluene (methylbenzene).

Heptane to toluene

\begin{aligned} C H_{3} (C H_{2})_{5} C H_{3} \\ \to_{o x i d e c a t a l y s t}^{773 K, 10 - 20 a t m} \\ C_{6} H_{5} C H_{3} + 4 H_{2} \end{aligned}

6. Reaction with steam. At 1273 K over nickel, methane and steam give carbon monoxide and hydrogen. This is the main industrial source of hydrogen, for example for ammonia in fertiliser plants.

Steam reforming

\begin{aligned} C H_{4} + H_{2} O \\ \to_{1273 K}^{N i} C O + 3 H_{2} \end{aligned}

7. Pyrolysis (cracking). Heated strongly in the absence of air, often over a catalyst, large alkanes break into smaller alkanes and alkenes. The reaction runs through free radicals, and the chain can snap at many points, so a mixture forms. Cracking turns heavy oil fractions into petrol and fuel gas.

One of many possible splits of decane

\begin{aligned} C_{10} H_{22} \overset{h e a t}{\to} \\ C_{8} H_{18} + C_{2} H_{4} \end{aligned}

Quick check

1 of 4

Which by-product of methane chlorination shows that methyl radicals are formed?

ethanehydrogenmethanolethene

Show answer

Answer (A) ethane

Ethane forms only when two methyl radicals meet in a termination step. Hydrogen is not formed, because chlorine atoms take hydrogen away as HCl.

9.2.4 Conformations

Hold a ball-and-stick model of ethane by one carbon and twist the other: it turns easily. Nothing resists, because the electron cloud of a σ bond is shaped like a cylinder around the C–C axis, and turning one end leaves the overlap unchanged. Each arrangement the molecule passes through is a conformation (also called a conformer or rotamer): an arrangement in space that changes into another by rotation about a single bond. There are infinitely many.

Definition

Conformations

Spatial arrangements of a molecule that change into one another by rotation about a single bond, without breaking any bond.

ExampleEclipsed and staggered ethane.

Strictly, the turn is not effortless. When C–H bonds on neighbouring carbons swing past one another, their electron clouds push apart. This torsional strain creates a small energy barrier, between 1 and 20 kJ mol⁻¹ in different molecules.

Conformations of ethane

Of all the arrangements of ethane, two stand out. In the eclipsed form each C–H bond on the front carbon hides one on the back carbon, so the hydrogens are as close as they can get. In the staggered form each front C–H bond points exactly between two back ones, as far apart as possible. Anything in between is a skew form. Bond lengths and angles stay the same throughout; only the angle of twist changes.

Chemists use two kinds of drawing (Fig. 9.6). A Sawhorse projection shows the C–C bond as a longish line seen at a slant, the front carbon at the lower end and the back carbon at the upper end, each with three bonds 120° apart. A Newman projection looks straight down the C–C axis: the front carbon is the point where three bonds meet, and the back carbon is a circle with three shorter bonds sticking out from its edge.

3/3Back C–H bonds midway between front ones: dihedral angle 60°.

Fig. 9.6Turning the back carbon by 60° changes eclipsed ethane into staggered ethane.NCERT Figs. 9.2, 9.3 · p. 305

Two drawings of ethane side by side: a Sawhorse projection on the left and a Newman projection on the right. Step 1 shows the eclipsed form, with the back hydrogens directly behind the front ones (drawn slightly offset so that they can be seen). Step 2 shows a curved arrow: the back carbon turns through 60 degrees while the front carbon stays still. Step 3 shows the staggered form, with each back hydrogen midway between two front ones and the dihedral angle labelled 60 degrees.

Seen along the axis, the angle between a front C–H bond and the nearest back C–H bond is the dihedral (torsional) angle: 0° for eclipsed ethane, 60° for staggered. Torsional strain peaks at 0° and is least at 60°, so the staggered form has the lowest energy and is the most stable. The gap is small, about 12.5 kJ mol⁻¹, and ordinary collisions at room temperature provide that much easily. Each ethane molecule therefore flips between forms billions of times a second and the conformers can never be isolated, although at any instant most molecules sit near the staggered form. That is why rotation about C–C is called almost, but not perfectly, free.

InteractiveEthane conformation explorer

Fig. 9.7Drag the slider (or press Play) to turn the back carbon and watch the energy change.

9.3 Alkenes

Alkenes contain at least one carbon–carbon double bond. Each double bond costs two hydrogens compared with the alkane, so alkenes with one C=C have the general formula $C_{n} H_{2 n}$ . The first stable member is ethene, $C_{2} H_{4}$ (common name ethylene); $C H_{2}$ itself exists only for an instant. The old name olefins, “oil-formers”, recalls that ethene and chlorine give an oily liquid, 1,2-dichloroethane.

Definition

Alkene

An unsaturated hydrocarbon with at least one C=C double bond; with one double bond and no ring, $C_{n} H_{2 n}$ .

ExamplePropene, $C H_{3} C H = C H_{2}$ .

9.3.1 Structure of Double Bond

Each carbon of a C=C bond is $s p^{2}$ hybridised. Head-on overlap of one $s p^{2}$ orbital from each carbon gives a strong σ bond (about 397 kJ mol⁻¹). The unhybridised 2p orbital left on each carbon stands at right angles to the plane of the molecule, and the two overlap sideways to give a π bond (about 284 kJ mol⁻¹) (Fig. 9.8). Sideways overlap is less effective than head-on overlap, so the π bond is the weaker partner.

Together they make the double bond stronger than a single bond (about 681 kJ mol⁻¹ in ethene against 348 kJ mol⁻¹ for C–C in ethane) and shorter (134 pm against 154 pm), but not twice as strong. All six atoms of ethene lie in one plane, with bond angles close to 120°.

The π electrons sit above and below that plane, loosely held and exposed. Reagents that seek electrons, called electrophiles, attack them readily. Swapping one weak π bond for two new σ bonds releases energy, so alkenes add reagents across the double bond and are far more reactive than alkanes.

Definition

Electrophile

An electron-seeking species, positively charged or electron-deficient, that attacks regions of high electron density.

Example $H^{+}$ , $B r^{+}$ , $N O_{2}^{+}$ .

3/4The p orbitals overlap sideways: electron cloud above and below.

Fig. 9.8A σ bond along the axis plus a π cloud above and below it.NCERT Figs. 9.4, 9.5 · p. 306

Ethene is drawn with its two carbons and four hydrogens in one horizontal plane. Step 1 shows the sigma framework: one C–C and four C–H sigma bonds made from sp2 orbitals, with angles of about 120 degrees. Step 2 adds an upright p orbital, a lobe above and a lobe below the plane, on each carbon. Step 3 shows the two p orbitals overlapping sideways into a pi cloud, one sausage above and one below the C–C axis. Step 4 labels the C=C length of 134 pm and the planar shape.

9.3.2 Nomenclature

To name an alkene, choose the longest chain that contains the double bond, number it from the end nearer the double bond, and replace -ane by -ene, giving the locant of the first doubly bonded carbon. So $C H_{2} = C H C H_{2} C H_{2} C H_{3}$ is pent-1-ene, $C H_{3} C H = C H C H_{2} C H_{3}$ is pent-2-ene and $(C H_{3})_{2} C = C H_{2}$ is 2-methylprop-1-ene. With two double bonds the ending is -diene: $C H_{2} = C H - C H = C H - C H_{3}$ is penta-1,3-diene.

Example 9.7NCERT Problem 9.7 · p. 307

Naming alkenes

Name these alkenes by the IUPAC rules: (i) $C H_{2} = C H C H (C H_{3}) C H_{2} C H_{3}$ , (ii) $C H_{3} C H = C H C H = C H C H_{3}$ , (iii) $(C H_{3})_{2} C = C (C H_{3}) C H_{2} C H_{3}$ , (iv) $C H_{3} C H = C H C H (C H_{3}) C H (C H_{3})_{2}$ .

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

Example 9.8NCERT Problem 9.8 · p. 307

Counting σ and π bonds

How many σ and π bonds are there in each molecule of Example 9.7?

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

9.3.3 Isomerism

Two kinds of isomerism are possible for alkenes: structural and geometrical.

Structural isomerism. Ethene and propene can each be built in only one way, but larger alkenes have isomers of two kinds. Pent-1-ene and pent-2-ene share a skeleton and differ only in where the double bond sits, so they are position isomers. 2-Methylbut-1-ene has a branched skeleton, so it is a chain isomer of both.

Example 9.9NCERT Problem 9.9 · p. 308

Unbranched hexenes

Write the structures and names of all alkenes $C_{6} H_{12}$ with an unbranched chain. What kind of isomers are they, and which of them can exist as cis and trans forms?

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

Geometrical isomerism

Suppose each carbon of a double bond carries two different groups, as in a molecule of type XYC=CXY. The two X groups can then sit together on one side of the double bond, or one on each side (Fig. 9.9). About a single bond these arrangements would interconvert by rotation. About a double bond, rotation would mean breaking the π bond, which needs far more energy than collisions supply at room temperature. The two arrangements are therefore separate compounds, called geometrical isomers or cis–trans isomers.

Definition

Geometrical (cis–trans) isomerism

Isomerism between compounds with the same structure but different arrangements of groups about a double bond, caused by restricted rotation.

Examplecis- and trans-but-2-ene.

cis (Latin, “on this side”) means the matching groups share a side of the double bond; trans (“across”) means they face opposite sides. The atoms are joined in the same order in both, so the structure is the same, but they are arranged differently in space, a difference of configuration. Such stereoisomers are distinct compounds that can be separated, and they differ in polarity, boiling point, melting point and solubility (see the table below).

3/3Same structure, different arrangement: cis is polar, trans is not.

Fig. 9.9Restricted rotation about C=C makes cis and trans isomers distinct compounds.NCERT p. 308

Step 1 shows a C–C single bond with a circular arrow: one end can spin freely. Step 2 shows a C=C bond with its pi lobes above and below; the circular arrow is crossed out because turning would break the pi bond. Step 3 shows cis-1,2-dichloroethene with both chlorines on the same side and a net dipole arrow pointing towards them (1.9 D), next to trans-1,2-dichloroethene with the chlorines on opposite sides, whose bond dipoles cancel (0 D).

Model it

The handshake model

You’ll need

a partner

Predict first

With both pairs of hands joined, can you turn one pair without letting go?

Yes, easilyOnly by letting go of one handYes, but only through half a turn

Grip your partner's right hand with your right hand, as in a handshake. Both of you turn your wrists: the joined hands rotate like a doorknob.
Now also hold their left hand with your left. Try the same turn.
Notice which of your partner's hands is on your left. Can you swap sides without letting go?

Table 9.3cis- and trans-1,2-dichloroethene, ClCH=CHCl

Property	cis	trans
Dipole moment / D	1.90	0
Boiling point / K	333	322
Melting point / K	192	224

The cis isomer is polar: its two C–Cl bond dipoles point partly the same way and add. In the trans isomer they point in opposite directions and cancel. The extra dipole–dipole attraction gives the cis isomer the higher boiling point. In the solid, the more symmetrical trans molecules pack together better, so the trans isomer usually melts higher. But-2-ene follows the same pattern: the cis form is slightly polar, the trans form non-polar.

Geometrical isomerism needs two different groups on each doubly bonded carbon. It occurs for XYC=CXY, XYC=CXZ and XYC=CZW, but not when either carbon carries two identical groups, as in $C H_{2} = C H C l$ or $(C H_{3})_{2} C = C H C H_{3}$ .

Example 9.10NCERT Problem 9.10 · p. 309

Drawing cis and trans forms

Sketch both geometrical isomers of each compound and label them cis or trans: (i) pent-2-ene, $C H_{3} C H = C H C H_{2} C H_{3}$ ; (ii) 1,2-diphenylethene, $C_{6} H_{5} C H = C H C_{6} H_{5}$ .

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

Example 9.11NCERT Problem 9.11 · p. 309

Which can be cis or trans?

Pick out the compounds that can exist as cis and trans forms: (i) $C H_{2} = C H C l$ (ii) $C H_{3} C H = C H C l$ (iii) $(C H_{3})_{2} C = C H C H_{3}$ (iv) $C H_{3} C H_{2} C H = C H C H_{2} C H_{3}$ (v) $C l_{2} C = C H C H_{3}$

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

9.3.4 Preparation

From alkynes

Hydrogen adds to an alkyne in two stages, and stopping after the first needs a measured amount of hydrogen and a weakened catalyst. Lindlar's catalyst, palladium on charcoal partly poisoned with sulfur compounds or quinoline, gives the cis alkene, because both hydrogen atoms arrive from the metal surface on the same side. Sodium in liquid ammonia gives the trans alkene instead.

Pent-2-yne gives either isomer

\begin{aligned} C H_{3} C \equiv C C_{2} H_{5} \\ \overset{H_{2}, L i n d l a r}{\to} cis \\ \overset{N a, l i q \cdot N H_{3}}{\to} trans \end{aligned}

For ethyne and propyne, partial hydrogenation over palladium on charcoal stops at ethene and propene, where cis or trans does not arise.

From alkyl halides

With alcoholic potash (KOH dissolved in ethanol), an alkyl halide loses H and X from neighbouring carbons. The halogen's carbon is called α and the next one β, so the reaction is a β-elimination; because HX is removed it is also called dehydrohalogenation. Speed depends on both partners: C–I breaks most easily and C–Cl least (iodide > bromide > chloride), and tertiary halides react fastest, then secondary, then primary.

1-Bromopropane to propene

\begin{aligned} C H_{3} C H_{2} C H_{2} B r \\ + K O H \to_{h e a t}^{e t h a n o l} \\ C H_{3} C H = C H_{2} \\ + K B r + H_{2} O \end{aligned}

From vicinal dihalides

Zinc dust strips both halogens from a vicinal dihalide (one whose halogens sit on adjacent carbons, from the Latin for neighbour), forming $Z n X_{2}$ and a C=C bond. This is dehalogenation.

2,3-Dibromobutane to but-2-ene

\begin{aligned} C H_{3} C H B r C H B r C H_{3} \\ + Z n \overset{}{\to} \\ C H_{3} C H = C H C H_{3} \\ + Z n B r_{2} \end{aligned}

From alcohols by acidic dehydration

Hot concentrated sulfuric acid pulls a water molecule out of an alcohol, R–OH: the –OH group and a hydrogen from the β-carbon leave together, so this acidic dehydration is one more β-elimination.

Propan-2-ol to propene

\begin{aligned} C H_{3} C H (O H) C H_{3} \\ \to_{h e a t}^{c o n c \cdot H_{2} S O_{4}} \\ C H_{3} C H = C H_{2} + H_{2} O \end{aligned}

Quick check

1 of 3

Which reagent converts but-2-yne into trans-but-2-ene?

Na in liquid

N H_{3}

H_{2}

with Lindlar's catalyst

H_{2}

with Ni, heatzinc dust

Show answer

Answer (A) Na in liquid $N H_{3}$

Sodium in liquid ammonia gives the trans alkene. Lindlar's catalyst gives the cis alkene, and nickel with excess hydrogen goes all the way to butane.

9.3.5 Properties

Physical properties

Physically, alkenes are close cousins of the alkanes; only their isomerism and slight polarity set them apart. Boiling points climb by roughly 20–30 K for each extra $C H_{2}$ , and branching lowers them, so ethene, propene and the butenes are gases, the alkenes from $C_{5}$ to about $C_{18}$ are liquids, and larger ones are solids. All are colourless, and only ethene has a faint, sweetish smell. Water will not dissolve them, but benzene and petroleum ether will.

Chemical properties

Most alkene chemistry is addition: two new groups attach to the doubly bonded carbons and the π bond disappears. Usually an electrophile starts the process; occasionally a free radical does (as with HBr and peroxide, below). Alkenes can also be oxidised or cut apart by ozone, and under special conditions they even undergo free-radical substitution.

1. Addition of hydrogen. Over finely divided nickel, palladium or platinum, alkenes add hydrogen to give alkanes (Section 9.2.2).

2. Addition of halogens. Shake a few drops of reddish-orange bromine in carbon tetrachloride with an alkene and the colour vanishes almost at once: bromine adds across the C=C bond to give a colourless vicinal dibromide. This makes a quick laboratory test for unsaturation. Chlorine adds the same way, but iodine does not add under ordinary conditions. The mechanism, an electrophilic addition through a ring-shaped halonium ion, comes in Class 12.

Propene to 1,2-dibromopropane (orange colour fades)

\begin{aligned} C H_{3} C H = C H_{2} + B r_{2} \\ \overset{C C l_{4}}{\to} \\ C H_{3} C H B r C H_{2} B r \end{aligned}

3. Addition of hydrogen halides. HCl, HBr and HI add to alkenes to give alkyl halides, with reactivity HI > HBr > HCl, the order in which the H–X bond gets weaker. A symmetrical alkene gives only one product:

Hex-3-ene to 3-chlorohexane

\begin{aligned} C_{2} H_{5} C H = C H C_{2} H_{5} \\ + H C l \overset{}{\to} \\ C_{2} H_{5} C H_{2} C H C l C_{2} H_{5} \end{aligned}

An unsymmetrical alkene such as propene could give either 1-bromopropane or 2-bromopropane. In 1869 the Russian chemist Vladimir Markovnikov summed up which one actually forms in a rule:

Definition

Markovnikov rule

When H–X adds to an unsymmetrical alkene, the hydrogen goes to the doubly bonded carbon that already holds more hydrogens, and X goes to the other one.

ExamplePropene + HBr gives mainly 2-bromopropane.

The mechanism explains the rule (Fig. 9.10). First, the electrophile $H^{+}$ from HBr attacks the π electrons and bonds to one carbon, leaving a positive charge, a carbocation, on the other. If $H^{+}$ joins the end $C H_{2}$ carbon, a secondary carbocation $C H_{3} - C H^{+} - C H_{3}$ results; if it joins the middle carbon, a primary carbocation $C H_{3} C H_{2} C H_{2}^{+}$ . The secondary carbocation is more stable, because two alkyl groups help spread its charge, so it forms faster and wins. Second, $B r^{-}$ bonds to the positive carbon, giving 2-bromopropane as the major product.

4/4 $B r^{∙}$ adds first, to the end carbon: 1-bromopropane (anti-Markovnikov).

Fig. 9.10Whichever species attacks first, the more stable intermediate decides the product.NCERT p. 311

Step 1 shows propene, CH3–CH=CH2, with H+ from HBr approaching the double bond. Step 2 shows the two possible carbocations: a secondary one, CH3–CH+–CH3, marked more stable, and a primary one, CH3CH2CH2+, marked less stable and faded. Step 3 shows bromide ion joining the secondary carbocation to give 2-bromopropane, the Markovnikov product. Step 4 shows the peroxide route: a bromine atom adds to the end carbon to give the secondary radical CH3–CH•–CH2Br, which takes H from HBr to give 1-bromopropane, the anti-Markovnikov product.

Peroxide effect. In 1933 Morris Kharasch and Frank Mayo at the University of Chicago found that HBr adds to propene the “wrong” way when an organic peroxide such as benzoyl peroxide is present, giving mainly 1-bromopropane. This peroxide (Kharasch) effect, or anti-Markovnikov addition, is seen only with HBr, not with HCl or HI.

Anti-Markovnikov addition of HBr

\begin{aligned} C H_{3} C H = C H_{2} + H B r \\ \overset{(C_{6} H_{5} C O)_{2} O_{2}}{\to} \\ C H_{3} C H_{2} C H_{2} B r \end{aligned}

The peroxide switches the reaction to a free-radical chain. Its weak O–O bond breaks into two radicals, which lose $C O_{2}$ to give phenyl radicals; a phenyl radical takes H from HBr and frees a bromine atom. Now $B r^{∙}$ , not $H^{+}$ , attacks first. Adding to the end $C H_{2}$ gives a secondary radical, more stable than the primary radical formed the other way. That radical takes H from another HBr, giving 1-bromopropane and a new $B r^{∙}$ to continue the chain:

Radical chain of the peroxide effect

\begin{aligned} (C_{6} H_{5} C O O)_{2} \overset{}{\to} \\ 2 C_{6} H_{5}^{∙} + 2 C O_{2} \\ C_{6} H_{5}^{∙} + H B r \overset{}{\to} \\ C_{6} H_{6} + B r^{∙} \\ C H_{3} C H = C H_{2} \\ + B r^{∙} \overset{}{\to} \\ C H_{3} C^{∙} H C H_{2} B r \\ + H B r \overset{}{\to} \\ C H_{3} C H_{2} C H_{2} B r \\ + B r^{∙} \end{aligned}

Why only HBr? The H–Cl bond (about 431 kJ mol⁻¹) is too strong for the carbon radical to break, so the chain cannot continue. The H–I bond (about 299 kJ mol⁻¹) breaks easily, but iodine atoms join up into $I_{2}$ rather than add to the double bond. H–Br (about 366 kJ mol⁻¹) lies in between, and both chain steps run smoothly.

Example 9.12NCERT Problem 9.12 · p. 312

HBr and 2-methylpropene

2-Methylpropene, $(C H_{3})_{2} C = C H_{2}$ , is treated with HBr. Predict and name the main product (i) of the plain reaction and (ii) of the reaction run with benzoyl peroxide present.

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

4. Addition of sulfuric acid. An alkene bubbled into cold concentrated sulfuric acid gives an alkyl hydrogen sulfate, with $- O S O_{3} H$ on the carbon that Markovnikov's rule predicts:

Markovnikov orientation (for but-1-ene, R = C₂H₅: butan-2-yl hydrogen sulfate)

\begin{aligned} R C H = C H_{2} \\ + H_{2} S O_{4} \overset{}{\to} \\ R C H (O S O_{3} H) C H_{3} \end{aligned}

5. Addition of water. Water alone does nothing, but a few drops of concentrated sulfuric acid catalyse its addition. H goes to the carbon with more hydrogens and –OH to the other, giving an alcohol:

2-Methylpropene gives 2-methylpropan-2-ol

\begin{aligned} (C H_{3})_{2} C = C H_{2} \\ + H_{2} O \overset{H^{+}}{\to} \\ (C H_{3})_{3} C O H \end{aligned}

6. Oxidation. (a) Baeyer's test. Purple, cold, dilute alkaline permanganate (Baeyer's reagent) loses its colour, and brown manganese dioxide appears, when shaken with an alkene. An –OH group has added to each doubly bonded carbon, giving a vicinal diol (glycol).

Propene gives propane-1,2-diol

\begin{aligned} C H_{3} C H = C H_{2} \\ + H_{2} O + [O] \\ \overset{c o l d d i l \cdot K M n O_{4}}{\to} \\ C H_{3} C H (O H) C H_{2} O H \end{aligned}

(b) Hot acidic permanganate or acidic dichromate goes further and breaks the double bond. Each doubly bonded carbon ends up in a C=O group: a carbon with two alkyl groups becomes a ketone, one with a hydrogen becomes a carboxylic acid, and a terminal $= C H_{2}$ is oxidised all the way to carbon dioxide and water.

2-Methylbut-2-ene gives propanone and ethanoic acid

\begin{aligned} (C H_{3})_{2} C = C H C H_{3} \\ \overset{K M n O_{4} / H^{+}}{\to} \\ (C H_{3})_{2} C = O \\ + C H_{3} C O O H \end{aligned}

7. Ozonolysis. Ozone adds across the double bond to give an unstable ozonide, which zinc and water then split into two carbonyl compounds (Fig. 9.11). The products are easy to predict: cut the C=C bond and put =O on each carbon. Working backwards, the products show exactly where the double bond was, which is why ozonolysis is used to locate double bonds in unknown compounds.

Pent-2-ene gives ethanal and propanal

\begin{aligned} C H_{3} C H = C H C_{2} H_{5} \\ \to_{(i i) Z n / H_{2} O}^{(i) O_{3}} \\ C H_{3} C H O \\ + C_{2} H_{5} C H O \end{aligned}

4/4Propanone from one carbon, ethanal from the other.

Fig. 9.11Cut the C=C bond and put =O on each carbon.NCERT p. 313

Step 1 shows 2-methylbut-2-ene, (CH3)2C=CH–CH3, with the double bond highlighted. Step 2 shows ozone added across the double bond as a five-membered ozonide ring of two carbons and three oxygens. Step 3 shows a dashed cutting line through the old double bond and the label Zn and H2O. Step 4 shows the two products: propanone, (CH3)2C=O, from the left carbon, and ethanal, O=CH–CH3, from the right carbon, each now with a C=O group where the double bond was.

Example

Decoding ozonolysis products

An alkene $C_{6} H_{12}$ gives propanone and propanal on ozonolysis. Identify the alkene.

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

8. Polymerisation. Under heat and pressure, and with a catalyst, ethene molecules join end to end by the thousand. This is polymerisation: the small units are monomers and the long-chain product, here polythene, is a polymer. Propene polymerises to polypropene in the same way.

Polythene and polypropene

\begin{aligned} n C H_{2} = C H_{2} \\ \to_{h e a t, p r e s s u r e}^{c a t a l y s t} \\ (- C H_{2} - C H_{2}^{-})_{n} \\ n C H_{3} C H = C H_{2} \\ \overset{c a t a l y s t}{\to} \\ (- C H (C H_{3}) - C H_{2}^{-})_{n} \end{aligned}

Polythene goes into carry bags, cling film, water tanks and cable insulation; polypropene into chairs, food containers, ropes and non-woven shopping bags. Their durability is also their drawback: discarded plastic lasts for centuries, which is one reason India banned many single-use plastic items in 2022.

InteractiveAlkene reaction bench

Two rows of buttons: six alkenes (ethene, propene, but-1-ene, but-2-ene, 2-methylpropene and 2-methylbut-2-ene) and eight reagents (hydrogen with nickel, bromine in CCl4, HBr, HBr with peroxide, water with acid, Baeyer's reagent, ozone then zinc and water, and hot acidic permanganate). The panel shows the reaction as reactant, reagent and product formulas, the IUPAC names of the products, any colour change, and one line explaining the rule, such as Markovnikov orientation or where the double bond is cut.

Fig. 9.12Choose an alkene and a reagent to see the product and the rule that decides it.

9.4 Alkynes

Alkynes contain at least one carbon–carbon triple bond. They have four hydrogens fewer than the alkane with the same number of carbons, so the general formula for one triple bond is $C_{n} H_{2 n - 2}$ . The first member is ethyne, $H C \equiv C H$ , better known as acetylene. Burned with pure oxygen in an oxyacetylene torch it gives a flame above 3000 °C, hot enough to cut and weld steel. Alkynes are also starting materials for a wide range of organic compounds.

Definition

Alkyne

An unsaturated hydrocarbon with at least one C≡C triple bond; with one triple bond and no ring, $C_{n} H_{2 n - 2}$ .

ExamplePropyne, $C H_{3} C \equiv C H$ .

9.4.1 Nomenclature and Isomerism

Common names treat alkynes as derivatives of acetylene: $C H_{3} C \equiv C H$ is methylacetylene and $C H_{3} C \equiv C C H_{3}$ is dimethylacetylene. IUPAC names replace -ane by -yne and give the locant of the first triply bonded carbon, so these two are propyne and but-2-yne, while $C H_{3} C H_{2} C \equiv C H$ is but-1-yne. Ethyne and propyne have one structure each. With four carbons the triple bond can start at C1 or C2, so but-1-yne and but-2-yne are position isomers; with five or more carbons, branched chain isomers appear as well.

Example 9.13NCERT Problem 9.13 · p. 315

Terminal alkynes C6H10

Write the structures and names of all the alkynes $C_{6} H_{10}$ that have the triple bond at the end of the chain. What isomerism do they show among themselves?

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

9.4.2 Structure of Triple Bond

Ethyne is a straight molecule, H–C≡C–H with 180° angles, because each carbon is $s p$ hybridised and its two $s p$ orbitals point in opposite directions (Fig. 9.13). One of them makes the C–C σ bond by head-on overlap with its partner; the other bonds to a hydrogen.

That leaves each carbon two spare 2p orbitals, at right angles to one another and to the molecular axis. Matching pairs on the two carbons overlap sideways, giving two π bonds in perpendicular planes. Ethyne therefore has three σ bonds (one C–C, two C–H) plus two π bonds between the carbons, and the two π clouds blend into a single sleeve of electron density wrapped evenly round the C–C axis.

Table 9.4Carbon–carbon bonds compared

Bond	Example	Carbon hybridisation	s character	Length / pm	Bond enthalpy / kJ mol⁻¹
C–C	ethane	$s p^{3}$	25 %	154	348
C=C	ethene	$s p^{2}$	33 %	134	681
C≡C	ethyne	$s p$	50 %	120	823

3/3Sideways overlap gives two π bonds: a cylinder of charge, C≡C 120 pm.

Fig. 9.13One σ bond and two perpendicular π bonds make a linear molecule.NCERT Fig. 9.6 · p. 315

Ethyne is drawn as a straight line H–C–C–H. Step 1 shows the sigma bonds made from sp orbitals and the 180 degree angle. Step 2 adds two p orbitals on each carbon: one pair standing up and down, one pair drawn as ellipses pointing towards and away from the viewer. Step 3 shows these overlapping sideways into two pi bonds, which merge into a cylinder of electron density around the C–C axis, labelled 120 pm.

9.4.3 Preparation

1. From calcium carbide. Ethyne for industry starts from limestone. Strong heating gives quicklime; quicklime and coke fused in an electric furnace give calcium carbide; and carbide plus water gives ethyne:

Limestone to ethyne

\begin{aligned} C a C O_{3} \overset{h e a t}{\to} \\ C a O + C O_{2} \\ C a O + 3 C \overset{}{\to} \\ C a C_{2} + C O \\ C a C_{2} + 2 H_{2} O \overset{}{\to} \\ C a (O H)_{2} + C_{2} H_{2} \end{aligned}

The same reaction once fuelled the hissing carbide lamps of miners and night markets. It is also behind an illegal practice: calcium carbide is sometimes hidden among unripe mangoes, because the ethyne it releases mimics the natural ripening gas, ethene.

2. From vicinal dihalides. Two eliminations in a row turn a vicinal dihalide into an alkyne. Alcoholic KOH takes off the first HX easily, leaving a vinylic (alkenyl) halide; the second HX, on a doubly bonded carbon, needs the much stronger base sodamide, $N a N H_{2}$ .

1,2-Dibromopropane to propyne

\begin{aligned} C H_{3} C H B r C H_{2} B r \\ \overset{a l c \cdot K O H}{\to} \\ C H_{3} C H = C H B r \\ \overset{N a N H_{2}}{\to} C H_{3} C \equiv C H \end{aligned}

9.4.4 Properties

Physical properties

Alkynes fit the pattern already seen: melting point, boiling point and density rise with molar mass, so the three smallest are gases, the next eight liquids and the rest solids. They are colourless, weakly polar, less dense than water and insoluble in it, but they dissolve in benzene, ether and carbon tetrachloride. Pure ethyne has almost no smell; the garlic odour of carbide acetylene comes from impurities such as phosphine.

Chemical properties

A. Acidic character. A terminal alkyne is a very weak acid, yet far stronger than an alkene or an alkane. Ethyne gives up a proton to strong bases such as sodium metal or sodamide, forming sodium ethynide; ethene and ethane show no such reaction. Hybridisation explains the difference. An $s p$ orbital is half s (50 %), against one-third for $s p^{2}$ and one-quarter for $s p^{3}$ , and s electrons stay close to the nucleus. So an $s p$ carbon is the most electronegative of the three, the C–H bonding pair sits nearer to it, and $H^{+}$ comes off most easily.

Terminal alkynes form sodium alkynides

\begin{aligned} 2 H C \equiv C H + 2 N a \overset{}{\to} \\ 2 H C \equiv C N a + H_{2} \\ C_{2} H_{5} C \equiv C H \\ + N a N H_{2} \overset{}{\to} \\ C_{2} H_{5} C \equiv C N a + N H_{3} \end{aligned}

This acidity belongs to the ≡C–H hydrogen alone; the methyl hydrogens of propyne are no more acidic than those of propane. So but-1-yne reacts with sodamide while but-2-yne, with no ≡C–H, does not, a handy way to tell terminal alkynes from alkanes, alkenes and internal alkynes. The acidity orders are $H C \equiv C H > H_{2} C = C H_{2} > C H_{3} C H_{3}$ and $H C \equiv C H > C H_{3} C \equiv C H$ ≫ $C H_{3} C \equiv C C H_{3}$ . Approximate pKa values: ethyne 25, ethene 44, ethane 50. Even ethyne is a far weaker acid than water (pKa about 16).

B. Addition reactions. A triple bond can take up two molecules of hydrogen, halogen or hydrogen halide, one after the other. Most of these additions are electrophilic. In an unsymmetrical alkyne the orientation follows Markovnikov's rule, decided by the stability of the vinylic cation (a positive charge on a doubly bonded carbon) formed in the first step.

(i) Hydrogen. Over Pt, Pd or Ni the alkene forms first and is quickly hydrogenated further to the alkane.

But-2-yne to butane

\begin{aligned} C H_{3} C \equiv C C H_{3} \\ \overset{H_{2}, N i}{\to} \\ [C H_{3} C H = C H C H_{3}] \\ \overset{H_{2}, N i}{\to} \\ C H_{3} C H_{2} C H_{2} C H_{3} \end{aligned}

(ii) Halogens. Bromine adds twice to give a tetrabromide, and the reddish-orange colour of bromine in $C C l_{4}$ fades, the same test for unsaturation as with alkenes.

Propyne gives 1,1,2,2-tetrabromopropane

\begin{aligned} C H_{3} C \equiv C H + 2 B r_{2} \\ \overset{}{\to} C H_{3} C B r_{2} C H B r_{2} \end{aligned}

(iii) Hydrogen halides. Two molecules of HX add. Markovnikov's rule sends both halogens to the same carbon, giving a geminal (gem) dihalide:

But-1-yne gives 2,2-dibromobutane

\begin{aligned} C_{2} H_{5} C \equiv C H + H B r \\ \overset{}{\to} C_{2} H_{5} C B r = C H_{2} \\ \overset{H B r}{\to} C_{2} H_{5} C B r_{2} C H_{3} \end{aligned}

(iv) Water. Water on its own leaves alkynes untouched. With mercury(II) ions as catalyst, mercury(II) sulfate in dilute sulfuric acid at about 333 K, one molecule of water adds. The first product, an enol (–OH on a doubly bonded carbon), rearranges at once into a carbonyl compound. Ethyne gives ethanal; every other alkyne gives a ketone.

Propyne gives propanone

\begin{aligned} C H_{3} C \equiv C H + H_{2} O \\ \to_{333 K}^{H g^{2 +}, H_{2} S O_{4}} \\ [C H_{3} C (O H) = C H_{2}] \\ \overset{}{\to} C H_{3} C O C H_{3} \end{aligned}

(v) Polymerisation. (a) Linear. Under suitable conditions ethyne links into polyacetylene (polyethyne), a long chain of alternating double and single bonds, $(- C H = C H - C H = C H^{-})_{n}$ . Specially treated films of it conduct electricity and have been tried as light, cheap electrodes for batteries.

(b) Cyclic. In a red-hot iron tube at 873 K, three ethyne molecules close into a single benzene ring, a direct link between open-chain (aliphatic) and aromatic chemistry.

Ethyne to benzene

\begin{aligned} 3 H C \equiv C H \\ \to_{873 K}^{r e d - h o t F e} C_{6} H_{6} \end{aligned}

Example 9.14NCERT Problem 9.14 · p. 318

From limestone to benzene

Using only limestone, coke and water (plus heat and an iron tube), outline how you would make benzene.

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

9.5 Aromatic Hydrocarbon

Toluene, naphthalene and their relatives are the arenes, or aromatic hydrocarbons. The name is historical: many early examples smelt pleasant (Greek aroma), but today the label describes structure, not smell. Most arenes contain a benzene ring, which keeps its unsaturation intact through most reactions; these are benzenoid arenes, such as toluene (methylbenzene), the xylenes, naphthalene (two fused rings) and biphenyl (two linked rings). A few aromatic species are built on another kind of ring, such as the cyclopentadienyl anion in Section 9.5.3; they are non-benzenoid.

Definition

Arene (aromatic hydrocarbon)

A hydrocarbon containing one or more aromatic rings, usually benzene rings.

ExampleToluene, naphthalene.

9.5.1 Nomenclature and Isomerism

Put one substituent on benzene and it makes no difference which carbon you choose, because all six are equivalent: there is only one chlorobenzene and one toluene. A second substituent can go next door, one carbon further away, or directly opposite, giving three position isomers: ortho (o-, 1,2), meta (m-, 1,3) and para (p-, 1,4). Counting the other way round, 1,6 is also ortho and 1,5 also meta. The dichlorobenzenes and the xylenes (dimethylbenzenes) each come in these three forms; Unit 8 covers their names.

9.5.2 Structure of Benzene

Benzene's formula, $C_{6} H_{6}$ , was known soon after Michael Faraday isolated it in 1825, but its structure took four decades to settle. Any proposal had to explain three awkward facts. The formula implies heavy unsaturation, yet benzene is unusually stable and resists the additions that alkenes undergo. It forms a triozonide, as if it had three C=C bonds. And it gives only one monosubstituted product, so all six C–H units must be alike.

August Kekulé's answer, in 1865, was a hexagon of carbons joined by alternate single and double bonds, each carbon carrying one hydrogen. Here was the snag: such a ring should give two 1,2-dibromobenzenes, one with the bromines across a double bond and one across a single bond, but only one exists. Kekulé suggested that the double bonds keep swapping places too fast for the two forms to be separated. Even this could not explain why benzene is so stable or why it prefers substitution to addition.

Resonance and stability of benzene

Today the two Kekulé structures are seen not as forms that flip back and forth but as resonance structures, neither of which exists on its own. The real molecule, the resonance hybrid, lies in between, with all six ring bonds identical. Chemists draw it as a hexagon with a circle inside (Fig. 9.14); the circle represents six π electrons spread evenly over the ring rather than confined to three double bonds.

Orbitals tell the same story. Every ring carbon is $s p^{2}$ hybridised, so the σ framework of six C–C and six C–H bonds lies flat, with 120° angles. Each carbon also keeps one p orbital standing upright from the ring. If these p orbitals paired off into three separate π bonds, there would be two equally good ways to do it: the two Kekulé structures. Instead each p orbital overlaps both of its neighbours equally, and the six merge into one continuous π system, a ring of electron density above the plane and a matching ring below.

Electrons spread over six nuclei in this way are delocalised, and being attracted by more nuclei lowers their energy. Two measurements back the picture. X-ray diffraction shows a flat ring whose six C–C bonds all measure 139 pm, between a single bond (154 pm) and a double bond (134 pm), whereas a fixed Kekulé ring would have two bond lengths. And benzene is markedly more stable than an imaginary cyclohexatriene with three ordinary double bonds (Exercise Q10 puts a number on this). With no isolated double bond to attack, benzene shrugs off most addition reagents.

4/4Delocalised π cloud; every C–C bond 139 pm.

Fig. 9.14Benzene is one hybrid with six delocalised π electrons, not two alternating structures.NCERT Fig. 9.7 · p. 320

Step 1 shows Kekulé structure A, a hexagon with double bonds at alternate edges. Step 2 shows Kekulé structure B, with the double bonds moved to the other three edges, and a double-headed resonance arrow between A and B. Step 3 shows six upright p orbitals, one on each carbon of the hexagon, each able to overlap with both neighbours. Step 4 shows the resonance hybrid: a hexagon with a circle inside, a pi cloud above and below the ring, and the label that all six C–C bonds are 139 pm, between 154 pm and 134 pm.

9.5.3 Aromaticity

Benzene was the original aromatic compound, but today the word describes any ring system, with or without a benzene ring, that has three features:

Planarity: all the ring atoms lie in one plane.
Complete delocalisation of the π electrons around the ring, which needs a p orbital on every ring atom.
$4 n + 2$ π electrons: the ring must hold 2, 6, 10, … π electrons ( $n = 0, 1, 2, \dots$ ). This is the Hückel rule, after the German physicist Erich Hückel (1931).

Definition

Aromaticity

The extra stability of a planar ring whose π electrons are completely delocalised and number $4 n + 2$ ( $n = 0, 1, 2, \dots$ ).

ExampleBenzene (6 π electrons, n = 1).

Table 9.5Testing rings with the Hückel rule

Ring	Planar, delocalised?	π electrons	Aromatic?
benzene	yes	6 (n = 1)	yes
cyclopropenyl cation, $C_{3} H_{3}^{+}$	yes	2 (n = 0)	yes
cyclopentadienyl anion, $C_{5} H_{5}^{-}$	yes	6 (n = 1)	yes
cyclooctatetraene dianion, $C_{8} H_{8}^{2 -}$	yes	10 (n = 2)	yes
naphthalene	yes	10 (n = 2)	yes
cyclobutadiene	yes	4	no (not 4n + 2)
cyclopentadiene	no ( $s p^{3}$ $C H_{2}$ )	4	no

9.5.4 Preparation of Benzene

Industry obtains benzene mainly from coal tar and from petroleum, by reforming (Section 9.2.3). In the laboratory it can be made in three ways:

by cyclic polymerisation of ethyne in a red-hot iron tube (Section 9.4.4);
by decarboxylation: heating sodium benzoate with soda lime;
by reduction of phenol: passing phenol vapour over heated zinc dust.

From sodium benzoate and from phenol

\begin{aligned} C_{6} H_{5} C O O N a \\ + N a O H \overset{C a O, h e a t}{\to} \\ C_{6} H_{6} + N a_{2} C O_{3} \\ C_{6} H_{5} O H + Z n \\ \overset{h e a t}{\to} C_{6} H_{6} + Z n O \end{aligned}

9.5.5 Properties

Physical properties

Arenes are non-polar and mostly colourless; the lighter ones are liquids, the heavier ones solids, and many have a distinctive smell. Like other hydrocarbons they do not mix with water but blend with organic solvents. Their high proportion of carbon makes them burn with a smoky, sooty flame, an easy way to tell an arene from an alkane in the laboratory.

Chemical properties

The signature reaction of arenes is electrophilic substitution: an electrophile takes the place of a ring hydrogen, and the aromatic ring comes through unchanged. Addition and oxidation happen only under special conditions. The common substitutions are:

Nitration. Heating benzene with a mixture of concentrated nitric and sulfuric acids (the nitrating mixture) gives nitrobenzene, $C_{6} H_{5} N O_{2}$ .
Halogenation. Chlorine or bromine with a Lewis acid such as anhydrous $A l C l_{3}$ , $F e C l_{3}$ or $F e B r_{3}$ gives a haloarene, for example bromobenzene, $C_{6} H_{5} B r$ .
Sulphonation. Heating with fuming sulfuric acid (oleum) replaces a hydrogen by $- S O_{3} H$ , giving benzenesulfonic acid.
Friedel–Crafts alkylation. An alkyl halide with anhydrous $A l C l_{3}$ attaches an alkyl group to the ring.
Friedel–Crafts acylation. An acyl chloride (or an acid anhydride) with anhydrous $A l C l_{3}$ attaches an acyl group, RCO–, giving an aryl ketone.

tert-Butylbenzene and 1-phenylpropan-1-one

\begin{aligned} C_{6} H_{6} + (C H_{3})_{3} C C l \\ \overset{a n h y d \cdot A l C l_{3}}{\to} \\ C_{6} H_{5} C (C H_{3})_{3} \\ + H C l \\ C_{6} H_{6} + C_{2} H_{5} C O C l \\ \overset{a n h y d \cdot A l C l_{3}}{\to} \\ C_{6} H_{5} C O C_{2} H_{5} \\ + H C l \end{aligned}

Substitution need not stop at one group. Given more reagent, the ring keeps swapping hydrogens: with excess chlorine and $A l C l_{3}$ all six can go, leaving $C_{6} C l_{6}$ .

Mechanism of electrophilic substitution

Electrophilic substitution ( $S_{E}$ ) runs in three steps (Fig. 9.15): (a) an electrophile is generated; (b) it attacks the ring, giving a carbocation intermediate; (c) the intermediate loses a proton.

Definition

Electrophilic substitution

A reaction in which an electrophile replaces an atom, usually hydrogen, on an aromatic ring by way of a σ-complex, so that the ring stays aromatic.

ExampleNitration of benzene.

(a) Generating the electrophile. A Lewis acid does this job in three of the reactions: anhydrous $A l C l_{3}$ grabs a chloride ion from chlorine, from an alkyl chloride or from an acyl chloride, leaving $C l^{+}$ , $R^{+}$ or the acylium ion $R C O^{+}$ , for instance $C l_{2} + A l C l_{3} \overset{}{\to} C l^{+} + [A l C l_{4}]^{-}$ . Nitration uses the nitronium ion, $N O_{2}^{+}$ , formed when sulfuric acid protonates nitric acid:

Making the nitronium ion

\begin{aligned} H N O_{3} + H_{2} S O_{4} \overset{}{⇌} \\ H_{2} N O_{3}^{+} + H S O_{4}^{-} \\ H_{2} N O_{3}^{+} \overset{}{⇌} \\ N O_{2}^{+} + H_{2} O \end{aligned}

Nitric acid, normally an acid, is acting here as a base towards the even stronger sulfuric acid.

(b) The σ-complex. The electrophile bonds to one ring carbon, which becomes $s p^{3}$ hybridised and carries both H and E. The positive charge left behind is shared by resonance among the other ring carbons; this carbocation is the arenium ion, or σ-complex. Because the $s p^{3}$ carbon has no p orbital, the ring's π system is broken at that point, so the intermediate is no longer aromatic and is high in energy.

(c) Loss of a proton. The anion left over from step (a), $H S O_{4}^{-}$ after nitration or $[A l C l_{4}]^{-}$ after an $A l C l_{3}$ -catalysed reaction, pulls the proton off the $s p^{3}$ carbon. Its two bonding electrons return to the ring and aromaticity is restored. Regaining that aromatic stability is why benzene substitutes rather than adds.

3/4The positive charge spreads over five carbons; aromaticity is lost.

Fig. 9.15Electrophile generated, σ-complex formed, proton lost: the ring stays aromatic.NCERT p. 323

Step 1 shows Br2 and FeBr3 forming Br+ and FeBr4−. Step 2 shows Br+ attacking a benzene ring and bonding to one carbon, which now carries both H and Br. Step 3 shows the sigma complex as a broken ring with a plus sign spread over the other five carbons, and labels that carbon sp3. Step 4 shows FeBr4− taking the proton, giving bromobenzene, HBr and FeBr3 back, with the full aromatic ring restored.

Addition. Benzene can be forced to add, but only under conditions far harsher than an alkene needs. Hydrogen over nickel at high temperature and pressure saturates the ring to cyclohexane, and chlorine in ultraviolet light adds three molecules to give benzene hexachloride (BHC), $C_{6} H_{6} C l_{6}$ ; one of its isomers was the insecticide lindane, also called gammaxane.

Addition needs forcing conditions

\begin{aligned} C_{6} H_{6} + 3 H_{2} \\ \to_{h i g h T, p}^{N i} C_{6} H_{12} \\ C_{6} H_{6} + 3 C l_{2} \\ \overset{U V}{\to} C_{6} H_{6} C l_{6} \end{aligned}

Combustion. Benzene burns in air with a sooty flame, $C_{6} H_{6} + \frac{15}{2} O_{2} \overset{}{\to} 6 C O_{2} + 3 H_{2} O$ . For any hydrocarbon:

Complete combustion of a hydrocarbon

\begin{aligned} C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \\ ⟶ x C O_{2} + \frac{y}{2} H_{2} O \end{aligned}

(9.83)

9.5.6 Directive influence of a functional group in monosubstituted benzene

Nitrate toluene and you get mostly 2- and 4-nitrotoluene; nitrate nitrobenzene and you get mostly 1,3-dinitrobenzene. The group already on the ring, not the arriving electrophile, decides where the second group goes and how fast. This steering is the directive influence of the substituent (Fig. 9.16).

Definition

Directive influence

The control a substituent already on a benzene ring has over where (ortho/para or meta) and how fast a new electrophile substitutes.

Example $- O H$ directs ortho/para; $- N O_{2}$ directs meta.

Ortho and para directors. In phenol, a lone pair on oxygen is shared with the ring. Drawing the resonance structures puts negative charge on C2, C4 and C6, the ortho and para carbons, but never on C3 or C5, so electrophiles go mainly to ortho and para. The ring as a whole also becomes richer in electrons than benzene and reacts faster, even though electronegative oxygen pulls some density back through the σ bond (–I effect): the resonance push wins. Such groups are activating. Alkyl groups (–CH₃, –C₂H₅), alkoxy groups (–OCH₃) and the nitrogen groups –NH₂, –NHR and –NHCOCH₃ behave the same way.

Halogens sit in between. Their resonance donation is weak, but it still favours ortho and para attack, as in chlorobenzene. Their strong –I effect wins the contest for overall electron density, however, so the ring is less reactive than benzene: halogens are deactivating ortho/para directors.

Meta directors. Groups such as –CHO, –COR, –COOH, –COOR, –CN, –SO₃H and –NO₂ pull electrons out of the ring, both through the σ bond and by resonance. The resonance structures of nitrobenzene place positive charge on the ortho and para carbons. The whole ring is deactivated, but the meta positions are the least electron-poor, so electrophiles attack there.

Compare

Three kinds of substituent

Activating, ortho/para

–OH, –OCH₃
–NH₂, –NHR, –NHCOCH₃
–CH₃, –C₂H₅

Deactivating, ortho/para

–F, –Cl, –Br, –I

Deactivating, meta

–CHO, –COR
–COOH, –COOR
–CN, –SO₃H, –NO₂

2/3Withdrawal: δ+ at ortho and para, so attack goes meta; ring deactivated.

Fig. 9.16The group already on the ring decides where the next one goes.NCERT p. 324

A benzene ring with a substituent at the top and its ortho, meta and para positions labelled. Step 1, with an OH group, marks the two ortho carbons and the para carbon with delta-minus signs and an arrow for donation into the ring: ortho/para, activating. Step 2, with an NO2 group, marks the ortho and para carbons with delta-plus signs and points the incoming electrophile to the two meta carbons: meta, deactivating. Step 3, with a Cl atom, shows a withdrawal arrow for the inductive effect but still marks ortho and para as the attack sites: ortho/para, deactivating.

9.6 Carcinogenicity and Toxicity

In 1775 the London surgeon Percivall Pott noticed that chimney sweeps often developed cancer. The culprits in soot were later identified as polynuclear aromatic hydrocarbons (PAHs), compounds with more than two benzene rings fused together. Many PAHs are carcinogenic (cancer-causing). They form whenever organic matter burns incompletely: in tobacco smoke, vehicle exhaust, coal, wood and dung smoke, and on charred food. Inside the body, enzymes convert them into reactive compounds that bind to DNA and damage it, which can start a cancer.

Well-known examples are benzo[a]pyrene, with five fused rings, present in cigarette smoke and soot; dibenz[a,h]anthracene, also with five rings; and 3-methylcholanthrene, used in laboratories to study tumours (Fig. 9.17). Benzene itself is toxic too: long exposure, for example to petrol vapour or solvent fumes at work, damages the bone marrow and can cause leukaemia.

Fig. 9.17Incomplete burning makes PAHs; once activated in the body they bind to DNA.NCERT p. 325

Three panels joined by arrows. On the left, a flame with smoke above it, labelled incomplete burning. In the middle, the five fused hexagonal rings of benzo[a]pyrene, labelled 5 fused rings. On the right, a DNA double helix with a cross marking where the reactive hydrocarbon binds.

Quick check

1 of 2

Polynuclear aromatic hydrocarbons mainly form by

complete combustion of methanehydrogenation of benzeneincomplete combustion of organic matterpolymerisation of ethene

Show answer

Answer (C) incomplete combustion of organic matter

Smoky, oxygen-starved burning of tobacco, coal, wood or fuel produces fused-ring hydrocarbons. Complete combustion gives only carbon dioxide and water.

Recap

Hydrocarbons: key points and summary

Hydrocarbons contain only carbon and hydrogen; they are our main fuels and the raw material for plastics, solvents and drugs. They are saturated, unsaturated or aromatic (Section 9.1).
Alkanes, $C_{n} H_{2 n + 2}$ , have $s p^{3}$ carbons; chain isomers multiply rapidly with size, and IUPAC names follow longest chain, lowest locants and alphabetical order (Section 9.2.1).
Alkanes are made by hydrogenation, by reducing alkyl halides (Zn/H⁺), by the Wurtz reaction, by decarboxylation with soda lime and by Kolbe electrolysis (Section 9.2.2).
Alkane boiling points rise with size and fall with branching. Alkanes undergo free-radical halogenation (3° > 2° > 1° H), combustion, controlled oxidation, isomerisation, aromatization, steam reforming and cracking (Section 9.2.3).
Rotation about a C–C bond gives conformations; staggered ethane is about 12.5 kJ mol⁻¹ more stable than eclipsed, shown in Sawhorse and Newman projections (Section 9.2.4).
A C=C bond is one σ bond plus one weaker π bond; the exposed π electrons attract electrophiles (Section 9.3.1).
Alkenes show chain, position and geometrical isomerism; cis–trans isomers need two different groups on each doubly bonded carbon (Section 9.3.3).
Alkenes come from alkynes (Lindlar: cis; Na/NH₃: trans), alkyl halides, vicinal dihalides and alcohols (Section 9.3.4).
Alkenes add $H_{2}$ , halogens, HX (Markovnikov via the more stable carbocation; anti-Markovnikov for HBr with peroxide), $H_{2} S O_{4}$ and water; Baeyer's reagent and bromine test for C=C; ozonolysis locates it; alkenes polymerise (Section 9.3.5).
Alkynes, $C_{n} H_{2 n - 2}$ , are linear ( $s p$ , one σ + two π). Terminal ≡C–H is acidic; alkynes add two molecules of reagent, hydrate to carbonyl compounds and trimerise to benzene (Section 9.4.4).
Benzene is a resonance hybrid with six delocalised π electrons and equal 139 pm bonds; aromatic rings are planar, fully delocalised and hold 4n + 2 π electrons (Section 9.5.3).
Arenes undergo electrophilic substitution (nitration, halogenation, sulphonation, Friedel–Crafts) through a σ-complex, and addition only under forcing conditions (Section 9.5.5).
Substituents decide orientation and rate: –OH, –NH₂ and –CH₃ activate and direct ortho/para; halogens deactivate but direct ortho/para; –NO₂, –CHO and –COOH deactivate and direct meta (Section 9.5.6).
Benzene and polynuclear hydrocarbons with more than two fused rings, formed by incomplete combustion, are toxic and carcinogenic (Section 9.6).

Practice · 25 questions

Practice questions with answers (NCERT “Exercises”)

Short answerEasyNCERT Exercises 9.1
When ethane is chlorinated in light, a little butane appears among the products. Explain how it forms.
Show solution
AnswerTwo ethyl radicals combine in a termination step.
Initiation: $C l_{2} \overset{}{\to} 2 C l^{∙}$ .
Propagation: $C l^{∙}$ takes H from ethane, giving HCl and an ethyl radical, $C_{2} H_{5}^{∙}$ , which takes Cl from $C l_{2}$ .
Termination: two ethyl radicals meet, $2 C_{2} H_{5}^{∙} \overset{}{\to} C_{4} H_{10}$ . Butane is the tell-tale by-product, just as ethane is for methane.
2
Short answerEasyNCERT Exercises 9.2
Name these compounds by the IUPAC rules:
1. $(C H_{3})_{2} C = C H C H_{2} C H_{3}$
  Show solution
  Answer2-methylpent-2-ene
  Five-carbon chain, C=C at C2 from the branched end, methyl on C2.
2. $H C \equiv C - C H = C H - C H_{3}$
  Show solution
  Answerpent-3-en-1-yne
  Numbering from the alkyne end gives the set {1, 3}; from the other end {2, 4}. The lower set wins; -ene is written before -yne.
3. $C_{6} H_{5} C H_{2} C H = C H C H_{3}$
  Show solution
  Answer1-phenylbut-2-ene
  Four-carbon chain with C=C at C2 from either end; the tie is broken by giving the phenyl group locant 1 rather than 4.
4. $C H_{3} C H_{2} C H (C H_{3})^{-}$ $C H_{2} C H (C_{2} H_{5}) C H_{2} C H_{3}$
  Show solution
  Answer3-ethyl-5-methylheptane
  Seven-carbon chain with substituents at 3 and 5 from either end; ethyl comes first in the alphabet, so it gets 3.
3
Short answerMediumNCERT Exercises 9.3
Write the structures and IUPAC names of all open-chain isomers of $C_{4} H_{6}$ that have:
1. one triple bond
  Show solution
  Answerbut-1-yne and but-2-yne
  $H C \equiv C C H_{2} C H_{3}$ (but-1-yne) and $C H_{3} C \equiv C C H_{3}$ (but-2-yne), which are position isomers.
2. two double bonds
  Show solution
  Answerbuta-1,3-diene and buta-1,2-diene
  $C H_{2} = C H - C H = C H_{2}$ (buta-1,3-diene) and $C H_{2} = C = C H - C H_{3}$ (buta-1,2-diene).
4
Short answerEasyNCERT Exercises 9.4
Name the products of ozonolysis of:
1. oct-4-ene
  Show solution
  Answerbutanal (2 mol)
  $C_{3} H_{7} C H = C H C_{3} H_{7}$ splits into two $C_{3} H_{7} C H O$ .
2. 2-methylpent-1-ene
  Show solution
  Answermethanal and pentan-2-one
  $C H_{2} =$ gives HCHO; the other carbon gives $C H_{3} C O C_{3} H_{7}$ .
3. 3-methylpent-2-ene
  Show solution
  Answerethanal and butan-2-one
  $C H_{3} C H =$ gives $C H_{3} C H O$ ; $= C (C H_{3}) C_{2} H_{5}$ gives $C H_{3} C O C_{2} H_{5}$ .
4. 1-phenylprop-1-ene
  Show solution
  Answerbenzaldehyde and ethanal
  $C_{6} H_{5} C H =$ gives $C_{6} H_{5} C H O$ ; $= C H C H_{3}$ gives $C H_{3} C H O$ .
Short answerMediumNCERT Exercises 9.5
Ozonolysis of alkene B produces propanone and butanal. Deduce B and name it.
Show solution
Answer2-methylhex-2-ene
Propanone supplies $(C H_{3})_{2} C =$ and butanal supplies $= C H C H_{2} C H_{2} C H_{3}$ .
$\begin{aligned} (C H_{3})_{2} C = C H^{-} \\ C H_{2} C H_{2} C H_{3} \end{aligned}$
The longest chain through C=C has six carbons; numbering from the branched end puts C=C and the methyl at C2: 2-methylhex-2-ene.
NumericalMediumNCERT Exercises 9.6
An alkene X has five C–C σ bonds, twelve C–H σ bonds and one C–C π bond. On ozonolysis it gives two moles of a single aldehyde of molar mass 58 u. Name X.
Show solution
Answerhex-3-ene
Twelve C–H bonds and one C=C in an open chain: $C_{6} H_{12}$ , with five C–C σ bonds along a six-carbon chain.
An aldehyde $C_{n} H_{2 n} O$ has $M = 14 n + 16 = 58$ , so $n = 3$ : propanal.
Two propanal units joined at their carbonyl carbons give $C_{2} H_{5} C H = C H C_{2} H_{5}$ : hex-3-ene.
Short answerMediumNCERT Exercises 9.7
Ozonolysis of an unknown alkene yields butanone and methanal. Deduce its structure.
Show solution
Answer2-methylbut-1-ene
Butanone, $C H_{3} C O C_{2} H_{5}$ , supplies $(C H_{3}) (C_{2} H_{5}) C =$ ; methanal supplies $= C H_{2}$ .
The alkene is $C H_{2} = C (C H_{3}) C_{2} H_{5}$ , 2-methylbut-1-ene.
8
Short answerEasyNCERT Exercises 9.8
Write balanced equations for the complete combustion of:
1. heptane
  Show solution
  $\begin{aligned} C_{7} H_{16} + 11 O_{2} \overset{}{\to} \\ 7 C O_{2} + 8 H_{2} O \end{aligned}$
2. propene
  Show solution
  $\begin{aligned} 2 C_{3} H_{6} + 9 O_{2} \overset{}{\to} \\ 6 C O_{2} + 6 H_{2} O \end{aligned}$
3. but-1-yne
  Show solution
  $\begin{aligned} 2 C_{4} H_{6} + 11 O_{2} \overset{}{\to} \\ 8 C O_{2} + 6 H_{2} O \end{aligned}$
4. ethylbenzene
  Show solution
  $\begin{aligned} 2 C_{8} H_{10} + 21 O_{2} \overset{}{\to} \\ 16 C O_{2} + 10 H_{2} O \end{aligned}$
Short answerMediumNCERT Exercises 9.9
Pent-2-ene exists as two geometrical isomers. Sketch them and predict, with a reason, which boils at the higher temperature.
Show solution
Answercis (slightly higher)
cis: $C H_{3}$ and $C_{2} H_{5}$ on the same side of C=C; trans: on opposite sides.
The cis isomer has a small net dipole moment, while in the trans isomer the bond dipoles nearly cancel. Extra dipole–dipole attraction gives cis-pent-2-ene the slightly higher boiling point (about 37 °C against 36 °C).
NumericalMediumNCERT Exercises 9.10
Hydrogenating cyclohexene (one C=C) releases 120 kJ mol⁻¹. Hydrogenating benzene to cyclohexane releases only 208 kJ mol⁻¹. By how much is benzene more stable than a ring with three isolated double bonds?
kJ/mol
Show solution
Three isolated C=C bonds would release about 360 kJ/mol. Benzene releases 152 kJ/mol less because its delocalised π electrons already sit at lower energy. This resonance (delocalisation) energy explains its unusual stability.
Short answerEasyNCERT Exercises 9.11
State the conditions for a ring to be aromatic, and use them to decide whether the cycloheptatrienyl cation, $C_{7} H_{7}^{+}$ , is aromatic.
Show solution
AnswerAromatic: planar, delocalised, 6 π electrons.
Conditions: a planar ring; a p orbital on every ring atom so the π electrons are completely delocalised; and 4n + 2 π electrons.
$C_{7} H_{7}^{+}$ : seven $s p^{2}$ carbons in a plane (the positive carbon has an empty p orbital), three C=C bonds giving 6 π electrons, so n = 1. It is aromatic.
12
Short answerMediumNCERT Exercises 9.12
None of these is aromatic. For each, say which condition fails.
1. cycloheptatriene, C₇H₈
  Show solution
  One ring carbon is an $s p^{3}$ $C H_{2}$ with no p orbital, so delocalisation is not complete.
2. cyclopentadienyl cation, C₅H₅⁺
  Show solution
  It is planar and fully conjugated but has only 4 π electrons, which is not 4n + 2.
3. cycloheptatrienyl anion, C₇H₇⁻
  Show solution
  It has 8 π electrons, which is not 4n + 2.
13
Short answerMediumNCERT Exercises 9.13
Outline a synthesis, starting from benzene, of:
1. m-bromonitrobenzene
  Show solution
  Nitrate first ( $H N O_{3}$ / $H_{2} S O_{4}$ ); –NO₂ then directs bromine ( $B r_{2}$ / $F e B r_{3}$ ) to the meta position.
2. p-bromotoluene
  Show solution
  Methylate first ( $C H_{3} C l$ / $A l C l_{3}$ ), then brominate ( $B r_{2}$ / $F e B r_{3}$ ); –CH₃ directs o/p, and the p isomer is separated.
3. m-nitroacetophenone
  Show solution
  Acylate first ( $C H_{3} C O C l$ / $A l C l_{3}$ ), then nitrate; –COCH₃ directs meta. (Nitrobenzene would not undergo Friedel–Crafts.)
4. benzophenone, C₆H₅COC₆H₅
  Show solution
  Friedel–Crafts acylation with benzoyl chloride, $C_{6} H_{5} C O C l$ , and anhydrous $A l C l_{3}$ .
Short answerEasyNCERT Exercises 9.14
In 2,2,4-trimethylhexane, $(C H_{3})_{3} C C H_{2} C H (C H_{3}) C_{2} H_{5}$ , how many 1°, 2°, 3° and 4° carbons are there, and how many H atoms are on each kind?
Show solution
Answer1°: 5 (15 H); 2°: 2 (4 H); 3°: 1 (1 H); 4°: 1 (0 H)
1°: the three methyls on C2, the methyl on C4 and C6 = 5 carbons, 3 H each = 15 H.
2°: C3 and C5 = 2 carbons, 4 H. 3°: C4, 1 H. 4°: C2, no H.
Check: 9 carbons and 20 hydrogens, $C_{9} H_{20}$ .
15
Short answerEasyNCERT Exercises 9.15
Arrange heptane, 2-methylhexane and 2,2-dimethylpentane in order of decreasing boiling point, and explain.
1. 2-methylhexane
2. 2,2-dimethylpentane
3. heptane
Show solution
All three are C₇H₁₆. More branching makes a molecule more compact, reducing the contact area and the dispersion forces, so the boiling point falls: heptane > 2-methylhexane > 2,2-dimethylpentane.
Long answerMediumNCERT Exercises 9.16
Without peroxide, but-1-ene and HBr give 2-bromobutane; with benzoyl peroxide the product is 1-bromobutane. Account for the switch and write the radical chain steps.
Show solution
AnswerCarbocation path gives 2-bromobutane; radical path gives 1-bromobutane.
Without peroxide: $H^{+}$ adds to C1, giving the secondary carbocation $C_{2} H_{5} C^{+} H C H_{3}$ ; $B r^{-}$ then gives 2-bromobutane (Markovnikov).
With peroxide, a radical chain runs instead:
$\begin{aligned} (C_{6} H_{5} C O O)_{2} \overset{}{\to} \\ 2 C_{6} H_{5}^{∙} + 2 C O_{2} \\ C_{6} H_{5}^{∙} + H B r \overset{}{\to} \\ C_{6} H_{6} + B r^{∙} \\ C H_{2} = C H C_{2} H_{5} \\ + B r^{∙} \overset{}{\to} \\ B r C H_{2} C^{∙} H C_{2} H_{5} \\ + H B r \overset{}{\to} \\ B r C H_{2} C H_{2} C_{2} H_{5} \\ + B r^{∙} \end{aligned}$
Br• adds to C1 so that the radical sits on the secondary C2, the more stable choice; the product is 1-bromobutane.
Long answerHardNCERT Exercises 9.17
Predict the products of ozonolysis of 1-ethyl-2-methylbenzene from each Kekulé structure. How would the observed mixture support Kekulé's idea of shifting double bonds?
Show solution
AnswerFour products, two of which can come only from different Kekulé structures.
Structure with C1=C2: fragments C2–C3, C4–C5, C6–C1 give 2-oxopropanal ( $C H_{3} C O C H O$ ), ethanedial ( $O H C C H O$ ) and 2-oxobutanal ( $C_{2} H_{5} C O C H O$ ).
Structure with C2=C3: fragments C1–C2, C3–C4, C5–C6 give pentane-2,3-dione ( $C_{2} H_{5} C O C O C H_{3}$ ) and two ethanedial.
Finding all four compounds shows that both arrangements contribute: the double bonds are not fixed, as Kekulé proposed (now explained by resonance).
18
Short answerEasyNCERT Exercises 9.18
Rank propane, propene and propyne from most to least acidic (consider each one's most acidic hydrogen) and explain the ranking.
1. propyne
2. propane
3. propene
Show solution
The most acidic hydrogens are on $s p$ (propyne, 50 % s), $s p^{2}$ (propene, 33 %) and $s p^{3}$ carbons (propane, 25 %). More s character holds the C–H electrons closer to carbon, so the proton leaves more easily: propyne > propene > propane.
Short answerMediumNCERT Exercises 9.19
Benzene reacts with $B r_{2}$ / $F e B r_{3}$ by substitution rather than addition, yet it does not react with nucleophiles such as $O H^{-}$ under ordinary conditions. Explain both observations.
Show solution
AnswerThe electron-rich π cloud attracts electrophiles and repels nucleophiles; substitution keeps the ring aromatic.
The six delocalised π electrons above and below the ring make benzene electron-rich, so it attracts electron-seeking electrophiles and repels electron-rich nucleophiles.
After the electrophile adds, losing H⁺ restores the stable aromatic ring, whereas addition would destroy it. So substitution wins.
20
Short answerEasyNCERT Exercises 9.20
How would you make benzene from:
1. calcium carbide
  Show solution
  Add water to get ethyne, then pass it through a red-hot iron tube at 873 K.
2. 1,2-dibromoethane
  Show solution
  Alcoholic KOH, then $N a N H_{2}$ , gives ethyne; then a red-hot iron tube gives benzene.
3. phenol
  Show solution
  Pass phenol vapour over heated zinc dust.
Short answerMediumNCERT Exercises 9.21
Several alkenes give 3-methylpentane when hydrogenated. Draw them all.
Show solution
Answer3 structural isomers (one with cis/trans forms)
3-Methylpent-1-ene, $C H_{2} = C H C H (C H_{3}) C_{2} H_{5}$ .
3-Methylpent-2-ene, $C H_{3} C H = C (C H_{3}) C_{2} H_{5}$ (cis and trans forms).
2-Ethylbut-1-ene, $C H_{2} = C (C_{2} H_{5})_{2}$ , with the double bond to the branch carbon.
22
Short answerMediumNCERT Exercises 9.22
Rank each trio from fastest to slowest in electrophilic substitution:
1. (a)
  benzene, phenol, chlorobenzene
  benzene
  phenol
  chlorobenzene
  Show answer
  Answer
  phenol
  benzene
  chlorobenzene
  –OH activates strongly; –Cl deactivates mildly.
2. (b)
  nitrobenzene, 1,3-dinitrobenzene, chlorobenzene
  1,3-dinitrobenzene
  chlorobenzene
  nitrobenzene
  Show answer
  Answer
  chlorobenzene
  nitrobenzene
  1,3-dinitrobenzene
  –Cl deactivates least; each –NO₂ deactivates strongly.
Short answerEasyNCERT Exercises 9.23
Benzene, ethylbenzene and benzoic acid are nitrated under the same conditions. Which reacts fastest? Explain.
Show solution
AnswerEthylbenzene
The ethyl group donates electron density to the ring (+I effect and hyperconjugation), activating it, so ethylbenzene nitrates fastest. –COOH withdraws electrons, so benzoic acid nitrates slowest.
Short answerEasyNCERT Exercises 9.24
Anhydrous $A l C l_{3}$ is not the only catalyst for Friedel–Crafts methylation with chloromethane. Name another Lewis acid that works and show how it generates the electrophile.
Show solution
AnswerAnhydrous $F e C l_{3}$
Anhydrous iron(III) chloride accepts a chloride ion: $C H_{3} C l + F e C l_{3}$ $\overset{}{\to} C H_{3}^{+} + [F e C l_{4}]^{-}$ . (In practice the $C H_{3}$ –Cl bond is strongly polarised rather than fully broken.)
Short answerMediumNCERT Exercises 9.25
Why is a mixture of 1-bromopropane and bromoethane with sodium a poor way to make pentane? Name the products.
Show solution
AnswerButane, pentane and hexane form together.
Sodium couples any two alkyl groups: ethyl + ethyl gives butane, ethyl + propyl gives pentane and propyl + propyl gives hexane.
Pentane is only part of the product and must be separated, so Wurtz is used only for symmetrical (even-carbon) alkanes.

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Quick check: MCQs on Hydrocarbons

9.1 Classification2 questions
9.2.1 Nomenclature and Isomerism3 questions
From carboxylic acids3 questions
Chemical properties4 questions
Conformations of ethane3 questions
Geometrical isomerism3 questions
From alcohols by acidic dehydration3 questions
Chemical properties4 questions
Chemical properties4 questions
Resonance and stability of benzene3 questions
Mechanism of electrophilic substitution4 questions
9.5.6 Directive influence of a functional group in monosubstituted benzene3 questions
9.6 Carcinogenicity and Toxicity2 questions

Glossary · 15

Key terms (English–Hindi)

alkaneऐल्केनailken: A saturated open-chain hydrocarbon, $C_{n} H_{2 n + 2}$ .First seen in the chapter →
alkeneऐल्कीनailkeen: A hydrocarbon with a C=C double bond; $C_{n} H_{2 n}$ for one double bond.First seen in the chapter →
alkyneऐल्काइनailkain: A hydrocarbon with a C≡C triple bond; $C_{n} H_{2 n - 2}$ for one triple bond.First seen in the chapter →
areneऐरीनaireen: A hydrocarbon containing an aromatic ring, usually a benzene ring.First seen in the chapter →
aromaticityऐरोमैटिकताaromaticata: Stability of a planar, fully delocalised ring with 4n + 2 π electrons.

conformationसंरूपणsanrupan: An arrangement of a molecule reached by rotation about a single bond, without breaking bonds.First seen in the chapter →

directive influence: The control a ring substituent has over where (o/p or m) and how fast the next substitution occurs.First seen in the chapter →

electrophileइलेक्ट्रॉनस्नेहीelectronsnehi: An electron-seeking species, such as $H^{+}$ or $N O_{2}^{+}$ .First seen in the chapter →
electrophilic substitutionइलेक्ट्रॉनस्नेही प्रतिस्थापनelectronsnehi pratisthapan: Replacement of a ring hydrogen by an electrophile through a σ-complex, keeping the ring aromatic.First seen in the chapter →

free radicalमुक्त मूलकmukt moolak: An atom or group with an unpaired electron, formed by homolysis of a bond.First seen in the chapter →

geometrical isomerismज्यामितीय समावयवताjyamitiya samavayavta: Isomerism from different arrangements of groups about a C=C bond, caused by restricted rotation.First seen in the chapter →

Markovnikov rule: In adding H–X to an unsymmetrical alkene, H goes to the doubly bonded carbon with more hydrogens.

ozonolysisओज़ोनी अपघटनozoni apghatan: Cleavage of a C=C bond by ozone and then Zn/H₂O, giving two carbonyl compounds.First seen in the chapter →

peroxide effectपरॉक्साइड प्रभावparoxide prabhav: Anti-Markovnikov addition of HBr to an unsymmetrical alkene in the presence of a peroxide, by a radical chain.First seen in the chapter →

torsional strain: The repulsion between bonds on adjacent atoms that raises the energy of eclipsed conformations.First seen in the chapter →

9.1 Classification

9.2 Alkanes

9.2.1 Nomenclature and Isomerism

9.2.2 Preparation

From unsaturated hydrocarbons

From alkyl halides

From carboxylic acids

9.2.3 Properties

Physical properties

Chemical properties

9.2.4 Conformations

Conformations of ethane

9.3 Alkenes

9.3.1 Structure of Double Bond

9.3.2 Nomenclature

9.3.3 Isomerism

Geometrical isomerism

9.3.4 Preparation

From alkynes

From alkyl halides

From vicinal dihalides

From alcohols by acidic dehydration

9.3.5 Properties

Physical properties

Chemical properties

9.4 Alkynes

9.4.1 Nomenclature and Isomerism

9.4.2 Structure of Triple Bond

9.4.3 Preparation

9.4.4 Properties

Physical properties

Chemical properties

9.5 Aromatic Hydrocarbon

9.5.1 Nomenclature and Isomerism

9.5.2 Structure of Benzene

Resonance and stability of benzene

9.5.3 Aromaticity

9.5.4 Preparation of Benzene

9.5.5 Properties

Physical properties

Chemical properties

Mechanism of electrophilic substitution

9.5.6 Directive influence of a functional group in monosubstituted benzene

9.6 Carcinogenicity and Toxicity

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