Chapter 15: Hydrocarbons
In the 1820s, the streets and shops of London were lit by gas. The gas was manufactured from coal and oils, squeezed into iron cylinders, and carted around the city. It worked well, but it left behind a nuisance: a small pool of oily liquid collecting at the bottom of the cylinders, which the gas companies treated as waste.
Michael Faraday, then working at the Royal Institution, decided that the waste deserved a proper look. He distilled it, separated the fractions, and analysed one of them carefully. The liquid contained only two elements, carbon and hydrogen, but with a strikingly high proportion of carbon. Faraday had no name for such a compound, so he described it by its composition and called it “bicarburet of hydrogen.” Chemists later renamed it benzene.
That single liquid, pulled out of an industrial leftover, went on to become one of the most important molecules in chemistry. Its structure puzzled the best minds in Europe for forty years, and the answer changed how chemists thought about bonding itself. Yet benzene belongs to the same broad family as the gas that burns on your kitchen stove, the polythene in a shopping bag, and the ethene that ripens mangoes: the hydrocarbons, compounds built from carbon and hydrogen alone. As Fig. 15.0 shows, that one family branches into several very different-looking groups.
Diagram
Fig. 15.0 – A 19th-century gas-lamp and a gas cylinder with a small pool of oily liquid at its base, arrow leading to a hexagonal benzene ring; below, four small icons labelled alkane (LPG cylinder), alkene (ripening mango), alkyne (welding torch), aromatic (benzene ring)
Keep these questions in mind as you read. This chapter will give you everything you need to reason out each one for yourself.
15.1 Alkanes
The Hydrocarbon Family
Organic chemistry contains millions of compounds, but a small group of them is built from just two elements. Compounds that contain carbon and hydrogen as the only elements are called hydrocarbons.
Before studying any one family in detail, it helps to see how the whole set is organised. Hydrocarbons are first divided by the shape of the carbon skeleton. If the carbon atoms are joined in a chain with two free ends, the compound is open chain (also called aliphatic). If the chain closes on itself to form a ring, the compound is cyclic.
The second division depends on the kind of carbon–carbon bonds present. A hydrocarbon in which every carbon–carbon bond is a single bond is saturated — it holds the maximum number of hydrogen atoms possible for that skeleton. A hydrocarbon that contains a carbon–carbon multiple bond, that is a double bond (\(\mathrm{C{=}C}\)) or a triple bond (\(\mathrm{C{\equiv}C}\)) or both, is unsaturated.
Cyclic hydrocarbons are divided one step further. A cyclic unsaturated hydrocarbon may be aromatic or nonaromatic, depending on whether its ring shows the special stability that benzene shows. Saturated cyclic hydrocarbons are called alicyclic. The complete scheme, along with the further split of open chain hydrocarbons into straight chain and branched chain, is drawn in Fig. 15.1.
Diagram
Fig. 15.1 – Classification of hydrocarbon: Hydrocarbon branching into Open chain (aliphatic) and Cyclic; Open chain into Saturated (alkane) and Unsaturated (alkene, alkyne), each into Straight chain and Branched chain; Cyclic into Saturated (alicyclic) and Unsaturated, the latter into Aromatic and Nonaromatic
What Makes an Alkane
The first family in this scheme is the alkanes. Alkanes are aliphatic saturated hydrocarbons containing carbon–carbon and carbon–hydrogen single covalent bonds. Every carbon in an alkane is \(sp^3\) hybridised, so every bond around it is a sigma bond and there is no multiple bond anywhere in the molecule.
You have learnt in Chapter 14 that alkanes have the general formula \(\mathrm{C_nH_{2n+2}}\), where ‘n’ stands for the number of carbon atoms in the alkane molecule. This single formula is a powerful tool. Put \(n = 1\) and you get \(\mathrm{CH_4}\) (methane); \(n = 2\) gives \(\mathrm{C_2H_6}\) (ethane); \(n = 6\) gives \(\mathrm{C_6H_{14}}\) (hexane). Because the formula fixes the hydrogen count for a given carbon count, any hydrocarbon that has fewer hydrogen atoms than \(2n+2\) must contain a multiple bond or a ring — a check you will use again and again in later sections.
15.1.1 Isomerism in Alkanes
Two compounds can share the same molecular formula and still be different substances, because the same atoms can be joined in more than one order. Such compounds are isomers. Alkanes with more than three carbon atoms generally exhibit structural isomerism, and in particular chain isomerism (refer to section 14.5.1), in which the carbon skeleton itself is arranged differently — in one isomer as an unbranched chain, in another as a shorter chain carrying a branch.
Methane, ethane and propane have no structural isomer, because with one, two or three carbon atoms there is only one way to join the skeleton. From four carbon atoms onward, branching becomes possible. Figure 15.2 shows the IUPAC and common names of structural isomers of \(\mathrm{C_4H_{10}}\) and \(\mathrm{C_5H_{12}}\).
Diagram
Fig. 15.2 – Structural isomers of butane and pentane: for C₄H₁₀, (i) full structural formula of Butane (n-Butane) and (ii) 2-Methylpropane (Isobutane); for C₅H₁₂, (i) Pentane (n-Pentane), (ii) CH₃–CH(CH₃)–CH₂–CH₃ labelled 2-Methylbutane (Isopentane) and (iii) C(CH₃)₄ drawn out, labelled 2,2-Dimethylpropane (Neopentane)
Notice how the count grows. Butane has two isomers, pentane three, hexane five. The number of possible structural isomers increases rapidly with the increase in the number of carbon atoms (see Table 15.1).
Table 15.1 : Alkanes and isomer number
| Number of Carbon | Alkane | Number of isomers |
|---|---|---|
| 1 | Methane | No structural isomer |
| 2 | Ethane | No structural isomer |
| 3 | Propane | No structural isomer |
| 4 | Butane | Two |
| 5 | Pentane | Three |
| 6 | Hexane | Five |
This rapid growth is one reason organic chemistry is so vast: a modest formula such as \(\mathrm{C_{10}H_{22}}\) already corresponds to seventy-five different alkanes, each a real substance with its own boiling point.
15.1.2 Conformations in Alkanes
Alkanes have only single bonded atoms. You have learnt that a single covalent bond is formed by coaxial overlap of orbitals and is called a sigma (\(\sigma\)) bond (refer to Chapter 5). “Coaxial” means the two orbitals overlap along the line joining the two nuclei.
That geometry has an important consequence. As a direct consequence of coaxial overlap of orbitals, a sigma bond is cylindrically symmetrical, and the extent of orbital overlap is unaffected by rotation about the single bond. Since rotating one carbon relative to the other does not weaken the overlap, nothing holds the two ends fixed, and therefore C–C bonds undergo rotation.
When such rotation happens, the atoms bonded to one carbon of a C–C single bond change their relative position with reference to the atoms on the other carbon of that bond. The resulting arrangements of the atoms in space about the C–C single bond are called conformations. Innumerable conformations result on complete rotation of a C–C single bond through 360°.
In Chapter 14 you have learnt about isomerism. The phenomenon of existence of conformation is a type of stereoisomerism — isomerism in which the atoms are joined in the same order but arranged differently in space. In conformational isomerism the conformations (or conformational isomers) interconvert by rotation about a C–C single bond.
Staggered and Eclipsed Ethane
Out of infinite conformations of the ethane molecule two are extreme and are called staggered and eclipsed conformation. In the staggered form, the hydrogen atoms on the front carbon sit exactly between the hydrogen atoms on the back carbon; in the eclipsed form, each front hydrogen lies directly in line with a back hydrogen.
These two extremes are represented by the Sawhorse formula and the Newman projection formula as shown in Fig. 15.3. In a sawhorse drawing the C–C bond is drawn as a slanting line, with three bonds at each end. In a Newman projection you look straight down the C–C bond: the front carbon is a point where three lines meet, and the back carbon is a circle carrying three lines.
Diagram
Fig. 15.3 – Representation of staggered and eclipsed conformation of ethane: (a) Sawhorse formula, staggered and eclipsed side by side; (b) Newman projection formula, staggered (back H’s bisecting front H’s) and eclipsed (back H’s hidden behind front H’s)
Conformational isomerism in other alkanes is more complex, because each additional C–C bond in the chain can rotate independently and the groups at the two ends are no longer three identical hydrogen atoms.
15.1.3 Industrial Preparation of Alkanes
Alkanes from Crude Oil
a. Industrial source : Crude petroleum and natural gas are the source of alkanes. Dead plants buried under the earth billions of years ago got converted into crude oil under the high temperature and pressure conditions. The crude oil collected in dome shaped cavities under the earth surface, which we call oil wells. Alkanes are obtained by fractional distillation of crude oil in oil refineries.
Fractional distillation works because the many alkanes in crude oil have different boiling points. When the hot vapour rises up a tall fractionating column, each group of alkanes condenses at the height where the temperature matches its boiling range, so the mixture separates into useful fractions instead of one shapeless liquid. The sequence is summarised in FC 15.1, and the column itself is drawn in Fig. 15.7.
Flowchart
FC 15.1 – Industrial source of alkanes: buried plant matter → high temperature and pressure over long time → crude oil trapped in dome-shaped cavities (oil wells) → pumped to refinery → fractional distillation → fractions: refinery gas → petrol → kerosene → diesel → lubricating oil → residue (bitumen/wax)
Diagram
Fig. 15.7 – A fractionating column of an oil refinery, crude oil entering the furnace at the base, with side outlets labelled (bottom to top) bitumen, lubricating oil, diesel, kerosene, petrol, refinery gas; arrow at side showing temperature decreasing upward
b. Methods of preparation of alkanes :
1. From Unsaturated Hydrocarbons
Catalytic hydrogenation of alkenes or alkynes with dihydrogen gas gives the corresponding alkane. The multiple bond is the weak point of the molecule: dihydrogen adds across it, and the unsaturated compound becomes saturated.
The catalyst decides how gentle the conditions can be. Finely divided powder of platinum (Pt) or palladium (Pd) catalyses the hydrogenation of alkenes and alkynes at room temperature. Relatively high temperature and pressure are required with finely divided nickel as the catalyst.
From alkenes
\[\mathrm{H_2C{=}CH_2} + \mathrm{H_2} \xrightarrow[\text{or Ni (high temp. and high pressure)}]{\ \mathrm{Pt}\ \text{or}\ \mathrm{Pd}\ \text{(room temp.)}\ } \mathrm{CH_3\!-\!CH_3}\]
\[\mathrm{H_3C\!-\!CH{=}CH_2} + \mathrm{H_2} \xrightarrow[\text{or Ni (high temp. and high pressure)}]{\ \mathrm{Pt}\ \text{or}\ \mathrm{Pd}\ \text{(room temp.)}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH_3}\]
Here ethene gives ethane and propene gives propane.
From alkynes
A triple bond has two \(\pi\) bonds, so two molecules of dihydrogen are needed to reach the alkane.
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} + 2\,\mathrm{H_2} \xrightarrow[\text{or Ni (high temp. and high pressure)}]{\ \mathrm{Pt}\ \text{or}\ \mathrm{Pd}\ \text{(room temp.)}\ } \mathrm{CH_3\!-\!CH_3}\]
Ethyne gives ethane. The two cases can be written in general form:
\[\mathrm{C_nH_{2n}} + \mathrm{H_2} \xrightarrow[\text{or Ni (high temp. \& high pressure)}]{\ \mathrm{Pt}\ \text{or}\ \mathrm{Pd}\ \text{(room temp.)}\ } \mathrm{C_nH_{2n+2}}\]
\[\mathrm{C_nH_{2n-2}} + 2\,\mathrm{H_2} \xrightarrow[\text{or Ni (high temp. \& high pressure)}]{\ \mathrm{Pt}\ \text{or}\ \mathrm{Pd}\ \text{(room temp.)}\ } \mathrm{C_nH_{2n+2}}\]
Alkene gives alkane, and alkyne gives alkane.
2. From Alkyl Halides
a. By reduction of alkyl halides : Alkyl halides on reduction with zinc and dilute hydrochloric acid form alkanes. The reduction of alkyl halides is due to nascent hydrogen obtained from the reducing agent Zn and dilute HCl. Nascent hydrogen means hydrogen at the moment of its formation, written as \([\mathrm{H}]\); it is far more reactive than ordinary \(\mathrm{H_2}\) gas because it has not yet paired up into a stable molecule.
\[\mathrm{CH_3\!-\!I} + 2[\mathrm{H}] \xrightarrow{\ \mathrm{Zn},\ \mathrm{HCl}\ } \mathrm{CH_4} + \mathrm{HI}\]
Methyl iodide and nascent hydrogen give methane.
\[\mathrm{CH_3\!-\!CH_2\!-\!Br} + 2[\mathrm{H}] \xrightarrow{\ \mathrm{Zn},\ \mathrm{HCl}\ } \mathrm{CH_3\!-\!CH_3} + \mathrm{HBr}\]
Ethyl bromide and nascent hydrogen give ethane.
b. By using reactive metal : In the 19th century alkyl halides were transformed to alkane having double the number of carbons by the Wurtz coupling reaction using the reactive metal sodium. Two alkyl groups are stitched together, so the product chain is twice as long as the chain in the halide.
\[\mathrm{CH_3\!-\!Br} + 2\,\mathrm{Na} + \mathrm{Br\!-\!CH_3} \xrightarrow{\ \text{dry ether}\ } \mathrm{CH_3\!-\!CH_3} + 2\,\mathrm{NaBr}\]
Later better methods using metals such as Mg and Li were developed. The Grignard reaction is one of the new methods widely employed for preparation of alkanes.
Alkyl magnesium halides (Grignard reagent) are obtained by treating alkyl halides with dry metal magnesium in the presence of dry ether. These on treatment with water give alkanes.
\[\mathrm{R\!-\!X} + \mathrm{Mg} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!Mg\!-\!X}\]
Alkyl halide gives alkyl magnesium halide.
\[\mathrm{R\!-\!Mg\!-\!X} + \mathrm{H\!-\!O\!-\!H} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!H} + \mathrm{MgX(OH)}\]
The alkyl magnesium halide gives the alkane.
\[\mathrm{CH_3\!-\!Mg\!-\!I} + \mathrm{H\!-\!O\!-\!H} \xrightarrow{\ \text{dry ether}\ } \mathrm{CH_3\!-\!H} + \mathrm{MgI(OH)}\]
Methyl magnesium iodide gives methane. Notice that unlike the Wurtz reaction, this route returns an alkane with the same number of carbon atoms as the halide, because the magnesium is simply replaced by hydrogen from water.
15.1.4 Physical Properties of Alkanes
i. Polarity : The electronegativity of carbon and hydrogen is nearly the same. Because neither atom pulls the shared pair appreciably towards itself, C–H and C–C bonds are non-polar covalent bonds and alkanes are, thus, nonpolar.
A nonpolar molecule has no permanent positive or negative end, so it cannot attract its neighbours electrostatically. The forces which hold non-polar molecules together are the van der Waals forces. Those are usually weak. They arise from the momentary, shifting distortions of the electron cloud, so they act only where two molecules are in close contact — and that makes contact area the deciding factor. Larger the surface area of molecules, stronger are such intermolecular van der Waals forces.
This is where the shape of the chain matters. A straight chain molecule can lie alongside its neighbour over its whole length, like two pencils side by side. A branched molecule is bunched up towards a ball shape and can touch its neighbour only over a small patch. Therefore the intermolecular forces are relatively stronger in straight chain alkanes than in branched alkanes (see Fig. 15.4).
Diagram
Fig. 15.4 – Relative surface area of straight chain and branched alkanes: two straight-chain pentane molecules drawn lying alongside each other with a bracket marked “large surface area”; below, two neopentane molecules drawn as ovals touching at a small region marked “small surface area”
Therefore straight chain alkanes have higher melting and boiling points. Branched alkanes have lower melting and boiling points. Melting and boiling differ only in how far the process goes: intermolecular forces are overcome partly during the melting process and completely during the boiling process.
Table 15.2 shows the melting and boiling points of some alkanes.
Table 15.2 : Melting and Boiling points in alkanes
| Molecular formula | Name | Molecular mass/u | b.p. (K) | m.p. (K) |
|---|---|---|---|---|
| \(\mathrm{CH_4}\) | Methane | 16 | 111.0 | 90.5 |
| \(\mathrm{C_2H_6}\) | Ethane | 30 | 184.4 | 101.0 |
| \(\mathrm{C_3H_8}\) | Propane | 44 | 230.9 | 85.3 |
| \(\mathrm{C_4H_{10}}\) | Butane | 58 | 272.4 | 134.6 |
| \(\mathrm{C_4H_{10}}\) | 2-Methylpropane | 58 | 261.0 | 114.7 |
| \(\mathrm{C_5H_{12}}\) | Pentane | 72 | 309.1 | 143.3 |
| \(\mathrm{C_5H_{12}}\) | 2-Methylbutane | 72 | 300.9 | 113.1 |
| \(\mathrm{C_5H_{12}}\) | 2,2-Dimethylpropane | 72 | 282.5 | 256.4 |
| \(\mathrm{C_6H_{14}}\) | Hexane | 86 | 341.9 | 178.5 |
Diagram
Fig. 15.8 – Two graphs side by side using Table 15.2 data: (a) boiling point (K) against number of carbon atoms for the straight-chain alkanes CH₄ to C₆H₁₄, smooth rising curve; (b) bar chart of b.p. and m.p. for the three C₅H₁₂ isomers — pentane, 2-methylbutane, 2,2-dimethylpropane
Colour, Odour and Physical State
Alkanes are colourless and odourless. At room temperature, the first four alkanes are gases, the next alkanes having 5 to 17 carbons are liquids and those with 18 carbons onwards are solids. The pattern follows directly from the trend above: the bigger the molecule, the stronger the total van der Waals attraction, and the more the molecules resist being pulled apart into a liquid or a gas.
15.1.5 Chemical Properties of Alkanes
Alkanes are relatively unreactive towards acids, bases, oxidizing and reducing agents. There are two reasons for this calm behaviour: the C–C and C–H bonds are strong, and they are almost non-polar, so a molecule of alkane offers no positively or negatively charged site for an acid, a base or an ion to attack. They undergo the following reactions under specified conditions — and each of those conditions is a way of supplying enough energy, or a reactive enough species, to get past that stability.
1. Halogenation of Alkanes
The reactions in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms are known as substitution reactions.
Substitution of H atoms of alkanes by X (halogen, X = Cl, Br, I and F) atom is called halogenation of alkanes.
Alkanes react with halogens in presence of UV light or diffused sunlight or at higher temperature (573 to 773 K) to give a mixture of alkyl halides. Light or heat is essential; in the dark and cold, an alkane and a halogen can be kept together with almost no reaction.
Chlorination of methane :
\[\mathrm{CH_4} + \mathrm{Cl\!-\!Cl} \xrightarrow{\ \text{UV light or }\Delta\ } \mathrm{CH_3\!-\!Cl} + \mathrm{H\!-\!Cl}\]
Methane and chlorine give chloromethane (methyl chloride) and hydrogen chloride. The product still has three hydrogen atoms left, and chlorine does not stop there.
\[\mathrm{CH_3\!-\!Cl} + \mathrm{Cl\!-\!Cl} \xrightarrow{\ \text{UV light or }\Delta\ } \mathrm{CH_2Cl_2} + \mathrm{H\!-\!Cl}\]
Chloromethane gives dichloromethane.
\[\mathrm{CH_2Cl_2} + \mathrm{Cl\!-\!Cl} \xrightarrow{\ \text{UV light or }\Delta\ } \mathrm{CHCl_3} + \mathrm{H\!-\!Cl}\]
Dichloromethane gives trichloromethane.
\[\mathrm{CHCl_3} + \mathrm{Cl\!-\!Cl} \xrightarrow{\ \text{UV light or }\Delta\ } \mathrm{CCl_4} + \mathrm{H\!-\!Cl}\]
Trichloromethane gives tetrachloromethane.
Since all four steps are possible, the proportions in the product mixture can be controlled by choosing which reactant is in excess. Tetrachloromethane is a major product when excess of chlorine is used. Chloromethane is obtained as major product when excess of methane is employed.
The reactivity of halogens toward alkanes follows the order
\[\mathrm{F_2} > \mathrm{Cl_2} > \mathrm{Br_2} > \mathrm{I_2}\]
The ease of replacement of hydrogen atoms from the carbon is in the order of
\[3^{\circ} > 2^{\circ} > 1^{\circ}\]
Here \(3^{\circ}\), \(2^{\circ}\) and \(1^{\circ}\) describe the carbon that carries the hydrogen: a tertiary carbon is joined to three other carbons, a secondary carbon to two, and a primary carbon to one. A hydrogen on a tertiary carbon is the easiest to remove.
For example,
\[\mathrm{H_3C\!-\!CH_2\!-\!CH_3} \xrightarrow[\text{UV light}]{\ \mathrm{Cl_2}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!Cl}\ (45\%)\ +\ \mathrm{H_3C\!-\!CHCl\!-\!CH_3}\ (55\%)\]
Propane gives 1-chloropropane (45%) and 2-chloropropane (55%).
\[\mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_3} \xrightarrow[\text{UV light}]{\ \mathrm{Cl_2}\ } \mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_2\!-\!Cl}\ (36\%)\ +\ \mathrm{(CH_3)_3C\!-\!Cl}\ (64\%)\]
2-Methylpropane gives 1-chloro-2-methylpropane (36%) and 2-chloro-2-methylpropane (64%).
Bromination gives corresponding bromides, but in different proportions. For example,
\[\mathrm{H_3C\!-\!CH_3} \xrightarrow[\text{UV light}]{\ \mathrm{Br_2}\ } \mathrm{H_3C\!-\!CH_2\!-\!Br}\]
Ethane gives bromoethane.
\[\mathrm{H_3C\!-\!CH_2\!-\!CH_3} \xrightarrow[\text{UV light}]{\ \mathrm{Br_2}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!Br}\ (3\%)\ +\ \mathrm{H_3C\!-\!CHBr\!-\!CH_3}\ (97\%)\]
Propane gives 1-bromopropane (3%) and 2-bromopropane (97%).
\[\mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!CH_3} \xrightarrow[\text{UV light}]{\ \mathrm{Br_2}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!Br}\ (2\%)\ +\ \mathrm{H_3C\!-\!CH_2\!-\!CHBr\!-\!CH_3}\ (98\%)\]
Butane gives 1-bromobutane (2%) and 2-bromobutane (98%).
\[\mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_3} \xrightarrow[\text{UV light}]{\ \mathrm{Br_2}\ } \mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_2\!-\!Br}\ (\text{trace})\ +\ \mathrm{(CH_3)_3C\!-\!Br}\ (\text{over } 99\%)\]
2-Methylpropane gives 1-bromo-2-methylpropane (trace) and 2-bromo-2-methylpropane (over 99%).
Compare the two sets of percentages. In bromination, there is a high degree of selectivity as to which hydrogen atoms are replaced. Chlorine attacks all kinds of hydrogen fairly indiscriminately, so both isomers appear in comparable amounts; bromine, being the less reactive of the two, is much fussier and picks out the most easily replaced hydrogen almost exclusively.
Halogenation of alkanes follows the free radical mechanism. Homolysis of the halogen molecule (\(\mathrm{X_2}\)) generates halogen atoms, that is, halogen free radicals. Homolysis means the bond breaks evenly, one electron going to each atom, which leaves each atom with a single unpaired electron. Such a species is a free radical, written with a dot.
The mechanism of the first step of chlorination of methane is shown below, and the electron movement is drawn in Fig. 15.9.
Initiation
\[\mathrm{Cl\!-\!Cl} \xrightarrow[\text{homolysis}]{\ \text{UV}\ } \mathrm{Cl^{\bullet}} + \mathrm{Cl^{\bullet}}\]
Propagation
\[\mathrm{CH_3\!-\!H} + \mathrm{Cl^{\bullet}} \rightarrow \mathrm{CH_3^{\bullet}} + \mathrm{H\!-\!Cl}\]
\[\mathrm{CH_3^{\bullet}} + \mathrm{Cl\!-\!Cl} \rightarrow \mathrm{CH_3\!-\!Cl} + \mathrm{Cl^{\bullet}}\]
Diagram
Fig. 15.9 – Free radical chlorination of methane: initiation, Cl–Cl bond with two single-headed (fishhook) arrows curving outward to each Cl showing one electron each; propagation step 1, fishhook arrows showing Cl• abstracting H from CH₄ to give •CH₃ + HCl; propagation step 2, •CH₃ attacking Cl–Cl to give CH₃Cl + Cl•
Look carefully at the second propagation step: it regenerates a chlorine radical, which can attack another methane molecule. That is why one photon of UV light can lead to the substitution of many molecules — the chain keeps itself running until the radicals combine with each other and the chain is terminated.
Flowchart
FC 15.2 – Free radical mechanism of chlorination: UV light → homolysis of Cl–Cl → 2 Cl• (initiation) → Cl• abstracts H from CH₄ → •CH₃ + HCl (propagation 1) → •CH₃ attacks Cl₂ → CH₃Cl + Cl• (propagation 2) → Cl• re-enters propagation 1 (chain continues); branch: two radicals combine → chain terminated
Two conclusions follow from all these examples. First, depending upon which hydrogen atom is replaced, a number of isomeric products can be formed from an alkane; for example, n-butane and isobutane yield two isomers each. Second, halogenation of an alkane yields a mixture of all possible isomeric products, indicating all the hydrogen atoms are susceptible to substitution.
2. Combustion
Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. That is why alkanes are good fuels.
\[\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O} \qquad \Delta_c H^0 = -890\ \text{kJ mol}^{-1}\]
Methane burns to give carbon dioxide and water.
\[\mathrm{C_4H_{10}} + \frac{13}{2}\,\mathrm{O_2} \rightarrow 4\,\mathrm{CO_2} + 5\,\mathrm{H_2O} \qquad \Delta_c H^0 = -2875.84\ \text{kJ mol}^{-1}\]
Butane burns to give carbon dioxide and water. The fractional coefficient \(13/2\) is perfectly acceptable here because the equation is written for exactly one mole of butane, which is what an enthalpy of combustion refers to.
A representative combustion equation for an alkane is
\[\mathrm{C_nH_{2n+2}} + \left(\frac{3n+1}{2}\right)\mathrm{O_2} \rightarrow n\,\mathrm{CO_2} + (n+1)\,\mathrm{H_2O} + \text{heat}\]
Read the coefficients as a bookkeeping statement: each carbon becomes one \(\mathrm{CO_2}\), each pair of hydrogen atoms becomes one \(\mathrm{H_2O}\), and the oxygen required is whatever those products need.
3. Pyrolysis
Alkanes on heating at higher temperature in absence of air decompose to lower alkanes, alkenes and hydrogen, etc. This is known as pyrolysis or cracking. Air must be absent, otherwise the alkane would simply burn; with no oxygen available, the only thing the heat can do is break the molecule apart.
\[\mathrm{C_6H_{14}} \xrightarrow{\ 773\ \text{K}\ } \mathrm{C_6H_{12}} + \mathrm{H_2}\]
\[\mathrm{C_6H_{14}} \xrightarrow{\ 773\ \text{K}\ } \mathrm{C_4H_8} + \mathrm{C_2H_6}\]
\[\mathrm{C_6H_{14}} \xrightarrow{\ 773\ \text{K}\ } \mathrm{C_3H_6} + \mathrm{C_2H_4} + \mathrm{CH_4}\]
The same starting alkane gives several different sets of products, because the chain can break at different C–C bonds. The three routes are set out together in FC 15.3.
Flowchart
FC 15.3 – Pyrolysis of hexane at 773 K: C₆H₁₄ branching into three routes → (1) C₆H₁₂ + H₂; (2) C₄H₈ + C₂H₆; (3) C₃H₆ + C₂H₄ + CH₄
4. Reforming
Straight chain alkanes containing 6 to 10 carbon atoms are converted to benzene and its homologues on heating under 10 to 20 atm pressure at about 773 K in the presence of \(\mathrm{V_2O_5}\), \(\mathrm{Cr_2O_3}\), \(\mathrm{Mo_2O_3}\) etc. supported over alumina. The reaction involves simultaneous dehydrogenation and cyclization. This reaction is known as aromatization or reforming.
\[\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!CH_3} \xrightarrow[\text{773 K, 10--20 atm}]{\ \mathrm{V_2O_5}\ } \mathrm{C_6H_6} + 4\,\mathrm{H_2}\]
n-Hexane gives benzene. Count the hydrogen atoms to see why four molecules of dihydrogen are released: the open chain \(\mathrm{C_6H_{14}}\) must lose eight hydrogen atoms to become the ring compound \(\mathrm{C_6H_6}\).
This process is used in refineries to produce high quality gasoline (the fuel used in automobiles).
15.1.6 Uses of Alkanes
Almost every use of alkanes traces back to one of two properties met in this section: they release a great deal of heat on burning, and they are non-polar, greasy and chemically unreactive.
- The first four alkanes are used as a fuel mainly for heating and cooking purpose, for example, LPG and CNG.
- CNG, petrol and diesel are used as fuels for automobiles.
- Lower liquid alkanes are used as solvents, since a non-polar liquid dissolves non-polar substances such as oils, fats and greases.
- Alkanes with more than 35 carbon atoms (tar) are used for road surfacing.
- Waxes are high molecular weight alkanes. They are used as lubricants. They are also used for the preparation of candles and carbon black, which is used in the manufacture of printing ink, shoe polish, etc.
15.2 Alkenes
Alkanes had no reactive site at all. Alkenes are the opposite: everything interesting about them happens at one place in the molecule.
Alkenes are unsaturated hydrocarbons containing at least one carbon–carbon double bond. Alkenes with one carbon–carbon double bond contain two hydrogen atoms less than the corresponding alkanes. They have the general formula \(\mathrm{C_nH_{2n}}\), where n = 2, 3, 4 … etc. You have also learnt about the IUPAC names of alkenes in Chapter 14.
Compare the two general formulae for the same carbon count. For \(n = 4\), the alkane is \(\mathrm{C_4H_{10}}\) but the alkene is \(\mathrm{C_4H_8}\) — two hydrogen atoms fewer, because a second bond between two carbon atoms uses up two positions that hydrogen would otherwise occupy. That double bond consists of one sigma bond and one pi (\(\pi\)) bond, and the loosely held \(\pi\) electrons are the source of nearly all alkene chemistry.
Dienes and Trienes
Nothing restricts a molecule to a single double bond. The aliphatic unsaturated hydrocarbons containing two or three carbon–carbon double bonds are called alkadienes and alkatrienes respectively. For example, \(\mathrm{H_2C{=}C(CH_3)\!-\!CH{=}CH_2}\) is methylbuta-1,3-diene (isoprene), the building unit of natural rubber, and \(\beta\)-phellandrene, a cyclic diene, is a constituent of oil of eucalyptus.
15.2.1 Isomerism in Alkenes
Alkenes with more than three carbon atoms show both structural isomerism and geometrical isomerism. The double bond adds a completely new possibility: not only can the skeleton be rearranged, the double bond itself can sit in different places, and the groups around it can be arranged differently in space.
i. Structural Isomerism
An alkene with molecular formula \(\mathrm{C_4H_8}\) is butene. The structural formulae for \(\mathrm{C_4H_8}\) can be drawn in three different ways:
\[\text{(I)}\quad \mathrm{\overset{1}{C}H_2{=}\overset{2}{C}H\!-\!\overset{3}{C}H_2\!-\!\overset{4}{C}H_3} \qquad \text{But-1-ene (chain of 4 carbon atoms)}\]
\[\text{(II)}\quad \mathrm{\overset{1}{C}H_3\!-\!\overset{2}{C}H{=}\overset{3}{C}H\!-\!\overset{4}{C}H_3} \qquad \text{But-2-ene (chain of 4 carbon atoms)}\]
\[\text{(III)}\quad \mathrm{\overset{1}{C}H_2{=}\overset{2}{C}(CH_3)\!-\!\overset{3}{C}H_3} \qquad \text{2-Methylprop-1-ene (chain of 3 carbon atoms)}\]
The three are compared in Fig. 15.10. Structure I and III along with II and III are the examples of chain isomerism. They differ in carbon chain length — the longest chain is four carbons in I and II, but only three in III. Structure I and II are the examples of position isomerism, because they differ in the position of the double bond in the same carbon chain.
Diagram
Fig. 15.10 – The three structural isomers of C₄H₈ drawn side by side with numbered carbons: (I) but-1-ene, (II) but-2-ene, (III) 2-methylprop-1-ene; brackets marking I–II as position isomers and I–III, II–III as chain isomers
ii. Geometrical Isomerism
The isomerism which arises due to the difference in spatial arrangement of atoms or groups about doubly bonded carbon (\(\mathrm{C{=}C}\)) atoms is called geometrical isomerism.
The reason this is possible at all is that a double bond cannot rotate. In an alkane, free rotation about the C–C sigma bond scrambles every arrangement into every other; in an alkene the \(\pi\) bond locks the two carbon atoms and their attached groups into one rigid plane. Whatever arrangement the groups have, they keep.
If the two atoms or groups bonded to each end of the \(\mathrm{C{=}C}\) double bond are different, then the molecule can be represented by two different spatial arrangements of the groups, shown as (A) and (B) in Fig. 15.11.
Diagram
Fig. 15.11 – Geometrical isomerism: (A) generalised alkene with X and X on the same side of the C=C and Y, Y on the other side; (B) the same alkene with X and Y on each side, i.e. the identical groups on opposite sides; below, cis-but-2-ene (both CH₃ on the same side, b.p. = 277 K) and trans-but-2-ene (CH₃ groups on opposite sides, b.p. = 274 K)
In structure (A), two identical atoms or groups lie on the same side of the double bond.
In structure (B), two identical atoms or groups lie on the opposite side of the double bond.
Due to the different arrangement of atoms or groups in space, these isomers differ in their physical properties like melting point, boiling point, solubility etc. Geometrical or cis–trans isomers of but-2-ene are represented in Fig. 15.11: cis-but-2-ene has b.p. = 277 K and trans-but-2-ene has b.p. = 274 K.
When cis–trans Isomerism Is Not Possible
Geometrical isomerism needs two different groups at each end of the double bond. If either carbon of the double bond carries two identical groups, flipping the molecule over gives back the same structure, and there is only one compound.
No terminal alkenes, i.e. those with a \(\mathrm{C{=}CH_2}\) unit, can exist as cis- and trans-isomers. No 1,1-disubstituted alkenes, i.e. those with a \(\mathrm{C{=}CR_2}\) unit, can exist as cis- and trans-isomers. Alkenes with the general formulae \(\mathrm{RCH{=}CHR}\), \(\mathrm{R_1R_2C{=}CR_1R_2}\), \(\mathrm{R_1CH{=}CR_1R_2}\), \(\mathrm{R_1CH{=}CR_2R_3}\), \(\mathrm{R_1CH{=}CHR_2}\) and \(\mathrm{R_1R_2C{=}CR_3R_4}\) exhibit cis–trans isomerism.
15.2.2 Preparation of Alkenes
a. Industrial Sources
The most important alkenes for the chemical industry are ethene, propene and buta-1,3-diene. Alkenes containing up to four carbon atoms can be obtained in pure form from the petroleum products. Ethene is produced from natural gas and crude oil by cracking — the same pyrolysis met in section 15.1.5, run deliberately on an industrial scale so that the alkene fragments, rather than the alkanes, become the valuable product.
b. Methods of Preparation of Alkenes
(I) Elimination reactions (1,2-elimination)
The reactions in which two atoms or groups are eliminated or removed from adjacent carbon atoms are called 1,2-elimination reactions. Two adjacent single-bonded carbon atoms each give up a substituent, and the electrons left behind form the second bond between them, as shown in Fig. 15.12.
Diagram
Fig. 15.12 – 1,2-Elimination: on the left, –C–C– with X on one carbon and Y on the adjacent carbon, both labelled sp³ hybridized; arrow marked “1,2-elimination, –XY”; on the right, –C=C– labelled sp² hybridized, alkene
Two points are worth fixing in memory. The carbon–carbon double bond can be generated from a carbon–carbon single bond by an elimination reaction. And during an elimination reaction the \(sp^3\) carbon atom changes to an \(sp^2\) carbon atom, because each carbon loses one substituent and is left with three groups around it instead of four.
Some elimination reactions useful to prepare alkenes are described below.
i. Dehydrohalogenation of alkyl halides
Dehydrohalogenation means removal of a hydrogen (H) atom and a halogen (X) atom from adjacent carbon atoms. To describe which carbon is which, chemists label them from the halogen: the carbon carrying X is called the \(\alpha\)-carbon atom, and the hydrogen atom from the adjacent carbon, called the \(\beta\)-carbon atom, is removed. The reaction is therefore known as \(\beta\)-elimination. The labelling is shown in Fig. 15.13.
Diagram
Fig. 15.13 – α and β carbon labelling: bromoethane with the C bearing Br marked α and the adjacent CH₃ carbon marked β, with the H to be removed circled; beside it 2-chlorobutane with the C bearing Cl marked α and both adjacent carbons marked β, showing two possible sites of elimination
Dehydrohalogenation suffers from the disadvantage that –H can be eliminated from the carbon on either side of the \(\alpha\)-carbon bearing the –X. When there are two different \(\beta\)-carbon atoms, two different alkenes can form, and a mixture results.
When an alkyl halide is boiled with a hot concentrated alcoholic solution of a strong base like KOH or NaOH, an alkene is formed with removal of a water molecule.
\[\mathrm{CH_3\!-\!CH_2\!-\!Br} + \mathrm{KOH}\ (\text{alcoholic}) \xrightarrow{\ \text{boil}\ } \mathrm{H_2C{=}CH_2} + \mathrm{KBr} + \mathrm{H_2O}\]
Bromoethane gives ethene. Here the \(\alpha\)-carbon has only one kind of neighbour, so only one alkene is possible. With 2-chlorobutane the situation is different:
\[\mathrm{CH_3\!-\!CH_2\!-\!CHCl\!-\!CH_3} + \mathrm{KOH}\ (\text{alcoholic}) \xrightarrow{\ \text{boil}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH{=}CH_2}\ (20\%) + \mathrm{CH_3\!-\!CH{=}CH\!-\!CH_3}\ (80\%) + \mathrm{KCl} + \mathrm{H_2O}\]
2-Chlorobutane gives but-1-ene (20%) and but-2-ene (80%).
In the dehydrohalogenation of 2-chlorobutane, but-2-ene (a disubstituted alkene) is the preferred product because it is formed faster than but-1-ene (a monosubstituted alkene). This is in accordance with Saytzeff rule.
The ease of dehydrohalogenation of alkyl halides accordingly is in the order
\[3^{\circ} > 2^{\circ} > 1^{\circ}\]
The ease of formation of alkenes :
\[\mathrm{R_2C{=}CR_2} > \mathrm{R_2C{=}CHR} > \mathrm{R_2C{=}CH_2},\quad \mathrm{RCH{=}CHR} > \mathrm{RCH{=}CH_2}\]
The stability of alkenes :
\[\mathrm{R_2C{=}CR_2} > \mathrm{R_2C{=}CHR} > \mathrm{R_2C{=}CH_2},\quad \mathrm{RCH{=}CHR} > \mathrm{RCH{=}CH_2} > \mathrm{CH_2{=}CH_2}\]
Read the two lists together: the alkene that is more stable is also the one that forms faster, which is exactly why Saytzeff’s rule works. The more alkyl groups sit on the doubly bonded carbon atoms, the more the double bond is stabilised, and the lower the energy barrier on the way to it.
ii. Dehydration of alcohols
Alcohols on heating with sulphuric acid form alkenes with elimination of a water molecule. The reaction is known as catalyzed dehydration of alcohols. The exact conditions of dehydration depend upon the alcohol — you will see below that the strength of the acid and the temperature both fall as the alcohol becomes more substituted.
Dehydration of an alcohol is an example of \(\beta\)-elimination, since the OH group from the \(\alpha\)-carbon along with an H atom from the \(\beta\)-carbon atom is removed.
\[\mathrm{{-}\overset{\beta}{C}H\!-\!\overset{\alpha}{C}(OH){-}} \xrightarrow[\Delta]{\ \text{conc. } \mathrm{H_2SO_4}\ } \mathrm{{-}C{=}C{-}} + \mathrm{H_2O}\]
An alcohol gives an alkene and water. The ease of dehydration of alcohols is in the order
\[3^{\circ} > 2^{\circ} > 1^{\circ}\]
\[\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!OH} \xrightarrow[413\ \text{K}]{\ 75\%\ \mathrm{H_2SO_4}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH{=}CH_2} + \mathrm{H_2O}\]
Butan-1-ol gives but-1-ene as the main product.
\[\mathrm{CH_3\!-\!CH_2\!-\!CH(OH)\!-\!CH_3} \xrightarrow[373\ \text{K}]{\ 60\%\ \mathrm{H_2SO_4}\ } \mathrm{H_3C\!-\!CH{=}CH\!-\!CH_3}\ (\text{major}) + \mathrm{H_3C\!-\!CH_2\!-\!CH{=}CH_2}\ (\text{minor}) + \mathrm{H_2O}\]
Here the secondary alcohol has two different \(\beta\)-carbon atoms, so two alkenes form, and once again the more substituted alkene dominates.
\[\mathrm{(CH_3)_3C\!-\!OH} \xrightarrow[363\ \text{K}]{\ 20\%\ \mathrm{H_2SO_4}\ } \mathrm{H_3C\!-\!C(CH_3){=}CH_2} + \mathrm{H_2O}\]
2-Methylpropan-2-ol gives 2-methylprop-1-ene. Compare the three sets of conditions: the primary alcohol needs 75% acid at 413 K, the secondary needs 60% acid at 373 K, and the tertiary needs only 20% acid at 363 K. That progression is the practical face of the \(3^{\circ} > 2^{\circ} > 1^{\circ}\) order.
iii. By dehalogenation of vicinal dihalides
Removal of two halogen atoms from adjacent carbon atoms is called dehalogenation. The dihalides of alkane in which two halogen atoms are attached to adjacent carbon atoms are called vicinal dihalides — “vicinal” simply means neighbouring. Vicinal dihalides on heating with zinc metal form alkenes; the zinc takes both halogen atoms away as zinc halide.
\[\mathrm{{-}C(X)\!-\!C(X){-}} + \mathrm{Zn} \rightarrow \mathrm{{-}C{=}C{-}} + \mathrm{ZnX_2}\]
\[\mathrm{CH_2Br\!-\!CH_2Br} + \mathrm{Zn} \rightarrow \mathrm{CH_2{=}CH_2} + \mathrm{ZnBr_2}\]
1,2-Dibromoethane gives ethene.
\[\mathrm{CH_3CHBr\!-\!CH_2Br} + \mathrm{Zn} \rightarrow \mathrm{CH_3CH{=}CH_2} + \mathrm{ZnBr_2}\]
1,2-Dibromopropane gives propene. The three elimination routes are collected in FC 15.4 so that you can see at a glance what leaves the molecule in each case.
Flowchart
FC 15.4 – Three elimination routes to an alkene: (1) alkyl halide R–CHX–CH₂R + alcoholic KOH, boil → removes H and X → alkene (Saytzeff product major); (2) alcohol R–CH(OH)–CH₂R + conc. H₂SO₄, heat → removes H and OH as H₂O → alkene; (3) vicinal dihalide R–CHX–CH₂X + Zn → removes X and X as ZnX₂ → alkene; all three arrows converge on “C=C formed, sp³ → sp²”
(II) Addition reactions : By partial reduction of alkynes (or controlled hydrogenation of alkynes)
The \(\mathrm{C{\equiv}C}\) triple bond of alkynes can be partially reduced to a \(\mathrm{C{=}C}\) double bond with a calculated quantity of dihydrogen in the presence of Lindlar’s catalyst (palladised charcoal deactivated partially with quinoline or a sulfur compound). The cis isomer is obtained by this method. Ordinary Pt or Pd would drive the reduction all the way to the alkane; deliberately poisoning the catalyst makes it stop at the alkene stage.
\[\mathrm{R\!-\!C{\equiv}C\!-\!R} + \mathrm{H\!-\!H} \xrightarrow[\text{quinoline}]{\ \mathrm{Pd\!-\!C}\ } \textit{cis}\text{-alkene}\]
Trans alkene is obtained from an alkyne on reduction with sodium in liquid ammonia.
\[\mathrm{R\!-\!C{\equiv}C\!-\!R} \xrightarrow[\text{liquid } \mathrm{NH_3}]{\ \mathrm{Na}\ } \textit{trans}\text{-alkene}\]
Both products are shown in Fig. 15.14. This pair of reactions is valuable precisely because it lets a chemist choose the geometry of the double bond, which no elimination reaction can do reliably.
Diagram
Fig. 15.14 – From one alkyne, two geometries: R–C≡C–R at the centre; upper arrow labelled “H₂, Pd-C/quinoline (Lindlar)” leading to cis-alkene with both R groups on the same side; lower arrow labelled “Na/liquid NH₃” leading to trans-alkene with R groups on opposite sides
15.2.3 Physical Properties of Alkenes
Table 15.3 : Melting point and boiling point of alkenes
| Formula | Name | Molecular mass (u) | b.p. (K) | m.p. (K) |
|---|---|---|---|---|
| \(\mathrm{CH_2{=}CH_2}\) | Ethene | 28 | 171 | 104 |
| \(\mathrm{CH_2{=}CHCH_3}\) | Propene | 42 | 225 | 88 |
| \(\mathrm{CH_2{=}CHCH_2CH_3}\) | But-1-ene | 56 | 267 | – |
| \(\mathrm{CH_2{=}CH(CH_2)_2CH_3}\) | Pent-1-ene | 70 | 303 | – |
| \(\mathrm{CH_2{=}CH(CH_2)_3CH_3}\) | Hex-1-ene | 84 | 337 | 135 |
| cis \(\mathrm{CH_3CH{=}CHCH_3}\) | cis-But-2-ene | 56 | 277 | 134 |
| trans \(\mathrm{CH_3CH{=}CHCH_3}\) | trans-But-2-ene | 56 | 274 | 167 |
| \(\mathrm{CH_2{=}C(CH_3)_2}\) | Isobutylene | 56 | 266 | 132 |
Alkenes are nonpolar or weakly polar compounds. The \(\pi\) bond makes the electron cloud a little more easily distorted than in an alkane, but not enough to give a real permanent dipole in a hydrocarbon. They are insoluble in water, and soluble in non-polar solvents like benzene, ether and chloroform, for the same “like dissolves like” reason met with the alkanes. They are less dense than water, so an alkene layer floats.
From Table 15.3, it is clear that the boiling point of alkene rises with increasing number of carbons. Branched alkenes have lower boiling point than the corresponding straight chain alkane. The boiling point of an alkene is very nearly the same as that of the alkane with the same carbon skeleton — a useful check that the weak van der Waals forces, not the double bond, are what boiling has to overcome.
Diagram
Fig. 15.15 – (a) Line graph of boiling point (K) against number of carbon atoms for the terminal alkenes ethene to hex-1-ene from Table 15.3; (b) bar chart comparing b.p. and m.p. of the three C₄H₈ isomers — cis-but-2-ene, trans-but-2-ene, isobutylene
15.2.4 Chemical Properties of Alkenes
Alkenes are more reactive than alkanes due to the presence of pi (\(\pi\)) electrons. The \(\pi\) electrons lie above and below the plane of the molecule, are loosely held, and are therefore easily offered to any electron-deficient species. Such a species — one that seeks electrons — is called an electrophile, and this is why alkenes undergo electrophilic addition reactions: the double bond attacks the electrophile, not the other way round. The different reactions of alkenes are given below.
1. Addition of dihydrogen / hydrogenation — see the preparation of alkanes from unsaturated hydrocarbons in section 15.1.3, where an alkene and dihydrogen over Pt, Pd or Ni give the corresponding alkane.
Alkenes are converted into the corresponding vicinal dihalides by addition of halogens (\(\mathrm{X_2} = \mathrm{Cl_2}\) or \(\mathrm{Br_2}\)). Note that this is the exact reverse of the dehalogenation used to make alkenes.
\[\mathrm{{-}C{=}C{-}} + \mathrm{X_2} \xrightarrow[\text{room temp.}]{\ \mathrm{CCl_4}\ } \mathrm{{-}C(X)\!-\!C(X){-}}\]
An alkene gives a vicinal dihalide. Iodine generally fails to react.
\[\mathrm{H_2C{=}CH_2} + \mathrm{Cl_2} \xrightarrow[\text{room temp.}]{\ \mathrm{CCl_4}\ } \mathrm{CH_2Cl\!-\!CH_2Cl}\]
Ethene gives 1,2-dichloroethane.
\[\mathrm{H_3C\!-\!CH{=}CH_2} + \mathrm{Br_2} \xrightarrow[\text{room temp.}]{\ \mathrm{CCl_4}\ } \mathrm{H_3C\!-\!CHBr\!-\!CH_2Br}\]
Propene and bromine, which is red brown, give 1,2-dibromopropane, which is colourless. That colour change is the whole basis of a laboratory test: addition of bromine is a useful test for detection of \(\mathrm{C{=}C}\) in unknown compounds. The red brown colour of bromine rapidly disappears in a carbon tetrachloride solution if a \(\mathrm{C{=}C}\) double bond is present in the compound.
Addition reaction of HBr to symmetrical alkenes : The addition of HBr to symmetrical alkenes yields only one product, because both carbon atoms of the double bond are identical, so it makes no difference which one receives the hydrogen.
\[\mathrm{H_2C{=}CH_2} + \mathrm{H\!-\!Br} \rightarrow \mathrm{H_3C\!-\!CH_2\!-\!Br}\]
\[\mathrm{H_2C{=}CH_2} + \mathrm{H\!-\!Br} \rightarrow \mathrm{Br\!-\!CH_2\!-\!CH_3}\]
Ethene gives bromoethane by either route — the two products are the same compound written from the two ends.
Addition reaction of HBr to unsymmetrical alkenes : Addition of HBr to unsymmetrical alkenes yields two isomeric products, along path A and path B.
\[\mathrm{H_3C\!-\!CH{=}CH_2} + \mathrm{H\!-\!Br} \rightarrow \underbrace{\mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!Br}}_{\text{1-Bromopropane (minor)}} + \underbrace{\mathrm{H_3C\!-\!CHBr\!-\!CH_3}}_{\text{2-Bromopropane (major)}}\]
Experimentally it has been found that 2-bromopropane is the major product. The Russian chemist Markovnikov studied hydrohalogenation of a number of unsymmetrical alkenes and formulated a general rule (1869) as follows.
In propene, C-2 carries one hydrogen and C-3 carries two, so bromine goes to C-2 and hydrogen to C-3, giving 2-bromopropane.
Anti-Markovnikov addition or peroxide effect or Kharasch–Mayo effect : In 1933, M. S. Kharasch and F. R. Mayo discovered that the addition of HBr to an unsymmetrical alkene in the presence of an organic peroxide (R–O–O–R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.
\[\mathrm{H_3C\!-\!CH{=}CH_2} + \mathrm{H\!-\!Br} \xrightarrow[\text{Benzoyl peroxide}]{\ (\mathrm{C_6H_5CO})_2\mathrm{O_2}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!Br}\]
Propene gives 1-bromopropane.
\[\mathrm{H_3C\!-\!C(CH_3){=}CH_2} + \mathrm{H\!-\!Br} \xrightarrow[\text{Benzoyl peroxide}]{\ (\mathrm{C_6H_5CO})_2\mathrm{O_2}\ } \mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_2\!-\!Br}\]
2-Methylprop-1-ene gives 1-bromo-2-methylpropane.
Two points must be remembered. The orientation of addition of HBr to an unsymmetrical alkene is determined by the presence or absence of peroxide. And the peroxide has no effect upon the addition of HCl and HI — the effect is specific to HBr.
The \(\mathrm{C{=}C}\) double bond gets cleaved on reaction with ozone followed by reduction. This overall process of formation of ozonide by reaction of ozone with an alkene in the first step, and then decomposing it to the carbonyl compounds by reduction in the second step, is called ozonolysis.
\[\mathrm{{-}\overset{1}{C}{=}\overset{2}{C}{-}} \xrightarrow{\ \mathrm{O_3}\ /\ \mathrm{CCl_4}\ } \text{alkene ozonide} \xrightarrow[\text{Zn dust}]{\ \mathrm{H_2O}\ } \overset{1}{\mathrm{C}}{=}\mathrm{O} + \mathrm{O}{=}\overset{2}{\mathrm{C}} + \mathrm{ZnO}\]
Ozone gas is passed into a solution of the alkene in an inert solvent like carbon tetrachloride, and an unstable alkene ozonide is obtained. This is subsequently treated with water in the presence of the reducing agent zinc dust to form carbonyl compounds, namely aldehydes and/or ketones. The ozonide structures are drawn in Fig. 15.16.
\[\mathrm{H_2C{=}CH_2} \xrightarrow{\ \mathrm{O_3}\ /\ \mathrm{CCl_4}\ } \text{Ethene ozonide} \xrightarrow[\text{Zn dust}]{\ \mathrm{H_2O}\ } \mathrm{HCHO} + \mathrm{HCHO} + \mathrm{ZnO}\]
Ethene gives formaldehyde and formaldehyde — two identical fragments, since the alkene itself was symmetrical.
\[\mathrm{H_3C\!-\!\overset{1}{C}H{=}\overset{2}{C}H_2} \xrightarrow{\ \mathrm{O_3}\ /\ \mathrm{CCl_4}\ } \text{Propene ozonide} \xrightarrow[\text{Zn dust}]{\ \mathrm{H_2O}\ } \mathrm{CH_3CHO} + \mathrm{HCHO} + \mathrm{ZnO}\]
Propene gives acetaldehyde and formaldehyde.
Diagram
Fig. 15.16 – Ozonide structures: (a) ethene ozonide, the five-membered ring with the –O–O– bridge and two CH₂ groups, formed from CH₂=CH₂ + O₃; (b) propene ozonide from CH₃–CH=CH₂ with carbons numbered 1 and 2; arrows to the carbonyl products showing which carbon becomes which product
Flowchart
FC 15.5 – Ozonolysis, and reading it backwards: alkene → step 1: O₃ in CCl₄ → unstable alkene ozonide → step 2: H₂O + Zn dust → carbonyl compounds (aldehyde and/or ketone) + ZnO; reverse branch for structure determination: given the carbonyl products → join the two carbonyl carbons by C=C → discard both oxygen atoms → original alkene
Four conclusions follow, and every one of them is examinable. In the cleavage products a carbonyl group (\(\mathrm{C{=}O}\)) is formed at each of the original doubly bonded carbon atoms. Knowing the number and arrangement of carbon atoms in these aldehydes and ketones produced, we can identify the structure of the original alkene. The role of zinc dust is to prevent the formation of hydrogen peroxide, which would oxidize the aldehydes to the corresponding acids. And this reaction is used to locate the position and determine the number of double bonds in alkenes.
Alkenes with diborane in tetrahydrofuran (THF) solvent undergo hydroboration to form trialkylborane, which on oxidation with alkaline peroxide forms a primary alcohol. The overall reaction gives the Anti-Markovnikov’s product from unsymmetrical alkenes — which makes it the natural partner to the sulphuric acid route, since between them they give either alcohol from the same alkene.
\[6\,(\mathrm{{-}C{=}C{-}}) + (\mathrm{BH_3})_2 \xrightarrow{\ \mathrm{THF}\ } 2\,\big(\mathrm{{-}C(H)\!-\!C(H){-}}\big)_3\mathrm{B}\]
An alkene and diborane give a trialkylborane.
\[\big(\mathrm{{-}C(H)\!-\!C(H){-}}\big)_3\mathrm{B} \xrightarrow[\mathrm{OH^{-}}]{\ \mathrm{HO\!-\!OH}\ } \mathrm{{-}C(H)\!-\!C(OH){-}} + \mathrm{B(OH)_3}\]
The trialkylborane gives the alcohol. Worked through with ethene:
\[6\,\mathrm{H_2C{=}CH_2} + (\mathrm{BH_3})_2 \xrightarrow[\text{solvent}]{\ \mathrm{THF}\ } 2\,(\mathrm{CH_3\!-\!CH_2\!-})_3\mathrm{B}\]
Ethene and diborane give triethylborane.
\[(\mathrm{CH_3\!-\!CH_2\!-})_3\mathrm{B} + 3\,\mathrm{HO\!-\!OH} \xrightarrow{\ \mathrm{NaOH}\ } 3\,\mathrm{CH_3\!-\!CH_2\!-\!OH} + \mathrm{B(OH)_3}\]
Triethylborane gives ethanol. The same sequence with propene:
\[6\,\mathrm{H_3C\!-\!CH{=}CH_2} + (\mathrm{BH_3})_2 \xrightarrow[\text{solvent}]{\ \mathrm{THF}\ } 2\,(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-})_3\mathrm{B}\]
Propene and diborane give tri-n-propylborane.
\[(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-})_3\mathrm{B} + 3\,\mathrm{HO\!-\!OH} \xrightarrow{\ \mathrm{NaOH}\ } 3\,\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!OH} + \mathrm{B(OH)_3}\]
Tri-n-propylborane gives propan-1-ol. Compare this with the sulphuric acid route: propene gave propan-2-ol there, but propan-1-ol here. The two stages are set out in FC 15.6.
Flowchart
FC 15.6 – Hydroboration–oxidation: alkene + (BH₃)₂ in THF → trialkylborane (hydroboration stage) → HO–OH / NaOH → primary alcohol + B(OH)₃ (oxidation stage); side note branch: OH ends up on the carbon with more hydrogen atoms → anti-Markovnikov product
The process in which a large number of small molecules join together and form very large molecules with repeating units is called polymerization. The compound having very large molecules made of a large number of repeating small units is called a polymer, and the simple compound forming the repeating units in the polymer is called a monomer.
For example, ethene at high temperature and under high pressure interacts with oxygen, and undergoes polymerization giving the high molecular weight polymer called polyethene.
\[n\,\mathrm{H_2C{=}CH_2} \xrightarrow[\text{polymerization}]{\ \mathrm{O_2},\ \text{heat, pressure}\ } \mathrm{({-}CH_2\!-\!CH_2{-})}_n\]
Ethene, the monomer, gives polyethene, the polymer. Here n represents the number of repeating units and is a large number. The double bond is what makes this possible: each ethene molecule opens its \(\pi\) bond and uses the freed electrons to bond to the next molecule, so a chain grows without anything being lost. The repeating unit is shown in Fig. 15.17.
Diagram
Fig. 15.17 – Polymerization of ethene: n molecules of CH₂=CH₂ on the left, arrow labelled O₂, heat, pressure; on the right the open chain –CH₂–CH₂–CH₂–CH₂– with wavy ends, and beside it the same shown as a bracketed repeating unit (–CH₂–CH₂–)ₙ labelled polyethene
Alkenes react with cold and dilute alkaline potassium permanganate to form glycols — compounds with two \(\mathrm{{-}OH}\) groups on adjacent carbon atoms.
\[\mathrm{H_2C{=}CH_2} + \mathrm{H\!-\!OH} + (\mathrm{O}) \xrightarrow{\ \text{alkaline } \mathrm{KMnO_4}\ } \mathrm{CH_2(OH)\!-\!CH_2(OH)}\]
Ethene gives ethane-1,2-diol.
\[\mathrm{H_3C\!-\!CH{=}CH_2} + \mathrm{H\!-\!OH} + (\mathrm{O}) \xrightarrow{\ \text{alkaline } \mathrm{KMnO_4}\ } \mathrm{H_3C\!-\!CH(OH)\!-\!CH_2(OH)}\]
Propene gives propane-1,2-diol.
Hydroxylation of alkenes is the most important method for the synthesis of 1,2-diols (glycols). During this reaction the purple colour of \(\mathrm{KMnO_4}\) disappears. Hence such a reaction serves as a qualitative test for detecting the presence of a double bond in the compound under test. This is known as Baeyer’s test.
Flowchart
FC 15.7 – Testing an unknown hydrocarbon for unsaturation: sample + Br₂ in CCl₄ → if red brown colour disappears → C=C or C≡C present; if colour persists → saturated (alkane); confirm with sample + cold dilute alkaline KMnO₄ → if purple colour disappears → unsaturated (Baeyer’s test positive); if purple colour persists → saturated
Alkenes on oxidation with acidic \(\mathrm{KMnO_4}\) or acidic potassium dichromate form ketones and/or acids. This is a far harsher treatment than Baeyer’s test: instead of adding two \(\mathrm{{-}OH}\) groups, the reagent cuts the molecule at the double bond and oxidises the fragments as far as it can.
\[\mathrm{H_3C\!-\!CH{=}CH_2} \xrightarrow[(\mathrm{O})]{\ \mathrm{KMnO_4},\ \mathrm{H^{\oplus}}\ } \mathrm{H_3C\!-\!COOH} + \mathrm{CO_2} + \mathrm{H_2O}\]
Prop-1-ene gives ethanoic acid, along with carbon dioxide and water.
\[\mathrm{H_3C\!-\!C(CH_3){=}CH\!-\!CH_3} \xrightarrow[(\mathrm{O})]{\ \mathrm{KMnO_4},\ \mathrm{H^{\oplus}}\ } \mathrm{H_3C\!-\!CO\!-\!CH_3} + \mathrm{H_3C\!-\!COOH}\]
2-Methylbut-2-ene gives propan-2-one and ethanoic acid. The rule behind the products is simple: a doubly bonded carbon carrying a hydrogen atom is oxidised further, to an acid or even to \(\mathrm{CO_2}\), while a doubly bonded carbon carrying two alkyl groups can only stop at a ketone.
15.2.5 Uses of Alkenes
- Alkenes are used as starting materials for preparation of alkyl halides, alcohols, aldehydes, ketones, acids etc. Every one of those conversions is one of the addition or oxidation reactions of section 15.2.4.
- Ethene and propene are used to manufacture polythene and polypropylene, which find use in bags, toys, bottles, etc.
- Ethene is used for artificial ripening of fruits, such as mangoes.
15.3 Alkynes
Alkynes are aliphatic unsaturated hydrocarbons containing at least one \(\mathrm{C{\equiv}C}\). The number of hydrogen atoms is still less in alkynes as compared to alkenes. Their general formula is \(\mathrm{C_nH_{2n-2}}\).
Follow the hydrogen count across the three families for the same two carbon atoms: ethane is \(\mathrm{C_2H_6}\), ethene is \(\mathrm{C_2H_4}\) and ethyne is \(\mathrm{C_2H_2}\). Each new bond between the two carbon atoms costs two hydrogen atoms. The triple bond is made of one sigma bond and two pi (\(\pi\)) bonds, and because the two carbon atoms are \(sp\) hybridised, the molecule is linear — every atom of ethyne lies on one straight line, as shown in Fig. 15.18. Table 15.4 shows names and formulae of lower alkynes.
Diagram
Fig. 15.18 – Bonding in ethyne: linear H–C≡C–H skeleton with bond angle 180° marked; the sigma framework drawn from sp hybrid orbitals, and the two π bonds drawn from two mutually perpendicular pairs of unhybridised p orbitals, shown as two overlapping clouds at right angles to each other
Table 15.4 : Lower alkynes
| n | Molecular formula | Structural formula and Common name | IUPAC name |
|---|---|---|---|
| 2 | \(\mathrm{C_2H_2}\) | \(\mathrm{H\!-\!C{\equiv}C\!-\!H}\), Acetylene | Ethyne |
| 3 | \(\mathrm{C_3H_4}\) | \(\mathrm{CH_3\!-\!C{\equiv}CH}\), Methyl acetylene | Propyne |
| 4 | \(\mathrm{C_4H_6}\) | \(\mathrm{CH_3\!-\!CH_2\!-\!C{\equiv}CH}\), Ethyl acetylene | But-1-yne |
| 4 | \(\mathrm{C_4H_6}\) | \(\mathrm{CH_3\!-\!C{\equiv}C\!-\!CH_3}\), Dimethyl acetylene | But-2-yne |
Notice from the table that the simplest alkyne is ethyne, which is known as acetylene, and that the carbon–carbon triple bond is a functional group — it is the site at which every reaction of the family takes place. As with the alkenes, more than one triple bond is possible: the aliphatic unsaturated hydrocarbons containing two or three carbon–carbon triple bonds are called alkadiynes and alkatriynes respectively.
\[\mathrm{HO\!-\!(CH_2)_3\!-\!C{\equiv}C\!-\!C{\equiv}C\!-\!(C_{10}H_{14})\!-\!OH}\]
Every part of the plant is dangerous, and it is easily mistaken for edible roots — a reminder that “natural” and “safe” are not the same statement in chemistry.
15.3.1 Isomerism in Alkynes
Alkynes show position isomerism, which is a type of structural isomerism. For example, 1-butyne and 2-butyne have the same molecular formula \(\mathrm{C_4H_6}\) and the same unbranched chain of four carbon atoms, and differ only in where the triple bond sits. 1-Alkynes are also called terminal alkynes, because the triple bond reaches the end of the chain and carries a hydrogen atom directly on a triply bonded carbon. That single hydrogen turns out to matter a great deal, as section 15.3.4 will show.
Note that a triple bond rules out geometrical isomerism. A triply bonded carbon has only one other group attached to it, and the whole unit is linear, so there is no “same side” or “opposite side” to speak of.
15.3.2 Preparation of Alkynes
a. Industrial Sources
i. Ethyne is industrially prepared by controlled, high temperature partial oxidation of methane. “Partial” is the key word — with enough oxygen methane would simply burn to carbon dioxide, so the oxygen supply is deliberately restricted.
\[6\,\mathrm{CH_4} + 2\,\mathrm{O_2} \xrightarrow{\ 1773\ \text{K}\ } 2\,\mathrm{H\!-\!C{\equiv}C\!-\!H} + 2\,\mathrm{CO_2} + 10\,\mathrm{H_2}\]
Methane gives ethyne.
ii. From calcium carbide : Industrially the alkyne ethyne is prepared by reaction of calcium carbide with water.
\[\mathrm{CaC_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{C_2H_2} + \mathrm{Ca(OH)_2}\]
b. Methods of Preparation of Alkynes
Removal of H and X from adjacent carbon atoms is called dehydrohalogenation. You met this reaction in section 15.2.2, where it stopped at the alkene. To reach an alkyne, it must be carried out twice — once to make the double bond, and again to make the triple bond.
Vicinal dihalides react with an alcoholic solution of potassium hydroxide to form an alkenyl halide, which on further treating with sodamide forms the alkyne.
\[\mathrm{{-}C(H)(X)\!-\!C(H)(X){-}} + \mathrm{KOH}\ (\text{alcoholic}) \xrightarrow[-\mathrm{H_2O}]{\ -\mathrm{KX}\ } \mathrm{{-}C(H){=}C(X){-}}\]
\[\mathrm{{-}C(H){=}C(X){-}} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Na^{+}NH_2^{-}},\ -\mathrm{NaX}\ } \mathrm{{-}C{\equiv}C{-}}\]
A vicinal dihalide gives an alkenyl halide, which gives the alkyne. Worked through with the two-carbon case:
\[\mathrm{CH_2Br\!-\!CH_2Br} + \mathrm{KOH}\ (\text{alcoholic}) \xrightarrow[-\mathrm{H_2O}]{\ -\mathrm{KBr}\ } \mathrm{CH_2{=}CHBr}\]
1,2-Dibromoethane gives bromoethene.
\[\mathrm{CH_2{=}CHBr} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Na^{+}NH_2^{-}},\ -\mathrm{NaBr}\ } \mathrm{H\!-\!C{\equiv}C\!-\!H}\]
Bromoethene gives ethyne. The three-carbon case runs in exactly the same way:
\[\mathrm{CH_3\!-\!CHCl\!-\!CH_2Cl} + \mathrm{KOH}\ (\text{alcoholic}) \xrightarrow[-\mathrm{H_2O}]{\ -\mathrm{KCl}\ } \mathrm{CH_3\!-\!CH{=}CHCl}\]
1,2-Dichloropropane gives 1-chloropropene.
\[\mathrm{CH_3\!-\!CH{=}CHCl} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Na^{+}NH_2^{-}},\ -\mathrm{NaCl}\ } \mathrm{H_3C\!-\!C{\equiv}C\!-\!H}\]
1-Chloropropene gives propyne. The full two-stage route is summarised in FC 15.8.
Flowchart
FC 15.8 – Vicinal dihalide to alkyne in two stages: R–CHX–CH₂X → stage 1: alcoholic KOH, boil (−KX, −H₂O) → alkenyl halide R–CH=CHX → stage 2: sodamide Na⁺NH₂⁻ (−NaX, −NH₃) → alkyne R–C≡CH; note on stage 2: a stronger base than alcoholic KOH is required to pull HX from a doubly bonded carbon
Terminal alkynes are the compounds in which a hydrogen atom is directly attached to a triply bonded carbon atom.
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} \qquad \mathrm{R\!-\!C{\equiv}C\!-\!H}\]
Such a hydrogen atom shows appreciable acidity and can be given away as a proton on reaction with a sufficiently strong base. This method turns that acidity into a synthetic tool: strip the hydrogen off a small alkyne, and the carbon left behind will attack an alkyl halide and build a longer chain.
In this method a smaller terminal alkyne first reacts with a very strong base like lithium amide to form a metal acetylide (lithium amide is easier to handle than sodamide). Higher alkynes are obtained by reacting metal acetylides (alkyn-1-yl lithium) with primary alkyl halides.
\[\mathrm{R\!-\!C{\equiv}C\!-\!H} + \mathrm{Li^{+}NH_2^{-}} \rightarrow \mathrm{R\!-\!C{\equiv}C\!-\!Li} + \mathrm{NH_3}\]
A terminal alkyne and lithium amide give alkyn-1-yl lithium.
\[\mathrm{R\!-\!C{\equiv}C\!-\!Li} + \mathrm{R^{1}\!-\!X} \rightarrow \mathrm{R\!-\!C{\equiv}C\!-\!R^{1}} + \mathrm{LiX}\]
Alkyn-1-yl lithium and a primary alkyl halide give the higher alkyne. Two examples:
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Li^{+}NH_2^{-}}\ } \mathrm{H\!-\!C{\equiv}C\!-\!Li} \xrightarrow{\ \mathrm{H_3C\!-\!CH_2\!-\!Br}\ } \mathrm{H\!-\!C{\equiv}C\!-\!CH_2\!-\!CH_3} + \mathrm{LiBr}\]
Ethyne gives ethynyllithium, which with bromoethane gives but-1-yne.
\[\mathrm{H_3C\!-\!C{\equiv}C\!-\!H} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Li^{+}NH_2^{-}}\ } \mathrm{H_3C\!-\!C{\equiv}C\!-\!Li} \xrightarrow{\ \mathrm{H_3C\!-\!CH_2\!-\!Br}\ } \mathrm{H_3C\!-\!C{\equiv}C\!-\!CH_2\!-\!CH_3} + \mathrm{LiBr}\]
Propyne gives prop-1-yn-1-yl lithium, which with bromoethane gives pent-2-yne. Count the carbon atoms in each case and you can see the logic of the method: the product chain is the alkyne’s chain plus the alkyl halide’s chain. FC 15.9 sets out how to plan such a synthesis.
Flowchart
FC 15.9 – Building a higher alkyne: target alkyne → decide the split, i.e. which part comes from the terminal alkyne and which from the primary alkyl halide → terminal alkyne R–C≡CH + Li⁺NH₂⁻ (−NH₃) → alkyn-1-yl lithium R–C≡C–Li → add primary alkyl halide R¹–X → higher alkyne R–C≡C–R¹ + LiX
15.3.3 Physical Properties of Alkynes
The physical properties of alkynes are similar to those of alkanes and alkenes. They are less dense than water. They are insoluble in water and quite soluble in less polar organic solvents like ether, benzene and carbon tetrachloride. The melting points and boiling points of alkynes increase with an increase in molecular mass (Table 15.5).
Table 15.5 : Melting point and Boiling point of alkynes
| Formula | Name | Molecular mass/u | Physical State | b.p. (K) | m.p. (K) |
|---|---|---|---|---|---|
| \(\mathrm{CH{\equiv}CH}\) | Ethyne | 26 | Gas | 198 | 191 |
| \(\mathrm{CH{\equiv}CCH_3}\) | Propyne | 40 | Gas | 250 | 171.5 |
| \(\mathrm{CH{\equiv}CCH_2CH_3}\) | But-1-yne | 54 | Gas | 282 | 151 |
| \(\mathrm{CH{\equiv}C(CH_2)_2CH_3}\) | Pent-1-yne | 68 | Liquid | 313 | 175 |
| \(\mathrm{CH{\equiv}C(CH_2)_3CH_3}\) | Hex-1-yne | 82 | Liquid | 345 | 149 |
Diagram
Fig. 15.20 – (a) Line graph of boiling point (K) against molecular mass for ethyne to hex-1-yne from Table 15.5, with the gas/liquid boundary marked by a dashed line; (b) the same five alkynes plotted for melting point, showing the irregular pattern
15.3.4 Chemical Properties of Alkynes
1. Acidity of Alkynes
The hydrogen bonded to a \(\mathrm{C{\equiv}C}\) triple bond has acidic character. Lithium amide (\(\mathrm{LiNH_2}\)) is a very strong base and it reacts with terminal alkynes to form lithium acetylides, indicating the acidic nature of terminal alkynes.
Why is it so? In terminal alkynes, the hydrogen atom is directly attached to an \(sp\) hybridized carbon atom. In an \(sp\) hybrid orbital, the percentage of s-character is 50%. An electron in an s-orbital is very close to the nucleus and is held tightly. The \(sp\) hybrid carbon atom in terminal alkynes is therefore more electronegative than the \(sp^2\) carbon in ethene or the \(sp^3\) carbon in ethane. Due to the high electronegative character of carbon in terminal alkynes, the hydrogen atom can be given away as a proton (\(\mathrm{H^{+}}\)) to a very strong base.
The chain of reasoning is worth reading slowly, because it links three ideas you already know. More s-character means the shared electrons sit closer to the carbon nucleus; electrons held closer to the nucleus mean a more electronegative carbon; a more electronegative carbon holds the C–H bonding pair more firmly and lets the hydrogen leave as a bare proton, keeping the electron pair for itself. The comparison is drawn in Fig. 15.19.
Diagram
Fig. 15.19 – s-character and acidity: three panels for ethane (sp³ carbon, 25% s-character), ethene (sp² carbon, 33% s-character) and ethyne (sp carbon, 50% s-character); in each the C–H bond is drawn with the bonding pair shown progressively closer to the carbon, and an arrow at the side labelled “increasing electronegativity of carbon, increasing acidity of the H”
Examples
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Li^{+}NH_2^{-}}\ } \mathrm{H\!-\!C{\equiv}C\!-\!Li}\]
Ethyne gives monolithium ethynide.
\[\mathrm{H\!-\!C{\equiv}C\!-\!Li} + \mathrm{Li^{+}NH_2^{-}} \rightarrow \mathrm{Li\!-\!C{\equiv}C\!-\!Li} + \mathrm{NH_3}\]
Monolithium ethynide gives dilithium ethynide — ethyne has two acidic hydrogen atoms, one at each end, and both can be removed in turn.
\[\mathrm{H_3C\!-\!C{\equiv}C\!-\!H} \xrightarrow[-\mathrm{NH_3}]{\ \mathrm{Li^{+}NH_2^{-}}\ } \mathrm{H_3C\!-\!C{\equiv}C\!-\!Li}\]
Prop-1-yne gives lithium propynide. Since propyne is a terminal alkyne with only one such hydrogen, only one lithium can be introduced.
The relative acidity of alkanes, alkenes and alkynes follows the order
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} > \mathrm{H_2C{=}CH_2} > \mathrm{H_3C\!-\!CH_3}\]
2. Addition of Dihydrogen
See the method of preparation of alkenes from unsaturated hydrocarbons in section 15.2.2, where a calculated quantity of dihydrogen over Lindlar’s catalyst stops at the cis-alkene, and sodium in liquid ammonia gives the trans-alkene; with Pt, Pd or Ni and excess dihydrogen the reduction goes all the way to the alkane, as in section 15.1.3.
3. Addition of Halogens
A triple bond has two \(\pi\) bonds, so it can accept halogen twice. The first molecule of halogen converts the alkyne to a dihaloalkene, and the second converts that to a tetrahaloalkane.
\[\mathrm{{-}C{\equiv}C{-}} + \mathrm{X\!-\!X} \xrightarrow{\ \mathrm{CCl_4}\ } \mathrm{{-}C(X){=}C(X){-}}\]
An alkyne, where \(\mathrm{X_2} = \mathrm{Cl_2}\) or \(\mathrm{Br_2}\), gives a 1,2-dihaloalkene.
\[\mathrm{{-}C(X){=}C(X){-}} + \mathrm{X\!-\!X} \rightarrow \mathrm{{-}C(X)_2\!-\!C(X)_2{-}}\]
The 1,2-dihaloalkene gives a 1,1,2,2-tetrahaloalkane. Ethyne reacts with bromine in an inert solvent such as carbon tetrachloride to give tetrabromoethane.
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} + \mathrm{Br\!-\!Br} \xrightarrow{\ \mathrm{CCl_4}\ } \mathrm{CHBr{=}CHBr}\]
Ethyne and bromine give 1,2-dibromoethene.
\[\mathrm{CHBr{=}CHBr} + \mathrm{Br\!-\!Br} \rightarrow \mathrm{CHBr_2\!-\!CHBr_2}\]
1,2-Dibromoethene gives 1,1,2,2-tetrabromoethane.
The red-brown colour of the solution of bromine in carbon tetrachloride disappears. This test can be used to detect the presence of unsaturation in a given compound. Note carefully what it can and cannot do: the bromine test says that a multiple bond is present, but it does not distinguish a double bond from a triple bond. For that you need the silver nitrate test, which responds only to a terminal alkyne.
4. Addition of Hydrogen Halides
\[\mathrm{{-}C{\equiv}C{-}} + \mathrm{H\!-\!X} \rightarrow \mathrm{{-}C(H){=}C(X){-}}\]
An alkyne and a hydrogen halide, where HX = HCl, HBr or HI, give a haloalkene.
\[\mathrm{{-}C(H){=}C(X){-}} + \mathrm{H\!-\!X} \rightarrow \mathrm{{-}CH_2\!-\!C(X)_2{-}}\]
The haloalkene gives a geminal dihalide. Hydrogen halides (HCl, HBr and HI) add to alkynes across the carbon–carbon triple bond in two steps to form geminal dihalides, in which two halogen atoms are attached to the same carbon atom. The addition of HX in both the steps takes place according to Markovnikov’s rule, which you met in section 15.2.4.
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} + \mathrm{H\!-\!Br} \rightarrow \mathrm{CH_2{=}CHBr}\]
Ethyne and hydrogen bromide give 1-bromoethene.
\[\mathrm{CH_2{=}CHBr} + \mathrm{H\!-\!Br} \rightarrow \mathrm{CH_3\!-\!CHBr_2}\]
1-Bromoethene gives 1,1-dibromoethane. Note where the second bromine lands: on the carbon that already carries a bromine and fewer hydrogen atoms, exactly as Markovnikov’s rule requires. That is why the product is geminal and not vicinal.
\[\mathrm{H_3C\!-\!C{\equiv}C\!-\!H} + \mathrm{H\!-\!Br} \rightarrow \mathrm{H_3C\!-\!C(Br){=}CH_2}\]
Propyne and hydrogen bromide give 2-bromopropene.
\[\mathrm{H_3C\!-\!C(Br){=}CH_2} + \mathrm{H\!-\!Br} \rightarrow \mathrm{H_3C\!-\!CBr_2\!-\!CH_3}\]
2-Bromopropene gives 2,2-dibromopropane. The order of reactivity of hydrogen halides is
\[\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\]
Flowchart
FC 15.10 – Two-step addition across a triple bond: alkyne –C≡C– → add one molecule of X₂ → 1,2-dihaloalkene → add second X₂ → 1,1,2,2-tetrahaloalkane; parallel branch: alkyne + one HX → haloalkene (Markovnikov) → + second HX (Markovnikov again) → geminal dihalide
5. Addition of Water
\[\mathrm{{-}C{\equiv}C{-}} + \mathrm{H\!-\!OH} \xrightarrow[1\%\ \mathrm{HgSO_4}]{\ 40\%\ \mathrm{H_2SO_4}\ } \big[\mathrm{{-}C(H){=}C(OH){-}}\big] \xrightarrow{\ \text{tautomerism}\ } \mathrm{{-}CH_2\!-\!C({=}O){-}}\]
An alkyne gives an unstable intermediate, which rearranges to a carbonyl compound. Alkynes react with water in the presence of 40% sulphuric acid and 1% mercuric sulphate to form aldehydes or ketones, i.e. carbonyl compounds.
The step marked “tautomerism” deserves a word, since the original product is never isolated. A compound with an \(\mathrm{{-}OH}\) group attached directly to a doubly bonded carbon is called an enol, and it is unstable. A hydrogen atom migrates from the oxygen to the neighbouring carbon while the double bond shifts from between the two carbon atoms to between carbon and oxygen. The result is the far more stable carbonyl compound. This rapid interconversion between the enol form and the carbonyl form is called tautomerism, and it is drawn in Fig. 15.21.
Diagram
Fig. 15.21 – Tautomerism of vinyl alcohol to ethanal: on the left CH₂=CH–OH with curly arrows showing the O–H hydrogen moving to the terminal carbon and the C=C π electrons shifting to become the C=O; on the right CH₃–CHO; double-headed arrow between the two labelled tautomerism
\[\mathrm{H\!-\!C{\equiv}C\!-\!H} + \mathrm{H\!-\!OH} \xrightarrow[1\%\ \mathrm{HgSO_4}]{\ 40\%\ \mathrm{H_2SO_4}\ } \big[\mathrm{CH_2{=}CH\!-\!OH}\big] \xrightarrow{\ \text{tautomerisation}\ } \mathrm{CH_3\!-\!CHO}\]
Ethyne gives vinyl alcohol, which tautomerises to ethanal.
\[\mathrm{H_3C\!-\!C{\equiv}C\!-\!H} + \mathrm{H\!-\!OH} \xrightarrow[1\%\ \mathrm{HgSO_4}]{\ 40\%\ \mathrm{H_2SO_4}\ } \big[\mathrm{H_3C\!-\!C(OH){=}CH_2}\big] \xrightarrow{\ \text{tautomerisation}\ } \mathrm{H_3C\!-\!CO\!-\!CH_3}\]
Propyne gives propanone. Notice the pattern: because the \(\mathrm{{-}OH}\) enters by Markovnikov’s rule at the more substituted carbon, ethyne is the only alkyne that gives an aldehyde, and every other alkyne gives a ketone.
Flowchart
FC 15.11 – Hydration of an alkyne: alkyne + H₂O with 40% H₂SO₄ and 1% HgSO₄ → unstable enol (–OH on a doubly bonded carbon) → tautomerism: H shifts from O to the adjacent C, C=C becomes C=O → stable carbonyl compound; branch: ethyne → ethanal (aldehyde); any other alkyne → ketone
15.3.5 Uses of Acetylene
- Ethyne (acetylene) is used in the preparation of ethanal (acetaldehyde), propanone (acetone) and ethanoic acid (acetic acid) — the first two by the hydration reaction you have just studied.
- It is used in the manufacture of polymers, synthetic rubber, synthetic fibre, plastic etc.
- For artificial ripening of fruits.
- In the oxy-acetylene flame — a mixture of oxygen and acetylene — for welding and cutting of metals.
15.4 Aromatic Hydrocarbons
Many organic compounds obtained from natural sources like resins, balsams, oil of wintergreen, etc. possessing pleasant fragrance (aroma = smell) are named as aromatic compounds. Aromatic hydrocarbons, also called arenes, contain only carbon and hydrogen. Benzene is the simplest aromatic compound.
The name came from smell, but chemists soon needed a definition based on behaviour rather than fragrance. Benzene and all compounds that have structures and chemical properties resembling benzene are classified as aromatic compounds. Examples are benzene, methylbenzene, bromobenzene, phenol and naphthalene.
Three points prevent a common misunderstanding. Aromatic compounds contain planar cyclic rings. Not all compounds that resemble benzene have a pleasant odour — anthracene, built from three fused benzene rings, is odourless. And many compounds having a pleasant odour do not resemble benzene at all: methyl acetate, ethyl acetate and trichloromethane (chloroform) all smell agreeable and none of them is aromatic in the chemical sense.
Table 15.6 : Difference between aromatic and aliphatic compounds
| Aromatic compounds | Aliphatic compounds |
|---|---|
| 1. Aromatic compounds contain higher percentage of carbon. | 1. Aliphatic compounds contain lower percentage of carbon. |
| 2. They burn with sooty flame. | 2. They burn with non-sooty flame. |
| 3. They are cyclic compounds with alternate single and double bonds. | 3. They are open chain compounds. |
| 4. They are not attacked by normal oxidizing and reducing agents. | 4. They are easily attacked by oxidizing and reducing agents. |
| 5. They do not undergo addition reactions easily. They do not decolourise dilute alkaline aqueous \(\mathrm{KMnO_4}\) and \(\mathrm{Br_2}\) in \(\mathrm{CCl_4}\), though double bonds appear in their structure. | 5. Unsaturated aliphatic compounds undergo addition reactions easily. They decolourise dilute aqueous alkaline \(\mathrm{KMnO_4}\) and \(\mathrm{Br_2}\) in \(\mathrm{CCl_4}\). |
| 6. They prefer substitution reactions. | 6. The saturated aliphatic compounds give substitution reactions. |
Row 5 of Table 15.6 is the one to dwell on. The two tests that identified a double bond in sections 15.2 and 15.3 both fail on benzene, even though its usual structure is drawn with three double bonds. Something about those bonds must be different, and the rest of this section is largely about what.
15.4.1 Benzene
The molecular formula of benzene is \(\mathrm{C_6H_6}\). Benzene is the parent compound of most of the aromatic compounds.
Coal-tar and petroleum are the two large-scale sources of benzene, and of aromatic compounds like toluene, phenol and naphthalene. It is a colourless liquid having a characteristic odour. Its boiling point is 353 K.
It was synthesized by Berthelot (1870) from acetylene — a striking result, since it showed that a simple two-carbon alkyne could be built up into the ring that had puzzled chemists for decades. You will meet the reaction itself in section 15.4.4.
15.4.2 Structure of Benzene
The structure of benzene was settled not by one experiment but by a chain of deductions, each ruling something out. FC 15.12 traces that chain, and the paragraphs below follow it step by step.
Flowchart
FC 15.12 – Deducing the structure of benzene: molecular formula C₆H₆ → high degree of unsaturation → test against alkene/alkyne behaviour (Table 15.7) → fails all three tests, so no open chain structure with double or triple bonds → only one monosubstituted product and hydrogenation to cyclohexane → cyclic ring of six carbons, one H each, three C=C → Kekulé structure (1865) → but only one o-disubstituted product exists → oscillating double bonds → still cannot explain stability and preference for substitution → resonance and delocalization of six π electrons
1. The molecular formula of benzene, \(\mathrm{C_6H_6}\), indicates a high degree of unsaturation. Compare it with the alkane formula: a saturated six-carbon compound would be \(\mathrm{C_6H_{14}}\), so benzene is short of eight hydrogen atoms.
2. Open chain structure NOT possible : An open chain or cyclic structure having double and triple bonds can be written for \(\mathrm{C_6H_6}\). But benzene does not behave like alkenes or alkynes (see Table 15.7). This indicates that benzene can not have the open chain structure.
Table 15.7 : Comparative reactivity of alkenes and benzene
| Reaction | Alkene | Benzene |
|---|---|---|
| With dil. alka. \(\mathrm{KMnO_4}\) | Decolourisation of purple colour of \(\mathrm{KMnO_4}\) | No decolourisation |
| With \(\mathrm{Br_2}\) in \(\mathrm{CCl_4}\) | Decolourisation of red brown colour of bromine | No decolourisation |
| With \(\mathrm{H_2O}\) in acidic medium | Addition of \(\mathrm{H_2O}\) molecule | No reaction |
3. Evidence of cyclic structure : Benzene yields only one, and no isomeric, monosubstituted bromobenzene (\(\mathrm{C_6H_5Br}\)) when treated with equimolar bromine in \(\mathrm{FeBr_3}\). This indicates that all the six hydrogen atoms in benzene are identical.
\[\mathrm{C_6H_6} + \mathrm{Br_2} \xrightarrow{\ \mathrm{FeBr_3}\ } \mathrm{C_6H_5Br} + \mathrm{HBr}\]
This is possible only if benzene has a cyclic structure of six carbons bound to one hydrogen atom each. In any open chain of six carbon atoms, the end carbons and the middle carbons would carry different numbers of hydrogen atoms, and replacing one hydrogen would give several different products.
Benzene on catalytic hydrogenation gives cyclohexane.
\[\mathrm{C_6H_6} + 3\,\mathrm{H_2} \xrightarrow[\Delta]{\ \mathrm{Ni}\ } \mathrm{C_6H_{12}}\]
This confirms the cyclic structure of benzene and three \(\mathrm{C{=}C}\) in it — three molecules of dihydrogen are absorbed, so three multiple bonds must be present, and the product is a ring.
4. Kekulé structure of benzene : August Kekulé in 1865 suggested the structure for benzene having a cyclic planar ring of six carbon atoms with alternate single and double bonds and a hydrogen atom attached to each carbon atom. It is usually written as a plain hexagon, as shown in Fig. 15.22.
The Kekulé structure indicates the possibility of two isomeric 1,2-dibromobenzenes. In one of the isomers, the bromine atoms would be attached to the doubly bonded carbon atoms, whereas in the other, they would be attached to single bonded carbons. However, benzene was found to form only one ortho-disubstituted benzene.
This problem was overcome by Kekulé by suggesting the concept of the oscillating nature of double bonds in benzene: the single and double bonds exchange places so rapidly that the two structures cannot be told apart.
Diagram
Fig. 15.22 – Kekulé structure of benzene: full structural formula with alternating single and double bonds and one H on each carbon, and beside it the shorthand hexagon; below, the two supposedly isomeric 1,2-dibromobenzenes with the C1–C2 bond drawn double in one and single in the other, marked with a query; at the bottom, two Kekulé hexagons linked by a ⇌ arrow illustrating oscillating double bonds
Even with this modification, the Kekulé structure of benzene failed to explain the unusual stability and the preference for substitution reactions rather than addition reactions, which was later explained by resonance.
5. Stability of benzene : Benzene is a hybrid of various resonance structures. The two structures, A and B, given by Kekulé are the main contributing structures. The resonance hybrid is represented by inserting a circle or a dotted circle inscribed in the hexagon as shown in (C). The circle represents six electrons delocalized over the six carbon atoms of the benzene ring. A double headed arrow between the resonance structures is used to represent the resonance phenomenon. All three are drawn in Fig. 15.23.
Diagram
Fig. 15.23 – Resonance in benzene: structure (A) and structure (B), the two Kekulé forms, joined by a double-headed arrow; structure (C) the resonance hybrid drawn as a hexagon with an inscribed dotted circle, and beside it the same with a solid circle; caption note that the circle stands for six delocalized electrons
A caution on notation: the double-headed arrow does not mean the molecule flips between A and B. There is only one benzene molecule, and its real electron distribution is the hybrid C. A and B are drawings we use because ordinary bond lines cannot show delocalization.
Stability of benzene : The actual molecule is more stable than any of its resonance structures. For benzene, the stability due to resonance is so high that the \(\pi\)-bonds of the molecule resist breaking. This explains the lack of reactivity of benzene towards addition. An addition reaction would have to localise the delocalized electrons into ordinary bonds, and that means giving up the extra stability — so benzene avoids addition and prefers substitution, which leaves the ring intact.
The Orbital Picture
The orbital overlap gives us a better picture of the structure of benzene. All six carbon atoms in benzene are \(sp^2\) hybridised. Two \(sp^2\) hybrid orbitals of each carbon overlap with those of its two neighbours and form carbon–carbon sigma (\(\sigma\)) bonds, and the remaining third \(sp^2\) hybrid orbital of each carbon overlaps with the s orbital of a hydrogen atom to form six C–H sigma bonds. Since \(sp^2\) hybridisation is trigonal planar, this framework is flat, as drawn in Fig. 15.24.
Diagram
Fig. 15.24 – Sigma framework of benzene: planar hexagonal ring of six sp² hybridised carbons with the twelve sigma bonds labelled σ (six C–C and six C–H), all bond angles marked 120°, all atoms shown lying in one plane
Three facts about benzene follow from this framework and are worth memorising. All carbon and hydrogen atoms lie in the same plane. The six sigma (\(\sigma\)) bonds of the ring lie in the same plane. And all bond angles are 120°.
The unhybrid p orbitals of the carbon atoms overlap laterally, forming \(\pi\) bonds. There are two possibilities of forming three \(\pi\) bonds by overlap of p orbitals — of \(\mathrm{C_1}\)-\(\mathrm{C_2}\), \(\mathrm{C_3}\)-\(\mathrm{C_4}\), \(\mathrm{C_5}\)-\(\mathrm{C_6}\), or of \(\mathrm{C_2}\)-\(\mathrm{C_3}\), \(\mathrm{C_4}\)-\(\mathrm{C_5}\), \(\mathrm{C_6}\)-\(\mathrm{C_1}\) — as shown in Fig. 15.5; both are equally probable. According to resonance theory (Chapter 5) these are the two resonance structures of benzene.
Diagram
Fig. 15.5 – Overlap of p orbitals in benzene: hexagonal ring with an unhybridised p orbital above and below each carbon, numbered 1 to 6; upper panel shows pairing C1–C2, C3–C4, C5–C6 with the corresponding Kekulé hexagon beside it; lower panel shows pairing C2–C3, C4–C5, C6–C1 with its Kekulé hexagon
According to molecular orbital (MO) theory (Chapter 5) the six p orbitals of the six carbons give rise to six molecular orbitals of benzene. The shape of the most stable MO is as shown in Fig. 15.6. Three of these \(\pi\) molecular orbitals lie above and the other three below those of the free carbon atom energies — which is exactly why the molecule is stabilised: the six electrons occupy the three orbitals of lower energy.
Diagram
Fig. 15.6 – Representative π molecular orbital in benzene: on the left the six p orbitals on the hexagonal ring, arrow to the right showing the merged doughnut-shaped π molecular orbital lying above and below the plane of the ring
The six electrons of the p orbitals cover all the six carbon atoms and are said to be delocalized. Delocalization of \(\pi\) electrons results in the stability of the benzene molecule.
6. Bond parameters of benzene : X-ray diffraction data indicate that all C–C bond lengths in benzene are equal (139 pm), which is intermediate between a C–C bond (154 pm) and a \(\mathrm{C{=}C}\) bond (133 pm). Thus the absence of a pure double bond in benzene accounts for its reluctance to addition reactions under normal conditions, which explains the unusual behaviour of benzene (refer to sec. 14.6.5).
Diagram
Fig. 15.25 – Bond length comparison on a single scale: C–C single bond 1.54 Å (154 pm), benzene C–C bond 1.39 Å (139 pm), C=C double bond 1.34 Å; the three drawn one below the other with the benzene value marked as intermediate, and the benzene hexagon shown with all six bonds identical
15.4.3 Aromatic Character (Huckel Rule)
Benzene undergoes substitution reactions rather than addition reactions. This property is common to all aromatic compounds and is referred to as aromaticity or aromatic character. The aromatic character of benzene is correlated to its structure.
Aromaticity is due to extensive cyclic delocalization of p electrons in planar ring structures. Since delocalization needs continuous overlap all the way round, the ring must satisfy strict geometric and electronic conditions — which is what the three rules below state.
The following three rules of aromaticity are useful in predicting whether a particular compound is aromatic or non-aromatic.
1. Aromatic compounds are cyclic and planar (all atoms in the ring are \(sp^2\) hybridized).
2. Each atom in the aromatic ring has a p-orbital. The p-orbitals must be parallel so that continuous overlap is possible around the ring.
3. Huckel Rule : The cyclic \(\pi\) molecular orbital formed by overlap of p-orbitals must contain \((4n + 2)\) p electrons, where n = integer 0, 1, 2, 3 … etc.
Note carefully how the rule is applied: count the \(\pi\) electrons, set the count equal to \(4n + 2\), and solve for n. If n comes out to be a whole number, the compound is aromatic; if it comes out as a fraction, it is not. All three rules must be satisfied together — the electron count alone is not enough, as the fourth example below shows. FC 15.13 puts the tests in order.
Flowchart
FC 15.13 – Testing a ring for aromaticity: is the ring cyclic and planar? → if no, non-aromatic → if yes, does every atom of the ring carry a parallel p-orbital? → if no (e.g. an sp³ carbon in the ring), non-aromatic → if yes, count the π electrons and solve 4n + 2 = number of π electrons → if n is an integer, aromatic → if n is a fraction, non-aromatic
Let us apply these rules to the following compounds, whose p-orbital pictures are drawn in Fig. 15.26.
Diagram
Fig. 15.26 – Applying the rules of aromaticity: (a) benzene hexagon with a p orbital on each carbon, continuous overlap above and below the ring; (b) naphthalene, two fused rings with p orbitals on every carbon; (c) pyridine, with p orbitals on five carbons and the nitrogen, and the nitrogen lone pair drawn in an sp² hybrid orbital lying in the ring plane, labelled “non bonding electron pair”; (d) cycloheptatriene, seven-membered ring with an arrow pointing to the one sp³ hybridized carbon that has no p orbital, breaking the overlap
1. Benzene : It is cyclic and planar. It has three double bonds and six \(\pi\) electrons. It has a p orbital on each carbon of the hexagonal ring. Hence a continuous overlap above and below the ring is possible. This compound is aromatic.
\[4n + 2 = \text{Number of }\pi\text{ electrons}\] \[4n + 2 = 6, \qquad \therefore 4n = 6 - 2 = 4\] \[n = 4/4 = 1\]
Here ‘n’ comes out to be an integer. Hence benzene is aromatic.
2. Naphthalene : It is cyclic and planar. It has 5 double bonds and 10 \(\pi\) electrons. It has a p orbital on each carbon atom of the ring. Hence a continuous overlap around the ring is possible. This is in accordance with the Huckel rule.
\[4n + 2 = \text{Number of }\pi\text{ electrons}\] \[4n + 2 = 10, \qquad \therefore 4n = 10 - 2 = 8\] \[n = 8/4 = 2\]
Here ‘n’ comes out to be an integer. Hence naphthalene is aromatic.
3. Pyridine : Pyridine has three double bonds and 6 \(\pi\) electrons. The six p orbitals containing six electrons form a delocalized \(\pi\) molecular orbital. The unused \(sp^2\) hybrid orbital of nitrogen containing two non-bonding electrons is as it is. Those two electrons lie in the plane of the ring, not in a p orbital, so they take no part in the delocalization and are not counted.
\[4n + 2 = \text{Number of p electrons}\] \[4n + 2 = 6, \qquad \therefore 4n = 6 - 2 = 4\] \[n = 4/4 = 1\]
Here ‘n’ comes out to be an integer. Hence pyridine is aromatic.
4. Cycloheptatriene : It is cyclic and planar. It has three double bonds and 6 \(\pi\) electrons. But one of the carbons is saturated (\(sp^3\) hybridized) and does not possess a p orbital. Hence a continuous overlap around the ring is not possible. Therefore, it is non-aromatic.
Cycloheptatriene is the important warning case. Its electron count of six would satisfy the Huckel rule perfectly, yet it is not aromatic, because rule 2 fails. Whenever an examination question gives you a ring, check the geometry and the p-orbitals first and only then count electrons.
15.4.4 Preparation of Aromatic Compounds
a. Industrial Source of Aromatic Compounds
Coal tar and petroleum are major sources of aromatic compounds.
b. Methods of Preparation of Benzene
1. From ethyne (by trimerization) : Alkynes when passed through a red hot iron tube at 873 K polymerize to form aromatic hydrocarbons. Ethyne when passed through a red hot iron tube at 873 K undergoes trimerization to form benzene. “Trimerization” means three molecules joining into one, and the arithmetic confirms it: three \(\mathrm{C_2H_2}\) units give \(\mathrm{C_6H_6}\) exactly, with nothing left over.
\[3\,\mathrm{C_2H_2} \xrightarrow[873\ \text{K}]{\ \text{red hot iron tube}\ } \mathrm{C_6H_6}\]
Diagram
Fig. 15.27 – Trimerization of ethyne: three molecules of H–C≡C–H arranged in a triangle with curved arrows showing them joining end to end, arrow labelled “red hot iron tube, 873 K”, giving the benzene hexagon and beside it the hexagon with an inscribed circle
2. From sodium benzoate (by decarboxylation) : When anhydrous sodium benzoate is heated with soda-lime it gives benzene. Decarboxylation means the loss of the carboxyl group as carbonate, and soda-lime — a mixture of sodium hydroxide and calcium oxide — supplies the alkali needed.
\[\mathrm{C_6H_5COONa} + \mathrm{NaOH} \xrightarrow[\Delta]{\ \mathrm{CaO}\ } \mathrm{C_6H_6} + \mathrm{Na_2CO_3}\]
Sodium benzoate gives benzene.
3. From phenol (by reduction) : When vapours of phenol are passed over heated zinc dust, it gives benzene. The zinc removes the oxygen of the \(\mathrm{{-}OH}\) group, being itself oxidised to zinc oxide.
\[\mathrm{C_6H_5OH} + \mathrm{Zn} \xrightarrow{\ \Delta\ } \mathrm{C_6H_6} + \mathrm{ZnO}\]
Phenol gives benzene.
15.4.5 Physical Properties of Benzene
- Benzene is a colourless liquid.
- Its boiling point is 353 K and melting point is 278.5 K.
- It is insoluble in water. It forms the upper layer when mixed with water, since it is less dense than water and non-polar.
- It is soluble in alcohol, ether, and chloroform.
- Benzene vapours are highly toxic, which on inhalation lead to unconsciousness.
15.4.6 Chemical Properties of Benzene
Aromatic compounds are characterised by electrophilic substitution reactions. However, they undergo addition and oxidation reactions under special conditions. Some reactions of benzene are discussed below.
I. Addition Reactions
Addition destroys the delocalization, so these reactions need forcing conditions — strong light, a catalyst and heat, or a powerful oxidant.
i. Addition of chlorine : Benzene when treated with chlorine in the presence of bright sunlight or UV light adds up three molecules of chlorine to give benzene hexachloride.
\[\mathrm{C_6H_6} + 3\,\mathrm{Cl_2} \xrightarrow{\ \text{UV light}\ } \mathrm{C_6H_6Cl_6}\]
Benzene gives benzene hexachloride (BHC), drawn in Fig. 15.28. The \(\gamma\)-isomer of benzene hexachloride is called gammexane or lindane, which is used as an insecticide.
Diagram
Fig. 15.28 – Addition of chlorine to benzene: benzene hexagon with the six H atoms shown, plus 3Cl₂ and the condition “UV light”, giving the non-planar saturated ring of benzene hexachloride with one H and one Cl on each carbon, labelled BHC
ii. Addition of hydrogen : When a mixture of benzene and hydrogen gas is passed over heated catalyst nickel at 453 K to 473 K, cyclohexane is formed.
\[\mathrm{C_6H_6} + 3\,\mathrm{H_2} \xrightarrow[453 - 473\ \text{K}]{\ \mathrm{Ni}\ } \mathrm{C_6H_{12}}\]
Benzene (\(\mathrm{C_6H_6}\)) gives cyclohexane (\(\mathrm{C_6H_{12}}\)). Compare the conditions with those in section 15.1.3, where an alkene was hydrogenated over Pt or Pd at room temperature. Benzene needs a much hotter catalyst, and that difference in conditions is a direct measure of the extra stability delocalization gives it.
iii. Addition of ozone : When benzene is treated with ozone in the presence of an inert solvent carbon tetrachloride, benzene triozonide is formed, which is then decomposed by zinc dust and water to give glyoxal.
\[\mathrm{C_6H_6} + 3\,\mathrm{O_3} \rightarrow \text{Benzene triozonide}\]
\[\text{Benzene triozonide} \xrightarrow{\ \mathrm{Zn}\ /\ \mathrm{H_2O}\ } 3\,\mathrm{OHC\!-\!CHO} + 3\,\mathrm{H_2O_2}\]
The product is ethanedial, or glyoxal. Three molecules of ozone are taken up and three identical two-carbon fragments are released, which is one more piece of evidence for a symmetrical six-carbon ring with three multiple bonds.
Diagram
Fig. 15.29 – Addition of ozone to benzene: benzene + 3O₃ giving the benzene triozonide with three –O–O–O– bridges drawn across the ring; second arrow labelled Zn/H₂O giving three molecules of OHC–CHO (ethanedial or glyoxal) and 3H₂O₂
II. Substitution Reactions
Benzene shows electrophilic substitution reactions, in which one or more hydrogen atoms of the benzene ring are replaced by electrophilic groups like \(\mathrm{{-}Cl}\), \(\mathrm{{-}Br}\), \(\mathrm{{-}NO_2}\), \(\mathrm{{-}SO_3H}\), \(\mathrm{{-}R}\) (alkyl group), \(\mathrm{{-}COR}\) (acyl group) etc. (see Chapter 14).
Every one of the five reactions below follows the same three-stage logic, set out in FC 15.14: a reagent and a catalyst first generate a positively charged electrophile, the electron-rich ring then supplies a pair of \(\pi\) electrons to it, and finally a hydrogen ion leaves so that the aromatic ring is restored. Because the last step gives the ring back, substitution costs the molecule nothing in delocalization energy — which is precisely why benzene prefers it.
Flowchart
FC 15.14 – General pattern of electrophilic substitution in benzene: reagent + catalyst → generation of the electrophile E⁺ → E⁺ attacked by the π electrons of the ring → intermediate carbocation with the ring temporarily disrupted → loss of H⁺ → substituted benzene with the aromatic ring restored + catalyst regenerated
i. Halogenation : In this reaction, a hydrogen atom of the benzene ring is replaced by a halogen atom.
Chlorination
\[\mathrm{C_6H_6} + \mathrm{Cl_2} \xrightarrow[\text{in dark}]{\ \mathrm{Fe}\ } \mathrm{C_6H_5Cl} + \mathrm{HCl}\]
Benzene gives chlorobenzene. Chlorine reacts with benzene in the dark in the presence of iron or ferric chloride or anhydrous aluminium chloride or red phosphorous as catalyst to give chlorobenzene. Note the contrast with the addition reaction above: bright light gives benzene hexachloride, darkness plus a catalyst gives chlorobenzene. The condition, not the reagent, decides the path.
Electrophile : \(\mathrm{Cl^{+}}\), chloronium ion.
Formation of the electrophile :
\[\mathrm{Cl\!-\!Cl} + \mathrm{FeCl_3} \rightarrow \mathrm{Cl^{+}} + [\mathrm{FeCl_4}]^{-}\]
Bromination of benzene is similar to chlorination :
\[\mathrm{C_6H_6} + \mathrm{Br_2} \xrightarrow{\ \mathrm{FeBr_3}\ } \mathrm{C_6H_5Br} + \mathrm{HBr}\]
Benzene gives bromobenzene.
Electrophile : \(\mathrm{Br^{+}}\).
Formation of the electrophile :
\[\mathrm{Br\!-\!Br} + \mathrm{FeBr_3} \rightarrow \mathrm{Br^{+}} + [\mathrm{FeBr_4}]^{-}\]
Iodination of benzene is not possible as it is a reversible process.
With excess of chlorine, benzene gives hexachlorobenzene.
\[\mathrm{C_6H_6} + 6\,\mathrm{Cl_2} \xrightarrow{\ \text{anhy. } \mathrm{AlCl_3}\ } \mathrm{C_6Cl_6} + 6\,\mathrm{HCl}\]
All six hydrogen atoms are replaced, one at a time, in six successive substitutions.
ii. Nitration :
\[\mathrm{C_6H_6} + \mathrm{HO\!-\!NO_2}\ (\text{conc.}) \xrightarrow[\text{Heat}]{\ \text{Conc. } \mathrm{H_2SO_4}\ } \mathrm{C_6H_5NO_2} + \mathrm{H_2O}\]
Benzene gives nitrobenzene. When benzene is heated with a mixture of concentrated nitric acid and concentrated sulfuric acid (nitrating mixture) at about 313 K to 333 K, it gives nitrobenzene.
Electrophile : \(\mathrm{NO_2^{+}}\), nitronium ion.
Formation of the electrophile :
\[\mathrm{HO\!-\!NO_2} + 2\,\mathrm{H_2SO_4} \rightleftharpoons 2\,\mathrm{HSO_4^{-}} + \mathrm{H_3O^{+}} + \mathrm{NO_2^{+}}\]
The role of the sulphuric acid is now clear: it is the stronger of the two acids, so it protonates nitric acid and forces it to give up water, leaving the nitronium ion that actually attacks the ring.
iii. Sulfonation :
\[\mathrm{C_6H_6} + \mathrm{HO\!-\!SO_3H} \xrightarrow[373\ \text{K}]{\ \text{fuming } \mathrm{H_2SO_4}\ } \mathrm{C_6H_5SO_3H} + \mathrm{H_2O}\]
Benzene gives benzenesulfonic acid. When benzene is heated with fuming sulfuric acid (oleum) at 373 K, it gives benzene sulfonic acid.
Electrophile : \(\mathrm{SO_3}\), free sulfur trioxide.
Formation of the electrophile :
\[2\,\mathrm{H_2SO_4} \rightarrow \mathrm{H_3O^{+}} + \mathrm{HSO_4^{-}} + \mathrm{SO_3}\]
Note that this electrophile carries no full positive charge. It does not need one — the sulfur atom of \(\mathrm{SO_3}\) is strongly electron-deficient because three oxygen atoms pull electron density away from it, and that is enough to make it electrophilic.
iv. Friedel–Craft’s alkylation reaction :
\[\mathrm{C_6H_6} + \mathrm{CH_3\!-\!Cl} \xrightarrow{\ \text{anhydrous } \mathrm{AlCl_3}\ } \mathrm{C_6H_5CH_3} + \mathrm{HCl}\]
Benzene and methyl chloride give methylbenzene (toluene). When benzene is treated with an alkyl halide like methyl chloride in the presence of anhydrous aluminium chloride, it gives toluene. The reaction is used to extend the chain outside the benzene ring — it is the only reaction in this section that adds carbon atoms to the molecule.
Electrophile : \(\mathrm{R^{+}}\).
Formation of the electrophile :
\[\mathrm{R\!-\!Cl} + \mathrm{AlCl_3} \rightarrow \mathrm{R^{+}} + \mathrm{AlCl_4^{-}}\]
v. Friedel–Craft’s acylation reaction :
\[\mathrm{C_6H_6} + \mathrm{CH_3\!-\!COCl} \xrightarrow[\text{heat}]{\ \text{anhy. } \mathrm{AlCl_3}\ } \mathrm{C_6H_5COCH_3} + \mathrm{HCl}\]
Benzene and acetyl chloride give 1-phenylethanone or acetophenone.
\[\mathrm{C_6H_6} + (\mathrm{CH_3\!-\!CO})_2\mathrm{O} \xrightarrow[\text{heat}]{\ \text{anhy. } \mathrm{AlCl_3}\ } \mathrm{C_6H_5COCH_3} + \mathrm{CH_3COOH}\]
Benzene and acetic anhydride also give 1-phenylethanone or acetophenone. When benzene is heated with an acyl halide or acid anhydride in the presence of anhydrous aluminium chloride, it gives the corresponding acyl benzene.
Electrophile : \(\mathrm{R\!-\!C^{+}{=}O}\), acylium ion.
Formation of the electrophile :
\[\mathrm{R\!-\!COCl} + \mathrm{AlCl_3} \rightarrow \mathrm{R\!-\!C^{+}{=}O} + \mathrm{AlCl_4^{-}}\]
15.4.7 Directive Influence of a Functional Group in Monosubstituted Benzene
Structure of Benzene
In benzene, all hydrogen atoms are equivalent. Therefore, only one product is possible when it undergoes electrophilic substitution reactions. Once one substituent is in place, however, the remaining positions are no longer all alike.
Monosubstituted Benzene
When one hydrogen is removed and a substituent S is put in its place, the ring positions can be numbered from the substituent, as shown in Fig. 15.30.
Positions of carbon atoms in monosubstituted benzene :
- The positions 2 and 6 are equivalent and give ortho (o-) products.
- The positions 3 and 5 are equivalent and give meta (m-) products.
- The position 4 is unique and gives the para (p-) product.
Now in benzene, five positions are available for electrophilic substitution, but because of the symmetry of the ring these five positions amount to only three distinct products.
Diagram
Fig. 15.30 – Positions in a monosubstituted benzene: hexagon with substituent S at C-1 and positions numbered 2 to 6, with 2 and 6 labelled ortho, 3 and 5 labelled meta, 4 labelled para; below, the same ring reacting with E⁺ (−H) branching into the three disubstituted products, each drawn twice to show the equivalent positions, labelled ortho-product, meta-product and para-product
When monosubstituted benzene is subjected to further electrophilic substitution, the second substituent i.e. the electrophile (see Chapter 14) or incoming group (E) can occupy any of these positions and give three disubstituted products. But these products are not formed in equal amounts.
Two types of behaviour are observed: either a. ortho- and para-products, or b. meta-products, are found as major products. This behaviour mainly depends on the nature of the substituent (S) already present in the benzene ring, and not on the nature of the second substituent (E), i.e. the incoming group.
That last statement is the heart of the section, and it is worth restating as a rule of thumb: the group already on the ring decides where the next group goes, whatever the next group happens to be.
Ortho and Para Directing Groups
Ortho and para directing groups :
\[\mathrm{{-}\ddot{C}l}\!:,\quad \mathrm{{-}\ddot{B}r}\!:,\quad \mathrm{{-}\ddot{O}\!-\!H},\quad \mathrm{{-}\ddot{O}\!-\!CH_3},\quad \mathrm{{-}\ddot{N}H_2},\quad \mathrm{{-}\ddot{N}HR},\quad \mathrm{{-}CH_3},\quad \mathrm{{-}C_2H_5},\quad \mathrm{{-}R}\ \text{etc.}\]
The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups.
Ortho and para directive influence of the -OH group : The resonance theory clearly explains why certain substituents are ortho/para or meta directing. Let us study the various resonance structures of phenol (see Chapter 14), drawn in Fig. 15.31.
Diagram
Fig. 15.31 – Resonance structures of phenol: the neutral structure with the oxygen lone pairs shown, then the structures in which a lone pair from oxygen has moved into the ring, placing a positive charge on oxygen and a negative charge in turn at the ortho, para and other ortho positions; curly arrows drawn on each step, all linked by double-headed arrows and labelled “resonating structures”
It is clear from the above resonance structures that the ortho and para positions have a greater electron density than the meta positions. Therefore, the \(\mathrm{{-}OH}\) group activates the benzene ring for the attack of the second substituent E at these electron-rich centres. An electrophile is attracted to negative charge, so it goes where the electron density has been pushed.
There is a complication worth noting. Due to the \(\mathrm{{-}I}\) effect of the \(\mathrm{{-}OH}\) group (see Chapter 14), the electron density on the ortho positions of the benzene ring gets slightly reduced. Thus the resonance effect and the inductive effect of the OH group act opposite to each other, but the \(\mathrm{{+}R}\) effect of \(\mathrm{{-}OH}\) is more powerful than the \(\mathrm{{-}I}\) effect — which is why the group is activating overall.
The pattern generalises: all ortho and para directing groups possess a nonbonding electron pair on the atom which is directly attached to the aromatic ring. That lone pair is what can be pushed into the ring by resonance.
Methyl group is an exception : The only exception to the above rule is the methyl or alkyl group. It is ortho and para directing, although it has no nonbonding electron pair on the key atom. This is explained on the basis of a special type of resonance called hyperconjugation or no bond resonance (see Chapter 14).
Halogens as a Special Case
In case of aryl halides, halogens are moderately deactivating. Because of their strong \(\mathrm{{-}I}\) effect, the overall electron density on the benzene ring decreases. It makes the electrophilic substitution difficult. However, due to resonance the electron density on the ortho and para positions is greater than at the meta positions. Halogens are therefore ortho and para directing. Let us study the various resonating structures of chlorobenzene, shown in Fig. 15.32.
Diagram
Fig. 15.32 – Resonance structures of chlorobenzene: the neutral structure with chlorine lone pairs, followed by the structures with a lone pair donated into the ring placing a positive charge on chlorine and a negative charge at the ortho and para positions in turn, curly arrows shown, linked by double-headed arrows and labelled “resonating structures”
Halogens therefore do two opposite things at once: they reduce the reactivity of the whole ring by the inductive effect, but they still steer the incoming group to the ortho and para positions by resonance. Deactivating and o/p-directing are not contradictory descriptions — one is about how fast, the other about where.
Meta Directing and Deactivating Groups
Meta directing and deactivating groups :
\[\mathrm{{-}NO_2},\quad \mathrm{{-}NH_3^{+}},\quad \mathrm{{-}C{\equiv}N},\quad \mathrm{{-}CHO},\quad \mathrm{{-}CO\!-\!R},\quad \mathrm{{-}COOH},\quad \mathrm{{-}SO_3H}\ \text{etc.}\]
All meta directing groups have a positive (or partial positive) charge on the atom which is directly attached to the aromatic ring. The groups which direct the incoming group to meta positions are called meta directing groups.
Metadirective influence of the \(\mathrm{{-}NO_2}\) group can be explained by resonance theory : A meta directing group withdraws electrons from the aromatic ring by resonance, making the ring electron-deficient. Therefore, meta groups are ring deactivating groups.
Due to the \(\mathrm{{-}I}\) effect, the \(\mathrm{{-}NO_2}\) group reduces electron density in the benzene ring at the ortho and para positions. So the attack of the incoming group becomes difficult at ortho and para positions. The incoming group can attack the meta positions more easily. Let us study the various resonance structures of nitrobenzene, given in Fig. 15.33.
Diagram
Fig. 15.33 – Resonance structures of nitrobenzene: the neutral structure with the –NO₂ group drawn with its formal charges, followed by structures in which ring π electrons are pulled towards the nitro group, placing a positive charge in turn at the ortho, para and other ortho positions; curly arrows shown, all linked by double-headed arrows and labelled “resonating structures”
It is clear from the above resonance structures that the ortho and para positions have comparatively less electron density than at the meta positions. Hence, the incoming group / electrophile attacks at the meta positions.
Note the logic carefully, because it is easy to state the wrong way round. The meta position is not especially activated. It is simply the least deactivated, so by elimination that is where the electrophile ends up.
Flowchart
FC 15.15 – Predicting where the second group goes: look at the substituent already on the ring → does the key atom attached to the ring carry a nonbonding electron pair? → if yes → o/p directing and activating (branch: if the atom is a halogen → o/p directing but moderately deactivating) → if no, is it an alkyl group? → if yes → o/p directing by hyperconjugation → if the key atom instead carries a positive or partial positive charge → m directing and deactivating
15.4.8 Carcinogenicity and Toxicity of Benzene
Benzene is both toxic and carcinogenic (cancer causing). In fact, it might be considered “the mother of all carcinogens”, as a large number of carcinogens have structures that include benzene rings.
Several polycyclic aromatic compounds, containing more than two fused benzene rings, are produced by incomplete combustion of tobacco, coal and petroleum. In the liver, benzene is oxidized to an epoxide. Benzopyrene is converted into an epoxy diol. These substances are carcinogenic and can react with DNA, which can induce mutation leading to uncontrolled growth of cancer cells.
Read this alongside section 15.4.6: the very stability that makes the aromatic ring survive so many reactions also means that once such a molecule enters the body it is not easily broken down and disposed of. Instead the liver tries to make it more water-soluble by oxidation, and the reactive intermediate formed on the way is what attacks DNA.
15.5 Summary
Hydrocarbons contain only carbon and hydrogen. They are classified first as open chain (aliphatic) or cyclic, then as saturated (only single C–C bonds) or unsaturated (containing \(\mathrm{C{=}C}\) and/or \(\mathrm{C{\equiv}C}\)). A cyclic unsaturated hydrocarbon may be aromatic or nonaromatic.
Alkanes are aliphatic saturated hydrocarbons containing carbon–carbon and carbon–hydrogen single covalent bonds, with the general formula \(\mathrm{C_nH_{2n+2}}\). They are called paraffins because of their low affinity for other chemicals.
Alkanes with more than three carbon atoms show structural isomerism, in particular chain isomerism. The number of isomers rises rapidly with carbon number: butane has two, pentane three and hexane five.
Rotation about a C–C sigma bond is free because the bond is cylindrically symmetrical, so the extent of overlap does not change on rotation. The arrangements of atoms in space that result are called conformations, of which the staggered and eclipsed forms of ethane are the two extremes, drawn as sawhorse and Newman projection formulae. Conformational isomerism is a type of stereoisomerism.
Alkanes are obtained industrially by fractional distillation of crude oil. In the laboratory they are prepared by catalytic hydrogenation of alkenes or alkynes over Pt or Pd at room temperature, or over Ni at high temperature and pressure; by reduction of alkyl halides with Zn and dilute HCl; by the Wurtz coupling reaction, which doubles the carbon number; and through the Grignard reagent, which keeps the carbon number the same.
\[\mathrm{R\!-\!X} + \mathrm{Mg} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!Mg\!-\!X} \qquad \mathrm{R\!-\!Mg\!-\!X} + \mathrm{H\!-\!O\!-\!H} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!H} + \mathrm{MgX(OH)}\]
C–H and C–C bonds are non-polar, so alkanes are non-polar and are held together only by weak van der Waals forces. Since these forces depend on contact area, straight chain alkanes have higher melting and boiling points than branched ones. Alkanes are colourless and odourless; the first four are gases, those with 5 to 17 carbons are liquids and those from 18 carbons onwards are solids.
Halogenation of alkanes is a substitution reaction needing UV light, diffused sunlight or 573 to 773 K, and it proceeds by a free radical mechanism with initiation and propagation steps. Halogen reactivity follows \(\mathrm{F_2} > \mathrm{Cl_2} > \mathrm{Br_2} > \mathrm{I_2}\), and hydrogen is replaced in the order \(3^{\circ} > 2^{\circ} > 1^{\circ}\). Bromination is far more selective than chlorination, but a mixture of all possible isomeric products is always obtained.
Alkanes burn in air to carbon dioxide and water with a large release of heat, which is why they are good fuels.
\[\mathrm{C_nH_{2n+2}} + \left(\frac{3n+1}{2}\right)\mathrm{O_2} \rightarrow n\,\mathrm{CO_2} + (n+1)\,\mathrm{H_2O} + \text{heat}\]
On strong heating in the absence of air, alkanes undergo pyrolysis or cracking to lower alkanes, alkenes and hydrogen. Straight chain alkanes of 6 to 10 carbon atoms undergo aromatization or reforming at about 773 K and 10 to 20 atm over \(\mathrm{V_2O_5}\), \(\mathrm{Cr_2O_3}\) or \(\mathrm{Mo_2O_3}\) on alumina, by simultaneous dehydrogenation and cyclization, giving benzene and its homologues for high quality gasoline.
Uses of alkanes: the first four as cooking and heating fuels (LPG, CNG); CNG, petrol and diesel as automobile fuels; lower liquid alkanes as solvents; alkanes above 35 carbon atoms as road tar; and waxes as lubricants, candles and carbon black.
Alkenes are unsaturated hydrocarbons with at least one \(\mathrm{C{=}C}\), general formula \(\mathrm{C_nH_{2n}}\), also called olefins. Those with two or three double bonds are alkadienes and alkatrienes.
Alkenes show chain and position isomerism, and also geometrical isomerism, which arises from the fixed spatial arrangement of groups about the doubly bonded carbon atoms. Identical groups on the same side give the cis-isomer and on opposite sides the trans-isomer. Geometrical isomerism is impossible if either doubly bonded carbon carries two identical groups, as in terminal and 1,1-disubstituted alkenes.
Alkenes are prepared by 1,2-elimination, in which two groups leave adjacent carbon atoms and an \(sp^3\) carbon becomes \(sp^2\): dehydrohalogenation of alkyl halides with hot concentrated alcoholic KOH or NaOH; dehydration of alcohols with sulphuric acid; and dehalogenation of vicinal dihalides with zinc. The ease of both dehydrohalogenation and dehydration follows \(3^{\circ} > 2^{\circ} > 1^{\circ}\).
Saytzeff rule: in dehydrohalogenation the preferred product is the alkene with the greater number of alkyl groups on the doubly bonded carbon atoms. The order of stability of alkenes runs \(\mathrm{R_2C{=}CR_2} > \mathrm{R_2C{=}CHR} > \mathrm{R_2C{=}CH_2}\) and \(\mathrm{RCH{=}CHR} > \mathrm{RCH{=}CH_2} > \mathrm{CH_2{=}CH_2}\), which is why the more substituted alkene forms faster.
An alkyne can also be partially reduced to an alkene: dihydrogen over Lindlar’s catalyst gives the cis-alkene, while sodium in liquid ammonia gives the trans-alkene.
Alkenes are non-polar or weakly polar, insoluble in water, soluble in non-polar solvents and less dense than water. Boiling point rises with carbon number and is very nearly that of the alkane with the same carbon skeleton.
Because of the loosely held \(\pi\) electrons, alkenes are more reactive than alkanes and undergo electrophilic addition. Halogens add in \(\mathrm{CCl_4}\) at room temperature to give vicinal dihalides, iodine generally failing to react; the disappearance of the red brown colour of bromine is a test for unsaturation.
Hydrogen halides add in the order of reactivity \(\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\). Markovnikov’s rule states that the negative part of an unsymmetrical reagent attaches to the carbon carrying fewer hydrogen atoms. In the presence of an organic peroxide the orientation of HBr addition is reversed — the peroxide effect or Kharasch–Mayo effect — while HCl and HI are unaffected.
Cold concentrated sulphuric acid adds by Markovnikov’s rule to give alkyl hydrogen sulfate, which on dilution and heating gives the corresponding alcohol; this is an important industrial route to alcohols. Direct hydration of reactive alkenes with water and \(\mathrm{H^{+}}\) also follows Markovnikov’s rule.
Ozonolysis cleaves the double bond in two steps: ozone in \(\mathrm{CCl_4}\) gives an unstable ozonide, and water with zinc dust decomposes it to aldehydes and/or ketones. A carbonyl group appears at each originally doubly bonded carbon, zinc dust prevents hydrogen peroxide from oxidising the aldehydes, and the products reveal the number and position of double bonds in the original alkene.
Hydroboration–oxidation with diborane in THF followed by alkaline peroxide gives the anti-Markovnikov product, that is the primary alcohol from a terminal alkene, along with \(\mathrm{B(OH)_3}\).
Polymerization joins many monomer molecules into a polymer; ethene with oxygen at high temperature and pressure gives polyethene.
\[n\,\mathrm{H_2C{=}CH_2} \xrightarrow[\text{polymerization}]{\ \mathrm{O_2},\ \text{heat, pressure}\ } \mathrm{({-}CH_2\!-\!CH_2{-})}_n\]
Hydroxylation with cold dilute alkaline \(\mathrm{KMnO_4}\) gives 1,2-diols (glycols) and discharges the purple colour; this is Baeyer’s test for a double bond. With acidic \(\mathrm{KMnO_4}\) or acidic potassium dichromate, alkenes are oxidised further to ketones and/or acids.
Uses of alkenes: as starting materials for alkyl halides, alcohols, aldehydes, ketones and acids; ethene and propene for polythene and polypropylene; and ethene for artificial ripening of fruits such as mangoes.
Alkynes are aliphatic unsaturated hydrocarbons with at least one \(\mathrm{C{\equiv}C}\), general formula \(\mathrm{C_nH_{2n-2}}\). The triple bond is a functional group; the carbon atoms are \(sp\) hybridised and the molecule is linear. Alkynes show position isomerism, and 1-alkynes are called terminal alkynes.
Ethyne is prepared industrially by controlled partial oxidation of methane at 1773 K, and from calcium carbide and water.
\[6\,\mathrm{CH_4} + 2\,\mathrm{O_2} \xrightarrow{\ 1773\ \text{K}\ } 2\,\mathrm{H\!-\!C{\equiv}C\!-\!H} + 2\,\mathrm{CO_2} + 10\,\mathrm{H_2} \qquad \mathrm{CaC_2} + 2\,\mathrm{H_2O} \rightarrow \mathrm{C_2H_2} + \mathrm{Ca(OH)_2}\]
In the laboratory, a vicinal dihalide with alcoholic KOH gives an alkenyl halide, which with sodamide gives the alkyne. Higher alkynes are built by treating a terminal alkyne with lithium amide to form an alkyn-1-yl lithium and then reacting it with a primary alkyl halide.
The hydrogen on a triply bonded carbon is acidic, because an \(sp\) hybrid orbital has 50% s-character, holds its electrons close to the nucleus and makes the carbon more electronegative than the \(sp^2\) carbon of ethene or the \(sp^3\) carbon of ethane. Hence acidity follows \(\mathrm{H\!-\!C{\equiv}C\!-\!H} > \mathrm{H_2C{=}CH_2} > \mathrm{H_3C\!-\!CH_3}\). Terminal alkynes give lithium acetylides with \(\mathrm{LiNH_2}\), and insoluble acetylides with \(\mathrm{Ag^{+}}\) or \(\mathrm{Cu^{+}}\), which distinguishes them from non-terminal alkynes.
Alkynes add halogen twice, giving first a 1,2-dihaloalkene and then a 1,1,2,2-tetrahaloalkane, and the disappearance of the bromine colour again indicates unsaturation. Hydrogen halides also add twice, by Markovnikov’s rule in both steps, giving geminal dihalides, with reactivity \(\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\).
With water in the presence of 40% \(\mathrm{H_2SO_4}\) and 1% \(\mathrm{HgSO_4}\), an alkyne gives an unstable enol which tautomerises to a carbonyl compound: ethyne gives ethanal and every other alkyne gives a ketone.
Uses of acetylene: preparation of ethanal, propanone and ethanoic acid; manufacture of polymers, synthetic rubber, synthetic fibre and plastics; artificial ripening of fruits; and the oxy-acetylene flame for welding and cutting metals.
Aromatic hydrocarbons, or arenes, contain only carbon and hydrogen; benzene is the simplest. Benzene and all compounds whose structures and chemical properties resemble benzene are aromatic. Aromatic compounds have a higher percentage of carbon, burn with a sooty flame, resist normal oxidising and reducing agents, do not decolourise alkaline \(\mathrm{KMnO_4}\) or bromine in \(\mathrm{CCl_4}\), and prefer substitution to addition.
Benzene is \(\mathrm{C_6H_6}\), a colourless liquid of characteristic odour, boiling point 353 K and melting point 278.5 K, insoluble in water but soluble in alcohol, ether and chloroform; its vapours are highly toxic. Coal-tar and petroleum are the large-scale sources, and Berthelot synthesized it from acetylene in 1870.
The structure of benzene was established step by step: the formula shows high unsaturation; failure of the alkene tests rules out an open chain; a single monosubstituted bromobenzene shows all six hydrogen atoms are identical; and hydrogenation to cyclohexane shows a ring with three \(\mathrm{C{=}C}\). Kekulé proposed the planar ring with alternate single and double bonds in 1865, and added oscillating double bonds to explain the single ortho-disubstituted product, but only resonance explains the stability and the preference for substitution.
All six carbon atoms of benzene are \(sp^2\) hybridised, all atoms lie in one plane, all bond angles are 120°, and the six p electrons are delocalized over all six carbon atoms. Delocalization is the source of benzene’s stability, and it makes every C–C bond length equal at 139 pm, intermediate between C–C at 154 pm and \(\mathrm{C{=}C}\) at 133 pm.
Aromaticity requires all three conditions together: the ring must be cyclic and planar with \(sp^2\) atoms; every ring atom must carry a parallel p-orbital allowing continuous overlap; and the ring must contain \((4n + 2)\) \(\pi\) electrons with n an integer, which is Huckel’s rule. Benzene (\(n = 1\)), naphthalene (\(n = 2\)) and pyridine (\(n = 1\)) are aromatic, while cycloheptatriene is not, because one ring carbon is \(sp^3\) and breaks the overlap.
Benzene is prepared from ethyne by trimerization through a red hot iron tube at 873 K, from anhydrous sodium benzoate by decarboxylation with soda-lime, and from phenol by passing its vapours over heated zinc dust.
Under special conditions benzene adds: three molecules of chlorine in UV light give benzene hexachloride, whose \(\gamma\)-isomer is gammexane or lindane; three molecules of dihydrogen over nickel at 453 to 473 K give cyclohexane; and three molecules of ozone followed by zinc and water give glyoxal.
Benzene is characterised by electrophilic substitution, in which a catalyst first generates an electrophile, the ring then supplies \(\pi\) electrons to it, and a hydrogen ion is lost so that the aromatic ring is restored: chlorination and bromination with \(\mathrm{Cl^{+}}\) and \(\mathrm{Br^{+}}\) using Fe or the ferric halide, nitration with \(\mathrm{NO_2^{+}}\) from the nitrating mixture at 313 to 333 K, sulfonation with \(\mathrm{SO_3}\) from fuming sulphuric acid at 373 K, Friedel–Craft’s alkylation with \(\mathrm{R^{+}}\), and Friedel–Craft’s acylation with the acylium ion \(\mathrm{R\!-\!C^{+}{=}O}\), both over anhydrous \(\mathrm{AlCl_3}\). Iodination is not possible because it is reversible.
Benzene burns in air with a sooty flame, and any hydrocarbon burns according to
\[\mathrm{C_xH_y} + \left(x + \frac{y}{4}\right)\mathrm{O_2} \rightarrow x\,\mathrm{CO_2} + \frac{y}{2}\,\mathrm{H_2O}\]
In a monosubstituted benzene, positions 2 and 6 are ortho, 3 and 5 are meta and 4 is para. Which of the three disubstituted products dominates is decided by the group already present, not by the incoming group.
Ortho and para directing groups carry a nonbonding electron pair on the atom attached to the ring — \(\mathrm{{-}OH}\), \(\mathrm{{-}OCH_3}\), \(\mathrm{{-}NH_2}\), \(\mathrm{{-}NHR}\), \(\mathrm{{-}Cl}\), \(\mathrm{{-}Br}\) — and push electron density into the ortho and para positions by resonance; alkyl groups are the exception, directing ortho and para through hyperconjugation. In aryl halides the strong \(\mathrm{{-}I}\) effect makes halogens moderately deactivating even though they remain ortho and para directing.
Meta directing groups carry a positive or partial positive charge on the atom attached to the ring — \(\mathrm{{-}NO_2}\), \(\mathrm{{-}NH_3^{+}}\), \(\mathrm{{-}CN}\), \(\mathrm{{-}CHO}\), \(\mathrm{{-}COR}\), \(\mathrm{{-}COOH}\), \(\mathrm{{-}SO_3H}\) — and withdraw electron density from the ring, deactivating it and leaving the meta positions comparatively richest in electrons.
Benzene is toxic and carcinogenic, and many carcinogens contain benzene rings. Polycyclic aromatic compounds from the incomplete combustion of tobacco, coal and petroleum, together with the epoxide formed when the liver oxidises benzene and the epoxy diol from benzopyrene, can react with DNA and induce mutations leading to uncontrolled growth of cancer cells.
The interconversions of the whole chapter are collected in FC 15.16.
Flowchart
FC 15.16 – Interconversions across the four families: alkane ⇄ alkene (alkane ← H₂/Pt, Pd or Ni from alkene; alkene ← alcoholic KOH from alkyl halide, ← conc. H₂SO₄ from alcohol, ← Zn from vicinal dihalide); alkene ⇄ alkyne (alkene ← H₂/Lindlar giving cis or Na/liq. NH₃ giving trans from alkyne; alkyne ← alcoholic KOH then sodamide from vicinal dihalide); alkane → aromatic (C₆–C₁₀ straight chain alkane → V₂O₅, 773 K, 10–20 atm → benzene + 4H₂, reforming); alkyne → aromatic (3 C₂H₂ → red hot iron tube, 873 K → benzene); side branches: alkene + X₂ → vicinal dihalide, alkene + HX → alkyl halide, alkene + O₃/Zn-H₂O → carbonyl compounds, alkene + alkaline KMnO₄ → glycol, alkyne + H₂O/H₂SO₄-HgSO₄ → carbonyl compound, benzene + electrophile/catalyst → substituted benzene
15.6 Check Your Concepts
Assertion : A branched alkane has a lower boiling point than its straight chain isomer. Reason : Branching reduces the surface area over which two molecules can be in contact. State whether the assertion and the reason are correct, and whether the reason correctly explains the assertion.
Alkanes are described as chemically unreactive, yet they are used as fuels in stoves and engines. Explain why there is no contradiction between these two statements.
Butane and 2-methylpropane can be kept in separate bottles as distinct compounds, but staggered and eclipsed ethane cannot. Explain the difference.
Bromination of 2-methylpropane gives over 99% of a single product, while chlorination of the same alkane gives 64% of one product and 36% of another. Account for this difference.
Halogenation of an alkane needs UV light or strong heating, whereas chlorination of benzene is carried out in the dark with a catalyst. Explain why the conditions are so different, and name the species that attacks the substrate in each case.
Why is the Wurtz reaction unsuitable for preparing methane from methyl iodide, while the Grignard route is suitable? Support your answer with equations.
Why must the preparation of a Grignard reagent be carried out in dry ether under dry and inert conditions?
Write the initiation and the two propagation steps for the chlorination of methane, and state which step regenerates the chain carrier.
Arrange butan-1-ol, butan-2-ol and 2-methylpropan-2-ol in decreasing order of ease of dehydration, and support your order by referring to the conditions each one needs.
In the dehydrohalogenation of 2-chlorobutane, but-2-ene is formed in far greater amount than but-1-ene. Name the rule that predicts this and explain it in terms of alkene stability.
Which of the following exhibit cis–trans isomerism, and why: pent-2-ene, hex-3-ene, 2-methylprop-1-ene, 2-methylbut-2-ene?
An organic peroxide reverses the orientation of HBr addition to propene, but has no effect when HCl or HI is used. Explain both observations.
Propene can be converted into propan-1-ol by one route and into propan-2-ol by another. Give the reagents for each route and state which one gives the anti-Markovnikov product.
State the role of zinc dust in the second step of ozonolysis, and explain what would go wrong if it were omitted.
Distinguish between a vicinal dihalide and a geminal dihalide. Give one example of each and state how each is prepared from a hydrocarbon.
Why does the addition of water to every alkyne except ethyne give a ketone rather than an aldehyde?
Ethyne reacts with lithium amide but ethene and ethane do not. Explain, referring to the hybridisation of the carbon atom holding the hydrogen.
Match the reagent in Column I with the product it gives from an alkene in Column II.
| Column I (reagent) | Column II (product) |
|---|---|
| a. Cold dilute alkaline \(\mathrm{KMnO_4}\) | i. Primary alcohol |
| b. \(\mathrm{O_3}\) in \(\mathrm{CCl_4}\), then \(\mathrm{H_2O}\)/Zn dust | ii. Vicinal dihalide |
| c. \((\mathrm{BH_3})_2\) in THF, then \(\mathrm{HO\!-\!OH}\)/\(\mathrm{OH^{-}}\) | iii. Glycol |
| d. \(\mathrm{Br_2}\) in \(\mathrm{CCl_4}\) | iv. Aldehyde and/or ketone |
You are given three unlabelled cylinders containing ethane, ethene and ethyne. Describe a scheme of chemical tests, with the reagent and the expected observation at each stage, that would identify all three.
Assertion : Benzene does not decolourise bromine dissolved in carbon tetrachloride. Reason : All six carbon–carbon bonds in benzene have the same length, 139 pm, intermediate between a single and a double bond. State whether each statement is correct and whether the reason explains the assertion.
List the three conditions a ring must satisfy to be aromatic, and use them to explain why cycloheptatriene is non-aromatic even though it has six \(\pi\) electrons.
Why is iodination of benzene not possible? Name the catalyst used for chlorination and for bromination of benzene, and write the equation showing how the electrophile is generated in each case.
Benzene burns with a sooty flame while hexane burns with a clean flame. Explain, using the information in Table 15.6.
Explain why the \(\mathrm{{-}NO_2}\) group is both meta directing and ring deactivating, while the \(\mathrm{{-}OH}\) group is ortho and para directing and ring activating.
Chlorine withdraws electron density from the benzene ring, yet chlorobenzene undergoes further substitution mainly at the ortho and para positions. Reconcile these two facts.
The methyl group has no nonbonding electron pair on the carbon attached to the ring, yet it is ortho and para directing. Name the effect responsible and state why the general rule for ortho/para directing groups does not apply to it.
Complete each conversion by naming the reagents and conditions, and write the balanced equation: a. n-hexane to benzene, b. ethyne to benzene, c. benzene to cyclohexane, d. 1,2-dibromoethane to ethyne.
Predict the major product and state whether the reaction will be faster or slower than the same reaction on benzene itself: a. nitration of methylbenzene, b. sulfonation of nitrobenzene.
Explain why ethene readily undergoes addition reactions while benzene prefers substitution, even though both are drawn with carbon–carbon double bonds.
Why is lindane, the \(\gamma\)-isomer of benzene hexachloride, no longer freely used as an insecticide? Relate your answer to the chemical stability of a saturated chlorinated ring.
The type of hydrocarbon that is used as lubricant.
Alkene used in the manufacture of polythene bags.
The hydrocarbon said to possess carcinogenic property.
What are the main natural sources of alkane?
Arrange the three isomers of alkane with molecular formula \(\mathrm{C_5H_{12}}\) in increasing order of boiling points and write their IUPAC names.
Write IUPAC names of the products obtained by the reaction of cold concentrated sulphuric acid followed by water with the following compounds.
- propene
- but-1-ene
Write the balanced chemical reaction for preparation of ethane from
- Ethyl bromide
- Ethyl magnesium iodide
How many monochlorination products are possible for
- 2-methylpropane?
- 2-methylbutane? Draw their structures and write their IUPAC names.
Write all the possible products for pyrolysis of butane.
Which of the following will exhibit geometrical isomerism?
- \(\mathrm{CH_3\!-\!CH_2\!-\!C({=}CH_2)\!-\!CH_3}\)
- \(\mathrm{(CH_3)_2C{=}CH_2}\)
- \(\mathrm{CH_3\!-\!C(C_2H_5){=}C(C_2H_5)\!-\!CH_3}\)
What is the action of following on ethyl iodide?
- alc. KOH
- Zn, HCl
An alkene ‘A’ on ozonolysis gives 2 moles of ethanal. Write the structure and IUPAC name of ‘A’.
Acetone and acetaldehyde are the ozonolysis products of an alkene. Write the structural formula of an alkene and give IUPAC name of it.
Write the reaction to convert
- propene to n-propyl alcohol
- propene to isopropyl alcohol.
What is the action of following on but-2-ene?
- dil. alkaline \(\mathrm{KMnO_4}\)
- acidic \(\mathrm{KMnO_4}\)
Complete the following reaction sequence: \[\mathrm{HC{\equiv}CH} \xrightarrow{\ \mathrm{LiNH_2}\ } \mathrm{A} \xrightarrow{\ \mathrm{CH_3CH_2Cl}\ } \mathrm{B} \xrightarrow{\ \mathrm{HBr}\ } \mathrm{C}\] Comment on the acidic nature of B.
Write the balanced chemical reactions to get benzene from
- Sodium benzoate.
- Phenol.
Predict the possible products of the following reaction.
- chlorination of nitrobenzene,
- sulfonation of chlorobenzene,
- bromination of phenol,
- nitration of toluene.
Identify the main product of the reaction
\[\mathrm{CH_3\!-\!CH_2\!-\!CH_3} \xrightarrow[\Delta]{\ \mathrm{O_2}\ }\]
b.\[\mathrm{CH_3\!-\!CH(OH)\!-\!CH_3} \xrightarrow[373\ \text{K}]{\ 60\%\ \mathrm{H_2SO_4}\ }\]
c.\[\mathrm{HC{\equiv}C\!-\!CH_3} \xrightarrow[\mathrm{Pd\!-\!C}/\text{quinoline}]{\ \mathrm{H_2}\ }\]
d.\[\mathrm{HC{\equiv}CH} + \mathrm{H_2O} \xrightarrow[1\%\ \mathrm{HgSO_4}]{\ 40\%\ \mathrm{H_2SO_4}\ }\]
Identify A, B, C in the following reaction sequence: \[\mathrm{CH_3\!-\!CH{=}CH_2} \xrightarrow[\text{room temperature}]{\ \mathrm{Br_2}/\mathrm{CCl_4}\ } \mathrm{A} \xrightarrow{\ \mathrm{Zn}\ } \mathrm{B} \xrightarrow{\ \text{dil. alka. } \mathrm{KMnO_4}\ } \mathrm{C}\]
Identify giving reason whether the following compounds are aromatic or not.
- cyclopenta-1,3-diene
- cyclobutadiene
- benzene
- pyridine (structures as drawn in Fig. 15.34)
Diagram
Fig. 15.34 – Four ring structures for the aromaticity question, labelled A to D: A, five-membered ring with two double bonds and one CH₂ carbon; B, four-membered ring with two double bonds; C, plain hexagon of benzene; D, six-membered ring with three double bonds and one N atom in the ring, marked N
- Name two reagents used for acylation of benzene.
15.7 Practice With Numericals
The enthalpy of combustion of methane is \(\Delta_c H^0 = -890\ \text{kJ mol}^{-1}\). Calculate the heat released when 11.2 L of methane measured at STP is burnt completely. (Molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).)
The enthalpy of combustion of butane is \(\Delta_c H^0 = -2875.84\ \text{kJ mol}^{-1}\). Calculate the heat released when 29 g of butane is burnt completely. (Molar mass of \(\mathrm{C_4H_{10}} = 58\ \text{g mol}^{-1}\).)
Using the general combustion equation for an alkane, calculate the number of moles of dioxygen required for the complete combustion of 0.25 mol of propane, and the mass of water formed. (Molar mass of \(\mathrm{H_2O} = 18\ \text{g mol}^{-1}\).)
Calculate the mass of ethane obtained by the complete catalytic hydrogenation of 14 g of ethene. (Molar masses: \(\mathrm{C_2H_4} = 28\ \text{g mol}^{-1}\), \(\mathrm{C_2H_6} = 30\ \text{g mol}^{-1}\).)
Calculate the volume of dihydrogen at STP required to convert 13 g of ethyne completely into ethane. (Molar mass of \(\mathrm{C_2H_2} = 26\ \text{g mol}^{-1}\); molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).)
Calculate the volume of ethyne obtained at STP when 32 g of calcium carbide reacts completely with water. (Molar mass of \(\mathrm{CaC_2} = 64\ \text{g mol}^{-1}\).)
Calculate the mass of 1,2-dibromopropane obtained when 8.4 g of propene reacts completely with bromine in carbon tetrachloride. (Molar masses: \(\mathrm{C_3H_6} = 42\ \text{g mol}^{-1}\), \(\mathrm{C_3H_6Br_2} = 202\ \text{g mol}^{-1}\).)
Calculate the mass of ethane-1,2-diol obtained when 5.6 g of ethene is treated with cold dilute alkaline \(\mathrm{KMnO_4}\), assuming complete conversion. (Molar masses: \(\mathrm{C_2H_4} = 28\ \text{g mol}^{-1}\), \(\mathrm{C_2H_6O_2} = 62\ \text{g mol}^{-1}\).)
Calculate the mass of methylbenzene obtained when 15.6 g of benzene undergoes Friedel–Craft’s alkylation with excess methyl chloride, assuming complete conversion. (Molar masses: \(\mathrm{C_6H_6} = 78\ \text{g mol}^{-1}\), \(\mathrm{C_7H_8} = 92\ \text{g mol}^{-1}\).)
Calculate the volume of dioxygen at STP required for the complete combustion of 15.6 g of benzene, and the mass of carbon dioxide formed. (Molar masses: \(\mathrm{C_6H_6} = 78\ \text{g mol}^{-1}\), \(\mathrm{CO_2} = 44\ \text{g mol}^{-1}\); molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).)
A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and its molar mass is \(56\ \text{g mol}^{-1}\). Determine its empirical formula and molecular formula, and write the structures and IUPAC names of the alkenes possible for that molecular formula.
An alkyne has molar mass \(54\ \text{g mol}^{-1}\). Using the general formula for alkynes, find the number of carbon atoms in the molecule, and write the structures and IUPAC names of its two position isomers.
A sample of polyethene has an average molar mass of \(28000\ \text{g mol}^{-1}\). Calculate the average number of ethene units, n, in one molecule of this polymer. (Molar mass of \(\mathrm{C_2H_4} = 28\ \text{g mol}^{-1}\).)
15.8 Case-Based Questions
- Read the following reaction and answer the questions given below. \[\mathrm{CH_3\!-\!C(CH_3){=}CH_2} + \mathrm{HBr} \xrightarrow{\ \text{benzoyl peroxide}\ } \mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2Br}\]
- Write IUPAC name of the product.
- State the rule that governs formation of this product.
- Read the following reaction and answer the questions given below. \[\mathrm{C_6H_6} + \mathrm{CH_3Cl} \xrightarrow{\ \text{anhydrous } \mathrm{AlCl_3}\ } \mathrm{C_6H_5CH_3} + \mathrm{HCl}\]
- Write the name of the reaction.
- Identify the electrophile in it.
- How is this electrophile generated?
15.9 Objective Questions
- Which of the following compound has highest boiling point?
- n-pentane
- iso-butane
- butane
- neopentane
- Acidic hydrogen is present in:
- acetylene
- ethane
- ethylene
- dimethyl acetylene
- Identify ‘A’ in the following reaction: \[\mathrm{CH_3\!-\!C(CH_3){=}CH_2} \xrightarrow{\ \mathrm{A}\ } \mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{CO_2} + \mathrm{H_2O}\]
- \(\mathrm{KMnO_4}/\mathrm{H^{+}}\)
- alkaline \(\mathrm{KMnO_4}\)
- dil. \(\mathrm{H_2SO_4}\)/1% \(\mathrm{HgSO_4}\)
- \(\mathrm{NaOH}/\mathrm{H_2O_2}\)
- Major product of chlorination of ethyl benzene is:
- m-chloroethyl benzene
- p-chloroethyl benzene
- chlorobenzene
- o-chloroethylbenzene
- 1-chloropropane on treatment with alc. KOH produces:
- propane
- propene
- propyne
- propyl alcohol