Hydrocarbons

Hydrocarbons are a class of organic compounds that consist of only carbon and hydrogen. They are found abundantly in petroleum and natural gas and are frequently used as domestic and industrial fuels. Unsaturated hydrocarbons, especially alkenes, are used as starting materials for many industrial products — most products in the form of medicines, plastics, perfumes and polymers have their origins in hydrocarbons.

Hydrocarbons are broadly divided into two classes:

• Aliphatic hydrocarbons — compounds that are not aromatic. They may be open-chain or cyclic, and saturated or unsaturated. If there are only carbon and hydrogen atoms in an aliphatic compound, it is called an aliphatic hydrocarbon. Alkanes are acyclic aliphatic hydrocarbons, whereas cycloalkanes are cyclic aliphatic hydrocarbons. Alkenes and alkynes are acyclic and unsaturated aliphatic hydrocarbons. Examples: methane, ethane, ethene, ethyne, cyclopropane, cyclobutane, cyclopentane, cyclohexane.
• Aromatic hydrocarbons — a special class of cyclic hydrocarbons that have a high carbon-to-hydrogen ratio and are based on benzene (C₆H₆) or resembling compounds. Benzene is the basic aromatic compound; other compounds such as toluene and phenol are its derivatives. Examples: benzene, methylbenzene (toluene), ethylbenzene.

📌 Interesting Information (book's sidebar): the term "aromatic" was derived from the Greek word aroma meaning "fragrance", and is used in organic chemistry for a special class of compounds (aromatic compounds) having characteristic odour.

Alkanes are saturated hydrocarbons with the general formula C_nH_2n+2. Methane (CH₄) is the simplest member of this family. The homologous series of alkanes with condensed structural formulae and molecular formulae is:

• Methane CH₄ — CH₄; Ethane C₂H₆ — CH₃CH₃; Propane C₃H₈ — CH₃CH₂CH₃; Butane C₄H₁₀ — CH₃CH₂CH₂CH₃; Pentane C₅H₁₂; Hexane C₆H₁₄; Heptane C₇H₁₆; Octane C₈H₁₈; Nonane C₉H₂₀; Decane C₁₀H₂₂.

What we obtain after the removal of one hydrogen atom from an alkane is called an alkyl group. Alkyl groups are named by replacing –ane of the alkane with –yl. Examples: methane → methyl (CH₃–, Me–); ethane → ethyl (CH₃–CH₂–, Et–).

For alkanes with more than two carbon atoms, more than one alkyl group can be derived. From propane: n-propyl (CH₃CH₂CH₂–) and iso-propyl (CH₃CH(CH₃)–, systematic name 1-methylethyl). From butane: n-butyl (butyl), sec-butyl (1-methylpropyl), iso-butyl (2-methylpropyl) and tert-butyl (1,1-dimethylpropyl).

📌 Did You Know? (book's sidebar): Butane is a fuel used in lighters. Isobutane is used as a propellant in products such as shaving gel.

Branched-chain alkanes are named according to the following rules:

1. Locate the longest continuous chain of carbon atoms; this chain determines the parent name. The longest chain may not always be obvious from the way the formula is written — a bent chain can still be the longest (e.g. a bent six-carbon chain is a hexane; a bent seven-carbon chain is a heptane).
2. Number the longest chain from the end nearest to the substituent. E.g. the same skeleton numbered one way gives 3-methylheptane and the other way 2-methylhexane — the correct numbering is the one that gives the substituent the lowest number.
3. Use the numbers from rule 2 to designate the location of the substituent. The parent name is placed last, and the substituent group, preceded by the number designating its location, is placed first. Example: 3-methylheptane; 2-methylhexane.
4. When two or more substituents are present, give each substituent a number corresponding to its location on the longest chain. Example: 4-ethyl-2-methylhexane. Substituent groups are listed alphabetically (ethyl before methyl); while deciding alphabetical order, ignore multiplying prefixes such as "di" and "tri".
5. When two substituents are present on the same carbon atom, use that number twice. Example: 3-ethyl-3-methylhexane.
6. When two or more substituents are identical, indicate this by the prefixes di, tri, tetra and so on, and make certain that each and every substituent has a number. Examples: 2,3-dimethylbutane; 2,3,4-trimethylpentane; 2,2,3,3-tetramethylbutane.
7. When two chains of equal length compete for selection as the parent chain, choose the chain with the greater number of substituents. Example: 2,3,5-trimethyl-4-propylheptane.
8. When branching first occurs at an equal distance from either end of the longest chain, choose the name that gives the lower number at the first point of difference. Example: 2,3,5-trimethylhexane (not 2,4,5-trimethylhexane).

Applying these eight rules allows us to name most of the alkanes that we shall encounter.

11.2.3 Nomenclature of cycloalkanes: Cycloalkanes are named by attaching the prefix "cyclo" to the names of the alkanes possessing the same number of carbon atoms. Examples: ethylcyclohexane; isopropylcyclohexane (IUPAC name: 1-methylethylcyclohexane); 1,1,4-trimethylcyclohexane; n-propylcyclohexane (IUPAC name: propylcyclohexane).

11.2.4 Shapes of alkanes and cycloalkanes: The carbon atom in an alkane has four sp³ hybridized orbitals and forms four single bonds that are equidistant from each other. Alkane molecules have a tetrahedral shape around carbon atoms. Carbon atoms in alkane molecules form sigma bonds with surrounding carbon and hydrogen atoms. As a result, carbon atoms in alkanes are surrounded by 4 pairs of bonding electrons which equally repel each other to form 109.5° bond angles and a C–C bond length of 1.54 Å. The shapes of organic molecules are generally represented using the ball and stick formula. Straight-chain alkanes: methane (CH₄), ethane (C₂H₆), propane (C₃H₈), butane (C₄H₁₀).

Cycloalkanes are cyclic, meaning that the carbon atoms of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbon atoms that make up the ring are singly bonded to other atoms. Structural shapes of cyclopropane (C₃H₆), cyclobutane (C₄H₈) and cyclopentane (C₅H₁₀) are shown in the book's Table 11.5.

📌 Quick Check 11.1 (book, p. 241): (a) structural formulae for 2-methylbutane and 2,2-dimethylpropane; (b) displayed formula of 2,3,3-trimethylhexane and methylcyclohexane; (c) two differences between cyclopentane and pentane; (d) eicosane is a straight-chain alkane whose molecules contain 20 carbon atoms — its molecular formula is C₂₀H₄₂ (from C_nH_2n+2).

1. Alkanes containing up to four carbon atoms are gases, while pentane to heptadecane (C₅ to C₁₇) are liquids. The higher members from C₁₈ onwards are waxy solids. Alkanes are colourless and odourless.
2. Alkanes are non-polar or very weakly polar and are insoluble in polar solvents like water, but soluble in non-polar solvents, such as hexane, benzene, ether, etc.
3. Their physical constants, e.g. boiling points, melting points, density, etc., increase with the increase in number of carbon atoms, whereas solubility decreases with the increase in molecular mass. The boiling points of alkanes having branched chain structures are lower than their isomeric normal chain alkanes, e.g. straight-chain butane has a higher boiling point, −0.50 °C, than 2-methylpropane (isobutane, −11.7 °C).

📌 Quick Check 11.2 (book, p. 243): (a) Branched alkanes have lower boiling points because branching makes the molecule more compact and spherical, so the surface area of contact between molecules falls and the weak intermolecular forces become weaker — less heat is needed to separate them. (b) Alkanes have a tetrahedral shape because each carbon is sp³ hybridized with four pairs of bonding electrons that repel each other equally (see sec4). (c) Cyclopropane is a three-membered ring (triangle) and cyclopentane a five-membered ring (pentagon) of CH₂ groups.

Reaction mechanism is the series of steps that take place in the course of the overall reaction. Like all chemical reactions, organic reactions involve the breaking of chemical bonds. There are two ways in which a covalent bond can break:

• (a) Homolytic fission: the covalent bond breaks evenly to produce free radicals. A free radical is a species that contains an unpaired electron and is very reactive. Free radicals are extremely reactive because of the tendency of unpaired electrons to become paired. Example: Cl–Cl —(hν)→ Cl• + Cl•. Homolytic fission is shown by half curly arrows (fish-hooks).
• (b) Heterolytic fission: the covalent bond breaks in such a way that the originally shared pair of electrons is gained by one atom only and oppositely charged ions are produced. This bond fission is shown by full curly arrows in mechanisms. Example: H–CH₂–Cl → H–CH₂⁺ + :Cl⁻ (carbocation + chloride ion).

11.4 Unreactive nature of alkanes towards polar reagents: Alkanes or paraffins (Latin: parum = little, affinis = affinity) are inert towards acids, alkalis, oxidizing and reducing agents under ordinary conditions. The lack of reactivity of alkanes under normal conditions may be explained on the basis of the non-polarity of the bonds forming them. The electronegativity values of carbon (2.5) and hydrogen (2.1) do not differ appreciably, and the bonding electrons between C–H and C–C are equally shared, making them almost non-polar and generally unreactive. The unreactive nature of alkanes can also be explained on the basis of inertness of a sigma bond: in a sigma bond the electrons are very tightly held between the nuclei, which makes it a very stable bond. A lot of energy is required to break it.

📌 Did You Know? (book's sidebar): Hydrocarbon combustion produces CO₂, a greenhouse gas contributing to climate change. Incomplete combustion can release carbon monoxide and particulate matter, which have serious health and environmental consequences.

Under high temperature or suitable conditions, alkanes undergo two types of reactions: (i) thermal and catalytic reactions, e.g. combustion and cracking; (ii) free radical substitution reactions — reactions that involve replacement of one atom, or group of atoms, by another through the free radical mechanism.

11.5.1 Free radical substitution reactions: Substitution reactions of alkanes take place in the presence of light energy through the formation of highly reactive free radicals. An example is the substitution of a halogen atom for a hydrogen of an alkane, called halogenation. Alkanes react with chlorine and bromine in the presence of sunlight or UV light, resulting in the successive replacement of hydrogen atoms with halogens. The extent of halogenation depends upon the amount of halogen used. Reaction of alkanes with fluorine is highly violent and results in a mixture of carbon, fluorinated alkanes and hydrofluoric acid. Iodine does not substitute directly because the reaction is too slow and reversible. The order of reactivity of halogens is: F₂ > Cl₂ > Br₂ > I₂.

Halogenation involves three steps:

• i) Initiation step: the first step in free radical substitution of alkanes by halogens. It involves breaking the halogen–halogen bond homolytically using energy from UV light: Cl–Cl —(hν)→ Cl• + Cl•.
• ii) Propagation step: the initially formed radicals attack an alkane molecule, generating more free radicals: CH₄ + Cl• → CH₃• + HCl; then CH₃• + Cl–Cl → CH₃Cl + Cl•. The new chlorine radical continues the chain.
• iii) Termination step: the final step, in which two radicals react together to form a product: CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → CH₃CH₃ (ethane, double the carbon atoms — rare, formed in traces, but its presence is evidence of the free radical mechanism); Cl• + Cl• → Cl₂.

By repetition of the propagation step, a mixture of halogen-substituted products (chloroalkanes) is obtained. The more chlorine gas in the reaction mixture to start with, the greater the proportions of chlorine atoms in the chloroalkane molecules formed: CH₃Cl (chloromethane) → CH₂Cl₂ (dichloromethane) → CHCl₃ (trichloromethane, also known as chloroform) → CCl₄ (tetrachloromethane, also known as carbon tetrachloride).

📌 Interesting Information (book's sidebar): Haloalkanes are anaesthetic in nature. The anaesthetic power of a haloalkane increases with the number of halogen atoms. Use of these as organic solvents in the lab is strictly monitored due to their toxic nature.

📌 Quick Check 11.3 (book, p. 246): (a) Alkanes have high stability because their C–C and C–H bonds are strong, non-polar sigma bonds (sec6). (b) Two types: thermal/catalytic (combustion, cracking) and free radical substitution. (c) Termination: two radicals combine (see step iii above). (d) A mixture of halogenoalkanes is obtained when excess halogen is present and propagation repeats — more halogen gives more highly substituted products. (e) Free radical chlorination of ethane gives chloroethane first, then dichloroethanes, trichloroethane and so on by successive substitution, exactly as for methane.

11.6.1 Nomenclature of alkenes: Alkenes are unsaturated hydrocarbons. Unsaturated hydrocarbons are compounds of hydrogen and carbon only whose molecules contain carbon-to-carbon double bonds (or triple bonds). Alkenes have two hydrogen atoms less than the corresponding saturated hydrocarbons, with general formula C_nH_2n. The IUPAC rules for naming alkenes are similar in many respects to those for naming alkanes:

1. Select the longest continuous chain that contains the C=C as the parent chain. Change the ending of the name of the alkane of identical length from –ane to –ene, e.g. 3-propyl-2-hexene.
2. Number the chain so as to include both carbon atoms of the double bond. Numbering begins from the end nearer to the double bond, e.g. 5-methyl-2-hexene.
3. Designate the location of the double bond by using the number of the first atom of the double bond as a prefix, e.g. 1-butene; 1-pentene.
4. Indicate the locations of the substituent groups by the numbers of the carbon atoms to which they are attached, e.g. 2-methyl-2-butene; 2,5,5-trimethyl-2-hexene.

11.6.2 Preparation of alkenes: Alkenes are prepared by the removal of small atoms (H, OH, X) from the adjacent carbon atoms of the saturated compounds, so as to create a double bond between carbon atoms. These reactions are known as elimination reactions. In such reactions, a small molecule such as H₂O or HCl is removed from adjacent carbons in an organic molecule. Preparation of ethene through various methods is discussed below:

• i) Dehydration of alcohols: Ethene is prepared by heating a mixture of ethanol and excess of concentrated sulphuric acid at 180 °C. The reaction involves the removal of a water molecule from the alcohol molecule: CH₃CH₂OH —(H₂SO₄, heat)→ CH₂=CH₂ + H₂O. In the first step, ethyl hydrogen sulphate is formed, which decomposes on heating to produce ethene: CH₃CH₂OH + H₂SO₄ —(180 °C)→ CH₃CH₂OSO₃H + H₂O; CH₃CH₂OSO₃H —(heat)→ H₂C=CH₂ + H₂SO₄.
• ii) Dehydrohalogenation of alkyl halides: On heating, ethyl bromide with alcoholic KOH gives ethene. Removal of hydrogen and halogen takes place from adjacent carbon atoms to create a double bond: H₃C–CH₂–Br + KOH_(alcoholic) —(heat)→ H₂C=CH₂ + KBr + H₂O.

11.6.3 Physical properties: (i) First three members, i.e. ethene, propene and butene, are gases at room temperature while C₅ to C₁₅ are liquids and the higher members are solids. (ii) They are insoluble in water but soluble in alcohol. (iii) They have characteristic smell and burn with luminous flame. (iv) Unlike alkanes, they show weakly polar properties because of sp² hybridization.

📌 Quick Check 11.4 (book, p. 248): (a) Propanol with H₂SO₄ undergoes dehydration like ethanol, giving propene: CH₃CH₂CH₂OH —(conc. H₂SO₄, heat)→ CH₃CH=CH₂ + H₂O (via propyl hydrogen sulphate). (b) Naming: (i) CH₃C(CH₃)=CH–CH₃ is 2-methyl-2-butene; (ii) CH₂=CHCH(CH₃)₂ is 3-methyl-1-butene. (c) Structures: (i) 3-methyl-2-butene is CH₃CH=C(CH₃)CH₃; (ii) 2-methyl-4-ethyl-2-pentene is CH₃C(CH₃)=CH–CH(C₂H₅)–CH₃. (d) Dehydrohalogenation of bromopropane: CH₃CH₂CH₂Br + KOH_(alcoholic) —(heat)→ CH₃CH=CH₂ + KBr + H₂O. (e) Yes, propanol can undergo dehydration — see (a).

11.7.1 Structure of alkenes: Ethene (C₂H₄) is the simplest alkene. Each carbon atom in ethene is sp² hybridized and forms three equivalent sp² hybrid orbitals which lie in the same plane. The unhybrid p orbital lies perpendicular (90°) to the plane of hybrid orbitals. These sp² orbitals are formed by mixing of the 2s orbital with two 2p orbitals, leaving one unhybrid p orbital. In ethene, each carbon uses two sp² orbitals to form two C–H bonds each. The remaining sp² orbital on each carbon is used to form a carbon–carbon σ bond by linear overlap.

The unhybrid p orbitals of each carbon atom, which are parallel to each other, overlap laterally to form another carbon–carbon bond. This bond formed by the lateral overlap of p orbitals is called a π bond. The two carbon atoms forming a double bond and the four atoms attached directly to them lie in the same plane. The sp² hybridization results in a trigonal planar arrangement around each carbon atom. The bond angles are approximately 120°, leading to a planar molecule.

11.7.2 Reactivity of π bond: Alkenes are more reactive than alkanes due to the presence of the double bond. The probability of finding electrons away from the line joining the two nuclei is high. Due to this reason, π-electrons are less firmly held between the nuclei. A π bond is, therefore, a weak bond as compared to a σ bond. During a reaction it breaks comparatively easily, rendering alkenes as reactive group of compounds. Moreover, the loosely held π-electrons are more exposed to attack by the electrophilic reagents. Alkenes, therefore, undergo electrophilic addition reactions very easily.

📌 Did You Know? (book's sidebar): Electrophile is a species that can act as electron pair acceptor; it is represented by E⁺. Alkenes act as nucleophile because they contain π electrons. Nucleophile is a species that can act as a donor of a pair of electrons; it is represented by Nu⁻.

📌 Interesting Information (book's sidebar): Plants produce ethene as a natural ripening hormone. For efficient harvesting and transportation to market, fruits and vegetables are often picked unripe and exposed to ethene so they could ripen in short time.

📌 Quick Check 11.5 (book, p. 250): (a) An alkane differs from an alkene in stability because the alkane has only strong σ bonds, while the alkene's π bond is weaker and breaks easily, making alkenes more reactive. (b) Nucleophile: OH⁻ (electron-pair donor); H₂ has no lone pair to donate and H⁺ accepts rather than donates. (c) Electrophile: H⁺ (electron-pair acceptor); OH⁻ donates rather than accepts, and H₂ is neutral with no electron deficiency.

The reactions of alkenes mostly proceed through the formation of carbocations. Therefore, before studying the addition reactions of alkenes, the concept of carbocations and their stability is very important.

11.7.3 Carbocation stability and inductive effect of alkyl group: A carbocation is an alkyl group with a single positive charge on one of its carbon atoms. Carbocations are classified into the following types:

• (i) A carbocation bonded to H only is named as methyl carbocation.
• (ii) If a carbocation is bonded to one C or alkyl substituent, it is named as primary (1°) carbocation.
• (iii) If it is bonded to 2 carbons (alkyl substituents) directly, it is named as secondary (2°) carbocation.
• (iv) If it is directly bonded to 3 carbons (alkyl substituents), it is a tertiary (3°) carbocation.

The inductive effect of alkyl groups plays an important role in the stability of carbocation. The polarization of a σ bond due to electron withdrawing or electron donating effect of adjacent groups or atoms is referred to as inductive effect. Alkyl groups have a slightly electron donating inductive effect. Electron donating species are said to have positive inductive effect (+I), whereas electron withdrawing species, such as a halogen atom, have a negative inductive effect (−I).

The alkyl groups attached to the positively charged carbon atoms are electron donating groups. Due to the positive charge on the carbon atom, carbocations are electrophile. Inductive effect is shown by the arrowheads on the bonds, to show the alkyl groups pushing electrons towards the positively charged carbon. As a result, the carbocation becomes less positively charged, which makes it energetically more stable. The stability of carbocation ions increases with the number of alkyl groups due to their +I effect. This means that tertiary carbocation is the most stable, as they have three electron-donating alkyl groups which energetically stabilize the carbocation. Stability order: 3° > 2° > 1° > methyl.

11.7.4 Electron withdrawing inductive effect: Electron withdrawing groups, such as a halogen atom, have a negative inductive effect (−I). Halogen withdraws bonded electrons from carbon to carbon bond and thus displays negative inductive effect. This causes a permanent dipole to arise in the molecule, wherein the halogen atom holds a negative charge and carbon becomes partial positive. The electron-deficient C₁ pulls electrons from the C₁–C₂ covalent bond, and the effect weakens along the chain (C₂ less positive than C₁).

📌 Quick Check 11.6 (book, p. 252): (a) Alkenes react with an electrophile because their loosely held π electrons attack the electrophile, forming a carbocation intermediate which is then attacked by a nucleophile (sec11). (b) The stability order is 3° > 2° > 1° because more alkyl groups donate more electrons by the +I effect, spreading and reducing the positive charge.

Alkenes are unsaturated hydrocarbons and they undergo electrophilic addition reactions mostly. Other minor types of reactions include oxidation, combustion and polymerization reactions.

11.8.1 Electrophilic addition reactions: The reactions in which an electrophile is added to the double bond of an alkene are called electrophilic addition reactions. Mechanism of electrophilic addition in alkenes: The C=C in alkenes is a region of high electron density, making it susceptible to attack by electrophiles. An alkene reacts with an electrophile to give a carbocation intermediate that then reacts with a nucleophile. In the generalized mechanism, X⁺ represents an electrophile and Y⁻ represents a nucleophile.

1. Halogenation: Halogenation means addition of halogens like chlorine or bromine. When bromine water (a solution of bromine in water having red-brown colour) is added to ethene in an inert solvent like carbon tetrachloride, its reddish brown colour is discharged at once. This reaction is used to identify the unsaturation of an organic compound: CH₂=CH₂ + Br₂ —(CCl₄)→ Br–CH₂–CH₂–Br (1,2-dibromoethane). Mechanism (three steps): (i) Bromine (Br₂) is a non-polar molecule; however, when a bromine molecule gets closer to the double bond of an alkene, the high electron density of the double bond repels the electron pair in Br–Br away from the Br atom. As a result, the Br atom closest to the double bond is slightly positive and acts as an electrophile (induced dipole). (ii) The δ⁺ bromine of Br₂ is attacked by the π electrons in the double bond, and the partial positive bromine accepts a pair of electrons from the C=C bond in the alkene. This results in the formation of a highly reactive carbocation intermediate. (iii) This carbocation intermediate reacts with the Br⁻ (nucleophile) to make the product.

2. Hydrohalogenation: When an alkene reacts with dry gaseous hydrogen halide at room temperature, the product is a haloalkane: CH₂=CH₂ + HBr → CH₃–CH₂–Br (bromoethane). Hydrogen halides are among the simplest examples of polar substances having permanent dipoles that add to alkenes. All hydrogen halides (HBr, HCl, HI, HF) give this reaction and add across C=C exactly the same way. But the rate of reaction increases from HF to HI. The actual product formed depends upon whether the alkene is symmetrical or unsymmetrical.

i) Addition to symmetrical alkenes: When the alkene (e.g. ethene) is symmetrical, only one product is theoretically possible. Mechanism of addition of HBr to ethene: (i) Nucleophilic π-bond attacks the hydrogen in hydrogen bromide. Hydrogen bromide gives a bromide ion (Br⁻). Ethene is converted to carbocation by adding hydrogen to one of the double bonded carbons. (ii) The bromide ion (nucleophile) attacks the carbocation to give the addition product, 1-bromoethane (alkyl halide).

ii) Addition to unsymmetrical alkenes — Markovnikov's Rule: When the alkene is unsymmetrical, two products are theoretically possible. For example, the addition of HBr to propene can give two products. But experimentally, it has been found that under these conditions, the major product is 2-bromopropane and the minor product is 1-bromopropane. Markovnikov's Rule is applicable for addition reactions in alkenes. According to this rule, "When a polar reagent adds to an unsymmetrical alkene, the negative part of the reagent attaches to the double bonded carbon which bears the lesser number of hydrogen." Mechanism: Propene reacts with HBr in two ways. (i) The π electrons of C=C attack the partial positive end of H–Br. Hydrogen bromide breaks heterolytically, creating bromide ion (electrophile) and hydrogen is added to double bonded carbon with more number of hydrogen, creating a 2° carbocation. (ii) The bromide ion attacks the secondary carbocation to give the major product. The minor product is formed by the attack on primary carbocation.

3. Hydrogenation: Hydrogenation means addition of molecular hydrogen to an alkene in the presence of a catalyst (Ni/Pt) to form a saturated compound at 250–300 °C: H₂C=CH₂ + H₂ —(Ni, 250–300 °C)→ H₃C–CH₃. On industrial scale, this reaction is used to convert vegetable oil into margarine (Banaspati ghee): Oil + H₂ —(Ni)→ Margarine.

4. Hydration: Gaseous alkenes react with steam at high temperature and pressure in the presence of concentrated acid (H₂SO₄) as the catalyst to form alcohols: H₂C=CH₂ + H₂O —(H₂SO₄)→ C₂H₅OH. Similar to hydrohalogenation, the alkene forms a carbocation intermediate, which is then attacked by OH⁻.

5. Halohydration: Halohydration is a reaction where an alkene reacts with a halogen (such as bromine or chlorine) in the presence of water to form a halohydrin. For ethene (C₂H₄), the reaction specifically produces a bromohydrin or chlorohydrin, depending on the halogen used: H₂C=CH₂ + Br₂ + H₂O → CH₃–CH(Br)–OH (1-bromoethanol, as drawn in the book) + HBr.

📌 Quick Check 11.7 (book, p. 254): (a) The relative rates of addition of HCl, HBr and HI to alkenes increase from HCl to HI — the book states the rate of hydrohalogenation increases from HF to HI, because the H–X bond becomes weaker and easier to break heterolytically down the group. (b) Addition vs elimination: addition adds atoms across the double bond (alkene → single product, saturation increases); elimination removes a small molecule from adjacent carbons and creates a double bond (the exact opposite). (c) An alkene reacts with an electrophile as in sec11's generalized mechanism: π electrons attack E⁺, a carbocation forms, then a nucleophile attacks it. (d) Markovnikov's rule applied to HBr + 2-pentene: 2-pentene (CH₃CH=CHCH₂CH₃) is unsymmetrical in the Markovnikov sense — H⁺ adds to give the more stable 2° carbocation, and Br⁻ attaches to the carbon bearing fewer hydrogens, so the major product is 3-bromopentane (with 2-bromopentane as the minor product). ⚠️ The book gives no worked example for 2-pentene; this answer applies the chapter's own Markovnikov mechanism, and both possible carbocations here are secondary, so the preference is weaker than in the propene case.

6. Epoxidation: The epoxidation of alkenes gives an oxygen-containing three-membered cyclic ether called epoxide. In the case of ethene (C₂H₄), this reaction can be accomplished using various oxidizing agents. Ethene can also be converted to ethylene oxide using molecular oxygen and metal catalysts. A transition metal catalyst facilitates the reaction, forming an epoxide directly from the alkene. The epoxides can be converted into diols by acid hydrolysis: ethene + O₂ —(Ag)→ epoxide —(H⁺/H₂O)→ ethane-1,2-diol.

7. Ozonolysis: Ozonolysis is a method of oxidative cleavage of alkenes using ozone (O₃), a reactive allotrope of oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Alkenes react with ozone (O₃) to form ozonides, which can be further reduced to carbonyl compounds (aldehydes or ketones), and ultimately to alcohols. Ethene: C₂H₄ + O₃ → C₂H₄O₃ → 2CH₂O (methanal). Propene: H₃C–CH=CH₂ —(O₃)→ ozonide —(Zn/H₂O)→ ethanal (CH₃CHO) + methanal (HCHO), which on reduction give ethanol and methanol.

8. Polymerization: Polymerization is the formation of extremely long molecules (polymers) from small reactive molecules that join together (monomers). Addition polymerization is one of the most important addition reactions of alkenes which form the basis of the plastics industry. Addition polymerization is the reaction in which many monomers containing at least one C=C double bond form long chains of polymers as the only product. Just like in other addition reactions of alkenes, the π-bond in each C=C bond breaks and then the monomers link together to form new C–C single bonds. A polymer is a long-chain molecule that is made up of many repeating units. Examples: ethene (monomer) → poly(ethene) (polymer, polyethylene, one of the world's most common plastics used in bags, bottles and packaging); chloroethene (monomer) → poly(chloroethene) (PVC, used e.g. in ball pens).

Deducing repeating units: A repeating unit is the smallest group of atoms that, when connected one after the other, make up the polymer chain. It is represented by square brackets in the displayed and general formula. In poly(alkenes) (such as poly(ethene)) and substituted poly(alkenes) (such as PVC) made of one type of monomer, the repeating unit is the same as the monomer except that the C=C double bond is changed to a C–C single bond — for poly(ethene), –CH₂CH₂–.

📌 Quick Check 11.8 (book, p. 256): (a) Ozonolysis of ethene: C₂H₄ + O₃ → ozonide —(Zn/H₂O)→ 2 methanal. (b) Epoxidation of 2-butene: CH₃CH=CHCH₃ + O₂ —(metal catalyst)→ 2,3-epoxybutane (the three-membered epoxide ring across C2–C3), which on acid hydrolysis gives butane-2,3-diol. (c) Ozonolysis locates a double bond because cleavage happens exactly at the C=C: each carbon of the double bond becomes a carbonyl carbon, so the carbonyl products reveal which carbons were joined — e.g. propene gives ethanal + methanal, proving the double bond was between C1 and C2.

📌 Quick Check 11.9 (book, p. 257): (a) (i) Ethene is suitable for polymerization because its C=C π bond breaks easily and each carbon can then form a new C–C single bond to the next monomer, extending the chain. (ii) During this reaction the hybridization changes from sp² (trigonal planar, ~120°) to sp³ (tetrahedral, ~109.5°) as the double bond becomes a single bond. (b) An addition reaction increases saturation because the π bond is lost and each double-bonded carbon gains one more single-bonded partner (H or another carbon), so the molecule becomes more saturated.

11.9 Conjugated dienes: As the name indicates, a diene is a molecule containing two double bonds (C=C). In a conjugated diene, the double bonds in the carbon chain are separated by a single bond. The adjacent double bonds in conjugated dienes allow for the overlap of p-orbitals on three or more adjacent atoms. This overlap leads to a delocalization of electrons, which can stabilize the molecule and affect its reactivity. In conjugated dienes, π-bond overlap extends over the entire system. Examples: but-1,3-diene CH₂=CH–CH=CH₂; 2-methylbut-1,3-diene (isoprene) CH₂=C(CH₃)–CH=CH₂. Conjugated alkenes are an important class of compounds in organic chemistry, featuring unique properties due to the arrangement of their double bonds. Conjugated dienes like 1,3-butadiene are used in the production of synthetic rubber. The conjugated structure allows for polymerization, leading to long chains that exhibit rubber-like properties.

📌 Quick Check 11.10 (book, p. 257): (a) Penta-1,3-diene: CH₂=CH–CH=CH–CH₃ (double bonds separated by one single bond — conjugated). (b) Delocalization in conjugated dienes: the p-orbitals of all four carbons overlap continuously, so the π electrons are spread over the whole system rather than confined to one double bond, which stabilizes the molecule.

11.10 Isomerism: The concept of isomerism is an important feature of organic compounds. Two or more compounds having the same molecular formula but different structural formulae and properties are said to be isomers and the phenomenon is called isomerism. The structural formula of a compound shows the arrangement of atoms and bonds present in it. The simplest hydrocarbon to have structural isomers is butane (C₄H₁₀). If we study the structural formula of butane or other higher hydrocarbons of the alkane family, we will observe that it is possible to arrange the atoms present in the molecule in more than one way to satisfy all valencies. This means that it is possible to have two or more different arrangements for the same molecular formula. For example, chlorobutane molecule can have two different arrangements as represented by 1-chlorobutane (CH₃CH₂CH₂CH₂Cl) and 2-chlorobutane (CH₃CHClCH₂CH₃). This fact has been supported by experimental evidence that there are two compounds with different physical properties but with the same molecular formula of C₄H₉Cl. Isomerism is not only possible but common if the compound contains more than three carbon atoms. As the number of carbon atoms in a hydrocarbon increases, the number of possible isomers increases very rapidly. The five carbon compound, pentane, has three isomers.

11.10.1 Types of isomerism — A. Structural isomerism: Structural isomerism is not confined to hydrocarbons only. In fact, all classes of organic compounds and their derivatives show the phenomenon of structural isomerism. Structural isomerism arises due to the difference in the arrangement of atoms within the molecule. Structural isomerism can be exhibited in five different ways:

• i) Chain isomerism: arises due to the difference in the nature of the carbon chain. Examples: butane vs 2-methylpropane (isobutane); pentane, 2-methylbutane (isopentane), 2,2-dimethylpropane (neopentane).
• ii) Position isomerism: arises due to the difference in the position of the same functional group on the carbon chain, while the arrangement of carbon atoms remains the same. Examples: 1-chloropropane vs 2-chloropropane; but-1-ene vs but-2-ene.
• iii) Functional group isomerism: compounds having the same molecular formula but different functional groups. Example: methoxy methane (H₃C–O–CH₃) vs ethanol (H₃C–CH₂–OH), both C₂H₆O.
• iv) Metamerism: arises due to the unequal distribution of carbon atoms on either side of the functional group. Such compounds belong to the same homologous series. Examples: ethoxy ethane (H₃C–H₂C–O–CH₂–CH₃) vs methoxy propane (H₃C–O–CH₂–CH₂–CH₃); for a ketonic compound C₅H₁₀O: 3-pentanone vs 2-pentanone.
• v) Tautomerism: arises due to shifting of a proton from one atom to another within the same molecule. This is commonly seen in amino acids. They exist as zwitterions, which are formed when a proton is shifted from the carboxylic acid end of an amino acid molecule to the amine group.

B. Stereoisomerism: Stereoisomers are such compounds which possess the same structural formula, but differ with respect to the positions of the identical groups in space. There are two types of stereoisomerism: (i) geometrical or cis–trans isomerism; (ii) optical isomerism. ⚠️ The book only names these two types here — it does not develop cis–trans or optical isomerism further in this chapter, so no further detail is added.

11.11 Organic redox reactions: Oxidation–reduction reactions that involve organic compounds are called organic redox reactions. These reactions are characterized by the addition or removal of atoms/bonds. Normally, during organic oxidation, oxygen is added or hydrogen is removed. During organic reduction, hydrogen is added or oxygen is removed. Some examples of oxidation and reduction reactions are given below:

• 11.11.1 Oxidation: Oxidation of organic compounds involving reaction with oxygen is usually the combustion reaction: 2C₂H₅ + 7O₂ → 4CO₂ + 6H₂O (as printed in the book; read as the combustion of ethane, C₂H₆). Reactions of organic compounds with oxidizing agents, e.g.: (i) Ozonolysis is a method of oxidative cleavage of alkenes using ozone (O₃) — propene gives ethanal (acetaldehyde) + methanal (formaldehyde), reducible to ethanol and methanol. (ii) The epoxidation of alkene is also an example of oxidation reaction — ethene + O₂ → epoxide —(H⁺/H₂O)→ ethane-1,2-diol. (iv) Reaction of primary alcohols with acidified K₂Cr₂O₇ to make carboxylic acids — detail will be discussed in class 12 syllabus: C₂H₅OH + 2[O] —(K₂Cr₂O₇/H₂SO₄)→ CH₃COOH + H₂O. ⚠️ The book numbers these (i), (ii), (iv) — point (iii) is missing in the printed book; nothing has been invented to fill it.
• 11.11.2 Reduction: Addition of hydrogen to a carbon–carbon double bond to form an alkane is an example of organic reduction reaction: ethene + H–H —(Pt)→ ethane. Various reducing agents are used in organic synthesis, e.g. LiAlH₄, NaBH₄, Sn + conc. HCl. Reduction of nitriles to amines, amides to amines, carboxylic acid to primary alcohols, nitrobenzene to phenylamine are all examples of reduction.

📌 Quick Check 11.11 (book, p. 261): (a) Positional isomers of C₄H₈: but-1-ene and but-2-ene (displayed formulae: CH₂=CH–CH₂–CH₃ and CH₃–CH=CH–CH₃). (b) Ethoxy methane (CH₃OC₂H₅) and propanol (C₃H₇OH) — both C₃H₈O — are functional group isomers (ether vs alcohol). (c) Different types of structural isomerism in alkenes: chain, position, functional group, metamerism and tautomerism (the five types of sec11.10.1, applied to alkenes e.g. chain isomers of pentene and position isomers but-1-ene/but-2-ene). (d) Isomers of C₄H₉NH₂ (butylamine): the four chain/position amine isomers — butan-1-amine, butan-2-amine, 2-methylpropan-1-amine and 2-methylpropan-2-amine — derived by applying the chapter's chain and position isomerism rules. ⚠️ The book does not list these four names; they are derived from the chapter's own isomerism rules, not quoted from a book passage.