Alkanes

What are alkanes?

In plain words: alkanes are chains of carbon atoms joined by single bonds, with hydrogen atoms on every bond that is left over. Methane, ethane, propane and butane are all alkanes. They have no double bonds and no lone pairs for a reagent to attack, so they react only under harsh conditions: heat, light or a flame. That is also why they make such good fuels: they keep safely until they are burnt.

Structure and the homologous series

NCERT builds the family from methane: replace one hydrogen of CH₄ by a –CH₃ group and you get ethane, C₂H₆; do it again and you get C₃H₈, then C₄H₁₀, and so on (NCERT Section 9.2, page 296). Each member differs from the next by CH₂, so the alkanes form a homologous series with the general formula C_nH_2n+2. Methane itself is a gas found in coal mines and marshy places.

Methane is tetrahedral, as VSEPR theory predicts for four bond pairs: carbon at the centre, a hydrogen at each corner of a regular tetrahedron, and every H–C–H angle 109.5°. In higher alkanes these tetrahedra are joined together. Each carbon is sp³ hybridised: the C–C bonds form by head-on overlap of sp³ orbitals of two carbons, and the C–H bonds by overlap of an sp³ orbital of carbon with the 1s orbital of hydrogen. The C–C bond is 154 pm long.

NCERT page 296 gives the C–H bond length in alkanes as 112 pm. That is too long: the C–H bond in alkanes is about 109 pm, the value NCERT itself gives for ethane in Unit 4 (Section 4.6.2, page 123).

Nomenclature and isomerism

The first three alkanes, methane, ethane and propane, have only one structure each. From butane onwards the carbons can be joined in more than one way (NCERT Section 9.2.1, pages 296–298):

• C₄H₁₀: butane, CH₃CH₂CH₂CH₃ (b.p. 272.4 K), and 2-methylpropane (isobutane), (CH₃)₂CHCH₃ (b.p. 261.0 K);
• C₅H₁₂: pentane (b.p. 309.1 K), 2-methylbutane (isopentane, b.p. 300.9 K) and 2,2-dimethylpropane (neopentane, b.p. 282.5 K).

The boiling points are those of NCERT Table 9.2 (page 302); the structures on pages 296–297 print them rounded, as 273 K, 261 K, 309 K and 301 K.

These compounds have the same molecular formula but different structures and properties, so they are structural isomers. Because they differ only in the carbon chain, continuous in one and branched in the other, they are called chain isomers. The number grows fast: C₆H₁₄ has five, C₇H₁₆ nine, and C₁₀H₂₂ as many as 75 (NCERT page 298).

Degree of a carbon atom. A carbon joined to no other carbon (as in methane) or to one carbon is primary (1°); terminal carbons are always primary. A carbon joined to two carbons is secondary (2°), to three tertiary (3°), and to four quaternary (4°, or neo) (NCERT page 297). Hydrogens take the name of their carbon: 2-methylpropane has nine primary hydrogens and one tertiary hydrogen.

Alkyl groups. Removing one hydrogen from an alkane gives an alkyl group, general formula C_nH_2n+1: –CH₃ (methyl), –C₂H₅ (ethyl), –C₃H₇ (propyl or isopropyl) (NCERT page 298). The C₅H₁₁ group alone has eight isomeric forms (NCERT Problem 9.2).

Naming. Alkanes are named by the IUPAC rules of Unit 8. Pick the longest chain (if two chains are equally long, the one with more substituents), number it so that the substituents get the lowest locants, and list the substituents alphabetically, ignoring di, tri, sec and tert. To go the other way, from a name to a structure, write the parent chain, number it, attach the substituents at the numbered carbons, and finally add hydrogens until every carbon has four bonds (NCERT page 299). NCERT’s Problems 9.3–9.5 practise both directions; Example 2 below works through all three.

Preparation

Petroleum and natural gas are the main sources of alkanes, but they can also be made in the laboratory (NCERT Section 9.2.2, pages 300–301).

1. From unsaturated hydrocarbons: hydrogenation. Dihydrogen adds to alkenes and alkynes in the presence of finely divided platinum, palladium or nickel. The metal adsorbs H₂ on its surface and activates the H–H bond. Platinum and palladium work at room temperature; nickel needs a higher temperature and pressure.

• CH₂=CH₂ + H₂ → CH₃–CH₃ (ethene to ethane)
• CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃ (propene to propane)
• CH₃C≡CH + 2H₂ → CH₃CH₂CH₃ (propyne to propane)

Three slips in NCERT’s equations 9.1–9.3 on page 300: the product of 9.1 is labelled “Propane” but is ethane; the reactant of 9.2 is labelled “Propane” but is propene; and 9.3 shows only one H₂, but an alkyne needs two molecules of H₂ (one for each π bond) to become an alkane.

2. From alkyl halides.

• Reduction. Alkyl halides, except fluorides, are reduced to alkanes by zinc and dilute hydrochloric acid: CH₃Cl + H₂ → CH₄ + HCl (Zn, H⁺). Chloroethane gives ethane and 1-chloropropane gives propane in the same way.
• Wurtz reaction. Alkyl halides react with sodium in dry ether (free from moisture) to give higher alkanes with an even number of carbon atoms, two alkyl groups joined together: 2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr, and 2C₂H₅Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr (butane). NCERT’s equation 9.7 names the reactant “Bromomenthane”; it is bromomethane (menthane is a different compound).

NCERT asks what happens if two different alkyl halides are used. Beyond the NCERT text: each alkyl group can pair with its own kind or with the other, so three alkanes form together, which are hard to separate (Example 4). This is why the Wurtz reaction is used only for symmetrical alkanes.

3. From carboxylic acids.

• Decarboxylation. The sodium salt of a carboxylic acid heated with soda lime (sodium hydroxide and calcium oxide) loses carbon dioxide and gives an alkane with one carbon atom fewer than the acid: CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO, heat).
• Kolbe’s electrolytic method. Electrolysis of an aqueous solution of the sodium or potassium salt of a carboxylic acid gives, at the anode, an alkane with an even number of carbon atoms: 2CH₃COONa + 2H₂O → CH₃–CH₃ + 2CO₂ + H₂ + 2NaOH.

The path NCERT gives for Kolbe’s electrolytic method (page 301): at the anode each acetate ion gives up an electron to become an acetate free radical, which loses CO₂ to give a methyl free radical; two methyl radicals join to form ethane. At the cathode, water takes electrons and hydrogen gas is set free. Because the product is always two alkyl groups joined, the smallest alkane Kolbe’s method can give is ethane: methane cannot be prepared this way.

Physical properties

Alkanes are almost non-polar, because the C–C and C–H bonds are covalent and carbon and hydrogen differ very little in electronegativity. Their molecules are held only by weak van der Waals forces (NCERT Section 9.2.3, pages 301–302). As a result:

• at 298 K, the straight-chain alkanes C₁ to C₄ are gases, C₅ to C₁₇ are liquids and those with 18 or more carbons are solids (2,2-dimethylpropane, b.p. 282.5 K, is a gas at 298 K);
• they are colourless and odourless;
• they do not dissolve in water but do dissolve in non-polar solvents: like dissolves like. Petrol and the lower fractions of petroleum remove grease stains in dry cleaning because grease, a mixture of higher alkanes, is also non-polar and hydrophobic.

Boiling points. The boiling point rises steadily with molecular mass, because the van der Waals forces grow with the size, or surface area, of the molecule. Among isomers, branching lowers the boiling point: pentane boils at 309.1 K, 2-methylbutane at 300.9 K and 2,2-dimethylpropane at 282.5 K (NCERT Table 9.2, page 302). As branches are added, the molecule becomes more nearly spherical, the area of contact between molecules shrinks, and the weaker forces are overcome at a lower temperature.

Data from NCERT Table 9.2 (page 302), which also lists the molecular masses (16 u for methane, rising by 14 u for each CH₂); eicosane is spelt icosane in current IUPAC names. *The melting point of ethane printed in the table, 101.0 K, is wrong: ethane melts at about 90 K (−182.8 °C).

Melting points do not rise as smoothly as boiling points, because they also depend on how well the molecules pack in a crystal: compact 2,2-dimethylpropane melts at 256.4 K, far above its isomers.

Chemical properties

Alkanes are generally inert towards acids, bases, oxidising agents and reducing agents, but they react under suitable conditions (NCERT Section 9.2.3, pages 302–304).

Substitution: halogenation

One or more hydrogen atoms of an alkane can be replaced by halogens, a nitro group or a sulphonic acid group; lower alkanes, however, do not undergo nitration or sulphonation. Halogenation takes place at a high temperature (573–773 K) or in diffused sunlight or ultraviolet light. Chlorination of methane in light gives a mixture of all four chloromethanes, one hydrogen at a time:

• CH₄ + Cl₂ → CH₃Cl + HCl (chloromethane)
• CH₃Cl + Cl₂ → CH₂Cl₂ + HCl (dichloromethane)
• CH₂Cl₂ + Cl₂ → CHCl₃ + HCl (trichloromethane)
• CHCl₃ + Cl₂ → CCl₄ + HCl (tetrachloromethane)

Ethane gives chloroethane in the same way: CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl.

Two orders to remember (NCERT page 303):

• the rate of reaction with halogens is F₂ > Cl₂ > Br₂ > I₂: fluorination is too violent to control, and iodination is very slow and reversible;
• hydrogens are replaced in the order 3° > 2° > 1° (NCERT page 303 prints this as 3 > 2 > 1, with the degree signs lost).

Iodination is carried out with an oxidising agent such as HIO₃ or HNO₃, which removes the HI formed:

• CH₄ + I₂ ⇌ CH₃I + HI
• HIO₃ + 5HI → 3I₂ + 3H₂O

Removing a product pulls the reversible reaction forward, as Le Chatelier’s principle predicts.

The free radical mechanism

Halogenation goes by a free radical chain mechanism in three steps (NCERT page 303):

1. Initiation. Light or heat breaks the Cl–Cl bond by homolysis, giving two chlorine free radicals: Cl₂ → 2Cl•. The Cl–Cl bond breaks first because it is weaker than the C–C and C–H bonds.
2. Propagation. (a) A chlorine radical takes a hydrogen from methane: CH₄ + Cl• → •CH₃ + HCl. (b) The methyl radical attacks a second chlorine molecule: •CH₃ + Cl₂ → CH₃Cl + Cl•. The new chlorine radical starts step (a) again, so (a) and (b) repeat as a chain. Similar steps give the more highly chlorinated products, for example CH₃Cl + Cl• → •CH₂Cl + HCl, then •CH₂Cl + Cl₂ → CH₂Cl₂ + Cl•.
3. Termination. The chain stops when the reactants are used up or when two radicals meet: Cl• + Cl• → Cl₂; •CH₃ + •CH₃ → CH₃–CH₃; •CH₃ + Cl• → CH₃Cl. The last step makes the product, but it uses up radicals and so ends the chain.

The second termination step explains why ethane appears as a by-product when methane is chlorinated.

Combustion

Heated in air or dioxygen, alkanes burn completely to carbon dioxide and water, with the evolution of a large amount of heat (NCERT page 303). The enthalpy of combustion is large and negative, which is why alkanes are used as fuels:

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); Δ_cH^⊖ = −890 kJ mol⁻¹
• C₄H₁₀(g) + 13/2 O₂(g) → 4CO₂(g) + 5H₂O(l); Δ_cH^⊖ = −2875.84 kJ mol⁻¹

NCERT’s Unit 5 gives −2658.0 kJ mol⁻¹ for the same equation, but that figure fits water formed as vapour (about −2657 kJ mol⁻¹). With liquid water, as written, −2875.84 kJ mol⁻¹ is close to the accepted −2877 kJ mol⁻¹ worked out on the enthalpy of combustion page linked above.

Each carbon needs one O₂ and each pair of hydrogens half an O₂, so a C_n alkane needs n + (n + 1)/2 = (3n + 1)/2 molecules of O₂.

With too little air or dioxygen, combustion is incomplete and carbon black forms: CH₄(g) + O₂(g) → C(s) + 2H₂O(l). Carbon black is used to make ink, printer ink and black pigments, and as a filler, for example in rubber (NCERT prints “filters”). In NCERT’s equations 9.17 and 9.18 on page 303 the symbol Δ_cH^⊖ prints with Ä for Δ and è for ⊖ (a font fault), 9.17 has no = sign before −890, and the water in 9.17, 9.18 and 9.20 is marked (1) where (l), liquid, is meant; the same fault prints Cl as “C1” in many equations on pages 300–303.

Other reactions

Controlled oxidation. Heated with a regulated supply of dioxygen or air at high pressure over a suitable catalyst, alkanes give a variety of oxidation products (NCERT page 304):

• 2CH₄ + O₂ → 2CH₃OH (Cu, 523 K, 100 atm; methanol)
• CH₄ + O₂ → HCHO + H₂O (Mo₂O₃, heat; methanal)
• 2CH₃CH₃ + 3O₂ → 2CH₃COOH + 2H₂O ((CH₃COO)₂Mn, heat; ethanoic acid)

Ordinarily alkanes resist oxidation, but an alkane with a tertiary hydrogen is oxidised by potassium permanganate to the corresponding alcohol: (CH₃)₃CH → (CH₃)₃COH, 2-methylpropane to 2-methylpropan-2-ol. NCERT’s equation 9.24 prints the product as “2-Methylpropane-2-01”, with the digits 0 and 1 for the letters “ol” and the final e of propane kept.

Isomerisation. n-Alkanes heated with anhydrous aluminium chloride and hydrogen chloride gas rearrange to branched alkanes. n-Hexane, CH₃(CH₂)₄CH₃, gives mainly 2-methylpentane and 3-methylpentane. NCERT’s equation 9.25 prints the formula of n-hexane as “CH₃(CH)₂)₄CH₃”; it should be CH₃(CH₂)₄CH₃.

Aromatization (reforming). n-Alkanes with six or more carbons, heated to 773 K at 10–20 atm over oxides of vanadium, molybdenum or chromium supported on alumina, lose hydrogen and close into a ring, giving benzene and its homologues: C₆H₁₄ → C₆H₆ + 4H₂. NCERT’s equation 9.26 leaves out the 4H₂. NCERT asks which alkane would give toluene, C₇H₈: n-heptane (Q10).

Reaction with steam. Methane and steam at 1273 K over a nickel catalyst give carbon monoxide and dihydrogen; this is the industrial preparation of hydrogen: CH₄ + H₂O → CO + 3H₂.

Pyrolysis (cracking). Higher alkanes heated to a high temperature break into smaller alkanes, alkenes and so on. n-Hexane at 773 K gives C₆H₁₂ + H₂, or C₄H₈ + C₂H₆, or C₃H₆ + C₂H₄ + CH₄. Pyrolysis is believed to be a free radical reaction; oil gas or petrol gas is made from kerosene or petrol this way. Dodecane, a constituent of kerosene, heated to 973 K over platinum, palladium or nickel, gives heptane and pentene: C₁₂H₂₆ → C₇H₁₆ + C₅H₁₀ + other products.

Conformations

The C–C σ bond is symmetrical about the internuclear axis, so turning one carbon about this axis does not disturb the overlap. This permits rotation about the C–C single bond, and the rotation produces different arrangements of the atoms in space (NCERT Section 9.2.4, page 305):

An alkane can have an infinite number of conformations. The rotation is not completely free, though: it is hindered by a small energy barrier of 1–20 kJ mol⁻¹, caused by weak repulsion between the bonds on adjacent carbons. This repulsion is called torsional strain.

Conformations of ethane. Hold one carbon still and turn the other about the C–C axis. Of the infinite number of arrangements, two are extreme (NCERT page 305):

• eclipsed: the hydrogens on the two carbons are as close together as possible;
• staggered: they are as far apart as possible;
• any arrangement in between is a skew conformation.

In every conformation the bond angles and bond lengths stay the same; only the angle of rotation changes.

Drawing them. In a Sawhorse projection the molecule is viewed obliquely, from one side of the C–C axis rather than straight down it, and the C–C bond is drawn as a longer straight line, tilted to the right or left at its upper end. The front carbon is at the lower end and the rear carbon at the upper end; each carries three lines for its three hydrogens, at 120° to each other. In a Newman projection the C–C bond is viewed head on: the front carbon is a point, with its three hydrogens on lines at 120° to each other, and the rear carbon is a circle, with its hydrogens on shorter lines from the edge of the circle, also at 120°.

Relative stability. In the staggered form the electron clouds of the C–H bonds are as far apart as possible: repulsion is least, energy lowest and stability greatest. Turning to the eclipsed form brings the C–H bond clouds closer, so repulsion and energy rise and stability falls (NCERT page 306). The size of the torsional strain depends on the angle of rotation about the C–C bond, called the dihedral or torsional angle: 60° in the staggered form and 0° in the eclipsed form. The staggered form has the least torsional strain and is the preferred conformation.

The energy difference between the two forms is only about 12.5 kJ mol⁻¹. Even at ordinary temperatures, ethane molecules gain enough energy from collisions to cross this barrier, so rotation about the C–C bond in ethane is almost free for all practical purposes, and the conformers of ethane have never been separated and isolated.

Solved examples

Common mistakes to avoid

Quick revision summary

• Alkanes: saturated open-chain hydrocarbons, C_nH_2n+2; sp³ carbons, C–C 154 pm, C–H about 109 pm (not NCERT’s 112 pm). Alkyl groups C_nH_2n+1.
• Chain isomers: C₄ 2, C₅ 3, C₆ 5, C₇ 9, C₁₀ 75. Carbons are 1°, 2°, 3° or 4°.
• Preparation: hydrogenation (Pt, Pd at room temperature; Ni hotter); Zn/HCl reduction of RX (not fluorides); Wurtz, 2RX + 2Na → R–R (even carbons); decarboxylation (one C fewer); Kolbe electrolysis (even carbons, no methane).
• Physical: non-polar, van der Waals forces; C₁–C₄ gases at 298 K; b.p. rises with size and falls with branching.
• Halogenation: free radical chain (initiation, propagation, termination); F₂ > Cl₂ > Br₂ > I₂; 3° > 2° > 1°; iodination needs HIO₃ or HNO₃; ethane is a by-product of methane chlorination.
• Combustion: Δ_cH^⊖(CH₄) = −890 kJ mol⁻¹; incomplete combustion gives carbon black. Also controlled oxidation, isomerisation (AlCl₃/HCl), aromatization (773 K, 10–20 atm), steam reforming (Ni, 1273 K) and pyrolysis.
• Conformations: staggered (60°, most stable) and eclipsed (0°); barrier in ethane 12.5 kJ mol⁻¹, so rotation is almost free.

Questions students ask

Why are alkanes unreactive, and why do they react by substitution?

Alkanes are unreactive because all their bonds are strong σ bonds, C–C and C–H, which are almost non-polar, so they offer no π electrons, lone pairs or partly charged atoms for an electrophile or a nucleophile to attack. Every carbon already holds four atoms, so nothing can add to an alkane without a bond breaking first: a new atom can only take the place of a hydrogen, which is why their typical reaction is substitution. Even that needs heat or light to split the halogen molecule into free radicals, as when methane is chlorinated at 573–773 K or in diffused sunlight.

Why is dry ether used in the Wurtz reaction?

Dry ether is used in the Wurtz reaction because sodium, and the alkylsodium thought to form from it as an intermediate, react with water: moisture would use up the sodium, 2Na + 2H₂O → 2NaOH + H₂, and turn the alkyl group into R–H, R–Na + H₂O → R–H + NaOH, instead of joining two alkyl groups into R–R. Diethyl ether, C₂H₅OC₂H₅, dissolves the alkyl halide but has no acidic hydrogen, so it does not react with sodium; it is in fact often dried over sodium wire. With 1-bromopropane, dry ether gives hexane, while moisture would give propane instead.

Why does the Wurtz reaction fail with tertiary alkyl halides?

The Wurtz reaction fails with tertiary alkyl halides because their bulky alkyl groups cannot easily join, and the strongly basic alkylsodium (or the alkyl radical) removes a hydrogen from a neighbouring carbon instead, so elimination gives an alkene and an alkane in place of the coupled product. 2-Bromo-2-methylpropane gives mainly 2-methylpropene and 2-methylpropane, 2(CH₃)₃CBr + 2Na → (CH₃)₂C=CH₂ + (CH₃)₃CH + 2NaBr, and hardly any 2,2,3,3-tetramethylbutane. Secondary halides also give some alkene in this way, so the Wurtz reaction works best with primary alkyl halides.

What is the difference between the Wurtz, Wurtz–Fittig and Fittig reactions?

The Wurtz reaction joins two alkyl groups, 2RX + 2Na → R–R + 2NaX in dry ether, while the Wurtz–Fittig reaction joins an alkyl group to an aryl group: an alkyl halide and an aryl halide with sodium in dry ether give an alkylbenzene, as in CH₃Br + C₆H₅Br + 2Na → C₆H₅CH₃ + 2NaBr (toluene). The Fittig reaction joins two aryl groups: 2C₆H₅Cl + 2Na → C₆H₅–C₆H₅ + 2NaCl (biphenyl). The Wurtz reaction is a Class 11 preparation of alkanes; it returns, with the Wurtz–Fittig and Fittig reactions, in the Class 12 unit Haloalkanes and Haloarenes.

What is the difference between Kolbe’s reaction and Kolbe electrolysis?

Kolbe electrolysis (Kolbe’s electrolytic method in NCERT, often just called the Kolbe reaction) makes alkanes: electrolysing an aqueous sodium or potassium carboxylate joins two alkyl groups at the anode, 2RCOONa + 2H₂O → R–R + 2CO₂ + H₂ + 2NaOH. Kolbe’s reaction, as NCERT Class 12 names it (the Kolbe–Schmitt reaction), is a reaction of phenol: sodium phenoxide is heated with carbon dioxide under pressure, and acidifying the product gives salicylic acid (2-hydroxybenzoic acid), the starting material for aspirin. Both are named after Hermann Kolbe; Kolbe electrolysis is in Class 11 Hydrocarbons, Kolbe’s reaction in Class 12 Alcohols, Phenols and Ethers.

What is the difference between chlorination and bromination of alkanes?

Chlorination and bromination of alkanes go by the same free radical chain, but chlorination is faster and much less selective: a chlorine atom removes 1°, 2° and 3° hydrogens at rates within a factor of about 5, while the less reactive bromine atom takes mainly the most easily removed one, so the order 3° > 2° > 1° matters far more in bromination. In standard textbook data (beyond NCERT), propane, with six 1° and two 2° hydrogens, gives about 45% 1-chloropropane and 55% 2-chloropropane in light at 298 K, but about 97% 2-bromopropane in light at about 400 K. Bromination is therefore the better choice when one product is wanted, while chlorination gives mixtures that must be separated.

What are the conformations of butane, and are they in the JEE Main 2027 and NEET (UG) 2026 syllabuses?

Rotation about the central C2–C3 bond of butane gives four named conformations: anti, with the two CH₃ groups 180° apart (most stable); gauche, 60° apart (about 3.8 kJ mol^⁻¹ higher); eclipsed, 120° apart, each CH₃ eclipsing a hydrogen (about 16 kJ mol^⁻¹ higher); and fully eclipsed, 0°, CH₃ on CH₃ (about 19 kJ mol^⁻¹ higher, least stable). The bulky methyl groups add steric strain to the torsional strain of ethane, yet the conformers still interconvert at room temperature. Both syllabuses name only ‘Sawhorse and Newman projections (of ethane)’ in Unit 15, and NCERT treats only ethane, so butane goes beyond both.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Write the molecular formula of the alkane with nine carbon atoms and of its alkyl group.

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   C₉H₂₀ (C_nH_2n+2) and –C₉H₁₉ (C_nH_2n+1).

2. Q2. How many primary, secondary, tertiary and quaternary carbons are there in 2,2,4-trimethylpentane, (CH₃)₃CCH₂CH(CH₃)CH₃?

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   Five primary (the five CH₃ groups), one secondary (CH₂), one tertiary (CH) and one quaternary (carbon 2): eight carbons in all.

3. Q3. Give the IUPAC name of CH₃CH(C₂H₅)CH₂CH(CH₃)CH₃.

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   The longest chain runs through the ethyl group: six carbons, with methyls on carbons 2 and 4 from one end (3 and 5 from the other). Name: 2,4-dimethylhexane.

4. Q4. Why can methane not be prepared by the Wurtz reaction or by Kolbe’s method?

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   Both methods join two alkyl groups, so the product has at least two carbons; the smallest is ethane. Methane can be made instead by decarboxylating sodium ethanoate with soda lime, or by reducing chloromethane with Zn and HCl.

5. Q5. Which alkane forms when sodium 2-methylpropanoate, (CH₃)₂CHCOONa, is heated with soda lime?

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   The –COONa group is replaced by H: (CH₃)₂CH₂, which is CH₃CH₂CH₃, propane. (CH₃)₂CHCOONa + NaOH → CH₃CH₂CH₃ + Na₂CO₃.

6. Q6. Why is ethane found among the products when methane is chlorinated in light (NCERT Exercise 9.1)?

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   Chlorination is a free radical chain: propagation, Cl• + CH₄ → •CH₃ + HCl, makes methyl radicals, and in termination two of them can combine: •CH₃ + •CH₃ → CH₃–CH₃.

7. Q7. Arrange in order of increasing boiling point: pentane, 2,2-dimethylpropane, butane, 2-methylbutane. What does this show about the effect of branching (NCERT Exercise 9.15)?

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   Butane (272.4 K) < 2,2-dimethylpropane (282.5 K) < 2-methylbutane (300.9 K) < pentane (309.1 K) (NCERT Table 9.2). The C₅ isomers boil above butane because they are heavier; among them, more branching means a lower boiling point: the molecule becomes more nearly spherical, the area of contact shrinks, and the weaker van der Waals forces are overcome at a lower temperature.

8. Q8. Write the balanced equation for the complete combustion of hexane and state the moles of O₂ needed per mole.

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   C₆H₁₄ + 19/2 O₂ → 6CO₂ + 7H₂O, so 9.5 mol of O₂ per mole of hexane ((3 × 6 + 1)/2 = 19/2).

9. Q9. In the staggered and eclipsed conformations of ethane, what are the dihedral angles? Do the H–C–H bond angles change between them?

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   60° (staggered) and 0° (eclipsed). No: bond angles and bond lengths stay the same; only the angle of rotation about the C–C bond changes.

10. Q10. Which alkane would you use to prepare toluene by aromatization? Write the equation.

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    n-Heptane: CH₃(CH₂)₅CH₃ → C₆H₅CH₃ + 4H₂ (773 K, 10–20 atm, oxides of V, Mo or Cr on alumina). The seven-carbon chain closes into a six-membered ring with one carbon left as the methyl group.