Alkanes are the saturated hydrocarbons $\mathrm{C_{n}H_{2n+2}}$: the fuels of the gas cylinder and the petrol tank. How hydrocarbons are classified, how alkanes are named and drawn, how they are made, how they burn and react with halogens by a free-radical chain, and why even ethane has conformations.
Natural gas, the LPG in a kitchen cylinder, petrol, diesel, kerosene, paraffin wax and the bitumen on a road are all mixtures of alkanes — chains of carbon atoms joined by single bonds and clothed in hydrogen. India burns more than $25$ million tonnes of LPG a year, and every kilogram of it releases about $50\ \mathrm{MJ}$ of heat. Alkanes are so unreactive towards acids, bases and most oxidising agents that they were once called paraffins, from the Latin for "little affinity". Yet they burn fiercely, and in sunlight they react with chlorine in a chain reaction that can run thousands of steps from a single spark of light.
This section starts the chapter by sorting hydrocarbons into families. It then looks at alkanes in detail: their structure, isomers and names; the reactions used to make them in the laboratory (hydrogenation, reduction of alkyl halides, the Wurtz reaction, decarboxylation and Kolbe's electrolysis); their physical properties; their reactions — free-radical halogenation, combustion, controlled oxidation, isomerisation, aromatisation, the reaction with steam and pyrolysis; and finally the different shapes, or conformations, that ethane takes as one end of the molecule turns relative to the other.
Hydrocarbons and the Alkane Family
Hydrocarbons differ in the kinds of carbon–carbon bond they contain and in whether their carbon atoms form chains or rings. The classification below is the map for the whole chapter.
📘 Definition 9.1 — Hydrocarbons and their classification
A hydrocarbon is a compound that contains only carbon and hydrogen. Hydrocarbons are classified by the carbon–carbon bonds they contain:
Saturated hydrocarbons contain only C–C and C–H single bonds. Open-chain ones are the alkanes ($\mathrm{C_{n}H_{2n+2}}$); ring ones are the cycloalkanes ($\mathrm{C_{n}H_{2n}}$, e.g. cyclohexane).
Unsaturated hydrocarbons contain at least one C=C double bond (alkenes, $\mathrm{C_{n}H_{2n}}$ for one double bond) or C≡C triple bond (alkynes, $\mathrm{C_{n}H_{2n-2}}$ for one triple bond).
Aromatic hydrocarbons (arenes) contain one or more benzene-type rings, e.g. benzene $\mathrm{C_{6}H_{6}}$, methylbenzene (toluene) $\mathrm{C_{6}H_{5}CH_{3}}$, naphthalene $\mathrm{C_{10}H_{8}}$.
Alkanes, alkenes, alkynes and cycloalkanes are together called aliphatic hydrocarbons, to distinguish them from the aromatic ones.
📘 Definition 9.2 — Alkanes
An alkane is an open-chain saturated hydrocarbon. Its general formula is $\mathrm{C_{n}H_{2n+2}}$: methane $\mathrm{CH_{4}}$, ethane $\mathrm{C_{2}H_{6}}$, propane $\mathrm{C_{3}H_{8}}$, butane $\mathrm{C_{4}H_{10}}$, … Successive members differ by $\mathrm{CH_{2}}$ (a homologous series). Every carbon atom is sp³ hybridised and tetrahedral, with bond angles close to $109.5^{\circ}$; the C–C bond length is $154\ \mathrm{pm}$ and the C–H bond about $109\ \mathrm{pm}$. Because the bonds around each carbon point to the corners of a tetrahedron, a "straight" chain is really a zig-zag.
A carbon atom in an alkane is called primary (1°), secondary (2°), tertiary (3°) or quaternary (4°) when it is bonded to one, two, three or four other carbon atoms; the hydrogen atoms on them are called 1°, 2° and 3° hydrogens.
Alkane names are built by the IUPAC rules of Chapter 8: find the longest chain, number it from the end nearer a branch, and name the branches (methyl, ethyl…) with their locants. From butane onwards the same molecular formula can belong to more than one compound — chain isomers, which differ in the branching of the carbon skeleton. The number of isomers grows rapidly: $\mathrm{C_{4}H_{10}}$ has 2, $\mathrm{C_{5}H_{12}}$ has 3, $\mathrm{C_{6}H_{14}}$ has 5, $\mathrm{C_{7}H_{16}}$ has 9 and $\mathrm{C_{8}H_{18}}$ has 18.
pentane (n-pentane) · b.p. 36 °C
2-methylbutane (isopentane) · b.p. 28 °C
2,2-dimethylpropane (neopentane) · b.p. 9.5 °C
✏️ Example 1 — Isomers, names and classes of carbon
(a) Draw and name the three isomers of $\mathrm{C_{5}H_{12}}$, and give the number of 1°, 2°, 3° and 4° carbon atoms in each. (b) 2,2,4-Trimethylpentane ("isooctane", the reference fuel for octane numbers) has the condensed formula $\mathrm{(CH_{3})_{3}CCH_{2}CH(CH_{3})_{2}}$. Confirm its molecular formula and count its 1°, 2° and 3° hydrogen atoms. (c) Why is "2-ethylbutane" not a correct name?
(b) formula Carbons: three $\mathrm{CH_{3}}$ on C2, the quaternary C2, $\mathrm{CH_{2}}$ (C3), $\mathrm{CH}$ (C4) and two $\mathrm{CH_{3}}$ on C4: $3+1+1+1+2=8$ C. Hydrogens: $5\times3+2+1=18$. $\mathrm{C_{8}H_{18}}=\mathrm{C_{n}H_{2n+2}}$ with $n=8$ ✓.
(b) classes Five $\mathrm{CH_{3}}$ groups carry the $15$ primary H; the $\mathrm{CH_{2}}$ carries $2$ secondary H; the CH carries $1$ tertiary H; the quaternary C2 has none. Total $15+2+1=18$ ✓.
(c) $\mathrm{CH_{3}CH_{2}CH(CH_{2}CH_{3})CH_{3}}$ written as "2-ethylbutane" has a longest chain of five carbons (through the ethyl group), so it is 3-methylpentane. Always find the longest chain first.
Preparing Alkanes
Petroleum and natural gas supply alkanes in bulk, but as mixtures. In the laboratory a particular alkane is made from a compound that already has the right carbon skeleton, or from one with half of it (two halves can be joined) or with one carbon too many (one can be removed as $\mathrm{CO_{2}}$).
🧭 Four routes to an alkane
Hydrogenation of alkenes and alkynes. Hydrogen adds across the multiple bond on a finely divided metal catalyst — platinum or palladium at room temperature, nickel at a higher temperature and pressure:
$$\mathrm{CH_{2}{=}CH_{2}(g)+H_{2}(g)\xrightarrow{\text{Pt}}CH_{3}CH_{3}(g)}$$
$$\mathrm{HC{\equiv}CH(g)+2H_{2}(g)\xrightarrow{\text{Pt}}CH_{3}CH_{3}(g)}$$
The carbon skeleton is unchanged; only the bonds become single.
Reduction of alkyl halides (except fluorides) by zinc and dilute hydrochloric acid, which replaces the halogen by hydrogen:
$$\mathrm{CH_{3}CH_{2}Cl+H_{2}\xrightarrow{\text{Zn, H}^{+}}CH_{3}CH_{3}+HCl}$$
The Wurtz reaction — two alkyl groups are joined (below).
From carboxylic acids — decarboxylation removes one carbon; Kolbe's electrolysis joins two alkyl groups (below).
📘 Definition 9.3 — Wurtz reaction, decarboxylation and Kolbe's electrolysis
In the Wurtz reaction an alkyl halide is heated with sodium metal in dry ether; two alkyl groups join to give an alkane with twice as many carbon atoms:
$$\mathrm{2R{-}X+2Na\xrightarrow{\text{dry ether}}R{-}R+2NaX}$$
e.g. $\mathrm{2CH_{3}Br+2Na\rightarrow CH_{3}CH_{3}+2NaBr}$. The ether must be dry because sodium reacts with water.
In decarboxylation the sodium salt of a carboxylic acid is heated with soda lime (sodium hydroxide mixed with calcium oxide); the $\mathrm{{-}COONa}$ group is replaced by H, giving an alkane with one carbon fewer:
$$\begin{aligned}&\mathrm{CH_{3}COONa(s)+NaOH(s)}\\&\quad\xrightarrow[\Delta]{\text{CaO}}\mathrm{CH_{4}(g)+Na_{2}CO_{3}(s)}\end{aligned}$$
In Kolbe's electrolysis a concentrated aqueous solution of the sodium or potassium salt of a carboxylic acid is electrolysed. At the anode the carboxylate ion loses an electron and then $\mathrm{CO_{2}}$, and the alkyl radicals pair up; hydrogen is formed at the cathode:
$$\begin{aligned}&\mathrm{2CH_{3}COONa(aq)+2H_{2}O(l)}\\&\quad\mathrm{\rightarrow CH_{3}CH_{3}(g)+2CO_{2}(g)}\\&\quad\quad\mathrm{{}+H_{2}(g)+2NaOH(aq)}\end{aligned}$$
Anode steps: $\mathrm{2CH_{3}COO^{-}\rightarrow 2CH_{3}COO^{\bullet}+2e^{-}}$, $\mathrm{2CH_{3}COO^{\bullet}\rightarrow 2{}^{\bullet}CH_{3}+2CO_{2}}$, $\mathrm{2{}^{\bullet}CH_{3}\rightarrow CH_{3}CH_{3}}$.
📐 Rule 9.1 — Counting carbons in alkane preparations
From a starting compound with an alkyl group of $n$ carbon atoms:
hydrogenation and reduction of $\mathrm{RX}$ keep the skeleton: product has $n$ carbons;
Wurtz ($\mathrm{R{-}X}$ with $n$ carbons) gives $\mathrm{R{-}R}$ with $2n$ carbons — always a symmetrical alkane with an even number of carbons when one halide is used;
decarboxylation of $\mathrm{RCOONa}$ (acid with $n+1$ carbons) gives $\mathrm{RH}$ with $n$ carbons;
Kolbe electrolysis of $\mathrm{RCOONa}$ gives $\mathrm{R{-}R}$ with $2n$ carbons.
Consequences: methane cannot be made by the Wurtz reaction or by Kolbe's method (both give at least two carbons), and an alkane with an odd number of carbons made by Wurtz needs two different halides, which gives a mixture of three alkanes that is hard to separate.
✏️ Example 2 — The Wurtz reaction
(a) Name the product when bromoethane is heated with sodium in dry ether, and write the equation. (b) A mixture of iodomethane and iodoethane is treated with sodium in dry ether. Which alkanes form? (c) Can 2-methylpropane be made cleanly by a Wurtz reaction from a single alkyl halide? (d) Why is methane never made by this method?
💡 Show solution▶
(a) Two ethyl groups join: $\mathrm{2CH_{3}CH_{2}Br+2Na\xrightarrow{\text{dry ether}}CH_{3}CH_{2}CH_{2}CH_{3}+2NaBr}$ — butane.
(b) Any two of the alkyl groups may pair: $\mathrm{CH_{3}{-}CH_{3}}$ (ethane), $\mathrm{CH_{3}{-}CH_{2}CH_{3}}$ (propane) and $\mathrm{CH_{3}CH_{2}{-}CH_{2}CH_{3}}$ (butane). Propane is only one of three products.
(c) No. $\mathrm{(CH_{3})_{3}CH}$ has four carbons, so it would have to be $\mathrm{R{-}R}$ with two-carbon R groups — but joining two ethyl groups gives butane, an unbranched chain. A symmetrical product $\mathrm{R{-}R}$ is needed, and 2-methylpropane cannot be cut into two identical halves.
(d) Wurtz joins two alkyl groups, so the product has at least two carbons ($\mathrm{CH_{3}{-}CH_{3}}$ from $\mathrm{CH_{3}X}$).
✏️ Example 3 — From sodium propanoate: one carbon fewer, or twice the alkyl group
$9.61\ \mathrm{g}$ of sodium propanoate, $\mathrm{CH_{3}CH_{2}COONa}$ ($M=96.06\ \mathrm{g\,mol^{-1}}$), is (a) heated with soda lime, or (b) dissolved in water and electrolysed (Kolbe). Name the alkane formed in each case, write the equation, and find the maximum mass of alkane. (c) Name every alkene (one C=C) and every alkyne (one C≡C) that gives 2-methylbutane on catalytic hydrogenation. ($M$: ethane $30.07$, butane $58.12\ \mathrm{g\,mol^{-1}}$)
💡 Show solution▶
Amount $n=9.61/96.06=0.1000\ \mathrm{mol}$ of salt; the alkyl group is $\mathrm{CH_{3}CH_{2}{-}}$ ($n=2$).
(a) $\mathrm{CH_{3}CH_{2}COONa+NaOH\xrightarrow[\Delta]{\text{CaO}}CH_{3}CH_{3}+Na_{2}CO_{3}}$: ethane, $1:1$, so $0.1000\times30.07=3.01\ \mathrm{g}$.
(b) $\mathrm{2CH_{3}CH_{2}COONa+2H_{2}O\rightarrow CH_{3}CH_{2}CH_{2}CH_{3}+2CO_{2}+H_{2}+2NaOH}$: butane, two salt units per molecule, so $0.0500\times58.12=2.91\ \mathrm{g}$ (at the anode, with $\mathrm{CO_{2}}$; $\mathrm{H_{2}}$ at the cathode).
(c) The skeleton $\mathrm{(CH_{3})_{2}CHCH_{2}CH_{3}}$ must be kept. Alkenes: 2-methylbut-1-ene $\mathrm{CH_{2}{=}C(CH_{3})CH_{2}CH_{3}}$, 2-methylbut-2-ene $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$ and 3-methylbut-1-ene $\mathrm{(CH_{3})_{2}CHCH{=}CH_{2}}$. Alkyne: 3-methylbut-1-yne $\mathrm{(CH_{3})_{2}CHC{\equiv}CH}$ — the only one, since C2 carries a branch and cannot form a triple bond.
Physical properties. C–C bonds are non-polar and C–H bonds almost so (Section 4.1), so alkane molecules are held together only by weak dispersion (van der Waals) forces. At $298\ \mathrm{K}$ the alkanes $\mathrm{C_{1}}$–$\mathrm{C_{4}}$ are gases, $\mathrm{C_{5}}$–$\mathrm{C_{17}}$ liquids and the higher members waxy solids. They are insoluble in water but dissolve in non-polar solvents such as benzene and ether — "like dissolves like". Their boiling points rise steadily along the series, because longer molecules have larger surfaces in contact and stronger dispersion forces; among isomers, branching lowers the boiling point, because a branched molecule is more compact (closer to a sphere) and touches its neighbours over a smaller area. Explore both trends in Interactive 1.
Interactive 1 — Boiling Points: Chain Length and Branching
In "chain length" mode the slider walks along the unbranched alkanes from methane to decane; in the isomer modes it steps through the isomers of $\mathrm{C_{5}H_{12}}$ or $\mathrm{C_{6}H_{14}}$ from least to most branched. The bar chart shows each normal boiling point (literature values), the dashed line marks room temperature, and the readouts give the state at $25\ ^{\circ}\mathrm{C}$. Press ▶ Play to step through automatically.
alkane—
boiling point—
change from previous—
state at 25 °C—
Halogenation. Alkanes do not react with acids, bases or aqueous oxidising agents at room temperature. In diffused sunlight or ultraviolet light, or on heating to about $573$–$773\ \mathrm{K}$, they react with chlorine and bromine, the halogen atoms replacing hydrogen atoms one at a time — a substitution:
$$\mathrm{CH_{4}(g)+Cl_{2}(g)\xrightarrow{h\nu}CH_{3}Cl(g)+HCl(g)}$$
With more chlorine the substitution continues: $\mathrm{CH_{3}Cl\rightarrow CH_{2}Cl_{2}\rightarrow CHCl_{3}\rightarrow CCl_{4}}$ (chloromethane, dichloromethane, trichloromethane (chloroform), tetrachloromethane). A large excess of methane favours chloromethane; excess chlorine drives the reaction to $\mathrm{CCl_{4}}$. The rate depends on the halogen, $\mathrm{F_{2}\gt Cl_{2}\gt Br_{2}\gt I_{2}}$: fluorination is explosive and must be moderated, and iodination is slow and reversible ($\mathrm{CH_{4}+I_{2}\rightleftharpoons CH_{3}I+HI}$), so it is carried out with an oxidising agent such as $\mathrm{HIO_{3}}$ or $\mathrm{HNO_{3}}$ that removes HI. For a given halogen, H atoms are replaced in the order 3° $\gt$ 2° $\gt$ 1°.
📐 Rule 9.2 — Free-radical chain mechanism of halogenation
Halogenation of an alkane goes by a free-radical chain (the radicals and homolytic fission of Chapter 8) in three stages. For methane and chlorine:
Propagation — each step uses up one radical and makes another, so the chain carries on:
$$\mathrm{Cl^{\bullet}+CH_{4}\rightarrow {}^{\bullet}CH_{3}+HCl}$$
$$\mathrm{{}^{\bullet}CH_{3}+Cl{-}Cl\rightarrow CH_{3}Cl+Cl^{\bullet}}$$
The $\mathrm{Cl^{\bullet}}$ made in the second step starts the first step again; one photon can lead to thousands of molecules of product.
Termination — two radicals meet and pair their electrons, ending a chain:
$$\mathrm{Cl^{\bullet}+Cl^{\bullet}\rightarrow Cl_{2}}$$
$$\mathrm{{}^{\bullet}CH_{3}+Cl^{\bullet}\rightarrow CH_{3}Cl}$$
$$\mathrm{{}^{\bullet}CH_{3}+{}^{\bullet}CH_{3}\rightarrow CH_{3}{-}CH_{3}}$$
Evidence: the reaction needs light or heat and does not occur in the dark at room temperature; a trace of ethane appears among the products (from the last termination step); and substances that destroy radicals, such as oxygen, slow it down.
Combustion. Alkanes burn in excess air or oxygen to carbon dioxide and water, releasing a large amount of heat — the reason they are fuels. The standard enthalpy of combustion of methane is $-890.3\ \mathrm{kJ\,mol^{-1}}$ (Section 5.3). In a limited supply of air, carbon monoxide and soot (carbon black, used in inks and tyres) are formed instead.
📐 Rule 9.3 — Combustion of an alkane
For complete combustion of any alkane,
$$\begin{aligned}&\mathrm{C_{n}H_{2n+2}+\tfrac{3n+1}{2}\,O_{2}}\\&\quad\mathrm{\rightarrow n\,CO_{2}+(n+1)\,H_{2}O}\end{aligned}$$
Check: carbon $n=n$; hydrogen $2n+2=2(n+1)$; oxygen $3n+1=2n+(n+1)$. When $n$ is even, $\tfrac{3n+1}{2}$ is a fraction; multiply the whole equation by 2, e.g. $\mathrm{2C_{4}H_{10}(g)+13O_{2}(g)\rightarrow 8CO_{2}(g)+10H_{2}O(l)}$.
📘 Definition 9.4 — Other reactions of alkanes
Controlled oxidation with a limited supply of oxygen and a catalyst gives useful oxygen compounds:
over copper at $523\ \mathrm{K}$ and $100\ \mathrm{atm}$, $\mathrm{2CH_{4}+O_{2}\rightarrow 2CH_{3}OH}$ (methanol); heated with $\mathrm{Mo_{2}O_{3}}$, $\mathrm{CH_{4}+O_{2}\rightarrow HCHO+H_{2}O}$ (methanal); heated with manganese(II) ethanoate, $\mathrm{2CH_{3}CH_{3}+3O_{2}\rightarrow 2CH_{3}COOH+2H_{2}O}$ (ethanoic acid).
Alkanes with a tertiary H are oxidised at that carbon by potassium permanganate: $\mathrm{(CH_{3})_{3}CH\xrightarrow{KMnO_{4}}(CH_{3})_{3}COH}$ (2-methylpropan-2-ol).
Isomerisation: an unbranched alkane heated with anhydrous aluminium chloride and hydrogen chloride gas rearranges into branched isomers, e.g. hexane → 2-methylpentane + 3-methylpentane.
Aromatisation: alkanes with six or more carbons in a chain, heated to $773\ \mathrm{K}$ at $10$–$20\ \mathrm{atm}$ over oxides of chromium, vanadium or molybdenum ($\mathrm{Cr_{2}O_{3}}$, $\mathrm{V_{2}O_{5}}$, $\mathrm{Mo_{2}O_{3}}$) on alumina, lose hydrogen and close into benzene rings: $\mathrm{C_{6}H_{14}\rightarrow C_{6}H_{6}+4H_{2}}$; heptane gives methylbenzene (toluene).
Reaction with steam: methane and steam over nickel at about $1273\ \mathrm{K}$ give carbon monoxide and hydrogen — the main industrial source of hydrogen:
$\mathrm{CH_{4}(g)+H_{2}O(g)\xrightarrow[\Delta]{\text{Ni}}CO(g)+3H_{2}(g)}$.
Pyrolysis (cracking): heating a higher alkane strongly, in the absence of air, breaks it into smaller alkanes, alkenes and hydrogen. Hexane at $773\ \mathrm{K}$ gives $\mathrm{C_{6}H_{12}+H_{2}}$, $\mathrm{C_{4}H_{8}+C_{2}H_{6}}$ and $\mathrm{C_{3}H_{6}+C_{2}H_{4}+CH_{4}}$; dodecane over Pt, Pd or Ni gives heptane and pentene among other products. Pyrolysis is believed to follow a free-radical mechanism; it turns heavy oil fractions into petrol and supplies ethene and propene for plastics.
✏️ Example 4 — Free-radical chlorination
(a) Write the mechanism of the monochlorination of ethane and explain why a little butane is found among the products. (b) How many structurally different monochloro products are formed from (i) propane, (ii) 2-methylpropane, (iii) 2-methylbutane, (iv) 2,2-dimethylpropane? Name those from (iii). (c) Which alkane of formula $\mathrm{C_{5}H_{12}}$ is the best choice for making a single monochloro compound in good yield?
💡 Show solution▶
(a) Initiation $\mathrm{Cl_{2}\xrightarrow{h\nu}2Cl^{\bullet}}$. Propagation $\mathrm{Cl^{\bullet}+CH_{3}CH_{3}\rightarrow CH_{3}CH_{2}^{\bullet}+HCl}$; $\mathrm{CH_{3}CH_{2}^{\bullet}+Cl_{2}\rightarrow CH_{3}CH_{2}Cl+Cl^{\bullet}}$. Termination: $\mathrm{2Cl^{\bullet}\rightarrow Cl_{2}}$, $\mathrm{CH_{3}CH_{2}^{\bullet}+Cl^{\bullet}\rightarrow CH_{3}CH_{2}Cl}$, $\mathrm{2CH_{3}CH_{2}^{\bullet}\rightarrow CH_{3}CH_{2}CH_{2}CH_{3}}$ — the last step explains the butane.
(b) idea Count the sets of equivalent hydrogen atoms: replacing any H of one set gives the same compound.
(b) (i) Propane: $\mathrm{CH_{3}}$ (6 H) and $\mathrm{CH_{2}}$ (2 H) → 2 (1-chloropropane, 2-chloropropane). (ii) 2-Methylpropane: 9 equivalent $\mathrm{CH_{3}}$ H and 1 tertiary H → 2. (iii) 2-Methylbutane $\mathrm{(CH_{3})_{2}CHCH_{2}CH_{3}}$: the two $\mathrm{CH_{3}}$ on C2, the CH, the $\mathrm{CH_{2}}$ and the C4 $\mathrm{CH_{3}}$ → 4: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane and 1-chloro-3-methylbutane. (iv) 2,2-Dimethylpropane: all 12 H equivalent → 1.
(c) 2,2-Dimethylpropane: it gives only 1-chloro-2,2-dimethylpropane, $\mathrm{(CH_{3})_{3}CCH_{2}Cl}$ (pentane gives 3 products and 2-methylbutane 4).
✏️ Example 5 — An LPG cylinder
Treat the $14.2\ \mathrm{kg}$ of fuel in a domestic LPG cylinder as pure butane ($M=58.12\ \mathrm{g\,mol^{-1}}$, $\Delta_{c}H^{\circ}=-2878\ \mathrm{kJ\,mol^{-1}}$). (a) Write the balanced equation for its complete combustion. (b) What mass of oxygen is used and what mass of $\mathrm{CO_{2}}$ is produced when the cylinder is used up? (c) How much heat is released, and what is the fuel value in $\mathrm{MJ\,kg^{-1}}$? (d) Write the equation for combustion to carbon monoxide, and say why a yellow, sooty flame on a gas stove is a warning sign. ($M$: $\mathrm{O_{2}}$ $32.00$, $\mathrm{CO_{2}}$ $44.01\ \mathrm{g\,mol^{-1}}$)
💡 Show solution▶
(a) Rule 9.3 with $n=4$: $\mathrm{C_{4}H_{10}+\tfrac{13}{2}O_{2}\rightarrow 4CO_{2}+5H_{2}O}$, i.e. $\mathrm{2C_{4}H_{10}(g)+13O_{2}(g)\rightarrow 8CO_{2}(g)+10H_{2}O(l)}$. Atoms: C $8=8$, H $20=20$, O $26=16+10$ ✓.
(b) $n(\mathrm{C_{4}H_{10}})=14\,200/58.12=244.3\ \mathrm{mol}$. $\mathrm{O_{2}}$: $6.5\times244.3=1588\ \mathrm{mol}$, $1588\times32.00=5.08\times10^{4}\ \mathrm{g}=50.8\ \mathrm{kg}$. $\mathrm{CO_{2}}$: $4\times244.3=977.3\ \mathrm{mol}$, $\times44.01=4.30\times10^{4}\ \mathrm{g}=43.0\ \mathrm{kg}$ — three times the mass of the fuel.
(c) $q=244.3\times2878=7.03\times10^{5}\ \mathrm{kJ}=703\ \mathrm{MJ}$; fuel value $703/14.2=49.5\ \mathrm{MJ\,kg^{-1}}$ ($=2878/58.12\ \mathrm{kJ\,g^{-1}}$).
(d) $\mathrm{2C_{4}H_{10}(g)+9O_{2}(g)\rightarrow 8CO(g)+10H_{2}O(l)}$ (O: $18=8+10$ ✓). A yellow, smoky flame means the air supply is too small: combustion is incomplete, producing soot and the poisonous gas CO, and less heat is obtained.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — predicting product ratios in chlorination
Not required by the CBSE syllabus. Halogenation of alkanes is listed by JEE Main, NEET and JEE Advanced (JEE Main and NEET name its mechanism); estimating product ratios this way is a typical JEE problem that no syllabus names.
At about $298\ \mathrm{K}$, a chlorine atom removes a 3°, 2° and 1° hydrogen with relative rates of roughly $5.0:3.8:1$ (per hydrogen atom). The share of each product is proportional to (number of H of that type) × (relative rate). For propane: 1-chloropropane $6\times1=6$, 2-chloropropane $2\times3.8=7.6$; so $6/13.6=44\%$ and $56\%$, close to the observed $45:55$. For 2-methylpropane: 1-chloro-2-methylpropane $9\times1=9$, 2-chloro-2-methylpropane $1\times5.0=5$, i.e. $64\%$ and $36\%$ — the many primary H atoms win on numbers even though each one reacts more slowly. Bromine atoms are far more selective (roughly $1600:80:1$), so bromination gives mainly the 3° product.
Conformations of Ethane
A σ bond is cylindrically symmetrical about the line joining the two nuclei (Section 4.3), so turning one end of the molecule does not break the overlap. In ethane the two $\mathrm{CH_{3}}$ groups can therefore rotate about the C–C bond. The molecule passes through an infinite number of arrangements in space; they all have the same bonds and are the same compound, but they are not quite equal in energy.
📘 Definition 9.5 — Conformations, projections and torsional strain
Conformations (conformers, rotamers) are the different spatial arrangements of the atoms of a molecule that can be interconverted by rotation about single bonds. They are described by the dihedral angle $\phi$: looking along the C–C bond, the angle between a C–H bond on the front carbon and a C–H bond on the back carbon.
Eclipsed conformation ($\phi=0^{\circ},120^{\circ},240^{\circ}$): each H on the back carbon is directly behind an H on the front carbon.
Staggered conformation ($\phi=60^{\circ},180^{\circ},300^{\circ}$): each back H lies midway between two front H atoms, as far from them as possible.
Any other arrangement is a skew conformation.
Two drawings are used. In a sawhorse projection the C–C bond is drawn as a slanting line, the front carbon at the lower left, and the three bonds on each carbon are drawn out from it. In a Newman projection the molecule is viewed along the C–C bond: the front carbon is a point where three bonds meet at $120^{\circ}$, and the back carbon is a circle from whose edge its three bonds start. Torsional strain is the extra energy of a conformation caused by repulsion between the electron clouds of bonds on adjacent carbon atoms; it is largest when the bonds are eclipsed.
sawhorse, staggered
sawhorse, eclipsed
Newman, staggered φ = 60°
Newman, eclipsed φ = 0° (drawn slightly turned)
📐 Rule 9.4 — Relative stability of the conformations of ethane
The staggered conformation of ethane has the lowest energy and is the most stable; the eclipsed conformation has the highest energy, about $12.5\ \mathrm{kJ\,mol^{-1}}$ higher, because of torsional strain; skew conformations lie in between. This energy barrier is so small that at room temperature the molecules collide often enough to rotate freely: the conformations interconvert millions of times per second and cannot be separated as different substances. Rotation about a C–C single bond is therefore described as "free", though strictly it is slightly hindered.
A good simple model of the torsional energy of ethane, measured from the staggered minimum, is
$$E(\phi)=\tfrac12E_{0}\,(1+\cos3\phi)$$
with $E_{0}=12.5\ \mathrm{kJ\,mol^{-1}}$, which is $E_{0}$ at $\phi=0^{\circ},120^{\circ},240^{\circ}$ and zero at $60^{\circ},180^{\circ},300^{\circ}$. Interactive 2 turns the back carbon for you and draws this curve.
✏️ Example 6 — Torsional energy of ethane
(a) Through what angle must the back $\mathrm{CH_{3}}$ group turn to go from one staggered conformation to the next, and how many eclipsed conformations are passed in one full turn? (b) Use the model above to find the torsional energy at $\phi=20^{\circ}$ and $\phi=45^{\circ}$. (c) At $298\ \mathrm{K}$, $RT=2.48\ \mathrm{kJ\,mol^{-1}}$. Express the barrier as a multiple of $RT$ and explain why the conformers of ethane cannot be isolated. (d) In the eclipsed conformation the nearest H atoms on the two carbons are about $227\ \mathrm{pm}$ apart; in the staggered one, about $249\ \mathrm{pm}$. Why do these distances alone not explain the barrier?
💡 Show solution▶
(a) The three C–H bonds are $120^{\circ}$ apart, so staggered conformations repeat every $120^{\circ}$ (at $60^{\circ}$, $180^{\circ}$, $300^{\circ}$). One full turn of $360^{\circ}$ passes three eclipsed conformations, at $0^{\circ}$, $120^{\circ}$ and $240^{\circ}$.
(c) $12.5/2.48=5.0$, so the barrier is only about $5RT$. Molecular collisions at room temperature easily supply this energy, so each molecule is turning all the time; no sample can be kept in one conformation.
(d) H atoms $227\ \mathrm{pm}$ apart are only slightly closer than the sum of their van der Waals radii ($2\times120=240\ \mathrm{pm}$): they barely touch, and the repulsion between two such small atoms is weak. The barrier comes mainly from the repulsion between the bonding electron pairs of the eclipsed C–H bonds (torsional strain), not from the H atoms colliding.
Interactive 2 — Newman-Projection Rotor for Ethane
Drag the slider to turn the back carbon (circle) relative to the front carbon (centre point), or press ▶ Play to spin it. The dot on the energy curve shows the torsional energy $E(\phi)=\tfrac12E_{0}(1+\cos3\phi)$ with $E_{0}=12.5\ \mathrm{kJ\,mol^{-1}}$, and the readouts give the name of the conformation and the distance between the closest pair of H atoms on the two carbons, calculated from the real geometry (C–C $154\ \mathrm{pm}$, C–H $109\ \mathrm{pm}$, tetrahedral angles).
60°
dihedral φ—
conformation—
torsional energy—
closest H···H—
Exercises
I
Classification, structure, isomers and names. Atomic masses: C $12.01$, H $1.008$.
1 Classify each as an alkane, alkene, alkyne, cycloalkane or arene: $\mathrm{CH_{3}CH{=}CH_{2}}$, cyclohexane, $\mathrm{CH_{3}C{\equiv}CH}$, $\mathrm{C_{6}H_{5}CH_{3}}$, $\mathrm{C(CH_{3})_{4}}$.
2 Which of these molecular formulae can belong to an alkane: $\mathrm{C_{7}H_{16}}$, $\mathrm{C_{8}H_{16}}$, $\mathrm{C_{10}H_{22}}$, $\mathrm{C_{5}H_{10}}$, $\mathrm{C_{20}H_{42}}$?
3 Give the molecular formula and molar mass of the alkane with 12 carbon atoms. Which alkane has a molar mass of $86.17\ \mathrm{g\,mol^{-1}}$?
4 Draw bond-line formulae for the five isomers of $\mathrm{C_{6}H_{14}}$ and name them.
5 Each name is wrong. Draw the structure and give the correct name: (a) 1-methylbutane; (b) 3-methylbutane; (c) 2-ethylpropane; (d) 4,4-dimethylpentane.
6 For 2,2,3-trimethylbutane, $\mathrm{(CH_{3})_{3}CCH(CH_{3})_{2}}$, count the 1°, 2°, 3° and 4° carbon atoms and the 1° and 3° hydrogen atoms, and check the molecular formula.
7 Why is an "unbranched" alkane chain zig-zag rather than straight? What is the C–C–C bond angle?
8 Which isomer of $\mathrm{C_{5}H_{12}}$ has only primary hydrogens? Which isomer of $\mathrm{C_{6}H_{14}}$ has exactly two kinds of hydrogen?
2 $\mathrm{C_{n}H_{2n+2}}$: $\mathrm{C_{7}H_{16}}$ ($n=7$), $\mathrm{C_{10}H_{22}}$ and $\mathrm{C_{20}H_{42}}$. $\mathrm{C_{8}H_{16}}$ and $\mathrm{C_{5}H_{10}}$ are $\mathrm{C_{n}H_{2n}}$ (an alkene or a cycloalkane).
3 $\mathrm{C_{12}H_{26}}$ (dodecane): $12(12.01)+26(1.008)=170.3\ \mathrm{g\,mol^{-1}}$. $14.03n+2.016=86.17$ gives $n=6$: hexane, $\mathrm{C_{6}H_{14}}$ (or any of its isomers).
5 (a) $\mathrm{CH_{3}{-}CH_{2}CH_{2}CH_{2}CH_{3}}$: a five-carbon chain, pentane. (b) $\mathrm{CH_{3}CH_{2}CH(CH_{3})CH_{3}}$ numbered from the wrong end: 2-methylbutane. (c) $\mathrm{CH_{3}CH(CH_{2}CH_{3})CH_{3}}$: longest chain 4, 2-methylbutane. (d) $\mathrm{CH_{3}CH_{2}CH_{2}C(CH_{3})_{2}CH_{3}}$ numbered from the wrong end: 2,2-dimethylpentane.
6 Five $\mathrm{CH_{3}}$ groups (1°), no $\mathrm{CH_{2}}$ (no 2°), one CH (3°, C3), one quaternary C (C2): $5+0+1+1=7$ C. Hydrogens: $15$ primary, $1$ tertiary, total $16$: $\mathrm{C_{7}H_{16}}$ ✓.
7 Each carbon is sp³ and tetrahedral, so consecutive C–C bonds meet at about $109.5^{\circ}$ (close to $112^{\circ}$ in long chains), not $180^{\circ}$.
8 2,2-Dimethylpropane (all 12 H on $\mathrm{CH_{3}}$ groups attached to the 4° carbon). 2,3-Dimethylbutane: 12 equivalent methyl H and 2 equivalent tertiary H.
II
Preparation of alkanes. $M$: sodium ethanoate $82.03$, ethane $30.07\ \mathrm{g\,mol^{-1}}$.
1 Name the product and write the equation: propene with hydrogen over nickel.
2 Name the product: 1-chloropropane with zinc and dilute hydrochloric acid.
3 Name the alkane formed when sodium in dry ether reacts with (a) 1-bromopropane, (b) 2-bromopropane.
4 A mixture of bromoethane and 1-bromopropane is treated with sodium in dry ether. Name all the alkanes formed.
5 Which alkane is obtained when sodium butanoate is heated with soda lime? Name two sodium salts that would give 2-methylpropane by this method.
6 An aqueous solution of $16.4\ \mathrm{g}$ of sodium ethanoate is electrolysed. Name the products at each electrode and find the maximum mass of ethane.
7 Why must the ether be dry in the Wurtz reaction? Why can methane not be made by Kolbe's electrolysis?
8 Suggest four different ways of preparing butane, naming the starting compound and the reagent in each.
5 $\mathrm{CH_{3}CH_{2}CH_{2}COONa\rightarrow CH_{3}CH_{2}CH_{3}}$: propane. For $\mathrm{(CH_{3})_{3}CH}$ the $\mathrm{{-}COONa}$ may replace any H: sodium 3-methylbutanoate $\mathrm{(CH_{3})_{2}CHCH_{2}COONa}$ or sodium 2,2-dimethylpropanoate $\mathrm{(CH_{3})_{3}CCOONa}$.
6 Anode: ethane and $\mathrm{CO_{2}}$; cathode: hydrogen (NaOH remains in solution). $n=16.4/82.03=0.200\ \mathrm{mol}$ salt → $0.100\ \mathrm{mol}$ ethane → $3.01\ \mathrm{g}$.
7 Sodium reacts with water ($\mathrm{2Na+2H_{2}O\rightarrow 2NaOH+H_{2}}$), which would use it up and stop the coupling. Kolbe joins two alkyl groups, so the smallest product is ethane, from ethanoate.
8 For example: but-1-ene or but-2-ene + $\mathrm{H_{2}}$/Pt; 1-chlorobutane + Zn/HCl; bromoethane + Na/dry ether (Wurtz); electrolysis of sodium propanoate (Kolbe); sodium pentanoate + soda lime (decarboxylation).
III
Properties and reactions. $M$: octane $114.22$, $\mathrm{CO_{2}}$ $44.01\ \mathrm{g\,mol^{-1}}$.
1 Arrange in order of increasing boiling point, with reasons: butane, 2-methylpropane, pentane, 2,2-dimethylpropane.
2 Why are alkanes insoluble in water but soluble in benzene?
3 Write the mechanism for the monobromination of methane in light, labelling the initiation, propagation and termination steps.
4 Methane is treated with excess chlorine in sunlight. Name the four chlorinated products in the order in which they form.
5 Write balanced equations for the complete combustion of propane and of octane.
6 What mass of $\mathrm{CO_{2}}$ is produced when $1.00\ \mathrm{kg}$ of octane burns completely?
7 Give the main product(s): (a) $\mathrm{CH_{4}+O_{2}}$ over $\mathrm{Mo_{2}O_{3}}$; (b) hexane with anhydrous $\mathrm{AlCl_{3}}$/HCl; (c) heptane over $\mathrm{Cr_{2}O_{3}}$ on alumina at $773\ \mathrm{K}$; (d) methane and steam over Ni at $1273\ \mathrm{K}$.
8 Decane, $\mathrm{C_{10}H_{22}}$, is cracked into octane and one other product. Identify it and say why cracking always gives some alkenes.
💡 Show solutions▶
1 2-Methylpropane ($-11.7\ ^{\circ}\mathrm{C}$) $\lt$ butane ($-0.5$) $\lt$ 2,2-dimethylpropane ($9.5$) $\lt$ pentane ($36.1\ ^{\circ}\mathrm{C}$). Four carbons below five (smaller dispersion forces); within each formula, more branching → more compact → smaller contact area → lower boiling point.
2 Water molecules are held by strong hydrogen bonds; non-polar alkane molecules cannot form such bonds, so mixing would cost energy. Benzene is non-polar like the alkane, with similar dispersion forces: "like dissolves like".
6 $n=1000/114.22=8.755\ \mathrm{mol}$; $\mathrm{CO_{2}}$ $=8\times8.755=70.04\ \mathrm{mol}$; mass $=70.04\times44.01=3.08\times10^{3}\ \mathrm{g}=3.08\ \mathrm{kg}$.
7 (a) Methanal, HCHO, and water. (b) 2-Methylpentane and 3-methylpentane. (c) Methylbenzene (toluene), $\mathrm{C_{7}H_{16}\rightarrow C_{6}H_{5}CH_{3}+4H_{2}}$. (d) $\mathrm{CO+3H_{2}}$.
8 $\mathrm{C_{10}H_{22}\rightarrow C_{8}H_{18}+C_{2}H_{4}}$: ethene. A molecule $\mathrm{C_{n}H_{2n+2}}$ split into two pieces cannot give two alkanes — there are not enough H atoms — so at least one fragment must be an alkene (or $\mathrm{H_{2}}$ must be given off).
IV
Conformations of ethane. Use $E(\phi)=\tfrac12E_{0}(1+\cos3\phi)$ with $E_{0}=12.5\ \mathrm{kJ\,mol^{-1}}$ where needed.
1 What are conformations? Are the conformations of ethane different compounds that could be separated?
2 Draw Newman projections of the staggered and eclipsed conformations of ethane.
3 Which conformation of ethane is more stable, and by about how much?
4 List all the dihedral angles between $0^{\circ}$ and $360^{\circ}$ at which ethane is (a) staggered, (b) eclipsed.
5 Find the torsional energy at $\phi=30^{\circ}$, $90^{\circ}$ and $100^{\circ}$.
6 What is torsional strain, and why is it greatest in the eclipsed conformation?
7 Sketch the graph of torsional energy against dihedral angle for one full rotation, marking the maxima and minima.
8 Rotation about the C–C bond of ethane is almost free. Predict, with a reason, whether the same is true of the C=C bond of ethene.
💡 Show solutions▶
1 The spatial arrangements of a molecule that interconvert by rotation about single bonds. No: the barrier is only $12.5\ \mathrm{kJ\,mol^{-1}}$, so they interconvert rapidly at room temperature and cannot be isolated.
2 As in the figure: front C–H bonds meeting at the centre, back bonds from the circle; staggered with back bonds midway between front ones, eclipsed with back bonds behind front ones.
3 Staggered, by about $12.5\ \mathrm{kJ\,mol^{-1}}$ (less torsional strain).
6 The extra energy due to repulsion between the electron clouds of C–H bonds on neighbouring carbons. When eclipsed, the bonds are as close together as they can be, so the repulsion is greatest.
7 A wave with three equal maxima ($12.5\ \mathrm{kJ\,mol^{-1}}$) at $0^{\circ}$, $120^{\circ}$, $240^{\circ}$ (and $360^{\circ}$) and three minima ($0$) at $60^{\circ}$, $180^{\circ}$, $300^{\circ}$ — the curve of Interactive 2.
8 No. Turning one end of a C=C bond twists the two p orbitals apart and breaks the π bond, which costs about $266\ \mathrm{kJ\,mol^{-1}}$ ($614-348$). Rotation is therefore restricted, which is why alkenes can show cis–trans isomerism (Section 9.2).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET
E1.NEET Which alkane gives only one monochlorination product? (A) propane (B) 2-methylbutane (C) 2,2-dimethylpropane (D) pentane
💡 Show solution▶
Idea Count sets of equivalent H atoms: propane 2, 2-methylbutane 4, pentane 3, 2,2-dimethylpropane 1 (all twelve H are on equivalent $\mathrm{CH_{3}}$ groups).
Answer(C).
E2.JEE Main An aqueous solution containing both sodium ethanoate and sodium propanoate is electrolysed. The hydrocarbons formed at the anode are (A) ethane only (B) ethane, propane and butane (C) propane only (D) methane and ethane
💡 Show solution▶
Idea Kolbe electrolysis forms methyl ($\mathrm{^{\bullet}CH_{3}}$) and ethyl ($\mathrm{CH_{3}CH_{2}^{\bullet}}$) radicals at the anode; any two may pair.
Pairs methyl + methyl → ethane; methyl + ethyl → propane; ethyl + ethyl → butane. Methane is not formed (D).
Answer(B).
E3.Board · 3 marks Explain, with the steps of the mechanism, why the chlorination of methane needs light, and why it is called a chain reaction.
💡 Show solution▶
1 mark Initiation: light supplies the energy to break Cl–Cl homolytically, $\mathrm{Cl_{2}\xrightarrow{h\nu}2Cl^{\bullet}}$; without it no radicals form, so there is no reaction in the dark at room temperature.
1 mark Propagation: $\mathrm{Cl^{\bullet}+CH_{4}\rightarrow {}^{\bullet}CH_{3}+HCl}$; $\mathrm{{}^{\bullet}CH_{3}+Cl_{2}\rightarrow CH_{3}Cl+Cl^{\bullet}}$. The $\mathrm{Cl^{\bullet}}$ regenerated restarts the cycle — a chain; one initiation produces many product molecules.
1 mark Termination: $\mathrm{2Cl^{\bullet}\rightarrow Cl_{2}}$, $\mathrm{{}^{\bullet}CH_{3}+Cl^{\bullet}\rightarrow CH_{3}Cl}$, $\mathrm{2{}^{\bullet}CH_{3}\rightarrow C_{2}H_{6}}$ remove radicals and stop the chain.
E4.JEE MainNumerical answer. Complete combustion of one mole of an alkane requires $9.5$ mol of oxygen. The number of structural isomers of this alkane is ____.
💡 Show solution▶
Formula Rule 9.3: $\tfrac{3n+1}{2}=9.5$, so $3n+1=19$, $n=6$: $\mathrm{C_{6}H_{14}}$.
E5.JEE AdvancedOne or more options correct. For ethane: (A) the staggered conformation is more stable than the eclipsed one by about $12.5\ \mathrm{kJ\,mol^{-1}}$; (B) the staggered and eclipsed conformers can be separated by fractional distillation at low temperature; (C) in the eclipsed conformation the dihedral angle between the nearest C–H bonds is $0^{\circ}$; (D) rotation about the C–C bond requires the σ bond to break and re-form.
💡 Show solution▶
(A) True (torsional strain in the eclipsed form).
(B) False: the barrier is only about $5RT$ at room temperature; the conformers interconvert far too quickly to be separated.
(C) True, by definition of eclipsed.
(D) False: a σ bond is cylindrically symmetrical, so its overlap is unchanged by rotation.
Answer(A), (C).
E6.Board · Case study A student has sodium ethanoate, soda lime, sodium metal, dry ether and 2-bromopropane. (i) Write the equation for the preparation of methane from sodium ethanoate and name the reaction. (2 marks) (ii) Why can the student not make methane by electrolysing aqueous sodium ethanoate? (1 mark) (iii) Name the alkane formed when 2-bromopropane is heated with sodium in dry ether. (1 mark)
💡 Show solution▶
(i) $\mathrm{CH_{3}COONa(s)+NaOH(s)\xrightarrow[\Delta]{\text{CaO}}CH_{4}(g)+Na_{2}CO_{3}(s)}$ — decarboxylation (heating with soda lime).
(ii) Kolbe electrolysis joins two methyl radicals, giving ethane, not methane.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Cities from Delhi to Bogotá have moved buses and taxis from petrol and diesel to compressed natural gas (CNG, mostly methane), and LPG (propane and butane) is used for cooking across Asia and Africa. For the same amount of heat, how much $\mathrm{CO_{2}}$ does each alkane fuel release — and why does methane come out best?
Why it matters. Fuel switching is one of the fastest ways to cut urban air pollution and carbon emissions. Policy makers compare fuels by "grams of $\mathrm{CO_{2}}$ per megajoule", a number you can calculate from a balanced equation and an enthalpy of combustion.
What to do.
Write balanced equations (Rule 9.3) for the complete combustion of methane, propane, butane and octane (a model for petrol).
Using $\Delta_{c}H^{\circ}$ ($\mathrm{kJ\,mol^{-1}}$, water as liquid): $\mathrm{CH_{4}}$ $-890.3$, $\mathrm{C_{3}H_{8}}$ $-2220.0$, $\mathrm{C_{4}H_{10}}$ $-2878$, $\mathrm{C_{8}H_{18}(l)}$ $-5471$, calculate the mass of $\mathrm{CO_{2}}$ released per $\mathrm{MJ}$ of heat for each fuel.
Plot $\mathrm{g\,CO_{2}\,MJ^{-1}}$ against the number of carbon atoms. Explain the trend using the ratio of H to C atoms in $\mathrm{C_{n}H_{2n+2}}$.
Methane is itself a powerful greenhouse gas. Find an estimate of how much leakage from pipelines would cancel its advantage over petrol.
Tools: Rule 9.3, Section 5.3, a spreadsheet, a national energy or environment agency report on CNG and LPG.
Hand in: A table and graph of $\mathrm{g\,CO_{2}\,MJ^{-1}}$ for the four fuels, and a 200-word note for a city council comparing CNG buses with petrol ones.
Going further. Each extra $\mathrm{CH_{2}}$ adds about $650\ \mathrm{kJ\,mol^{-1}}$ to $|\Delta_{c}H^{\circ}|$. What value does $\mathrm{g\,CO_{2}\,MJ^{-1}}$ approach for a very long alkane, and how does hydrogen ($\Delta_{c}H^{\circ}=-285.8\ \mathrm{kJ\,mol^{-1}}$) compare?
Instructor note: $\mathrm{g\,CO_{2}\,MJ^{-1}}=1000\,n\times44.01/|\Delta_{c}H^{\circ}|$: $\mathrm{CH_{4}}$ $49.4$; $\mathrm{C_{3}H_{8}}$ $59.5$; $\mathrm{C_{4}H_{10}}$ $61.2$; $\mathrm{C_{8}H_{18}}$ $64.4$. The H : C ratio $(2n+2)/n$ falls from 4 towards 2, so a smaller share of the heat comes from forming water. Long-chain limit $\approx44.01/0.65\approx68\ \mathrm{g\,MJ^{-1}}$; hydrogen releases none. Studies commonly put the break-even leakage for CNG versus petrol vehicles at a few per cent of the gas.
🧪 Self-check quiz
0/5 correct
Q1. Which molecular formula belongs to an alkane?
Q2. The correct order of boiling points is
Q3. Which is a chain-propagation step in the chlorination of methane?
Q4. The eclipsed conformation of ethane is less stable than the staggered one by about
Q5. Which alkane cannot be obtained as the only organic product of a Wurtz reaction?