Hydrocarbons

The chapter moves through four hydrocarbon families in increasing order of unsaturation. It opens with classification (9.1) before building alkanes (9.2) from methane upward: nomenclature, chain isomerism, preparation methods, physical and chemical properties culminating in the free-radical halogenation mechanism, and the three-dimensional idea of conformation. Section 9.3 covers alkenes: the double bond's structure, structural and geometrical isomerism, preparation, and the electrophilic addition chemistry that produces Markovnikov's rule and its peroxide-effect exception. Section 9.4 covers alkynes in parallel — the triple bond, the acidity of the terminal C-H, preparation from calcium carbide, and addition and polymerisation, including the cyclic trimerisation of ethyne to benzene that bridges aliphatic and aromatic chemistry. Section 9.5 is benzene and the arenes: the Kekulé structure and its resonance resolution, aromaticity (Hückel's rule), the electrophilic substitution mechanism, and the directive influence of ring substituents. The chapter closes (9.6) with the carcinogenicity of polynuclear aromatic hydrocarbons. Two threads run throughout: identify the reactive intermediate (free radical, carbocation or arenium ion) before predicting a product, and check whether a reaction is addition (alkenes, alkynes, benzene only under force) or substitution (alkanes, benzene under ordinary conditions).

Classifying hydrocarbons

Hydrocarbons contain only carbon and hydrogen, and the type of carbon-carbon bond present sorts them into three families. Saturated hydrocarbons have only single C-C and C-H bonds: an open chain gives the alkanes, while a closed ring of single-bonded carbons gives the cycloalkanes. Unsaturated hydrocarbons carry one or more carbon-carbon multiple bonds — a double bond defines the alkenes, a triple bond the alkynes. Aromatic hydrocarbons are a special class of cyclic compounds built around benzene's uniquely stable, delocalised ring, and behave chemically unlike ordinary alkenes despite containing formal double bonds. Every family obeys carbon's tetravalency and hydrogen's monovalency, which is why a model built from toothpicks and plasticine balls always closes every valency exactly.

Alkane structure, hybridisation and nomenclature

Methane, $CH_4$, is the first alkane; replacing one hydrogen by a $-CH_3$ group generates ethane, $C_2H_6$, and repeating this generates the whole family, $C_nH_{2n+2}$. Each carbon is $sp^3$ hybridised, giving methane a tetrahedral shape with H-C-H angles of $109.5\degree$; the C-C bond length is 154 pm and C-H about 112 pm, both formed by head-on sigma overlap ($sp^3$-$sp^3$ for C-C, $sp^3$-1s for C-H). Because alkanes resisted acids, bases and most reagents, they were once called paraffins (Latin parum affinis, little affinity).

where: $n$ is the number of carbon atoms in the chain. IUPAC naming picks the longest continuous chain, numbers it for the lowest locants to substituents (ties broken alphabetically), and names branches as alkyl groups ($C_nH_{2n+1}$, an alkane minus one hydrogen). Carbons are also classified by how many other carbons they touch: primary touches one (or none, as in methane), secondary touches two, tertiary touches three, quaternary touches four — terminal carbons are always primary.

Isomerism in alkanes

Methane, ethane and propane have only one possible structure each, but from butane onward more than one arrangement of the same formula is possible. Butane ($C_4H_{10}$) has two chain isomers — straight-chain n-butane (b.p. 273 K) and branched isobutane (b.p. 261 K) — differing only in skeleton branching, not formula. Pentane ($C_5H_{12}$) has three: n-pentane (309.1 K), isopentane (301 K) and neopentane (282.5 K). The count grows fast: hexane has five, heptane nine, decane 75 — hand-counting becomes impractical well before ten carbons.

Preparation of alkanes

Petroleum and natural gas are the industrial sources of alkanes, but four laboratory routes matter for mechanism. Hydrogenation adds $H_2$ across an alkene or alkyne over finely divided Pt, Pd or Ni (Pt/Pd at room temperature, Ni needs heat and pressure): $CH_2{=}CH_2 + H_2 \xrightarrow{\text{Pt/Pd/Ni}} CH_3{-}CH_3$. Alkyl halides (except fluorides) are reduced by zinc and dilute HCl to the alkane. The Wurtz reaction couples two molecules of an alkyl halide with sodium in dry ether, $2R{-}X + 2Na \xrightarrow{\text{dry ether}} R{-}R + 2NaX$, useful only for even-carbon alkanes; two different halides give a statistical mixture of three products. Sodium salts of carboxylic acids undergo decarboxylation on heating with soda lime (NaOH + CaO), losing $CO_2$ to give an alkane with one fewer carbon — sodium ethanoate gives methane, sodium butanoate gives propane. Kolbe's electrolytic method electrolyses a carboxylate solution; at the anode the carboxylate loses an electron and $CO_2$ to form a radical, and two radicals combine, so sodium acetate gives ethane, not methane — coupling always joins two fragments, which is why Kolbe electrolysis cannot make methane.

Physical and chemical properties of alkanes

Alkanes are almost non-polar, held together only by weak van der Waals forces: $C_1$-$C_4$ are gases, $C_5$-$C_{17}$ liquids, $C_{18}$ and above solids, all colourless, odourless and insoluble in water but soluble in non-polar solvents ("like dissolves like", why petrol removes grease). Boiling point rises with molecular mass but falls with branching, since a spherical molecule has less surface area to grip — the three pentane isomers show this (309.1, 301, 282.5 K). Chemically, alkanes undergo substitution (H replaced by halogen, nitro or sulphonic groups, mechanism below), combustion (complete oxidation to $CO_2$ and $H_2O$ with a large heat release, or incomplete combustion to carbon black when oxygen is limited), controlled oxidation (regulated $O_2$ over a catalyst gives methanol, methanal or ethanoic acid; tertiary C-H is oxidised by $KMnO_4$ to a tertiary alcohol), isomerisation (rearranging to branched isomers over anhydrous $AlCl_3$/HCl), aromatization ($C_6$+ n-alkanes dehydrogenate and cyclise to benzene homologues over V/Mo/Cr oxides at 773 K, 10-20 atm), reaction with steam ($CH_4 + H_2O \xrightarrow{\text{Ni}, 1273\,\text{K}} CO + 3H_2$, industrially making dihydrogen), and pyrolysis (cracking), breaking higher alkanes into smaller alkanes and alkenes by heat alone.

where: this is the general combustion equation for any alkane; methane's combustion releases about 890 kJ per mole and butane's about 2875.84 kJ per mole, which is why alkane-rich fuels like LPG and CNG are prized as high-energy, comparatively clean-burning sources of heat.

Conformations of ethane

Rotation about a C-C sigma bond is essentially free because the sigma orbital is symmetric about the internuclear axis, so rotating one carbon does not break the bond. Hindered only by a small energy barrier of roughly 1-20 kJ per mole from weak repulsion between bonds on adjacent carbons (torsional strain), this generates infinite spatial arrangements called conformations or conformers, of which the two extremes are the eclipsed conformation (hydrogens as close together as possible) and the staggered conformation (hydrogens as far apart as possible); any in-between arrangement is a skew conformation. These are drawn using Sawhorse projections (the C-C bond as a tilted line) or Newman projections (viewed down the C-C axis, front carbon a point, rear carbon a circle).

In the staggered form, the C-H electron clouds on the two carbons are as far apart as possible, giving minimum repulsion, minimum energy and maximum stability; the eclipsed form has maximum torsional strain. The energy gap between the extremes is only about 12.5 kJ per mole — small enough that ordinary thermal collisions at room temperature let ethane pass over the barrier constantly, so rotation is "almost free" in practice, and the individual conformers have never been isolated even though staggered is always the more populated form.

Alkenes: the structure of the double bond and nomenclature

Alkenes are unsaturated hydrocarbons with at least one carbon-carbon double bond, general formula $C_nH_{2n}$; also called olefins ("oil-forming") since ethene forms an oily liquid with chlorine. Each doubly bonded carbon is $sp^2$ hybridised: a strong sigma bond (about 397 kJ per mole) forms by head-on $sp^2$ overlap, and a weaker pi bond (about 284 kJ per mole) by sideways overlap of the two unhybridised $2p$ orbitals, giving a total C=C enthalpy of about 681 kJ per mole — built from one strong and one comparatively fragile component. The C=C bond length, 133 pm, is shorter than the 154 pm C-C single bond. Because the pi electrons are loosely held and mobile, alkenes are attacked readily by electron-seeking electrophilic reagents, the basis of their addition chemistry. IUPAC nomenclature selects the longest chain containing the double bond, numbers from the end nearer it, and replaces "-ane" with "-ene": propene, but-1-ene, but-2-ene, 2-methylprop-1-ene.

Isomerism in alkenes

Ethene and propene have only one structure, but $C_4H_8$ already shows both chain isomerism (but-1-ene/but-2-ene versus branched 2-methylprop-1-ene) and position isomerism (but-1-ene versus but-2-ene). Restricted rotation about C=C gives a third kind: when each doubly bonded carbon carries two different groups, two spatial arrangements exist, called geometrical or cis-trans isomers. The cis isomer has two identical groups on the same side, the trans on opposite sides; cis-but-2-ene has a dipole moment of 0.33 D while trans-but-2-ene is essentially non-polar (its C-CH3 dipoles cancel). The trans isomer of a solid alkene generally has the higher melting point, its symmetric shape packing more efficiently.

Preparation of alkenes

Alkynes undergo partial reduction with calculated $H_2$ over palladised charcoal deactivated by sulphur compounds or quinoline (Lindlar's catalyst), giving the cis-alkene; the same alkyne with sodium in liquid ammonia gives the trans-alkene. Alkyl halides heated with alcoholic KOH undergo dehydrohalogenation (a $\beta$-elimination); rate depends on halogen (I greater than Br greater than Cl) and alkyl group (tertiary greater than secondary greater than primary). Vicinal dihalides lose $ZnX_2$ with zinc dust, a dehalogenation regenerating the double bond. Alcohols heated with concentrated $H_2SO_4$ lose water in an acidic dehydration, another $\beta$-elimination.

Chemical properties of alkenes and Markovnikov's rule

Alkenes are rich sources of loosely held pi electrons and undergo electrophilic addition: $H_2$ over Pt/Pd/Ni gives the alkane; $Br_2$ or $Cl_2$ (not $I_2$) gives a vicinal dihalide, decolourising reddish-orange bromine in $CCl_4$ — a standard unsaturation test; cold concentrated $H_2SO_4$ gives an alkyl hydrogen sulphate; water, catalysed by a trace of concentrated $H_2SO_4$, gives an alcohol. For an unsymmetrical alkene, HX addition follows Markovnikov's rule: the negative part of the addendum attaches to the carbon with fewer hydrogens (derived below); under peroxides, HBr alone reverses this (the Kharash or peroxide effect, also derived below). Cold, dilute, alkaline $KMnO_4$ (Baeyer's reagent) oxidises the double bond to a vicinal diol, another decolourisation test; hot, acidic $KMnO_4$ or $K_2Cr_2O_7$ cleave it oxidatively to ketones and/or acids. Ozonolysis forms an ozonide that Zn/$H_2O$ cleaves reductively to two carbonyl compounds, precisely locating the double bond. High temperature, pressure and a catalyst drive polymerisation: ethene to polythene, propene to polypropene.

Alkynes: the structure of the triple bond

Alkynes contain at least one carbon-carbon triple bond, general formula $C_nH_{2n-2}$; the first stable member, ethyne (acetylene), burns in an oxyacetylene flame hot enough for metal welding. Each triple-bonded carbon is $sp$ hybridised: one $sp$-$sp$ sigma bond joins the carbons, each carbon's other $sp$ orbital bonds to hydrogen (linear H-C-C angle, $180\degree$), and the two remaining, perpendicular unhybridised $p$ orbitals on each carbon overlap sideways to form two pi bonds, giving a cylindrically symmetric electron cloud. The triple bond (823 kJ per mole, 120 pm) is stronger and shorter than the double bond (681 kJ per mole, 133 pm) and single bond (348 kJ per mole, 154 pm). IUPAC names replace "-ane" with "-yne"; but-1-yne and but-2-yne are position isomers.

Preparation and acidic character of alkynes

Industrially, ethyne is made from calcium carbide and water, $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$, the carbide itself coming from heating quicklime with coke, $CaO + 3C \rightarrow CaC_2 + CO$. In the laboratory, vicinal dihalides undergo a double dehydrohalogenation with alcoholic KOH (via an alkenyl halide, sometimes finished with sodamide). Alkynes show a behaviour alkanes and alkenes do not: terminal alkynes ($\equiv C{-}H$) are weakly acidic. Sodium metal or sodamide liberates $H_2$ from ethyne, forming sodium or disodium acetylide; propyne likewise forms sodium propynide with $NaNH_2$. The reason is hybridisation: the $sp$ orbital housing the C-H electrons has 50% s-character (versus 33% for $sp^2$, 25% for $sp^3$), and since s-orbitals hold electrons closer to the nucleus, the sp carbon pulls the shared pair more strongly, releasing the proton more easily. Acidity falls as $HC{\equiv}CH > CH_2{=}CH_2 > CH_3{-}CH_3$; internal alkynes like but-2-yne have no acidic hydrogen and do not react with Na or $NaNH_2$.

Addition reactions and polymerisation of alkynes

Like alkenes, alkynes undergo electrophilic addition, but can add two molecules of the same reagent since the triple bond accepts addition twice: $H_2$ over Pt/Pd/Ni gives first the alkene, then the alkane; $X_2$ gives first a dihaloalkene, then a tetrahalide; $HX$ (two molecules) gives a gem-dihalide (both halogens on one carbon), again by Markovnikov's rule via a resonance-stabilised vinylic cation. Water adds only with a $HgSO_4$/dilute $H_2SO_4$ catalyst at 333 K, giving (after enol-to-keto tautomerisation) a carbonyl compound — ethyne uniquely gives ethanal, other terminal alkynes give methyl ketones. Ethyne also polymerises: linear linkage gives conducting polyacetylene, used in battery electrodes, while a red-hot iron tube at 873 K gives cyclic trimerisation, three molecules combining to form benzene — the classic bridge from aliphatic to aromatic chemistry.

Aromatic hydrocarbons: nomenclature and the structure of benzene

Aromatic hydrocarbons ("arenes") were named for the pleasant odour many possess. All six hydrogens of benzene, $C_6H_6$, are chemically identical, so mono-substitution gives only one product, while di-substitution gives three position isomers: ortho (1,2), meta (1,3) and para (1,4) — toluene, o-, m- and p-xylene illustrate the pattern. Benzene's structure puzzled chemists for decades: it forms a triozonide (suggesting three double bonds) yet gives only one 1,2-disubstituted product, not the two a fixed alternating-bond pattern should allow. Kekulé's 1865 hexagonal structure, even amended with "oscillating" bonds, could not fully resolve this. The modern resolution is resonance: all six carbons are $sp^2$ hybridised, forming a planar ring of six C-C and six C-H sigma bonds, leaving one unhybridised $p$ orbital per carbon, overlapping sideways with both neighbours equally. This delocalises six pi electrons over all six carbons at once (two doughnut-shaped clouds above and below the ring) rather than confining them to three fixed double bonds. X-ray diffraction confirms every C-C bond in benzene is identical at 139 pm — intermediate between a single bond (154 pm) and a double bond (133 pm) — proof that no localised double bond exists and the delocalised structure beats either Kekulé form or the hypothetical fixed-bond cyclohexatriene.

Aromaticity, preparation and the electrophilic substitution mechanism

A ring system, whether or not it contains a benzene ring, is aromatic if it satisfies three conditions: planarity, complete delocalisation of pi electrons around the ring, and a pi-electron count of $(4n+2)$ for some whole number $n = 0, 1, 2, \ldots$ — Hückel's rule.

where: for benzene, six delocalised pi electrons fit $n=1$; ring systems whose pi-electron count fails this rule, or that are non-planar, are not aromatic, whatever their formal double bonds might suggest. Beyond isolation from coal tar, benzene can be made by cyclic polymerisation of ethyne, decarboxylation of sodium benzoate with soda lime, or reduction of phenol vapour over heated zinc dust. Chemically, arenes are dominated by electrophilic substitution: nitration (conc. $HNO_3$/conc. $H_2SO_4$, giving nitrobenzene), halogenation ($X_2$ with anhydrous $FeX_3$ or $AlCl_3$), sulphonation (fuming $H_2SO_4$, giving benzenesulphonic acid), Friedel-Crafts alkylation (R-X with anhydrous $AlCl_3$ — 1-chloropropane surprisingly gives isopropylbenzene, since the initial primary carbocation rearranges to the more stable secondary one first), and Friedel-Crafts acylation (an acyl halide or anhydride with $AlCl_3$, giving an aryl ketone such as acetophenone). Only under forcing conditions does benzene add reagents: $H_2$ with Ni at high temperature/pressure gives cyclohexane, $Cl_2$ under UV light gives benzene hexachloride ($C_6H_6Cl_6$, "BHC"). Combustion follows the general pattern, $C_6H_6 + \tfrac{15}{2}O_2 \rightarrow 6CO_2 + 3H_2O$, burning with a sooty flame from the high carbon-to-hydrogen ratio.

Carcinogenicity and toxicity

Benzene and polynuclear aromatic hydrocarbons (two or more fused benzene rings) are toxic, and many are carcinogenic. They form during incomplete combustion of tobacco, coal and petroleum, enter the body, undergo biochemical transformation, and can damage DNA, causing cancer.

Board level - Chain isomers of hexane and classifying its carbon atoms (NCERT Problem 9.1, extended)

Write the structures of all chain isomers of $C_6H_{14}$, give their IUPAC names, and for the most branched isomer state how many primary, secondary, tertiary and quaternary carbon atoms it contains.

Step 1. Confirm the formula fits the alkane family: $C_nH_{2n+2}$ with $n = 6$ gives $C_6H_{14}$, so every isomer is a fully saturated, open-chain hydrocarbon with only single bonds.

Step 2. Build the unbranched chain first: $CH_3{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}CH_3$, n-hexane. Then introduce one methyl branch on the five-carbon chain at each distinct position: 2-methylpentane and 3-methylpentane (branching at carbon 4 repeats 2-methylpentane by symmetry).

Step 3. Introduce two methyl branches on a four-carbon chain: both on carbon 2 gives 2,2-dimethylbutane, one each on carbons 2 and 3 gives 2,3-dimethylbutane. No further distinct skeleton exists, giving five isomers total.

Step 4. Classify the carbons of 2,2-dimethylbutane, $CH_3{-}C(CH_3)_2{-}CH_2{-}CH_3$: the central carbon (bonded to four carbons) is quaternary; the two methyl branches and terminal ethyl carbon are primary; the $CH_2$ carbon is secondary; there is no tertiary carbon here. Answer: five chain isomers — n-hexane, 2-methylpentane, 3-methylpentane, 2,3-dimethylbutane and 2,2-dimethylbutane; 2,2-dimethylbutane has four primary carbons, one secondary carbon and one quaternary carbon.

Board level - Choosing the correct longest chain when naming a branched alkane (NCERT Problem 9.5 style)

The name "2-Ethylpentane" is proposed for a compound drawn as $CH_3{-}CH(C_2H_5){-}CH_2{-}CH_2{-}CH_3$. Explain why this name is incorrect and give the correct IUPAC name.

Step 1. Draw the full carbon skeleton: a five-carbon "pentane" chain with a two-carbon ethyl branch on carbon 2. Counting every carbon regardless of which chain was originally intended gives $5 + 2 = 7$ carbons total.

Step 2. Search for the longest continuous chain through those seven carbons, not just the assumed one. Tracing through the ethyl branch and along the main chain reveals an unbroken run of six carbons — one longer than the five-carbon chain "2-Ethylpentane" was built on.

Step 3. IUPAC rules require the parent chain to be the longest available, so the parent must be hexane, not pentane; renumbering from the end nearer the substituent places a single methyl branch at carbon 3. Answer: the correct IUPAC name is 3-methylhexane; "2-Ethylpentane" is wrong because an ethyl branch on a five-carbon chain always hides a six-carbon (or longer) chain that must be chosen instead.

Competitive level (JEE / NEET) - Addition of HBr to hex-1-ene, with and without peroxide

Hex-1-ene, $CH_2{=}CH{-}CH_2{-}CH_2{-}CH_2{-}CH_3$, is treated with HBr (i) in the absence of peroxide and (ii) in the presence of benzoyl peroxide. Give the IUPAC name of the major product in each case.

Step 1. Without peroxide, the mechanism is ionic: $H^+$ adds to whichever carbon generates the more stable carbocation. Protonating the terminal carbon (C1) leaves the positive charge on C2, a secondary carbocation; protonating C2 would leave a far less stable primary carbocation on C1.

Step 2. The secondary carbocation at C2 is therefore favoured, and is captured by $Br^-$, placing bromine on C2 — the carbon with fewer hydrogens, as Markovnikov's rule predicts. Result (i): 2-bromohexane.

Step 3. With peroxide, the mechanism switches to free-radical (Kharash effect): a bromine radical adds first to the terminal, less hindered carbon (C1), generating the more stable secondary radical at C2, which then abstracts H from another HBr molecule.

Step 4. This places bromine permanently at C1, the terminal carbon. Answer: (i) without peroxide, 2-bromohexane (Markovnikov); (ii) with peroxide, 1-bromohexane (anti-Markovnikov, Kharash effect). Timing note: under three minutes. Spot the phrase "in the presence of peroxide" and immediately flip your expected product from internal to terminal halogen placement, without re-deriving the full mechanism unless asked. The common slip is applying the peroxide effect to a question that actually specifies HCl or HI — check the halogen first, since the effect is real only for HBr.

Competitive level (JEE / NEET) - Identifying an alkene from its ozonolysis products

An alkene A, on ozonolysis followed by Zn/H2O workup, gives a mixture of ethanal and pentan-3-one as the only two products. Deduce the structure and IUPAC name of A.

Step 1. Recall what ozonolysis does: it cleaves a C=C double bond completely, replacing it with two C=O double bonds, one on each fragment — so the two carbonyl products, rejoined at their carbonyl carbons with a double bond, must reconstruct the original alkene exactly.

Step 2. Mark each product's carbonyl carbon: ethanal, $CH_3{-CHO}$, has it bonded to one $CH_3$ and one H; pentan-3-one, $CH_3{-}CH_2{-CO-}CH_2{-}CH_3$, has it bonded to two ethyl groups.

Step 3. Rejoin the two carbonyl carbons with a double bond in place of the two oxygens, keeping every other substituent unchanged: $CH_3{-CH=}$ from ethanal plus $=C(C_2H_5)_2$ from pentan-3-one gives $CH_3{-}CH{=}C(C_2H_5)_2$.

Step 4. Number the chain: the longest chain through the double bond is five carbons (through one ethyl group), with the double bond starting at carbon 2 and an ethyl branch at carbon 3. Answer: A is 3-ethylpent-2-ene, $CH_3{-}CH{=}C(C_2H_5){-}CH_2{-}CH_3$, which cleaves on ozonolysis to ethanal and pentan-3-one. Timing note: aim for under four minutes. Draw both carbonyl fragments with the carbonyl carbon inside, delete the oxygens, and join those two carbons with a double bond — faster and more reliable than guessing the alkene and verifying it forward. Double-check by mentally re-cleaving: you should recover exactly the two given products.

The most frequent slip is applying Markovnikov's rule mechanically without checking for "peroxide" — the Kharash effect reverses regiochemistry for HBr only, never HCl or HI. A second error treats chain, position and geometrical isomerism as interchangeable: chain isomers differ in skeleton branching, position isomers differ in where a multiple bond or substituent sits on an identical chain, and geometrical isomers differ only in spatial arrangement about a restricted C=C bond, with identical connectivity. In naming, students often accept a plausible chain without searching for a longer one hidden through a branch — any alkyl substituent longer than methyl should trigger this check. With the Wurtz reaction, it only gives even-carbon alkanes reliably, and mixing two different alkyl halides gives a mixture of three products, not one. With ozonolysis, remember the reductive Zn/H2O workup stops the reaction at the carbonyl stage. Finally, halogens on benzene are a favourite exception: ortho/para directing (lone-pair resonance) yet still net deactivating (inductive withdrawal) — a combination students often "simplify" incorrectly.

• Classify any hydrocarbon as an alkane, alkene, alkyne or arene from its formula and bonding, stating the general formula for each family.
• Name a hydrocarbon by IUPAC rules, choosing the longest chain and lowest locant set, and reverse the process to draw a structure from a name.
• Count and draw the chain, position and geometrical isomers of a molecular formula, and explain what distinguishes each type.
• Explain why C-C rotation is hindered rather than free, and identify the staggered and eclipsed conformations of ethane by relative stability.
• State Markovnikov's rule, derive it from carbocation stability, and predict when and why the peroxide effect reverses it for HBr specifically.
• Write the free-radical chain mechanism (initiation, propagation, termination) for alkane halogenation, and explain byproducts such as ethane from methane chlorination.
• Explain, using resonance and bond-length evidence, why benzene is more stable than a hypothetical fixed-bond cyclohexatriene, and apply Huckel's rule to judge aromaticity.
• Write the three-step electrophilic aromatic substitution mechanism and predict whether a ring substituent is ortho/para or meta directing, activating or deactivating.
• Deduce an alkene's structure from its ozonolysis products, and vice versa.
• Distinguish the preparation methods for alkanes, alkenes and alkynes and state the reagents and conditions each requires.

Ask the tutor: Want to work through drawing a free-radical or carbocation mechanism step by step for a specific hydrocarbon reaction, or practice predicting ortho/para versus meta products for a substituent of your choice?