Hydrocarbons

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 9.

One reagent, two products: the puzzle that runs this chapter

Add HBr to propene and you get 2-bromopropane. Add the same HBr to the same propene with a pinch of peroxide and you get 1-bromopropane instead. Same atoms, same bond to break, and the product flips. Why?

The answer is that organic reactions are not decided by what is in the flask but by the path the electrons take. In one case the path goes through a carbocation, in the other through a free radical, and each path favours a different carbon. Hydrocarbons are the simplest compounds on which you can watch this happen, which is why almost every later organic chapter (haloalkanes, alcohols, aldehydes, amines) borrows its ideas from here.

A hydrocarbon contains only carbon and hydrogen. Petroleum and natural gas are the main sources. Natural gas is mostly methane and, compressed, is sold as CNG, while LPG is mainly butane with some propane. Everything else in this chapter is about how the arrangement of bonds decides how a hydrocarbon behaves.

The family and the one idea that sorts it

Cycloalkanes are saturated rings with the formula CₙH₂ₙ. The idea that sorts the whole table is how tightly the electrons are held. In an alkane every electron sits in a strong σ bond, so there is nothing for an attacker to grab, and only an aggressive free radical can force a hydrogen out. In an alkene the π bond is a cloud of electrons above and below the plane, held more loosely than the σ bond, so electron-hungry reagents (electrophiles) are drawn to it and simply add across it. In benzene the π electrons are spread round the whole ring, which makes the ring so stable that the molecule would rather swap a hydrogen (and keep the ring) than add anything.

More s character means the carbon holds its bonding electrons closer to the nucleus. That single fact will explain why ethyne is acidic, further down.

Naming and isomerism

Find the longest chain (for alkenes and alkynes, the longest chain that contains the multiple bond), number from the end that gives the lowest locants (the multiple bond gets priority over substituents), name the substituents alphabetically, and finish with the suffix: -ane, -ene or -yne.

Isomerism. Alkanes show chain isomerism only: C₄H₁₀ has 2 isomers, C₅H₁₂ has 3 (pentane, 2-methylbutane, 2,2-dimethylpropane), C₆H₁₄ has 5 and C₁₀H₂₂ has 75. Alkenes add position isomerism (but-1-ene and but-2-ene) and geometrical (cis–trans) isomerism. Alkynes show chain isomerism from C₅H₈ onwards and position isomerism from C₄H₆.

Cis–trans isomerism needs each carbon of the double bond to carry two different groups, because the double bond cannot rotate. But-2-ene has it. But-1-ene does not (one carbon has two hydrogens) and neither does 2-methylbut-2-ene.

Alkanes: strong bonds, patient chemistry

Preparation. The methods differ in what happens to the carbon count, which is exactly what CUET tests.

Three consequences follow. Wurtz reaction with two different alkyl halides gives a mixture of three alkanes (R–R, R′–R′ and R–R′), so it is only useful for symmetrical, even carbon products. Kolbe electrolysis can never give methane, since two R groups must join and R has at least one carbon. And decarboxylation is the route to methane (from sodium acetate), because it removes one carbon rather than joining two.

Worked example: carbon counting. Predict the alkane from (a) sodium propanoate with sodalime, heated, and (b) aqueous sodium propanoate electrolysed.

The common wrong step: applying the same rule to both. Decarboxylation subtracts one carbon, while Kolbe doubles R.

Pause and predict: you heat a mixture of CH₃Cl and C₂H₅Cl with sodium in dry ether. How many different alkanes can form?

Check your answer: three. CH₃Cl + CH₃Cl gives ethane, C₂H₅Cl + C₂H₅Cl gives butane, and the cross coupling gives propane. This is why Wurtz is a poor way to make an alkane such as propane.

Physical properties. Alkanes are non-polar, held together only by weak van der Waals forces. C₁–C₄ are gases, C₅–C₁₇ are liquids and C₁₈ onwards are solids. Boiling point rises with molecular mass, because bigger molecules have more surface to attract each other. Branching lowers the boiling point, because a compact molecule has less surface in contact with its neighbours.

Chemical reactions. Alkanes are called paraffins ("little affinity") because they resist most reagents. They react under forcing conditions.

Halogens rival each other in reactivity: F₂ > Cl₂ > Br₂ > I₂. Fluorination is violent, and iodination is so reversible that it needs an oxidising agent to remove the HI. For the hydrogen being replaced, the rate order is 3° > 2° > 1°, because the tertiary radical is the most stable.

How chlorination really happens. Halogenation is a free radical chain reaction, and the three-stage pattern appears in many exam questions.

Conformations of ethane. The C–C σ bond allows free rotation, so ethane has infinitely many arrangements called conformations. The two extremes are staggered (hydrogens on the two carbons as far apart as possible) and eclipsed (hydrogens directly in line). In the eclipsed form the electron clouds of the C–H bonds repel each other, which is called torsional strain. The staggered form is the most stable, and the eclipsed form is higher in energy by about 12.5 kJ mol⁻¹. That barrier is so small that molecules rotate freely at room temperature, so you can never isolate one conformer. In a Newman projection you look straight down the C–C bond, with the front carbon as a point and the rear carbon as a circle; a Sawhorse projection views the bond from slightly above.

Beyond the textbook. For the exam: branching lowers the boiling point, and isomerisation and aromatisation turn straight chain alkanes into branched and aromatic ones. In reality: petrol quality is rated by octane number, which is defined using 2,2,4-trimethylpentane (100) and n-heptane (0). Refineries run exactly these reactions to convert straight chain alkanes into branched and aromatic ones that burn smoothly instead of knocking. Why it matters: the dull-looking reactions in the table above are what sits behind the fuel in a car tank.

Alkenes: where the π bond does the talking

The C=C bond is one σ bond (head-on overlap of sp² orbitals, 397 kJ mol⁻¹) and one π bond (sideways overlap of 2p orbitals, 284 kJ mol⁻¹). The π bond is the weaker, more exposed bond, and that is where reagents attack.

Preparation.

Saytzeff rule. When elimination can go two ways, the more substituted (more stable) alkene is the major product. 2-Bromobutane with alcoholic KOH gives mainly but-2-ene, with but-1-ene as the minor product.

Reactions of alkenes.

Markovnikov's rule and why it holds. When HX adds to an unsymmetrical alkene, the negative part (X) goes to the carbon with fewer hydrogens. A rule you can memorise is fine, but the reason is what lets you handle new alkenes. The reaction goes in two steps, and the first one decides everything.

So the rule is really "the H goes to the carbon that already has more H and Br lands on the better carbocation". For 2-methylpropene, (CH₃)₂C=CH₂ + HBr gives (CH₃)₃CBr (2-bromo-2-methylpropane) through a tertiary carbocation.

Beyond the textbook. For the exam: addition of HX follows Markovnikov's rule and the product is the one from the more stable carbocation. In reality: a carbocation will rearrange if a more stable one is nearby. 3-Methylbut-1-ene with HCl gives mainly 2-chloro-2-methylbutane, because the first (secondary) carbocation takes a hydrogen with its bonding pair from the neighbouring carbon and becomes tertiary. Why it matters: the same rearrangement explains why Friedel–Crafts alkylation of benzene with n-propyl chloride gives isopropylbenzene, a favourite NTA twist.

The peroxide (Kharasch) effect. With HBr and a peroxide, the addition runs through free radicals, not carbocations, and the orientation reverses.

So Br ends up on the terminal carbon: 1-bromopropane. The effect is seen only with HBr. The H–Cl bond (430.5 kJ mol⁻¹) is too strong for the radical step to occur, and the H–I bond (296.8 kJ mol⁻¹) is so weak that iodine atoms just pair up to give I₂ instead of adding to the double bond.

Pause and predict: propene is treated with HCl in the presence of peroxide. What is the product?

Check your answer: 2-chloropropane, the ordinary Markovnikov product. HCl cannot sustain the radical chain, so the polar carbocation path takes over and the peroxide changes nothing.

Ozonolysis: reading a molecule by cutting it. Ozone cleaves the C=C bond, and Zn / H₂O replaces each end of the old double bond with a carbonyl group (C=O). A carbon with an H becomes an aldehyde, and a carbon with two alkyl groups becomes a ketone. Because the pieces tell you where the double bond was, ozonolysis is used to locate it.

Worked example: working backwards. An alkene on ozonolysis (O₃, then Zn / H₂O) gives propanal and pentan-3-one. Find the alkene.

The common wrong step: joining the oxygens instead of the carbons. The carbonyl carbons are the ones that were double bonded.

For the forward case, 2-methylbut-2-ene, (CH₃)₂C=CHCH₃, gives propanone (from the (CH₃)₂C= end) and ethanal (from the =CHCH₃ end). With hot acidified KMnO₄ the aldehyde end is oxidised further, so the same alkene gives propanone and ethanoic acid.

Beyond the textbook. For the exam: ethene is a starting material for polythene and a gas that decolourises bromine water. In reality: ethene is also a plant hormone. Plants make it themselves, and it triggers fruit ripening, which is why a ripe banana in a bag speeds up the green ones next to it. Why it matters: the simplest alkene is both an industrial feedstock and a biological signal.

Alkynes: the acidic hydrocarbon

The triple bond is one σ bond and two π bonds at right angles, so the π cloud forms a cylinder around the carbon–carbon axis. The molecule is linear (H–C≡C–H, 180°). Ethyne is the only common hydrocarbon you will meet that is a weak acid, and that comes straight from the s character figure above.

Why terminal alkynes are acidic. An sp carbon is 50% s, and s electrons sit nearer the nucleus. So the sp carbon is the most electronegative of the three types, holds the C–H pair more tightly, and the H comes off as H⁺ more easily. The acidity order is HC≡CH > H₂C=CH₂ > CH₃–CH₃.

But-2-yne has no hydrogen on a triple-bonded carbon, so none of these tests work. That is the usual distractor.

Preparation.

Reactions.

Hydration, the step students miss. The first product, an enol (a molecule with C=C–OH), is unstable and flips (tautomerises) to the carbonyl form. Ethyne gives ethanal (CH₃CHO), and propyne gives propanone (CH₃COCH₃) because water follows Markovnikov's rule. Ethyne is the only alkyne that gives an aldehyde. Every other alkyne gives a ketone.

Pause and predict: which of propyne and but-2-yne will give a white precipitate with ammoniacal silver nitrate?

Check your answer: propyne only. It has the terminal ≡C–H. But-2-yne, CH₃C≡CCH₃, has no acidic hydrogen.

Beyond the textbook. For the exam: ethyne burns with a sooty flame and is made from calcium carbide and water. In reality: burned in pure oxygen, ethyne gives a flame hotter than 3000 °C, which is why oxyacetylene torches cut and weld steel. Why it matters: the C≡C bond holds a lot of energy, and the flame shows that energy being released.

Benzene: the molecule that breaks the rules

Benzene, C₆H₆, was isolated by Faraday in 1825. It should be wildly unsaturated, yet it does not decolourise bromine water or Baeyer's reagent and does not do addition under mild conditions. Why?

Structure. Kekulé (1865) proposed a hexagon with alternating single and double bonds that swap places rapidly. We now say benzene is a resonance hybrid of the two Kekulé structures, so all six C–C bonds are identical at 139 pm, between a C–C single bond (154 pm) and a C=C double bond (134 pm). Orbital theory explains the rest. Each carbon is sp² hybridised and uses its three sp² orbitals for σ bonds to two carbons and one hydrogen. The leftover p orbital on each carbon points up and down, and all six overlap sideways to form two doughnut shaped clouds above and below the ring. The six π electrons are delocalised over all six carbons, which is what gives benzene its unusual stability.

Aromaticity (Hückel's rule). A molecule is aromatic if it is planar, cyclic, fully conjugated and has $(4n + 2)$ π electrons, with $n = 0, 1, 2$ and so on:

$$N_{\pi} = 4n + 2$$

Benzene has 6 π electrons ($n = 1$). Naphthalene has 10 ($n = 2$) and is aromatic too.

Beyond the textbook. For the exam: benzene is stable because of resonance and delocalised π electrons. In reality: you can measure that stability. Hydrogenating one C=C in cyclohexene releases about 120 kJ mol⁻¹, so a hypothetical ring with three isolated double bonds should release about 360 kJ mol⁻¹. Benzene releases only about 208 kJ mol⁻¹, which means it is about 150 kJ mol⁻¹ more stable than the alternating structure predicts. Why it matters: that missing 150 kJ mol⁻¹ is the energy a reagent would have to pay to destroy the aromatic ring, and that is why benzene substitutes rather than adds.

Preparation. Three routes: cyclic polymerisation of ethyne (above), decarboxylation of sodium benzoate with sodalime (C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃), and reduction of phenol vapour over heated zinc dust (C₆H₅OH + Zn → C₆H₆ + ZnO).

Electrophilic aromatic substitution (EAS). This is the reaction family of benzene. An electrophile replaces H and the ring survives.

With excess chlorine and anhydrous AlCl₃ in the dark, all six hydrogens can be replaced: C₆H₆ + 6Cl₂ → C₆Cl₆ + 6HCl. The electrophile in nitration comes from HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻, with the sulphuric acid acting as a stronger acid than nitric acid and forcing it to lose water.

Addition reactions (forcing conditions only). Under vigorous conditions benzene will add: C₆H₆ + 3H₂ → C₆H₁₂ (cyclohexane) over Ni at high temperature and pressure, and C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (benzene hexachloride, BHC or gammaxane) in UV light. Benzene burns with a smoky, sooty flame because it is carbon rich: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O.

Directive influence. When a ring already has a substituent, the second incoming group goes to a predictable place. The substituent either pushes electrons into the ring (activating, usually o/p-directing) or pulls them out (deactivating, usually m-directing).

For a meta director, ortho and para carbons lose electron density the most, so the meta carbon is the "least poor" position and the electrophile attacks there.

Worked example: order of steps matters. Starting from benzene, prepare m-bromonitrobenzene (1-bromo-3-nitrobenzene).

The common wrong step: brominating first. The –Br group directs to ortho and para, so nitration afterwards gives mainly o- and p-bromonitrobenzene, not the meta isomer.

Pause and predict: which is nitrated faster, toluene or chlorobenzene?

Check your answer: toluene. The –CH₃ group activates the ring, while –Cl deactivates it by its –I effect. Both direct to ortho and para, but toluene reacts faster than benzene and chlorobenzene reacts slower.

Toxicity. Benzene and polynuclear aromatic hydrocarbons with more than two fused benzene rings (for example benzpyrene) are toxic and cancer causing, because they damage DNA. They form during incomplete combustion of tobacco, coal and petroleum.

Reagent to product: quick recognition

Exam traps

60-second revision

Where this chapter connects

• Before this: Organic Chemistry: Some Basic Principles and Techniques gives you the inductive effect, hyperconjugation, resonance and carbocation stability used in every mechanism here.
• Next: Haloalkanes and Haloarenes shows what the alkyl halides made here do next, and Alcohols, Phenols and Ethers uses the same hydration and dehydration ideas in reverse.
• Forward link: Aldehydes, Ketones and Carboxylic Acids picks up the carbonyl compounds that ozonolysis and alkyne hydration produce, and Amines starts from nitrobenzene.