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Hydrocarbons is the chapter where organic chemistry either starts to make sense or turns into a list of reactions to memorise. The difference is whether you see the pattern. Each class of hydrocarbon has one main kind of reaction, and most of the named reactions are variations on it. This post gives you that map, works through the questions that appear most often, and ends with a practice plan.
What does the syllabus include?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus share a hydrocarbons unit. It covers conformations of ethane and the mechanism of halogenation of alkanes; for alkenes, geometrical isomerism, electrophilic addition, Markovnikov's rule, the peroxide effect, ozonolysis and polymerisation; for alkynes, acidic character and addition reactions; and for benzene, aromaticity, electrophilic substitution, Friedel-Crafts reactions and the directive influence of substituents. The NCERT Class 11 chapter on hydrocarbons covers all of it. Check the current year's syllabus.
What is the one reaction type for each class?
Alkanes: free-radical substitution. With only strong, non-polar C–H and C–C bonds, alkanes react with chlorine or bromine in light by a chain reaction of radicals: initiation, propagation, termination.
Alkenes: electrophilic addition. The π bond is a source of electrons. An electrophile such as H⁺ or Br⁺ adds first, then a nucleophile completes the addition.
Alkynes: addition, plus acidity. Two π bonds take up two molecules of H₂, X₂ or HX. Terminal alkynes are also weakly acidic.
Benzene: electrophilic substitution. Addition would destroy the aromatic ring, so benzene swaps an H for the electrophile and keeps its six π electrons.
The map: how the classes connect
Alkyne → alkene: H₂ with Lindlar's catalyst gives the cis alkene.
Alkene → alkane: H₂ with Ni, Pt or Pd.
Alkene → haloalkane: HX by Markovnikov's rule, or HBr with peroxide anti-Markovnikov.
Alkene → 1,2-diol: cold, dilute alkaline KMnO₄ (Baeyer's reagent). The purple colour disappears, which is a test for unsaturation.
Alkene → aldehydes and ketones: ozonolysis, then Zn and water.
Haloalkane → alkane: Wurtz reaction with sodium in dry ether; sodium salts of carboxylic acids also give alkanes by Kolbe's electrolysis.
Benzene → nitrobenzene, bromobenzene, toluene or acetophenone: nitration, halogenation, Friedel-Crafts alkylation and acylation.
Worked example 1: Markovnikov or peroxide effect?
What is the major product when propene reacts with (a) HBr, (b) HBr with benzoyl peroxide, and (c) HCl with peroxide?
(a) H⁺ adds to the end carbon so the more stable secondary carbocation forms on the middle carbon. Br⁻ attacks there: the product is 2-bromopropane.
(b) Peroxide starts a free-radical chain. A bromine radical adds to the end carbon so the more stable secondary radical forms. The product is 1-bromopropane.
(c) The peroxide effect does not work with HCl. NCERT explains that the H–Cl bond (430.5 kJ/mol) is stronger than H–Br (363.7 kJ/mol) and is not broken by the free radical. The product is 2-chloropropane, by Markovnikov's rule.
The common mistake: applying the peroxide effect to every hydrogen halide. It is an HBr-only effect.
Worked example 2: working backwards from ozonolysis
An alkene C₅H₁₀ gives propanone and ethanal on ozonolysis. What is it?
Ozonolysis cuts the C=C and turns each end into C=O.
Write the products with their C=O carbons facing each other: (CH₃)₂C=O and O=CH–CH₃.
Remove both oxygens and join the two carbons with a double bond: (CH₃)₂C=CH–CH₃.
Check the formula: five carbons and ten hydrogens. The alkene is 2-methylbut-2-ene.
Name it carefully; our guide to IUPAC naming of organic compounds covers the rules.
Why are terminal alkynes acidic?
The carbon in C≡C–H is sp hybridised, with 50% s character, so it holds the electron pair of the C–H bond close to the nucleus. NCERT gives the order HC≡CH > H₂C=CH₂ > CH₃–CH₃, and among alkynes HC≡CH > CH₃–C≡CH >> CH₃–C≡C–CH₃. Ethyne reacts with sodium or sodamide to form sodium acetylide; ethene and ethane do not. But-2-yne has no hydrogen on a triple-bonded carbon, so it gives no acetylide.
How do substituents direct benzene substitution?
A group already on the ring decides where the next one goes.
Ortho- and para-directing: –OH, –NH₂, –OCH₃, –CH₃ and halogens. Most of these also activate the ring; halogens direct ortho and para but deactivate it.
Meta-directing: –NO₂, –CN, –CHO, –COOH and –SO₃H. These withdraw electrons and deactivate the ring.
Worked example 3: make m-bromonitrobenzene from benzene. Nitrate first, because –NO₂ sends the bromine to the meta position. If you brominate first, the bromine directs ortho and para, and you get o- and p-bromonitrobenzene instead. Order of steps is a common JEE question.
The reasons behind these effects are in our post on inductive, resonance and hyperconjugation effects.
Why do students drop mechanisms, and why should you keep them?
Grove, Cooper and Rush (2012) followed second-year organic chemistry students through a year and found that a large proportion chose not to use the curved-arrow mechanism convention in their work. Skipping mechanisms feels faster, but the map above only works if you can say why each product forms. When you cannot, you are back to memorising. Our post on learning organic mechanisms instead of lists explains how to practise them.
When a conversion question stumps you, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see which reaction connects each step.
A practice routine for hydrocarbons
Draw the map from memory on one page, with reagents on every arrow.
Do ten "predict the product" questions, writing the intermediate (carbocation or radical) each time.
Do five reverse ozonolysis problems.
Do five two-step benzene syntheses where the order of steps matters.
Redraw the map a week later and compare it with the first version.
Today, write the products of propene with HBr, with HBr and peroxide, and with Baeyer's reagent, naming each.
