Why mechanism families beat compound classes
Learning alkene reactions as a list means memorising hydration, hydrohalogenation, hydroboration, halogenation and ozonolysis separately. Learning them as electrophilic addition means one idea: the pi bond is electron-rich, an electrophile attacks it, and the outcome is decided by which intermediate is more stable.
That reframing is what makes Markovnikov's rule predictable rather than arbitrary, and it explains immediately why peroxides reverse the outcome - the radical route has a different intermediate with a different stability order. Preparing by mechanism converts a list into a small set of principles.
How JEE actually asks Hydrocarbons
NTA does not publish chapter-wise weightage, so figures online are coaching estimates from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
The recurring types are: predict the major product of an addition, identify an alkene from its ozonolysis products, decide where a substituent directs on a benzene ring, and distinguish compounds by a chemical test. Reaction-condition questions - which reagent gives which product - are also common and reward knowing the small set of distinguishing conditions.
Key concepts, compressed
- ✓Alkanes are saturated and relatively unreactive, undergoing free-radical substitution under light or heat.
- ✓Alkenes and alkynes are electron-rich at the multiple bond and undergo electrophilic addition.
- ✓Markovnikov's rule follows from carbocation stability: the electrophile adds to give the more stable intermediate.
- ✓Aromatic compounds resist addition because it would destroy aromatic stabilisation, so they undergo substitution instead.
- ✓Substituents on a benzene ring direct incoming groups to specific positions and change the ring's reactivity.
- ✓Chemical tests distinguish the classes: bromine water and Baeyer's reagent for unsaturation, ammoniacal silver nitrate for terminal alkynes.
Results you need before attempting the questions
| Item | Result | Note |
|---|---|---|
| Alkane formula | CₙH₂ₙ₊₂ | sp³, tetrahedral |
| Alkene formula | CₙH₂ₙ | sp², planar, 120° |
| Alkyne formula | CₙH₂ₙ₋₂ | sp, linear, 180° |
| Degree of unsaturation | (2C + 2 + N − H − X)/2 | |
| Markovnikov | electrophile adds to give the more stable carbocation | |
| Peroxide effect | anti-Markovnikov, HBr only | radical mechanism |
| Hydroboration-oxidation | anti-Markovnikov, syn addition | no rearrangement |
| Ozonolysis | cleaves C=C, oxygen on each carbon | |
| Baeyer's reagent | cold dilute alkaline KMnO₄, decolourised by C=C | |
| Terminal alkyne test | white precipitate with ammoniacal AgNO₃ | |
| Lindlar catalyst | alkyne to cis-alkene | |
| Na in liquid NH₃ | alkyne to trans-alkene | |
| Activating groups | electron-donating, ortho-para directing | |
| Deactivating groups | electron-withdrawing, meta directing | halogens are the exception |
The five mistakes that cost the most marks
- ✓Applying the peroxide effect to HCl or HI. Only HBr gives anti-Markovnikov addition with peroxides.
- ✓Treating internal alkynes as acidic. The acidic hydrogen must be attached to the sp carbon, so only terminal alkynes react with sodium amide or ammoniacal silver nitrate.
- ✓Classifying halogens as meta directors because they deactivate. They deactivate by induction but direct ortho-para by resonance - the standard exception.
- ✓Breaking the wrong bond in ozonolysis. It cleaves the carbon-carbon double bond and places an oxygen on each of the two carbons.
- ✓Checking only the 4n + 2 electron count for aromaticity, without confirming the ring is cyclic, planar and fully conjugated.
Practice set 1: alkanes
1. What is the general formula of an alkane, and what is its hybridisation?
CₙH₂ₙ₊₂, with every carbon sp³ hybridised and tetrahedral at about 109.5°. Alkanes are saturated - all bonds are single - which is why they are relatively unreactive and why their reactions require harsh conditions such as light or high temperature.
2. Why are alkanes relatively unreactive?
Because C-C and C-H sigma bonds are strong and only slightly polar, so there is no electron-rich or electron-poor site for a reagent to attack. Without a pi bond or a significant dipole, alkanes offer nothing for nucleophiles or electrophiles to engage with, and only radicals will react with them.
3. Describe the mechanism of free-radical substitution.
Three stages. Initiation: light or heat homolyses a chlorine molecule into two radicals. Propagation: a chlorine radical abstracts a hydrogen to give an alkyl radical, which attacks another chlorine molecule and regenerates a chlorine radical. Termination: two radicals combine. The chain nature of propagation is what makes the reaction efficient.
4. Why does the chlorination of methane give a mixture of products?
Because the chloromethane produced still has hydrogens available for further substitution, and it is at least as reactive as methane. So dichloromethane, trichloromethane and tetrachloromethane form alongside it. Controlling the outcome requires a large excess of methane, which makes free-radical substitution poor for selective synthesis.
5. What is the Wurtz reaction?
Two alkyl halide molecules coupled by sodium in dry ether to give a symmetrical alkane with twice the carbon count. Its limitation is that using two different alkyl halides gives a mixture of three products, so it is practical only for symmetrical alkanes.
6. Which conformation of ethane is more stable and why?
The staggered conformation, by about 12 kJ/mol. In the eclipsed conformation the C-H bonds on adjacent carbons align, causing torsional strain from the repulsion of their bonding electrons. Staggered maximises the separation. The barrier is low enough that rotation is rapid at room temperature, so the conformers cannot be separated.
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Practice set 2: alkenes
7. What is the general formula of an alkene and its geometry at the double bond?
CₙH₂ₙ, with both carbons sp² hybridised, planar, and bond angles near 120°. The double bond consists of one sigma bond from hybrid-orbital overlap and one pi bond from sideways overlap of unhybridised p orbitals. The pi bond prevents rotation, which is what makes geometrical isomerism possible.
8. Why do alkenes undergo electrophilic addition?
Because the pi bond is a region of high electron density above and below the molecular plane, exposed and loosely held. That makes it attractive to electron-deficient species. The pi bond is also weaker than a sigma bond, so breaking it to form two new sigma bonds is energetically favourable.
9. State Markovnikov's rule and its underlying reason.
In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon already bearing more hydrogens. The reason is carbocation stability: the proton adds so as to generate the more substituted and therefore more stable carbocation, which the halide then attacks. Stating the rule as a fact is far less useful than deriving it from stability.
10. What is the peroxide effect and which reagent does it apply to?
In the presence of peroxides, HBr adds anti-Markovnikov because the mechanism switches from ionic to free-radical, and the bromine radical adds first to give the more stable carbon radical. It applies to HBr only - HCl has too strong a bond to propagate the chain, and HI too weak a one. Applying it to HCl or HI is a standard error.
11. What does ozonolysis do and how is it used?
It cleaves the carbon-carbon double bond and places a carbonyl oxygen on each carbon, giving aldehydes or ketones depending on the substitution. Its main exam use is backwards: given the carbonyl products, rejoin them at the carbonyl carbons to identify the original alkene. Reductive workup gives aldehydes, oxidative workup gives carboxylic acids.
12. What is distinctive about hydroboration-oxidation?
It adds water across the double bond with anti-Markovnikov orientation and syn stereochemistry, without any carbocation intermediate - so no rearrangement occurs. Acid-catalysed hydration gives the Markovnikov product and can rearrange. The two reactions are the standard complementary pair for making either alcohol from the same alkene.
13. What is Baeyer's reagent and what does it test for?
Cold dilute alkaline potassium permanganate, which is decolourised from purple by an alkene or alkyne while producing a brown precipitate of manganese dioxide. It tests for unsaturation, and it also converts the alkene into a vicinal diol by syn addition. Bromine water serves the same diagnostic purpose.
14. What conditions are required for geometrical isomerism in an alkene?
Restricted rotation about the double bond, and two different groups on each of the two doubly-bonded carbons. If either carbon bears two identical groups, swapping them changes nothing and no isomerism exists. The second condition is the one most often overlooked.
Practice set 3: alkynes
15. What is the general formula of an alkyne and its geometry?
CₙH₂ₙ₋₂, with both carbons sp hybridised and a linear arrangement at 180°. The triple bond is one sigma and two pi bonds. The linear geometry means alkynes show no geometrical isomerism, unlike alkenes.
16. Why is the C-H bond in a terminal alkyne acidic?
Because the hydrogen sits on an sp carbon with 50% s character, which holds the electron pair closer to the nucleus and stabilises the resulting carbanion. Acidity therefore increases from sp³ to sp² to sp, making terminal alkynes far more acidic than alkenes or alkanes - though still far weaker acids than water.
17. How do you distinguish a terminal alkyne from an internal one?
By reaction with ammoniacal silver nitrate, which gives a white precipitate with a terminal alkyne and no reaction with an internal one. Ammoniacal cuprous chloride gives a red precipitate similarly. Both tests depend on the acidic terminal hydrogen, which internal alkynes lack.
18. What is the product of the acid-catalysed hydration of propyne?
Propanone, a ketone. Water adds according to Markovnikov's rule to give an enol, which tautomerises to the more stable keto form. Terminal alkynes other than ethyne always give ketones by this route; ethyne itself is the sole exception, giving acetaldehyde.
19. How do you reduce an alkyne to a cis-alkene?
With hydrogen over Lindlar's catalyst - palladium poisoned with quinoline and barium sulphate. The syn addition of both hydrogens to the same face gives the cis product, and the poisoning stops the reduction at the alkene rather than continuing to the alkane.
20. How do you reduce an alkyne to a trans-alkene?
With sodium in liquid ammonia, sometimes called a dissolving-metal reduction. It proceeds through radical anion intermediates that adopt the less hindered trans arrangement, giving anti addition. The Lindlar and sodium-ammonia pair is the standard way to control alkene geometry, and questions frequently ask which to use.
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Practice set 4: aromatic hydrocarbons
21. Describe the structure and stability of benzene.
A planar regular hexagon with all six carbon-carbon bonds of equal length, intermediate between single and double. All carbons are sp² hybridised and the six p orbitals form a delocalised pi system above and below the ring. That delocalisation gives roughly 150 kJ/mol of extra stability, which is why benzene resists addition.
22. State the conditions for aromaticity.
Cyclic, planar, fully conjugated around the ring, and containing 4n + 2 pi electrons. All four are required. Cyclooctatetraene has 8 pi electrons and adopts a non-planar tub shape, failing on two counts, which is why it behaves as an ordinary polyene rather than an aromatic compound.
23. Why does benzene undergo substitution rather than addition?
Because addition would destroy the delocalised pi system and lose the aromatic stabilisation. Substitution replaces a hydrogen and leaves the ring intact, so the stabilisation is preserved. This is the single fact that governs the entire chemistry of aromatic compounds.
24. How do activating and deactivating groups differ?
Activating groups donate electron density to the ring, making it more nucleophilic and speeding up electrophilic substitution - alkyl, hydroxyl and amino groups are examples. Deactivating groups withdraw density and slow the reaction - nitro, carbonyl and cyano groups. Reactivity and directing effect usually go together, with one important exception.
25. Which positions do activating and deactivating groups direct to?
Activating groups direct ortho and para; deactivating groups direct meta. The reason is which intermediate is better stabilised: an electron-donating group can stabilise the positive charge when attack occurs ortho or para, but not meta. Electron-withdrawing groups make ortho and para attack worse, leaving meta as the least bad option.
26. Why are halogens an exception to that pattern?
Because they act through two opposing effects. Their strong electron-withdrawing inductive effect deactivates the ring, but their lone pairs can donate by resonance specifically at the ortho and para positions. Induction dominates the rate while resonance dominates the orientation, so halogens are deactivating yet ortho-para directing.
27. What are the Friedel-Crafts reactions and what limits them?
Alkylation attaches an alkyl group and acylation attaches an acyl group, both using an aluminium chloride catalyst. Alkylation suffers two problems: the product is more activated than the starting material, so polyalkylation occurs, and the carbocation intermediate can rearrange. Acylation avoids both, which is why acylation followed by reduction is the preferred route to an alkylbenzene.
Practice set 5: mixed
28. How is the degree of unsaturation calculated and what does it tell you?
(2C + 2 + N − H − X)/2, where C, N, H and X are the counts of carbon, nitrogen, hydrogen and halogen. Each unit corresponds to one ring or one pi bond. A value of 4 usually signals a benzene ring, since that accounts for three double bonds and one ring - which is a useful shortcut in structure-determination questions.
29. How many structural isomers does C₄H₁₀ have?
Two: butane and 2-methylpropane. With only four carbons the skeleton is either a straight chain or a chain of three with one branch. Isomer counts grow rapidly with carbon number, and the systematic method is to enumerate carbon skeletons first and then place substituents.
30. How do boiling points vary among hydrocarbons?
They rise with increasing molecular size, because larger molecules have stronger van der Waals attractions. Among isomers, branching lowers the boiling point, since a more compact molecule has less surface contact with its neighbours. So butane boils higher than 2-methylpropane despite identical molecular formulas.
How to study this chapter efficiently
- ✓Organise your notes by mechanism - free-radical substitution, electrophilic addition, electrophilic substitution - rather than by compound class.
- ✓For every addition reaction, write the intermediate before the product. Markovnikov and anti-Markovnikov both follow from which one is more stable.
- ✓Learn the reagent pairs that control outcome: Lindlar versus sodium-ammonia, and acid hydration versus hydroboration.
- ✓Make a single table of directing effects, and mark halogens explicitly as the exception.
- ✓Practise ozonolysis backwards - given products, reconstruct the alkene. That is how it is usually asked.
- ✓Learn the three distinguishing tests together: bromine water, Baeyer's reagent, and ammoniacal silver nitrate.
Turn this into active practice
The trap in this chapter is that reading a reaction list feels like progress. Recognition of a reaction you have seen is not the same as predicting the product of one you have not, and only the second is tested.
The JEE Hydrocarbons quiz on QUFF generates fresh questions across alkanes, alkenes, alkynes and aromatics, marks them instantly and explains each answer. Do mixed sets so the compound class is not signposted, and for each error note whether you misidentified the mechanism or the intermediate.