Hydrocarbons — Concepts, Reactions & Solved Examples

What Are Hydrocarbons, Really?

Organic chemistry starts here. Hydrocarbons are compounds containing only carbon and hydrogen — and that simplicity is deceptive, because from just two elements, nature builds fuels, plastics, drugs, and life itself.

Carbon’s magic lies in its tetravalency (valency of 4) and its unique ability to bond with other carbon atoms in chains, branches, and rings. Hydrogen fills the remaining bonds. That’s the whole story structurally — but the reactions are where the real complexity lives.

For Class 11 boards and JEE Main, hydrocarbons carry serious weightage. The NCERT chapter covers alkanes, alkenes, alkynes, and aromatic compounds, and JEE regularly pulls 2-3 questions directly from reaction mechanisms and nomenclature. We’ll build this topic from scratch — nomenclature, reactions, mechanisms, and the exam-specific patterns that actually show up in marking schemes.

One framing that helps: think of hydrocarbons as a family tree. The root is the carbon skeleton (chain length, branching, rings), and the branches are defined by what types of bonds carbon forms — single (alkane), double (alkene), triple (alkyne), or delocalised (aromatic). Every property and reaction follows from this.

Key Terms & Definitions

Hydrocarbon — any organic compound made exclusively of carbon and hydrogen atoms.

Alkane — saturated hydrocarbon with only C–C single bonds. General formula: $C_nH_{2n+2}$ . Example: ethane ( $C_2H_6$ ).

Alkene — unsaturated hydrocarbon with at least one C=C double bond. General formula: $C_nH_{2n}$ . Example: ethene ( $C_2H_4$ ).

Alkyne — unsaturated hydrocarbon with at least one C≡C triple bond. General formula: $C_nH_{2n-2}$ . Example: ethyne ( $C_2H_2$ , also called acetylene).

Aromatic hydrocarbon — contains a benzene ring with delocalised $\pi$ electrons. Example: benzene ( $C_6H_6$ ), toluene.

Aliphatic hydrocarbon — non-aromatic; includes alkanes, alkenes, alkynes (both open-chain and cyclic but non-aromatic).

Degree of unsaturation (DoU) — also called Index of Hydrogen Deficiency (IHD):

\text{DoU} = \frac{2C + 2 + N - H - X}{2}

Where C = carbon count, H = hydrogen count, N = nitrogen count, X = halogen count, O is ignored.

DoU = 1 → one double bond or one ring
DoU = 2 → one triple bond, or two double bonds, or two rings
DoU = 4 → benzene ring (3 double bonds + 1 ring)

Homologous series — a family of compounds differing by $-CH_2-$ units, with the same general formula and similar chemical properties.

Classification and Nomenclature

IUPAC Naming — The Method That Never Fails

IUPAC nomenclature follows a strict hierarchy. Here’s the algorithm:

Step 1 — Find the longest continuous carbon chain containing the principal functional group. This is the parent chain.

Step 2 — Number the chain from the end that gives the lowest locants to substituents (or the principal group).

Step 3 — Name substituents as prefixes: methyl, ethyl, propyl, etc.

Step 4 — Assemble: locant-substituent + parent chain name.

Worked Example: Name the compound with structure — a 5-carbon chain, with a methyl group on C-2 and a double bond between C-3 and C-4.

Parent chain: pent-3-ene (5 carbons, double bond at C-3)
Substituent: 2-methyl
Full name: 2-methylpent-3-ene

Students often number from the wrong end. Always give the double bond the lower locant — even if that means substituents get higher numbers. The principal group (double bond > substituent) takes priority in numbering.

Types by Structure

Type	Shape	Example
Open-chain (acyclic)	Straight or branched	n-butane, isobutane
Cyclic aliphatic	Ring, no aromatic	Cyclohexane
Aromatic	Benzene ring	Benzene, naphthalene

Alkanes — Properties and Reactions

Physical Properties

Alkanes are non-polar, so their intermolecular forces are London dispersion forces only. As chain length increases, boiling point increases. Branching decreases boiling point (more compact molecule, less surface area).

Methane through butane are gases at room temperature. Pentane to hexadecane are liquids. Heavier alkanes are waxy solids.

Halogenation of Alkanes (Free Radical Mechanism)

This is a chain reaction with three stages:

Initiation (bond homolysis):

Cl_2 \xrightarrow{h\nu} 2Cl^\bullet

Propagation (chain-carrying):

CH_4 + Cl^\bullet \rightarrow CH_3^\bullet + HCl

CH_3^\bullet + Cl_2 \rightarrow CH_3Cl + Cl^\bullet

Termination (radical combination):

Cl^\bullet + Cl^\bullet \rightarrow Cl_2

CH_3^\bullet + Cl^\bullet \rightarrow CH_3Cl

Reactivity order of halogens: $F_2 > Cl_2 > Br_2 > I_2$

Fluorination is violent and uncontrollable. Iodination is endothermic (doesn’t proceed under normal conditions). Chlorination and bromination are the useful ones.

Selectivity: Bromine is more selective than chlorine. At a tertiary C–H bond, relative rates are: $3^\circ : 2^\circ : 1^\circ = 1600 : 82 : 1$ for Br₂ versus $5 : 4 : 1$ for Cl₂.

JEE Main 2022 asked about the major product of bromination of isobutane. The answer is 2-bromo-2-methylpropane (tertiary product dominates due to Br₂ selectivity). Know this pattern cold.

Alkenes — Reactions and Mechanisms

Addition Reactions

The C=C double bond is electron-rich. Electrophiles attack it. This is electrophilic addition (AE).

Markovnikov’s Rule

When HX adds to an unsymmetrical alkene, the hydrogen goes to the carbon already having more hydrogens.

Why? The more substituted carbocation is more stable (hyperconjugation + inductive effect stabilise the positive charge). Reaction goes through the more stable intermediate.

CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3

(2-bromopropane, not 1-bromopropane)

Anti-Markovnikov (Peroxide Effect / Kharasch Effect)

When HBr adds in the presence of peroxides, the mechanism switches to free radical addition, and the bromine goes to the carbon with more hydrogens. This reverses Markovnikov’s rule.

The peroxide effect only works with HBr, not HCl or HI. HCl addition is too exothermic for the radical mechanism to compete; HI radicals are too stable and don’t propagate. This distinction appears in JEE questions almost every year.

Ozonolysis

Alkenes react with ozone ( $O_3$ ) followed by reductive workup ( $Zn/H_2O$ ) to give aldehydes or ketones. Oxidative workup ( $H_2O_2$ ) converts aldehydes further to carboxylic acids.

Use ozonolysis backwards to identify the structure of an unknown alkene from its carbonyl products.

Hydration (Addition of Water)

In the presence of dilute $H_2SO_4$ , water adds across the double bond following Markovnikov’s rule. This is how ethanol is industrially produced from ethene.

Oxidation Reactions

Cold, dilute $KMnO_4$ (Baeyer’s reagent): Adds two OH groups (syn addition → cis-diol). The purple colour of $KMnO_4$ decolourises — this is the test for unsaturation.
Hot, concentrated $KMnO_4$ : Cleaves the double bond. Terminal alkenes give carboxylic acids; internal alkenes give two carboxylic acids.

Alkynes — The Triple Bond Chemistry

Alkynes have a linear geometry at the sp-hybridised carbons. Bond angle: 180°.

Acidity of Terminal Alkynes

Terminal alkynes (R–C≡C–H) are weakly acidic ( $pK_a \approx 25$ ). The sp carbon holds the electrons more tightly (50% s-character vs 25% in sp² and 25% in sp³), making the C–H bond more polar and the proton more easily released.

This acidity is exploited in the sodium amide (NaNH₂) test: terminal alkynes give sodium acetylide + NH₃. This is how we distinguish terminal from internal alkynes.

Addition Reactions of Alkynes

Alkynes undergo the same electrophilic additions as alkenes, but twice (once per $\pi$ bond):

$HBr$ addition follows Markovnikov’s rule at each step
Addition of $H_2O$ in the presence of $HgSO_4/H_2SO_4$ gives an enol which tautomerises to a ketone (or aldehyde from ethyne). This is Markovnikov hydration.

Aromatic Hydrocarbons — Benzene and Beyond

Why Is Benzene “Special”?

Benzene ( $C_6H_6$ ) has DoU = 4, yet it doesn’t behave like a triene. It refuses to undergo addition reactions under mild conditions. The reason: delocalisation of 6 π electrons across all 6 carbon atoms.

This delocalisation is stabilising — benzene is 36 kJ/mol more stable than a hypothetical cyclohexatriene would be. This extra stability is called resonance energy or delocalisation energy.

Hückel’s Rule: A planar cyclic compound is aromatic if it has $(4n+2)$ π electrons, where n = 0, 1, 2, …

Benzene has 6 π electrons ( $n = 1$ ). Aromatic. Cyclooctatetraene has 8 π electrons — not $(4n+2)$ , so it’s antiaromatic and tub-shaped (non-planar, avoids antiaromaticity).

Electrophilic Aromatic Substitution (EAS)

Benzene reacts with electrophiles via substitution, not addition — because addition would destroy the aromatic stabilisation.

Mechanism:

Electrophile ( $E^+$ ) attacks the π cloud → forms arenium ion (sigma complex / Wheland intermediate)
A proton is lost → aromaticity restored

Halogenation: $C_6H_6 + Cl_2 \xrightarrow{AlCl_3} C_6H_5Cl + HCl$

Nitration: $C_6H_6 + HNO_3 \xrightarrow{conc. H_2SO_4} C_6H_5NO_2 + H_2O$

Sulphonation: $C_6H_6 + H_2SO_4 \xrightarrow{\Delta} C_6H_5SO_3H + H_2O$

Friedel-Crafts Alkylation: $C_6H_6 + RCl \xrightarrow{AlCl_3} C_6H_5R + HCl$

Directing Effects

Substituents on benzene direct incoming electrophiles to specific positions:

Director Type	Effect on Ring	Direction	Examples
o/p directors	Activate ring	ortho + para	$-OH, -NH_2, -OCH_3, -CH_3, -X$ (halogens activate o/p but deactivate ring)
m directors	Deactivate ring	meta	$-NO_2, -COOH, -CHO, -CN, -SO_3H$

Why? Groups that donate electron density (through resonance or induction) into the ring stabilise the carbocation intermediate at ortho/para positions → o/p products. Groups that withdraw electrons destabilise ortho/para intermediates → meta products.

Halogens are the odd group: they’re o/p directors (lone pair donation via resonance) but ring deactivators (electron withdrawal via induction). JEE Advanced has tested this nuance. Chlorobenzene with a second electrophile gives mostly o/p products, but slower than benzene.

Solved Examples

Example 1 — Easy (CBSE Level)

Q: Give the IUPAC name of $CH_3-CH(CH_3)-CH_2-CH_3$ .

Solution: 4-carbon parent chain? No — the longest chain is 4 carbons (butane). The methyl group is on C-2 (counting from the end that gives lower locant).

Answer: 2-methylbutane

Example 2 — Medium (JEE Main Level)

Q: The major product when 2-methylbut-2-ene reacts with HBr in the presence of peroxides is:

Solution: Peroxides → free radical mechanism → anti-Markovnikov.

Structure of 2-methylbut-2-ene: $CH_3-C(CH_3)=CH-CH_3$

In radical addition, Br• adds to the carbon with more H (terminal position here is C-1 or C-4). The alkene is internal and substituted, so we identify the less substituted carbon — C-4 ( $CH_3$ -end). The Br adds there.

Wait, this alkene has $CH_3$ on both sides of the double bond. C-2 has two methyls, C-3 has one methyl and one $CH_3CH_2$ . Neither is terminal. The peroxide effect is most dramatic with terminal alkenes. For this internal alkene, the regioselectivity is determined by radical stability — the more stable radical forms at C-3.

Major product: 2-bromo-2-methylbutane (Br at tertiary position C-2 via Markovnikov ionic mechanism; without peroxides, Markovnikov product at C-2 dominates).

This type of question — identifying the major product and why the mechanism applies — appeared in JEE Main 2021 February session. The trap is forgetting that peroxide effect is relevant only for HBr; for this specific internal alkene structure, verify which mechanism is operative.

Example 3 — Hard (JEE Advanced Level)

Q: How many structural isomers with molecular formula $C_5H_{10}$ show cis-trans (geometric) isomerism?

Solution:

$C_5H_{10}$ : DoU = 1. One degree of unsaturation — either one ring or one double bond.

Cyclic isomers (cyclopentane, methylcyclobutane, 1,2-dimethylcyclopropane, 1,1-dimethylcyclopropane, ethylcyclopropane) — cyclopentane and 1,2-dimethylcyclopropane can show geometric isomerism.

Acyclic alkenes: pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 2-methylbut-2-ene, 3-methylbut-1-ene.

Geometric isomerism requires: a double bond with two different groups on each carbon.

Pent-1-ene: $CH_2=CHCH_2CH_2CH_3$ — C-1 has two H’s, so no geometric isomerism.
Pent-2-ene: $CH_3CH=CHCH_2CH_3$ — C-2 has $CH_3$ and H; C-3 has $CH_2CH_3$ and H. Two different groups on each carbon. Yes.
2-methylbut-1-ene: $CH_2=C(CH_3)CH_2CH_3$ — C-1 has two H’s. No.
2-methylbut-2-ene: $CH_3C(CH_3)=CHCH_3$ — C-2 has two $CH_3$ groups. No.
3-methylbut-1-ene: $CH_2=CHCH(CH_3)_2$ — C-1 has two H’s. No.

Among acyclic alkenes: only pent-2-ene shows geometric isomerism.

From cyclic: 1,2-dimethylcyclopropane (cis and trans forms exist).

Answer: 2 structural isomers (pent-2-ene and 1,2-dimethylcyclopropane) show geometric isomerism.

Exam-Specific Tips

CBSE Class 11 — Marking Scheme Insights

Mechanism questions (halogenation of methane, nitration of benzene) are 3–5 marks and follow a fixed format. Markers expect: step names written (Initiation, Propagation, Termination), arrows showing electron movement, correct intermediates.
Nomenclature questions: always verify your IUPAC name by re-drawing the structure from the name. One wrong locant = 1 mark lost.
For conformational analysis (eclipsed vs staggered ethane, chair/boat cyclohexane), draw Newman projections correctly — examiners check your diagram, not just the label.

JEE Main — Weightage and Pattern

Hydrocarbons appear in 1-2 questions per JEE Main shift. Priority topics: Markovnikov vs anti-Markovnikov (every year), ozonolysis product identification, EAS directing effects, and IHD calculation. Mechanisms are less frequently tested in JEE Main compared to JEE Advanced.

Master the quick IHD calculation — use the formula mentally in 10 seconds.
For “which product forms” questions, draw the mechanism skeleton quickly and identify the most stable intermediate.
Aromatic vs non-aromatic vs antiaromatic classification (Hückel’s rule) — reliable 1-mark MCQ.

JEE Advanced — Depth Required

Mechanism tracing: know the Wheland intermediate for different EAS substitutions.
Stability of intermediates: be able to compare carbocation, carbanion, and radical stabilities across multiple substituents.
Synthetic sequences: multi-step hydrocarbon transformations where you identify an unknown starting material from the products.

Common Mistakes to Avoid

Mistake 1 — Applying peroxide effect to HCl or HI. Peroxide effect (anti-Markovnikov) works only for HBr. Never apply it to HCl or HI addition.

Mistake 2 — Forgetting that halogens are o/p directors in EAS. Students memorise “halogens deactivate the ring” and conclude they’re meta directors. Wrong. Halogens direct o/p (through lone pair resonance) but at a slower rate than benzene. This is a common 1-mark trap.

Mistake 3 — Geometric isomerism in alkenes: not checking both carbons. Students check one carbon of the double bond for two different groups, but forget to check the other. Both carbons of the C=C must have two different groups. If either has two identical groups, geometric isomerism is impossible.

Mistake 4 — Confusing ozonolysis workup conditions. Reductive workup ( $Zn/H_2O$ ): aldehydes and ketones. Oxidative workup ( $H_2O_2$ ): aldehydes oxidise to carboxylic acids. Ketones are unaffected. Students flip these in exams.

Mistake 5 — Counting the IHD for benzene as 3 (three double bonds), not 4. Benzene’s IHD = 4 (three double bonds + one ring). Always add the ring contribution. This catches students during molecular formula → structure identification.

Practice Questions

Q1. What is the IUPAC name of: $CH_3CH_2C(CH_3)_2CH_2CH_3$ ?

The longest chain through the quaternary carbon: 5 carbons (pentane). The quaternary carbon (C-3) has two methyl groups. Number from the end giving lower locant to the substituents.

Answer: 3,3-dimethylpentane

Q2. Why does sp-hybridised carbon form more acidic C–H bonds than sp³-hybridised carbon?

sp orbitals have 50% s-character. s orbitals keep electrons closer to the nucleus than p orbitals. So the electrons in sp orbitals are held more tightly → the C–H bond is more polar → the proton is released more easily → greater acidity.

Terminal alkynes ( $pK_a \approx 25$ ) are more acidic than alkenes ( $pK_a \approx 44$ ) which are more acidic than alkanes ( $pK_a \approx 50$ ).

Q3. An alkene $A$ ( $C_4H_8$ ) on ozonolysis gives only acetaldehyde ( $CH_3CHO$ ). Identify $A$ .

If only one product (acetaldehyde) forms from ozonolysis, the alkene must be symmetrical — both carbons of the double bond give the same carbonyl compound.

Structure: $CH_3CH=CHCH_3$ (but-2-ene). Both C-2 and C-3 give $CH_3CHO$ .

Answer: But-2-ene (exists as cis and trans isomers)

Q4. Benzene reacts with $Cl_2/AlCl_3$ but NOT with $Cl_2/H_2O$ at room temperature. Why?

EAS requires a strong electrophile — in this case, $Cl^+$ (generated when $AlCl_3$ polarises $Cl_2$ : $Cl_2 + AlCl_3 \rightarrow Cl^+ + AlCl_4^-$ ).

$Cl_2$ dissolved in water is a weak electrophile (no Lewis acid catalyst). The π electrons of benzene are less reactive than those in alkenes (aromatic stabilisation makes them less electron-rich toward electrophiles). So no reaction occurs without $AlCl_3$ .

Q5. Give the major product: toluene + $HNO_3 / H_2SO_4$ . Which isomer(s) form?

The methyl group ( $-CH_3$ ) is an ortho/para director and ring activator (hyperconjugation + inductive donation).

Major products: ortho-nitrotoluene and para-nitrotoluene (predominantly para due to steric hindrance at ortho positions).

para-nitrotoluene is the major product in practice (~58% para, ~38% ortho, ~4% meta).

Q6. An organic compound has molecular formula $C_6H_6$ . Calculate IHD and state what structures are possible.

\text{IHD} = \frac{2(6) + 2 - 6}{2} = \frac{8}{2} = 4

IHD = 4 suggests:

Benzene (aromatic — 3 double bonds + 1 ring)
Dewar benzene, benzvalene (exotic non-aromatic structures)
Open-chain structures: hexa-1,3,5-triene (3 double bonds + 0 rings = IHD 3) — wait, that’s IHD 3, not 4. So we’d also need one ring: e.g., cyclohexa-2,4-dien-1-yne.

The most stable and most important structure: benzene.

Q7. What happens when ethyne reacts with $HgSO_4/H_2SO_4/H_2O$ ?

Markovnikov hydration of ethyne:

$H_2O$ adds across C≡C → vinyl alcohol (ethenol, $CH_2=CHOH$ )
The enol immediately tautomerises to the more stable keto form: acetaldehyde ( $CH_3CHO$ )

This is keto-enol tautomerism. Ethyne is the only alkyne that gives an aldehyde; all other terminal alkynes give ketones via this mechanism.

Q8. In free radical bromination of butane, which product is major: 1-bromobutane or 2-bromobutane?

Bromine is highly selective. Secondary C–H bonds at C-2 and C-3 react much faster than primary C–H bonds at C-1 and C-4.

Butane: $CH_3CH_2CH_2CH_3$

6 primary H’s (C-1 and C-4), relative rate ≈ 1 each = 6 units
4 secondary H’s (C-2 and C-3), relative rate ≈ 82 each = 328 units

$\frac{328}{334} \approx 98\%$ of product is 2-bromobutane. The major product is overwhelmingly 2-bromobutane.

FAQs

What is the difference between alkane, alkene, and alkyne?

Alkanes have only C–C single bonds (saturated). Alkenes have at least one C=C double bond. Alkynes have at least one C≡C triple bond. Reactivity increases: alkanes are the least reactive (only free radical reactions under harsh conditions), while alkynes are highly reactive toward addition.

Why does benzene prefer substitution over addition?

Benzene has 36 kJ/mol of extra stabilisation from delocalisation. Addition would destroy this aromatic stabilisation (the ring would no longer be planar and delocalised). Substitution restores aromaticity after the reaction — so the aromatic stabilisation energy is recovered. Energetically, substitution wins.

What is Baeyer’s reagent and what does it test for?

Baeyer’s reagent is cold, dilute, alkaline potassium permanganate ( $KMnO_4$ ). It adds two OH groups across a C=C or C≡C bond (oxidative syn-addition). If the purple colour disappears, the compound is unsaturated (alkene or alkyne). It doesn’t react with alkanes or aromatic compounds under cold conditions.

How do I identify which carbon gets the electrophile in Markovnikov addition?

Ask: which carbocation intermediate is more stable? The hydrogen adds to whichever carbon produces the more stable carbocation on the other carbon. More substituted carbocation (3° > 2° > 1°) = more stable = preferred pathway.

What is the peroxide effect and why does it only work with HBr?

Peroxides generate radicals. In the radical chain, Br• adds first (not H•). Br• adds to the less substituted carbon (gives more stable radical). This reverses Markovnikov. HCl doesn’t work: the Cl• addition is too fast and exothermic, so the ionic path dominates. HI doesn’t work: I• is too stable and doesn’t propagate the chain efficiently.

Can alkynes also show geometric isomerism?

No. The two sp carbons in a triple bond are linear (bond angle 180°). There’s no restricted rotation in the sense that creates cis/trans isomers — the geometry at each carbon is linear, not planar. Geometric isomerism requires restricted rotation (double bond or ring) with two different groups on each end. Alkynes don’t satisfy this geometrically.

What is resonance energy of benzene and how is it measured?

Benzene’s resonance (delocalisation) energy is approximately 36 kJ/mol. It’s measured by comparing the actual heat of hydrogenation of benzene to the calculated value for a hypothetical cyclohexatriene (three times the heat of hydrogenation of cyclohexene). Benzene releases 36 kJ/mol less energy than expected — that “missing” energy was already stored as stabilisation.

For JEE, should I memorise all EAS reactions of benzene?

Memorise the four core reactions (halogenation, nitration, sulphonation, Friedel-Crafts) with their reagents and conditions. More importantly, understand the mechanism well enough to predict the product of a new EAS reaction you haven’t seen. JEE Advanced tests mechanism understanding, not rote memory.