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HydrocarbonsNEET Chemistry · Class 11 · NCERT Chapter 9

Classification of Hydrocarbons

Hydrocarbons are organic compounds that contain only carbon and hydrogen. They are the simplest class of organic compounds and form the basis for understanding all of organic chemistry. NEET asks about their reactions every year.

TypeFunctional GroupExample
AlkaneC-C (single bond only)CH₄, C₂H₆, propane
AlkeneC=C (double bond)CH₂=CH₂ (ethylene)
AlkyneC≡C (triple bond)HC≡CH (acetylene)
AromaticBenzene ring (delocalized π)C₆H₆, toluene
CycloalkaneRing of C-C bondsCyclohexane (C₆H₁₂)

General formula: alkane CₙH₂ₙ₊₂; alkene CₙH₂ₙ; alkyne CₙH₂ₙ₋₂; benzene series CₙH₂ₙ₋₆.

Alkanes: Conformations and Reactions

Rotate the dihedral angle to see how conformation affects relative energy. Key: staggered conformations (60° apart) are more stable than eclipsed.

Dihedral angle

60°

Staggered

H atoms of back carbon bisect the H-H angles of front carbon (60°, 180°, 300°). Minimum energy — most stable conformation.

Relative energy:

12.0 kJ/mol

Newman projection (schematic)

H
H
H
H
H
H

Solid lines = front carbon bonds. Dashed = back carbon bonds. Dihedral = 60°

Stability order for ethane

Staggered (0 kJ) >> Eclipsed (12 kJ)
Torsional (Pitzer) strain in eclipsed = repulsion between electron clouds of C-H bonds.

Alkanes are saturated hydrocarbons (all single bonds). Their main reactions are combustion and free radical halogenation. They are chemically less reactive than alkenes or alkynes.

Conformational Isomers

In ethane, the two CH₃ groups can rotate around the C-C bond. Different spatial arrangements arising from rotation around single bonds are called conformations. Two extreme conformations of ethane:

  • Staggered conformation: H atoms on adjacent carbons are as far apart as possible. This has minimum torsional strain and is the most stable conformation.
  • Eclipsed conformation: H atoms on adjacent carbons are directly aligned. This has maximum torsional strain and is less stable.

Conformations are not permanent structures (they interconvert rapidly at room temperature) and are not structural isomers.

Halogenation of Alkanes: Free Radical Mechanism

Alkanes react with chlorine or bromine in the presence of UV light (hν) via a free radical chain mechanism. For methane + Cl₂:

1. Initiation: UV light breaks the Cl-Cl bond to form chlorine radicals.

2. Propagation: The chain is carried forward.

3. Termination: Two radicals combine to form a stable molecule.

Reactivity of C-H bonds toward halogenation: 3° > 2° > 1° (due to stability of the radical intermediate formed). Chlorination is less selective than bromination (bromine is more selective because radical formation by Br• is more endothermic, making the reaction more sensitive to radical stability).

Fluorination is violent and uncontrollable; iodination is too slow (thermodynamically unfavourable).

Alkenes: Reactions and Markovnikov's Rule

Markovnikov's addition
Free radical halogenation
Baeyer's test
Ozonolysis
Friedel-Crafts alkylation
Combustion of hydrocarbons

Electrophilic addition

Markovnikov's addition

Reactant

CH₃CH=CH₂ (propene)

+

Reagent

HBr

Product

CH₃CHBrCH₃ (2-bromopropane)

Mechanism

H⁺ adds to the carbon with MORE H atoms (Markovnikov's rule). Br⁻ adds to the carbon with the more stable (more substituted) carbocation intermediate.

Rich get richer: H goes to the C that already has more H.

Quick questions

Addition of HBr to propene gives mainly:

1-bromopropane
2-bromopropane
propyl bromide (equal mixture)
1,2-dibromopropane

The carbocation intermediate in Markovnikov addition to propene is:

CH₃CH⁺CH₃ (secondary)
CH₃CH₂CH₂⁺ (primary)
CH₂=CHCH₃ itself
CH₃CH(Br)CH₃

Alkenes contain a C=C double bond (one σ and one π bond). The π bond is the site of most reactions. Alkenes undergo electrophilic addition reactions because the π electrons act as a Lewis base and attract electrophiles.

Addition of HX (Hydrogen Halides)

HX adds across the double bond. For unsymmetrical alkenes, Markovnikov's rule applies:

The H goes to the carbon with more H atoms; X goes to the carbon with fewer H atoms. The reason: the more substituted carbocation intermediate (formed when H⁺ adds first) is more stable.

Anti-Markovnikov Addition (Peroxide Effect)

In the presence of peroxides (ROOR or H₂O₂), HBr adds via a free radical mechanism to give the anti-Markovnikov product:

This is called the Kharasch effect or peroxide effect. It only works with HBr (not HCl or HI). HCl: Cl• too reactive (indiscriminate). HI: I• too stable (unreactive toward the alkene).

Addition of H₂O (Hydration)

Alkenes react with water in the presence of an acid catalyst (H₃PO₄ or H₂SO₄) to give alcohols. Markovnikov's rule applies: OH goes to the more substituted carbon.

Addition of Cl₂ / Br₂ (Halogenation)

Halogens add across the double bond. The reaction with Br₂ in CCl₄ is used as a test for unsaturation (the orange-brown colour of Br₂ decolourises).

Ozonolysis

Treatment with ozone (O₃) followed by zinc and water (reductive workup) cleaves the C=C bond to give carbonyl compounds:

  • Terminal =CH₂ gives formaldehyde (HCHO).
  • =CHR gives an aldehyde (RCHO).
  • =CR₂ gives a ketone (R₂C=O).

Ozonolysis is used to determine the position of the double bond in an unknown alkene.

Oxidation with KMnO₄

Cold dilute KMnO₄ (Baeyer's reagent) adds two OH groups across the double bond (gives a diol). The purple colour of KMnO₄ decolourises — another test for unsaturation. Hot concentrated KMnO₄ cleaves the double bond completely (similar to oxidative ozonolysis).

Combustion

Master hydrocarbon reactions step by step

Practice Markovnikov's rule, ozonolysis, and EAS reactions with our AI tutor. It explains every mechanism in simple, clear language.

Alkynes: Acidic Character and Reactions

Alkynes have a C≡C triple bond (one σ and two π bonds). The sp hybridised carbon holds electrons closer to the nucleus, giving the terminal C-H bond a slightly acidic character.

Acidic Character of Terminal Alkynes

Terminal alkynes (HC≡CR) react with strong bases (Na, NaNH₂) to release H⁺ and form acetylide ions (RC≡C⁻). This makes alkynes more acidic than alkenes or alkanes.

Acidic order: alkyne (pKa ≈ 25) >> alkene (pKa ≈ 44) > alkane (pKa ≈ 50). Water has pKa = 15.7 and is more acidic than alkynes in the common sense, but among hydrocarbons, alkynes are the most acidic.

Addition Reactions of Alkynes

Alkynes undergo electrophilic addition, but less readily than alkenes (the triple bond is less reactive toward electrophiles because the sp carbons are more electronegative and hold the π electrons more tightly).

  • HX addition follows Markovnikov's rule. For acetylene + HCl: HC≡CH + HCl → CH₂=CHCl (vinyl chloride, first addition) → CH₃CHCl₂ (second HCl addition).
  • Addition of H₂O (in presence of Hg²⁺/H₂SO₄) gives an enol which immediately tautomerises to a carbonyl compound. HC≡CH + H₂O → [CH₂=CHOH] → CH₃CHO (acetaldehyde).
  • Hydrogenation: partial (Lindlar's catalyst, poisoned Pd) gives cis-alkene. Complete H₂ + Pt gives alkane.

Benzene: Structure and Stability

Benzene (C₆H₆) has a cyclic structure with three alternating double bonds (Kekulé structure). However, the actual structure is a resonance hybrid — all six C-C bond lengths are equal (1.40 Å, between single 1.54 Å and double 1.34 Å), and all bond angles are 120°. The six π electrons are fully delocalised over the ring.

Evidence for Delocalisation

  • All C-C bonds in benzene are equal in length (1.40 Å).
  • Benzene does not decolourise Br₂ in CCl₄ (unlike alkenes) — it does not undergo addition readily.
  • The heat of hydrogenation of benzene (−208 kJ/mol) is much less than the expected value for cyclohexatriene (−360 kJ/mol). The difference (≈ 152 kJ/mol) is the resonance energy.

Hückel's Rule

A monocyclic, planar compound with (4n + 2) π electrons (where n = 0, 1, 2, ...) is aromatic. For benzene: 6 π electrons, n = 1 → aromatic. Cyclooctatetraene (8 π electrons, 4n with n = 2) is anti-aromatic and non-planar in practice.

Electrophilic Aromatic Substitution (EAS)

Benzene undergoes substitution (not addition) because the delocalized ring is too stable to add across. An electrophile attacks the ring, and a proton is lost to restore aromaticity. General mechanism:

  1. Formation of the electrophile (E⁺) using a Lewis acid catalyst.
  2. The electrophile attacks the benzene ring, forming a non-aromatic sigma complex (arenium ion, also called a Wheland intermediate or sigma complex). Aromaticity is temporarily lost.
  3. A proton is lost from the carbon that bonded to E⁺, restoring aromaticity. This step is fast.

Important EAS Reactions of Benzene

ReactionReagent / ConditionsProduct
Nitrationconc. HNO₃ + conc. H₂SO₄Nitrobenzene (PhNO₂)
HalogenationCl₂ or Br₂ + Lewis acid (FeCl₃, FeBr₃, AlCl₃)Chlorobenzene / Bromobenzene
SulphonationFuming H₂SO₄ (oleum)Benzenesulphonic acid (PhSO₃H)
Friedel-Crafts alkylationAlkyl halide + AlCl₃Alkylbenzene (e.g., toluene)
Friedel-Crafts acylationAcyl halide (RCOCl) + AlCl₃Aryl ketone (e.g., acetophenone)

Electrophiles formed in these reactions: NO₂⁺ (nitronium) in nitration; Cl⁺ or FeCl₄⁻Cl complex in halogenation; R⁺ in alkylation.

Directive Influence of Substituents

When a substituent is already on the benzene ring, it directs an incoming electrophile to specific positions.

Ortho/Para Directors (Ring-Activating Groups)

Groups that donate electrons to the ring (by +M effect or +I effect) activate it toward EAS and direct the electrophile to the ortho and para positions. These positions become relatively electron-rich due to resonance donation.

  • Examples: −OH, −OR, −NH₂, −NHR, −NR₂, −NHCOCH₃, alkyl groups (−CH₃).
  • Halogens (−F, −Cl, −Br, −I) are ortho/para directors but ring-deactivating (−I effect dominates over +M effect for rate, but position still ortho/para).

Meta Directors (Ring-Deactivating Groups)

Groups that withdraw electrons from the ring (by −M effect) deactivate it and direct the electrophile to the meta position. The ortho and para positions are made more electron-poor by the withdrawing group's resonance, leaving meta as the preferred site.

  • Examples: −NO₂, −COOH, −SO₃H, −CHO, −COR (ketone), −CN, −CF₃, −NR₃⁺.

A useful memory tip: groups that contain a positive charge or a highly electronegative atom directly attached to the ring (especially through a double bond toward oxygen or nitrogen) are meta directors.

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Worked NEET Problems

1

NEET-style problem · Markovnikov's Rule

Question

What is the major product of the reaction of 2-methylpropene with HBr?

Solution

2-methylpropene: (CH₃)₂C=CH₂. Apply Markovnikov's rule: H⁺ adds to the less substituted carbon (=CH₂, which has 2 H), giving the more stable tertiary carbocation (CH₃)₂C⁺-CH₃. Then Br⁻ attacks the carbocation.

Major product: (CH₃)₂CBr-CH₃ = 2-bromo-2-methylpropane.

2

NEET-style problem · Ozonolysis

Question

An alkene on ozonolysis (reductive workup) gives CH₃CHO and CH₂O. Identify the alkene.

Solution

Ozonolysis cleaves the double bond. CH₃CHO (acetaldehyde) comes from a =CHCH₃ end. CH₂O (formaldehyde) comes from a =CH₂ end. Join the two fragments at the double bond: CH₃CH=CH₂ = propene (prop-1-ene). The alkene is propene.

3

NEET-style problem · Electrophilic Aromatic Substitution

Question

Where will a second nitro group be introduced in nitrobenzene? Explain.

Solution

−NO₂ is a meta director because it withdraws electron density from the ring by −M effect. The ortho and para positions become electron-poor relative to the meta position.

The second −NO₂ group will attach at the meta position to give 1,3-dinitrobenzene as the major product.

4

NEET-style problem · Acidic Character of Alkynes

Question

Arrange the following in decreasing order of acidity: CH₄, C₂H₄, C₂H₂, H₂O.

Solution

Acidity increases with increasing s-character of the C-H bond (sp carbon holds electrons tightest, stabilising the conjugate carbanion). H₂O is the most acidic of the set (pKa ≈ 15.7). C₂H₂ (sp carbon, pKa ≈ 25) > C₂H₄ (sp² carbon, pKa ≈ 44) > CH₄ (sp³ carbon, pKa ≈ 50).

Decreasing acidity: H₂O > C₂H₂ > C₂H₄ > CH₄.

Summary Cheat Sheet

HydrocarbonKey Reactions
AlkaneFree radical halogenation (UV light); combustion; reactivity: 3° > 2° > 1°
AlkeneElectrophilic addition (HX, H₂O, X₂); Markovnikov's rule; ozonolysis; KMnO₄ oxidation
Anti-Markovnikov (HBr)Peroxide effect, free radical mechanism; Br goes to less substituted C
AlkyneMore acidic than alkenes/alkanes (sp C-H); reacts with Na/NaNH₂; hydration gives carbonyl
Benzene EASNitration (NO₂⁺), halogenation (Cl⁺/Br⁺ with Lewis acid), sulphonation, Friedel-Crafts
Ortho/para directors−OH, −NH₂, −OCH₃, alkyl, −X (halogen); +M or +I effect; activate ring
Meta directors−NO₂, −COOH, −CHO, −CN, −SO₃H; −M effect; deactivate ring
Resonance energy of benzene≈ 152 kJ/mol; makes benzene prefer substitution over addition
Hückel's rule4n + 2 π electrons for aromaticity; benzene has 6 π (n=1)
Baeyer's testCold dil. KMnO₄ decolourises with alkene/alkyne (not with benzene)

Frequently asked questions

What is Markovnikov's rule and when does it apply?

Markovnikov's rule applies to the electrophilic addition of unsymmetrical reagents (like HX or H₂O) across a double bond. It states that the hydrogen atom (or the positive part of the reagent) adds to the carbon of the double bond that already has more hydrogen atoms. The underlying reason is that the more substituted (more stable) carbocation intermediate is formed preferentially. Example: propene + HBr gives 2-bromopropane (Markovnikov product), not 1-bromopropane.

What is anti-Markovnikov addition (peroxide effect)?

When HBr adds to an alkene in the presence of peroxides (e.g., H₂O₂ or ROOR), the reaction proceeds via a free radical mechanism instead of ionic. The bromine atom (not HBr as an ionic species) adds first to give the more stable radical intermediate. This produces the anti-Markovnikov product (Br goes to the less substituted carbon). This effect is specific to HBr; HCl and HI do not show this effect with peroxides.

Why is benzene more stable than expected?

A hypothetical cyclohexatriene (three alternating double bonds, like Kekulé's structure) would have a heat of hydrogenation of about −360 kJ/mol (3 × −120 kJ). But the actual benzene has a heat of hydrogenation of only −208 kJ/mol. The difference (about 152 kJ/mol) is the resonance energy. This extra stability comes from the complete delocalisation of the 6 pi electrons over all 6 carbon atoms in a ring, following Hückel's rule (4n + 2 π electrons with n = 1).

What are ortho-para directors and meta directors in EAS?

Groups attached to a benzene ring influence where the incoming electrophile attacks. Ortho/para directors (also called ring-activating groups) have a lone pair that can donate electron density to the ring via resonance (+M effect). They activate the ring and direct the electrophile to the ortho and para positions. Examples: −OH, −NH₂, −OCH₃, −alkyl. Meta directors (ring-deactivating groups) withdraw electron density from the ring by −M or −I effects, leaving the meta positions relatively electron-rich. Examples: −NO₂, −COOH, −SO₃H, −CHO, −CN.

Why are alkynes more acidic than alkenes and alkanes?

Acidity depends on how well the conjugate base (the carbanion) is stabilised. The sp carbon in alkynes has more s-character (50%) than sp² (33%) or sp³ (25%) carbon. More s-character means electrons are held closer to the nucleus, stabilising the negative charge on carbon. So the sp carbanion (from an alkyne) is more stable → alkyne is more acidic than alkene or alkane. Acidic order: alkyne (pKa ≈ 25) >> alkene (pKa ≈ 44) > alkane (pKa ≈ 50).

What are the products of ozonolysis of an alkene?

Ozonolysis cleaves the double bond completely. The double bond is replaced by two C=O groups. Reductive ozonolysis (with Zn/H₂O or Me₂S) gives aldehydes from terminal carbons (=CH₂ → HCHO) and ketones from internal carbons (=CR₂ → R₂C=O). Oxidative ozonolysis (with H₂O₂ or KMnO₄) converts aldehydes further to carboxylic acids. This reaction is used to determine the position of a double bond in an unknown alkene.

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