Chemistry of Hydrocarbons

Hydrocarbons — compounds containing only carbon and hydrogen — are the parent skeletons of all organic chemistry. The PMDC MDCAT 2026 syllabus expects fluency with alkanes (C_nH_2n+2), alkenes (C_nH_2n), alkynes (C_nH_2n-2) and benzene, including their nomenclature, preparation, and characteristic mechanisms. This chapter typically delivers 4-6 MCQs on the paper.

PMC Table of Specifications. Hydrocarbons covers Alkanes, Alkenes, Alkynes, Benzene and the substitution-vs-addition contrast. Master the mechanisms — about half of all hydrocarbon MCQs are mechanism based.

Alkanes — Nomenclature & Free Radical Mechanism

Alkanes are saturated hydrocarbons with only single C−C and C−H bonds. General formula C_nH_2n+2. Each carbon is sp³ hybridised, tetrahedral, with bond angles ~109.5°. They are non-polar, almost unreactive towards ionic reagents, but undergo combustion and free-radical halogenation.

IUPAC nomenclature

Find the longest continuous carbon chain — this is the parent.
Number from the end giving the lowest locants to substituents.
List substituents alphabetically with their locants, separated by hyphens.
Use prefixes di-, tri-, tetra- for repeated identical substituents (ignored in alphabetising).

Example: CH₃−CH(CH₃)−CH₂−CH₃ is 2-methylbutane, not 3-methylbutane.

Free radical chain mechanism (halogenation)

Alkanes react with Cl₂ or Br₂ in UV light to give haloalkanes. The mechanism is a three-step free-radical chain:

Initiation: Cl₂ absorbs UV light and undergoes homolytic fission: Cl−Cl → 2 Cl•
Propagation: Cl• + CH₄ → CH₃• + HCl, then CH₃• + Cl₂ → CH₃Cl + Cl•. The cycle repeats — one photon can produce thousands of product molecules.
Termination: Two radicals combine: Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → C₂H₆.

Common trap. Halogenation of alkanes is a substitution reaction, not addition. Examiners often pair this with the alkene contrast to test whether you can tell them apart.

Alkenes — Nomenclature, Shape, Reactivity, Preparation

Alkenes are unsaturated hydrocarbons containing one C=C double bond. General formula C_nH_2n. The double bond consists of one σ and one π bond; both carbons are sp² hybridised, planar, with bond angles ~120°.

Nomenclature

Use the longest chain that contains the double bond. The suffix "-ane" of the alkane becomes "-ene". Number from the end giving the C=C the lowest locant. Example: CH₂=CH−CH₂−CH₃ is but-1-ene.

Preparation

Dehydration of alcohols with conc. H₂SO₄ (or Al₂O₃, 350°C).
Dehydrohalogenation of alkyl halides with alc. KOH.
Dehalogenation of vicinal dihalides with Zn dust.

Reactivity — addition reactions

Hydrogenation (H₂/Ni or Pt) gives alkanes.
Halogenation with Br₂/CCl₄ gives vicinal dihalide; decolourisation of bromine water is the classical test for unsaturation.
Hydrohalogenation (HX) follows Markovnikov's rule: H goes to the carbon bearing more hydrogens. With peroxides, HBr adds anti-Markovnikov (Kharasch effect, free-radical mechanism).
Hydration (H₂O/H⁺) gives an alcohol, also Markovnikov.
Ozonolysis (O₃, then Zn/H₂O) cleaves C=C into two carbonyl fragments — useful for locating the position of the double bond.

Markovnikov mnemonic. "The rich get richer." The carbon already bearing more hydrogens (richer in H) gets the H; the other carbon gets the X.

Alkynes — IUPAC Naming, Preparation, Acidity, Reactions

Alkynes contain a C≡C triple bond (one σ + two π). Both alkyne carbons are sp hybridised, linear, with bond angles of 180°. General formula C_nH_2n-2.

IUPAC naming

Replace "-ane" with "-yne". Lowest locant goes to the triple bond. CH≡C−CH₂−CH₃ is but-1-yne.

Preparation

From CaC₂ + H₂O → ethyne (acetylene) — classic lab/industrial route.
Double dehydrohalogenation of vicinal or geminal dihalides with alc. KOH (or NaNH₂).

Acidity of terminal alkynes

The terminal ≡C−H bond is weakly acidic (pKa ≈ 25) because the conjugate base sits on an sp carbon (50% s character) which holds the lone pair tightly. Reaction with NaNH₂/liq. NH₃ generates the acetylide ion R−C≡C⁻, which is a powerful nucleophile in C−C bond formation.

Reactions

Hydrogenation (H₂/Ni) gives alkane; with Lindlar's catalyst stops at cis alkene.
Hydrohalogenation: two equivalents of HX add Markovnikov to give a geminal dihalide.
Hydration (HgSO₄/H₂SO₄) gives a ketone via enol tautomerisation (acetylene gives acetaldehyde — the only exception).
Acetylide formation with Na, AgNO₃/NH₄OH (white precipitate) and Cu₂Cl₂/NH₄OH (red precipitate) — tests for terminal alkynes.

Benzene — MOT, Resonance, Reactivity, Electrophilic Substitution

Benzene C₆H₆ is the prototypical aromatic compound. All six carbons are sp², planar hexagonal, with C−C bond length intermediate between single (154 pm) and double (134 pm) at 139 pm.

Molecular orbital treatment

Each sp² carbon contributes one unhybridised p-orbital perpendicular to the ring; these six p-orbitals overlap sideways to form a delocalised π system above and below the ring plane. The six π electrons satisfy Hückel's rule (4n + 2) with n = 1, accounting for the pronounced stability.

Resonance & aromaticity

Benzene is a resonance hybrid of two Kekulé structures. Resonance energy ≈ 150 kJ/mol. Aromaticity criteria: cyclic, planar, every ring atom sp² (or with a lone pair in a p-orbital), and (4n + 2) π electrons.

Reactivity — electrophilic aromatic substitution (EAS)

The electron-rich π cloud attracts electrophiles, but benzene preserves aromaticity by undergoing substitution rather than addition. The five exam-relevant EAS reactions are:

Halogenation: C₆H₆ + Br₂ with FeBr₃ → bromobenzene + HBr.
Nitration: conc. HNO₃ + conc. H₂SO₄ at 50°C generates NO₂⁺; product is nitrobenzene.
Sulphonation: fuming H₂SO₄ (SO₃) gives benzenesulphonic acid; reversible.
Friedel-Crafts alkylation: RCl + AlCl₃ gives an alkylbenzene. Suffers from rearrangement and polyalkylation.
Friedel-Crafts acylation: RCOCl + AlCl₃ gives an aryl ketone — cleaner than alkylation, no rearrangement.

Directing effects of substituents

Ortho/para directors (activating): −OH, −OR, −NH₂, −NHR, alkyl groups. Halogens direct o/p but are weakly deactivating.
Meta directors (deactivating): −NO₂, −CN, −COOH, −SO₃H, −CHO, −COR.

High-yield rule. Benzene undergoes substitution while alkenes undergo addition — aromaticity must be preserved. Expect at least one MCQ on this contrast every year.

Substitution vs Addition

Two of the most fundamental organic reaction classes — and a favourite MDCAT contrast.

Alkane vs Alkene vs Alkyne — structural & reactivity comparison
Property	Alkane	Alkene	Alkyne
General formula	C_nH_2n+2	C_nH_2n	C_nH_2n−2
Bonding at carbon	All single (C–C, C–H)	One C=C double bond	One C≡C triple bond
Hybridisation	sp³	sp²	sp
Bond angle	109.5°	120°	180°
Geometry	Tetrahedral	Trigonal planar	Linear
Bond length C–C	1.54 Å	1.34 Å	1.20 Å
Saturation	Saturated	Unsaturated	Unsaturated
Acidity (terminal C–H)	Inert	Inert	Weakly acidic (pK_a ~25); reacts with Na, NaNH₂
Characteristic reaction	Substitution (free-radical halogenation)	Addition (H₂, X₂, HX, H₂O)	Addition + acidic substitution
Bromine water test	No decolourisation	Decolourises (rapid)	Decolourises (rapid)
Baeyer's test (KMnO₄)	No decolourisation	Decolourises (purple → brown MnO₂)	Decolourises
Examples	CH₄, C₂H₆, propane, butane	Ethene CH₂=CH₂, propene	Ethyne (acetylene) HC≡CH, propyne

Substitution vs Addition reactions
Property	Substitution	Addition
What happens	One atom/group replaced by another	Atoms add across a multiple bond → bond becomes single
Saturation preserved?	Yes	No (decreases — multiple bond is consumed)
Substrate	Alkanes, aromatic rings, alcohols, alkyl halides	Alkenes, alkynes, carbonyls
Common examples	CH₄ + Cl₂ → CH₃Cl (UV); benzene + Br₂/FeBr₃ → C₆H₅Br	CH₂=CH₂ + H₂ → CH₃CH₃; alkene + HBr → alkyl bromide (Markovnikov)

Worked MCQs

Five MCQs that capture the high-yield testing patterns for hydrocarbons. Read every explanation — the deeper concept lives there.

Q1. The chlorination of methane in UV light proceeds through which mechanism?

Electrophilic addition
Nucleophilic substitution
Free-radical substitution
Electrophilic substitution

UV light homolytically cleaves Cl−Cl into chlorine radicals (initiation), which abstract H from CH₄ producing CH₃• (propagation). Radical recombination terminates the chain. The overall pathway is free-radical substitution.

Q2. Addition of HBr to propene in the absence of peroxides yields predominantly:

1-bromopropane
2-bromopropane
1,2-dibromopropane
Propan-1-ol

Markovnikov's rule: the H adds to the carbon already bearing more hydrogens (C1 of propene), so Br ends up on C2. The intermediate 2° carbocation is more stable than the 1° alternative.

Q3. The reagent that distinguishes a terminal alkyne from a non-terminal alkyne is:

Br₂/CCl₄
cold dilute KMnO₄
Ammoniacal AgNO₃
conc. H₂SO₄

The acidic terminal ≡C−H proton is replaced by silver to give a white silver acetylide precipitate. Internal alkynes lack this acidic hydrogen and give no precipitate. Bromine water is decolourised by all alkynes.

Q4. Benzene undergoes electrophilic substitution rather than addition because:

The ring carbons are sp³ hybridised
Benzene has no π electrons
Substitution preserves aromatic stability while addition would destroy it
Electrophiles cannot approach the ring

Addition would rupture the cyclic 6-π-electron system and forfeit the ~150 kJ/mol resonance energy. Substitution restores aromaticity in the product, which is energetically far more favourable.

Q5. Which of the following is a meta-directing, deactivating group on a benzene ring?

−OH
−CH₃
−NH₂
−NO₂

−NO₂ withdraws electrons through both inductive and resonance effects, deactivating the ring and destabilising the o/p arenium intermediates. The meta product is left as the least destabilised, hence dominant. Hydroxyl, methyl and amino groups are all activating o/p directors.

Quick Recap

Alkanes C_nH_2n+2, sp³, saturated — undergo free-radical substitution.
Alkenes C_nH_2n, sp², planar — undergo electrophilic addition (Markovnikov; anti-Markovnikov with peroxides).
Alkynes C_nH_2n-2, sp, linear — terminal ≡C−H is acidic (pKa ~25); detected by Ag/Cu acetylide tests.
Ozonolysis cleaves C=C into two carbonyl fragments — locates the double bond.
Benzene = aromatic, 6 π electrons (4n+2 with n = 1), prefers substitution to preserve aromaticity.
EAS reactions: halogenation, nitration, sulphonation, Friedel-Crafts alkylation/acylation.
o/p directors: −OH, −NH₂, −OR, alkyl. m-directors: −NO₂, −COOH, −CN, −SO₃H.

Test yourself. Take a timed practice test or browse the topic-wise MCQs to lock these concepts in.