Hydrocarbons and Aromaticity (AAMC 5D)

Low-yield for the MCAT: Alkene and alkyne addition reactions and electrophilic aromatic substitution are not listed in the AAMC content outline. What is tested: naming, hybridization and stability, resonance and aromaticity, and degrees of unsaturation. Spend reaction-mechanism time on alcohols, aldehydes and ketones and carboxylic acids instead.

Hydrocarbons, compounds containing only carbon and hydrogen, are the parent skeletons of all organic chemistry. The AAMC MCAT content outline 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 is one of the highest-yield organic chapters.

AAMC content category 5D. 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 hybridized, 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 hybridized, 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; decolorization 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 hybridized, 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: Molecular Orbital Theory, 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 unhybridized p-orbital perpendicular to the ring; these six p-orbitals overlap sideways to form a delocalized 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.
Sulfonation: fuming H₂SO₄ (SO₃) gives benzenesulfonic 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. This contrast is a frequent testing point.

Substitution vs Addition

Two of the most fundamental organic reaction classes, and a favorite exam 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
Hybridization	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 decolorization	Decolorizes (rapid)	Decolorizes (rapid)
Baeyer's test (KMnO₄)	No decolorization	Decolorizes (purple → brown MnO₂)	Decolorizes
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)

Degrees of Unsaturation

The degree of unsaturation (index of hydrogen deficiency, IHD) counts the number of rings plus π bonds in a molecule, a fast way to narrow down a structure from a molecular formula on the MCAT.

Formula for a C_cH_hN_nX_x compound

DoU = (2c + 2 + n − h − x) / 2, where c = carbons, h = hydrogens, n = nitrogens, x = halogens. Oxygen (and sulfur) are ignored. They do not change the count.
· Each ring or double bond = 1 degree; each triple bond = 2 degrees.
· Example: benzene C₆H₆ → (12 + 2 − 6)/2 = 4 (one ring + three π bonds). Hexane C₆H₁₄ → 0 (saturated).

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 decolorized by all alkynes.

Q4. Benzene undergoes electrophilic substitution rather than addition because:

The ring carbons are sp³ hybridized
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 favorable.

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 destabilizing the o/p arenium intermediates. The meta product is left as the least destabilized, 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, sulfonation, Friedel-Crafts alkylation/acylation.
o/p directors: −OH, −NH₂, −OR, alkyl. m-directors: −NO₂, −COOH, −CN, −SO₃H.
Degrees of unsaturation = (2c + 2 + n − h − x)/2 = rings + π bonds (O ignored).

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