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Chemistry: Organic Chemistry Basics

Part of the HAT-UG (HEC Aptitude Test - Undergraduate) study roadmap. Subject Knowledge topic sk-6 of Subject Knowledge.

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Chemistry: Organic Chemistry Basics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your HAT-UG Subject Knowledge paper.

Organic chemistry basics covers carbon compounds, their bonding, nomenclature, isomerism, and reaction types. Carbon is tetravalent and forms sp³, sp², or sp hybrid orbitals, giving rise to single (σ), double (σ + π), and triple (σ + 2π) bonds. The Degree of Unsaturation formula (2C + 2 + N − H − X)/2 tells you how many rings or π bonds sit in an unknown molecule. IUPAC naming hinges on the longest carbon chain, lowest locants, and the principal functional group chosen as the suffix (priority: −oic acid > −al > −ol > −ene > −yne). HAT-UG tests these directly.

  • Functional groups to recognise: −OH (alcohol), −CHO (aldehyde), C=O (ketone), −COOH (acid), −NH₂ (amine), −X (halo).
  • Isomerism split: structural (chain, position, functional) vs stereoisomerism (cis-trans, optical/R-S).
  • Markovnikov rule: H adds to the carbon with more H’s in HX addition to alkenes; reversed with peroxides.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Hybridization and Bonding

Carbon uses sp³ in alkanes (tetrahedral, 109.5°), sp² in alkenes (trigonal, 120°, leaves one p-orbital for a π bond), and sp in alkynes (linear, 180°). Every single bond is a σ bond; the second bond of a double bond and both extra bonds of a triple bond are π bonds. Pi bonds restrict rotation, which is exactly why cis-trans isomerism exists around C=C but never around a tetrahedral C−C single bond.

Degree of Unsaturation

A quick pre-screening tool: DoU = (2C + 2 + N − H − X)/2. Each ring or π bond consumes one degree. Oxygen and sulfur are ignored because they don’t change the hydrogen count for a neutral molecule.

Functional-Group Priority for Naming

Priority (high → low)SuffixPrefix
Carboxylic acid−oic acidcarboxy-
Aldehyde−alformyl-/oxo-
Ketone−oneoxo-
Alcohol−olhydroxy-
Alkene−ene—
Alkyne−yne—
  • The highest-priority group becomes the suffix and gets the lowest locant.
  • Substituents are listed alphabetically (chloro before methyl).
  • The parent chain is the longest continuous chain, not necessarily the one drawn straight.

Reaction Types

Substitution (SN1/SN2), electrophilic addition to C=C, elimination (E1/E2, follow Zaitsev for major alkene), and rearrangement. Electrophiles accept electrons (H⁺, Br⁺); nucleophiles donate them (OH⁻, NH₃, CN⁻).

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Electronic Effects and Carbocation Stability

Three effects govern reactivity in organic molecules: the inductive effect (σ-bond electron shift through atoms, +I from alkyl groups, −I from −X, −OH, −NO₂), the mesomeric/resonance effect (π-bond delocalisation, +M from −OH, −NH₂; −M from −NO₂, C=O), and hyperconjugation (σ C−H overlap with empty p-orbital, stabilising carbocations and free radicals). Stability order for carbocations: 3° > 2° > 1° > methyl. This explains why Markovnikov addition to an unsymmetrical alkene puts H on the less-substituted carbon: the intermediate carbocation forms at the more-substituted site.

Worked Micro-Example

Compute the degree of unsaturation for C₄H₆O₂ (a possible methyl acrylate skeleton): DoU = (2·4 + 2 + 0 − 6 − 0)/2 = (8 + 2 − 6)/2 = 2. Two degrees → one C=C plus one C=O, consistent with an ester.

Common Traps and Exam Strategy

MistakeCorrection
Cis-trans on tetrahedral COnly C=C or rings allow cis-trans
Markovnikov with HBr/peroxidePeroxides reverse it (anti-Markovnikov, free radical)
Choosing −ol over −al as suffixAldehyde outranks alcohol in IUPAC priority
Ranking CIP by atomic sizeUse atomic number, not mass or radius

HAT-UG Subject Knowledge contributes roughly 4% of the aggregate, typically one short question (5–8 marks) on naming, functional-group identification, isomerism type, or reaction product. Spend ≤6 minutes, write the parent chain, mark the principal group, then assign locants — three deliberate steps beat intuition.

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