Difficulty: Intermediate | Prerequisites: CHEM 2301 Exams 1–3 material (functional groups, hybridisation, stereochemistry basics)
This covers the tail end of Organic Chemistry I: naming compounds with multiple functional groups (alkenes, alkynes, alcohols), predicting physical and chemical properties from structure, and connecting molecular structure to reactivity. You should already be comfortable with Lewis structures, hybridisation (sp, sp², sp³), basic stereochemistry (R/S, E/Z), and the idea that bond strength and acidity are tied to orbital character. If those feel shaky, revisit your Exam 2 and 3 notes first.
TL;DR
IUPAC naming of alkenes, alkynes and alcohols requires correct identification of the longest chain containing the functional group, lowest-locant numbering, and E/Z or R/S stereodescriptors where applicable. Structure determines reactivity: hybridisation controls bond length and acidity, substitution controls alkene stability, and radical stability governs selectivity in radical halogenation.
IUPAC nomenclature
The systematic method for naming organic compounds endorsed by the International Union of Pure and Applied Chemistry. Names encode the parent chain, substituents, stereochemistry and functional groups in a standardised format.
In simple terms, it is the universal "address system" for molecules so that any chemist reading the name can draw the exact structure.
Alkyne
A hydrocarbon containing a carbon–carbon triple bond (C≡C). The suffix is "-yne" in IUPAC naming.
Think of it as the most unsaturated version of a two-carbon unit: three bonds shared between two carbons, with sp hybridisation.
Alkene
A hydrocarbon containing a carbon–carbon double bond (C=C). The suffix is "-ene" in IUPAC naming.
In simple terms, a double bond means the carbons are sp²-hybridised and the geometry around them is trigonal planar.
(E)/(Z) designation
A system for specifying the configuration of groups around a C=C double bond, based on Cahn–Ingold–Prelog priority rules. (Z) = higher-priority groups on the same side; (E) = higher-priority groups on opposite sides.
Think of Z as "zusammen" (together) and E as "entgegen" (opposite).
(R)/(S) designation
Absolute configuration labels for a chiral centre, assigned using CIP priority rules viewed from the side opposite the lowest-priority group. Clockwise = R, anticlockwise = S.
Bond dissociation energy (BDE)
The energy required to homolytically break a specific bond in a molecule, producing two radicals. Lower BDE means a weaker, more easily broken bond.
Benzylic position
The carbon directly attached to a benzene ring. A radical or cation at this position is stabilised by resonance delocalisation into the aromatic ring.
In simple terms, this is the carbon "next door" to the ring, and anything reactive sitting there gets extra stability from the ring's electron cloud.
Enol
The tautomer of a carbonyl compound in which the α-carbon bears a hydroxyl group and a C=C double bond replaces the C=O. Enols are typically less stable than their keto forms but appear as intermediates in acid-catalysed reactions of alkynes.
Tautomerisation (keto–enol)
The equilibrium interconversion between a keto form (C=O with adjacent C–H) and an enol form (C=C–OH). In most cases the keto form is heavily favoured.
Naming procedure
Identify the longest continuous carbon chain that includes the highest-priority functional group. Priority order for suffixes: alcohol ("-ol") > alkyne ("-yne") > alkene ("-ene").
Number the chain to give the functional group the lowest possible locant.
When both a double bond and a triple bond are present, both are indicated in the name (e.g. "pent-3-en-1-yne"). If there is a tie in numbering, the double bond gets the lower number.
Assign E/Z descriptors at double bonds and R/S at chiral centres.
Substituents are listed alphabetically as prefixes with their locants.
Worked example from the exam: (E)-pent-3-en-1-yne
Five-carbon chain = "pent."
Triple bond at C1 = "-1-yne."
Double bond at C3 = "-3-en."
The double-bond geometry is (E): higher-priority groups are on opposite sides.
Full structure: HC≡C–CH=CH–CH₃, with the E configuration at the C3=C4 bond.
Worked example: (S)-but-3-yn-2-ol
Four-carbon chain = "but."
Triple bond at C3 = "-3-yn."
Hydroxyl at C2 = "-2-ol" (alcohol has naming priority, so C2 gets the lowest locant).
The chiral centre at C2 has S configuration.
Naming a branched alkyne: 3,3-dimethylpent-1-yne
Five-carbon parent chain containing the triple bond.
Triple bond at C1, two methyl groups at C3.
This is answer (A) from the exam. Common traps include mislabelling the parent chain length or confusing "-yne" with "-ene."
Acidity and conjugate base stability
The most acidic hydrogen in a molecule is the one whose removal produces the most stable conjugate base. Stability of the conjugate base depends on:
Electronegativity of the atom bearing the charge
Resonance delocalisation of the negative charge
Hybridisation (more s-character = electrons held closer to the nucleus = more stable anion)
Inductive effects from nearby electronegative atoms
From the exam: in a structure with Hₐ (on a terminal alkyne carbon), Hᵇ and Hᶜ, the answer is Hₐ. The sp-hybridised C–H has the most s-character (50%), so the conjugate base is most stabilised. Additionally, the alkynide anion can be stabilised by resonance with adjacent conjugated bonds.
Hybridisation and bond length/strength
The relationship between hybridisation, bond length and bond strength is central:
sp³ C–H bonds are the longest and weakest (25% s-character).
sp² C–H bonds are intermediate.
sp C–H bonds are the shortest and strongest (50% s-character).
The same logic applies to C–C bonds:
C(sp³)–C(sp³) single bonds are longest.
C=C double bonds are shorter.
C≡C triple bonds are shortest.
From the exam (structure with numbered bonds):
Longest C–H bond: bond 6 (sp³ carbon)
Longest C–C bond: bond 3 (single bond between sp³ carbons)
Shortest C–C bond: bond 2 (the double or triple bond in the chain)
The heat of hydrogenation is the energy released when H₂ adds across a double bond over a metal catalyst (Pd, Pt or Ni). A more stable alkene releases less energy upon hydrogenation, because it starts from a lower energy state.
Stability order for alkenes (most to least stable):
Tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted
Trans (E) disubstituted > cis (Z) disubstituted (less steric strain in the trans isomer)
Conjugated alkenes are more stable than isolated alkenes
From the exam: given a set of cyclic and acyclic alkenes, the one requiring the least energy to reduce is the most substituted (or most stable) alkene. The tetrasubstituted cycloalkene was the correct answer.
Why substitution stabilises alkenes:
Alkyl groups are weakly electron-donating (hyperconjugation), which stabilises the electron-poor π system. More substituents = more hyperconjugation = lower energy.
Radical halogenation overview
When a hydrocarbon is treated with Br₂ and light (hν) or heat, a radical chain reaction occurs:
Initiation: Br₂ → 2 Br· (homolytic cleavage by light)
Propagation: Br· abstracts an H from the substrate to form HBr + a carbon radical; the carbon radical reacts with Br₂ to form the product + a new Br·
Termination: two radicals combine
Selectivity in radical bromination
Bromine radicals are highly selective. They preferentially abstract the hydrogen that produces the most stable radical intermediate. Radical stability follows the same trend as carbocation stability:
3° > 2° > 1° > methyl
Benzylic and allylic radicals are especially stable due to resonance delocalisation
Exam example: bromination of a molecule with formula C₁₀H₁₂
The starting material has a benzylic C–H (a hydrogen on the carbon directly attached to a benzene ring). This hydrogen is abstracted because:
The benzylic C–H has the lowest BDE (weakest bond).
The resulting benzylic radical is the most stable radical that can form.
Either argument (lowest BDE or most stable radical) earns full credit.
The product is C₁₀H₁₁Br, with bromine replacing the benzylic hydrogen.
Real-world connection: Radical halogenation selectivity is why industrial synthesis of certain brominated intermediates targets benzylic positions. The predictability of radical stability makes it a reliable tool for selective functionalisation.
Students often number the chain starting from the wrong end, giving the functional group a higher locant than necessary. Always start numbering from the end nearest the highest-priority group.
Students confuse E/Z with cis/trans. E/Z is based on CIP priority, not on whether the same groups are on the same side. They sometimes give different answers.
Students assume the most acidic hydrogen is on the most electronegative atom. In organic chemistry, hybridisation and resonance stabilisation of the conjugate base are often more important than raw electronegativity.
Students think all C–H bonds are the same length. They vary meaningfully with hybridisation: an sp C–H is roughly 1.08 Å, an sp³ C–H is roughly 1.09 Å. This matters for predicting bond strength and acidity.
⚠️ Nomenclature questions are free points if you practise. Pomerantz awards partial credit for the correct parent chain, correct chain length, and correct stereodescriptor, separately.
⚠️ Acidity ranking by hybridisation is a recurring exam topic. Know the order: sp > sp² > sp³ for acidity, and be ready to explain why using conjugate base stability.
⚠️ Bond-length questions require you to identify the hybridisation of each carbon. Label each carbon's hybridisation on the structure before answering.
⚠️ Radical bromination questions test two things at once: product prediction and mechanistic reasoning (why that hydrogen is abstracted). Both the stability argument and the BDE argument are accepted.
True or false: In IUPAC naming, when a molecule has both an alkene and an alkyne, the alkyne always gets the lower locant.
False. The double bond gets the lower number when there is a tie.
Fill in the blank: An sp-hybridised carbon has ____% s-character, making its C–H bond the ______ and ______ of all C–H bonds.
50%, shortest, strongest
True or false: The most substituted alkene releases the most heat upon catalytic hydrogenation.
False. It releases the least heat because it starts at a lower energy.
Fill in the blank: In radical bromination, selectivity is governed by the stability of the __________ intermediate.
radical (carbon radical)
True or false: A benzylic radical is stabilised by hyperconjugation.
False. It is stabilised by resonance delocalisation into the aromatic ring. Hyperconjugation stabilises alkyl-substituted radicals/cations.
Q: Draw the structure of (E)-pent-3-en-1-yne. How many degrees of unsaturation does it have?
A: HC≡C–CH=CH–CH₃ with E geometry at the C3=C4 double bond. Three degrees of unsaturation (one triple bond = 2, one double bond = 1).
Q: Which is more acidic: a terminal alkyne C–H (sp) or a vinyl C–H (sp²)? Explain.
A: The terminal alkyne C–H is more acidic. The sp carbon has 50% s-character vs 33% for sp². Greater s-character means the bonding electrons are held closer to the carbon nucleus, stabilising the conjugate base (the alkynide anion).
Q: Rank these alkenes from most to least stable: cyclohexene, 1-methylcyclohexene, 2,3-dimethylbut-2-ene, ethene.
A: 2,3-Dimethylbut-2-ene (tetrasubstituted) > 1-methylcyclohexene (trisubstituted) > cyclohexene (disubstituted) > ethene (unsubstituted).
Q: A molecule C₁₀H₁₂ with a benzylic C–H is treated with Br₂/hν. What is the molecular formula of the product, and why is the benzylic hydrogen selectively abstracted?
A: C₁₀H₁₁Br. The benzylic hydrogen is abstracted because it has the lowest BDE and produces the most stable (resonance-stabilised) radical intermediate.
Q: In a reaction coordinate diagram for a two-step reaction, how do you distinguish intermediates from transition states?
A: Intermediates sit in energy minima (valleys) between two transition states. Transition states sit at energy maxima (peaks). Intermediates have finite lifetimes; transition states do not.
Nomenclature links directly to every other section of this exam. You cannot write correct reagents in a synthesis roadmap or identify NMR splitting patterns if you cannot name and draw the molecule first. Hybridisation and bond character reappear in the mechanisms section, where understanding orbital overlap determines whether you can draw correct curly arrows.
Acidity and conjugate base stability connect forward to Organic Chemistry II topics: enolate chemistry, nucleophilic addition, and base-mediated eliminations all depend on predicting which proton is removed and why.
Radical bromination is the simplest radical chain mechanism you will see. The selectivity principles (radical stability, BDE) return in more complex radical reactions and in rationalising polymerisation.
Related Terms / Search Tags
IUPAC naming, organic nomenclature, alkene naming, alkyne naming, E/Z configuration, R/S configuration, CIP priority rules, Cahn-Ingold-Prelog, stereodescriptors, sp hybridisation, sp2 hybridisation, sp3 hybridisation, bond dissociation energy, BDE, radical bromination, benzylic radical, allylic radical, heat of hydrogenation, alkene stability, substitution and stability, catalytic hydrogenation, conjugate base stability, acidity in organic chemistry, CHEM 2301, Pomerantz, Exam 4