Difficulty: Intermediate | Prerequisites: IUPAC naming basics, Lewis structures, orbital hybridisation
Alkenes contain C=C double bonds (sp² carbons), alkynes contain C≡C triple bonds (sp carbons), and naming them correctly means finding the longest chain that includes the functional group, numbering to give the lowest locant, and assigning E/Z or R/S stereochemistry where applicable. Stability of both alkenes and carbon radicals follows substitution patterns: more substituted = more stable, and understanding this hierarchy is tested heavily on Exam 4.
Alkene
A hydrocarbon containing at least one carbon-carbon double bond (C=C). The carbons of the double bond are sp² hybridised. In simple terms, it is an unsaturated molecule with a flat, trigonal planar region around the double bond.
Alkyne
A hydrocarbon containing at least one carbon-carbon triple bond (C≡C). The carbons of the triple bond are sp hybridised. Think of it as a linear, rod-like bond that is shorter and stronger than a double bond.
Terminal alkyne
An alkyne where the triple bond is at the end of the carbon chain (C≡C-H). The terminal hydrogen is mildly acidic (pKa ~25) because the electrons sit in an sp orbital with more s character.
Internal alkyne
An alkyne where the triple bond is flanked by carbon atoms on both sides (R-C≡C-R). Internal alkynes are more stable than terminal alkynes, much the same way that more substituted alkenes are more stable.
IUPAC nomenclature
The systematic naming convention maintained by the International Union of Pure and Applied Chemistry. For alkenes, the suffix is "-ene"; for alkynes, "-yne". When both are present, the compound is an "enyne" and numbering gives the lowest set of locants to the multiple bonds, with the triple bond getting the lower number in a tie.
E/Z stereochemistry
A system for naming the configuration of groups around a C=C double bond, based on Cahn-Ingold-Prelog priority rules. Z (zusammen) = higher-priority groups on the same side. E (entgegen) = higher-priority groups on opposite sides. In simple terms, Z roughly corresponds to "cis" and E to "trans" when using CIP priorities.
R/S configuration
A system for designating the absolute configuration of a chiral centre (a carbon with four different substituents). Assign CIP priorities 1 through 4, orient the lowest priority away from you, then trace 1→2→3: clockwise = R, anticlockwise = S.
Degree of unsaturation (index of hydrogen deficiency, IHD)
Calculated as (2C + 2 + N - H - X) / 2 for a molecular formula CₙHₙNₙOₙXₙ. Each degree corresponds to one ring or one pi bond. A double bond = 1 degree; a triple bond = 2 degrees; a ring = 1 degree.
Hyperconjugation
The stabilising interaction in which electrons in a C-H or C-C sigma bond adjacent to a p orbital (or a radical/carbocation) donate electron density into that orbital. This is the main reason more substituted alkenes, radicals, and carbocations are more stable.
Carbon radical
A carbon atom bearing an unpaired electron. Radicals are sp² hybridised (planar) and follow the same substitution-based stability trend as carbocations: methyl < primary < secondary < tertiary. Additional stabilisation comes from resonance (e.g. allylic or benzylic radicals).
Find the longest continuous chain that includes the C≡C triple bond.
Number the chain to give the triple bond the lowest possible locant.
Change the parent alkane ending from "-ane" to "-yne" and place the locant before "-yne" (e.g. 1-butyne, 2-pentyne).
Name and number all substituents as prefixes (e.g. 3-methyl-1-butyne means a four-carbon chain with a triple bond at C1 and a methyl group at C3).
When the molecule also contains a hydroxyl group, the "-ol" suffix takes naming priority for the lowest locant, and the triple bond locant appears as an infix: 4-heptyn-3-ol means a seven-carbon chain, triple bond between C4 and C5, hydroxyl on C3.
Find the longest chain containing the C=C double bond.
Number to give the double bond the lowest locant.
Replace "-ane" with "-ene" and indicate the position.
The compound is named as an "-en-...-yne" (the "-ene" comes first alphabetically in the name).
Number the chain to give the lowest set of locants to the combination of multiple bonds.
If there is a tie, give the double bond the lower number.
Example: (Z)-4-hexen-1-yne is a six-carbon chain with a triple bond starting at C1 and a double bond starting at C4, with Z configuration at the double bond.
A chiral centre adjacent to a triple bond is common in exam questions (e.g. (R)-2-chloro-3-hexyne).
Assign CIP priorities to the four groups on the chiral carbon. Remember that a triple bond counts as three bonds to the other carbon (the "phantom atom" expansion): C≡C is treated as C bonded to C, C, C.
Orient lowest priority (4) away from you, then trace 1→2→3.
Identify the two groups on each carbon of the C=C.
Use CIP priority rules to rank the two substituents on each carbon.
Higher-priority groups on the same side = Z. Opposite sides = E.
Watch for cases where the substituents are not simply "bigger" or "smaller" by size. CIP rules go atom by atom: Cl > C > H at the first point of difference.
Carbon radicals follow a substitution-based stability order:
Methyl radical (least stable) < primary < secondary < tertiary (most stable)
Allylic and benzylic radicals gain additional stability through resonance delocalisation of the unpaired electron across the pi system.
On this exam, you may be asked to rank radicals attached to ring systems. A radical on a carbon adjacent to a double bond (allylic position) is more stable than a secondary radical that lacks resonance.
The underlying reason is hyperconjugation: adjacent C-H and C-C sigma bonds can overlap with the half-filled p orbital on the radical carbon, spreading out the electron deficiency.
Alkene stability increases with substitution:
Ethylene (no substituents, least stable)
Monosubstituted
Disubstituted (cis < trans for acyclic alkenes, because trans has less steric strain)
Trisubstituted
Tetrasubstituted (most stable)
This ordering is confirmed experimentally by heats of hydrogenation: more substituted alkenes release less heat when hydrogenated, meaning they started at a lower energy (more stable).
For disubstituted alkenes specifically:
Trans (E) is more stable than cis (Z) in acyclic systems because the two larger substituents are farther apart, reducing steric strain.
In cyclic systems, this can reverse depending on ring strain.
Hyperconjugation is the primary explanation. Alkyl groups donate electron density into the pi system of the double bond (or the half-filled orbital of a radical), which stabilises the species. More alkyl groups = more hyperconjugative donors = greater stabilisation.
Degree of unsaturation (IHD)
IHD = (2C + 2 + N - H - X) / 2
Where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Oxygen does not appear in the formula (it does not change the IHD).
For C₇H₁₄: IHD = (2(7) + 2 - 14) / 2 = 1. One degree of unsaturation, which means one ring or one double bond.
CIP priority assignment (quick reference)
Higher atomic number = higher priority (Cl > O > N > C > H).
If tied at the first atom, move outward to the next atoms attached and compare.
Double and triple bonds use the phantom-atom expansion: C=C means each carbon is bonded to an extra phantom carbon; C≡C means each is bonded to two phantom carbons.
Heats of hydrogenation trend
More substituted alkene → lower heat of hydrogenation → more stable starting material.
Students often think that cis/trans and Z/E are interchangeable labels. They are not. Cis/trans is an informal system based on whether "similar" groups are on the same side, while Z/E uses strict CIP priority rules. A cis alkene is not always Z.
Students sometimes number the chain to give substituents (not the functional group) the lowest locant. The double or triple bond always gets naming priority for the lowest locant, not the substituent.
A common error is forgetting that a triple bond counts as two degrees of unsaturation, not one. If a molecular formula gives IHD = 2, it could be one triple bond, two double bonds, two rings, one double bond and one ring, or other combinations.
When ranking radical stability, students often confuse radical stability with carbocation stability. While the trends are similar (more substituted = more stable), the reasoning differs. Radicals are stabilised primarily by hyperconjugation, not by inductive effects alone. Resonance stabilisation (allylic, benzylic) matters for both.
⚠️ Nomenclature questions (drawing structures from IUPAC names) appeared as Question 1 on this practice exam and are worth 8 points. You must draw the correct structure and clearly show stereochemistry (R/S, E/Z, wedge/dash).
⚠️ Stability ranking of radicals and alkenes appeared as Question 2 and is worth 11 points. Be ready to rank three or four species in order of increasing stability.
⚠️ The IHD formula is not directly tested as a standalone question here, but it is essential for the spectral analysis question (Question 6). If you cannot calculate degrees of unsaturation, you cannot deduce structures from molecular formulas.
⚠️ CIP priority assignment underpins both R/S and E/Z questions. Practice with substituents that include halogens, alkyl groups of different sizes, and groups with multiple bonds.
True or false: A tetrasubstituted alkene is more stable than a trisubstituted alkene. (True)
Fill in the blank: The IUPAC suffix for an alkyne is ____. (-yne)
True or false: In (Z)-2-butene, the two methyl groups are on opposite sides of the double bond. (False, Z = same side)
Fill in the blank: The degree of unsaturation for C₅H₈ is ____. (2)
True or false: A tertiary radical is less stable than a secondary radical. (False, tertiary is more stable)
Q: Draw the structure of (R)-2-chloro-3-hexyne. What functional groups and stereocentres are present?
A: The parent chain is hexane (six carbons). The triple bond runs between C3 and C4. A chlorine substituent is on C2, which is the chiral centre. Assign CIP priorities to the four groups on C2 (Cl, the C3≡C4-C5-C6 chain, the CH₃ at C1, and H), orient H away from you, and trace 1→2→3 clockwise for R configuration. Draw with a wedge or dash on C2 to show the R arrangement.
Q: Place the following in order of increasing stability: ethylene, trans-2-butene, 2-methylpropene, 2,3-dimethyl-2-butene.
A: Ethylene (unsubstituted, least stable) < trans-2-butene (disubstituted) < 2-methylpropene (disubstituted, but note that trans-disubstituted and gem-disubstituted are close; this ordering may vary by source) < 2,3-dimethyl-2-butene (tetrasubstituted, most stable). The key principle: more alkyl groups on the double bond = more hyperconjugation = lower energy.
Q: Rank in order of increasing stability: primary radical, allylic radical, tertiary radical.
A: Primary radical (least stable) < tertiary radical < allylic radical (most stable, assuming the allylic radical benefits from resonance delocalisation across the pi system). If the allylic radical is also secondary or tertiary, it is even more stabilised.
Q: What is the degree of unsaturation for (Z)-4-hexen-1-yne (C₆H₈)?
A: IHD = (2(6) + 2 - 8) / 2 = 3. One triple bond (2 degrees) + one double bond (1 degree) = 3 total. This matches.
Q: A student names a compound "2-methyl-3-butyne." What error have they made?
A: The triple bond should receive the lowest possible locant. The correct name is 3-methyl-1-butyne (number from the end nearest the triple bond, placing it at C1 rather than C3).
Nomenclature and stability are foundational for every reaction topic that follows. The regiochemistry of additions to alkenes (Markovnikov vs anti-Markovnikov) depends on which intermediate (carbocation or radical) is more stable, and that stability follows the same substitution trends covered here. Stereochemistry assignments (R/S, E/Z) return in every reaction where a new chiral centre forms or where syn/anti addition matters.
The spectral analysis question on this exam (Question 6) requires you to work backwards from an NMR spectrum and a molecular formula to deduce a structure, which starts with calculating the degree of unsaturation.
Alkene, alkyne, enyne, IUPAC nomenclature, E/Z configuration, R/S configuration, CIP priority rules, Cahn-Ingold-Prelog, radical stability, alkene stability, hyperconjugation, degree of unsaturation, index of hydrogen deficiency, IHD, heats of hydrogenation, terminal alkyne, internal alkyne, sp hybridisation, sp2 hybridisation, organic chemistry I, CHEM 2301, Exam 4, stereochemistry, chiral centre, wedge-dash notation