Hydrocarbons · Lesson 14 of 14
Chapter Summary and Practice
“Connect bonding, isomerism, reaction mechanisms, and structural evidence across every hydrocarbon family.”
• Relate hydrocarbon classification and formulas to carbon bonding and geometry. • Distinguish structural, conformational, and geometrical changes. • Select reaction conditions and predict products through appropriate intermediates. • Use oxidation and ozonolysis to infer carbon skeletons. • Explain aromaticity, substitution direction, and hydrocarbon applications and hazards. • Solve integrated naming, conversion, and reasoning problems across the chapter.
Begin with structure, then ask what can change
A hydrocarbon problem becomes manageable when approached in a consistent order. Identify the carbon skeleton and its rings or multiple bonds. Check valence and formula, then decide which reaction site and mechanism the conditions make available. The same molecular formula can describe different skeletons or configurations, so do not allow formula alone to replace a structural drawing.
| Family | Bonding and geometry to inspect | Typical chemical pattern |
|---|---|---|
| Open-chain alkane | Tetrahedral carbon, sigma bonds, rotatable single bonds | Radical substitution; combustion; high-temperature transformations |
| Alkene | One sigma and one pi component at a double bond; planar local geometry | Electrophilic addition; oxidation; polymerisation |
| Alkyne | One sigma and two pi components; linear sp centres | Sequential addition; terminal-hydrogen acidity; polymerisation |
| Arene | Cyclic conjugated delocalised pi system | Electrophilic substitution with aromaticity restored |
For single open-chain members, the formulas progress from to to . These are family relationships with stated assumptions, not universal structure identifiers. A ring can reduce the hydrogen count too. Carbon hybridisation connects shape with acidity: a terminal alkyne leaves a relatively stabilised anion at carbon, whereas an internal alkyne lacks that particular hydrogen.
Problem
A compound has formula C4H8. What additional evidence is needed before treating it as but-1-ene?
- 1.The formula is consistent with an acyclic monoalkene, but it also allows saturated cyclic frameworks. Even within the alkene category, but-1-ene, but-2-ene, and 2-methylpropene are possible.
- 2.Evidence such as addition behaviour and oxidation fragments can establish the multiple bond and its position. Geometrical information is a further question if the structure allows it.
- 3.Thus family identification, connectivity, and configuration are separate levels of information. One formula does not settle all three.
Problem
Arrange ethyne, benzene, and n-hexane in decreasing acidity.
- 1.Compare the carbon directly bonded to the hydrogen being removed. It is sp in ethyne, sp2 in benzene, and sp3 in n-hexane.
- 2.Greater s character generally stabilises the negative charge left after proton removal in this comparison. The order is therefore ethyne, benzene, then n-hexane.
- 3.This does not make benzene or hexane strong aqueous acids. The comparison is relative, and benzene losing a ring proton is not the same event as an electrophile temporarily attacking its pi cloud.
Compare the different meanings of an isomer
A chain or position isomer changes connectivity. An ethane conformer changes spatial arrangement through single-bond rotation without changing connectivity. Cis/trans configurations arise when both carbons of a restricted double bond carry different groups. Rotating the page is merely a change of viewpoint; rotating one end of ethane is a physical conformational change. Neither act by itself changes the carbon skeleton.
Names provide another way to check structures. Trace the parent containing the required multiple bond, assign locants, place branches, and complete hydrogen valence. At a quaternary alkane carbon there is no hydrogen. At a triple-bond carbon only one remaining valence connection is available beyond its triple-bond partner. These simple checks catch many plausible-looking but impossible drawings.
Let conditions choose the mechanism
Methane chlorination begins with chlorine homolysis and proceeds through radicals. Ordinary HBr addition to an alkene proceeds through a carbocation in the present model. HBr with peroxide instead proceeds through a carbon radical after bromine adds first. Benzene substitution involves a temporary arenium ion and final proton removal. Each explanation needs its own electron movement; the word “reaction” does not identify one universal intermediate.
Problem
You need to convert propene into either 1-bromopropane or 2-bromopropane. Choose and justify conditions.
- 1.For 2-bromopropane, use ordinary HBr addition without peroxide. Protonation leading to the secondary carbocation is favoured, and bromide attaches at the middle carbon.
- 2.For 1-bromopropane, use HBr with an appropriate peroxide initiator. Bromine radical attaches at the terminal carbon to leave a secondary radical, then that radical takes hydrogen from HBr.
- 3.Do not transfer the peroxide instruction to HCl or HI. A sustainable radical chain needs favourable propagation energetics, not just the presence of a radical initiator.
Preparation methods can be selected by carbon accounting. Hydrogenation preserves the skeleton; Wurtz coupling joins alkyl fragments; soda-lime heating removes one carboxyl carbon; Kolbe electrolysis removes a carboxyl carbon from each partner before radical coupling. Cracking divides a skeleton, while aromatisation converts an appropriate straight chain into an aromatic ring with hydrogen loss.
Use fragments and atom balance as independent evidence
Reductive ozonolysis preserves the substituents attached to each original double-bond carbon and makes two carbonyl centres. Cold dilute permanganate instead forms a vicinal diol, while vigorous oxidation can take cleavage fragments to acids or carbon dioxide. Keep these work-ups separate. Reconstructing an alkene is the reverse of ozonolysis: join the two carbonyl carbons after conceptually removing their oxygens.
Problem
An acyclic alkene has three carbon–carbon sigma bonds, eight carbon–hydrogen bonds, and one pi bond. Ozonolysis gives two molecules of an aldehyde of relative molecular mass 44. Identify it.
- 1.Three carbon–carbon connections in an open connected skeleton mean four carbons. The eight C–H bonds give C4H8, consistent with one double bond.
- 2.The aldehyde of relative molecular mass 44 is ethanal. Two identical ethanal fragments require each double-bond carbon to carry methyl and hydrogen.
- 3.Joining them gives CH3CH=CHCH3, but-2-ene. These clues establish connectivity but do not distinguish its cis and trans configurations.
Complete combustion can be checked independently of mechanism: gives and , requiring . A fractional coefficient is valid; multiply the entire equation if whole-number coefficients are preferred. Incomplete combustion is chemically different and can produce soot or carbon monoxide.
Aromatic stability changes the preferred outcome
An aromaticity claim needs a cyclic planar conjugated system before the electron count is tested. Benzene has six delocalised pi electrons and equal carbon–carbon distances. Substitution temporarily interrupts this system but restores it after proton loss; ordinary addition would retain the loss of aromatic stabilisation in the final product. This explains why benzene cannot be treated as three independent ordinary alkene double bonds.
Problem
Compare the three connectivity sketches above.
- 1.In A, one saturated ring carbon interrupts a continuous cyclic p-orbital path. An exocyclic double bond does not automatically repair that interrupted circuit.
- 2.Neutral cyclopentadiene in B also has a saturated CH2 carbon and is not continuously conjugated around its ring. Removing a proton to form the cyclopentadienyl anion changes its orbital and electron-count situation.
- 3.Cyclooctatetraene in C has eight pi electrons and adopts a non-planar conformation, so it does not satisfy the planar 4n + 2 aromatic conditions. A flat connectivity sketch is not evidence of actual planarity.
Problem
What carbonyl fragments are associated with ozonolysis of ortho-xylene, and how can they be related to the Kekulé contributors?
- 1.Track the adjacent methyl-substituted ring carbons. Cutting alternating pi connections in the two contributing drawings gives pairs of adjacent ring carbons with either zero, one, or two methyl substituents.
- 2.Those fragments correspond to ethanedial (glyoxal), 2-oxopropanal (methylglyoxal), and butane-2,3-dione (diacetyl). For one contributor, two one-methyl fragments and one unsubstituted fragment arise; for the other, one two-methyl fragment and two unsubstituted fragments arise.
- 3.This connects the fragment pattern with the two standard structural representations. It must not be interpreted as benzene switching between two isolable localised compounds: the actual ring is one delocalised molecule.
An existing substituent then changes how readily the ring undergoes another substitution and which positions it favours. Donation usually activates and directs ortho/para. Strong withdrawal usually deactivates and directs meta. Halogens deactivate overall but direct ortho/para. Reaction order therefore matters in planning disubstituted products, and preferred direction does not guarantee a pure single isomer.
Connect useful materials with their consequences
Hydrocarbons supply fuels, solvents, synthesis feedstocks, and polymers. Their size and shape affect dispersion attractions, boiling, solubility, and crystal packing. Addition polymerisation can turn reactive small molecules into durable materials; durability also affects waste persistence. Incomplete combustion can produce harmful mixtures, including certain carcinogenic fused-ring compounds. Knowing the family of a molecule does not determine all of its uses or hazards.
Check the three most common shortcuts: a matching formula does not prove a unique structure; reagent names without conditions do not prove a unique product; and ortho/para direction does not necessarily mean activation. Use a drawing, a mechanism, and an atom count together.
Quiz
Which preparation joins two alkyl fragments without deliberately losing a carboxyl carbon?
Which pair illustrates conformations of one compound?
Reductive ozonolysis gives two ethanal molecules. Which starting connectivity fits?
Which structure lacks a terminal alkyne hydrogen?
What is essential before applying the aromatic electron count?
Which group deactivates but directs ortho/para?
Which change normally preserves the entire carbon skeleton?
What distinguishes ultraviolet chlorine addition to benzene from exhaustive substitution?
Practice Problems
- Explain the ethane byproduct in methane chlorination by writing the relevant radical steps.
- Name all of these structures: (a) ; (b) ; (c) ; (d) ; (e) a benzene ring bearing adjacent hydroxyl and methyl groups; (f) ; (g) .
- Draw and name all acyclic structural isomers of C4H8 containing one double bond and C5H8 containing one triple bond. Identify any additional geometrical configurations separately.
- Name all reductive ozonolysis products of pent-2-ene, 3,4-dimethylhept-3-ene, 2-ethylbut-1-ene, and 1-phenylbut-1-ene.
- Reconstruct and name an alkene that gives ethanal and pentan-3-one on reductive ozonolysis.
- An alkene has three C–C sigma bonds, eight C–H bonds, and one pi bond. Its ozonolysis gives two molecules of an aldehyde with relative molecular mass 44. Identify the alkene and state what remains unknown about its configuration.
- Reconstruct the alkene whose reductive ozonolysis gives propanal and pentan-3-one, and verify the total carbon count.
- Balance complete combustion equations for butane, a pentene, a hexyne, and toluene.
- Draw cis- and trans-hex-2-ene. Explain the expected boiling-point difference in terms of molecular polarity.
- Explain benzene’s unusual stability despite the three formal double bonds in a Kekulé drawing.
- State and explain every condition required for the simple aromaticity test used in this chapter.
- Use the three connectivity sketches above to explain failure of aromaticity in each neutral system.
- Plan benzene conversions to para-nitrobromobenzene, meta-nitrochlorobenzene, para-nitrotoluene, and acetophenone. Give reagents, reaction order, and any likely isomer-mixture qualification.
- For , identify primary, secondary, tertiary, and quaternary carbons and count the hydrogens attached to each category.
- Explain how branching changes the boiling point of alkane isomers and why the same simple rule need not predict their melting points.
- Give mechanisms explaining why propene with HBr produces different major products with and without benzoyl peroxide.
- Name the carbonyl fragments from ortho-xylene ozonolysis and relate their pattern to the conventional Kekulé representations without treating resonance as an equilibrium.
- Arrange benzene, n-hexane, and ethyne in decreasing acidity and explain the role of carbon hybridisation.
- Explain why benzene is susceptible to electrophilic substitution but ordinary nucleophilic substitution is difficult without appropriate activating features.
- Plan conversions of ethyne, ethene, and n-hexane into benzene. Specify reagents and the change in carbon framework at each stage.
- Draw and name all acyclic monoalkenes that yield 2-methylbutane on complete hydrogenation. Explain how removing a hydrogen from each end of a possible single bond helps generate the candidates.
- Arrange each set in decreasing reactivity towards an electrophile: (a) chlorobenzene, para-nitrochlorobenzene, 2,4-dinitrochlorobenzene; (b) toluene, para-nitrotoluene, para-dinitrobenzene. Explain both comparisons.
- Compare ease of nitration of benzene, meta-dinitrobenzene, and toluene, separating activation from positional direction.
- Suggest a suitable Lewis acid other than aluminium chloride for benzene ethylation and explain its activating role.
- Explain why a mixed Wurtz reaction is inconvenient for preparing a pure odd-carbon alkane. Illustrate all three carbon-coupling possibilities for methyl and ethyl halides.
Key Takeaways
• Use skeleton, bond type, and geometry together to classify and name a hydrocarbon. • Connectivity, conformation, and configuration describe different structural information. • Intermediate stability and conditions explain product patterns more reliably than isolated reaction memorisation. • Fragment analysis and atom balance provide independent checks on a proposed structure. • Aromaticity, substituent effects, applications, and toxicity connect molecular electron distribution to observable behaviour.
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Predicting Substitution on Aromatic Rings
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