Hydrocarbon classification, IUPAC naming, and structural isomerism
Question 1
Which naming rule is correct for an enyne?
(a)Number from the nearest substituent even if the multiple bond gets a larger locant
(b)Choose the longest chain even when it omits the multiple bond
(c)Choose a parent chain containing both multiple bonds and give them the lowest possible locants.
(d)Turning a drawing of one structure around makes it a different structural isomer.
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(c) Choose a parent chain containing both multiple bonds and give them the lowest possible locants.
Hydrocarbons are classified as aliphatic or aromatic and as saturated or unsaturated. A defensible IUPAC name follows the longest chain containing the maximum number of multiple bonds, lowest locants, and alphabetised substituents; constitutional isomers differ in connectivity, not merely drawing orientation.
Question 2
What distinguishes chain isomers?
(a)Turning a drawing of one structure around makes it a different structural isomer.
(b)They have the same molecular formula, but the carbon atoms are joined together in a different way.
(c)Number from the nearest substituent even if the multiple bond gets a larger locant
(d)Choose the longest chain even when it omits the multiple bond
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(b) They have the same molecular formula, but the carbon atoms are joined together in a different way.
Hydrocarbons are classified as aliphatic or aromatic and as saturated or unsaturated. A defensible IUPAC name follows the longest chain containing the maximum number of multiple bonds, lowest locants, and alphabetised substituents; constitutional isomers differ in connectivity, not merely drawing orientation.
Substituent effects, orientation, reactivity, and hydrocarbon toxicity
Question 3
Where does methyl direct nitration of toluene?
(a)Mainly to the ortho and para positions.
(b)Every deactivating substituent is meta directing
(c)A methyl group directs incoming electrophiles meta
(d)A nitro group activates benzene toward electrophiles
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(a) Mainly to the ortho and para positions.
A substituent changes ring electron density and directs further electrophilic substitution. Electron-releasing groups such as alkyl activate and direct ortho/para; strongly electron-withdrawing groups such as nitro deactivate and usually direct meta. Halogens are deactivating yet ortho/para directing because their inductive withdrawal and resonance donation affect rate and orientation differently. Many fused aromatic hydrocarbons are toxic or carcinogenic.
Question 4
Where does nitro direct further substitution?
(a)Every deactivating substituent is meta directing
(b)To the meta position.
(c)A methyl group directs incoming electrophiles meta
(d)A nitro group activates benzene toward electrophiles
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(b) To the meta position.
A substituent changes ring electron density and directs further electrophilic substitution. Electron-releasing groups such as alkyl activate and direct ortho/para; strongly electron-withdrawing groups such as nitro deactivate and usually direct meta. Halogens are deactivating yet ortho/para directing because their inductive withdrawal and resonance donation affect rate and orientation differently. Many fused aromatic hydrocarbons are toxic or carcinogenic.
Alkyne structure, preparation, acidity, and additions
Question 5
What bonds constitute C≡C?
(a)A carbon-carbon triple bond contains three sigma bonds
(b)Every alkyne has an acidic hydrogen on its triple bond
(c)Alkyne carbons are sp2 hybridised and bent
(d)One sigma and two pi bonds.
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(d) One sigma and two pi bonds.
Alkynes contain a linear sp–sp C≡C bond with one sigma and two pi bonds. They are prepared by double dehydrohalogenation of vicinal or geminal dihalides. Terminal alkynes are weakly acidic because the conjugate-base charge resides in an orbital with 50% s character; additions occur stepwise and complete hydrogenation consumes two equivalents of H2.
Question 6
What is the geometry at an alkyne carbon?
(a)Alkyne carbons are sp2 hybridised and bent
(b)A carbon-carbon triple bond contains three sigma bonds
(c)Linear, with sp hybridisation.
(d)Every alkyne has an acidic hydrogen on its triple bond
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(c) Linear, with sp hybridisation.
Alkynes contain a linear sp–sp C≡C bond with one sigma and two pi bonds. They are prepared by double dehydrohalogenation of vicinal or geminal dihalides. Terminal alkynes are weakly acidic because the conjugate-base charge resides in an orbital with 50% s character; additions occur stepwise and complete hydrogenation consumes two equivalents of H2.
Preparation of alkanes and limits of coupling methods
Question 7
Why does mixed Wurtz coupling fail as a selective synthesis?
(a)Two different alkyl halides in Wurtz reaction give only the cross-coupled alkane
(b)It forms two symmetrical alkanes as well as the cross-coupled product.
(c)Wurtz reaction is preferred for pure odd-carbon alkanes
(d)Soda-lime decarboxylation retains every carbon atom
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(b) It forms two symmetrical alkanes as well as the cross-coupled product.
Alkanes are prepared by hydrogenation of unsaturated hydrocarbons, reduction of alkyl halides, Wurtz coupling, decarboxylation, and Kolbe electrolysis. Each method has a characteristic carbon-count consequence and substrate limitation; mixed Wurtz coupling gives mixtures and is unsuitable for a pure odd-carbon alkane.
Question 8
What carbon-count change occurs in soda-lime decarboxylation?
(a)The alkane contains one carbon fewer than the carboxylate.
(b)Soda-lime decarboxylation retains every carbon atom
(c)Two different alkyl halides in Wurtz reaction give only the cross-coupled alkane
(d)Wurtz reaction is preferred for pure odd-carbon alkanes
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(a) The alkane contains one carbon fewer than the carboxylate.
Alkanes are prepared by hydrogenation of unsaturated hydrocarbons, reduction of alkyl halides, Wurtz coupling, decarboxylation, and Kolbe electrolysis. Each method has a characteristic carbon-count consequence and substrate limitation; mixed Wurtz coupling gives mixtures and is unsuitable for a pure odd-carbon alkane.
Alkane properties, combustion, pyrolysis, and radical halogenation
Question 9
Why does branching lower boiling point?
(a)Greater branching raises an alkane's boiling point
(b)Ethane in methane chlorination is the main propagation product
(c)A compact branched molecule has less surface contact and weaker dispersion forces.
(d)UV light is required in every propagation step
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(c) A compact branched molecule has less surface contact and weaker dispersion forces.
Alkanes are non-polar, water-insoluble fuels whose boiling points rise with molar mass and fall with branching at fixed formula. Their reactions include combustion, controlled oxidation, cracking and free-radical halogenation. Chlorination proceeds by initiation, propagation, and termination; radical recombination explains ethane during methane chlorination.
Question 10
How is ethane formed during methane chlorination?
(a)UV light is required in every propagation step
(b)Greater branching raises an alkane's boiling point
(c)Ethane in methane chlorination is the main propagation product
(d)Two methyl radicals combine in a termination step.
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(d) Two methyl radicals combine in a termination step.
Alkanes are non-polar, water-insoluble fuels whose boiling points rise with molar mass and fall with branching at fixed formula. Their reactions include combustion, controlled oxidation, cracking and free-radical halogenation. Chlorination proceeds by initiation, propagation, and termination; radical recombination explains ethane during methane chlorination.
Alkene bonding and geometrical isomerism
Question 11
What bonds constitute C=C?
(a)A C=C double bond contains two sigma bonds
(b)Every alkene shows cis-trans isomerism
(c)One sigma bond and one pi bond.
(d)Free rotation occurs about C=C without breaking the bond
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(c) One sigma bond and one pi bond.
An alkene double bond contains one sigma and one pi bond between trigonal-planar sp2 carbons. Restricted rotation permits cis/trans or E/Z isomerism only when each double-bond carbon bears two different substituents. Cis isomers can be more polar and may boil higher, while trans isomers are often more symmetrical and may pack better.
Conformations of ethane and torsional strain
Question 12
Which ethane conformation is most stable?
(a)Eclipsed ethane is the most stable conformation
(b)A Newman projection views along a C–H bond
(c)The staggered conformation.
(d)Interconversion of ethane conformers requires breaking a C–C bond
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(c) The staggered conformation.
Rotation about the ethane C–C sigma bond produces conformations represented by sawhorse or Newman projections. The staggered form is lowest in energy and the eclipsed form highest because eclipsing C–H bonds create torsional strain. Conformers interconvert without bond breaking and are not constitutional isomers.
Alkene preparation, electrophilic addition, peroxide effect, and ozonolysis
Question 13
What is the major product of propene plus HBr without peroxide?
(a)Ozonolysis keeps the C=C carbon skeleton intact
(b)Markovnikov addition places halide on the less substituted carbon
(c)The peroxide effect applies equally to HCl, HBr, and HI
(d)2-bromopropane by Markovnikov addition.
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(d) 2-bromopropane by Markovnikov addition.
Alkenes arise by dehydration of alcohols and dehydrohalogenation of haloalkanes. They undergo electrophilic additions, oxidation and ozonolysis. Markovnikov addition places H on the carbon already bearing more H; HBr with peroxide follows a radical anti-Markovnikov path. Ozonolysis cleaves C=C to carbonyls and can be reversed to reconstruct the alkene.
Benzene structure, resonance, and aromaticity
Question 14
Why are all benzene C–C bonds equal?
(a)A (4n+2) electron count alone guarantees aromaticity
(b)Kekule structures rapidly interconvert in equilibrium
(c)Benzene has three short double bonds and three long single bonds
(d)The six pi electrons are delocalised over the ring.
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(d) The six pi electrons are delocalised over the ring.
Benzene is a planar cyclic conjugated system with six delocalised pi electrons and equal C–C bonds. Its resonance hybrid is more stable than any localised Kekule form. Huckel aromaticity requires a planar, cyclic, continuously conjugated system with (4n+2) pi electrons; electron count alone is insufficient if planarity or conjugation is absent.
Preparation and electrophilic substitution of benzene
Question 15
What electrophile attacks benzene during nitration?
(a)Friedel–Crafts alkylation needs no Lewis acid
(b)Nitric acid alone generates the nitrating electrophile efficiently
(c)The nitronium ion, NO2+.
(d)Benzene undergoes nucleophilic substitution more readily than electrophilic substitution
Show answer
(c) The nitronium ion, NO2+.
Benzene is prepared by cyclic polymerisation of ethyne, decarboxylation of sodium benzoate, and aromatisation of alkanes. Its electron-rich pi cloud reacts with electrophiles in nitration, sulfonation, halogenation and Friedel–Crafts reactions. Electrophile generation requires the stated acid or Lewis-acid catalyst; substitution restores aromaticity after sigma-complex formation.
Does a benzene ring react like three independent double bonds?
Wrong
Benzene should readily add bromine like an alkene.
Why it feels right. Its alternating-bond drawing resembles three alkene bonds.
Right
Benzene has a delocalised aromatic system and commonly undergoes substitution under suitable conditions.
Worked example
Alkene
Br₂ adds across C=C→Two new C–Br bonds
Benzene
Br₂ with catalyst→H replaced by Br→Aromatic ring retained
Bromine adds across an alkene double bond. Under suitable catalytic conditions benzene instead exchanges H for Br.
An alkene can add bromine across C=C.
Benzene bromination with a suitable catalyst replaces one hydrogen and preserves aromaticity in the product.
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