Classification, nomenclature and isomerism of hydrocarbons
Link to Std 9–10 : You learnt that compounds of carbon and hydrogen are called hydrocarbons, with names like methane and ethene; now we classify them into families and give them proper names.
LPG in a kitchen cylinder is mostly propane and butane, petrol is a mixture of liquid hydrocarbons, and candle wax is a solid hydrocarbon. All are made of only carbon and hydrogen.
Hydrocarbons contain only carbon and hydrogen. Saturated hydrocarbons (alkanes) have only single bonds. Unsaturated ones have a double bond (alkenes) or a triple bond (alkynes). Aromatic hydrocarbons (arenes) contain a benzene ring.
In the figure : Hydrocarbons divide into saturated alkanes and unsaturated alkenes, alkynes and arenes, with their general formulae.
Rule 1: General formulae — Alkanes CₙH₂ₙ₊₂, alkenes CₙH₂ₙ (and cycloalkanes), alkynes CₙH₂ₙ₋₂.
- Easy : An alkane with 6 carbons is C₆H₁₄.
- Medium : An alkene with 5 carbons is C₅H₁₀ and an alkyne with 5 carbons is C₅H₈.
- Exam-type : A hydrocarbon C₇H₁₂ fits the alkyne formula CₙH₂ₙ₋₂ (n = 7), so it can be an alkyne.
Rule 2: Naming — Longest chain with the multiple bond, numbered to give it the lowest locant: -ane, -ene, -yne.
- Easy : CH₃CH₂CH₃ is propane; CH₂=CHCH₃ is propene; CH₃C≡CH is propyne.
- Medium : CH₃CH₂CH=CH₂ is but-1-ene; CH₃CH=CHCH₃ is but-2-ene.
- Exam-type : (CH₃)₂C=CHCH₃ is 2-methylbut-2-ene (chain of 4 carbons, double bond at 2, methyl at 2).
Rule 3: Types of carbon — A carbon bonded to one other carbon is primary (1°), two is secondary (2°), three is tertiary (3°), four is quaternary (4°).
- Easy : In 2-methylbutane, CH₃CH(CH₃)CH₂CH₃, the CH carbon is tertiary.
- Medium : In the same molecule, the CH₂ carbon is secondary and the three CH₃ carbons are primary.
- Exam-type : In 2,2-dimethylpropane the central carbon is quaternary.
Rule 4: Isomers — Chain isomers exist from C₄H₁₀ for alkanes; alkenes also show position and geometrical isomers.
- Easy : C₄H₁₀ has two isomers; C₅H₁₂ has three.
- Medium : C₄H₈ has the position isomers but-1-ene and but-2-ene, and also 2-methylpropene.
- Exam-type : But-2-ene shows cis–trans isomerism while but-1-ene does not.
Each extra double bond takes away 2 hydrogens; each ring does the same.
Numbering from the wrong end. The multiple bond must get the lowest number.
Memory clue: "ane, ene, yne = single, double, triple".
Spot the difference
| Point | Alkane | Alkene | Alkyne |
|---|---|---|---|
| Bond | Single | Double | Triple |
| Formula | CₙH₂ₙ₊₂ | CₙH₂ₙ | CₙH₂ₙ₋₂ |
| Example | Ethane | Ethene | Ethyne |
Practice questions
- 1Define a hydrocarbon.
- 2Write the molecular formula of the alkane with 6 carbon atoms.
- 3What is the suffix for a triple bond?
- 4Name CH₃CH₂CH=CH₂.
- 5Write the formula of the alkene with 5 carbon atoms.
- 6Name the type of the central carbon in 2,2-dimethylpropane.
- 7Name (CH₃)₂C=CHCH₃.
- 8Name the three isomers of C₄H₈ that are alkenes.
Show answers (try them yourself first)
- 1A compound containing only carbon and hydrogen.
- 2C₆H₁₄.
- 3-yne.
- 4But-1-ene.
- 5C₅H₁₀.
- 6Quaternary.
- 72-Methylbut-2-ene.
- 8But-1-ene, but-2-ene (cis and trans forms) and 2-methylpropene.
Alkanes: preparation, halogenation and combustion
Link to Std 9–10 : You know that methane burns and that cooking gas is an alkane; now we see how alkanes are made and why they react only under special conditions.
A gas stove burns LPG with a blue flame and releases heat. Alkanes are good fuels but do not react easily with acids, bases or most reagents, which is why LPG is safe to store in a steel cylinder.
Alkanes are saturated hydrocarbons with only C–C and C–H sigma bonds. They are fairly unreactive; their main reactions are combustion and free-radical halogenation, in light or heat.
In the figure : The chain reaction of methane with chlorine: initiation, two propagation steps and termination.
Rule 1: Preparation — Wurtz reaction: 2R–X + 2Na → R–R + 2NaX in dry ether. Hydrogenation of alkenes or alkynes with Ni, Pt or Pd also gives alkanes.
- Easy : 2CH₃Br + 2Na → C₂H₆ + 2NaBr (ethane).
- Medium : CH₂=CH₂ + H₂ → CH₃–CH₃ with Ni at 473–573 K.
- Exam-type : Wurtz reaction with two different halides gives a mixture of three alkanes, so it is useful mainly with a single halide.
Rule 2: Physical properties — Boiling point rises with the number of carbons. For the same carbon count, branching lowers the boiling point because the molecule becomes more compact.
- Easy : The first four alkanes (C₁ to C₄) are gases at room temperature.
- Medium : For C₅H₁₂ the boiling points are pentane 36 °C, 2-methylbutane 28 °C and 2,2-dimethylpropane 9.5 °C.
- Exam-type : Alkanes are non-polar, so they dissolve in non-polar solvents and not in water.
Rule 3: Halogenation by free radicals — Light or heat splits Cl₂ into Cl• radicals, which start a chain. Reactivity order of halogens: F₂ > Cl₂ > Br₂ > I₂.
- Easy : CH₄ + Cl₂ → CH₃Cl + HCl in sunlight.
- Medium : Ethane forms as a small by-product, when two CH₃• radicals combine at the termination step.
- Exam-type : With excess chlorine the reaction continues to CH₂Cl₂, CHCl₃ and CCl₄.
Rule 4: Combustion and conformations — Complete combustion gives CO₂ and H₂O with heat: CₙH₂ₙ₊₂ + (3n + 1)/2 O₂ → n CO₂ + (n + 1) H₂O. Ethane has staggered (more stable) and eclipsed conformations, differing by about 12.5 kJ mol⁻¹.
- Easy : CH₄ + 2O₂ → CO₂ + 2H₂O; 16 g of methane gives 44 g of CO₂.
- Medium : C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; burning 22 g (0.5 mol) of propane needs 2.5 mol = 80 g of O₂ and gives 1.5 mol = 66 g of CO₂.
- Exam-type : The staggered form of ethane is more stable than the eclipsed form by about 12.5 kJ mol⁻¹, because of less repulsion between C–H bonds.
Alkanes react by substitution, not by addition.
Saying that the halogenation of alkanes needs no light. It needs light or heat to make radicals.
Memory clue: "Light starts, radicals carry, radicals meet to stop" — initiation, propagation, termination.
Spot the difference
| Point | Staggered | Eclipsed |
|---|---|---|
| H atoms | As far apart as possible | Directly in line |
| Stability | More stable | Less stable |
| Energy | Lower | About 12.5 kJ mol⁻¹ higher |
Practice questions
- 1Write the Wurtz reaction for bromomethane.
- 2Why are alkanes insoluble in water?
- 3Name the products of complete combustion of an alkane.
- 4Write the balanced equation for the combustion of propane.
- 5What mass of CO₂ forms when 16 g of methane burns completely?
- 6Arrange pentane, 2-methylbutane and 2,2-dimethylpropane by increasing boiling point.
- 7How much O₂ is needed to burn 22 g of propane completely?
- 8Write the three steps of the chlorination of methane.
Show answers (try them yourself first)
- 12CH₃Br + 2Na → CH₃CH₃ + 2NaBr.
- 2Alkanes are non-polar and cannot form hydrogen bonds with water.
- 3Carbon dioxide and water, with heat.
- 4C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
- 51 mol of CH₄ (16 g) gives 1 mol of CO₂, which is 44 g.
- 62,2-dimethylpropane (9.5 °C) < 2-methylbutane (28 °C) < pentane (36 °C).
- 722 g = 0.5 mol; O₂ = 0.5 × 5 = 2.5 mol = 80 g.
- 8Initiation: Cl₂ → 2Cl•. Propagation: CH₄ + Cl• → CH₃• + HCl, then CH₃• + Cl₂ → CH₃Cl + Cl•. Termination: two radicals combine, for example CH₃• + CH₃• → C₂H₆.
Alkenes
Link to Std 9–10 : You know that ethene is used to ripen fruit and to make polythene; now we see why a double bond makes alkenes so reactive.
Plastic bags and pipes are made of polythene, which comes from ethene. The double bond lets many small molecules join into one long chain.
Alkenes contain a C=C double bond (one σ bond and one π bond). The weaker π bond is exposed, so alkenes mainly undergo addition reactions. Markovnikov's rule: in the addition of HX to an unsymmetrical alkene, hydrogen goes to the carbon that already has more hydrogens.
In the figure : Propene with HBr gives 2-bromopropane (Markovnikov) normally and 1-bromopropane in the presence of peroxide (anti-Markovnikov).
Rule 1: Preparation — Dehydration of alcohols (conc. H₂SO₄, 443 K) and dehydrohalogenation of alkyl halides (alcoholic KOH). When two products are possible, the more substituted alkene is the major product (Saytzeff rule).
- Easy : CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 443 K).
- Medium : CH₃CH₂CH₂Br + alcoholic KOH → CH₃CH=CH₂ + KBr + H₂O.
- Exam-type : 2-Bromobutane with alcoholic KOH gives mainly but-2-ene (more substituted) and a little but-1-ene.
Rule 2: Addition reactions — H₂ (Ni, Pt) gives alkanes; X₂ gives dihalides; HX gives alkyl halides; water (with acid) gives alcohols.
- Easy : CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br; the reddish-brown colour of bromine fades (test for unsaturation).
- Medium : CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (Markovnikov).
- Exam-type : With peroxide, CH₃CH=CH₂ + HBr gives CH₃CH₂CH₂Br (anti-Markovnikov, a free-radical path, only for HBr).
Rule 3: Oxidation and ozonolysis — Cold dilute KMnO₄ (Baeyer's test) turns pink-purple to colourless and gives a diol. Ozonolysis cuts the C=C bond to give carbonyl compounds.
- Easy : Ethene with cold dilute KMnO₄ gives ethane-1,2-diol and the purple colour disappears.
- Medium : But-2-ene on ozonolysis gives 2 molecules of ethanal.
- Exam-type : 2-Methylbut-2-ene on ozonolysis gives propanone and ethanal.
Rule 4: Geometrical isomers and polymerisation — cis-Alkenes have the same groups on the same side of the double bond; trans have them on opposite sides. Many alkene molecules join to give a polymer.
- Easy : But-2-ene exists as cis and trans.
- Medium : The trans isomer often has a higher melting point than the cis isomer because it packs better.
- Exam-type : n CH₂=CH₂ → (–CH₂–CH₂–)ₙ is polythene; the product is an addition polymer.
Alkenes give addition reactions; alkanes give substitution reactions.
Applying anti-Markovnikov addition to HCl or HI. It works only for HBr with peroxide.
Memory clue: "The rich get richer" — hydrogen adds to the carbon with more hydrogens already.
Spot the difference
| Point | Markovnikov | Anti-Markovnikov |
|---|---|---|
| Condition | No peroxide | Peroxide |
| Product with propene + HBr | 2-Bromopropane | 1-Bromopropane |
| Mechanism | Ionic (carbocation) | Free radical |
Practice questions
- 1How do you test for a double bond with bromine water?
- 2Write the product of CH₂=CH₂ + H₂ (Ni).
- 3What does ethene form on polymerisation?
- 4State Markovnikov's rule.
- 5Give the product of propene + HBr in the absence of peroxide.
- 6What does but-2-ene give on ozonolysis?
- 7Predict the products of ozonolysis of 2-methylbut-2-ene.
- 8Why does peroxide change the product of propene + HBr?
Show answers (try them yourself first)
- 1The reddish-brown colour of bromine disappears.
- 2Ethane, CH₃CH₃.
- 3Polythene, (–CH₂–CH₂–)ₙ.
- 4In HX addition to an unsymmetrical alkene, the negative part goes to the carbon with fewer hydrogens (hydrogen goes to the carbon with more hydrogens).
- 52-Bromopropane.
- 6Two molecules of ethanal.
- 7Propanone and ethanal.
- 8Peroxide starts a free-radical chain in which Br• adds first, so bromine ends up on the terminal carbon, giving 1-bromopropane.
Alkynes
Link to Std 9–10 : You know that welders use a very hot flame from acetylene gas; now we see its structure and chemistry.
A gas welder burns ethyne (acetylene) with oxygen to cut and join metals; the flame reaches over 3000 °C. Ethyne is made from calcium carbide and water.
Alkynes contain a C≡C triple bond (one σ and two π bonds). The carbon atoms are sp hybridised and the molecule is linear (180°). Alkynes with a hydrogen on the triple-bond carbon are weakly acidic.
In the figure : Ethyne is linear with 3σ and 2π bonds. Acidity of C–H falls in the order ethyne > ethene > ethane.
Rule 1: Preparation — CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂. Elimination from vicinal dihalides with alcoholic KOH also gives alkynes.
- Easy : Ethyne forms when calcium carbide reacts with water.
- Medium : 6.4 g of CaC₂ (M = 64 g mol⁻¹) is 0.1 mol and gives 0.1 mol of ethyne, which is 2.6 g.
- Exam-type : CH₂Br–CH₂Br with alcoholic KOH, then sodamide, gives ethyne in two eliminations.
Rule 2: Acidic character — An sp carbon (50% s-character) holds its bonding electrons closer, so the H is lost more easily as H⁺ than from sp² or sp³ carbon.
- Easy : Ethyne reacts with sodium: 2HC≡CH + 2Na → 2HC≡CNa + H₂.
- Medium : Order of acidity: HC≡CH > CH₂=CH₂ > CH₃–CH₃.
- Exam-type : But-2-yne has no hydrogen on a triple-bond carbon, so it does not react with sodium.
Rule 3: Addition reactions — Addition goes in two stages. H₂ gives alkene and then alkane; HX adds twice (Markovnikov); water (HgSO₄, H₂SO₄) gives a carbonyl compound.
- Easy : HC≡CH + H₂ (Pd, poisoned) → CH₂=CH₂, and with excess H₂ and Ni → CH₃CH₃.
- Medium : Propyne with water and HgSO₄ gives propanone, CH₃COCH₃.
- Exam-type : Ethyne + water (HgSO₄) gives ethanal, the only alkyne that gives an aldehyde.
Rule 4: Polymerisation and uses — Three ethyne molecules pass through a red-hot iron tube at 873 K to give benzene. Ethyne is used in welding and as a raw material.
- Easy : 3C₂H₂ → C₆H₆ (benzene) in a red-hot iron tube.
- Medium : The oxy-acetylene flame is much hotter than a flame from ordinary fuel gases.
- Exam-type : Ethyne burns with a sooty flame in air because of its high percentage of carbon.
Terminal alkynes (R–C≡C–H) are acidic; internal alkynes are not.
Saying that all alkynes react with sodium. Only those with ≡C–H do.
Memory clue: "sp is sour" — more s-character means more acidic.
Spot the difference
| Point | Ethyne | Ethene |
|---|---|---|
| Bond | Triple | Double |
| Hybridisation | sp | sp² |
| Acidity | Higher | Lower |
Practice questions
- 1How is ethyne prepared from calcium carbide?
- 2What is the shape of ethyne?
- 3How many σ and π bonds does ethyne have?
- 4Why is ethyne more acidic than ethene?
- 5What does propyne give with water and HgSO₄?
- 6Which will react with sodium: but-1-yne or but-2-yne?
- 7How much ethyne forms from 6.4 g of CaC₂?
- 8Write the reaction in which ethyne forms benzene.
Show answers (try them yourself first)
- 1CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂.
- 2Linear, with bond angle 180°.
- 33 σ bonds and 2 π bonds.
- 4Its carbon is sp hybridised (50% s), which holds the C–H pair more tightly, so the H leaves more easily.
- 5Propanone.
- 6But-1-yne, because it has a hydrogen on the triple-bond carbon.
- 76.4 ÷ 64 = 0.1 mol of CaC₂ gives 0.1 mol of C₂H₂, which is 0.1 × 26 = 2.6 g.
- 83C₂H₂ → C₆H₆ through a red-hot iron tube at 873 K.
Benzene and aromaticity
Link to Std 9–10 : You know that carbon compounds burn and that some have a pleasant smell; benzene and its relatives are called aromatic because many early examples were fragrant.
Mothballs, dyes and many medicines contain benzene rings. Benzene itself is a colourless liquid; because it is harmful, only closed systems use it in industry.
Benzene (C₆H₆) is a planar ring of six sp² carbons with six π electrons shared equally around the ring. A compound is aromatic if it is cyclic, planar, fully conjugated and has (4n + 2) π electrons (Hückel's rule).
In the figure : All C–C bonds in benzene are 139 pm, between single (154) and double (134). Three isolated double bonds would release 360 kJ on hydrogenation, but benzene releases 208, so it is about 150 kJ mol⁻¹ more stable.
Rule 1: Structure and resonance — Benzene is a resonance hybrid of two Kekulé structures. Each carbon is sp² with a p-orbital whose π cloud lies above and below the ring.
- Easy : Benzene has 6 carbons and 6 hydrogens; all six C–C bonds are equal.
- Medium : It has 12 σ bonds and 3 π bonds in the Kekulé picture, and the π electrons are spread over the whole ring.
- Exam-type : Because of resonance, benzene does not show the usual addition reactions expected for three separate double bonds.
Rule 2: Resonance energy — Compare the expected and the actual heat of hydrogenation.
- Easy : Cyclohexene releases about 120 kJ mol⁻¹ on hydrogenation; three such double bonds would release 3 × 120 = 360 kJ mol⁻¹.
- Medium : Benzene actually releases only 208 kJ mol⁻¹.
- Exam-type : Extra stability = 360 − 208 = 152 kJ mol⁻¹, the resonance (delocalisation) energy.
Rule 3: Hückel's rule — Count the π electrons; if 2, 6, 10 … (4n + 2) and the ring is planar and conjugated, it is aromatic.
- Easy : Benzene has 6 π electrons (n = 1): aromatic.
- Medium : Naphthalene has 10 π electrons (n = 2): aromatic.
- Exam-type : Cyclobutadiene (4 π electrons) is anti-aromatic and cyclooctatetraene (8, non-planar) is not aromatic; the cyclopentadienyl anion and the tropylium cation (6 each) are aromatic.
Rule 4: Properties — Benzene is a non-polar liquid that burns with a sooty flame (92.3% carbon) and prefers substitution to addition.
- Easy : Benzene is insoluble in water but dissolves in organic solvents.
- Medium : % carbon in C₆H₆ = 72 ÷ 78 × 100 = 92.3%, higher than ethane (80%), hence the sooty flame.
- Exam-type : Benzene does not decolourise bromine water or cold dilute KMnO₄, which shows that it has no ordinary C=C bonds.
Aromatic means (4n + 2) π electrons in a planar, cyclic, conjugated ring.
Treating benzene as three separate double bonds that react like an alkene.
Memory clue: "2, 6, 10 are aromatic; 4 and 8 are not".
Spot the difference
| Point | Benzene | Cyclohexene |
|---|---|---|
| C–C bond | All equal (139 pm) | Single and double |
| Reaction with Br₂ water | None | Decolourises |
| Typical reaction | Substitution | Addition |
Practice questions
- 1State Hückel's rule.
- 2How many π electrons does benzene have?
- 3What is the C–C bond length in benzene?
- 4Is naphthalene aromatic? Why?
- 5Calculate the resonance energy from the hydrogenation values 120 (cyclohexene), and 208 kJ mol⁻¹ (benzene).
- 6Why does benzene burn with a sooty flame?
- 7Why is cyclobutadiene not aromatic?
- 8Why does benzene not decolourise bromine water?
Show answers (try them yourself first)
- 1A planar, cyclic, conjugated system with (4n + 2) π electrons is aromatic.
- 2Six.
- 3139 pm.
- 4Yes; it is planar, cyclic and conjugated, with 10 π electrons (n = 2).
- 53 × 120 − 208 = 152 kJ mol⁻¹.
- 6Benzene has a high percentage of carbon (92.3%), so the carbon is not completely burnt.
- 7It has 4 π electrons, which does not fit 4n + 2, so it is anti-aromatic.
- 8Its π electrons are delocalised over the ring, which gives extra stability, so it does not behave like an ordinary alkene.
Electrophilic substitution and directing effects
Link to Std 9–10 : You know that some reagents attack where electrons are plentiful; benzene's electron-rich ring attracts electron-seeking reagents, so the ring reacts by substitution.
Nitrobenzene smells like almonds, but it is toxic. Aromatic nitro compounds are made in industries and are used for dyes and explosives.
In electrophilic substitution an electrophile (E⁺) replaces one hydrogen of the ring. A group already on the ring directs the next substituent to the ortho and para positions (activating groups) or to the meta position (deactivating groups).
In the figure : Four substitutions of benzene: nitration, halogenation, sulphonation and Friedel–Crafts alkylation.
Rule 1: Common reactions — Nitration (conc. HNO₃ + conc. H₂SO₄ gives NO₂⁺), halogenation (X₂ with FeX₃), sulphonation (fuming H₂SO₄) and Friedel–Crafts alkylation or acylation (R–Cl or RCOCl with AlCl₃).
- Easy : C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (nitrobenzene).
- Medium : C₆H₆ + Br₂ → C₆H₅Br + HBr with FeBr₃.
- Exam-type : C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl with AlCl₃ (acetophenone). 7.8 g (0.1 mol) of benzene would give 0.1 mol (12 g) of acetophenone.
Rule 2: Mechanism in three steps — Formation of the electrophile; attack of the ring on the electrophile to give a carbocation; loss of H⁺ to restore aromaticity.
- Easy : H₂SO₄ + HNO₃ → NO₂⁺ + HSO₄⁻ + H₂O makes the nitronium ion.
- Medium : The ring is attacked by NO₂⁺ to give a carbocation intermediate.
- Exam-type : The carbocation loses H⁺ to form nitrobenzene, since regaining aromaticity is strongly favourable.
Rule 3: Directing groups — Activating o/p-directors: –OH, –NH₂, –OCH₃, –CH₃. Deactivating m-directors: –NO₂, –CN, –CHO, –COOH. Halogens are deactivating but o/p-directing.
- Easy : Nitration of toluene (–CH₃) gives o-nitrotoluene and p-nitrotoluene.
- Medium : Nitration of nitrobenzene gives m-dinitrobenzene.
- Exam-type : Phenol reacts much faster than benzene because –OH releases electrons into the ring.
Rule 4: Toxicity — Some aromatic compounds with several fused rings are carcinogenic (cancer-causing); benzene itself is a toxic and carcinogenic liquid.
- Easy : Benzene vapour is toxic and must be avoided.
- Medium : Polynuclear hydrocarbons such as benzo[a]pyrene form when organic matter burns incompletely.
- Exam-type : Cigarette smoke and exhaust gases contain such polynuclear hydrocarbons.
Ortho, para-directors push electrons into the ring; meta-directors pull electrons out.
Thinking that halogens activate the ring because they are o/p-directing. They deactivate it.
Memory clue: "Electron givers send o/p; electron takers send m".
Spot the difference
| Point | Activating group | Deactivating group |
|---|---|---|
| Effect on ring | Electron-rich | Electron-poor |
| Position of next group | Ortho and para | Meta (except halogens) |
| Example | –OH, –CH₃ | –NO₂, –COOH |
Practice questions
- 1Name the electrophile in the nitration of benzene.
- 2What catalyst is used in the bromination of benzene?
- 3What does benzene give with CH₃Cl and AlCl₃?
- 4Name the products of nitration of toluene.
- 5What is the product of nitrating nitrobenzene?
- 6Why is the –OH group an activating group?
- 7How much acetophenone can be obtained from 7.8 g of benzene (100% yield)?
- 8Why do halogens deactivate the ring yet direct to o/p positions?
Show answers (try them yourself first)
- 1The nitronium ion, NO₂⁺.
- 2FeBr₃ (or iron).
- 3Toluene (methylbenzene).
- 4o-Nitrotoluene and p-nitrotoluene.
- 5m-Dinitrobenzene.
- 6It releases electrons into the ring by resonance (+R), making the ring more reactive.
- 77.8 g is 0.1 mol of benzene (78 g mol⁻¹); it gives 0.1 mol of acetophenone (120 g mol⁻¹), so 12 g.
- 8Halogens pull electrons by the inductive effect (deactivating) but their lone pairs release electrons by resonance at the o/p positions (directing).