ChemistryISC Class 11 · Hydrocarbons

Alkynes and Benzene: Acidity, Aromaticity and Electrophilic Substitution

Updated 17 October 2026 · Fact-checked

Terminal alkynes are weakly acidic (sp carbon, 50% s-character) and add reagents twice; benzene is aromatic (planar, cyclic, 6 delocalised π electrons, 4n + 2), stabilised by resonance, and undergoes electrophilic substitution rather than addition. Substituents direct the incoming group: –CH₃, –Cl, –OH to ortho/para; –NO₂, –CHO, –COOH to meta.

Benzene keeps its aromatic ring and substitutes; –CH₃, –Cl, –OH send the new group ortho/para, while –NO₂, –CHO, –COOH send it meta.

Illustration drawn for QuestionBuddy.

Understand Alkynes and Benzene: Aromaticity and Substitution

Alkynes. Ethyne is prepared by CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂ and by dehydrohalogenation of vicinal dihalides with alcoholic KOH then NaNH₂. The carbon atoms are sp hybridised (linear, 50% s-character), so the C–H bond is more acidic than in alkenes (33%) and alkanes (25%): terminal alkynes react with sodium or sodamide to give sodium acetylide (HC≡CH + NaNH₂ → HC≡CNa + NH₃), and with ammoniacal silver nitrate (white precipitate) or ammoniacal cuprous chloride (red precipitate). This distinguishes ethyne from ethene and ethane.

Reactions of alkynes. Addition of H₂ over Ni gives the alkane; with Lindlar's catalyst it gives a cis-alkene; with Na/liquid NH₃ it gives a trans-alkene. Addition of two moles of X₂ gives the tetrahaloalkane (e.g. CHBr₂CHBr₂ from ethyne with Br₂), and addition of two moles of HX follows Markovnikov's rule. Hydration with dilute H₂SO₄/HgSO₄ gives ethanal from ethyne (enol → aldehyde) and propanone from propyne. Trimerisation of ethyne through a red-hot iron tube (873 K) gives benzene.

Structure of benzene. C₆H₆ is a planar hexagon of sp² carbons, with all C–C bonds equal (139 pm, between typical single 154 pm and double 134 pm) and a delocalised cloud of six π electrons above and below the ring; it is represented as a resonance hybrid of two Kekulé structures. The resonance energy is about 150 kJ mol⁻¹. Aromaticity requires a planar, cyclic, fully conjugated ring with (4n + 2) π electrons (Hückel's rule).

Reactions of benzene. Electrophilic substitution (retains the aromatic ring): nitration (conc. HNO₃ + conc. H₂SO₄, 323 to 333 K, electrophile NO₂⁺) gives nitrobenzene; halogenation (Cl₂ or Br₂ + FeCl₃ or FeBr₃) gives chlorobenzene; sulphonation (fuming H₂SO₄) gives benzenesulphonic acid; Friedel-Crafts alkylation (CH₃Cl, anhydrous AlCl₃) gives toluene; Friedel-Crafts acylation (CH₃COCl, AlCl₃) gives acetophenone. Addition reactions need drastic conditions: H₂/Ni gives cyclohexane; Cl₂ in sunlight gives C₆H₆Cl₆. Benzene burns with a sooty flame. Preparation: decarboxylation of sodium benzoate with soda lime.

Directing effect. Activating, ortho/para-directing groups: –OH, –NH₂, –CH₃, –OR (halogens are o/p-directing but deactivating). Deactivating, meta-directing groups: –NO₂, –CN, –CHO, –COOH, –SO₃H.

Alkynes and Benzene: Aromaticity and Substitution: formulas and key facts

Hückel's rule: $π e l ec t r o n s = 4 n + 2 (n = 0, 1, 2...)$; Planar, cyclic, conjugated: benzene has 6.
Alkyne hydration: $H C \equiv C H + H_{2} O \to (H g S O_{4}, H_{2} S O_{4}) C H_{3} C H O$; Propyne gives CH₃COCH₃.
Nitrating mixture: $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$; NO₂⁺ is the electrophile.
Hybridisation and acidity: $s p (50$; More s-character, more acidic C–H.

Alkynes and Benzene: Aromaticity and Substitution in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Telling ethyne from ethene

Suppose a technician has two unlabelled gas cylinders, one of ethene and one of ethyne.

Describe a test that distinguishes them and explain.

Try it first, then show the solutionSolution

Both decolourise bromine water, so that test is not enough.
Pass each gas through ammoniacal silver nitrate: ethyne gives a white precipitate of silver acetylide, ethene gives none.
The terminal sp C–H of ethyne is acidic enough to be replaced by Ag⁺.

Answer: Ethyne gives a white precipitate with ammoniacal AgNO₃; ethene does not.

Memory hook: Only terminal alkynes (RC≡CH) give this test.

Tricks to remember Alkynes and Benzene: Aromaticity and Substitution

Short memory aids. Say each one aloud once, then use it on a question the same day.

sp is acidic
The more s-character, the more electronegative the carbon and the more acidic its C–H bond.
Use it when: Comparing acidity of ethane, ethene and ethyne.
Nitrate first for meta
Put the meta-director on first if you want a meta product; the o/p-director first for para.
Use it when: Multi-step synthesis of disubstituted benzenes.

Alkynes and Benzene: Aromaticity and Substitution: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Hybridisation in ethyne · 1 mark · MCQBoard level

The hybridisation of each carbon in ethyne is:
(a) sp
(b) sp²
(c) sp³
(d) sp³d

Try it first, then show the solutionSolution

Each carbon forms two σ bonds (one C–H, one C–C) and two π bonds: sp.

Answer: (a) sp

Example 2: Distinguishing hydrocarbons · 3 marks · Short answerBoard level

Distinguish between ethane, ethene and ethyne by chemical tests.

Try it first, then show the solutionSolution

Bromine water (or Baeyer's reagent) is decolourised by ethene and ethyne but not by ethane. (1 mark)
Ammoniacal silver nitrate gives a white precipitate with ethyne only. (1 mark)
The remaining gas, which gives no reaction with either, is ethane. (1 mark)

Answer: Ethane: no reaction; ethene: decolourises bromine water only; ethyne: decolourises bromine water and gives a white precipitate with ammoniacal AgNO₃.

Example 3: Reactions of ethyne · 4 marks · Short answerHarder board level

Write the equations and conditions for (i) hydration of ethyne, (ii) hydration of propyne, (iii) conversion of ethyne to benzene, and (iv) the formation of sodium acetylide.

Try it first, then show the solutionSolution

(i) HC≡CH + H₂O →(HgSO₄, dil. H₂SO₄, 333 K) CH₃CHO. (1 mark)
(ii) CH₃C≡CH + H₂O → CH₃COCH₃ (Markovnikov addition, then enol-keto change). (1 mark)
(iii) 3HC≡CH →(red-hot iron tube, 873 K) C₆H₆. (1 mark)
(iv) HC≡CH + NaNH₂ → HC≡C⁻Na⁺ + NH₃. (1 mark)

Answer: (i) ethanal; (ii) propanone; (iii) benzene; (iv) sodium acetylide.

Example 4: Aromaticity · 4 marks · Short answerHarder board level

Using Hückel's rule, decide whether benzene, cyclobutadiene (C₄H₄, planar), the cyclopentadienyl anion (C₅H₅⁻) and cyclohexane are aromatic.

Try it first, then show the solutionSolution

Benzene: 6 π electrons = 4(1) + 2, planar and cyclic: aromatic. (1 mark)
Cyclobutadiene: 4 π electrons, which is not 4n + 2: not aromatic. (1 mark)
C₅H₅⁻: 6 π electrons, planar and fully conjugated: aromatic. (1 mark)
Cyclohexane: no π electrons and not planar: not aromatic. (1 mark)

Answer: Benzene and C₅H₅⁻ are aromatic; cyclobutadiene and cyclohexane are not.

Example 5: Synthesis with directing effects · 5 marks · Long answerHOTS / competency

Starting from benzene, outline the preparation of (i) m-chloronitrobenzene and (ii) p-chloronitrobenzene (the main product), giving reagents and reasons for the order of steps.

Try it first, then show the solutionSolution

Nitro group is a meta-director; chlorine is an ortho/para-director. (1 mark)
(i) Benzene →(conc. HNO₃/conc. H₂SO₄) nitrobenzene →(Cl₂, FeCl₃) m-chloronitrobenzene (Cl enters meta to NO₂). (2 marks)
(ii) Benzene →(Cl₂, FeCl₃) chlorobenzene →(conc. HNO₃/conc. H₂SO₄) a mixture of o- and p-chloronitrobenzene, of which the para isomer is the major product. (1.5 marks)
The order matters: the group put on first controls where the second group enters. (0.5 mark)

Answer: (i) nitrate first, then chlorinate; (ii) chlorinate first, then nitrate.

Where students lose marks in Alkynes and Benzene: Aromaticity and Substitution

The slips examiners see most often, and what to write instead.

Mistake: Wrong order of steps in synthesis
Chlorinating nitrobenzene to get the para isomer.
Do this: Choose the order so that the first group directs the second to the required position.
Usually costs: 2 marks.
Mistake: Writing Kekulé structures as separate molecules
Saying benzene has alternate single and double bonds.
Do this: Benzene is a resonance hybrid; all bonds are equal.
Usually costs: 1 to 2 marks.
Mistake: Missing catalyst
Writing Friedel-Crafts without AlCl₃ or halogenation without FeCl₃.
Do this: Write the Lewis acid catalyst above the arrow.
Usually costs: 1 mark.

Exam tips

State Hückel's rule when asked about aromaticity and count π electrons.
Give the electrophile for each substitution.
Describe the tests in terms of observations: colour, precipitate.
Write the reagent and conditions above each arrow.

Practice questions on Alkynes and Benzene: Aromaticity and Substitution

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The reaction of benzene with CH₃Cl in the presence of anhydrous AlCl₃ is called:
(a) nitration
(b) Friedel–Crafts alkylation
(c) sulphonation
(d) Wurtz reaction
Hint
The product is toluene.
Answer
(b) Friedel–Crafts alkylation
Q2 · 2 marks · Short answerBoard level
Write the equation for the nitration of benzene and name the electrophile.
Hint
Use the nitrating mixture.
Answer
C₆H₆ + HNO₃ →(conc. H₂SO₄, 323–333 K) C₆H₅NO₂ + H₂O. The electrophile is the nitronium ion, NO₂⁺.
Q3 · 3 marks · Short answerHarder board level
Why is ethyne more acidic than ethene and ethane? Give the s-character of the carbon in each.
Hint
Higher s-character means higher electronegativity.
Answer
The carbon atoms are sp (50% s), sp² (33% s) and sp³ (25% s) respectively. More s-character holds the C–H electrons closer to carbon, so the proton is released more easily: acidity ethyne > ethene > ethane.
Q4 · 3 marks · Short answerHarder board level
Complete the sequence: HC≡CH →(NaNH₂) A →(CH₃Br) B →(H₂, Lindlar's catalyst) C. Identify A, B and C.
Hint
The acetylide ion is a nucleophile.
Answer
A = HC≡C⁻Na⁺ (sodium acetylide); B = HC≡CCH₃ (propyne); C = CH₃CH=CH₂ (propene).
Q5 · 3 marks · Short answerHarder board level
Write the equations to show how the electrophile NO₂⁺ is formed in the nitration of benzene, and give the product of the nitration of nitrobenzene with a reason.
Hint
The nitro group deactivates the ring.
Answer
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Nitration of nitrobenzene gives m-dinitrobenzene, because –NO₂ is a deactivating meta-director.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the heats of hydrogenation are cyclohexene −120 kJ mol⁻¹ and benzene −208 kJ mol⁻¹ (to cyclohexane).
(i) Predict the heat of hydrogenation of a hypothetical 1,3,5-cyclohexatriene with three isolated double bonds. (1 mark)
(ii) Find the resonance energy of benzene. (1 mark)
(iii) Explain why benzene prefers substitution to addition. (1 mark)
(iv) Compare the C–C bond length in benzene with typical single and double bonds (154 pm and 134 pm). (1 mark)
Hint
Resonance energy = predicted − observed.
Answer
(i) 3 × (−120) = −360 kJ mol⁻¹. (ii) 360 − 208 = 152 kJ mol⁻¹. (iii) Substitution keeps the stable delocalised ring; addition would destroy the aromatic stabilisation. (iv) All C–C bonds are 139 pm, between 154 pm and 134 pm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Hybridisation in benzene?Show answer
sp².
2.Hückel's rule?Show answer
4n + 2 π electrons.
3.Test for terminal alkynes?Show answer
White precipitate with ammoniacal AgNO₃.
4.Electrophile in nitration?Show answer
NO₂⁺.

Alkynes and Benzene: Aromaticity and Substitution: frequently asked questions

Why does benzene not decolourise bromine water?

The delocalised π cloud makes benzene far less reactive to addition than an ordinary alkene; it reacts by substitution with Br₂ only with a catalyst such as FeBr₃.

What is resonance energy?

The extra stability of the real molecule over the most stable contributing structure; for benzene it is about 150 kJ mol⁻¹.

Why does AlCl₃ work in the Friedel-Crafts reaction?

As a Lewis acid it removes Cl⁻ from CH₃Cl, generating the electrophile CH₃⁺ (as a complex) that attacks the benzene ring.