This MCQ module is based on: Alkynes Aromatic
TOPIC 11 OF 13
Alkynes Aromatic
🎓 Class 11
Chemistry
CBSE
Theory
Ch 9 – Hydrocarbons
⏱ ~14 min
🌐 Language: [gtranslate]
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Alkynes and Aromatic Hydrocarbons
9.5 Alkynes
Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond (C≡C). General formula CₙH₂ₙ₋₂. Ethyne (HC≡CH), commonly called acetylene, is the first member.
9.5.1 Structure of the Triple Bond
Each carbon of C≡C is sp-hybridised: two sp orbitals point in opposite directions (180°) and form two σ bonds. Each carbon has two unhybridised p-orbitals (pₓ and p_y, perpendicular to the bond axis) that overlap laterally with the corresponding orbitals on the other carbon to form two π bonds. So C≡C = 1 σ + 2 π. The two π bond clouds together form a cylindrical electron sleeve around the C–C axis. Bond length C≡C = 120 pm (shorter than C=C, 134 pm, and C–C, 154 pm). C–H in ethyne is only 106 pm because the s-character of the bonding orbital is high (50% s in sp).
9.5.2 Nomenclature
Same as alkenes, but the suffix is '-yne'. Examples: HC≡CH = ethyne; CH₃C≡CH = propyne; CH≡C–CH₂–CH₃ = but-1-yne; CH₃C≡CCH₃ = but-2-yne.
9.5.3 Isomerism
Alkynes show chain isomerism (e.g., pent-1-yne and 3-methylbut-1-yne) and position isomerism (e.g., but-1-yne vs but-2-yne). They do NOT show geometrical isomerism because the sp carbons have only one substituent each.
9.5.4 Preparation of Alkynes
(a) From calcium carbide
Industrial method for ethyne. Calcium carbide reacts with water:
CaC₂ + 2 H₂O → HC≡CH + Ca(OH)₂CaC₂ is itself made by heating CaO with coke at 2275 K: CaO + 3 C → CaC₂ + CO.
(b) From vicinal dihalides
Vicinal dihalides on heating with alcoholic KOH (2 equivalents) lose two HX:
CH₂Br–CH₂Br →KOH(alc) CH₂=CHBr →NaNH₂ HC≡CH9.5.5 Physical Properties
Ethyne is a colourless gas with a faint sweet smell (the garlic smell of commercial acetylene is from impurities). C₁–C₃ are gases, C₄–C₈ liquids, and higher members solids. Insoluble in water, soluble in organic solvents.
9.5.6 Chemical Reactions
(i) Acidic Character of Terminal Alkyne
The H attached to the sp carbon of a terminal alkyne (≡C–H) is weakly acidic (pKa ≈ 25) because the sp orbital has 50% s character → the bonding electrons are pulled closer to C → C–H bond becomes more polar → H⁺ leaves more readily. Compare:
| Compound | Hybridisation of C | % s character | pKa |
|---|---|---|---|
| CH₃–CH₃ (alkane) | sp³ | 25% | ~50 (essentially non-acidic) |
| CH₂=CH₂ (alkene) | sp² | 33% | ~44 |
| HC≡CH (alkyne) | sp | 50% | ~25 (weakly acidic) |
| H₂O | — | — | 15.7 |
Reactions confirming acidity:
HC≡CH + Na →liq NH₃ HC≡C⁻Na⁺ + ½ H₂ (sodium acetylide)HC≡CH + AgNO₃/NH₄OH → AgC≡CAg ↓ (white silver acetylide)
HC≡CH + Cu₂Cl₂/NH₄OH → CuC≡CCu ↓ (red copper acetylide)
The white precipitate with Tollens' reagent and red precipitate with Cu₂Cl₂ are characteristic tests for terminal alkynes (≡CH). Internal alkynes such as but-2-yne do NOT respond.
(ii) Addition of H₂
HC≡CH + H₂ →Pt/Pd/Ni CH₂=CH₂ →H₂ CH₃CH₃With Lindlar's catalyst (Pd–CaCO₃, Pb(OAc)₂) only one H₂ adds → cis-alkene; with Na/liq NH₃ → trans-alkene.
(iii) Addition of halogens
HC≡CH + Br₂ → CHBr=CHBr → CHBr₂–CHBr₂ (1,1,2,2-tetrabromoethane)Decolourisation of red-brown Br₂ in CCl₄ — confirms unsaturation, like with alkenes.
(iv) Addition of HX (Markovnikov)
HC≡CH + HBr → CH₂=CHBr → CH₃–CHBr₂ (1,1-dibromoethane)CH₃C≡CH + HBr → CH₃–CBr=CH₂ → CH₃–CBr₂–CH₃ (2,2-dibromopropane)
(v) Addition of water (industrial — synthesis of acetaldehyde)
HC≡CH + H₂O →HgSO₄/40% H₂SO₄, 333 K [CH₂=CHOH] → CH₃CHOThe unstable enol tautomerises to acetaldehyde. With unsymmetrical alkynes, water addition follows Markovnikov rule.
(vi) Polymerisation — cyclic trimerisation to benzene
Three molecules of ethyne polymerise on passing through a red-hot iron tube at 873 K to give benzene:
3 HC≡CH →red-hot Fe tube, 873 K C₆H₆ (benzene)Linear polymerisation can also occur to give polyacetylene, –(CH=CH)–ₙ — a conducting polymer (Nobel Prize 2000 to Heeger, MacDiarmid, Shirakawa).
Hybridisation–Acidity Connection
Pick a C–H type — see how %-s character drives the acidity (lower pKa = stronger acid).
% s character: 25% | pKa ≈ 50
Alkanes are essentially non-acidic. Their C–H bonds use sp³ orbitals (only 25% s character), so the bonding electrons are far from C and H⁺ does not leave easily.
9.6 Aromatic Hydrocarbons
Benzene (C₆H₆) was first isolated by Michael Faraday in 1825. Despite its unsaturation (formula C₆H₆ implies 4 degrees of unsaturation), benzene shows remarkable stability and undergoes substitution rather than addition reactions — defining a whole new class called aromatic compounds.
9.6.1 Kekulé's Structure of Benzene (1865)
August Kekulé proposed a six-membered cyclohexatriene structure with alternating C=C double and C–C single bonds. To explain why only one (and not three) ortho-disubstituted product is observed, he proposed dynamic equilibrium between two oscillating Kekulé structures.
9.6.2 Resonance & Stability of Benzene
Modern view: benzene is a resonance hybrid of two equivalent Kekulé structures. The six π electrons are delocalised over all six carbons, forming two doughnut-shaped π clouds above and below the molecular plane. Consequences:
- All six C–C bonds equivalent (length 139 pm) — between C–C single (154 pm) and C=C double (134 pm).
- All bond angles 120°; molecule perfectly planar.
- Heat of hydrogenation (benzene + 3 H₂ → cyclohexane) is –208 kJ/mol — much less than 3 × cyclohexene's –120 kJ/mol = –360 kJ/mol. Difference (~150 kJ/mol) = resonance energy — the extra stability of benzene over a hypothetical cyclohexatriene.
9.6.3 Aromaticity — Hückel's Rule
Hückel's (4n+2) Rule: A compound is aromatic if it satisfies ALL three criteria:
- Planarity — the cyclic system must be flat so that p-orbitals can overlap continuously.
- Complete delocalisation of π electrons (cyclic conjugation) — every ring atom must have a p-orbital perpendicular to the plane.
- The number of delocalised π electrons must be (4n + 2) where n = 0, 1, 2, …
For benzene, n = 1: (4×1 + 2) = 6 π electrons → aromatic. Naphthalene (10 π, n=2) and anthracene (14 π, n=3) are also aromatic.
9.6.4 Preparation of Benzene
- From ethyne (cyclic trimerisation): 3 HC≡CH →red-hot Fe, 873 K C₆H₆.
- From phenol (industrial reduction with Zn dust): C₆H₅OH + Zn →Δ C₆H₆ + ZnO.
- Decarboxylation of sodium benzoate with soda lime: C₆H₅COONa + NaOH →CaO/Δ C₆H₆ + Na₂CO₃.
9.6.5 Physical Properties
Aromatic hydrocarbons are colourless liquids or solids with a characteristic aroma. They are non-polar (hexane-like) and immiscible with water but soluble in organic solvents. They burn with a sooty (luminous) flame because of the high C/H ratio and incomplete combustion of the aromatic ring. Benzene melts at 5.5 °C and boils at 80.1 °C; toluene boils at 110 °C.
Activity 9.3 — Test for terminal alkyne
Setup: Three test tubes contain colourless gases dissolved in solvent: but-1-yne, but-2-yne, and pentene.
Predict: Using ammoniacal silver nitrate (Tollens' reagent) and bromine water, how would you distinguish all three?
Step 1 (Tollens' / AgNO₃·NH₃):
- but-1-yne (terminal ≡CH) → white precipitate of silver butynide. Identified.
- but-2-yne (internal ≡) → no precipitate.
- pentene → no precipitate.
Step 2 (Br₂ / CCl₄ on the remaining two):
- but-2-yne → decolourises Br₂ rapidly (×2 since two π bonds).
- pentene → also decolourises Br₂ but less consumption per mole. Quantitative bromination would consume 2 mol Br₂ for the alkyne but only 1 mol for the alkene.
Combination of Tollens' (acidity) and bromine consumption (degree of unsaturation) cleanly distinguishes terminal alkyne, internal alkyne, and alkene.
Worked Example 1: Sodium acetylide synthesis
Suggest a synthesis of pent-2-yne starting from acetylene and any alkyl halide.
Plan: pent-2-yne = CH₃–C≡C–CH₂CH₃. Build by SN2 alkylation of an acetylide ion.
Step 1: HC≡CH + NaNH₂ → HC≡C⁻Na⁺ (sodium acetylide).
Step 2: HC≡C⁻Na⁺ + CH₃Br → HC≡C–CH₃ (propyne) + NaBr.
Step 3: HC≡C–CH₃ + NaNH₂ → CH₃–C≡C⁻Na⁺ (methylacetylide).
Step 4: CH₃–C≡C⁻Na⁺ + CH₃CH₂Br → CH₃–C≡C–CH₂CH₃ (pent-2-yne).
This sequence converts the acidic ≡CH proton (pKa 25) into a useful nucleophilic carbon.
Step 1: HC≡CH + NaNH₂ → HC≡C⁻Na⁺ (sodium acetylide).
Step 2: HC≡C⁻Na⁺ + CH₃Br → HC≡C–CH₃ (propyne) + NaBr.
Step 3: HC≡C–CH₃ + NaNH₂ → CH₃–C≡C⁻Na⁺ (methylacetylide).
Step 4: CH₃–C≡C⁻Na⁺ + CH₃CH₂Br → CH₃–C≡C–CH₂CH₃ (pent-2-yne).
This sequence converts the acidic ≡CH proton (pKa 25) into a useful nucleophilic carbon.
Worked Example 2: Why benzene is exceptionally stable
Calculate the resonance energy of benzene given that ΔH(hydrogenation, cyclohexene) = –120 kJ/mol and ΔH(hydrogenation, benzene) = –208 kJ/mol.
Hypothetical cyclohexatriene with three isolated C=C bonds would release 3 × 120 = 360 kJ/mol on hydrogenation to cyclohexane.
Actual benzene releases only 208 kJ/mol on the same reduction.
Difference = 360 – 208 = 152 kJ/mol — this is the resonance (delocalisation) energy of benzene.
Interpretation: benzene is 152 kJ/mol more stable than the hypothetical cyclohexatriene with localised double bonds. This is why benzene resists addition (which would destroy aromaticity) and prefers substitution (which preserves the aromatic ring).
Actual benzene releases only 208 kJ/mol on the same reduction.
Difference = 360 – 208 = 152 kJ/mol — this is the resonance (delocalisation) energy of benzene.
Interpretation: benzene is 152 kJ/mol more stable than the hypothetical cyclohexatriene with localised double bonds. This is why benzene resists addition (which would destroy aromaticity) and prefers substitution (which preserves the aromatic ring).
Worked Example 3: Hückel's rule application
Predict whether the following are aromatic, anti-aromatic, or non-aromatic: (a) cyclobutadiene (C₄H₄), (b) cyclopentadienyl anion (C₅H₅⁻), (c) cycloheptatrienyl cation (C₇H₇⁺).
(a) Cyclobutadiene: 4-membered, planar (forced), 4 π electrons. n = ½ ⟹ not (4n+2). It has 4n with n=1 (4 π) → anti-aromatic, very unstable.
(b) Cyclopentadienyl anion (C₅H₅⁻): 5-membered ring, sp² carbons, 6 π electrons (4 from two C=C + 2 from the lone pair on the carbanion). 6 = 4(1)+2 ⟹ n=1 → aromatic. This is why cyclopentadiene (pKa ~16) is more acidic than typical hydrocarbons.
(c) Cycloheptatrienyl cation (tropylium, C₇H₇⁺): 7-membered ring, sp² carbons, 6 π electrons (the empty orbital at C contributes 0 e). 6 = (4n+2), n=1 → aromatic. This is why tropylium salts are stable solids.
(b) Cyclopentadienyl anion (C₅H₅⁻): 5-membered ring, sp² carbons, 6 π electrons (4 from two C=C + 2 from the lone pair on the carbanion). 6 = 4(1)+2 ⟹ n=1 → aromatic. This is why cyclopentadiene (pKa ~16) is more acidic than typical hydrocarbons.
(c) Cycloheptatrienyl cation (tropylium, C₇H₇⁺): 7-membered ring, sp² carbons, 6 π electrons (the empty orbital at C contributes 0 e). 6 = (4n+2), n=1 → aromatic. This is why tropylium salts are stable solids.
Competency-Based Questions
Q1. The general formula of an alkyne is: L1 Remember
Answer: (c) CₙH₂ₙ₋₂. Each triple bond reduces hydrogen count by 4 compared to alkane.
Q2. Why does ethyne react with sodium to release H₂, but ethene does not? L4 Analyse
Answer: The C–H bond in ethyne uses an sp orbital with 50% s-character; in ethene the C–H uses an sp² orbital (33% s). Higher s-character pulls bonding electrons closer to C, increasing the polarity of C–H and the acidity of H. Hence ethyne (pKa ~25) reacts with strong bases like Na to release H₂; ethene (pKa ~44) is far less acidic and does not react under the same conditions.
Q3. Identify which of the following are aromatic by Hückel's rule: cyclopropenyl cation (C₃H₃⁺, 2 π electrons), benzene (6 π), cyclooctatetraene (8 π, planar). L3 Apply
Answer:
(i) Cyclopropenyl cation: 2 π electrons = 4(0)+2 → aromatic (n = 0). Stable, isolable salt.
(ii) Benzene: 6 π = 4(1)+2 → aromatic (n = 1). Most familiar example.
(iii) Planar cyclooctatetraene: 8 π = 4(2) → anti-aromatic (does NOT satisfy 4n+2). In reality, it adopts a non-planar tub shape and behaves as a normal alkene.
(i) Cyclopropenyl cation: 2 π electrons = 4(0)+2 → aromatic (n = 0). Stable, isolable salt.
(ii) Benzene: 6 π = 4(1)+2 → aromatic (n = 1). Most familiar example.
(iii) Planar cyclooctatetraene: 8 π = 4(2) → anti-aromatic (does NOT satisfy 4n+2). In reality, it adopts a non-planar tub shape and behaves as a normal alkene.
Q4. Compare the bond lengths and bond strengths of C–C, C=C and C≡C. Comment on the trend. L5 Evaluate
Bond length: C–C (154 pm) > C=C (134 pm) > C≡C (120 pm).
Bond enthalpy: C–C (348 kJ/mol) < C=C (681 kJ/mol) < C≡C (839 kJ/mol).
Trend: as the bond order increases (1 → 2 → 3), atoms are pulled closer (length decreases) and the total bonding energy increases. However, the bond enthalpy is less than additive: a triple bond is NOT 3× a single bond because the σ bond is the strongest, the first π adds less, and the second π even less (poorer orbital overlap). This is why alkenes/alkynes still readily undergo addition — opening a π is exothermic.
Bond enthalpy: C–C (348 kJ/mol) < C=C (681 kJ/mol) < C≡C (839 kJ/mol).
Trend: as the bond order increases (1 → 2 → 3), atoms are pulled closer (length decreases) and the total bonding energy increases. However, the bond enthalpy is less than additive: a triple bond is NOT 3× a single bond because the σ bond is the strongest, the first π adds less, and the second π even less (poorer orbital overlap). This is why alkenes/alkynes still readily undergo addition — opening a π is exothermic.
Q5. HOT (Create): Suggest a reasonable synthesis of methyl ethyl ketone (CH₃COC₂H₅) starting from but-1-yne. L6 Create
Plan: but-1-yne is CH≡C–CH₂CH₃ (terminal). Hydration would give methyl ketone via Markovnikov:
CH≡C–CH₂CH₃ + H₂O →HgSO₄/H₂SO₄ CH₂=C(OH)–CH₂CH₃ → tautomerises → CH₃–CO–CH₂CH₃ (butan-2-one = methyl ethyl ketone, MEK).
Markovnikov adds H to the terminal C (more H), OH to the internal C, generating the more stable enol which tautomerises to a methyl ketone — never to an aldehyde (in contrast, hydration of acetylene itself gives acetaldehyde because the symmetry forces both carbons equivalent).
CH≡C–CH₂CH₃ + H₂O →HgSO₄/H₂SO₄ CH₂=C(OH)–CH₂CH₃ → tautomerises → CH₃–CO–CH₂CH₃ (butan-2-one = methyl ethyl ketone, MEK).
Markovnikov adds H to the terminal C (more H), OH to the internal C, generating the more stable enol which tautomerises to a methyl ketone — never to an aldehyde (in contrast, hydration of acetylene itself gives acetaldehyde because the symmetry forces both carbons equivalent).
Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Terminal alkynes give a white precipitate with ammoniacal AgNO₃.
R: The H attached to the sp-hybridised carbon is acidic and is replaced by Ag to form a silver acetylide.
Answer: (A). Both true; R explains A. The white silver acetylide is a sensitive test for ≡C–H.
A: Benzene undergoes substitution rather than addition reactions.
R: Substitution preserves the (4n+2) π electron aromatic system, while addition would destroy it.
Answer: (A). Both true; R explains A. The 150 kJ/mol resonance stabilisation strongly favours retention of the aromatic ring.
A: Cyclooctatetraene (C₈H₈) is aromatic.
R: It contains 8 π electrons in a fully conjugated cycle.
Answer: (D). Assertion is FALSE — cyclooctatetraene is non-aromatic. It adopts a tub-shaped (non-planar) geometry, breaking conjugation; even if forced planar it would have 8 π = 4n electrons → anti-aromatic. Reason is TRUE about its π count, but the molecule avoids planarity precisely to escape anti-aromaticity.
Frequently Asked Questions — Alkynes and Aromatic Hydrocarbons
What are alkynes and what is their general formula?
Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond (C≡C). General formula: CₙH₂ₙ₋₂. Each carbon of the triple bond is sp-hybridised with linear geometry (180°). The C≡C consists of one σ and two π bonds, making it shorter (~120 pm) and stronger than C=C or C-C. NCERT Class 11 Chemistry Chapter 9 covers ethyne (acetylene, C₂H₂) as the simplest alkyne and propyne (C₃H₄). Alkynes are more reactive than alkanes but slightly less reactive than alkenes in many addition reactions due to the cylindrical π-electron cloud.
How is acetylene prepared?
NCERT Class 11 Chemistry Chapter 9 lists preparation methods of acetylene (ethyne): (1) from calcium carbide and water — CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂; this is the major industrial method using CaC₂ from CaO + C in an electric furnace; (2) dehydrohalogenation of vicinal dihalides — CH₂Br-CH₂Br + 2KOH (alc.) → HC≡CH + 2KBr + 2H₂O; (3) from sodium acetylide RC≡C⁻Na⁺ + R'X; (4) Kolbe electrolysis of sodium salt of dicarboxylic acid. Acetylene is used in oxy-acetylene welding (3000°C flame), in vinyl chloride and PVC manufacture, and as an intermediate for many chemicals.
Why are terminal alkynes acidic?
Terminal alkynes (≡C-H) are weakly acidic (pK_a ≈ 25) because the C-H bond involves an sp-hybridised carbon. The sp orbital has 50% s character (compared to 33% in sp² and 25% in sp³), making it more electronegative and holding the bonding electrons more tightly. The conjugate base — the acetylide anion (RC≡C⁻) — is stabilised by the high s-character orbital. NCERT Class 11 Chemistry Chapter 9 shows that terminal alkynes react with strong bases like sodamide (NaNH₂) to give sodium acetylides: RC≡CH + NaNH₂ → RC≡C⁻Na⁺ + NH₃. Internal alkynes (R-C≡C-R') are not acidic.
What is aromaticity and Hückel's rule?
Aromaticity is a special stability associated with certain cyclic, planar, fully conjugated molecules with delocalised π electrons. Hückel's rule (1931) states that an organic molecule is aromatic if it satisfies: (1) cyclic, (2) planar, (3) every atom in the ring has a p-orbital perpendicular to the ring, (4) it contains (4n + 2) π electrons where n = 0, 1, 2, 3,…. NCERT Class 11 Chemistry Chapter 9 examples: benzene (n = 1, 6 π electrons), naphthalene (n = 2, 10 π electrons), cyclopentadienyl anion. Anti-aromatic compounds have 4n π electrons and are unusually unstable.
What is the structure of benzene?
Benzene (C₆H₆) is a planar, regular hexagonal molecule with all C-C bond lengths equal at 139 pm (intermediate between single 154 pm and double 134 pm). Each carbon is sp²-hybridised, forming three σ bonds (two to adjacent C, one to H) in a plane. The remaining p-orbitals overlap above and below the plane forming a delocalised π system of 6 electrons. NCERT Class 11 Chemistry Chapter 9 describes benzene as a resonance hybrid of two Kekulé structures, often represented as a hexagon with an inscribed circle. The delocalisation provides ~150 kJ/mol of resonance stabilisation energy.
What are electrophilic aromatic substitution reactions?
Electrophilic aromatic substitution (EAS) is the characteristic reaction of aromatic compounds in which an electrophile replaces a hydrogen atom on the aromatic ring, preserving the aromaticity. NCERT Class 11 Chemistry Chapter 9 lists five common EAS reactions of benzene: (1) nitration — benzene + HNO₃/H₂SO₄ → nitrobenzene + H₂O; (2) halogenation — benzene + Cl₂/FeCl₃ → chlorobenzene + HCl; (3) sulphonation — benzene + H₂SO₄ → benzenesulphonic acid; (4) Friedel-Crafts alkylation — benzene + RCl/AlCl₃ → R-benzene + HCl; (5) Friedel-Crafts acylation — benzene + RCOCl/AlCl₃ → R-CO-benzene + HCl. The mechanism involves an arenium ion intermediate.
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