CBSE Class 11 & JEE Mains • Module 07 of 20 • The sp Hybrid — Acidic, Linear, Reactive
Topics Covered: General Formula CₙH₂ₙ₋₂ · sp Hybridization · Linear Geometry · Acidic Nature of Terminal Alkynes · Preparation (dehydrohalogenation, from calcium carbide) · Chemical Reactions (addition of H₂, HX, X₂, H₂O-Markovnikov to ketone via Wacker, Lindlar/Na/NH₃ reductions, Ozonolysis) · Identification Tests
General formula: CₙH₂ₙ₋₂ (one triple bond)
Triple bond = 1 σ + 2 π bonds
sp hybridization: Each triple-bond carbon forms 2 sp hybrid orbitals (180° apart → linear). Two unhybridized p orbitals per carbon remain (perpendicular to each other and to the σ-bond axis) → these overlap sideways to form TWO π bonds.
Bond angle: 180° (linear). C≡C bond length: 1.20 Å (shorter than C=C 1.34 Å or C–C 1.54 Å). C≡C bond strength: 839 kJ/mol (strongest C–C bond type).
Terminal alkynes (R–C≡C–H) are weakly acidic — the ≡C–H hydrogen can be removed by strong bases.
Why? The sp carbon is the most electronegative type of carbon (50% s-character → holds electrons close to nucleus). The C–H bond is more polarized (Cδ⁺–Hδ⁺) → H is more "acidic".
Acidity order: HC≡CH (pKa ~25) > H₂C=CH₂ (pKa ~44) > CH₄ (pKa ~50)
Key reactions showing acidic nature:
CaC₂ + H₂O → HC≡CH (ethyne / acetylene) + Ca(OH)₂
Calcium carbide is made by: CaO + 3C →(electric furnace, 2000°C)→ CaC₂ + CO
The carbide ion C₂²⁻ reacts with water to give acetylene. This is the main industrial source of ethyne.
Two moles of KOH/alcohol required:
R–CHBr–CHBr–R' + 2KOH (alc.) → R–C≡C–R' + 2KBr + 2H₂O
Mechanism: First elimination gives vinyl halide (R–CBr=CH–R'). Second KOH/alc. removes another HBr from vinyl halide → alkyne. (Vinyl halides are much less reactive in E2 — need strong base and higher temperature.)
R–CHBr₂ + 2KOH (alc.) → R–C≡CH + 2KBr + 2H₂O (two HBr removed from same C and adjacent C)
Gem-dihalides have both halogens on same carbon.
The remarkable feature of alkynes: you can stop reduction at the alkene stage (do NOT reduce all the way to alkane) using specific catalysts.
Full reduction → Alkane: R–C≡C–R' + 2H₂ →(Ni/Pt, heat)→ R–CH₂–CH₂–R'
Lindlar's Catalyst → cis-Alkene (Z):
Both H atoms delivered from the same face of the Pd surface → syn addition → cis product.
Birch Reduction (Na/NH₃) → trans-Alkene (E):
Mechanism: electron addition to alkyne → vinyl radical → vinyl anion → protonation. Anti addition → trans product (bulky groups go to same side for less steric interaction in the vinyl anion step).
Step 1: R–C≡CH + HX → R–C(X)=CH₂ (haloalkene — Markovnikov addition)
Step 2 (1 more HX): R–C(X)=CH₂ + HX → R–CX₂–CH₃ (gem-dihalide — Markovnikov again)
Both halogen atoms end up on the SAME carbon (gem-dihalide). This is called gem-addition.
Example: HC≡CH + 2HCl → CH₃–CHCl₂ (1,1-dichloroethane)
R–C≡C–R' + Br₂ → R–CBr=CBr–R' (trans adduct, 1,2-dibromo alkene) [first addition]
+ Br₂ again → R–CBr₂–CBr₂–R' (1,1,2,2-tetrabromoalkane)
Alkynes also decolorise bromine water (like alkenes) → test for unsaturation.
Condition: H₂O + H₂SO₄ + HgSO₄ catalyst (or Wacker process with PdCl₂/CuCl₂)
R–C≡CH + H₂O →(H₂SO₄, HgSO₄)→ [R–C(OH)=CH₂] (enol) → tautomerises → R–CO–CH₃ (ketone)
The enol form (vinyl alcohol) is unstable → immediately tautomerises to the more stable keto form.
Markovnikov applies: OH adds to the more substituted C → ketone (not aldehyde, except for acetylene).
Exception — Acetylene: HC≡CH + H₂O →(H₂SO₄, HgSO₄)→ [CH₂=CHOH] (vinyl alcohol) → CH₃CHO (acetaldehyde, ethanal — an aldehyde not a ketone, because the molecule is symmetric)
Linear trimerization: 3 HC≡CH →(600°C, charcoal)→ C₆H₆ (benzene!) — Berthelot reaction
Polymerization of vinyl acetylene: 2 HC≡CH → CH₂=CH–C≡CH (vinylacetylene) → further addition of HCl → gives neoprene (chloroprene polymer) — a synthetic rubber.
R–C≡C–R' + O₃ → 2 RCOOH (two carboxylic acids by full oxidation)
HC≡CH → 2 CO₂ + H₂O (acetylene gives CO₂)
R–C≡CH → RCOOH + CO₂ (terminal alkyne gives acid + CO₂)
Useful for identifying the structure of unknown alkynes (same backward-reasoning as alkene ozonolysis).
E Q1. Why is ethyne more acidic than ethene? What property of the sp carbon makes it so?
E Q2. Write the reaction of ethyne with silver nitrate (Tollens' reagent, Ag(NH₃)₂⁺). What does a positive test for this look like?
M Q3. An alkyne gives a white precipitate with AgNO₃/NH₃ and on ozonolysis gives CH₃COOH and CO₂. Identify the alkyne and write all reactions.
M Q4. Convert but-1-yne to (a) (Z)-but-2-ene and (b) (E)-but-2-ene using appropriate reagents.
H Q5. How would you convert ethyne to (a) acetaldehyde, (b) acetone, (c) benzene? Write all steps and conditions.
H Q6. Explain why gem-dihalides (not vicinal dihalides) are the product of double HX addition to alkynes, using Markovnikov's rule for each step.
| Reaction | Condition | Product |
|---|---|---|
| H₂ (full) | Ni/Pt/Pd, heat | Alkane (−2 degrees of unsaturation) |
| H₂ (half, syn) | Lindlar's catalyst | cis-Alkene (Z) |
| H₂ (half, anti) | Na/liquid NH₃ (Birch) | trans-Alkene (E) |
| HX (1st mole) | Markovnikov, HX | Haloalkene (vinyl halide) |
| HX (2nd mole) | Markovnikov, HX | gem-Dihalide (both X on same C) |
| H₂O | H₂SO₄/HgSO₄ | Ketone (via enol tautomerism); acetylene → acetaldehyde |
| AgNO₃/NH₃ | Terminal alkyne only | White AgC≡CR precipitate (test) |
| Ozonolysis | O₃, then H₂O | Carboxylic acids (or CO₂ for ≡CH) |