The Hook: The Flame That Cuts Steel
Ever seen welders cutting through thick steel beams with an incredibly hot flame? That’s an oxyacetylene torch - burning acetylene (C₂H₂) with oxygen produces a flame reaching 3,300°C!
But here’s what makes acetylene special in chemistry: it’s the only hydrocarbon that can act as an acid. How can a hydrocarbon give up a proton? The answer lies in the triple bond!
JEE Question: Why can ethyne (acetylene) react with sodium metal while ethene and ethane cannot?
The Core Concept
What are Alkynes?
Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond (C≡C).
$$\boxed{\text{General formula: C}_n\text{H}_{2n-2}}$$In simple terms: The triple bond has one σ bond and two π bonds, making alkynes even more unsaturated than alkenes!
The Triple Bond: σ + π + π
A C≡C triple bond consists of:
- One σ (sigma) bond - sp-sp overlap (strong)
- Two π (pi) bonds - p-p overlap (weaker)
Hybridization: sp (50% s-character, 50% p-character)
Bond characteristics:
- Bond length: C≡C (120 pm) < C=C (134 pm) < C-C (154 pm)
- Bond energy: C≡C (839 kJ/mol) > C=C (611 kJ/mol) > C-C (347 kJ/mol)
- Bond strength: Triple > Double > Single
Key concept for JEE Advanced:
The sp hybridized carbon has 50% s-character (compared to 33% in sp² and 25% in sp³)
Higher s-character → Electrons held closer to nucleus → More electronegative C → Can stabilize negative charge better
$$\boxed{\text{R-C≡C-H} + \text{NaNH}_2 \rightarrow \text{R-C≡C}^- \text{Na}^+ + \text{NH}_3}$$This makes terminal alkynes weakly acidic (pKa ≈ 25)
Related: Acid-Base Chemistry
Alkynes: 2-3 questions in JEE Main, 1-2 in JEE Advanced High-yield topics:
- Acidic nature of terminal alkynes
- Addition reactions (electrophilic and nucleophilic)
- Polymerization and industrial applications
Preparation of Alkynes
Method 1: From Calcium Carbide (Industrial)
Historical importance - commercial production of acetylene
$$\boxed{\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{HC≡CH} + \text{Ca(OH)}_2}$$Calcium carbide preparation:
$$\text{CaO} + 3\text{C} \xrightarrow{2000°\text{C}} \text{CaC}_2 + \text{CO}$$Uses of acetylene:
- Oxyacetylene welding (3,300°C flame)
- Starting material for many organic syntheses
- Production of PVC (via vinyl chloride)
Method 2: Dehydrohalogenation of Vicinal or Geminal Dihalides
From Vicinal Dihalides (X on adjacent carbons)
$$\boxed{\text{R-CHX-CHX-R} + 2\text{KOH (alc.)} \xrightarrow{\Delta} \text{R-C≡C-R} + 2\text{KX} + 2\text{H}_2\text{O}}$$Mechanism: Two successive eliminations
Step 1: First elimination → alkene
$$\text{R-CHBr-CHBr-R} + \text{KOH} \rightarrow \text{R-CH=CHBr-R}$$Step 2: Second elimination → alkyne
$$\text{R-CH=CHBr-R} + \text{KOH} \rightarrow \text{R-C≡C-R}$$From Geminal Dihalides (both X on same carbon)
$$\boxed{\text{R-CHX}_2\text{-CH}_3 + 2\text{KOH (alc.)} \xrightarrow{\Delta} \text{R-C≡CH} + 2\text{KX} + 2\text{H}_2\text{O}}$$Example:
$$\text{CH}_3\text{-CHBr}_2 + 2\text{KOH (alc.)} \rightarrow \text{CH}_3\text{-C≡CH}$$“Vicinity is Next door, Gemini is Twin”
Vicinal: Halogens on adjacent carbons (neighbors)
- CHBr-CHBr (1,2-dibromo)
Geminal: Halogens on same carbon (twins on same spot)
- CHBr₂ (1,1-dibromo)
JEE Tip: Both give alkynes with excess KOH, but geminal dihalides are preferred (easier to prepare from aldehydes/ketones)
Related: Alkyl Halides
Method 3: Alkylation of Terminal Alkynes
Building longer carbon chains using sodium acetylide
$$\boxed{\text{HC≡C}^- \text{Na}^+ + \text{R-X} \rightarrow \text{R-C≡CH} + \text{NaX}}$$Full sequence:
Step 1: Form acetylide ion
$$\text{HC≡CH} + \text{NaNH}_2 \rightarrow \text{HC≡C}^- \text{Na}^+ + \text{NH}_3$$Step 2: Nucleophilic substitution
$$\text{HC≡C}^- + \text{CH}_3\text{Br} \rightarrow \text{CH}_3\text{-C≡CH}$$Step 3: Can repeat for longer chain
$$\text{CH}_3\text{-C≡CH} + \text{NaNH}_2 \rightarrow \text{CH}_3\text{-C≡C}^- \text{Na}^+$$ $$\text{CH}_3\text{-C≡C}^- + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{-C≡C-CH}_2\text{CH}_3$$Question: Why use NaNH₂ and not NaOH to form acetylide?
Wrong answer: “Both are bases, so both work”
Correct answer:
pKa comparison:
- HC≡CH: pKa ≈ 25
- NH₃: pKa ≈ 35
- H₂O: pKa ≈ 15.7
For complete deprotonation: Conjugate acid of base must have higher pKa than substrate
$$\text{pKa (NH}_3\text{) = 35} > \text{pKa (HC≡CH) = 25} \quad ✓$$ $$\text{pKa (H}_2\text{O) = 15.7} < \text{pKa (HC≡CH) = 25} \quad ✗$$Conclusion: NaNH₂ is strong enough, NaOH is too weak!
JEE One-liner: “For acetylide formation, Need NaNH₂ (Not NaOH)”
Acidic Character of Alkynes
Why Terminal Alkynes are Acidic
Acidity order:
$$\boxed{\text{HC≡CH} > \text{H}_2\text{C=CH}_2 > \text{H}_3\text{C-CH}_3}$$pKa values:
- Ethyne (HC≡CH): 25
- Ethene (H₂C=CH₂): 44
- Ethane (H₃C-CH₃): 50
Reason: Stability of carbanion
When H⁺ is removed:
$$\text{R-C≡C-H} \rightarrow \text{R-C≡C}^- + \text{H}^+$$The acetylide ion (R-C≡C⁻) is stabilized by:
- High s-character (50% in sp) → electrons closer to nucleus
- sp hybridized carbon is more electronegative → holds negative charge better
Electronegativity:
$$\text{sp (50\% s)} > \text{sp}^2\text{ (33\% s)} > \text{sp}^3\text{ (25\% s)}$$More s-character → More electronegative → Better stabilization of (-)ve charge
Q: Arrange in order of increasing acidity: (a) CH₃-CH₃ (b) CH₂=CH₂ (c) HC≡CH
Answer: (a) < (b) < (c)
Explanation:
| Compound | Hybridization | s-character | pKa | Acidity |
|---|---|---|---|---|
| Ethane | sp³ | 25% | 50 | Least acidic |
| Ethene | sp² | 33% | 44 | Weak acid |
| Ethyne | sp | 50% | 25 | Strongest (of these) |
JEE Shortcut: More s-character → More acidic
Reactions Showing Acidic Nature
Reaction 1: With Sodium metal
$$\boxed{2\text{HC≡CH} + 2\text{Na} \rightarrow 2\text{HC≡C}^- \text{Na}^+ + \text{H}_2↑}$$Reaction 2: With NaNH₂ (stronger base, more common)
$$\boxed{\text{HC≡CH} + \text{NaNH}_2 \rightarrow \text{HC≡C}^- \text{Na}^+ + \text{NH}_3}$$Reaction 3: With Grignard reagent
$$\boxed{\text{HC≡CH} + \text{CH}_3\text{MgBr} \rightarrow \text{HC≡C}^- \text{Mg}^{2+}\text{Br}^- + \text{CH}_4}$$Test for terminal alkynes:
With ammoniacal AgNO₃ or ammoniacal Cu₂Cl₂:
$$\boxed{\text{R-C≡CH} + \text{AgNO}_3 + \text{NH}_3 \rightarrow \text{R-C≡C-Ag}↓ + \text{NH}_4\text{NO}_3}$$White/yellow precipitate forms (silver acetylide)
Similarly with copper:
$$\text{R-C≡CH} + \text{Cu}_2\text{Cl}_2 + \text{NH}_3 \rightarrow \text{R-C≡C-Cu}↓$$Red/brown precipitate (copper acetylide)
Metal acetylides (Ag-C≡C-R, Cu-C≡C-R) are explosive when dry!
Handle with care in laboratory.
JEE won’t ask about explosions, but will test:
- Which alkynes give this test? (Only terminal alkynes)
- Internal alkynes (R-C≡C-R) → No precipitate
Reactions of Alkynes
Type 1: Addition Reactions
Alkynes can add two molecules of reagent (two π bonds to break!)
Reaction 1: Hydrogenation
Complete Hydrogenation (excess H₂)
$$\boxed{\text{R-C≡C-R} + 2\text{H}_2 \xrightarrow{\text{Pt/Pd/Ni}} \text{R-CH}_2\text{-CH}_2\text{-R}}$$Triple bond → Single bond (alkane)
Partial Hydrogenation to Alkene
Method A: Lindlar’s Catalyst (gives cis-alkene)
$$\boxed{\text{R-C≡C-R} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4\text{, quinoline}} \text{R-CH=CH-R (cis)}}$$- Syn addition (both H from same side)
- Catalyst is “poisoned” with quinoline to prevent over-reduction
Method B: Na in liquid NH₃ (gives trans-alkene)
$$\boxed{\text{R-C≡C-R} + 2\text{Na/liq. NH}_3 \rightarrow \text{R-CH=CH-R (trans)}}$$- Anti addition (H from opposite sides)
“Lindlar is Like cis, Liquid gives trans”
- Lindlar → Like → close together → cis
- Liquid NH₃ → Long way apart → trans
JEE loves asking: “How will you prepare cis-2-butene from 2-butyne?” Answer: Lindlar’s catalyst!
Related: Alkenes
Reaction 2: Addition of Halogens
With one mole X₂ → Dihaloalkene
$$\boxed{\text{R-C≡C-R} + \text{X}_2 \rightarrow \text{R-CX=CX-R}}$$With excess X₂ → Tetrahaloalkane
$$\boxed{\text{R-C≡C-R} + 2\text{X}_2 \rightarrow \text{R-CX}_2\text{-CX}_2\text{-R}}$$Example:
$$\text{HC≡CH} + \text{Br}_2 \rightarrow \text{HCBr=CHBr}$$ $$\text{HCBr=CHBr} + \text{Br}_2 \rightarrow \text{HCBr}_2\text{-CHBr}_2$$Test for unsaturation: Br₂/CCl₄ decolorized (same as alkenes)
Reaction 3: Addition of Hydrogen Halides (HX)
Follows Markovnikov’s Rule
With one mole HX:
$$\boxed{\text{R-C≡CH} + \text{HX} \rightarrow \text{R-CX=CH}_2}$$Mechanism:
Step 1: Protonation → vinylic carbocation
$$\text{R-C≡CH} + \text{H}^+ \rightarrow \text{R-C}^+\text{=CH}_2$$(more stable)
Not: R-CH=C⁺H (less stable due to sp carbon bearing +ve charge)
Step 2: X⁻ attacks
$$\text{R-C}^+\text{=CH}_2 + \text{X}^- \rightarrow \text{R-CX=CH}_2$$With excess HX: (adds to alkene product)
$$\boxed{\text{R-CX=CH}_2 + \text{HX} \rightarrow \text{R-CX}_2\text{-CH}_3}$$Overall:
$$\text{R-C≡CH} \xrightarrow{2\text{HX}} \text{R-CX}_2\text{-CH}_3$$Question type: “What is the product when propyne reacts with excess HBr?”
Common mistake: Writing CH₃-CHBr-CHBr₂ (vicinal)
Correct answer: CH₃-CBr₂-CH₃ (geminal)
Why?
- First addition: CH₃-C≡CH + HBr → CH₃-C(Br)=CH₂ (Markovnikov)
- Second addition: CH₃-C(Br)=CH₂ + HBr → CH₃-CBr₂-CH₃ (Markovnikov again!)
Both halogens end up on the same carbon → Geminal dihalide
Memory: “Markovnikov Makes geMs” (geminal dihalides)
Reaction 4: Hydration
Using H₂SO₄/HgSO₄ - Gives carbonyl compounds
$$\boxed{\text{R-C≡CH} \xrightarrow{\text{H}_2\text{SO}_4, \text{HgSO}_4} \text{R-CO-CH}_3}$$Mechanism:
Step 1: Addition of H₂O (Markovnikov)
$$\text{R-C≡CH} \xrightarrow{\text{H}^+, \text{H}_2\text{O}} \text{R-C(OH)=CH}_2$$(enol form)
Step 2: Keto-enol tautomerism
$$\text{R-C(OH)=CH}_2 \rightarrow \text{R-CO-CH}_3$$(keto form - stable)
Important: Enols are unstable and spontaneously convert to ketones!
Special case: Acetylene
$$\boxed{\text{HC≡CH} \xrightarrow{\text{H}_2\text{SO}_4, \text{HgSO}_4} \text{CH}_3\text{-CHO}}$$Gives acetaldehyde (not ketone, since it’s terminal)
Enol (unstable) ⇌ Keto (stable)
OH O
| ||
R-C=CH₂ ←→ R-C-CH₃
Equilibrium strongly favors keto form (more stable)
Ratio: Keto : Enol ≈ 10⁶ : 1
Why keto is more stable:
- C=O bond (740 kJ/mol) stronger than C=C bond (611 kJ/mol)
- C-H bond (413 kJ/mol) stronger than O-H bond (463 kJ/mol)… wait, that’s wrong!
Actually: C=O stronger + C-H bond compensates for loss of O-H and C=C
Net result: Keto form has lower energy
Related: Carbonyl Compounds
Hydroboration-Oxidation (Anti-Markovnikov)
$$\boxed{\text{R-C≡CH} \xrightarrow[\text{2. H}_2\text{O}_2/\text{OH}^-]{\text{1. (sia-BH)}_2} \text{R-CH}_2\text{-CHO}}$$Gives aldehyde (anti-Markovnikov hydration)
Type 2: Polymerization
Linear Polymerization
Ethyne → Benzene
$$\boxed{3\text{HC≡CH} \xrightarrow{\text{Red hot Fe tube, 873 K}} \text{C}_6\text{H}_6}$$Mechanism: Cyclic trimerization
HC≡CH + HC≡CH + HC≡CH → Benzene ring
Polymerization to PVC
Step 1: Ethyne → Vinyl chloride
$$\text{HC≡CH} + \text{HCl} \xrightarrow{\text{HgCl}_2, 333\text{ K}} \text{H}_2\text{C=CHCl}$$Step 2: Polymerization
$$n\text{H}_2\text{C=CHCl} \xrightarrow{\text{peroxide}} (\text{-CH}_2\text{-CHCl-})_n$$(PVC)
Uses: Pipes, insulation, flooring
Related: Polymers
Distinction: Alkanes vs Alkenes vs Alkynes
| Test | Alkane | Alkene | Alkyne (Terminal) | Alkyne (Internal) |
|---|---|---|---|---|
| Br₂/CCl₄ | No reaction (brown color persists) | Decolorizes | Decolorizes | Decolorizes |
| KMnO₄ | No reaction | Decolorizes (purple → colorless) | Decolorizes | Decolorizes |
| Ammoniacal AgNO₃ | No reaction | No reaction | White ppt | No reaction |
| NaNH₂ | No reaction | No reaction | Forms acetylide | No reaction |
Given: Unknown hydrocarbon
Test 1: Br₂/CCl₄
- Decolorizes → Unsaturated (alkene or alkyne)
- No change → Saturated (alkane)
Test 2: If unsaturated, add ammoniacal AgNO₃
- White ppt → Terminal alkyne
- No ppt → Alkene or internal alkyne
Test 3: If no ppt, check molecular formula
- CₙH₂ₙ → Alkene
- CₙH₂ₙ₋₂ → Internal alkyne
JEE loves: Multi-step identification problems!
Common Mistakes to Avoid
Wrong: Hydration of alkynes gives alcohols
Correct: Hydration gives carbonyl compounds (aldehydes/ketones) via enol intermediate
Example:
$$\text{CH}_3\text{-C≡CH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{-CO-CH}_3 \quad \text{(NOT alcohol!)}$$The initially formed enol immediately tautomerizes to ketone.
Wrong: HC≡CH + 2HBr → CHBr-CHBr₂ (vicinal)
Correct: HC≡CH + 2HBr → CHBr₂-CH₃ (geminal)
Reason: Markovnikov’s rule applies at EACH step
- Both additions follow Markovnikov
- Result: Both Br on same carbon
Wrong: “Internal alkynes (R-C≡C-R) don’t react with bases”
Correct: They DO react with strong bases for addition reactions, but:
- Can’t form acetylide ions (no acidic H)
- Won’t give precipitate with AgNO₃/Cu₂Cl₂
- Won’t undergo alkylation reactions
JEE distinction: Terminal vs internal alkynes - know which reactions need terminal H!
Practice Problems
Level 1: Foundation (NCERT)
Q: Write the IUPAC name of CH₃-C≡C-CH₂-CH₃
Solution:
- Longest chain with triple bond: 5 carbons
- Number from end closest to triple bond
- Triple bond between C-2 and C-3
Answer: Pent-2-yne
Structure:
CH₃-C≡C-CH₂-CH₃
1 2 3 4 5
Triple bond starts at C-2, so “pent-2-yne”
Q: Why is ethyne more acidic than ethene?
Solution:
Key factor: Stability of conjugate base (carbanion)
When H⁺ is lost:
- Ethyne → HC≡C⁻ (acetylide ion)
- Ethene → H₂C=CH⁻ (vinyl anion)
Stability comparison:
HC≡C⁻:
- Carbon is sp hybridized (50% s-character)
- Electrons in orbital closer to nucleus
- More stable
H₂C=CH⁻:
- Carbon is sp² hybridized (33% s-character)
- Electrons farther from nucleus
- Less stable
Rule: More stable conjugate base → Stronger acid
$$\text{pKa (ethyne) = 25 < pKa (ethene) = 44}$$Answer: Higher s-character in sp carbon stabilizes negative charge better
Level 2: JEE Main
Q: What is the major product when propyne reacts with HBr (1 mole)?
Solution:
Structure: CH₃-C≡CH
Step 1: Identify possible carbocations
Addition to which carbon?
Option 1: CH₃-C⁺=CH₂ (vinyl carbocation, more substituted)
Option 2: CH₃-CH=C⁺H (vinyl carbocation, less substituted)
Markovnikov’s rule: More substituted carbocation forms
Step 2: Br⁻ attacks carbocation
CH₃-C⁺=CH₂ + Br⁻ → CH₃-C(Br)=CH₂
Answer: CH₃-C(Br)=CH₂ (2-bromopropene)
Common mistake: Don’t write CH₃-CH=CHBr (this violates Markovnikov)
Q: How will you prepare cis-2-butene from 2-butyne?
Solution:
Need: CH₃-CH=CH-CH₃ (cis configuration)
Starting: CH₃-C≡C-CH₃
Method: Partial hydrogenation with Lindlar’s catalyst
$$\text{CH}_3\text{-C≡C-CH}_3 + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} \text{CH}_3\text{-CH=CH-CH}_3 \text{ (cis)}$$Why Lindlar?
- Gives syn addition → cis product
- Catalyst poisoned to prevent over-reduction
Wrong answer: Na/liq. NH₃ (gives trans, not cis)
JEE Tip: Lindlar = cis, Na/NH₃ = trans (memorize this!)
Level 3: JEE Advanced
Q: Starting from acetylene, how will you prepare 2-butanone?
Solution:
Target: CH₃-CO-CH₂-CH₃
Starting: HC≡CH
Strategy: Need 4 carbons and a ketone group
Step 1: Alkylation (add first methyl)
$$\text{HC≡CH} + \text{NaNH}_2 \rightarrow \text{HC≡C}^- \text{Na}^+$$ $$\text{HC≡C}^- + \text{CH}_3\text{Br} \rightarrow \text{CH}_3\text{-C≡CH}$$Step 2: Second alkylation (add ethyl)
$$\text{CH}_3\text{-C≡CH} + \text{NaNH}_2 \rightarrow \text{CH}_3\text{-C≡C}^- \text{Na}^+$$ $$\text{CH}_3\text{-C≡C}^- + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{-C≡C-CH}_2\text{CH}_3$$Step 3: Hydration (convert triple bond to ketone)
$$\text{CH}_3\text{-C≡C-CH}_2\text{CH}_3 \xrightarrow{\text{H}_2\text{SO}_4/\text{HgSO}_4} \text{?}$$Wait! Which ketone forms?
Internal alkyne hydration:
- Can give mixture
- Markovnikov still applies but either direction possible
Actually for symmetric internal alkynes: Only one ketone possible!
$$\text{CH}_3\text{-C≡C-CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{-CO-CH}_2\text{CH}_3$$or
$$\text{CH}_3\text{-CH}_2\text{-CO-CH}_3$$Both are same compound! (2-butanone)
Answer:
- HC≡CH + NaNH₂ → HC≡C⁻Na⁺
- CH₃Br → CH₃-C≡CH
- NaNH₂ → CH₃-C≡C⁻Na⁺
- C₂H₅Br → CH₃-C≡C-C₂H₅
- H₂SO₄/HgSO₄ → CH₃-CO-C₂H₅
JEE Advanced loves: Multi-step synthesis with alkylation + functional group interconversion!
Q: Explain why hydration of 1-butyne gives 2-butanone and not butanal.
Solution:
Structure: CH₃-CH₂-C≡CH
Hydration conditions: H₂SO₄/HgSO₄
Mechanism:
Step 1: Electrophilic addition of H₂O (follows Markovnikov)
Possible enols:
Option 1: CH₃-CH₂-C(OH)=CH₂ (more substituted C gets OH)
Option 2: CH₃-CH₂-CH=C(OH)H (less substituted)
Markovnikov: OH adds to more substituted carbon → Option 1
Step 2: Keto-enol tautomerism
CH₃-CH₂-C(OH)=CH₂ → CH₃-CH₂-CO-CH₃
(enol) (2-butanone)
Why not butanal?
For butanal (CH₃-CH₂-CH₂-CHO), we’d need:
CH₃-CH₂-CH=C(OH)H → CH₃-CH₂-CH₂-CHO
But this violates Markovnikov’s rule! (OH on less substituted C)
Answer: Markovnikov’s rule ensures OH adds to internal carbon, giving ketone (2-butanone) not aldehyde.
JEE Insight: For terminal alkynes:
- If triple bond at end (like ethyne) → aldehyde
- If triple bond internal (like 1-butyne) → ketone
Actually, 1-butyne IS terminal! Let me reconsider…
Correction:
1-butyne: CH₃-CH₂-C≡CH (terminal)
Hydration:
$$\text{CH}_3\text{-CH}_2\text{-C≡CH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{-CH}_2\text{-CO-CH}_3$$Forms enol: CH₃-CH₂-C(OH)=CH₂ → Tautomerizes to CH₃-CH₂-CO-CH₃
Why ketone and not aldehyde?
Because H and OH add across triple bond following Markovnikov:
- H adds to terminal C (has more H already)
- OH adds to next C (more substituted)
Result: Ketone
For aldehyde: Would need anti-Markovnikov addition (not possible with H₂SO₄)
To get aldehyde: Use hydroboration-oxidation (anti-Markovnikov)
$$\text{CH}_3\text{-CH}_2\text{-C≡CH} \xrightarrow{\text{BH}_3/\text{H}_2\text{O}_2} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CHO}$$JEE Takeaway:
- H₂SO₄/HgSO₄ → Markovnikov → Ketone (from terminal alkyne)
- Hydroboration-oxidation → Anti-Markovnikov → Aldehyde
Quick Revision Table
| Property/Reaction | Details | Key Points |
|---|---|---|
| General Formula | CₙH₂ₙ₋₂ | Two H less than alkenes |
| Hybridization | sp | 50% s-character → acidic |
| Bond | 1σ + 2π | Triple bond |
| Acidity (pKa) | HC≡CH: 25 | Can form acetylide with NaNH₂ |
| Hydrogenation | + 2H₂ → Alkane | Pt/Pd/Ni catalyst |
| Partial reduction | Lindlar → cis | Na/NH₃ → trans |
| HX addition | Markovnikov | Geminal dihalide with excess |
| Hydration | H₂SO₄/HgSO₄ → Ketone | Via enol intermediate |
| Metal acetylide test | AgNO₃/NH₃ → ppt | Only terminal alkynes |
| Alkylation | HC≡C⁻ + R-X → R-C≡CH | Build longer chains |
Quick Decision Tree
Working with alkynes?
│
├─ Need to identify?
│ ├─ Br₂/CCl₄ → Decolorizes (unsaturated)
│ └─ AgNO₃/NH₃ → ppt (terminal) or no ppt (internal)
│
├─ Need alkene from alkyne?
│ ├─ Want cis → Lindlar's catalyst
│ └─ Want trans → Na/liquid NH₃
│
├─ Need carbonyl from alkyne?
│ ├─ Want ketone → H₂SO₄/HgSO₄ (Markovnikov)
│ └─ Want aldehyde → Hydroboration-oxidation (anti-Markovnikov)
│
└─ Need longer chain?
└─ Alkylation: 1. NaNH₂ 2. R-X
Connection to Other Topics
Prerequisites:
- Alkenes - Understanding double bonds and addition reactions
- Chemical Bonding - sp hybridization
- Acid-Base Chemistry - pKa and conjugate bases
Related Topics:
- Alkyl Halides - Alkylation reactions
- Carbonyl Compounds - Hydration products
- Reaction Mechanisms - Addition mechanisms
Applications:
- Polymers - PVC from acetylene
- Benzene Chemistry - Trimerization of acetylene
- Industrial Chemistry - Oxyacetylene welding
Teacher’s Summary
1. Alkynes have C≡C triple bond (1σ + 2π) - Formula: CₙH₂ₙ₋₂
2. Unique Property: Acidic Character
- Terminal alkynes (R-C≡C-H) are weakly acidic (pKa ≈ 25)
- sp hybridization → 50% s-character → stabilizes negative charge
- React with NaNH₂ to form acetylide ions (R-C≡C⁻)
3. Preparation Methods:
- CaC₂ + H₂O → Acetylene (industrial)
- Dehydrohalogenation: R-CHX-CHX-R + 2KOH → R-C≡C-R
- Alkylation: HC≡C⁻ + R-X → R-C≡CH (chain building!)
4. Addition Reactions - Two moles can add!
| Reagent | Product | Special Notes |
|---|---|---|
| H₂/Pt (excess) | Alkane | Complete reduction |
| H₂/Lindlar | cis-Alkene | Syn addition, poisoned catalyst |
| Na/NH₃ | trans-Alkene | Anti addition |
| HX (excess) | Geminal dihalide | Both X on same C (Markovnikov) |
| H₂O/H₂SO₄ | Ketone/Aldehyde | Via enol, Markovnikov |
| Hydroboration | Aldehyde | Anti-Markovnikov |
5. Tests for Terminal Alkynes:
- AgNO₃/NH₃ → White precipitate (silver acetylide)
- Cu₂Cl₂/NH₃ → Red precipitate (copper acetylide)
- Internal alkynes (R-C≡C-R) → No precipitate
6. JEE Strategy:
High-Yield Concepts:
- Acidic nature and acetylide formation (unique to alkynes!)
- Markovnikov addition → geminal dihalides
- Stereochemistry: Lindlar (cis) vs Na/NH₃ (trans)
- Multi-step synthesis using alkylation
Common Traps:
- Hydration gives carbonyl, NOT alcohol
- HX addition gives geminal, NOT vicinal dihalides
- Only NaNH₂ (not NaOH) can form acetylides
- Peroxide effect doesn’t work with alkynes
7. Key Difference from Alkenes:
- Can add TWO molecules of reagent
- Terminal alkynes are acidic (alkenes are not)
- Used in alkylation to build carbon chains
“Alkynes are the acid rebels of hydrocarbons - the only ones that can donate a proton!”
Master the acidic character and addition patterns, and you’ll ace alkyne questions! Next, explore benzene to see how three molecules of acetylene form aromatic rings!
Interactive Demo: Visualize Addition Reactions
Explore step-by-step mechanisms of electrophilic and nucleophilic addition reactions to alkynes.