Noesis
4.2.7Hydrocarbons

Alkynes — preparation, acidity of terminal alkynes, addition reactions, hydration to ketones

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🇮🇳 Yeh note Hinglish mein padho →

1. What is an alkyne? (WHAT)

The triple-bond carbon is ==spsp hybridised==, bond angle 180∘180^\circ, linear geometry. The C≡CC\equiv C bond length (120120 pm) is shorter than C=CC=C (134134 pm) and C−CC-C (154154 pm).

2. Preparation (HOW)

2.1 From calcium carbide (industrial ethyne)

CaC2+2H2O⟶C2H2↑+Ca(OH)2CaC_2 + 2H_2O \longrightarrow C_2H_2\uparrow + Ca(OH)_2 Why? The carbide ion C22−C_2^{2-} is the conjugate base of ethyne; water (a stronger acid) protonates it twice.

2.2 By dehydrohalogenation of vicinal / geminal dihalides

A vic-dihalide loses two HX (needs two eliminations → needs strong base): CH3−CHBr−CH2Brvic-dihalide→ΔKOH (alc.)CH3−CBr=CH2→ΔNaNH2CH3−C≡CH\underset{\text{vic-dihalide}}{CH_3-CHBr-CH_2Br} \xrightarrow[\Delta]{\text{KOH (alc.)}} CH_3-CBr=CH_2 \xrightarrow[\Delta]{NaNH_2} CH_3-C\equiv CH

2.3 By alkylation of a terminal alkyne (builds bigger alkynes)

HC≡CH→NaNH2HC≡C−Na+→R−XHC≡C−RHC\equiv CH \xrightarrow{NaNH_2} HC\equiv C^-Na^+ \xrightarrow{R-X} HC\equiv C-R Why this works: The acetylide ion is a strong nucleophile that does SN2S_N2 on a primary alkyl halide. (Secondary/tertiary R–X → elimination dominates instead.)

3. Acidity of terminal alkynes (the WHY star of the chapter)

4. Electrophilic addition reactions (HOW)

Two π\pi bonds → can add one or two equivalents of reagent.

4.1 Addition of HX (Markovnikov)

CH3−C≡CH→HBrCH3−CBr=CH2→HBrCH3−CBr2−CH3CH_3-C\equiv CH \xrightarrow{HBr} CH_3-CBr=CH_2 \xrightarrow{HBr} CH_3-CBr_2-CH_3 Why Markovnikov? H⁺ adds to give the more stable (secondary) vinyl cation; X⁻ then adds to the cationic carbon. So Br ends up on the more substituted carbon.

4.2 Addition of H2OH_2O → see hydration (Section 5).

4.3 Catalytic hydrogenation (control the product)

R−C≡C−R→(Lindlar)H2/Pd−CaCO3R−CH=CH−RcisR-C\equiv C-R \xrightarrow[\text{(Lindlar)}]{H_2/Pd-CaCO_3} \underset{cis}{R-CH=CH-R} R−C≡C−R→Na/liq. NH3R−CH=CH−RtransR-C\equiv C-R \xrightarrow{Na/liq.\,NH_3} \underset{trans}{R-CH=CH-R} Why? Lindlar (poisoned Pd) delivers both H's from the same face → cis. Dissolving-metal reduction goes through a trans-radical anion → trans alkene.

Figure — Alkynes — preparation, acidity of terminal alkynes, addition reactions, hydration to ketones

5. Hydration to ketones (the headline reaction)

5.1 Markovnikov hydration (H2O/H2SO4/HgSO4H_2O / H_2SO_4 / HgSO_4)

R−C≡CH→H2SO4, HgSO4H2O[R−COH=CH2]→tautomeriseR−C∥O−CH3R-C\equiv CH \xrightarrow[H_2SO_4,\ HgSO_4]{H_2O} \big[R-\underset{OH}{C}=CH_2\big] \xrightarrow{\text{tautomerise}} R-\overset{O}{\overset{\|}{C}}-CH_3

Why a ketone (not aldehyde)? OH adds Markovnikov → onto the more substituted carbon → enol carbon bearing R → on tautomerisation gives a methyl ketone.

5.2 Keto–enol tautomerism (the rearrangement)

C=C−OHenol⇌C∥O−C−Hketo\underset{\text{enol}}{C=C-OH}\rightleftharpoons \underset{\text{keto}}{\overset{O}{\overset{\|}{C}}-C-H} Why keto is favoured: the C=OC=O π\pi bond (≈ 360\,360 kJ/mol) is much stronger than C=CC=C (≈ 270\,270) + the O−HO-H vs C−HC-H swap; net energy drops → keto dominates.

6. Quick recall

Recall Active recall — cover the right side
  • General formula of alkyne ::: CnH2n−2C_nH_{2n-2}
  • Hybridisation & geometry ::: spsp, linear, 180∘180^\circ
  • Why terminal ≡C−H\equiv C-H acidic ::: 50% s-character → anion held close to nucleus → stable
  • Acidity order ::: sp>sp2>sp3sp > sp^2 > sp^3 (ethyne > ethene > ethane)
  • Base to make acetylide ::: NaNH2NaNH_2 (not NaOH)
  • Markovnikov hydration product ::: methyl ketone (ethyne → acetaldehyde)
  • Lindlar gives ::: cis alkene; Na/NH₃ gives ::: trans alkene
Recall Feynman: explain to a 12-year-old

Imagine three rubber bands joining two atoms — that's the triple bond, packed with extra electrons. Those electrons are like sticky candy; anything "positive" wants to grab them, so alkynes react eagerly. Also, the H at the very end is loosely held — like a hat that blows off in a strong wind. A very strong wind (NaNH2NaNH_2) takes the hat; a weak breeze (NaOHNaOH) can't. When you splash water on the triple bond with a special mercury helper, it first makes a wobbly "enol" that instantly flips into a steady ketone, because the new oxygen double bond is super strong and comfy.

Connections

  • Alkenes — electrophilic addition (alkynes add twice, same logic)
  • Hybridisation and s-character (root of acidity)
  • Keto–enol tautomerism (drives hydration product)
  • Aldehydes and Ketones — carbonyl chemistry (the products)
  • SN2 reactions (acetylide alkylation)
  • Markovnikov rule and carbocation stability
General formula of an open-chain alkyne
CnH2n−2C_nH_{2n-2}
Hybridisation and bond angle at triple-bond carbon
spsp, 180∘180^\circ, linear
Why is a terminal alkyne C–H acidic?
Conjugate-base lone pair is in an spsp orbital (50% s-character), held close to nucleus → stable carbanion
Order of acidity: ethyne, ethene, ethane
ethyne (spsp) > ethene (sp2sp^2) > ethane (sp3sp^3)
Why does NaOH fail to deprotonate ethyne but NaNH₂ works?
pKapK_a(water)=15.7 < pKapK_a(ethyne)=25 < pKapK_a(NH₃)=38; only a base whose conjugate acid is weaker (NH₃) deprotonates
Reagents for Markovnikov hydration of an alkyne
H2OH_2O, dil. H2SO4H_2SO_4, HgSO4HgSO_4 catalyst
Product of Markovnikov hydration of propyne
propan-2-one (a methyl ketone)
Why does ethyne give an aldehyde, not a ketone, on hydration
Both carbons equivalent; enol can only tautomerise to acetaldehyde
Product of anti-Markovnikov (hydroboration–oxidation) of a terminal alkyne
an aldehyde (RCH2CHORCH_2CHO)
Lindlar's catalyst converts an alkyne to
a cis-alkene
Na in liquid NH₃ converts an internal alkyne to
a trans-alkene
Test reagent for terminal alkynes
Ammoniacal AgNO3AgNO_3 (white ppt) or ammoniacal CuClCuCl (red ppt)
Two HX needed: which base for the 2nd elimination to form the triple bond
NaNH2NaNH_2 (sodium amide)
What intermediate forms first in alkyne hydration before the carbonyl
an enol (C=C−OHC=C-OH)

Concept Map

has

two types

two types

two weak pi bonds

attack electrophiles

acidic equiv C-H

explained by

stabilises

needs stronger base than OH-

acetylide + primary R-X SN2

second elimination

preparation

hydration

Alkyne CnH2n-2 triple bond

sp carbon linear 180

Terminal alkyne

Internal alkyne no acidic H

Pi electrons nucleophile-rich

Addition reactions

Terminal H acidity

sp orbital 50% s-character

Acetylide carbanion

NaNH2 deprotonates

Alkylation builds bigger alkynes

Dehydrohalogenation of dihalides

Ketones

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, alkyne ka funda do baaton pe tikta hai. Pehla: triple bond (C≡CC\equiv C) me do π\pi bonds hote hain jo electron se bharpoor hain, isliye koi bhi "positive" cheez (electrophile, jaise H+H^+) inhe attack karti hai — bilkul alkene jaisa, bas do baar. Doosra: terminal alkyne ka end wala ≡C−H\equiv C-H thoda acidic hota hai. Kyun? Kyunki H nikalne ke baad jo carbanion banta hai, uska lone pair spsp orbital me baithta hai jisme 50% s-character hota hai — yaani electron nucleus ke paas, tightly held, stable anion. Isliye acidity order: ethyne > ethene > ethane. Yaad rakho — zyada s-character matlab zyada acidic.

Ab base ka chakkar: NaOH se ethyne deprotonate nahi hota, kyunki paani (pKa 15.7) ethyne (pKa 25) se zyada acidic hai. Acetylide banane ke liye chahiye strong base NaNH2NaNH_2 (kyunki NH3NH_3 ka pKa 38 hai — ethyne se zyada). Ye acetylide ek powerful nucleophile hai jo primary alkyl halide pe SN2S_N2 karke bada alkyne bana deta hai — yeh ek important prep hai.

Hydration sabse important reaction hai. Jab H2OH_2O, dil. H2SO4H_2SO_4 aur HgSO4HgSO_4 ke saath water add karte ho, to OH Markovnikov rule se zyada substituted carbon pe lagta hai, pehle ek enol (C=C−OHC=C-OH) banta hai jo unstable hai, aur turant tautomerise hoke ketone ban jata hai. Sirf ethyne exception hai — wo acetaldehyde (CH3CHOCH_3CHO) deta hai kyunki dono carbon same hain. Yaad rakho contrast: hydroboration–oxidation (anti-Markovnikov) terminal alkyne ko aldehyde deta hai. Toh "Mark makes Ketones, Boron makes Aldehydes". Reduction me Lindlar = cis, Na/liq. NH₃ = trans. Bas yeh patterns pakad lo, exam me sab questions inhi se ghoom ke aayenge.

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Connections

Aldehydes and ketones — preparation; nucleophilic addition; aldol, Cannizzaro, Wittig, Claisen-Schmidt, Mannich, ReformaChemistry · 4.3.7