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CBSE • कक्षा XI • Chemistry • अध्याय 9
अनुमानित समय: 45 Mins
प्रगति: अध्ययनरत

हाइड्रोकार्बन (Hydrocarbons)

In Class 11 Chemistry, "Hydrocarbons" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

🛢️ Have You Ever Wondered?

Why does natural gas (methane) burn with a clean, invisible blue flame while welding torches burn acetylene with blazing white heat hot enough to slice through solid steel? The chemistry of Alkanes, Alkenes, Alkynes, and Aromatic Hydrocarbons.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Chemistry, "Hydrocarbons" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Hydrocarbons from Class 10.
  • IUPAC naming from Chapter 8.
  • Combustion reactions.

इस अध्याय के लक्ष्य

  • Explain Conformations of Ethane: Sawhorse and Newman projections (Staggered vs Eclipsed stability).
  • Explain preparation and reactions of Alkenes: Markovnikov's Rule and Anti-Markovnikov's (Peroxide) Effect.
  • Analyze Alkynes: Acidic nature of terminal alkynes and cyclic polymerization to Benzene.
  • State Huckel's Rule of Aromaticity ($(4n + 2)\pi$ electrons).
  • Explain Electrophilic Aromatic Substitution mechanism in Benzene: Nitration, Sulphonation, Halogenation, and Friedel-Crafts Alkylation/Acylation.

अध्याय रूपरेखा एवं प्रगति

1 1. Conformations & Markovnikov's Ru...
2 2. Alkynes & Acidity
3 3. Huckel's Rule & Benzene Electrop...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Conformations & Markovnikov's Rule

  • Ethane Conformations: Rotation around $C-C$ single bond yields infinite spatial forms. Staggered is most stable (dihedral angle $60^\circ$, minimal torsional strain); Eclipsed is least stable ($0^\circ$, maximum electron repulsion).
  • Markovnikov's Rule: In the addition of an unsymmetrical reagent ($HX$) to an unsymmetrical alkene, the negative part adds to the double-bonded carbon containing fewer hydrogen atoms (e.g. $\text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \to \text{CH}_3\text{CH(Br)}\text{CH}_3$, 2-bromopropane!).
  • Peroxide Effect (Kharasch): With $\text{HBr}$ in the presence of organic peroxides, addition reverses to give 1-bromopropane!

2. Alkynes & Acidity

Terminal alkynes ($R-C\equiv CH$) are acidic because the $sp$-hybridized carbon has $50\%$ $s$-character, making it highly electronegative and allowing easy release of $H^+$ (reacts with sodium to release $H_2$ gas). Cyclic polymerization: $3\text{CH}\equiv\text{CH} \xrightarrow{\text{red-hot Fe}, 873\text{K}} \text{Benzene (C}_6\text{H}_6)$!

3. Huckel's Rule & Benzene Electrophilic Substitution

Huckel's Rule ($4n+2$): A cyclic, planar, fully conjugated ring system possessing $\mathbf{(4n + 2)\pi}$ electrons ($n=0, 1, 2\dots$; e.g. 6 in benzene, 10 in naphthalene) is extraordinarily stable (Aromatic!).
Electrophilic Substitution:
• Nitration: Conc. $\text{HNO}_3 + \text{H}_2\text{SO}_4$ generates nitronium electrophile $\text{NO}_2^+$.
• Friedel-Crafts Alkylation: $\text{CH}_3\text{Cl} + \text{anhyd. AlCl}_3$ generates $\text{CH}_3^+$, forming Toluene.

Hydrocarbons - Key Conceptual Architecture & Molecular Model

Hydrocarbons - Molecular Architecture Thermodynamic & Kinetic Foundations Equilibrium laws & state transformations Orbital & Electronic Mechanisms VSEPR, hybridization & MOT electron density Industrial Synthesis & Competitive Analysis CBSE board problem frameworks, JEE/NEET diagnostic applications & lab benchmarks

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
Newman Projections: Staggered conformation minimizing torsional strain by $12.5\text{ kJ/mol}$.
मुख्य बिंदु 2
Markovnikov Rule: Electrophilic addition guided by intermediate carbocation stability.
मुख्य बिंदु 3
Anti-Markovnikov Radical Addition: Peroxide effect operating exclusively with HBr via free-radical mechanism.
मुख्य बिंदु 4
Huckel's $(4n+2)$ Law: The quantum criterion governing aromatic planar electronic stability.
मुख्य बिंदु 5
Friedel-Crafts Synthesis: Anhydrous $\text{AlCl}_3$ catalyzed electrophilic carbon alkylation on benzene.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
State Markovnikov's Rule. Predict the major product formed when HBr is added to Propene.
उत्तर एवं व्याख्या देखें
उत्तर: Markovnikov's rule states that when an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the reagent attaches to that double-bonded carbon which carries the lesser number of hydrogen atoms. Propene ($ ext{CH}_3 ext{CH}= ext{CH}_2$) + $ ext{HBr}$ yields 2-Bromopropane ($ ext{CH}_3 ext{CH(Br)} ext{CH}_3$) as the major product via the more stable secondary carbocation.
Major product is 2-Bromopropane.
2
Why is the Staggered conformation of ethane more stable than the Eclipsed conformation?
उत्तर एवं व्याख्या देखें
उत्तर: In the staggered conformation, the $C-H$ bonds on adjacent carbon atoms are as far apart as possible (dihedral angle $60^\circ$), resulting in minimum electron-cloud repulsion (minimum torsional strain). In eclipsed ($0^\circ$), repulsions are maximized.
Minimal torsional strain and maximum spatial separation.
3
State Huckel's Rule of Aromaticity. Show that Cyclopentadienyl anion is aromatic.
उत्तर एवं व्याख्या देखें
उत्तर: Huckel's rule states that a cyclic, planar, fully conjugated system possessing $(4n + 2)\pi$ electrons (where $n = 0, 1, 2\dots$) exhibits aromatic stability. Cyclopentadienyl anion has a planar 5-membered conjugated ring with 2 double bonds (4 electrons) + 1 lone pair on carbanion (2 electrons) = $6\pi$ electrons ($n=1$), satisfying Huckel's rule.
Planar ring with (4n + 2) pi electrons = 6 pi electrons.
4
What is the electrophile in the Nitration of Benzene? Write the reaction for its generation.
उत्तर एवं व्याख्या देखें
उत्तर: The electrophile is the Nitronium ion ($ ext{NO}_2^+$). Generated by concentrated sulfuric acid protonating nitric acid: $ ext{HNO}_3 + 2 ext{H}_2 ext{SO}_4 ightleftharpoons ext{NO}_2^+ + ext{H}_3 ext{O}^+ + 2 ext{HSO}_4^-$.
Nitronium ion (NO2+).
5
Explain why terminal alkynes like Ethyne are acidic in nature, while ethene and ethane are not.
उत्तर एवं व्याख्या देखें
उत्तर: The carbon in ethyne is $sp$-hybridized with 50% $s$-character, making it significantly more electronegative than $sp^2$ (33%) in ethene or $sp^3$ (25%) in ethane. High electronegativity attracts shared $C-H$ electrons strongly, allowing easy release of proton ($H^+$).
sp-hybridized carbon has 50% s-character, making it highly electronegative.
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