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JEE Chemistry Focus Reading Time: 14 min 2 Key Comparisons

Acidity & Basicity in Hydrocarbons: From Alkynes to Arenes

Master the acidity trends in hydrocarbons and basicity differences in nitrogen heterocycles with mechanisms and JEE applications.

2
Key Comparisons
100%
JEE Relevance
4+
Mechanisms
18min
Practice Time

Why Acidity & Basicity Trends Matter in JEE

Understanding acidity and basicity in hydrocarbons is crucial for solving complex organic chemistry problems in JEE. These concepts appear in:

  • Reaction mechanism prediction and arrow-pushing
  • Compound stability and reactivity comparisons
  • Acid-base equilibrium problems
  • Synthesis planning and reagent selection
Concept 1 Essential

Acidity of Terminal Alkynes vs Alkanes/Alkenes

📊 Acidity Trend:

Terminal Alkyne > Alkane ≈ Alkene
pKₐ ≈ 25      pKₐ ≈ 50      pKₐ ≈ 44

🔬 Why Terminal Alkynes are Acidic:

1. Hybridization Effect:

• Terminal alkyne H is attached to sp-hybridized carbon (50% s-character)

• Alkane H is attached to sp³-hybridized carbon (25% s-character)

• Higher s-character → electrons closer to nucleus → more acidic H

2. Stability of Conjugate Base:

HC≡C:⁻ (Acetylide ion) vs R-CH₂⁻ (Carbanion)

• Acetylide ion is stabilized by sp-hybridization and triple bond

• Carbanions from alkanes are highly unstable

🎯 JEE Application Example:

Problem: Which hydrogen is most acidic and why?

CH₃-CH₃
CH₂=CH₂
HC≡CH

Solution: HC≡CH (terminal alkyne) is most acidic because:

• sp-hybridization (50% s-character)

• Conjugate base HC≡C:⁻ is stabilized by triple bond

• pKₐ ≈ 25 compared to ≈44-50 for others

Concept 2 Important

Basicity: Pyridine vs Pyrrole

📊 Basicity Trend:

Pyridine > Pyrrole
pKₐ ≈ 5.2      pKₐ ≈ 0.4

⚖️ Comparative Analysis:

Pyridine (More Basic):
  • Nitrogen is sp² hybridized
  • Lone pair in sp² orbital
  • Lone pair not involved in aromaticity
  • Available for protonation
  • pKₐ ≈ 5.2
N: (lone pair available)
Pyrrole (Less Basic):
  • Nitrogen is sp² hybridized
  • Lone pair in p orbital
  • Lone pair involved in aromatic sextet
  • Protonation destroys aromaticity
  • pKₐ ≈ 0.4
N: (lone pair in aromatic system)

🔬 Key Factor: Aromaticity Preservation

Pyridine Protonation:

• Lone pair not part of aromatic system

• Protonation doesn't affect aromaticity

• Conjugate acid remains aromatic

Pyrrole Protonation:

• Lone pair essential for aromatic sextet

• Protonation destroys aromaticity

• High energy penalty makes it less basic

🚀 Problem-Solving Strategies

For Acidity Comparisons:

  • Check hybridization (sp > sp² > sp³)
  • Look at conjugate base stability
  • Consider resonance and induction effects
  • Remember: pKₐ ↓ = acidity ↑

For Basicity Comparisons:

  • Check if lone pair is available
  • Consider aromaticity preservation
  • Look at hybridization effects
  • Remember: pKₐ ↑ = basicity ↑

💡 Memory Aid

Acidity Mnemonic

S.P.A.R.K.

S - s-character

P - Proton removal

A - Anion stability

R - Resonance effects

K - Kₐ value

Basicity Rule

A.L.P.

A - Available lone pair

L - Localized electrons

P - Preserve aromaticity

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. Arrange in decreasing acidity: CH₄, HC≡CH, NH₃, H₂O

2. Explain why pyrrole is less basic than pyridine but more acidic

3. Which is stronger base: aniline or ammonia? Justify

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