**Hydrocarbons** are organic compounds containing only carbon and hydrogen atoms. They are classified based on the types of carbon-carbon bonds present:
**Three Main Categories:**
1. **Saturated Hydrocarbons** - contain only C-C and C-H single bonds
2. **Unsaturated Hydrocarbons** - contain C=C (double bonds) or C≡C (triple bonds)
3. **Aromatic Hydrocarbons** - special cyclic compounds with delocalized π electrons
**Importance**: Hydrocarbons are primary sources of energy (LPG, CNG, petrol, diesel, kerosene), used in polymer manufacturing (polythene, polypropene), as solvents for paints, and as starting materials for dyes and drugs.
---
**Alkanes** are saturated open-chain hydrocarbons containing only C-C and C-H single bonds. Methane (CH₄) is the first member of this homologous series.
**General Formula**: **CₙH₂ₙ₊₂** (where n = number of carbon atoms)
**Structure of Methane (CH₄)**:
**Homologous Series Properties**:
---
#### IUPAC Nomenclature of Alkanes
**Rules for IUPAC Nomenclature**:
1. **Identify the longest continuous chain of carbon atoms** - this is the parent alkane
2. **Number the chain** from the end that gives substituents the lowest numbers
3. **Name substituents** (alkyl groups) with their position numbers
4. **Arrange substituents alphabetically** (ignoring prefixes di-, tri-, etc.)
5. **Combine**: [position-number: substituent name]-parent alkane
**Alkyl Groups** (General formula CₙH₂ₙ₊₁):
**Example Nomenclature**:
```
CH₃-CH₂-CH(CH₃)-CH₂-CH₃
Numbering: 1 2 3 4 5
IUPAC Name: 3-Methylpentane
```
#### Types of Structural Isomerism in Alkanes
**1. Chain Isomerism** - difference in carbon chain arrangement
**2. Carbon Atom Classification** (based on number of carbon atoms attached):
**Example**:
```
CH₃ (1°)
|
CH₃-CH₂-C-CH₃ (4°)
|
CH₃ (1°)
In 2,2-dimethylpropane: central carbon is quaternary
```
**Writing Structures from IUPAC Names**:
Steps:
1. Write the longest parent chain of carbons
2. Number the carbons from appropriate end
3. Attach substituents at correct positions
4. Satisfy valence by adding hydrogen atoms
**Example**: 3-ethyl-2-methylhexane
```
Step 1: C-C-C-C-C-C (hexane)
Step 2: C₁-C₂-C₃-C₄-C₅-C₆
Step 3: Attach CH₃ at C2, C₂H₅ at C3
Step 4: CH₃-CH(CH₃)-CH(C₂H₅)-CH₂-CH₂-CH₃
```
---
#### 1. From Unsaturated Hydrocarbons (Hydrogenation)
Dihydrogen gas adds to alkenes and alkynes in the presence of **catalysts** (Pt, Pd, or Ni) to form alkanes.
**Reactions**:
```
CH₂=CH₂ + H₂ --Pt/Pd/Ni--> CH₃-CH₃ (Ethane)
Ethene
CH₃-CH=CH₂ + H₂ --Pt/Pd/Ni--> CH₃-CH₂-CH₃ (Propane)
Propene
CH₃-C≡C-H + 2H₂ --Pt/Pd/Ni--> CH₃-CH₂-CH₃ (Propane)
Propyne
```
**Mechanism**: Hydrogen gas is adsorbed on catalyst surface and activates H-H bond, facilitating addition to C=C or C≡C bonds.
#### 2. From Alkyl Halides
**a) Reduction with Zn and dilute HCl**:
```
R-Cl + H₂ --Zn/H⁺--> R-H + HCl
Alkyl halide Alkane
```
Examples:
```
CH₃-Cl + H₂ --Zn/H⁺--> CH₄ + HCl
C₂H₅-Cl + H₂ --Zn/H⁺--> C₂H₆ + HCl
```
**Note**: Fluorides cannot be reduced by this method because C-F bond is too strong.
**b) Wurtz Reaction** - treatment with sodium metal in dry ethereal solution (gives alkanes with even number of carbon atoms):
```
R-Hal + Na --dry ether--> R-R + NaHal
(2 moles)
```
Examples:
```
2 CH₃-Br + 2Na --dry ether--> CH₃-CH₃ + 2NaBr
Bromomethane Ethane
2 C₂H₅-Br + 2Na --dry ether--> C₂H₅-C₂H₅ + 2NaBr
Bromoethane n-Butane
```
**Mechanism**: Alkyl carbanion (R⁻) is formed first, which acts as nucleophile, attacking another alkyl halide molecule.
**Mixed Wurtz Reaction**: Two different alkyl halides give mixture of three alkanes (not used for pure products).
#### 3. From Carboxylic Acids
**a) Decarboxylation (with soda lime)**:
```
R-COO⁻Na⁺ + NaOH --Δ,CaO--> R-H + Na₂CO₃
Sodium salt of Alkane
carboxylic acid
```
Product has **one less carbon atom** than the carboxylic acid.
Example:
```
CH₃-CH₂-COO⁻Na⁺ + NaOH --Δ,CaO--> CH₃-CH₃ + Na₂CO₃
Sodium propanoate Ethane
```
**Problem**: To prepare propane, use sodium salt of **butanoic acid**:
```
CH₃-CH₂-CH₂-COO⁻Na⁺ --heat,soda lime--> CH₃-CH₂-CH₃ + CO₂ + NaOH
```
**b) Kolbe's Electrolytic Method**:
```
2 R-COO⁻Na⁺ + 2H₂O --electrolysis--> R-R + 2CO₂ + H₂ + 2NaOH
Sodium salt of Alkane
carboxylic acid (even number of C)
```
**Example**:
```
2 CH₃-COO⁻Na⁺ + 2H₂O --electrolysis--> CH₃-CH₃ + 2CO₂ + H₂ + 2NaOH
Sodium acetate Ethane
```
**Mechanism** (at anode):
1. Acetate ion formation: 2CH₃COO⁻Na⁺ → 2CH₃-C(=O)-O⁻Na⁺
2. Loss of electron: 2CH₃-C(=O)-O⁻ - e⁻ → 2CH₃•C(=O)• (acetyl radical)
3. Decarboxylation: 2CH₃•C(=O)• → 2CH₃• + 2CO₂
4. Coupling: 2CH₃• → CH₃-CH₃ (ethane)
At cathode: 2H⁺ + 2e⁻ → H₂
**Note**: Methane cannot be prepared by Kolbe's method because methanoate (formate) doesn't have R group for coupling.
**Industrial Source**: Petroleum and natural gas are the main sources of alkanes.
---
**General Characteristics**:
**Physical State at 298 K**:
**Boiling Point Trends**:
```
Methane (CH₄): 111.5 K
Ethane (C₂H₆): 184.5 K
Propane (C₃H₈): 231.0 K
Butane (C₄H₁₀): 272.6 K
```
**Factors Affecting Boiling Point**:
1. **Molecular Mass**: b.p. increases with increase in molecular mass due to increased van der Waals forces
2. **Surface Area/Branching**: Isomers with continuous chains have higher b.p. than branched isomers
```
Pentane (n-pentane): 309.1 K (continuous chain)
2-Methylbutane (isopentane): 301.0 K
2,2-Dimethylpropane (neopentane): 282.5 K (most branched)
```
**Reason**: Branched molecules are more spherical, reducing contact surface area, leading to weaker intermolecular forces and lower boiling points.
3. **Melting Point**: Increases with molecular weight, but less regular trend than b.p. due to crystal packing efficiency.
**Practical Applications**:
---
**General Reactivity**: Alkanes are relatively **inert** under normal conditions because:
They do not react with:
**Reactions under specific conditions**:
#### 1. Substitution Reactions (Free Radical Mechanism)
**Halogenation**: One or more hydrogen atoms replaced by halogens (F, Cl, Br, I)
**Chlorination of Methane**:
```
CH₄ + Cl₂ --hν--> CH₃Cl + HCl (Chloromethane/Methyl chloride)
CH₃Cl + Cl₂ --hν--> CH₂Cl₂ + HCl (Dichloromethane)
CH₂Cl₂ + Cl₂ --hν--> CHCl₃ + HCl (Trichloromethane/Chloroform)
CHCl₃ + Cl₂ --hν--> CCl₄ + HCl (Tetrachloromethane/Carbon tetrachloride)
```
**Chlorination of Ethane**:
```
C₂H₆ + Cl₂ --hν--> CH₃-CH₂Cl + HCl
Ethane Chloroethane
```
**Conditions Required**:
**Reactivity Order of Halogens**:
```
F₂ > Cl₂ > Br₂ > I₂
```
**Iodination with Oxidizing Agent**:
```
CH₄ + I₂ --HIO₃--> CH₃I + HI
HIO₃ + 5HI --> 3I₂ + 3H₂O (I₂ regenerated)
```
**Reactivity of Different Hydrogen Atoms**:
```
Relative reactivity: Primary (1°) : Secondary (2°) : Tertiary (3°) = 1 : 1.2 : 1.5 (approximately 1:2:3 for bromination)
```
This is because tertiary radicals are more stable than secondary, which are more stable than primary radicals (due to hyperconjugation and inductive effects).
**Free Radical Chain Mechanism** (for chlorination of methane):
**Step 1: Initiation** (formation of free radicals):
```
Cl₂ --hν--> 2Cl• (chlorine free radical)
```
**Step 2: Propagation** (chain reaction):
```
a) Cl• + CH₄ --> CH₃• + HCl
(chlorine radical abstracts H from methane)
b) CH₃• + Cl₂ --> CH₃Cl + Cl•
(methyl radical attacks Cl₂)
```
Steps 2a and 2b repeat many times, making it a chain reaction.
**Step 3: Termination** (destruction of free radicals):
```
a) CH₃• + Cl• --> CH₃Cl
b) Cl• + Cl• --> Cl₂
c) CH₃• + CH₃• --> CH₃-CH₃
```
**Note**:
#### 2. Other Reactions
**Combustion**: (Exothermic reaction)
```
CₙH₂ₙ₊₂ + [(3n+1)/2]O₂ --> nCO₂ + (n+1)H₂O
CH₄ + 2O₂ --> CO₂ + 2H₂O (ΔH = -890.7 kJ/mol)
```
**Oxidation** (at high temperature or with oxidizing agents):
**Cracking** (thermal decomposition):
**Isomerization**: Alkanes with branched chains convert to unbranched forms or vice versa in presence of catalysts and heat.
---
**[Detailed content would continue with Alkenes section covering structure, nomenclature, preparation, physical/chemical properties, addition reactions, and mechanisms per NCERT requirements]**
**General Formula**: **CₙH₂ₙ**
**Key Characteristics**:
**IUPAC Naming Rules**:
1. Identify longest carbon chain containing **C=C double bond**
2. Number chain from end that gives **lowest number to C=C**
3. Indicate double bond position by number of first carbon of double bond
4. Use suffix **-ene** instead of -ane
5. Position of substituents: Number from end giving lowest numbers
**Examples**:
```
CH₂=CH-CH₃ : Prop-1-ene (Propylene)
CH₃-CH=CH-CH₃ : But-2-ene
CH₂=C(CH₃)₂ : 2-Methylprop-1-ene (Isobutylene)
```
**Common Names**:
**1. Chain Isomerism**: Different carbon skeleton
```
But-1-ene: CH₂=CH-CH₂-CH₃
2-Methylprop-1-ene: CH₂=C(CH₃)₂
```
**2. Position Isomerism**: Double bond at different positions
```
Pent-1-ene: CH₂=CH-CH₂-CH₂-CH₃
Pent-2-ene: CH₃-CH=CH-CH₂-CH₃
```
**3. Geometric (cis-trans) Isomerism**: Due to restricted rotation around C=C bond
```
But-2-ene exists as:
(Z)-But-2-ene (cis): CH₃-CH=CH-CH₃ (both CH₃ on same side)
(E)-But-2-ene (trans): CH₃-CH=CH-CH₃ (CH₃ groups on opposite sides)
```
**Condition for Geometric Isomerism**: Each carbon of C=C must have **two different groups attached**.
```
Pent-2-ene shows geometric isomerism
Pent-1-ene does NOT (one carbon has two H atoms)
```
#### 1. Dehydration of Alcohols
```
R-CH₂-CH(OH)-R' --H₂SO₄, Δ--> R-CH=CH-R' + H₂O
Alcohol Alkene
```
**Example**:
```
CH₃-CH₂-OH --conc. H₂SO₄, 170°C--> CH₂=CH₂ + H₂O
Ethanol Ethene
```
**Mechanism** (E1 or E2 depending on conditions):
**Zaitsev's Rule**: When multiple alkenes can form, the **more substituted alkene** (more stable) is the major product.
```
(CH₃)₂CH-CH₂-OH --H₂SO₄, Δ--> (CH₃)₂C=CH₂ (Major, more substituted)
+ CH₃-CH=CH-CH₃ (Minor)
```
#### 2. Dehydrohalogenation of Alkyl Halides
```
R-CHX-CH₂-R' + KOH/alcoholic --> R-CH=CH-R' + KX + H₂O
Alkyl halide Alkene
```
**Example**:
```
CH₃-CHBr-CH₃ + KOH(alc.) --Δ--> CH₂=CH-CH₃ + KBr + H₂O
2-Bromopropane Propene
```
**Conditions**:
#### 3. From Alkynes (Partial Reduction)
```
RC≡CR' + H₂ --Lindlar's catalyst--> R-CH=CH-R' (cis-alkene)
Alkyne (Z-isomer)
```
**Note**: Lindlar's catalyst is Pd/CaCO₃ (selective reduction of triple bond to double bond only).
**Alkenes are much more reactive than alkanes** due to presence of loosely held π electrons in C=C bond, which are easily accessible to electrophiles.
#### 1. Addition Reactions (Most Important)
The π bond of C=C is broken and two atoms/groups add across the double bond.
**General Reaction**:
```
R-CH=CH-R' + A-B --> R-CH(A)-CH(B)-R'
Alkene Reagent Addition product
```
**a) Addition of Hydrogen (Hydrogenation)**:
```
CH₂=CH₂ + H₂ --Pt/Pd/Ni, Δ--> CH₃-CH₃
Ethene Ethane
```
**b) Addition of Halogens**:
```
CH₂=CH₂ + Br₂ --CCl₄--> CH₂Br-CH₂Br
Ethene 1,2-Dibromoethane
(yellow/orange color disappears)
```
**Test for Unsaturation**: Bromine water decolorizes in presence of alkenes (brown color disappears to colorless).
**Mechanism**: Electrophilic addition (formation of cyclic bromonium ion, then Br⁻ attacks)
**c) Addition of Hydrogen Halides (HX = HCl, HBr, HI)**:
**Markovnikov's Rule**: When an unsymmetrical alkene reacts with HX, the hydrogen of HX adds to the carbon having **more hydrogen atoms** (more substituted carbon of C=C), and halogen adds to the **less hydrogenated carbon**.
```
CH₃-CH=CH₂ + HBr --> CH₃-CHBr-CH₃ (Markovnikov product)
Propene 2-Bromopropane (Major, ~90%)
+ CH₃-CH₂-CH₂Br (Anti-Markovnikov, ~10%)
1-Bromopropane (Minor)
```
**Mechanism** (Electrophilic Addition - E.A.):
```
Step 1: Protonation of alkene (formation of carbocation)
CH₃-CH=CH₂ + H⁺ --> CH₃-CH⁺-CH₃ (secondary carbocation, more stable)
+ CH₃-CH₂-CH₂⁺ (primary carbocation, less stable)
Step 2: Attack of halide nucleophile
CH₃-CH⁺-CH₃ + Br⁻ --> CH₃-CHBr-CH₃
```
**Carbocation Stability Order**:
```
Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (CH₃⁺)
Stability order: 3° > 2° > 1° > CH₃⁺
```
**Rearrangement**: If primary or secondary carbocation formed can rearrange to more stable tertiary or secondary carbocation, rearrangement occurs.
```
(CH₃)₂C=CH₂ + HCl --> (CH₃)₃C-Cl (tertiary, Markovnikov)
But if rearrangement possible:
H Cl
| |
C - C - with hydride or methyl shift
```
**d) Addition of Sulfuric Acid (H₂SO₄)**:
```
CH₃-CH=CH₂ + H₂SO₄ --25°C--> CH₃-CH(HSO₄)-CH₃
Propene (intermediate, unstable)
Then hydrolysis:
CH₃-CH(HSO₄)-CH₃ + H₂O --> CH₃-CH(OH)-CH₃ + H₂SO₄
Propan-2-ol (secondary alcohol)
```
**e) Addition of Water (Hydration)**:
```
R-CH=CH₂ + H₂O --H⁺ (catalyst)--> R-CH(OH)-CH₃
Alkene Alcohol (secondary)
```
Follows **Markovnikov's rule**: OH adds to more substituted carbon.
```
CH₃-CH=CH₂ + H₂O --H₂SO₄--> CH₃-CH(OH)-CH₃ (Major)
Propene Propan-2-ol (secondary)
+ CH₃-CH₂-CH₂OH (Minor)
Propan-1-ol (primary)
```
#### 2. Oxidation Reactions
**a) Oxidation with Dilute Permanganate (KMnO₄)**:
```
CH₂=CH₂ + KMnO₄ (dilute) --cold, neutral--> CH₂OH-CH₂OH
Ethene 1,2-Ethanediol (Ethylene glycol)
Or: [O]
CH₃-CH=CH-CH₃ --KMnO₄, Δ--> CH₃-CHOH-CHOH-CH₃
But-2-ene Butane-2,3-diol
```
**Brown color of KMnO₄** is discharged (disappears), producing **colorless or pale pink solution** with brown MnO₂ precipitate.
**Mechanism**: Electrophilic addition of MnO₄⁻, formation of cyclic manganate ester intermediate, hydrolysis produces diol.
**b) Oxidation with Hot/Dilute Permanganate**:
```
CH₃-CH=CH-CH₃ + KMnO₄ (hot, conc.) --Δ--> CH₃COOH + CH₃COOH
But-2-ene Acetic acid (mixture)
CH₃-CH=CH₂ + KMnO₄ --Δ--> CH₃COOH + HCOOH
Propene (mixture of carboxylic acids)
```
C=C bond is broken at the double bond; products depend on substitution pattern.
**c) Oxidation with Ozone (Ozonolysis)**:
```
R-CH=CH-R' + O₃ --> R-CH-O-O-CH-R' (Primary ozonide)
\ /
O
Then: Reduction with Zn/H₂O:
R-CHO + O=CHR' (aldehydes/ketones)
Or: Oxidation (with H₂O₂):
R-COOH + HOOC-R' (carboxylic acids)
```
**Application**: Ozonolysis used to **determine structure** of alkenes (breaking at C=C, identifying carbonyl products).
#### 3. Polymerization (Addition Polymerization)
```
nCH₂=CH₂ --catalyst, Δ, pressure--> (-CH₂-CH₂-)ₙ
Ethene Polyethene (Polythene)
nCH₃-CH=CH₂ --catalyst--> (-CH₃-CH=CH-)ₙ or (-CH(CH₃)-CH₂-)ₙ
Propene Polypropene (Polypropylene)
```
Industrial production of plastics and polymers.
**Conditions**:
---
**General Formula**: **CₙH₂
Q1. Which of the following is the correct IUPAC name for the compound with structure CH₃–CH(CH₃)–CH₂–CH₃?
Answer: A — The longest chain contains 4 carbons (butane); the methyl branch is at position 2 from the nearest end, giving 2-methylbutane.
Q2. The general molecular formula for alkanes is CₙH₂ₙ₊₂. What is the formula for an alkane with 8 carbon atoms?
Answer: B — Using CₙH₂ₙ₊₂ with n = 8: C₈H₂(₈)₊₂ = C₈H₁₈.
Q3. Which statement correctly describes the difference between chain isomers and positional isomers?
Answer: B — Chain isomers result from different ways to arrange carbon atoms (e.g., n-butane vs isobutane); positional isomers have the same skeleton but different functional group locations.
Q4. A carbon atom bonded to exactly two other carbon atoms in an alkane is classified as:
Answer: B — A secondary (2°) carbon is attached to exactly two carbon atoms; it is not a terminal carbon.
Q5. Consider the structural isomers of C₆H₁₄. How many chain isomers are possible?
Answer: C — The five chain isomers of C₆H₁₄ are: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane.
Q6. Assertion: Alkanes are unreactive under normal conditions. Reason: Alkanes contain only single σ bonds, which are very strong and require high energy to break.
Answer: A — Alkanes are inert (paraffinic) because their C–C and C–H σ bonds are very strong, requiring high activation energy for reaction.
Q7. Which of the following is NOT a correct characteristic of alkanes?
Answer: A — Alkanes do NOT react with dilute acids under normal conditions; this is why they are classified as inert or paraffinic.
Q8. The bond angle in methane (CH₄) is approximately 109.5°. This is best explained by:
Answer: C — The 109.5° angle in methane arises from sp³ hybridisation of carbon, resulting in a tetrahedral geometry as predicted by VSEPR theory.
Q9. Among the following, identify the pair that represents chain isomers: (i) CH₃CH₂CH₂CH₃ and CH₃CH(CH₃)CH₃ (ii) n-pentane and 2-pentene. Which is correct?
Answer: A — Pair (i) both have formula C₄H₁₀ with different carbon skeletons (chain isomers). Pair (ii) are n-pentane (C₅H₁₂) and 2-pentene (C₅H₁₀), different formulas—not isomers.
Q10. Given that the C–H bond length in alkanes is 112 pm and the C–C bond length is 154 pm, which inference can be drawn about the relative strength of these bonds?
Answer: C — Shorter bond length generally correlates with higher bond dissociation energy; thus, the 112 pm C–H bond is stronger than the 154 pm C–C bond.
What is the general molecular formula for alkanes?
CₙH₂ₙ₊₂, where n is the number of carbon atoms.
Define chain isomers with one example.
Structural isomers that differ in the arrangement of the carbon chain skeleton; example: n-butane and 2-methylpropane both have formula C₄H₁₀.
What is the bond angle in a tetrahedral alkane molecule like methane?
109.5° (degrees), determined by VSEPR theory for sp³ hybridised carbon.
Why are alkanes called 'paraffins'?
From Latin 'parum affinis' (little affinity) because alkanes are unreactive with acids, bases, and most reagents under normal conditions.
Distinguish between primary and tertiary carbon atoms.
Primary carbon is bonded to one other carbon atom (or none, as in methane); tertiary carbon is bonded to three other carbon atoms.
What does LPG stand for and what is its source?
Liquified Petroleum Gas; it is obtained from petroleum and natural gas deposits and used as a domestic fuel with minimal pollution.
Name the three main classes of hydrocarbons based on bonding.
Saturated (single bonds only), unsaturated (double or triple bonds), and aromatic (cyclic with delocalised electrons).
What are the bond lengths of C–C and C–H in alkanes?
C–C bond: 154 pm; C–H bond: 112 pm.
How many structural isomers does C₅H₁₂ have?
Three chain isomers: n-pentane, 2-methylbutane, and 2,2-dimethylpropane.
What orbitals overlap to form C–C and C–H bonds in alkanes?
sp³ hybrid orbitals of carbon atoms overlap with each other (for C–C) or with 1s orbitals of hydrogen (for C–H) via head-on σ bonding.
Define chain isomers and write the structures and IUPAC names of all chain isomers of C₅H₁₂. [2 marks]
Chain isomers differ in the arrangement of the carbon skeleton. Draw three possible arrangements of 5 carbons and identify longest chains; assign position numbers to any methyl branches.
Explain why alkanes are called 'paraffins' and justify their unreactivity under normal conditions using concepts of bonding. What type of bond is present between carbon atoms in alkanes, and why does this contribute to their low reactivity? [5 marks]
Paraffins = low affinity (from Latin). Discuss σ bonds formed by sp³ hybridisation, high bond dissociation energy, and lack of polar character. Show that strong σ bonds require high activation energy to break.
For the molecular formula C₆H₁₄, identify and draw all possible chain isomers. For each isomer, classify each carbon atom as primary (1°), secondary (2°), tertiary (3°), or quaternary (4°). Explain why branched isomers have lower boiling points than the straight-chain isomer despite having the same molecular formula. [6 marks]
Five chain isomers exist for C₆H₁₄. For each structure, count carbons bonded to each carbon atom to classify. Lower boiling points in branched isomers result from decreased surface area and weaker intermolecular van der Waals forces compared to extended linear chains.
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