JEE Main & Advanced

Hydrocarbons

Hydrocarbons for JEE Main & Advanced

1
Module 1

Alkanes and Alkenes

Alkanes — Preparation, Properties, ReactionsTopic 1

Alkanes: Saturated hydrocarbons, general formula $C_nH_{2n+2}$. $sp^3$ hybridized C. Single bonds only.

Examples: $CH_4$ methane, $C_2H_6$ ethane, $C_3H_8$ propane, $C_4H_{10}$ butane.

Preparation Methods:

1. From Unsaturated Hydrocarbons (Hydrogenation): $$CH_2=CH_2 + H_2 \xrightarrow{\text{Ni}/\text{Pt}/\text{Pd}, \Delta} CH_3CH_3$$

  • Catalyst: Ni (most common), Pt, Pd
  • Heterogeneous catalysis

2. From Alkyl Halides:

  • Wurtz Reaction: $2RX + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$
  • Two molecules of alkyl halide; Na in dry ether
  • Best with primary RX; gives symmetrical alkanes
  • $2CH_3CH_2Br + 2Na \to CH_3CH_2CH_2CH_3 + 2NaBr$ (butane)
  • Reduction with Zn/HCl: $RX + 2[H] \to RH + HX$

3. From Carboxylic Acids (Decarboxylation):

  • Sodium salt heated with NaOH/CaO (soda-lime):

$$RCOONa + NaOH \xrightarrow{CaO, \Delta} RH + Na_2CO_3$$

  • e.g., $CH_3COONa + NaOH \to CH_4 + Na_2CO_3$
  • Kolbe's electrolysis: Electrolysis of sodium carboxylate solution gives alkanes:

$2RCOO^- \to R-R + 2CO_2 + 2e^-$ (anode)

Physical Properties:

  • C1-C4: gases at RT; C5-C17: liquids; C18 and above: solids
  • BP increases with chain length (more London forces)
  • Branched alkanes have lower BP than straight chain (less surface area)
  • Insoluble in water; soluble in non-polar solvents
  • Density < $1$ g/mL

Chemical Properties:

1. Combustion: $$C_nH_{2n+2} + \frac{3n+1}{2}O_2 \to nCO_2 + (n+1)H_2O$$

  • Highly exothermic
  • Major use as fuel

2. Halogenation (Free Radical Substitution): $$CH_4 + Cl_2 \xrightarrow{h\nu \text{ or } \Delta} CH_3Cl + HCl$$

  • Photochemical or thermal initiation
  • Further substitution: $CH_3Cl \to CH_2Cl_2 \to CHCl_3 \to CCl_4$
  • Order of reactivity of halogens: F₂ > Cl₂ > Br₂ > I₂
  • Selectivity: $3° H > 2° H > 1° H > CH_3$ (radical stability)
  • F is too reactive (explosive); I doesn't react (endothermic)

Mechanism (3 steps):

  • Initiation: $Cl_2 \xrightarrow{h\nu} 2Cl^\bullet$
  • Propagation:
  • $CH_4 + Cl^\bullet \to CH_3^\bullet + HCl$
  • $CH_3^\bullet + Cl_2 \to CH_3Cl + Cl^\bullet$
  • Termination: $Cl^\bullet + Cl^\bullet \to Cl_2$; $CH_3^\bullet + Cl^\bullet \to CH_3Cl$; etc.

3. Pyrolysis (Cracking):

  • $C_{10}H_{22} \to C_5H_{12} + C_5H_{10}$ (mix of smaller alkanes and alkenes)
  • Used in petroleum cracking

4. Isomerization:

  • $n$-Butane $\to$ isobutane (anhydrous $AlCl_3 + HCl$, heat)
  • Used in petroleum refining

5. Aromatization:

  • Higher alkanes ($C_6+$) heated over $Cr_2O_3/V_2O_5$ on alumina at ~$770$ K, 10-20 atm → aromatic compounds
  • e.g., $n$-hexane → benzene

6. Reaction with steam:

  • $CH_4 + H_2O \xrightarrow{Ni, 1273K} CO + 3H_2$ (syn gas — used for methanol)
Worked Examples
1

Identify product of Wurtz reaction with $C_2H_5Br$.

Show solution

$2C_2H_5Br + 2Na \xrightarrow{ether} C_2H_5-C_2H_5 + 2NaBr$ $= CH_3CH_2CH_2CH_3$ (n-butane)

Final Answer: n-Butane.

2

Predict product when soda-lime is heated with $CH_3CH_2COONa$.

Show solution

$CH_3CH_2COONa + NaOH \xrightarrow{CaO, \Delta} CH_3CH_3 + Na_2CO_3$

Final Answer: Ethane.

✎ Self-Check — 5 questions0 / 5
Q1.

General formula of alkanes:

Q2.

Wurtz reaction is for:

Q3.

Chlorination of methane is initiated by:

Q4.

Selectivity of halogenation: $3°$ vs $1°$ H:

Q5.

Cracking is used in:

Alkenes — Preparation and Electrophilic AdditionTopic 2

Alkenes: Unsaturated hydrocarbons with one C=C; general formula $C_nH_{2n}$. $sp^2$ C; planar around C=C.

Examples: $CH_2=CH_2$ ethene, $CH_3CH=CH_2$ propene.

Preparation Methods:

1. From Alkyl Halides (Dehydrohalogenation): $$RCH_2CH_2X \xrightarrow{\text{alc. KOH}, \Delta} RCH=CH_2 + HX$$

  • $\beta$-elimination
  • Follows Saytzeff's rule: more substituted alkene (more stable) is major
  • Example: $CH_3CH_2CH(Br)CH_3 \to CH_3CH=CHCH_3$ (major, but-2-ene) over $CH_2=CHCH_2CH_3$

2. From Alcohols (Dehydration): $$RCH_2CH_2OH \xrightarrow{conc.\ H_2SO_4, \Delta} RCH=CH_2 + H_2O$$

  • Acid-catalyzed; conc. $H_2SO_4$ at $443$ K, or alumina at $623$ K
  • Saytzeff rule applies
  • Reactivity: $3° > 2° > 1°$ alcohols

3. From Vicinal Dihalides (Dehalogenation): $$RCH(X)CH(X)R' + Zn \to RCH=CHR' + ZnX_2$$

  • $CH_2BrCH_2Br + Zn \to CH_2=CH_2 + ZnBr_2$

4. From Alkynes (Partial Reduction):

  • $RC \equiv CR' + H_2 \xrightarrow{Lindlar} \text{cis-alkene}$
  • $RC \equiv CR' + Na/NH_3(liq) \to \text{trans-alkene}$ (Birch reduction)

Physical Properties:

  • Similar to alkanes; slightly higher BPs due to greater polarizability of C=C
  • C2-C4 gases; C5-C17 liquids
  • Insoluble in water

Chemical Properties (Electrophilic Addition Reactions):

1. Addition of Hydrogen (Catalytic Hydrogenation): $CH_2=CH_2 + H_2 \xrightarrow{Ni/Pd/Pt} CH_3CH_3$

2. Addition of Halogens: $CH_2=CH_2 + Br_2 \to BrCH_2CH_2Br$ (1,2-dibromoethane)

  • Decolorization of bromine water = test for unsaturation
  • Mechanism: $\pi$ → cyclic bromonium ion → trans-addition

3. Addition of HX (Hydrogen halides): $CH_2=CH_2 + HBr \to CH_3CH_2Br$ (ethyl bromide)

Markovnikov's Rule: In addition of HX to unsymmetrical alkene, H adds to the C with more H atoms; X to C with fewer H. $CH_3CH=CH_2 + HBr \to CH_3CHBrCH_3$ (2-bromopropane, not 1-)

Mechanism: Protonation gives more stable carbocation (Markovnikov product follows carbocation stability: $3° > 2° > 1°$).

Anti-Markovnikov (Peroxide Effect, Kharasch): In presence of peroxides, HBr addition follows opposite: $CH_3CH=CH_2 + HBr \xrightarrow{ROOR} CH_3CH_2CH_2Br$ (1-bromopropane)

  • Only HBr exhibits this (not HCl, HI)
  • Free radical mechanism

4. Addition of H₂SO₄: $CH_2=CH_2 + H_2SO_4 \to CH_3CH_2OSO_3H$ (ethyl hydrogen sulfate) $CH_3CH_2OSO_3H + H_2O \xrightarrow{\Delta} CH_3CH_2OH + H_2SO_4$ (hydration)

5. Addition of Water (Acid-catalyzed): $CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$ (follows Markovnikov)

6. Oxidation:

  • Cold dilute KMnO₄ (Baeyer's reagent): $CH_2=CH_2 + H_2O \xrightarrow{KMnO_4} CH_2(OH)CH_2(OH)$ (glycol; cis-diol)
  • Hot KMnO₄/acidic: Cleavage of C=C → carboxylic acids/ketones/CO₂
  • OsO₄: gives syn-diol; followed by NaHSO₃

7. Ozonolysis: $RCH=CHR' + O_3 \to \text{ozonide} \xrightarrow{H_2O, Zn} RCHO + R'CHO$

  • Cleaves C=C to give aldehydes (or ketones if substituted)
  • Used to determine position of double bond

8. Polymerization: $nCH_2=CH_2 \xrightarrow{cat} -[CH_2CH_2]_n-$ (polyethylene)

Worked Examples
1

Product of $CH_3CH=CH_2 + HBr$ (i) without peroxide (ii) with peroxide.

Show solution

(i) Markovnikov: H goes to terminal $CH_2$, Br to inner C: $CH_3CHBrCH_3$ (2-bromopropane) (ii) Anti-Markovnikov (peroxide effect): reverse: $CH_3CH_2CH_2Br$ (1-bromopropane)

Final Answer: (i) 2-bromopropane; (ii) 1-bromopropane.

2

Ozonolysis of 2-butene gives:

Show solution

$CH_3CH=CHCH_3 + O_3 \to \text{ozonide} \xrightarrow{H_2O, Zn} 2CH_3CHO$ (two molecules of acetaldehyde)

Final Answer: $2CH_3CHO$ (acetaldehyde).

✎ Self-Check — 5 questions0 / 5
Q1.

General formula of alkenes:

Q2.

Markovnikov rule:

Q3.

Anti-Markovnikov is observed in:

Q4.

Decolorization of Br₂/CCl₄ tests for:

Q5.

Ozonolysis of $CH_3CH=CH_2$ gives:

2
Module 2

Alkynes and Aromatic Hydrocarbons

Alkynes — Preparation, Acidic Character, ReactionsTopic 1

Alkynes: Hydrocarbons with at least one C≡C triple bond; general formula $C_nH_{2n-2}$. $sp$ hybridized C at triple bond; linear geometry, bond angle $180°$.

Examples: $HC \equiv CH$ ethyne (acetylene), $CH_3 C \equiv CH$ propyne.

Preparation:

1. From Calcium Carbide: $CaC_2 + 2H_2O \to Ca(OH)_2 + C_2H_2$ (acetylene)

  • $CaC_2$ from $CaO + 3C \xrightarrow{\text{electric furnace}} CaC_2 + CO$

2. From Vicinal Dihalides: $RCH(X)CH(X)R' + 2KOH(alc) \xrightarrow{\Delta} RC \equiv CR' + 2KX + 2H_2O$

  • Double dehydrohalogenation
  • e.g., $CH_3CHBrCH_2Br + 2KOH \to CH_3C\equiv CH$

3. From Tetrahalides: $RCX_2CX_2R' + 2Zn \to RC \equiv CR' + 2ZnX_2$

4. From Sodium Acetylide: $HC \equiv C-Na + RX \to HC \equiv C-R + NaX$ (extends chain)

Acidic Character (Distinctive!):

Terminal alkynes (with $\equiv C-H$) are weakly acidic due to high s-character of $sp$ C-H bond (50% s → more EN).

Compound$pK_a$
$HC \equiv CH$ (ethyne)$\sim 25$
$CH_2 = CH_2$$\sim 44$
$CH_4$$\sim 50$

Order: Alkyne > Alkene > Alkane in acidity. Reason: more s-character → e⁻ closer to nucleus → more stable C⁻.

Acidic reactions:

  • $HC \equiv CH + Na \to HC \equiv C^-Na^+ + \frac{1}{2}H_2$
  • $HC \equiv CH + NaNH_2 \to HC \equiv C^-Na^+ + NH_3$
  • $RC \equiv CH + [Cu(NH_3)_2]^+ \to RC \equiv C-Cu \downarrow$ (red ppt — test for terminal alkyne)
  • $RC \equiv CH + [Ag(NH_3)_2]^+ \to RC \equiv C-Ag \downarrow$ (white-grey ppt — test)

Note: Only terminal alkynes show this acidity.

Addition Reactions (similar to alkenes but two steps):

1. Hydrogenation:

  • Complete: $RC \equiv CR' + 2H_2 \xrightarrow{Pt} RCH_2CH_2R'$
  • Partial: $RC \equiv CR' + H_2 \xrightarrow{Lindlar (Pd/CaCO_3/Pb)} cis$-alkene
  • Partial trans: $Na/liq NH_3$ → trans-alkene

2. Halogenation: Two molar of X₂; gives tetrahalide. $HC \equiv CH + 2Br_2 \to CHBr_2CHBr_2$

3. Addition of HX: Two steps (each follows Markovnikov): $CH_3C \equiv CH + HBr \to CH_3CBr=CH_2 \xrightarrow{HBr} CH_3CBr_2CH_3$

4. Addition of Water (Hydration): $RC \equiv CH + H_2O \xrightarrow{H_2SO_4, HgSO_4} \text{enol} \to RCOCH_3$ (ketone)

  • $CH \equiv CH + H_2O \to CH_3CHO$ (acetaldehyde — only alkyne giving aldehyde)
  • Other terminal alkynes give methyl ketones (Markovnikov)

5. Polymerization:

  • $3HC \equiv CH \xrightarrow{Cu \text{ tube, } 873K} C_6H_6$ (benzene)
  • Trimerization

6. Ozonolysis: $RC \equiv CR' + O_3 \to RCOOH + R'COOH$ (carboxylic acids)

Worked Examples
1

Why is acetylene acidic?

Show solution

The $\equiv C-H$ bond has high s-character (50% in $sp$ vs 25% in $sp^3$). More s-character → electrons closer to C nucleus → C-H bond more polar → H easily removed as $H^+$, leaving stable acetylide ion. Thus $pK_a \sim 25$ vs alkanes' $\sim 50$.

Final Answer: High s-character of $sp$ C → polar C-H → acidic.

2

Predict product of hydration of propyne.

Show solution

$CH_3C \equiv CH + H_2O \xrightarrow{H_2SO_4, HgSO_4} CH_3COCH_3$ (acetone, by Markovnikov)

Mechanism: $H_2O$ adds with H going to terminal C (more H) and -OH to internal C ($sp$ C with R), giving enol $CH_3C(OH)=CH_2$ which tautomerizes to ketone.

Final Answer: Acetone ($CH_3COCH_3$).

✎ Self-Check — 5 questions0 / 5
Q1.

Acetylene from CaC₂:

Q2.

$HC \equiv CH$ has hybridization:

Q3.

Most acidic:

Q4.

Lindlar catalyst gives:

Q5.

Trimerization of acetylene gives:

Aromatic Hydrocarbons — Aromaticity, Electrophilic SubstitutionTopic 2

Aromatic compounds have special stability beyond ordinary unsaturated compounds — "aromatic character" or aromaticity.

Hückel's Rule for Aromaticity (4n + 2 rule): A compound is aromatic if:

  1. Planar cyclic structure
  2. Conjugated $\pi$ system (alternate double-single bonds)
  3. All atoms in ring contribute to $\pi$ system ($p$ orbital)
  4. $(4n+2)$ $\pi$ electrons in ring, $n = 0, 1, 2, 3, \ldots$

Examples:

  • Benzene ($C_6H_6$): 6 $\pi$ electrons ($n = 1$)
  • Pyridine, pyrrole, furan, thiophene: all aromatic
  • Naphthalene: 10 $\pi$ ($n=2$)
  • Cyclopentadienyl anion ($C_5H_5^-$): 6 $\pi$
  • Cycloheptatrienyl cation ($C_7H_7^+$, tropylium): 6 $\pi$

Antiaromatic ($4n$ $\pi$ electrons, planar): Less stable; cyclobutadiene, cyclooctatetraene dianion is aromatic.

Benzene Structure (Kekulé):

  • Two resonance structures (alternating double-single bonds)
  • Real molecule: hybrid; all C-C bonds equal length ($1.39$ Å, intermediate between $C-C$ $1.54$ and $C=C$ $1.34$)
  • 6 $\pi$ electrons in delocalized $\pi$ molecular orbital above/below plane
  • Resonance energy ≈ $150$ kJ/mol (more stable than expected)

Electrophilic Aromatic Substitution ($S_E^{Ar}$):

General mechanism: $E^+$ attacks $\pi$ system → forms $\sigma$-complex (carbocation intermediate; loss of aromaticity) → loss of $H^+$ → restored aromatic ring.

Major Reactions:

1. Nitration: $C_6H_6 + HNO_3 \xrightarrow{conc.\ H_2SO_4} C_6H_5NO_2 + H_2O$

  • $E^+ = NO_2^+$ (formed by $HNO_3 + H_2SO_4 \to NO_2^+ + HSO_4^- + H_2O$)

2. Halogenation: $C_6H_6 + Br_2 \xrightarrow{FeBr_3 \text{ or } AlBr_3} C_6H_5Br + HBr$

  • $E^+ = Br^+$ (Lewis acid activates)

3. Sulfonation: $C_6H_6 + H_2SO_4 \to C_6H_5SO_3H$

  • Reversible; can desulfonate with steam
  • $E^+ = SO_3$ or $HSO_3^+$

4. Friedel-Crafts Alkylation: $C_6H_6 + RCl \xrightarrow{AlCl_3} C_6H_5R + HCl$

  • $E^+ = R^+$ (carbocation)
  • Issues: polyalkylation, rearrangement of carbocation
  • Doesn't work on strongly deactivated rings (e.g., nitrobenzene)

5. Friedel-Crafts Acylation: $C_6H_6 + RCOCl \xrightarrow{AlCl_3} C_6H_5COR + HCl$

  • $E^+ = RCO^+$ (acylium ion)
  • No polyacylation (deactivating product); no rearrangement

Directing Effects in Disubstituted Benzene:

GroupDirectingActivating/DeactivatingExamples
Ortho/Para directors, activatingo/pActivating$-OH, -OR, -NH_2, -NR_2, -NHCOR, -CH_3, -OR$
Ortho/Para directors, deactivatingo/pDeactivating$-X$ (halogens: F, Cl, Br, I)
Meta directors, deactivatingmDeactivating$-NO_2, -CN, -COOH, -COR, -CHO, -SO_3H, -NR_3^+$

Reasons:

  • o/p directors: Donate $e^-$ to ring (via +R or +I); stabilize intermediate at o/p positions where +ve charge can resonate onto substituted carbon
  • m directors: Withdraw $e^-$; destabilize o/p (carbocation at carbon bearing EWG); m attack avoids this

Important Note on Halogens: Despite being o/p directors (due to +R donation from lone pair), they deactivate ring overall (strong -I from electronegativity).

Worked Examples
1

Predict product: $C_6H_5CH_3 + HNO_3/H_2SO_4 \to ?$

Show solution

$-CH_3$ is o/p director, activator. Nitration gives ortho and para products as major: $o$-Nitrotoluene + $p$-Nitrotoluene (mainly p, since less steric).

Final Answer: Major: o-nitrotoluene and p-nitrotoluene.

2

Predict product: $C_6H_5NO_2 + Br_2/FeBr_3 \to ?$

Show solution

$-NO_2$ is meta director, strongly deactivating. Bromination gives: $m$-bromonitrobenzene as major product. Reaction is slower than for benzene.

Final Answer: m-Bromonitrobenzene.

✎ Self-Check — 5 questions0 / 5
Q1.

Hückel's rule: aromatic needs:

Q2.

Number of $\pi$ electrons in benzene:

Q3.

Nitration of benzene gives nitrobenzene; electrophile is:

Q4.

-OH group on benzene is:

Q5.

Friedel-Crafts alkylation does NOT work on:

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