Hydrocarbons
Hydrocarbons for JEE Main & Advanced
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)
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.
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.
General formula of alkanes:
Wurtz reaction is for:
Chlorination of methane is initiated by:
Selectivity of halogenation: $3°$ vs $1°$ H:
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)
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.
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).
General formula of alkenes:
Markovnikov rule:
Anti-Markovnikov is observed in:
Decolorization of Br₂/CCl₄ tests for:
Ozonolysis of $CH_3CH=CH_2$ gives:
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)
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.
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 ($CH_3COCH_3$).
Acetylene from CaC₂:
$HC \equiv CH$ has hybridization:
Most acidic:
Lindlar catalyst gives:
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:
- Planar cyclic structure
- Conjugated $\pi$ system (alternate double-single bonds)
- All atoms in ring contribute to $\pi$ system ($p$ orbital)
- $(4n+2)$ $\pi$ 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:
| Group | Directing | Activating/Deactivating | Examples |
|---|---|---|---|
| Ortho/Para directors, activating | o/p | Activating | $-OH, -OR, -NH_2, -NR_2, -NHCOR, -CH_3, -OR$ |
| Ortho/Para directors, deactivating | o/p | Deactivating | $-X$ (halogens: F, Cl, Br, I) |
| Meta directors, deactivating | m | Deactivating | $-NO_2, -CN, -COOH, -COR, -CHO, -SO_3H, -NR_3^+$ |
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).
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.
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.
Hückel's rule: aromatic needs:
Number of $\pi$ electrons in benzene:
Nitration of benzene gives nitrobenzene; electrophile is:
-OH group on benzene is:
Friedel-Crafts alkylation does NOT work on:
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