Hydrocarbons Formulas for JEE 2027, Download PDF

Classification of Hydrocarbons

• Saturated — contain only C–C single bonds → Alkanes (C$$_n$$H$$_{2n+2}$$)
• Unsaturated — contain C=C or C≡C bonds:
  • Alkenes (C$$_n$$H$$_{2n}$$) — contain at least one C=C double bond
  • Alkynes (C$$_n$$H$$_{2n-2}$$) — contain at least one C≡C triple bond
• Aromatic — contain a benzene ring or similar cyclic conjugated system → Arenes

Methods of Preparing Alkanes

• Wurtz Reaction: Two molecules of an alkyl halide react with sodium metal to form an alkane with double the carbon chain. $$$2\text{R--X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R--R} + 2\text{NaX}$$$ E.g., $$2\text{CH}_3\text{Br} + 2\text{Na} \rightarrow \text{CH}_3\text{CH}_3 + 2\text{NaBr}$$
• Decarboxylation: The sodium salt of a carboxylic acid is heated with soda lime (NaOH + CaO) to remove CO$$_2$$. $$$\text{RCOONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{R--H} + \text{Na}_2\text{CO}_3$$$
• Kolbe's Electrolysis: Electrolysis of an aqueous solution of sodium/potassium salt of a carboxylic acid. $$$2\text{RCOO}^- \xrightarrow{\text{electrolysis}} \text{R--R} + 2\text{CO}_2 + 2e^- \quad \text{(at anode)}$$$

Alkanes react with halogens (Cl$$_2$$ or Br$$_2$$) in the presence of UV light or heat. One H atom is replaced by a halogen atom.

$$$\text{R--H} + \text{X}_2 \xrightarrow{h\nu \text{ or } \Delta} \text{R--X} + \text{HX}$$$

Mechanism (3 steps):

• Initiation: $$\text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}\cdot$$ (homolytic cleavage by UV light)
• Propagation:
  $$\text{Cl}\cdot + \text{R--H} \rightarrow \text{R}\cdot + \text{HCl}$$
  $$\text{R}\cdot + \text{Cl}_2 \rightarrow \text{R--Cl} + \text{Cl}\cdot$$
• Termination: Two radicals combine.
  $$\text{Cl}\cdot + \text{Cl}\cdot \rightarrow \text{Cl}_2$$ or $$\text{R}\cdot + \text{Cl}\cdot \rightarrow \text{R--Cl}$$

Reactivity of H atoms: $$3° > 2° > 1°$$ (more substituted C–H is easier to break because the resulting radical is more stable).

Alkanes burn in excess oxygen to produce CO$$_2$$ and H$$_2$$O, releasing a large amount of heat.

$$$\text{C}_n\text{H}_{2n+2} + \frac{3n+1}{2}\text{O}_2 \longrightarrow n\text{CO}_2 + (n+1)\text{H}_2\text{O} + \text{heat}$$$

• Staggered conformation: The H atoms on the front carbon are positioned between the H atoms on the back carbon. This is the most stable conformation because atoms are as far apart as possible. Dihedral angle = $$60°$$.
• Eclipsed conformation: The H atoms on both carbons are directly aligned. This is the least stable conformation due to maximum torsional strain. Dihedral angle = $$0°$$.
• Energy difference: The eclipsed form is about 12.5 kJ/mol higher in energy than the staggered form.

Methods of Preparing Alkenes

• Dehydration of Alcohols: An alcohol loses water (H$$_2$$O) when heated with a strong acid catalyst. $$$\text{R--CH}_2\text{--CH}_2\text{--OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta} \text{R--CH=CH}_2 + \text{H}_2\text{O}$$$ Ease of dehydration: $$3° > 2° > 1°$$ alcohol (because more substituted carbocations are more stable).
• Dehydrohalogenation of Alkyl Halides: An alkyl halide loses HX when treated with a strong base (alcoholic KOH). $$$\text{R--CH}_2\text{--CHX--R'} \xrightarrow{\text{alc. KOH}} \text{R--CH=CH--R'} + \text{KX} + \text{H}_2\text{O}$$$ Saytzeff's Rule: The more substituted alkene is the major product.

Markovnikov's Rule

When an unsymmetrical reagent (like HBr, HCl, H$$_2$$O) adds to an unsymmetrical alkene:

"The hydrogen adds to the carbon that already has more hydrogen atoms" (i.e., the less substituted carbon).

Reason: The more substituted carbocation intermediate is more stable.

$$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \longrightarrow \text{CH}_3\text{CHBrCH}_3 \quad \text{(major, Markovnikov)}$$$

Anti-Markovnikov Addition (Peroxide Effect / Kharash Effect)

When HBr is added to an alkene in the presence of organic peroxides (R$$_2$$O$$_2$$), the addition follows anti-Markovnikov orientation:

$$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \quad \text{(anti-Markovnikov)}$$$

Mechanism: The reaction proceeds via a free radical mechanism (not carbocation). The Br radical adds first to the less substituted carbon because the resulting radical on the more substituted carbon is more stable.

Note: This peroxide effect works only with HBr, not with HCl or HI.

Ozonolysis of Alkenes

Alkenes react with ozone (O$$_3$$) followed by reductive workup (using Zn/H$$_2$$O or dimethyl sulfide) to cleave the double bond completely, producing aldehydes and/or ketones.

$$$\text{R}_1\text{R}_2\text{C=CR}_3\text{R}_4 \xrightarrow{1.\, \text{O}_3} \xrightarrow{2.\, \text{Zn/H}_2\text{O}} \text{R}_1\text{R}_2\text{C=O} + \text{O=CR}_3\text{R}_4$$$

• If a C of the double bond has 2 alkyl groups → ketone
• If a C of the double bond has 1 alkyl group and 1 H → aldehyde
• If a C of the double bond has 2 H atoms → formaldehyde (HCHO)

Methods of Preparing Alkynes

• From Calcium Carbide: Calcium carbide reacts with water to produce acetylene (ethyne). $$$\text{CaC}_2 + 2\text{H}_2\text{O} \longrightarrow \text{CH}\equiv\text{CH} + \text{Ca(OH)}_2$$$
• Dehydrohalogenation (Double): A vicinal dihalide or geminal dihalide loses 2 molecules of HX with strong base. $$$\text{R--CHX--CHX--R'} \xrightarrow{2 \text{ equiv. NaNH}_2} \text{R--C}\equiv\text{C--R'} + 2\text{HX}$$$

Terminal alkynes (R–C≡C–H) have an acidic hydrogen. The C–H bond in alkynes is more acidic than in alkenes or alkanes because the carbon is $$sp$$ hybridized (50% s-character), holding the bonding electrons closer to carbon.

Acidity order: $$sp\text{-C–H} > sp^2\text{-C–H} > sp^3\text{-C–H}$$ (Alkynes > Alkenes > Alkanes)

Reactions showing acidity:

• $$\text{R--C}\equiv\text{C--H} + \text{NaNH}_2 \longrightarrow \text{R--C}\equiv\text{C}^-\text{Na}^+ + \text{NH}_3$$
• $$\text{R--C}\equiv\text{C--H} + [\text{Ag(NH}_3\text{)}_2]^+ \longrightarrow \text{R--C}\equiv\text{C--Ag}\downarrow + 2\text{NH}_3 + \text{H}^+$$
  (white precipitate of silver acetylide — used as a test for terminal alkynes)

Since a triple bond has two $$\pi$$ bonds, alkynes can add two equivalents of the reagent.

• Addition of H$$_2$$:
  $$\text{R--C}\equiv\text{C--R'} \xrightarrow{\text{H}_2/\text{Pd}} \text{R--CH=CH--R'} \xrightarrow{\text{H}_2/\text{Pd}} \text{R--CH}_2\text{CH}_2\text{--R'}$$
  To stop at the alkene stage, use Lindlar's catalyst (Pd/CaCO$$_3$$ with quinoline) which gives the cis-alkene.
• Addition of HX: Follows Markovnikov's rule for each step.
  $$\text{CH}\equiv\text{CH} + \text{HCl} \rightarrow \text{CH}_2\text{=CHCl} \xrightarrow{\text{HCl}} \text{CH}_3\text{CHCl}_2$$ (geminal dihalide)
• Addition of H$$_2$$O: In the presence of Hg$$^{2+}$$ and H$$_2$$SO$$_4$$ (hydration). Follows Markovnikov's rule.
  $$\text{CH}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}, \text{H}_2\text{SO}_4} \text{CH}_3\text{CHO}$$ (acetaldehyde, via tautomerism of enol)

A compound is aromatic if it satisfies all of these conditions:

• Cyclic — the atoms form a ring
• Planar — all atoms in the ring lie in the same plane
• Fully conjugated — every atom in the ring has a p orbital (continuous overlap)
• $$(4n+2)$$ $$\pi$$ electrons — where $$n = 0, 1, 2, 3, \ldots$$

So the magic numbers of $$\pi$$ electrons for aromaticity are: 2, 6, 10, 14, ...

Benzene has 6 $$\pi$$ electrons ($$n = 1$$) → aromatic.

General Mechanism of EAS

Step 1: Generation of the electrophile (E$$^+$$).

Step 2: Electrophilic attack — E$$^+$$ attacks the $$\pi$$ electron cloud, forming a carbocation intermediate (arenium ion / sigma complex). Aromaticity is temporarily lost.

Step 3: Proton loss — a proton (H$$^+$$) is removed from the carbon bearing E, restoring aromaticity.

Important EAS Reactions of Benzene

Ortho/Para Directors vs. Meta Directors

Ortho/para directors (new group enters at ortho and para positions):

• Activating (increase reaction rate): –OH, –NH$$_2$$, –OR, –NHCOR, –R (alkyl)
• Halogens (deactivating, but still o/p directing): –F, –Cl, –Br, –I

Meta directors (new group enters at meta position):

• All are deactivating ($$-$$M effect): –NO$$_2$$, –CN, –CHO, –COOH, –COR, –SO$$_3$$H, –CF$$_3$$

Why? Ortho/para directors have lone pairs or +I effect that stabilize the arenium ion at o/p positions. Meta directors withdraw electrons ($$-$$M), destabilizing the intermediate at o/p, so meta attack is preferred.