Organic Chemistry Principles and Reaction Mechanisms

Posted on Jun 22, 2026 in Chemistry

Electrophilic Substitution Reaction

Answer: An electrophilic substitution reaction is a reaction in which an electrophile replaces a hydrogen atom in an aromatic compound.

Example: Nitration of benzene.

Nitration of Benzene

Answer: Nitration is the reaction in which a nitro group (NO₂) is introduced into benzene.

It is carried out using concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄).

Product formed: Nitrobenzene.

Friedel-Crafts Alkylation Reaction

Answer: Friedel-Crafts alkylation is a reaction in which an alkyl group is introduced into a benzene ring in the presence of an AlCl₃ catalyst.

Example: Benzene + CH₃Cl → Toluene

Classification of Alcohols with Examples

Answer: Alcohols are classified into three types:

• Primary alcohol: Ethanol (CH₃CH₂OH)
• Secondary alcohol: Isopropyl alcohol (CH₃CHOHCH₃)
• Tertiary alcohol: tert-butyl alcohol ((CH₃)₃COH)

Acidity of Phenol

Answer: Phenol is acidic in nature because it easily releases an H⁺ ion.

The formed phenoxide ion is stable due to resonance, which increases acidity.

Aldehydes vs. Ketones

Answer: Aldehydes contain a carbonyl group attached to at least one hydrogen atom.

Example: Formaldehyde (HCHO)

Ketones contain a carbonyl group attached to two carbon atoms.

Example: Acetone (CH₃COCH₃)

Aldol Condensation

Answer: Aldol condensation is a reaction in which two molecules of an aldehyde or ketone combine in the presence of a dilute base to form a larger molecule containing a β-hydroxy carbonyl compound, which may further dehydrate.

SN1 and SN2 Reactions

Answer:

• SN1 reaction: It is a substitution reaction that occurs in two steps via carbocation formation.
• SN2 reaction: It is a substitution reaction that occurs in one step (a single concerted step).

Electrophiles and Nucleophiles

Answer: An electrophile is a species that accepts electrons. It is electron-deficient or positively charged.

Example: H⁺, NO₂⁺

A nucleophile is a species that donates electrons. It is electron-rich.

Example: OH^–, NH₃

Homolytic and Heterolytic Bond Cleavage

Answer: In homolytic cleavage, the bond breaks equally and each atom gets one electron, forming free radicals.

Example: Cl₂ → Cl• + Cl•

In heterolytic cleavage, the bond breaks unevenly and one atom takes both electrons, forming ions.

Example: H–Cl → H⁺ + Cl^–

Carbocations, Carbanions, and Free Radicals

Answer:

• Carbocation is a positively charged carbon species. Example: CH₃⁺
• Carbanion is a negatively charged carbon species. Example: CH₃^–
• Free radical is a neutral species having an unpaired electron. Example: CH₃•

Inductive Effect

Answer: The inductive effect is the permanent displacement of electrons through sigma bonds due to an electronegativity difference.

Example: Cl shows the –I effect.

Resonance Effect

Answer: The resonance effect is the delocalization of electrons within a molecule.

Example: Benzene shows resonance.

Markovnikov’s Rule

Answer: According to Markovnikov’s rule, in the addition of HX to an unsymmetrical alkene, hydrogen attaches to the carbon having more hydrogen atoms and the halogen attaches to the other carbon.

Example: Propene + HBr → 2-bromopropane

Aromaticity and Hückel’s Rule

Answer: Aromaticity is the special stability shown by cyclic compounds having delocalized π electrons.

According to Hückel’s rule, a compound must have (4n + 2) π electrons to be aromatic.

Example: Benzene has 6 π electrons.

Basicity of Amines

Q: Discuss the basicity of amines

Answer:

Definition: Amines are basic in nature because the nitrogen atom in amines has a lone pair of electrons which can accept a proton (H⁺). Therefore, amines act as Lewis bases / Brønsted bases.

General reaction: R–NH₂ + H⁺ → R–NH₃⁺

Reason for basicity: The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom. A more available lone pair means a more basic nature.

Factors affecting basicity of amines:

1. Inductive effect (+I effect): Alkyl groups increase electron density on nitrogen and increase basicity. Order in gas phase: 3° amine > 2° amine > 1° amine > NH₃
2. Solvation effect: In aqueous solution, the stability of the ammonium ion due to hydration is important. Order in water: 2° amine > 1° amine > 3° amine > NH₃ (3° amines are less hydrated due to steric hindrance).
3. Resonance effect: If the lone pair is involved in resonance, basicity decreases. Example: Aniline (C₆H₅NH₂) is less basic because the lone pair is delocalized into the benzene ring.
4. Electron withdrawing groups: Groups like –NO₂, –CN, and –COOH decrease basicity by pulling electron density away from nitrogen.
5. Hybridization: sp³ nitrogen is more basic than sp² nitrogen because the lone pair is more available.

Acidity of Carboxylic Acids

Q: Discuss the acidity of carboxylic acids and the effect of substituents

Acidity of Carboxylic Acids

Definition: Carboxylic acids (R–COOH) are weak acids because they can donate a proton (H⁺) from the –COOH group in water.

Reason for Acidity: When a carboxylic acid loses H⁺, it forms a carboxylate ion (R–COO^–).

R–COOH ⇌ R–COO^– + H⁺

Stability of carboxylate ion: Very stable due to resonance.

Resonance structures: R–C(=O)O^– ↔ R–C(O^–)=O

Because the conjugate base is stable, acidity increases.

Key Point: Carboxylic acids are more acidic than alcohols because alcohol ions (RO^–) are not resonance stabilized, while carboxylate ions are resonance stabilized.

Effect of Substituents on Acidity

Substituents attached to carboxylic acids affect acidity by electron-withdrawing or electron-donating effects.

• (A) Electron Withdrawing Groups (–I effect): Increase acidity. They stabilize the carboxylate ion by pulling electron density. Examples: –Cl, –NO₂, –F, –CN. Example: ClCH₂COOH > CH₃COOH. Chloroacetic acid is a stronger acid than acetic acid because the –Cl group stabilizes the COO^– ion by the –I (inductive) effect.
• (B) Electron Donating Groups (+I effect): Decrease acidity. They destabilize the carboxylate ion. Examples: –CH₃, –C₂H₅, –OH (weak +I effect). Example: CH₃COOH > CH₃CH₂COOH (slightly weaker acid).
• (C) Effect of Distance of Substituent: The closer the substituent to the –COOH group, the stronger the effect. The effect decreases with distance. Example: ClCH₂COOH > ClCH₂CH₂COOH.
• (D) Multiple Substituents: More electron-withdrawing groups lead to more acidity. Example: CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH.
• (E) Resonance effect (+R / –R): The –NO₂ group (–R effect) increases acidity strongly. The –OH or –OCH₃ (+R effect) decreases acidity (generally in aromatic acids).

Aromatic Carboxylic Acids (Important)

Example: Benzoic acid > Toluic acid (CH₃ substituted). –NO₂ increases acidity; –CH₃ decreases acidity.

Structure and Uses of Common Organic Compounds

Q: Write a short note on the structure and uses of Formaldehyde, Benzaldehyde, Vanillin, Acetic acid, Citric acid, and Salicylic acid.

Formaldehyde

Structure: H–CHO

Uses: Used as a disinfectant and preservative (formalin solution); used in the preparation of resins like Bakelite and urea-formaldehyde resins; used for the preservation of biological tissues and specimens; used in organic synthesis.

Benzaldehyde

Structure: C₆H₅–CHO

Uses: Used in perfumes (almond-like smell); used in the synthesis of dyes and pharmaceuticals; used in Benzoin and Perkin condensation reactions; used as a flavoring agent.

Vanillin

Structure: HO–C₆H₃(OCH₃)–CHO (4-hydroxy-3-methoxybenzaldehyde)

Uses: Used as a flavoring agent (vanilla flavor in food); used in perfumes and cosmetics; used in pharmaceutical preparations; used in chemical synthesis.

Acetic Acid

Structure: CH₃–COOH

Uses: Used in vinegar as a food preservative; used in the manufacture of esters and acetic anhydride; used in the pharmaceutical industry; used as a solvent in chemical reactions.

Citric Acid

Structure: HOOC–CH₂–C(OH)(COOH)–CH₂–COOH

Uses: Used as a food preservative and acidity regulator; used in soft drinks and beverages; used in pharmaceutical effervescent tablets; used in cleaning agents (removes metal deposits).

Salicylic Acid

Structure: HO–C₆H₄–COOH

Uses: Used in the treatment of acne and skin diseases; used in the preparation of aspirin (acetylsalicylic acid); used as a keratolytic agent (removes dead skin); used in antiseptic and medicinal ointments.

Benzoin and Perkin Condensation Mechanisms

Q: Discuss the mechanism and reaction of Benzoin and Perkin condensation

Benzoin Condensation

Definition: Benzoin condensation is a reaction in which two molecules of an aromatic aldehyde (benzaldehyde) in the presence of KCN (or NaCN) combine to form an α-hydroxy ketone (benzoin).

Reaction: 2 C₆H₅CHO —KCN→ C₆H₅CH(OH)COC₆H₅ (Product: Benzoin)

Mechanism:

• Step 1: Cyanide ion attack: CN^– attacks benzaldehyde: C₆H₅CHO + CN^– → C₆H₅CH(OH)CN (cyanohydrin formation).
• Step 2: Formation of carbanion (umpolung): The CN group helps reverse polarity; the carbon becomes nucleophilic.
• Step 3: Attack on second benzaldehyde: C₆H₅CH^– + C₆H₅CHO → C₆H₅CH(OH)CH(OCN)C₆H₅.
• Step 4: Proton transfer and elimination of CN^–: Final product formed: C₆H₅CH(OH)COC₆H₅ (benzoin).

Key Points: Catalyst: CN^– (KCN/NaCN); Product: α-hydroxy ketone; Only aromatic aldehydes work well.

Perkin Condensation

Definition: Perkin condensation is a reaction between an aromatic aldehyde and an acid anhydride in the presence of a sodium salt of the acid (like sodium acetate) to form an α,β-unsaturated carboxylic acid.

Reaction (Example): C₆H₅CHO + (CH₃CO)₂O —NaOAc→ C₆H₅CH=CHCOOH (Product: Cinnamic acid)

Mechanism:

• Step 1: Formation of carbanion: The acetate ion removes α-hydrogen from the anhydride: CH₃COO^– → CH₂^–COO.
• Step 2: Nucleophilic attack: The carbanion attacks benzaldehyde: C₆H₅CHO + CH₂^–COO → C₆H₅CH(OH)CH₂COO.
• Step 3: Elimination of acetate: Water elimination occurs: C₆H₅CH=CHCOO^–.
• Step 4: Acidification: C₆H₅CH=CHCOOH (cinnamic acid).

Key Points: Reagents: aromatic aldehyde + acid anhydride; Catalyst: sodium salt of acid; Product: α,β-unsaturated acid.

Cannizzaro and Crossed Cannizzaro Reactions

Q: Discuss the mechanism and reaction of Cannizzaro and Crossed Cannizzaro reaction

Cannizzaro Reaction

Definition: The Cannizzaro reaction is a disproportionation reaction in which aldehydes having no α-hydrogen react with concentrated alkali (NaOH/KOH) to give a mixture of alcohol and carboxylic acid salt.

Reaction (Example: Formaldehyde): 2 HCHO + NaOH → CH₃OH + HCOONa

Mechanism:

• Step 1: Nucleophilic attack of OH^–: OH^– attacks the carbonyl carbon: HCHO + OH^– → HO–CH₂O^–.
• Step 2: Hydride ion transfer (key step): One molecule acts as a reducing agent, the other as an oxidizing agent: HO–CH₂O^– + HCHO → HCOO^– + CH₃O^–.
• Step 3: Formation of products: Formate ion (HCOO^–) → carboxylic acid salt; Alkoxide ion → alcohol after protonation: CH₃O^– + H₂O → CH₃OH.

Key Points: Requires aldehydes without α-H; Strong base needed; One molecule oxidized, one reduced.

Crossed Cannizzaro Reaction

Definition: The Crossed Cannizzaro reaction occurs between two different non-enolizable aldehydes, usually Formaldehyde + an aromatic aldehyde.

Reaction (Example): HCHO + C₆H₅CHO + NaOH → CH₃OH + C₆H₅COONa

Mechanism:

• Step 1: OH^– attack on formaldehyde: HCHO + OH^– → HO–CH₂O^–.
• Step 2: Hydride transfer to aromatic aldehyde: Formaldehyde donates a hydride (H^–) to benzaldehyde: HO–CH₂O^– + C₆H₅CHO → HCOO^– + C₆H₅CH₂O^–.
• Step 3: Formation of final products: C₆H₅CH₂O^– + H₂O → C₆H₅CH₂OH (benzyl alcohol + carboxylate salt).

Key Points: Between two different aldehydes; One usually is formaldehyde (more reactive); Produces alcohol + salt of acid.

Aldol and Crossed Aldol Condensation

Q: Discuss the mechanism and reaction of Aldol and Crossed Aldol Condensation

Aldol Condensation

Aldol condensation is a reaction in which aldehydes or ketones having α-hydrogen react in the presence of a dilute base (NaOH/KOH) to form a β-hydroxy aldehyde/ketone (aldol), which on heating gives an α,β-unsaturated carbonyl compound.

Reaction (Example: Acetaldehyde): 2 CH₃CHO —dil. NaOH→ CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O (Product: Crotonaldehyde)

Mechanism:

• Step 1: Formation of enolate ion: Base removes α-hydrogen: CH₃CHO + OH^– → CH₂^–CHO (enolate ion).
• Step 2: Nucleophilic attack: Enolate attacks another aldehyde molecule: CH₂^–CHO + CH₃CHO → CH₃CH(OH)CH₂CHO (β-hydroxy aldehyde formed = aldol).
• Step 3: Dehydration (on heating): Loss of water: CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O.

Key Points: Forms β-hydroxy carbonyl compound first; Then gives α,β-unsaturated compound; Requires α-hydrogen.

Crossed Aldol Condensation

Definition: Crossed aldol condensation occurs between two different aldehydes or ketones (both having α-hydrogen or one having no α-H).

Reaction (Example: Acetaldehyde + Benzaldehyde): CH₃CHO + C₆H₅CHO —NaOH→ CH₃CH(OH)CH₂C₆H₅ → CH₃CH=CHC₆H₅ (Product: Cinnamaldehyde-type compound)

Mechanism:

• Step 1: Enolate formation: CH₃CHO + OH^– → CH₂^–CHO.
• Step 2: Attack on second aldehyde: Enolate attacks benzaldehyde: CH₂^–CHO + C₆H₅CHO → CH₃CH(OH)CH₂C₆H₅.
• Step 3: Dehydration: CH₃CH(OH)CH₂C₆H₅ → CH₃CH=CHC₆H₅ + H₂O.

Important Note: If both compounds have α-H, a mixture of products forms. To avoid a mixture, one compound should have no α-H (like benzaldehyde).

Key Differences (Exam point): Aldol: same aldehyde reacts; Crossed aldol: two different carbonyl compounds react.

Structure and Uses of Alcohols and Halides

Q: Write a short note on the structure and uses of glycerol, dichloromethane, benzyl alcohol, iodoform, ethyl alcohol, and chloroform.

Glycerol

Structure: CH₂OH–CHOH–CH₂OH

Uses: Used as a moisturizer (humectant) in creams and cosmetics; used in pharmaceutical syrups; used in nitroglycerin (explosives) preparation; used as a sweetening agent in medicines.

Dichloromethane (CH₂Cl₂)

Structure: CH₂Cl₂

Uses: Used as a solvent in pharmaceuticals and labs; used in paint removers and degreasing agents; used in extraction processes.

Benzyl Alcohol

Structure: C₆H₅–CH₂OH

Uses: Used as a preservative in injections; used in cosmetics and perfumes; used as a mild local anesthetic; used as a solvent for drugs.

Iodoform

Structure: CHI₃

Uses: Used as an antiseptic and disinfectant; used in wound treatment (external use); used in the iodoform test in organic chemistry.

Ethyl Alcohol (Ethanol)

Structure: CH₃–CH₂OH

Uses: Used as an antiseptic (sanitizer); used as a solvent in medicines; used in tinctures and syrups; used as fuel (bioethanol).

Chloroform

Structure: CHCl₃

Uses: Used as a solvent in pharmaceuticals; previously used as an anesthetic (now limited due to toxicity); used in organic extraction processes.

Chemical Tests to Differentiate Alcohols

Q: Define various chemical tests for alcohols to differentiate between primary, secondary, and tertiary alcohols

Alcohols can be distinguished as 1° (primary), 2° (secondary), and 3° (tertiary) using specific chemical tests based on their oxidation behavior and reaction speed.

Lucas Test (Very Important)

Reagent: Conc. HCl + anhydrous ZnCl₂ (Lucas reagent)

Principle: Formation of alkyl chloride causes turbidity (cloudiness).

Observation:

• Tertiary alcohol (3°): Immediate turbidity
• Secondary alcohol (2°): Turbidity in 5–10 minutes
• Primary alcohol (1°): No turbidity at room temperature (only on heating)

Example: (CH₃)₃COH → immediate turbidity; CH₃CH(OH)CH₃ → slow turbidity; CH₃CH₂OH → no reaction (cold).

Victor Meyer Test

Principle: Conversion into nitro compounds gives different colors depending on the type of alcohol.

Observation: 1° alcohol: Red color; 2° alcohol: Blue color; 3° alcohol: No color.

Oxidation Test

Reagent: Acidified KMnO₄ or K₂Cr₂O₇

Principle: Alcohols are oxidized differently.

Observation:

• 1° alcohol: Oxidized to aldehyde → carboxylic acid
• 2° alcohol: Oxidized to ketone
• 3° alcohol: No oxidation (under mild conditions)

Example: 1°: CH₃CH₂OH → CH₃CHO → CH₃COOH; 2°: CH₃CHOHCH₃ → CH₃COCH₃; 3°: (CH₃)₃COH → no reaction.

Iodoform Test (Specific for some alcohols)

Reagent: I₂ + NaOH

Positive result: Yellow precipitate of CHI₃ (iodoform).

Observation: Positive: Ethanol (1°) and 2° alcohols with CH₃–CHOH group; Negative: Most other 1° and all 3° alcohols.

Chemical Tests to Differentiate Alcohols (Repeated)

Q: Define various chemical tests for alcohols to differentiate between primary, secondary and tertiary alcohols

Ans: Alcohols can be distinguished as 1° (primary), 2° (secondary), and 3° (tertiary) using specific chemical tests based on their oxidation behavior and reaction speed.

Lucas Test (Very Important)

Reagent: Conc. HCl + anhydrous ZnCl₂ (Lucas reagent)

Principle: Formation of alkyl chloride causes turbidity (cloudiness).

Observation: Tertiary alcohol (3°): Immediate turbidity; Secondary alcohol (2°): Turbidity in 5–10 minutes; Primary alcohol (1°): No turbidity at room temperature (only on heating).

Example: (CH₃)₃COH → immediate turbidity; CH₃CH(OH)CH₃ → slow turbidity; CH₃CH₂OH → no reaction (cold).

Victor Meyer Test

Principle: Conversion into nitro compounds gives different colors depending on the type of alcohol.

Observation: 1° alcohol: Red color; 2° alcohol: Blue color; 3° alcohol: No color.

Oxidation Test

Reagent: Acidified KMnO₄ or K₂Cr₂O₇

Principle: Alcohols are oxidized differently.

Observation: 1° alcohol: Oxidized to aldehyde → carboxylic acid; 2° alcohol: Oxidized to ketone; 3° alcohol: No oxidation (under mild conditions).

Example: 1°: CH₃CH₂OH → CH₃CHO → CH₃COOH; 2°: CH₃CHOHCH₃ → CH₃COCH₃; 3°: (CH₃)₃COH → no reaction.

Iodoform Test (Specific for some alcohols)

Reagent: I₂ + NaOH

Positive result: Yellow precipitate of CHI₃ (iodoform).

Observation: Positive: Ethanol (1°) and 2° alcohols with CH₃–CHOH group; Negative: Most other 1° and all 3° alcohols.

sp3 Hybridization in Alkanes

Q: Discuss sp³ hybridization in alkanes

Definition: In alkanes, each carbon atom undergoes sp³ hybridization, in which one 2s orbital and three 2p orbitals mix to form four equivalent sp³ hybrid orbitals. These orbitals are used to form four single (sigma) bonds (σ bonds).

Formation of sp³ Hybridization:

• Carbon electronic configuration (ground state): 1s² 2s² 2p²
• Excited state: 1s² 2s¹ 2p³
• Now, one 2s + three 2p orbitals combine → four sp³ hybrid orbitals

Geometry of sp³ Hybridized Carbon:

• Shape: Tetrahedral
• Bond angle: 109.5°
• All four orbitals are equivalent in energy and shape

Structure Example (Methane – CH₄):

    H
    |
H — C — H
    |
    H

Carbon is sp³ hybridized, forming 4 sigma bonds with hydrogen.

Bonding in Alkanes: All bonds are sigma (σ) bonds; strong, single covalent bonds; free rotation around C–C bond possible.

Example (Ethane): CH₃–CH₃ → both carbons are sp³ hybridized.

Halogenation (Chlorination) in Alkenes

Q: Define Halogenation (Chlorination) in Alkenes

Definition: Halogenation of alkenes is an electrophilic addition reaction in which an alkene reacts with chlorine (Cl₂) to form a 1,2-dihaloalkane (vicinal dihalide). In this reaction, the C=C double bond breaks and two halogen atoms add across it.

General Reaction: RCH=CHR′ + Cl₂ → RCHCl–CHClR′

Example (Ethene): CH₂=CH₂ + Cl₂ → CH₂Cl–CH₂Cl

Product Name: 1,2-dichloroethane

Structure:

CH2 = CH2 + Cl–Cl
      ↓
   CH2–CH2
   |   |
   Cl  Cl

Mechanism (Very Important):

• Step 1: Formation of cyclic chloronium ion: Alkene has π electrons (electron-rich). It attacks Cl₂, causing polarization of the Cl–Cl bond. One Cl attaches to both carbon atoms forming a three-membered chloronium ion: CH₂=CH₂ + Cl₂ → [CH₂–CH₂]⁺ (chloronium ion) + Cl^–.
• Step 2: Attack by chloride ion (Cl^–): The Cl^– attacks from the opposite side (backside attack). This opens the ring structure and results in anti addition of chlorine atoms.

Ozonolysis of Alkenes

Q: Define Ozonolysis

Ans: Definition: Ozonolysis is the reaction in which ozone (O₃) cleaves a C=C double bond of alkenes to form carbonyl compounds (aldehydes or ketones) after workup.

General Reaction (Must write in exam): RCH=CHR’ + O₃ → Ozonide → (Zn/H₂O or (CH₃)₂S) → Aldehydes / Ketones

Structure Example (Ethene): CH₂ = CH₂ + O₃ → Ozonide → HCHO + HCHO

Product structure: H–C=O + H–C=O (formaldehyde)

Key Point: Used to identify the position of the double bond in alkenes.

Markovnikov and Anti-Markovnikov Mechanisms

Question: Illustrate mechanism of Markovnikov’s and Anti-Markovnikov’s orientation of alkene with suitable example

Markovnikov’s Rule (Orientation)

Definition: When an unsymmetrical reagent (like HX) is added to an unsymmetrical alkene, the hydrogen (H) attaches to the carbon having more hydrogen atoms, while the halide (X) attaches to the carbon having fewer hydrogen atoms.

Example: Addition of HBr to Propene

Reaction: Plain text CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (Product: 2-bromopropane major product)

Mechanism (Markovnikov Addition):

• Step 1: Protonation of Alkene (Carbocation Formation): Double bond attacks H⁺: Plain text CH₃-CH=CH₂ + H⁺ → CH₃-CH⁺-CH₃. A more stable secondary carbocation is formed.
• Step 2: Nucleophilic Attack by Br^–: Plain text CH₃-CH⁺-CH₃ + Br^– → CH₃-CHBr-CH₃.

Key Point: Reaction proceeds via carbocation intermediate; more stable carbocation forms (3° > 2° > 1°); this controls orientation.

Anti-Markovnikov’s Rule (Peroxide Effect)

Definition: In the presence of peroxides (ROOR), addition of HBr to unsymmetrical alkenes occurs in a reverse manner: Br attaches to the carbon with more H atoms, and H attaches to the more substituted carbon. This effect is seen only with HBr (not HCl or HI).

Example: Propene + HBr (in presence of peroxide)

Reaction: Plain text CH₃-CH=CH₂ + HBr (ROOR) → CH₃-CH₂-CH₂Br (Product: 1-bromopropane)

Mechanism (Free Radical Mechanism):

• Step 1: Formation of Free Radicals (Initiation): Plain text RO-OR → 2RO•; RO• + HBr → ROH + Br•.
• Step 2: Chain Propagation: (a) Br• adds to alkene: Plain text CH₃-CH=CH₂ + Br• → CH₃-CH•-CH₂Br (Formation of more stable secondary radical). (b) Radical reacts with HBr: Plain text CH₃-CH•-CH₂Br + HBr → CH₃-CH₂-CH₂Br + Br•. Chain continues.

Classification and IUPAC Nomenclature

Q: Classify Organic Compounds. Define IUPAC Nomenclature for Naming of Organic Compounds.

Ans: Classification of Organic Compounds

Organic compounds are chemical compounds that contain carbon. They are classified as follows:

Acyclic (Open Chain) Compounds

These compounds have carbon atoms arranged in straight or branched chains.

• (a) Saturated Compounds (Alkanes): Contain only single bonds between carbon atoms. Example: CH₃–CH₂–CH₃ (Propane)
• (b) Unsaturated Compounds: Contain one or more double or triple bonds. Examples: CH₂=CH₂ (Ethene), HC≡CH (Ethyne)

Cyclic (Closed Chain) Compounds

These compounds contain carbon atoms arranged in ring structures.

• (a) Homocyclic Compounds: The ring contains only carbon atoms.
  • Alicyclic Compounds: Non-aromatic cyclic compounds such as Cyclohexane.
  • Aromatic Compounds: Contain benzene rings such as Benzene.
• (b) Heterocyclic Compounds: The ring contains one or more hetero atoms like oxygen, nitrogen, or sulfur along with carbon atoms. Examples: Pyridine, Furan, Thiophene.

IUPAC Nomenclature

Definition: IUPAC (International Union of Pure and Applied Chemistry) nomenclature is a systematic method used for naming organic compounds according to internationally accepted rules.

Rules of IUPAC Nomenclature:

1. Select the longest continuous carbon chain as the parent chain.
2. Number the carbon atoms from the end nearest to the substituent or functional group.
3. Identify and name all substituents attached to the parent chain.
4. Indicate the position of substituents by appropriate numbers.
5. If more than one substituent is present, arrange them alphabetically.
6. Name the principal functional group using the appropriate suffix.

Examples:

• CH₄ → Methane
• CH₃CH₃ → Ethane
• CH₃CH₂OH → Ethanol
• CH₃CHO → Ethanal
• CH₃COOH → Ethanoic Acid
• CH₃–CH(CH₃)–CH₂–CH₃ → 2-Methylbutane