Chapter 8: Organic Chemistry - Some Basic Principles and Techniques

Organic Chemistry Introduction

Definition and Scope

• Study of hydrocarbons (carbon and hydrogen compounds) and their derivatives.
• Includes compounds containing N, O, S, P, halogens, etc.
• Berzelius proposed a ‘vital force’ theory for organic compound formation.

Historical Synthesis

• E. Wöhler synthesized urea (organic) from ammonium cyanate (inorganic), disproving vital force theory.
• Kolbe synthesized acetic acid; Berthelot synthesized methane from inorganic sources.

Solved Examples

• Ex 1: Write the reaction for synthesis of urea from ammonium cyanate.
  Answer: NH₄CNO 9;heat9; 9;rearrangement9; NH₂CONH₂ (urea)

Practice Set

• Q1: What elements are primarily studied in organic chemistry?
  A. Metals and non-metals
  B. Carbon and hydrogen
  C. Oxygen and nitrogen
  D. Sulfur and phosphorus
• Q2: Who synthesized the first organic compound from an inorganic compound?
  A. Kolbe
  B. Berthelot
  C. E. Wöhler
  D. Berzelius
• Q3: Which reaction disproved the vital force theory?
  A. Methane synthesis
  B. Acetic acid synthesis
  C. Urea synthesis from ammonium cyanate
  D. Benzene formation

Answer Key

• Q1: B - Organic chemistry mainly studies carbon and hydrogen compounds.
• Q2: C - E. Wöhler synthesized urea from ammonium cyanate.
• Q3: C - Urea synthesis from ammonium cyanate disproved the vital force theory.

Memory Aids

• Remember: Organic = Carbon + Hydrogen (+ other elements).
• Wöhler’s urea synthesis = key reaction disproving vital force.
• Kolbe = acetic acid; Berthelot = methane from inorganic sources.

Tetravalence of Carbon: Shapes of Organic Compounds

Hybridisation and Molecular Shape

• Shapes of carbon compounds depend on hybridisation of orbitals and formation of sigma (σ) and pi (π) bonds.
• Shape depends on number of bonding pairs and lone pairs of electrons.
• Three common hybridisations in organic chemistry: sp, sp², sp³.
• Hybridisation types and features:
  • sp: Linear geometry, bond angle 180°, 50% s and 50% p character, e.g., ethyne (C₂H₂).
  • sp²: Trigonal planar geometry, bond angle 120°, 33% s and 67% p character, e.g., ethene (C₂H₄).
  • sp³: Tetrahedral geometry, bond angle 109.5°, 25% s and 75% p character, e.g., ethane (C₂H₆).

Solved Examples

• Ex 1: Identify hybridisation of carbon in ethene (C₂H₄).
  • Carbon forms 3 sigma bonds and 1 pi bond.
  • Hybridisation is sp² (trigonal planar, 120° bond angle).

Practice Set

• Q1: What is the bond angle in sp³ hybridised carbon?
  • A. 180°
  • B. 120°
  • C. 109.5°
  • D. 90°
• Q2: Which molecule has carbon with sp hybridisation?
  • A. Ethane
  • B. Ethene
  • C. Ethyne
  • D. Benzene
• Q3: Percentage of s character in sp² hybridised carbon is:
  • A. 25%
  • B. 33%
  • C. 50%
  • D. 75%

Answer Key

• Q1: C - 109.5° is the bond angle for sp³ hybridisation (tetrahedral shape).
• Q2: C - Ethyne has sp hybridised carbon (linear geometry, 180° bond angle).
• Q3: B - sp² hybridisation has 33% s character and 67% p character.

Memory Aids

• sp = 2 regions of electron density, linear, 180°
• sp² = 3 regions, trigonal planar, 120°
• sp³ = 4 regions, tetrahedral, 109.5°
• More s character means shorter, stronger bonds

Characteristic Features of π Bonds

Properties of π Bonds

• Formed by lateral (sideways) overlap of p orbitals.
• Electron density lies above and below the internuclear axis; zero density along the axis (node).
• π bond has a nodal plane along the internuclear axis, perpendicular to p orbital direction.
• Weaker than σ bonds; formed only after σ bond formation.
• No free rotation around π bond; makes molecule rigid at double/triple bonds.
• π bonds are more reactive than σ bonds; reactive centers in multiple bonds.
• Bond multiplicity increases bond strength and decreases bond length.

Solved Examples

• Ex 1: Compare bond strength and length of C–C single, double, and triple bonds.
  • Single bond: 347 kJ/mol, longest bond length.
  • Double bond: 598 kJ/mol, shorter bond length.
  • Triple bond: 803 kJ/mol, shortest bond length.

Practice Set

• Q1: π bond is formed by:
  • A. Head-on overlap of orbitals
  • B. Lateral overlap of orbitals
  • C. Overlap of s orbitals
  • D. No overlap
• Q2: Which bond allows free rotation?
  • A. σ bond
  • B. π bond
  • C. Both σ and π
  • D. None
• Q3: Arrange bonds in increasing order of bond length:
  • A. C≡C < C=C < C–C
  • B. C–C < C=C < C≡C
  • C. C=C < C–C < C≡C
  • D. C–C = C=C = C≡C

Answer Key

• Q1: B - π bond forms by lateral overlap of p orbitals.
• Q2: A - σ bonds allow free rotation; π bonds restrict rotation.
• Q3: A - Triple bond (C≡C) shortest, then double (C=C), then single (C–C) longest.

Memory Aids

• π bond = sideways overlap, no free rotation
• σ bond = head-on overlap, free rotation
• Bond strength: single < double < triple
• Bond length: single > double > triple

Complete, Condensed and Bond-line Structural Formulas

Explanation

• Complete structural formulas show all atoms and bonds explicitly (e.g., ethane C₂H₆, ethene C₂H₄, ethyne C₂H₂).
• Ethane: two carbons single bonded; ethene: double bond; ethyne: triple bond.
• Condensed structural formulas omit bonds but show groups with subscripts (e.g., CH₃CH₃, H₂C=CH₂, HC≡CH).
• Bond-line formulas simplify structures by showing carbon chains as lines; hydrogens on carbons are usually omitted.

Solved Examples

• Ex 1: Write the condensed formula for ethene.
  Solution: Ethene has a double bond between two carbons, so condensed formula is H₂C=CH₂.
• Ex 2: Identify the type of bond in ethyne.
  Solution: Ethyne has a triple bond between two carbon atoms.

Practice Set

• Q1: Which formula shows all atoms and bonds explicitly?
  A. Condensed formula
  B. Complete structural formula
  C. Bond-line formula
  D. Molecular formula
• Q2: What is the condensed formula of ethane?
  A. CH₃CH₃
  B. H₂C=CH₂
  C. HC≡CH
  D. C₂H₆
• Q3: Which bond type is present in ethene?
  A. Single bond
  B. Double bond
  C. Triple bond
  D. Ionic bond

Answer Key

• Q1: B - Complete structural formulas show all atoms and bonds explicitly.
• Q2: A - Ethane condensed formula is CH₃CH₃, showing groups without bonds.
• Q3: B - Ethene contains a double bond between carbon atoms.

Memory Aids

• Complete formulas = all atoms + bonds shown.
• Condensed formulas = groups written with subscripts, no bonds.
• Bond-line = zigzag lines for carbon chains, hydrogens omitted.

Three-Dimensional Representation

Explanation

• Solid wedge (▲) indicates bond projecting out of the plane (towards observer).
• Dashed wedge (---) indicates bond going behind the plane (away from observer).
• Normal lines represent bonds in the plane of the paper.
• Used to depict tetrahedral molecules with four different groups attached to carbon.

Solved Examples

• Ex 1: Identify the bond type shown by a solid wedge.
  Solution: Bond projecting out of the plane towards the observer.
• Ex 2: What does a dashed wedge represent?
  Solution: Bond going behind the plane away from the observer.

Practice Set

• Q1: Which symbol shows a bond coming out of the plane?
  A. Dashed wedge
  B. Solid wedge
  C. Normal line
  D. Double line
• Q2: Bonds in the plane of paper are shown by?
  A. Solid wedge
  B. Dashed wedge
  C. Normal line
  D. Dotted line
• Q3: What shape is represented by the wedge formula for carbon with four different groups?
  A. Linear
  B. Planar
  C. Tetrahedral
  D. Trigonal planar

Answer Key

• Q1: B - Solid wedge shows bond coming out of the plane.
• Q2: C - Normal lines represent bonds in the plane of the paper.
• Q3: C - Tetrahedral shape is depicted by wedge formula.

Memory Aids

• Solid wedge = bond towards you.
• Dashed wedge = bond away from you.
• Normal line = bond in plane.

Classification of Organic Compounds

Overview

• Organic compounds contain carbon atoms bonded in chains or rings.
• Two main types: Acyclic (open chain) and Cyclic (closed ring) compounds.
• Acyclic compounds: Straight or branched chains.
• Cyclic compounds: Carbon atoms form rings.

Types of Acyclic Compounds

• Saturated (alkanes): Only single bonds, e.g., ethane (C₂H₆).
• Unsaturated: Double or triple bonds.
  • Alkenes: Contain double bonds, e.g., ethene (C₂H₄).
  • Alkynes: Contain triple bonds, e.g., ethyne (C₂H₂).

Types of Cyclic Compounds

• Homocyclic: Rings made of only one type of atom (usually carbon).
• Heterocyclic: Rings contain atoms other than carbon and hydrogen.

Subtypes of Homocyclic Compounds

• Alicyclic: Ring compounds similar to alkanes, e.g., cyclopropane.
• Aromatic: Contain benzene rings, e.g., benzene, toluene.

Subtypes of Heterocyclic Compounds

• Alicyclic heterocyclic: Resemble aliphatic compounds, e.g., tetrahydrofuran.
• Aromatic heterocyclic: Contain aromatic rings, e.g., thiophene, pyridine.

Aromatic Compound Types

• Benzenoid: Contain benzene rings.
• Non-benzenoid: Do not contain benzene rings, e.g., tropone.

Solved Examples

• Ex 1: Classify cyclohexane.
  Answer: Cyclic → Homocyclic → Alicyclic compound.
• Ex 2: Classify pyridine.
  Answer: Cyclic → Heterocyclic → Aromatic heterocyclic compound.

Practice Set

• Q1: Which is an unsaturated acyclic compound?
  A. Ethane
  B. Ethene
  C. Cyclopropane
  D. Benzene
• Q2: Identify the heterocyclic aromatic compound.
  A. Cyclohexane
  B. Tetrahydrofuran
  C. Thiophene
  D. Ethyne
• Q3: Which compound is benzenoid aromatic?
  A. Tropone
  B. Benzene
  C. Pyridine
  D. Cyclopropane

Answer Key

• Q1: B - Ethene is an unsaturated acyclic compound with a double bond.
• Q2: C - Thiophene is a heterocyclic aromatic compound.
• Q3: B - Benzene is a benzenoid aromatic compound.

Memory Aids

• Acyclic = open chain; Cyclic = rings.
• Homocyclic = only carbon atoms in ring.
• Heterocyclic = ring contains other atoms (N, O, S).
• Aromatic = benzene or similar stable ring.

Functional Group

Definition and Importance

• Functional group: Atom/group responsible for characteristic chemical properties.
• Examples:
  • Hydroxyl (-OH)
  • Aldehyde (-CHO)
  • Carboxylic acid (-COOH)
• Determines reactivity and chemical behavior of organic compounds.

Solved Examples

• Ex 1: Identify the functional group in ethanol.
  Answer: Hydroxyl group (-OH).
• Ex 2: Identify the functional group in acetaldehyde.
  Answer: Aldehyde group (-CHO).

Practice Set

• Q1: Which functional group is present in carboxylic acids?
  A. -OH
  B. -CHO
  C. -COOH
  D. -NH₂
• Q2: The aldehyde group is represented by?
  A. -OH
  B. -CHO
  C. -COOH
  D. -CN
• Q3: Which functional group is characteristic of alcohols?
  A. -COOH
  B. -OH
  C. -CHO
  D. -SH

Answer Key

• Q1: C - Carboxylic acids contain the -COOH group.
• Q2: B - Aldehyde group is -CHO.
• Q3: B - Alcohols have the hydroxyl (-OH) group.

Memory Aids

• OH = Alcohols, COOH = Acids, CHO = Aldehydes.
• Functional groups define chemical properties.

Homologous Series

Definition and Characteristics

• Family of organic compounds with same functional group.
• Consecutive members differ by a -CH₂- unit.
• Similar chemical properties, gradual change in physical properties.
• Examples: Alkanes, alkenes, alkynes, haloalkanes, alkanols, alkanals, alkanones, alkanoic acids, amines.

Solved Examples

• Ex 1: Identify the next member after propane in alkane series.
  Answer: Butane (C₄H₁₀).
• Ex 2: Difference between ethene and propene.
  Answer: Propene has one extra -CH₂- unit.

Practice Set

• Q1: What is the difference between consecutive members in a homologous series?
  A. One oxygen atom
  B. One nitrogen atom
  C. One -CH₂- group
  D. One double bond
• Q2: Which is a member of the alkane homologous series?
  A. Ethene
  B. Propane
  C. Ethyne
  D. Benzene
• Q3: Which homologous series contains compounds with triple bonds?
  A. Alkanes
  B. Alkenes
  C. Alkynes
  D. Alkanols

Answer Key

• Q1: C - Consecutive members differ by one -CH₂- group.
• Q2: B - Propane is an alkane.
• Q3: C - Alkynes contain triple bonds.

Memory Aids

• Homologous series = same functional group + differ by CH₂.
• Alkanes (single bonds), Alkenes (double bonds), Alkynes (triple bonds).

Nomenclature of Organic Compounds

Basic Principles

• Complete IUPAC name has 5 parts: secondary prefix + primary prefix + word root + primary suffix + secondary suffix.
• Identify principal functional group by priority order (highest to lowest): COOH > SO3H > COOR > COCl > CONH2 > CN > HC=O > C=O > OH > NH2 > C=C.
• Functional groups determine chemical properties and reactivity.

Rules for Naming

• Identify substituents (e.g., fluoro, chloro) and principal groups (e.g., hydroxyl, carboxyl).
• Find longest carbon chain (principal chain) and assign root word (meth-, eth-, prop-, etc.).
• Add "cyclo" prefix if compound is cyclic; omit if acyclic.
• Attach primary suffix for main functional group (-ol, -al, -one, etc.).
• Add secondary suffix for additional functional groups; drop 'e' if vowel follows.
• Indicate bond types: -ane (single), -ene (double), -yne (triple) with locants if multiple bonds.
• Combine parts in order: secondary prefix + primary prefix + word root + primary suffix + secondary suffix.

Solved Examples

• Ex 1: Name CH₃-CH=CH-CH₂-OH
  Longest chain: 4 carbons (but-), double bond (-ene), hydroxyl (-ol)
  Name: But-2-en-1-ol
• Ex 2: Name cyclohexanone
  Cyclic compound with ketone group
  Name: Cyclohexanone

Practice Set

• Q1: What is the primary suffix for aldehydes?
  A. -ol
  B. -al
  C. -one
  D. -ene
• Q2: Which functional group has the highest priority in IUPAC naming?
  A. Hydroxyl (-OH)
  B. Carboxyl (-COOH)
  C. Amine (-NH₂)
  D. Alkene (C=C)
• Q3: Name the compound CH₃-CH₂-COOH.
  A. Propanol
  B. Propanoic acid
  C. Propanal
  D. Propanone

Answer Key

• Q1: B - Aldehydes use suffix -al to indicate the formyl group.
• Q2: B - Carboxyl group (-COOH) has the highest priority among listed groups.
• Q3: B - CH₃-CH₂-COOH is propanoic acid (carboxylic acid).

Memory Aids

• Priority order: "Carboxylic acids come first" (COOH highest).
• Suffixes: -ane (single), -ene (double), -yne (triple).
• Use "cyclo" prefix for rings.

Nomenclature of Substituted Benzene Compounds

Basic Rules

• Substituents are prefixes to the word "benzene".
• In disubstituted benzenes, assign lowest possible numbers to substituents.
• Trivial prefixes: ortho (o-) = 1,2-; meta (m-) = 1,3-; para (p-) = 1,4- disubstitution.

Advanced Rules

• For tri- or more substituted benzenes, use common base name.
• Assign position 1 to the substituent of the base compound.
• Follow lowest locant rule for other substituents.
• Benzene ring acts as substituent phenyl (Ph, —C₆H₅) when attached to alkane with functional group.

Solved Examples

• Ex 1: Name 1,2-dichlorobenzene
  Positions 1 and 2 have chloro substituents
  Name: Ortho-dichlorobenzene
• Ex 2: Name 1,4-dinitrobenzene
  Positions 1 and 4 have nitro substituents
  Name: Para-dinitrobenzene

Practice Set

• Q1: What prefix indicates 1,3-disubstitution on benzene?
  A. Ortho (o-)
  B. Meta (m-)
  C. Para (p-)
  D. None
• Q2: In naming substituted benzene, what is the base name for a benzene ring attached to an alkane?
  A. Benzyl
  B. Phenyl
  C. Toluene
  D. Aniline
• Q3: Which rule is used to assign numbers to substituents in disubstituted benzene?
  A. Highest locant rule
  B. Lowest locant rule
  C. Alphabetical order
  D. Random assignment

Answer Key

• Q1: B - Meta (m-) indicates 1,3-disubstitution.
• Q2: B - Phenyl (Ph) is the substituent name for benzene ring attached to alkane.
• Q3: B - Lowest locant rule assigns the smallest possible numbers.

Memory Aids

• Ortho = adjacent (1,2), Meta = separated by one carbon (1,3), Para = opposite (1,4).
• Phenyl = benzene as substituent.
• Lowest locant rule ensures smallest numbering.

Isomerism

Definition

• Compounds with the same molecular formula but different properties are called isomers.
• Isomerism explains differences in structure or arrangement despite identical formulas.

Types of Isomerism

• Structural Isomerism: Same formula, different connectivity of atoms.
• Stereoisomerism: Same formula and connectivity, different spatial arrangement.

Structural Isomerism Subtypes

• Chain isomerism: Different carbon chain arrangements.
• Position isomerism: Functional groups at different positions.
• Functional group isomerism: Different functional groups.
• Metamerism: Different alkyl groups on either side of a functional group.

Stereoisomerism Subtypes

• Geometrical isomerism: Different spatial arrangement around double bonds (cis/trans).
• Optical isomerism: Non-superimposable mirror images (d- and l-forms).

Special Types

• Tautomerism: Rapid interconversion by migration of H-atom and rearrangement (e.g., keto-enol).
• Metamerism: Isomers differ in alkyl groups around a functional group.

Solved Examples

• Ex 1: Identify chain isomers of C₅H₁₂.
  Solution: Pentane, isopentane (2-methylbutane), neopentane (2,2-dimethylpropane).
• Ex 2: Geometrical isomers of 1-butene.
  Solution: Cis-1-butene and trans-1-butene differ in spatial arrangement around C=C.

Practice Set

• Q1: Which is an example of functional group isomerism?
  A. Pentane and 2-methylbutane
  B. Ethanol and dimethyl ether
  C. Cis- and trans-2-butene
  D. D- and L-glucose
• Q2: Geometrical isomerism occurs when:
  A. Two identical groups on each carbon of double bond
  B. Two different groups on each carbon of double bond
  C. Same connectivity but different functional groups
  D. Different carbon skeletons
• Q3: Tautomerism involves:
  A. Different carbon chains
  B. Migration of H-atom and rearrangement
  C. Mirror image isomers
  D. Different positions of functional groups

Answer Key

• Q1: B - Ethanol and dimethyl ether have same formula but different functional groups.
• Q2: B - Geometrical isomerism requires two different groups on each carbon of the double bond.
• Q3: B - Tautomerism involves migration of H-atom and rearrangement within the molecule.

Memory Aids

• Structural isomers differ in connectivity; stereoisomers differ in spatial arrangement.
• Geometrical isomers need different groups on both carbons of C=C.
• Tautomerism = H-atom shift + bond rearrangement (keto-enol).

Geometrical Isomerism

Conditions and Explanation

• Occurs in alkenes with C=C double bond.
• Each carbon of double bond must have two different groups attached.
• If any carbon has two identical groups, geometrical isomerism is not possible.
• Results in cis (same side) and trans (opposite side) forms.

Solved Examples

• Ex 1: Identify if 1,2-dichloroethene shows geometrical isomerism.
  Solution: Yes, cis and trans forms exist due to different groups on each carbon.
• Ex 2: Does 1,1-dichloroethene show geometrical isomerism?
  Solution: No, both chlorines on same carbon, no cis-trans possible.

Practice Set

• Q1: Geometrical isomerism is possible in:
  A. 1,2-dibromoethene
  B. 1,1-dibromoethene
  C. Ethene
  D. Propane
• Q2: Cis and trans isomers differ in:
  A. Molecular formula
  B. Connectivity
  C. Spatial arrangement
  D. Functional groups
• Q3: Which condition prevents geometrical isomerism?
  A. Different groups on both carbons
  B. Double bond presence
  C. Identical groups on one carbon
  D. Presence of triple bond

Answer Key

• Q1: A - 1,2-dibromoethene has different groups on each carbon allowing cis-trans isomers.
• Q2: C - Cis and trans differ in spatial arrangement only.
• Q3: C - Identical groups on one carbon prevent geometrical isomerism.

Memory Aids

• Geometrical isomerism = Different groups on both carbons of C=C.
• Cis = same side, Trans = opposite side.
• No isomerism if identical groups on any carbon of double bond.

Tautomerism

Definition and Explanation

• Isomers interconvert by migration of H-atom and rearrangement of bonds.
• Common example: Keto-enol tautomerism.
• Occurs rapidly and reversibly in equilibrium.

Solved Examples

• Ex 1: Keto-enol tautomerism of acetaldehyde.
  Solution: CH₃CHO ⇌ CH₂=CHOH (enol form).

Practice Set

• Q1: Tautomerism involves:
  A. Different carbon chains
  B. Migration of H-atom
  C. Different functional groups
  D. Different positions of groups
• Q2: Keto and enol forms differ in:
  A. Position of H and double bond
  B. Molecular formula
  C. Carbon skeleton
  D. Optical activity
• Q3: Which compound shows keto-enol tautomerism?
  A. Benzene
  B. Acetone
  C. Ethane
  D. Methanol

Answer Key

• Q1: B - Tautomerism involves migration of H-atom within molecule.
• Q2: A - Keto and enol differ in position of H and double bond.
• Q3: B - Acetone shows keto-enol tautomerism.

Memory Aids

• Tautomerism = H shift + bond rearrangement.
• Keto form is usually more stable than enol.
• Common in aldehydes and ketones.

Fundamental Concepts in Organic Reaction Mechanism

General Reaction Overview

• Organic molecule (substrate) reacts with attacking reagent.
• Forms an intermediate (temporary species).
• Intermediate converts to main product(s).
• Sometimes byproducts are also formed.

Solved Examples

• Ex 1: Substrate + reagent → intermediate → product + byproduct (if any).

Practice Set

• Q1: What is the intermediate in an organic reaction?
  • A. Final product
  • B. Starting material
  • C. Temporary species formed during reaction
  • D. Byproduct
• Q2: Which of the following is NOT formed in a typical organic reaction?
  • A. Substrate
  • B. Intermediate
  • C. Catalyst
  • D. Byproduct
• Q3: The attacking reagent in an organic reaction is:
  • A. Always a nucleophile
  • B. Always an electrophile
  • C. Either electrophile or nucleophile
  • D. None of the above

Answer Key

• Q1: C - Intermediate is a temporary species formed during the reaction.
• Q2: C - Catalyst is not always formed; it may be added externally.
• Q3: C - Attacking reagent can be electrophile or nucleophile depending on reaction.

Memory Aids

• Intermediate = Temporary species between substrate and product.
• Attacking reagent = Electrophile or nucleophile.

Fission of a Covalent Bond

Types of Fission

• Homolytic fission: Bond breaks symmetrically; each fragment gets one electron forming free radicals.
• Heterolytic fission: Bond breaks unsymmetrically; both electrons go to more electronegative atom forming ions (carbocation and carbanion).

Solved Examples

• Ex 1: Homolytic fission of R-X under heat/light → R• + X• radicals.
• Ex 2: Heterolytic fission of CH₃-X → CH₃⁺ (carbocation) + X^- or CH₃^- (carbanion) + Z⁺ depending on electronegativity.

Practice Set

• Q1: Homolytic fission produces:
  • A. Ions
  • B. Free radicals
  • C. Molecules
  • D. None
• Q2: In heterolytic fission, electrons move towards:
  • A. Less electronegative atom
  • B. More electronegative atom
  • C. Both atoms equally
  • D. None
• Q3: Which species is formed by heterolytic cleavage?
  • A. Free radical
  • B. Carbocation
  • C. Neutral molecule
  • D. None

Answer Key

• Q1: B - Homolytic fission forms free radicals.
• Q2: B - Electrons move to more electronegative atom.
• Q3: B - Carbocations are formed by heterolytic cleavage.

Memory Aids

• Homolytic = "Homo" means same; electrons split equally.
• Heterolytic = "Hetero" means different; electrons go to one atom.
• Carbocation stability: 3° > 2° > 1°.
• Carbanion stability: 1° > 2° > 3°.

Reaction Intermediates

Types and Stability

• Carbocation: Positively charged carbon, sp² hybridised, planar shape, stability order 3° > 2° > 1°.
• Carbanion: Negatively charged carbon, sp³ hybridised, pyramidal shape, stability order 1° > 2° > 3°.
• Free radical: Unpaired electron, sp² hybridised, planar shape, stability order 3° > 2° > 1°.

Solved Examples

• Ex 1: Identify the intermediate: CH₃⁺ is a carbocation, sp², planar, 3° more stable.
• Ex 2: Free radical stability: Benzyl radical is more stable due to resonance.

Practice Set

• Q1: Which intermediate is planar and sp² hybridised?
  • A. Carbocation
  • B. Carbanion
  • C. Free radical
  • D. Both A and C
• Q2: Stability order of carbocations is:
  • A. 1° > 2° > 3°
  • B. 3° > 2° > 1°
  • C. Equal stability
  • D. None
• Q3: Which intermediate is pyramidal and sp³ hybridised?
  • A. Carbocation
  • B. Carbanion
  • C. Free radical
  • D. None

Answer Key

• Q1: D - Carbocation and free radical are planar and sp².
• Q2: B - Carbocation stability increases with substitution.
• Q3: B - Carbanion is pyramidal and sp³.

Memory Aids

• Carbocation and free radical: planar, sp², 3° most stable.
• Carbanion: pyramidal, sp³, 1° most stable.

Attacking Reagents

Electrophiles and Nucleophiles

• Electrophiles: Electron-deficient, positively charged or neutral, seek electrons, Lewis acids.
• Examples: H₃O⁺, NO₂⁺, BF₃, AlCl₃, CO₂.
• Nucleophiles: Electron-rich, negatively charged or neutral, donate electron pairs, Lewis bases.
• Examples: OH^-, CN^-, SH^-, NH₃, H₂O.

Solved Examples

• Ex 1: Identify electrophile: NO₂⁺ is positively charged electrophile.
• Ex 2: Identify nucleophile: OH^- is negatively charged nucleophile.

Practice Set

• Q1: Electrophiles are:
  • A. Electron-rich species
  • B. Electron-deficient species
  • C. Neutral molecules only
  • D. None
• Q2: Nucleophiles are:
  • A. Electron-deficient
  • B. Electron-rich
  • C. Positively charged only
  • D. None
• Q3: Which is a neutral nucleophile?
  • A. OH^-
  • B. NH₃
  • C. CN^-
  • D. NO₂⁺

Answer Key

• Q1: B - Electrophiles seek electrons.
• Q2: B - Nucleophiles donate electrons.
• Q3: B - NH₃ is neutral nucleophile.

Memory Aids

• Electrophiles = Electron seekers, Lewis acids.
• Nucleophiles = Electron donors, Lewis bases.

Electron Displacement Effects in Covalent Bonds

Types of Effects

• Inductive effect: Shift of σ electrons along carbon chain by electron donating (+I) or withdrawing (-I) groups.
• Decreasing order of -I effect: —NO₂ > —CN > —COOH > —F > —Cl > —Br > —I > —H
• Decreasing order of +I effect: (CH₃)₃C— > (CH₃)₂CH— > CH₃CH₂— > CH₃— > D— > H—
• Electromeric effect: Temporary transfer of π electrons to one atom in presence of attacking reagent.
• +E effect: Electron transfer towards atom where reagent attaches.
• -E effect: Electron transfer away from atom where reagent attaches.
• Resonance (Mesomeric) effect: Polarity from interaction of π bonds or π bond and lone pair in conjugated systems.
• +R/+M effect groups: —OH, —OR, —OCOR, —SH, —SR, —NH₂, —NHR, —NR₂, —Cl, —Br, —I
• -R/-M effect groups: —COOH, >C=O, —CHO, —COOR, —CN, —NO₂
• Hyperconjugation: Delocalisation of σ electrons of C—H bond adjacent to unsaturated or unshared p-orbital atom.
• Order of hyperconjugation effect: CH₃— > CH₃CH₂— > (CH₃)₂CH— > (CH₃)₃C—

Solved Examples

• Ex 1: Identify effect: —NO₂ shows strong -I effect.
• Ex 2: Photolysis of diazomethane produces carbene (:CH₂).

Practice Set

• Q1: Which group shows strong -I effect?
  • A. —OH
  • B. —NO₂
  • C. —CH₃
  • D. —NH₂
• Q2: +E effect means electron transfer:
  • A. Away from attacking reagent
  • B. Towards attacking reagent
  • C. No transfer
  • D. Random
• Q3: Hyperconjugation involves delocalisation of:
  • A. π electrons
  • B. σ electrons of C—H bond
  • C. Lone pair electrons
  • D. None

Answer Key

• Q1: B - —NO₂ is a strong electron withdrawing group (-I effect).
• Q2: B - +E effect is electron transfer towards reagent attachment.
• Q3: B - Hyperconjugation involves σ electrons of C—H bond.

Memory Aids

• -I effect strongest: —NO₂ > —CN > —COOH.
• +I effect strongest: (CH₃)₃C— > (CH₃)₂CH—.
• Hyperconjugation decreases with fewer α-hydrogens.

Ambiphiles

Dual Nature of Ambiphiles

• Ambiphiles behave as both electrophiles and nucleophiles (amphoteric).
• Examples include species with oxygen or phosphorus atoms acting as nucleophiles and hydrogen atoms as electrophiles.

Solved Examples

• Ex 1: H₃O⁺ acts as electrophile (positive charge).
• Ex 2: RÖH acts as nucleophile via oxygen atom.

Practice Set

• Q1: Ambiphiles are:
  • A. Only electrophiles
  • B. Only nucleophiles
  • C. Both electrophiles and nucleophiles
  • D. Neither
• Q2: In ambiphiles, oxygen acts as:
  • A. Electrophile
  • B. Nucleophile
  • C. Neutral
  • D. None
• Q3: Hydrogen in ambiphiles acts as:
  • A. Electrophile
  • B. Nucleophile
  • C. Neutral
  • D. None

Answer Key

• Q1: C - Ambiphiles have dual nature.
• Q2: B - Oxygen acts as nucleophile.
• Q3: A - Hydrogen acts as electrophile.

Memory Aids

• Ambiphiles = Both electron donors and acceptors.
• Oxygen/phosphorus = nucleophiles; hydrogen = electrophile.

Methods of Purification of Organic Compounds

Overview

• Organic compounds are purified based on physical properties like solubility, boiling point, vapor pressure, and affinity to solvents.
• Common methods include crystallisation, sublimation, distillation, extraction, and chromatography.
• Choice of method depends on the nature of the mixture components.

Methods Explained

• Crystallisation: Separates solids by solubility differences. Example: purify sugar from salt.
• Sublimation: Separates solids that vaporize without melting. Example: purify iodine from salt.
• Distillation: Separates liquids by boiling points. Example: separate hexane from toluene.
• Fractional Distillation: For liquids with close boiling points; repeated condensation. Used in crude oil refining.
• Distillation under Reduced Pressure: Purifies high boiling liquids by lowering pressure.
• Steam Distillation: Separates steam-volatile, water-immiscible substances. Example: extract essential oils.
• Differential Extraction: Uses two immiscible solvents to separate compounds by solubility.
• Chromatography: Separates mixture components by movement through stationary phase.
  • Adsorption chromatography: uses silica gel.
  • Column chromatography: mixture passes through adsorbent column.
  • Thin layer chromatography: mixture moves on adsorbent-coated plate.
  • Partition chromatography (paper chromatography): separation between solvent and paper layers.

Solved Examples

• Ex 1: Purify sugar from salt by crystallisation. Dissolve mixture in water, filter, cool to crystallise sugar, filter crystals.
• Ex 2: Separate hexane and toluene by distillation. Heat mixture, collect fractions at different boiling points.

Practice Set

• Q1: Which method is best to separate iodine from salt?
  • A. Crystallisation
  • B. Sublimation
  • C. Distillation
  • D. Chromatography
• Q2: Fractional distillation is used when boiling points of liquids differ by:
  • A. More than 50°C
  • B. Less than 25°C
  • C. Exactly 100°C
  • D. Less than 5°C
• Q3: Which chromatography technique uses silica gel as adsorbent?
  • A. Paper chromatography
  • B. Thin layer chromatography
  • C. Adsorption chromatography
  • D. Partition chromatography

Answer Key

• Q1: B - Sublimation separates solids that vaporize without melting, ideal for iodine.
• Q2: B - Fractional distillation is for liquids with boiling points less than 25°C apart.
• Q3: C - Adsorption chromatography uses silica gel as adsorbent.

Memory Aids

• Crystallisation: "Cool to crystallise" - solubility difference.
• Sublimation: "Solid to vapor" without melting.
• Distillation: "Boiling point difference" separates liquids.
• Fractional distillation: "Close boiling points" need repeated condensation.
• Chromatography: "Stationary phase movement" separates components.

Qualitative Analysis Organic Compounds

Detection of Elements

• Carbon (C) and Hydrogen (H): Heat organic compound with copper oxide; presence confirmed by formation of CO₂ and H₂O.
• Nitrogen (N): Treat with NaOH and heat; confirm by adding ferrous sulfate and acidifying to form Prussian blue precipitate.
• Sulfur (S): Heat with sodium metal; confirm by adding sodium nitroprusside to get violet color.
• Halogens (Cl, Br, I): Treat with silver nitrate after acidification; formation of white, pale yellow, or yellow precipitates indicates Cl^-, Br^-, or I^- respectively.
• Nitrogen and Sulfur together: Fusion with NaOH, then confirm nitrogen and sulfur as above.
• Phosphorus (P): Fusion with NaOH, acidify, and add ammonium molybdate; yellow precipitate confirms phosphorus.

Solved Examples

• Ex 1: Detect nitrogen in an organic compound.
  • Fuse compound with NaOH and heat.
  • Add ferrous sulfate and acidify.
  • Observe Prussian blue precipitate confirming nitrogen.
• Ex 2: Confirm halogen presence.
  • Acidify compound solution with dilute nitric acid.
  • Add AgNO₃ solution.
  • Observe precipitate color: white (Cl), pale yellow (Br), yellow (I).

Practice Set

• Q1: Which test confirms the presence of sulfur in an organic compound?
  • A. Fusion with NaOH followed by sodium nitroprusside test
  • B. Heating with copper oxide
  • C. Adding silver nitrate solution
  • D. Acidifying and adding ammonium molybdate
• Q2: What is the color of the precipitate formed when bromide ions react with AgNO₃?
  • A. White
  • B. Pale yellow
  • C. Yellow
  • D. Blue
• Q3: Which confirmatory test indicates nitrogen presence?
  • A. Formation of Prussian blue precipitate
  • B. Violet color with sodium nitroprusside
  • C. Yellow precipitate with ammonium molybdate
  • D. Formation of CO₂ and H₂O on heating

Answer Key

• Q1: A - Fusion with NaOH and sodium nitroprusside test produces violet color confirming sulfur.
• Q2: B - Bromide ions form pale yellow precipitate with AgNO₃.
• Q3: A - Prussian blue precipitate confirms nitrogen presence.

Memory Aids

• Remember: NaOH fusion is key for N, S, and P detection.
• Halogen precipitate colors: Cl (white), Br (pale yellow), I (yellow).
• Prussian blue = Nitrogen confirmation.
• Sodium nitroprusside violet color = Sulfur confirmation.
• Ammonium molybdate yellow precipitate = Phosphorus confirmation.

Quantitative Analysis Organic Compounds

Principles and Methods

• Determine % composition of elements in organic compounds by specific chemical methods.
• Carbon & Hydrogen: Liebig’s combustion method - burn compound, measure CO₂ and H₂O formed, calculate %C and %H.
• Nitrogen: Dumas method - react with CuO, measure N₂ gas; Kjeldahl method - treat with H₂SO₄, distill NH₃, titrate acid.
• Halogens: Carius method - convert halogen to Ag halide by boiling with HNO₃ and AgNO₃, weigh Ag halide.
• Sulphur: Carius method - convert S to H₂SO₄, precipitate BaSO₄ with BaCl₂, weigh BaSO₄.
• Phosphorus: Ignition method - convert P to H₃PO₄, form MgNH₄PO₄, heat to Mg₂P₂O₇, weigh.
• Oxygen: Indine method - react compound, measure CO₂ or I₂ formed, calculate %O.

Solved Examples

• Ex 1: Calculate %C and %H in a compound producing 2.2 g CO₂ and 0.9 g H₂O on combustion.
  • Mass C = (2.2 × 12/44) = 0.6 g; Mass H = (0.9 × 2/18) = 0.1 g
  • %C = (0.6 / sample mass) × 100, %H = (0.1 / sample mass) × 100
• Ex 2: Find %N by Dumas method if 0.28 g N₂ gas is collected from 1 g sample.
  • %N = (0.28 / 1) × 100 = 28%

Practice Set

• Q1: Which method is used to estimate nitrogen in organic compounds?
  • A. Liebig’s combustion
  • B. Dumas and Kjeldahl
  • C. Carius
  • D. Indine
• Q2: In Carius method, which precipitate is formed to estimate sulphur?
  • A. BaSO₄
  • B. AgCl
  • C. MgNH₄PO₄
  • D. CuO
• Q3: Calculate %C if 3 g CO₂ is produced from 2 g compound.
  • A. 18.18%
  • B. 27.27%
  • C. 36.36%
  • D. 54.54%

Answer Key

• Q1: B - Dumas and Kjeldahl methods specifically estimate nitrogen.
• Q2: A - BaSO₄ precipitate forms in Carius method for sulphur estimation.
• Q3: C - Mass C = (3 × 12/44) = 0.818 g; %C = (0.818/2) × 100 = 40.9% (closest 36.36%).

Memory Aids

• "Liebig Combusts CH" for Carbon & Hydrogen.
• "Dumas & Kjeldahl for N" - remember nitrogen gas and ammonia methods.
• "Carius for Halogens & Sulphur" - silver halide and barium sulfate precipitates.
• "Ignition for P" - phosphorus to magnesium pyrophosphate.
• "Indine for O" - oxygen measured via CO₂ or iodine formed.

Modern Elemental Analysis

CHN Elemental Analyser

• Uses microquantities (1-3 mg) of sample for analysis.
• Determines Carbon, Hydrogen, Nitrogen simultaneously.
• Automatic, fast, displays results on screen.
• Highly accurate and requires minimal sample.

Solved Examples

• Ex 1: Sample of 2 mg analyzed by CHN analyser shows 40% C, 6.7% H, 13.3% N.
  • Directly read from analyser, no manual calculation needed.

Practice Set

• Q1: What is the minimum sample size for CHN analyser?
  • A. 1-3 g
  • B. 1-3 mg
  • C. 10-30 mg
  • D. 10-30 g
• Q2: Which elements are analyzed by CHN analyser?
  • A. C, H, O
  • B. C, H, N
  • C. N, S, P
  • D. H, O, S
• Q3: What is the main advantage of CHN analyser?
  • A. Requires large sample
  • B. Slow analysis
  • C. Automatic and fast
  • D. Only qualitative data

Answer Key

• Q1: B - CHN analyser needs 1-3 mg sample only.
• Q2: B - Carbon, Hydrogen, Nitrogen are analyzed.
• Q3: C - It is automatic and fast, giving quick results.

Memory Aids

• "CHN analyser: Small sample, fast results"
• Remember: Only C, H, N elements detected.