Module 2 · NEET Chemistry

Organic Chemistry

Hydrocarbons, functional groups, reaction mechanisms, biomolecules, polymers.
Hydrocarbons · Functional Groups · Mechanisms · Biomolecules
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Learning Objectives

  • Understand hybridization (sp³, sp², sp) and tetravalency of carbon in organic compounds
  • Apply IUPAC nomenclature rules to name complex organic molecules systematically
  • Analyse electron displacement effects — inductive, resonance, hyperconjugation, electromeric
  • Distinguish between structural and stereoisomerism with solved NEET examples
  • Predict products of hydrocarbon reactions: Markovnikov addition, ozonolysis, electrophilic substitution
  • Compare SN1/SN2 and E1/E2 mechanisms with stereochemical outcomes
  • Master functional group interconversions: alcohols to aldehydes to acids and beyond
  • Identify biomolecules, polymers, and drugs relevant to NEET and everyday chemistry

1. Fundamental Concepts & Nomenclature

Tetravalency & Hybridization

Carbon has electronic configuration 1s² 2s² 2p². It exhibits tetravalency by promoting one 2s electron to the 2p orbital, forming four equivalent sp³ hybrid orbitals in methane. The type of hybridization determines the geometry and bond angles of the molecule.

Carbon can adopt three hybridization states depending on the number of sigma bonds and lone pairs.

HybridizationOrbitals UsedGeometryBond AngleExamples
sp³1s + 3pTetrahedral109.5°CH₄, C₂H₆, C₂H₅OH
sp²1s + 2pTrigonal planar120°C₂H₄, C₆H₆, HCHO
sp1s + 1pLinear180°C₂H₂, HCN, CO₂
Trick: s-Character & Acidity
Greater the s-character, more electronegative the carbon and more acidic the C—H bond. sp > sp² > sp³. Hence acetylene (sp) is acidic while ethane (sp³) is not.
Trick: Bond Length & s-Character
More s-character means shorter and stronger bond. C—H bond lengths: sp³ (1.09 Å) > sp² (1.08 Å) > sp (1.06 Å). C—C bond lengths: C≡C (1.20 Å) < C=C (1.34 Å) < C—C (1.54 Å).

Sigma (σ) & Pi (π) Bonds

All covalent bonds are classified as σ or π bonds. A σ bond is formed by the head-on (end-to-end) overlap of atomic orbitals along the internuclear axis. It is strong, cylindrically symmetrical, and allows free rotation. A π bond is formed by the lateral (sideways) overlap of unhybridised p-orbitals. It is weaker than a σ bond and restricts rotation (giving rise to geometrical isomerism).

Single bonds (C—C) consist of one σ bond. Double bonds (C=C) consist of one σ + one π bond. Triple bonds (C≡C) consist of one σ + two π bonds (the second π bond is perpendicular to the first). The bond strength order: C≡C > C=C > C—C. The π electrons are more exposed and more polarizable than σ electrons, making alkenes and alkynes more reactive than alkanes towards electrophilic addition.

Bond TypeOrbital OverlapStrengthRotationElectron Location
σ (single)Head-on (s-s, s-p, p-p)StrongFree rotationBetween nuclei (internuclear axis)
π (in double bond)Sideways (p-p)Weaker than σRestrictedAbove and below the internuclear axis
π (in triple bond)Two perpendicular sideways overlapsTwo π bonds, one σRestrictedAround the axis in two planes
Example — Bond Length & Bond Order
Arrange the following in increasing order of C—C bond length: C₂H₂, C₂H₄, C₂H₆
Solution: Bond order: C≡C (3) > C=C (2) > C—C (1). Higher bond order means shorter bond. Increasing bond length: C₂H₂ (1.20 Å) < C₂H₄ (1.34 Å) < C₂H₆ (1.54 Å).

IUPAC Nomenclature

The International Union of Pure and Applied Chemistry (IUPAC) system provides a standard method for naming organic compounds. The process involves three steps: 1. Identify the longest continuous carbon chain (parent chain). 2. Number the chain to give the lowest locants to substituents. 3. Name substituents in alphabetical order.

Alkanes, Alkenes, Alkynes

Alkanes end in -ane (methane, ethane, propane). Alkenes end in -ene (ethene, propene) and alkynes end in -yne (ethyne, propyne). For alkenes and alkynes, the position of the multiple bond is indicated by the lowest possible locant.

Functional Groups

When multiple functional groups are present, priority determines the suffix. The principal functional group is used as the suffix; all others are treated as prefixes.

PriorityFunctional GroupFormulaSuffixPrefix
1Carboxylic acid—COOH-oic acidcarboxy
2Sulfonic acid—SO₃H-sulfonic acidsulfo
3Ester—COOR-oatealkoxycarbonyl
4Acyl halide—COX-oyl halidehalocarbonyl
5Amide—CONH₂-amidecarbamoyl
6Nitrile—C≡N-nitrilecyano
7Aldehyde—CHO-alformyl / oxo
8Ketone—CO—-oneoxo
9Alcohol—OH-olhydroxy
10Amine—NH₂-amineamino

Priority Rules

When numbering, the principal functional group gets the lowest number. If there is a tie, the first point of difference rule is applied. Substituents are listed alphabetically (ignoring prefixes di-, tri-, tetra-).

Common Names vs IUPAC Names

Many organic compounds are still widely known by their common (trivial) names. NEET often tests both. Important examples: HCHO — Formaldehyde (methanal); CH₃CHO — Acetaldehyde (ethanal); C₆H₅CHO — Benzaldehyde; HCOOH — Formic acid (methanoic acid); CH₃COOH — Acetic acid (ethanoic acid); C₆H₅OH — Phenol (benzenol); CHCl₃ — Chloroform (trichloromethane); CCl₄ — Carbon tetrachloride (tetrachloromethane); CH₂=CH₂ — Ethylene (ethene); CH≡CH — Acetylene (ethyne); C₆H₆ — Benzene; CH₃COCH₃ — Acetone (propanone); CH₃CH₂OH — Ethyl alcohol (ethanol).

Common NameIUPAC NameFormula
FormaldehydeMethanalHCHO
AcetaldehydeEthanalCH₃CHO
AcetonePropanoneCH₃COCH₃
Formic acidMethanoic acidHCOOH
Acetic acidEthanoic acidCH₃COOH
ChloroformTrichloromethaneCHCl₃
EthyleneEtheneCH₂=CH₂
AcetyleneEthyneHC≡CH
GlycerolPropane-1,2,3-triolHOCH₂CH(OH)CH₂OH
AnilineBenzenamineC₆H₅NH₂
Example 1 — IUPAC Nomenclature
Give the IUPAC name of CH₃—CH(Cl)—CH₂—CH(OH)—COOH.

a) 3-chloro-2-hydroxypentanoic acid
b) 3-chloro-2-hydroxypentanoic acid
c) 2-hydroxy-3-chloropentanoic acid
d) 3-chloro-4-hydroxypentanoic acid
Solution: The principal group is COOH (pentanoic acid). Number from COOH carbon. OH on C-2, Cl on C-3. Alphabetically "chloro" comes before "hydroxy". Hence 3-chloro-2-hydroxypentanoic acid. Option (a).
Example 2 — Naming with Multiple Bonds
The IUPAC name of CH₂=C(CH₃)—CH=CH₂ is:

a) 3-methylbuta-1,3-diene
b) 2-methylbuta-1,3-diene
c) Isoprene
d) Both b and c
Solution: Four-carbon chain with two double bonds (diene). Number to give lowest locants to the double bonds: C1=C2—C(CH₃)=C3—C4. Methyl on C-2. IUPAC: 2-methylbuta-1,3-diene. Common name: Isoprene. Option (d).
Mnemonic: Functional Group Priority
"Crazy Students Eat Acidic Apples Near A Kitchen Area" — Carboxylic acid > Sulfonic > Ester > Acyl halide > Amide > Nitrile > Aldehyde > Ketone > Amine

Electron Displacement Effects

These effects govern the electron density distribution in a molecule, influencing reactivity, acidity, basicity, dipole moment, and orientation in substitution reactions. Understanding these effects is essential for predicting reaction outcomes in NEET organic chemistry. There are four main types: inductive, resonance (mesomeric), hyperconjugation, and electromeric.

Inductive Effect (I-effect)

Permanent polarization along a sigma bond due to electronegativity difference. It is a through-bond effect that diminishes with distance (practically negligible after 3-4 bonds). Groups that withdraw electrons via sigma bonds are -I groups: —NO₂, —CN, —COOH, —F, —Cl, —Br, —I, —OH, —OR, —NH₃⁺, —CHO, —COR. Groups that donate electrons via sigma bonds are +I groups: —CH₃, —C₂H₅, —CH(CH₃)₂, —C(CH₃)₃ (alkyl groups), —O⁻, —COO⁻.

Consequences of Inductive Effect: (1) Acidity of carboxylic acids: Electron-withdrawing groups increase acidity by stabilizing the conjugate base. Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH. (2) Basicity of amines: Electron-donating alkyl groups increase electron density on nitrogen, enhancing basicity. (3) Dipole moment: —I groups create a dipole with the positive end on the group.

Resonance / Mesomeric Effect (M-effect)

Delocalization of pi electrons or lone pairs through conjugated systems. It is represented by drawing multiple contributing structures (resonance structures) connected by double-headed arrows (↔). The actual molecule is a hybrid of all contributing structures and is more stable than any individual contributing structure. Rules for resonance: (1) Only π electrons and lone pairs move; σ bonds are never broken. (2) All contributing structures must have the same atomic framework. (3) The most stable structure has the maximum number of covalent bonds and minimal charge separation. +M groups donate electrons via resonance: —OH, —NH₂, —OR, —NHR, —NR₂, —X (F, Cl, Br, I). -M groups withdraw electrons via resonance: —NO₂, —CN, —CHO, —COR, —COOH, —COOR, —CONH₂.

Note: The mesomeric effect is stronger than the inductive effect. For example, —OH is strongly electron-donating by resonance (+M) but weakly electron-withdrawing by induction (-I). The net effect determines the overall electron density distribution.

Hyperconjugation (Baker-Nathan Effect)

The interaction between sigma (σ) electrons of a C—H (or C—C) bond with an adjacent empty or partially filled p-orbital or pi (π*) system. Also called no-bond resonance because it can be represented by resonance structures where the C—H bond is broken. The number of α-hydrogen atoms directly correlates with the extent of hyperconjugation: more α-H means greater delocalization and more stabilization. Applications: (1) Carbocation stability: (CH₃)₃C⁺ (9 α-H) > (CH₃)₂CH⁺ (6 α-H) > CH₃CH₂⁺ (3 α-H) > CH₃⁺ (0 α-H). (2) Alkene stability: more highly substituted alkenes are more stable (R₂C=CR₂ > R₂C=CHR > R₂C=CH₂ > RHC=CH₂). (3) Free radical stability: 3° > 2° > 1° > CH₃·.

Electromeric Effect (E-effect)

The complete transfer of a pi-electron pair to one of the atoms at the demand of an attacking reagent. It is a temporary, polarizable effect — it occurs only at the moment of attack by a reagent and ceases once the reagent is removed. +E effect: The π-electron pair moves towards the attacking reagent (observed in alkenes during electrophilic addition). -E effect: The π-electron pair moves away from the attacking reagent (observed in carbonyl compounds during nucleophilic addition). The electromeric effect is represented by a curved arrow showing the direction of electron movement.

EffectTypePermanent / TemporaryTransmission
Inductiveσ-electron displacementPermanentThrough sigma bonds
Resonanceπ-electron delocalizationPermanentThrough conjugated pi system
Hyperconjugationσ → π interactionPermanentThrough overlapping orbitals
Electromericπ-electron transferTemporaryAt the point of attack
Example 2 — Electron Displacement
Arrange the following in decreasing order of acidic strength:
CH₃COOH, ClCH₂COOH, Cl₂CHCOOH, Cl₃CCOOH
Solution: Chlorine is -I (electron withdrawing). More Cl atoms means stronger -I effect, more stabilized conjugate base, hence stronger acid. Order: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH.
Trick: Acidity and Inductive Effect
EWG (—NO₂, —CN, —X) increase acidity by stabilizing conjugate base. EDG (—CH₃, —C₂H₅) decrease acidity. More electronegative the substituent and closer to COOH, stronger the acid.

Reactive Intermediates

Carbocations, Carbanions, Free Radicals

Carbocations (R₃C⁺) are electron-deficient, planar (sp²), and stabilized by resonance and hyperconjugation. Stability order: 3° > 2° > 1° > methyl. Carbanions (R₃C⁻) are electron-rich, pyramidal (sp³), and stability is reverse: 1° > 2° > 3°. Free radicals (R₃C·) are neutral species with an unpaired electron; stability: 3° > 2° > 1° > methyl.

IntermediateHybridizationGeometryStability Trend
Carbocationsp²Trigonal planar3° > 2° > 1° > CH₃⁺
Carbanionsp³PyramidalCH₃⁻ > 1° > 2° > 3°
Free radicalsp² (approx.)Trigonal planar3° > 2° > 1° > CH₃·
Example 3 — Carbocation Stability
Which of the following carbocations is most stable?

a) CH₃⁺   b) CH₃CH₂⁺   c) (CH₃)₂CH⁺   d) (CH₃)₃C⁺
Solution: Tertiary carbocation (3°) has the greatest hyperconjugation (9 α-H) and is most stabilized. Order: (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺. Hence option (d).

2. Isomerism

Isomers are different compounds that have the same molecular formula but differ in the arrangement of atoms. Isomerism is broadly classified into structural isomerism (different connectivity) and stereoisomerism (same connectivity, different spatial arrangement). This topic is highly scoring in NEET, with frequent questions on chiral centres, geometrical isomerism, and conformational analysis.

Structural Isomerism

Chain, Position, Functional, Metamerism

Chain isomerism: Different carbon skeleton (n-butane vs isobutane). Branched isomers have lower boiling points due to reduced surface area. Position isomerism: Same skeleton, different position of substituent or functional group (1-propanol vs 2-propanol). Position isomers often have different chemical reactivity — 1° alcohols oxidize to aldehydes/acids while 2° alcohols oxidize to ketones. Functional group isomerism: Same molecular formula but different functional group (C₂H₅OH alcohol vs CH₃OCH₃ ether; C₂H₄O₂ can be acetic acid, methyl formate, or glycolaldehyde). Metamerism: Different alkyl groups on either side of a polyvalent functional group such as —O—, —CO—, —NH— (C₂H₅—O—C₂H₅ vs CH₃—O—C₃H₇). Tautomerism: A special type of functional group isomerism where two isomers (tautomers) exist in dynamic equilibrium, differing in the position of a proton and a double bond. Keto-enol tautomerism is the most common type. The keto form is generally more stable except when the enol form is stabilized by resonance (phenol) or intramolecular hydrogen bonding (β-diketones).

TypeExample PairFormulaKey Feature
Chainn-butane / isobutaneC₄H₁₀Different carbon skeleton
Position1-chloropropane / 2-chloropropaneC₃H₇ClSame skeleton, different substituent position
Functional groupEthanol / Dimethyl etherC₂H₆OAlcohol vs ether functional groups
MetamerismDiethyl ether / Methyl propyl etherC₄H₁₀ODifferent alkyl groups around O
TautomerismKeto / Enol (acetoacetic ester)C₄H₆O₂Proton and double bond shift
Ring-chainCyclopropane / PropeneC₃H₆Cyclic vs open-chain structure
Example 4 — Structural Isomerism
How many structural isomers (including stereoisomers) are possible for C₄H₁₀O?

a) 4   b) 7   c) 8   d) 6
Solution: Alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol (4). Ethers: CH₃OC₃H₇, C₂H₅OC₂H₅, CH₃OCH(CH₃)₂ (3). Total 7. Option (b).

Stereoisomerism

Stereoisomers have the same structural formula (same connectivity) but differ in the three-dimensional arrangement of atoms. They are classified into geometrical (cis-trans/E-Z), optical (enantiomers/diastereomers), and conformational isomerism.

Geometrical Isomerism (Cis-Trans / E-Z Isomerism)

Arises due to restricted rotation about a double bond (C=C) or in cyclic compounds. For a compound to exhibit geometrical isomerism, each of the two doubly bonded carbons must have two different substituents. In the cis-trans system, identical groups on the same side = cis; opposite sides = trans. The E-Z system (Cahn-Ingold-Prelog priority rules) assigns priority based on atomic number: if higher-priority groups are on the same side → Z (zusammen, German for \"together\"); opposite sides → E (entgegen, \"opposite\"). The E-Z system is preferred when the cis-trans system is ambiguous (e.g., 1-bromo-1-chloropropene has Br and Cl on one carbon and CH₃ and H on the other). Geometrical isomers differ in physical properties (melting point, boiling point, dipole moment) and sometimes in chemical reactivity.

Example — Geometrical Isomerism Counting
The number of geometrical isomers possible for CH₃—CH=CH—CH=CH—CH₃ is:

a) 2   b) 3   c) 4   d) 6
Solution: The molecule has two double bonds, each capable of cis-trans isomerism. Number of geometrical isomers = 2ⁿ = 2² = 4 (cis-cis, cis-trans, trans-cis, trans-trans). Option (c).

Optical Isomerism

Compounds that rotate plane-polarized light are optically active. A carbon with four different substituents is a chiral centre (stereocenter, asymmetric carbon). Enantiomers are non-superimposable mirror-image molecules. They have identical physical properties (melting/boiling point, density, solubility in achiral solvents) but differ in the direction of rotation of plane-polarized light — one is dextrorotatory (+), the other levorotatory (−). A 1:1 mixture of enantiomers is a racemic mixture (racemate), which is optically inactive. The R/S system (Cahn-Ingold-Prelog) assigns absolute configuration: assign priority (1→4) based on atomic number, orient the lowest priority group (4) away from view, then check the direction of 1→2→3; clockwise = R (rectus), anticlockwise = S (sinister). Diastereomers are stereoisomers that are not mirror images; they differ in physical properties (melting point, boiling point, solubility). Meso compounds: Achiral compounds with multiple chiral centres but an internal plane of symmetry (e.g., meso-tartaric acid). They are optically inactive. Number of optical isomers: For a compound with n chiral centres, the maximum number of optical isomers = 2ⁿ (if no meso form exists). If meso forms are possible, the number is less than 2ⁿ.

Conformational Isomerism

Different spatial arrangements achieved by rotation about single (σ) bonds. Conformers interconvert rapidly at room temperature. Newman projections visualize conformations by looking along a C—C bond axis. Ethane: Staggered conformation (more stable, torsional strain minimized, 0° dihedral angle between bonds) and eclipsed conformation (less stable, torsional strain ~12 kJ/mol due to bond-bond repulsion). Butane: Anti (most stable, dihedral angle 180°), gauche (less stable, 60°, steric strain from gauche butane interaction), and fully eclipsed (least stable). Cyclohexane: Chair conformation (most stable, all C—C bond angles ~109.5°, no angle strain), boat (less stable due to flagpole interactions), and twist-boat. Axial bonds (↑↓, parallel to the axis of the ring) and equatorial bonds (↗↘, pointing outward) interchange via ring flip. For substituted cyclohexanes, the bulkier substituent prefers the equatorial position to minimize 1,3-diaxial interactions. The preference is quantified by the A-value (energy difference between axial and equatorial). For methyl, A ≈ 1.74 kcal/mol; for tert-butyl, A ≈ 5.0 kcal/mol (essentially locks the ring).

Example 5 — Optical Isomerism
Which of the following compounds is optically active?

a) CH₃CH₂OH   b) CH₃CH(OH)COOH   c) CH₃COOH   d) CCl₄
Solution: Lactic acid, CH₃CH(OH)COOH, has a chiral carbon (C with —H, —OH, —CH₃, —COOH). Hence option (b) is optically active.
Mnemonic: Optical Activity
"CARB — Carbon with All Radicals Be different" = Chiral centre. Check if any carbon has four different substituents.

3. Hydrocarbons

Hydrocarbons are compounds composed exclusively of carbon and hydrogen. They are classified as aliphatic (alkanes, alkenes, alkynes) and aromatic (benzene and its derivatives). Hydrocarbons form the fundamental building blocks of organic chemistry and are a core topic in the NEET syllabus, with questions on preparation, reactions, and mechanisms.

Alkanes (CₙH₂ₙ₊₂)

Preparation of Alkanes

1. Catalytic hydrogenation (Sabatier-Senderens): Alkene/alkyne + H₂ (Ni/Pt/Pd, 200-300°C) → alkane. This is a syn-addition of H₂. 2. Wurtz reaction: 2R—X + 2Na (dry ether) → R—R + 2NaX. Suitable only for symmetrical alkanes. Cross-coupling gives a mixture of three products. 3. Reduction of alkyl halides: R—X + Zn/HCl → R—H + ZnX₂ (or LiAlH₄/NaBH₄). 4. Decarboxylation (soda lime): R—COONa + NaOH/CaO (heat) → R—H + Na₂CO₃. The product alkane has one carbon less than the parent carboxylic acid. 5. Kolbe's electrolysis: Electrolysis of concentrated aqueous solution of sodium/potassium salt of carboxylic acid yields alkane at the anode: 2RCOO⁻ → R—R + 2CO₂ + 2e⁻. The alkane formed has double the number of carbon atoms. 6. Frankland reaction: 2R—X + Zn (in presence of water) → 2R—H + ZnX₂. 7. Grignard reagent with water: RMgX + H₂O → R—H + MgX(OH).

Reactions of Alkanes

1. Free radical halogenation: Alkanes + Cl₂/Br₂ (UV light or heat, 300-400°C) → haloalkane + HX. The reactivity of halogens: F₂ > Cl₂ > Br₂ > I₂ (I₂ does not react). The ease of H-abstraction: 3° > 2° > 1°. Selectivity increases from Cl₂ (low selectivity, almost statistical ratio) to Br₂ (high selectivity, predominantly 3° substitution). The reaction proceeds via free radical chain mechanism (initiation, propagation, termination). 2. Combustion: Alkane + O₂ → CO₂ + H₂O + heat. Incomplete combustion gives CO and soot. 3. Pyrolysis (cracking): High-temperature (500-700°C) decomposition in the absence of air yields smaller alkanes, alkenes, and H₂. 4. Isomerization: n-Alkanes → branched alkanes using AlCl₃/HCl catalyst. 5. Aromatization: Alkanes with 6+ carbons (Pt, 500°C) → aromatic compounds. Hexane → benzene + 4H₂.

ReactionReagent / ConditionProductKey Feature
Wurtz reaction2RX + 2Na (dry ether)R—R + 2NaXSymmetrical alkanes; cross-coupling gives mixture
Sabatier-SenderensAlkene + H₂ (Ni, 200-300°C)AlkaneCatalytic hydrogenation, syn addition
DecarboxylationRCOONa + NaOH/CaO, heatR—H + Na₂CO₃One carbon less than parent acid
Kolbe's electrolysis2RCOOK + H₂O (electrolysis)R—R + 2CO₂ + H₂ + 2KOHDoubles the number of carbon atoms
Frankland reactionRX + Zn + H₂O (trace)R—H + ZnX(OH)Reduction of alkyl halide
HalogenationCl₂/Br₂ + hv or heatR—X + HXFree radical chain mechanism
Example 6 — Wurtz Reaction
What is the product when a mixture of CH₃Br and C₂H₅Br is treated with sodium in dry ether?

a) C₂H₆   b) C₄H₁₀   c) C₃H₈   d) Mixture of C₂H₆, C₄H₁₀, and C₃H₈
Solution: Cross-coupling gives three products: CH₃—CH₃, CH₃—C₂H₅ (propane), and C₂H₅—C₂H₅. Option (d).

Alkenes (CₙH₂ₙ)

Preparation of Alkenes

1. Dehydration of alcohols: R—CH₂—CH₂OH + conc. H₂SO₄ (170°C) or Al₂O₃ (350°C) → R—CH=CH₂ + H₂O. Follows E1 (for 2°/3° alcohols) or E2 (for 1° alcohols) mechanism. Zaitsev product (more substituted alkene) is favoured. 2. Dehydrohalogenation of alkyl halides: R—CH₂—CHX—R' + alc. KOH (heat) → R—CH=CH—R' + KX + H₂O. E2 elimination with strong base. Favours more substituted alkene (Zaitsev) except with bulky bases (K-tBuO) that favour Hoffmann product. 3. Dehalogenation of vicinal dihalides: R—CHX—CHX—R' + Zn (alcohol) → R—CH=CH—R' + ZnX₂. 4. Partial reduction of alkynes: Lindlar's catalyst (Pd/CaCO₃ + quinoline, H₂) → cis-alkene. Na/NH₃ (Birch reduction) → trans-alkene.

Addition Reactions — Markovnikov & Peroxide Effect

Markovnikov's rule: In the addition of HX (HCl, HBr, HI) to an unsymmetrical alkene, the hydrogen (the electropositive part) adds to the carbon with more hydrogen atoms (the less substituted carbon), and the halogen adds to the carbon with fewer hydrogen atoms (the more substituted carbon). This is because the reaction proceeds through the more stable carbocation intermediate. Peroxide effect (Kharasch effect): In the presence of organic peroxides (ROOR), HBr adds against Markovnikov's rule via a free radical mechanism. The Br· radical (formed by homolytic cleavage of HBr initiated by the peroxide) adds to the terminal (less substituted) carbon to form a more stable carbon radical. HCl and HI do not show the peroxide effect (HCl has a bond too strong to cleave homolytically; HI is easily oxidized by peroxides).

Other addition reactions: Hydration (H₂O/dil. H₂SO₄ → Markovnikov alcohol via carbocation, may involve rearrangement). Oxymercuration-demercuration [Hg(OAc)₂/H₂O then NaBH₄] → Markovnikov alcohol without rearrangement. Hydroboration-oxidation [BH₃/THF then H₂O₂/OH⁻] → anti-Markovnikov alcohol (syn addition of H and OH). Halogenation (Br₂/CCl₄ → vicinal dibromide; brown colour of Br₂ is discharged — test for unsaturation). Ozonolysis (O₃ then Zn/H₂O → aldehydes/ketones; structure of alkene deduced from ozonolysis products). Hydrogenation (H₂/Ni → alkane, syn addition). Anti-dihydroxylation (cold KMnO₄ → vicinal diol, Baeyer's test; or OsO₄ → syn dihydroxylation). Hot KMnO₄ cleaves the double bond.

ReactionReagent / ConditionsProductStereochemistry / Orientation
Markovnikov additionHX (HCl, HBr, HI), no peroxideAlkyl halideH⁺ to C with more H (via carbocation)
Anti-MarkovnikovHBr + peroxide (ROOR)Alkyl bromideBr· to C with more H (free radical mechanism)
HydrationH₂O + dil. H₂SO₄AlcoholMarkovnikov; may involve rearrangement
Oxymercuration-demercurationHg(OAc)₂/H₂O then NaBH₄AlcoholMarkovnikov, no rearrangement; syn addition
Hydroboration-oxidationBH₃/THF then H₂O₂/OH⁻AlcoholAnti-Markovnikov; syn addition
HalogenationBr₂ in CCl₄ (room temp)vic-DibromideAnti addition (trans); brown colour discharged
Ozonolysis (reductive)O₃ then Zn/H₂OAldehydes / KetonesCleaves C=C; useful for structural elucidation
Syn-dihydroxylationCold, dilute KMnO₄ (OH⁻) or OsO₄vic-Diol (glycol)Syn addition of two —OH groups (Baeyer test)
Catalytic hydrogenationH₂ + Ni/Pt/PdAlkaneSyn addition of H₂
Trick: Markovnikov vs Anti-Markovnikov
"Rich get richer" — In Markovnikov, the carbon with more H atoms gets the H⁺ (H becomes the plus part). In anti-Markovnikov (only HBr + peroxide), Br goes to the carbon with more H.
Example 7 — Markovnikov Rule
The major product of addition of HBr to propene in the absence of peroxide is:

a) 1-bromopropane   b) 2-bromopropane   c) 1,2-dibromopropane   d) Propane
Solution: H⁺ adds to C-1 (more H atoms) forming a 2° carbocation. Br⁻ attacks C-2 → 2-bromopropane. Option (b).

Alkynes (CₙH₂ₙ₋₂)

Acidic Nature & Metal Acetylides

Terminal alkynes (R—C≡C—H) are weakly acidic due to the high s-character (50%) of the sp-hybridized carbon. The pKa of acetylene is ~25 (compared to ~44 for ethene, ~50 for ethane). Terminal alkynes react with strong bases (NaNH₂, NaH, RMgX) to form metal acetylides: HC≡CH + NaNH₂ → HC≡C⁻ Na⁺ + NH₃. Silver acetylide (white precipitate with ammoniacal AgNO₃) and cuprous acetylide (red precipitate with ammoniacal Cu₂Cl₂) are used to distinguish terminal alkynes from internal alkynes (distinctive test). Internal alkynes (R—C≡C—R') lack the acidic H and do not form these precipitates.

Addition Reactions of Alkynes

Alkynes undergo two-step addition (adding one mole of reagent, then a second). 1. Hydrogenation: Lindlar's catalyst (Pd/CaCO₃ + quinoline, H₂) → cis-alkene. Na/NH₃ (Birch) → trans-alkene. Excess H₂/Ni → alkane. 2. Halogenation: Br₂ (1 eq, CCl₄) → trans-dibromoalkene; Br₂ (2 eq) → tetrahaloalkane. 3. Addition of HX: HX (1 eq) → vinyl halide (Markovnikov); HX (2 eq) → geminal dihalide (two halogens on the same carbon). 4. Hydration (Kucherov reaction): Alkyne + H₂O + HgSO₄/H₂SO₄ → enol (unstable) → tautomerizes to ketone. For HC≡CH, the product is CH₃CHO (acetaldehyde). For R—C≡CH, the product is R—CO—CH₃ (methyl ketone). For internal alkynes, the product is a mixture of ketones. 5. Ozonolysis: Alkynes + O₃ then H₂O → carboxylic acids + CO₂. 6. Polymerization: Passing acetylene through a red-hot tube gives benzene (cyclic trimerization). Linear polymerization gives polyacetylene (conducting polymer).

Example — Kucherov Hydration
The product of hydration of propyne (CH₃C≡CH) with HgSO₄/H₂SO₄ is:

a) Propanal   b) Propanone   c) Propanoic acid   d) Propan-1-ol
Solution: Propyne (CH₃C≡CH) undergoes Markovnikov hydration to form an enol [CH₃C(OH)=CH₂] which tautomerizes to propanone (CH₃COCH₃, acetone). Option (b).
Example 8 — Alkyne Acidity
Which of the following is most acidic?

a) CH₃CH₃   b) CH₂=CH₂   c) HC≡CH   d) CH₄
Solution: Acidity increases with s-character: sp > sp² > sp³. HC≡CH (sp) is most acidic. Option (c).

Aromatic Hydrocarbons

Benzene, Huckel's Rule & Aromaticity

Benzene (C₆H₆) is the parent aromatic compound — a planar, hexagonal molecule with 6 sp²-hybridized carbons and a delocalized π-electron system (6 π electrons, 3 double bonds). Each carbon contributes one p-orbital perpendicular to the ring; these p-orbitals overlap to form a continuous π-cloud above and below the plane. Hückel's rule: A planar, monocyclic, fully conjugated molecule with (4n+2) π electrons is aromatic (n = 0, 1, 2, 3...). n=1 → 6π (benzene); n=2 → 10π (naphthalene). Non-aromatic: Does not satisfy Hückel's rule (e.g., cyclooctatetraene, C₈H₈, 8π — tub-shaped, non-planar, non-aromatic). Antiaromatic: Planar, cyclic, fully conjugated with 4n π electrons (e.g., cyclobutadiene, 4π — extremely unstable). Aromatic ions: Cyclopentadienyl anion (C₅H₅⁻, 6π) is aromatic. Cyclopropenyl cation (C₃H₃⁺, 2π) is aromatic. Tropylium cation (C₇H₇⁺, 6π) is aromatic. Heterocyclic aromatics: Pyridine (C₅H₅N, 6π), Pyrrole (C₄H₅N, 6π with lone pair on N), Furan (C₄H₄O, 6π), Thiophene (C₄H₄S, 6π).

Electrophilic Aromatic Substitution (EAS)

Benzene's characteristic reaction is electrophilic substitution, not addition (addition would destroy the stable aromatic ring). The mechanism involves two steps: Step 1 (Slow, RDS): The electrophile (E⁺) attacks the π-electron cloud, forming a resonance-stabilized arenium ion (σ-complex, Wheland intermediate). The positive charge is delocalized over three carbon atoms of the ring. Step 2 (Fast): Loss of H⁺ from the arenium ion restores aromaticity. Base (e.g., HSO₄⁻) abstracts the proton.

EAS ReactionReagentsElectrophile (E⁺)Product
NitrationConc. HNO₃ + Conc. H₂SO₄NO₂⁺ (nitronium ion)Nitrobenzene
Halogenation (Cl₂)Cl₂ + FeCl₃/AlCl₃ (anhydrous)Cl⁺ (or polarized Cl—FeCl₃)Chlorobenzene
Halogenation (Br₂)Br₂ + FeBr₃/AlBr₃Br⁺ (or polarized Br—FeBr₃)Bromobenzene
SulphonationFuming H₂SO₄ (SO₃ + H₂SO₄)SO₃ (or SO₃H⁺)Benzene sulphonic acid
Friedel-Crafts alkylationR—Cl + AlCl₃ (anhydrous)R⁺ (carbocation, may rearrange)Alkylbenzene
Friedel-Crafts acylationR—CO—Cl + AlCl₃R—C≡O⁺ (acylium ion, no rearrangement)Phenyl alkyl ketone

Directing Effects of Substituents

Activating groups (ortho-para directing): —OH, —NH₂, —NHR, —NR₂, —OCH₃, —CH₃, —C₂H₅, —C₆H₅. These donate electrons (+M and/or +I), increasing electron density at ortho and para positions. Deactivating groups (meta directing): —NO₂, —CN, —CHO, —COR, —COOH, —COOR, —SO₃H, —NH₃⁺. These withdraw electrons (-M and/or -I), decreasing overall ring electron density, with the meta positions being least deactivated. Halogens (—F, —Cl, —Br, —I): Ortho-para directing but deactivating. The +M effect (lone pair donation) directs ortho-para, but the stronger -I effect reduces overall ring electron density, making them deactivating. This is a unique case that NEET frequently tests.

Mnemonic: EAS Reactions
"Nitrogen Has Some Fancy Atoms" — Nitration, Halogenation, Sulphonation, Friedel-Crafts, ... (the rest are miscellaneous).
Example 9 — Aromaticity
Which of the following is non-aromatic?

a) Benzene   b) Cyclopentadienyl anion   c) Cyclopentadiene   d) Pyridine
Solution: Cyclopentadiene has a saturated CH₂ and lacks full conjugation; it is non-aromatic. Benzene (6π), cyclopentadienyl anion (6π), and pyridine (6π) are all aromatic. Option (c).

4. Haloalkanes & Haloarenes

Haloalkanes (alkyl halides) and haloarenes (aryl halides) are compounds containing halogen atoms (F, Cl, Br, I) bonded to sp³ and sp² hybridized carbons respectively. The polarity of the C—X bond (δ⁺ on C, δ⁻ on X) makes the carbon electrophilic, enabling nucleophilic substitution. The reactivity difference between alkyl and aryl halides is a key NEET concept.

SN1 — Unimolecular Nucleophilic Substitution

Mechanism: Two-step process. Step 1 (slow, RDS): Heterolytic cleavage of C—X bond forms a planar carbocation intermediate (R⁺) and X⁻. Step 2 (fast): The nucleophile attacks the carbocation from either face. Kinetics: Rate = k[RX] (first order). The rate depends only on the concentration of the alkyl halide, not on the nucleophile. Stereochemistry: Racemization — the planar carbocation is attacked from the top or bottom with equal probability, giving a 50:50 mixture of R and S. However, some net inversion is often observed due to ion-pairing (the leaving group partially blocks one face before complete dissociation). Factors favouring SN1: (1) Tertiary alkyl halides (3° > 2° > 1°; methyl does not undergo SN1). (2) Weak nucleophiles (H₂O, ROH, CH₃COOH). (3) Polar protic solvents (H₂O, ROH) stabilize the carbocation intermediate by solvation. (4) Good leaving groups (I⁻, Br⁻, OTs⁻). Rearrangement: Carbocation intermediates frequently undergo 1,2-hydride shifts or 1,2-alkyl shifts to form more stable carbocations, leading to rearranged products (a common NEET exam trick).

SN2 — Bimolecular Nucleophilic Substitution

Mechanism: One-step, concerted process. The nucleophile attacks the electrophilic carbon from the back side (opposite to the leaving group) while the C—X bond breaks simultaneously. The transition state is pentacoordinated (sp²-like) with the nucleophile and leaving group partially bonded. Kinetics: Rate = k[RX][Nu⁻] (second order). The rate depends on both the alkyl halide and the nucleophile. Stereochemistry: Complete inversion of configuration (Walden inversion). (R)-configuration reacts to give (S)-product and vice versa. This is a defining characteristic of SN2. Factors favouring SN2: (1) Primary alkyl halides (CH₃X > 1° > 2° > 3°; 3° is extremely slow due to steric hindrance). (2) Strong nucleophiles (OH⁻, CN⁻, OR⁻, I⁻, HS⁻). (3) Polar aprotic solvents (DMSO, DMF, acetone, CH₃CN) — the cation is solvated but the nucleophile anion is not strongly solvated, making it more reactive. (4) Good leaving groups. Steric hindrance: The rate decreases sharply with increasing substitution on the α-carbon because bulky groups hinder the backside attack.

FeatureSN1SN2
Number of stepsTwo (with carbocation intermediate)One (concerted, transition state only)
KineticsFirst order: Rate ∝ [RX]Second order: Rate ∝ [RX][Nu⁻]
IntermediateCarbocation (planar, sp²)None (pentacoordinated transition state)
StereochemistryRacemization (mostly)Complete inversion (Walden inversion)
NucleophileWeak (neutral: H₂O, ROH, CH₃COOH)Strong (anionic: OH⁻, CN⁻, OR⁻, I⁻)
Alkyl halide preference3° > 2° > 1° (stable carbocation needed)CH₃ > 1° > 2° > 3° (steric hindrance prevents)
SolventPolar protic (H₂O, ROH)Polar aprotic (DMSO, DMF, acetone, CH₃CN)
RearrangementPossible (hydride/alkyl shift to more stable carbocation)Not possible (no intermediate to rearrange)
Leaving group requirementGood LG (I⁻ > Br⁻ > Cl⁻ > F⁻; OTs⁻, OMs⁻)Same requirement
Example 10 — SN2 Stereochemistry
The reaction of (R)-2-bromooctane with NaOH gives (S)-2-octanol. This is an example of:

a) Retention   b) Inversion   c) Racemization   d) Epimerization
Solution: SN2 involves backside attack, inverting the configuration (Walden inversion). Hence (R) → (S). Option (b).
Trick: SN1 vs SN2
"SN1Slow No 1 (tertiary)" and "SN2 — Steric hindrance kills it (tertiary is slow)". Primary → SN2; Tertiary → SN1; Secondary → either, depending on nucleophile/solvent.

Elimination Reactions (E1 & E2)

Elimination reactions compete with substitution reactions. A strong base and high temperature favour elimination over substitution. E1 mechanism: Two-step process via carbocation intermediate (similar to SN1). The leaving group departs first to form a carbocation, followed by loss of a β-proton to a weak base. Favours 3° RX. Gives the more substituted alkene (Zaitsev product). E2 mechanism: One-step, concerted process. A strong base (OH⁻, RO⁻) abstracts a β-proton while the leaving group departs simultaneously. Requires anti-periplanar geometry of the H and the leaving group (the H and X must be on opposite sides of the C—C bond, 180° apart). Favours 1° and 2° RX. Zaitsev vs Hoffmann: With a normal base (e.g., alc. KOH), the more substituted alkene (Zaitsev) predominates. With a bulky base (e.g., K-tert-butoxide), the less substituted alkene (Hoffmann) predominates because the bulky base cannot access the sterically hindered β-proton. E2 is stereospecific: The anti-periplanar requirement means that the stereochemistry of the reactant determines the stereochemistry of the alkene product.

FeatureE1E2
MechanismTwo-step (carbocation intermediate)One-step (concerted)
Base requirementWeak base (solvent acts as base)Strong base (OH⁻, RO⁻, NH₂⁻)
KineticsFirst order: Rate ∝ [RX]Second order: Rate ∝ [RX][Base]
Alkyl halide3° > 2° > 1°1° > 2° > 3°
RearrangementPossible (carbocation rearrangement)Not possible (concerted)
StereochemistryNot stereospecificStereospecific (anti-periplanar required)
Product selectivityZaitsev (more substituted alkene)Zaitsev (normal base) or Hoffmann (bulky base)
Example — SN1 vs SN2 vs E1 vs E2
2-Bromo-2-methylpropane [(CH₃)₃CBr] reacts with NaOH (aq) at room temperature. The major product is:

a) (CH₃)₃COH   b) (CH₃)₂C=CH₂   c) (CH₃)₃CH   d) Mixture of a and b
Solution: The alkyl halide is tertiary. At room temperature with aqueous NaOH, SN1 (and possibly E1) dominate. The major product is (CH₃)₃COH via SN1. Some elimination product (CH₃)₂C=CH₂ may also form. Option (a) is the major product.
Example — Hoffmann Elimination
Treatment of 2-bromobutane with a bulky base like potassium tert-butoxide gives:

a) But-1-ene (major)   b) But-2-ene (major)   c) Butane   d) 2-methylpropene
Solution: Bulky base abstracts the less hindered β-hydrogen, giving the less substituted alkene (Hoffmann product). Hence but-1-ene is major. Option (a).

Haloarenes (Aryl Halides)

Aryl halides (Ar—X) are much less reactive towards nucleophilic substitution than alkyl halides. The reasons are: (1) Resonance stabilization: The lone pair of the halogen is delocalized into the aromatic ring, giving the C—X bond partial double-bond character (making it stronger and more difficult to break). (2) sp² hybridization: The C—X bond is shorter and stronger (more s-character) than in alkyl halides. (3) No backside attack possible: The planar aromatic ring prevents the backside approach required for SN2. Haloarenes undergo nucleophilic substitution only under harsh conditions via addition-elimination (SNAr) or benzyne mechanisms. Electron-withdrawing groups (—NO₂, —CN) at ortho and para positions activate the ring towards nucleophilic substitution by stabilizing the Meisenheimer complex intermediate.

5. Alcohols, Phenols & Ethers

Alcohols (R—OH) are classified as 1° (e.g., ethanol), 2° (e.g., isopropyl alcohol), or 3° (e.g., tert-butyl alcohol) based on the number of alkyl groups attached to the carbon bearing the —OH group. Phenols have —OH directly attached to an aromatic ring. Ethers (R—O—R') have an oxygen bridging two alkyl/aryl groups. Alcohols have significantly higher boiling points than corresponding alkanes due to intermolecular hydrogen bonding. Methanol, ethanol, and propanol are miscible with water; solubility decreases as the alkyl chain length increases.

Preparation of Alcohols

1. Hydration of alkenes: Alkene + H₂O/dil. H₂SO₄ → alcohol (Markovnikov addition, via carbocation). May involve rearrangement. 2. Hydroboration-oxidation: Alkene + BH₃/THF then H₂O₂/OH⁻ → alcohol (anti-Markovnikov, syn addition, no rearrangement). 3. Reduction of carbonyl compounds: Aldehydes → 1° alcohols; Ketones → 2° alcohols (NaBH₄ or LiAlH₄). Carboxylic acids and esters → 1° alcohols (requires LiAlH₄; NaBH₄ is not strong enough). 4. Grignard synthesis: RMgX + HCHO → 1° alcohol (after hydrolysis). RMgX + R'CHO → 2° alcohol. RMgX + R'COR" → 3° alcohol. This is a versatile method that forms new C—C bonds. 5. Fermentation: C₆H₁₂O₆ (glucose) → 2C₂H₅OH + 2CO₂ (yeast, anaerobic conditions).

Reactions of Alcohols

1. Dehydration: Conc. H₂SO₄ (170°C) or Al₂O₃ (350°C) → alkene + H₂O. Reactivity: 3° > 2° > 1°. Follows E1 for 2°/3°, E2 for 1°. 2. Oxidation: 1° alcohols (PCC/CH₂Cl₂) → aldehyde; (K₂Cr₂O₇/H⁺, KMnO₄) → carboxylic acid. 2° alcohols → ketone. 3° alcohols are resistant to mild oxidation (no α-H at the —OH carbon). 3. Lucas test: R—OH + ZnCl₂ + HCl (conc.) → R—Cl (insoluble, appears as turbidity/cloudiness). 3°: immediate turbidity. 2°: turbidity in 5-10 min. 1°: no turbidity at RT (requires heating). 4. Esterification (Fischer): R—OH + R'COOH (H₂SO₄, heat) → R'COOR + H₂O. Reversible reaction. 5. Reaction with metals: 2R—OH + 2Na → 2R—ONa + H₂↑ (sodium alkoxide). 6. Conversion to alkyl halides: R—OH + HX (ZnCl₂ catalyst for 1°) → R—X + H₂O. Also using PX₃, PX₅, or SOCl₂ (SOCl₂ + pyridine is the best method for 1° alcohols — high yield, no rearrangement).

Phenols

Acidity: Phenols (pKa ≈ 10) are more acidic than alcohols (pKa ≈ 16) but less acidic than carboxylic acids (pKa ≈ 4-5). The phenoxide ion (C₆H₅O⁻) is stabilized by resonance delocalization of the negative charge into the aromatic ring. Effects on acidity: Electron-withdrawing groups (—NO₂) at ortho and para positions significantly enhance acidity (picric acid, 2,4,6-trinitrophenol, is a strong acid with pKa ≈ 0.3). Electron-donating groups (—CH₃) decrease acidity. Kolbe-Schmitt reaction: Phenol + CO₂ (NaOH, 125°C, 4-7 atm) → sodium salicylate → salicylic acid (aspirin precursor). Reimer-Tiemann reaction: Phenol + CHCl₃ + NaOH (50-70°C) → o-hydroxybenzaldehyde (salicylaldehyde) as the major product. Electrophilic substitution: —OH is strongly activating (+M), so phenol undergoes EAS readily even under mild conditions. Bromination with Br₂/H₂O gives 2,4,6-tribromophenol (white precipitate) — a characteristic test for phenol. Coupling with diazonium salts: Phenol couples with benzene diazonium chloride in alkaline medium to give a brightly coloured azo dye (p-hydroxyazobenzene).

Ethers

Williamson's synthesis: R—O⁻Na⁺ + R'—X → R—O—R' + NaX. This is the best method for preparing both symmetrical and unsymmetrical ethers. The alkoxide should be used as the nucleophile, and the alkyl halide should be primary (secondary/tertiary alkyl halides undergo elimination). For preparing aryl alkyl ethers (e.g., anisole, C₆H₅OCH₃): sodium phenoxide + CH₃I → anisole + NaI. Cleavage of ethers: R—O—R' + HI (hot, conc.) → R—I + R'—OH. If excess HI is used, both products are converted to alkyl iodides. The cleavage mechanism is SN2 (for 1° alkyl groups) or SN1 (for 3° alkyl groups). The order of cleavage efficacy: HI > HBr > HCl. Anisole: Methyl phenyl ether undergoes EAS at ortho and para positions (the —OCH₃ group is strongly activating, strong +M effect). This is used in the preparation of p-methoxyacetophenone via Friedel-Crafts acylation.

Trick: Lucas Test
"3, 2, 1 — Cloudy, then run!" 3° alcohols give immediate turbidity; 2° in 5-10 min; 1° no turbidity at room temperature.
Example 12 — Alcohol Oxidation
Which alcohol on oxidation gives a ketone?

a) CH₃OH   b) CH₃CH₂CH₂OH   c) (CH₃)₂CHOH   d) (CH₃)₃COH
Solution: 1° alcohols give aldehydes → acids. 2° alcohols give ketones. 3° alcohols are not easily oxidized. (CH₃)₂CHOH is 2° → gives acetone. Option (c).

6. Aldehydes, Ketones & Carboxylic Acids

The carbonyl group (C=O) is one of the most versatile functional groups in organic chemistry. The carbon is sp² hybridized, and the strong π-bond is polarized (δ⁺ on C, δ⁻ on O), making the carbonyl carbon highly electrophilic. Aldehydes are more reactive than ketones towards nucleophilic addition due to less steric hindrance and greater electrophilicity of the carbonyl carbon (ketones have two +I alkyl groups that stabilize the carbonyl carbon). Formaldehyde (HCHO) is the most reactive aldehyde.

Aldehydes & Ketones — Nucleophilic Addition

1. Addition of HCN: Aldehyde/ketone + HCN (catalytic amount of base) → cyanohydrin (α-hydroxy nitrile). The CN⁻ nucleophile attacks the carbonyl carbon. Cyanohydrins are useful synthetic intermediates — the —CN group can be hydrolysed to —COOH or reduced to —CH₂NH₂. 2. Addition of NaHSO₃: Aldehyde/ketone + saturated NaHSO₃ solution → white crystalline bisulphite addition product. This is a reversible reaction used for the purification of carbonyl compounds. Methyl ketones and cyclic ketones react readily; the product can be regenerated with dil. Na₂CO₃ or dil. HCl. 3. Addition of RMgX (Grignard reagents): HCHO → 1° alcohol (after hydrolysis); RCHO → 2° alcohol; RCOR' → 3° alcohol. This is a powerful C—C bond-forming reaction. 4. Addition of alcohols: Aldehyde + ROH (dry HCl gas) → hemiacetal (unstable) → acetal (stable). Ketones form ketals. Acetals/ketals are used as protecting groups for the carbonyl function. 5. Addition of ammonia derivatives (condensation): NH₂OH → oxime; NH₂NH₂ → hydrazone; 2,4-DNPH (2,4-dinitrophenylhydrazine) → orange/red precipitate of 2,4-dinitrophenylhydrazone (used for identification/characterization of carbonyl compounds); NH₂—NH—CS—NH₂ (thiosemicarbazide) → thiosemicarbazone.

Reduction of Aldehydes & Ketones

1. Metal hydrides: NaBH₄ (mild, selective for C=O, works in aqueous/methanolic solution) or LiAlH₄ (strong, reduces C=O, C≡N, COOH, esters). Aldehydes → 1° alcohols; Ketones → 2° alcohols. 2. Catalytic hydrogenation: H₂/Ni (Pt, Pd, Ru) → alcohol. 3. Wolff-Kishner reduction: NH₂NH₂ (hydrazine) + KOH (heat, ethylene glycol) → R—CH₃ (alkane from aldehyde) or R₂CH₂ (alkane from ketone). The C=O is reduced to CH₂. 4. Clemmensen reduction: Zn(Hg) + HCl (conc., heat) → alkane. Used for acid-sensitive carbonyl compounds (the Wolff-Kishner uses strong base, so Clemmensen is preferred for base-sensitive compounds).

Oxidation of Aldehydes & Ketones

Aldehydes are easily oxidized to carboxylic acids by mild oxidizing agents. Tollens' test: Aldehyde + ammoniacal AgNO₃ → Ag mirror (Ag⁺ reduced to Ag⁰). Fehling's test: Aldehyde + Cu²⁺/tartrate (in NaOH) → red precipitate of Cu₂O. Aromatic aldehydes do not react with Fehling's solution. Benedict's test: Similar to Fehling's, used clinically to detect glucose in urine. Schiff's test: Aldehyde + Schiff's reagent (rosaniline decolourized by SO₂) → pink/magenta colour. Ketones do not give any of these tests. Ketones require strong oxidizing conditions (hot KMnO₄, hot HNO₃) and undergo C—C bond cleavage, giving a mixture of carboxylic acids.

Aldol & Cannizzaro Reactions

Aldol condensation: Carbonyl compounds with at least one α-hydrogen undergo condensation in the presence of dilute base (NaOH) to form β-hydroxy carbonyl compounds (aldols). On heating, the aldol dehydrates to form an α,β-unsaturated carbonyl compound. Crossed aldol (Claisen-Schmidt reaction) between two different carbonyl compounds gives a mixture of four products unless one of them has no α-H (e.g., benzaldehyde + acetaldehyde gives a single product — cinnamaldehyde). Cannizzaro reaction: Aldehydes without α-H (HCHO, C₆H₅CHO, (CH₃)₃CCHO) undergo disproportionation in concentrated base: 2 molecules → 1 alcohol (reduced) + 1 carboxylic acid (oxidized). In a crossed Cannizzaro reaction, HCHO (which is more electrophilic) is always oxidized to formic acid (or formate), while the other aldehyde is reduced to the corresponding alcohol.

ReagentDetectsObservation
Tollens' reagent [Ag(NH₃)₂]⁺OH⁻AldehydeSilver mirror
Fehling's solution (Cu²⁺ + tartrate)AldehydeRed precipitate of Cu₂O
Benedict's solutionAldehydeRed precipitate
Iodoform test (I₂ + NaOH)CH₃CO— or CH₃CH(OH)—Yellow precipitate of CHI₃
2,4-DNPAldehyde / KetoneOrange precipitate
Schiff's reagentAldehydePink / magenta colour
Example 13 — Aldol Condensation
The aldol condensation product of acetaldehyde (CH₃CHO) in the presence of dil. NaOH is:

a) CH₃CH(OH)CH₂CHO   b) CH₃COCH₃   c) CH₃CH=CHCHO   d) CH₃COOCH₃
Solution: α-H of one molecule adds to carbonyl of another → β-hydroxy aldehyde (3-hydroxybutanal). On heating it dehydrates to crotonaldehyde (CH₃CH=CHCHO). The initial product is (a).
Mnemonic: Aldehyde Tests
"Tollens, Fehling, Benedict — All Aldehydes give colour!" and "Iodoform — only for compounds with CH₃CO— or CH₃CH(OH)— group."

Carboxylic Acids

Acidity: Carboxylic acids (pKa ~4-5) are stronger than phenols due to resonance stabilization of the carboxylate anion. Derivatives: Acid chlorides (SOCl₂), esters (alcohol + H⁺), amides (NH₃), anhydrides (P₂O₅). Hell-Volhard-Zelinsky (HVZ) reaction: α-halogenation using X₂/P. Decarboxylation: Heat with soda lime → alkane.

Carboxylic Acids (R—COOH)

Carboxylic acids are characterized by the carboxyl group (—COOH). They are the most acidic class of neutral organic compounds (pKa ~4-5). The acidity is due to resonance stabilization of the conjugate base (carboxylate anion, RCOO⁻), where the negative charge is delocalized over two equivalent oxygen atoms. Preparation: (1) Oxidation of 1° alcohols/aldehydes (KMnO₄, K₂Cr₂O₇/H⁺). (2) Hydrolysis of nitriles (RCN + H₂O/H⁺ → RCOOH + NH₄⁺). (3) Carbonation of Grignard reagents (RMgX + CO₂ → RCOOMgX → RCOOH). (4) Hydrolysis of esters and amides. Reactions: (1) Esterification (Fischer, ROH/H⁺, reversible). (2) Formation of acid chlorides (SOCl₂, PCl₅, PCl₃). (3) Formation of amides (NH₃ → ammonium salt → heat → amide). (4) Reduction (LiAlH₄ → 1° alcohol; BH₃/THF selectively reduces COOH without affecting NO₂ or ester groups). (5) HVZ reaction (Hell-Volhard-Zelinsky): Br₂ + catalytic red P → α-bromo acid (then with NH₃ → α-amino acid). (6) Decarboxylation: RCOONa + NaOH/CaO (soda lime, heat) → R—H + Na₂CO₃. (7) Arndt-Eistert synthesis: RCOOH → RCH₂COOH (increases chain by one carbon).

Trick: Comparative Acidity
Order of acidity: Mineral acids > Carboxylic acids (pKa 4-5) > Phenols (pKa 10) > Alcohols (pKa 16) > Alkanes (pKa ~50). Within carboxylic acids: EWG increase acidity (Cl₃CCOOH strong), EDG decrease it (CH₃COOH weaker than HCOOH).
Example — Decarboxylation
The product obtained when benzoic acid is heated with soda lime is:

a) Benzene   b) Toluene   c) Benzaldehyde   d) Phenol
Solution: Decarboxylation using soda lime (NaOH + CaO) removes CO₂, giving benzene. Option (a).

Distinction Tests — Aldehydes vs Ketones

Test / ReagentTargetPositive ResultNotes
Tollens' [Ag(NH₃)₂]⁺OH⁻AldehydesSilver mirror on test tubeAg⁺ → Ag⁰; ketones do not react
Fehling's (Cu²⁺ + tartrate in NaOH)Aldehydes (aliphatic)Red Cu₂O precipitateAromatic aldehydes do not react
Benedict's (Cu²⁺ + citrate)Aldehydes / reducing sugarsRed Cu₂O precipitateClinical test for glucose in urine
Iodoform (I₂ + NaOH)CH₃CO— or CH₃CH(OH)—Yellow CHI₃ precipitateAcetaldehyde: +; other aldehydes: —; Methyl ketones: +
2,4-DNPHAldehydes & KetonesOrange/red precipitateGeneral test for carbonyl compounds
Schiff's reagentAldehydesPink/magenta colourKetones do not restore colour

7. Amines

Amines are derivatives of NH₃ where one or more hydrogen atoms are replaced by alkyl (aliphatic amines) or aryl (aromatic amines) groups. Classified as 1° (primary, RNH₂), 2° (secondary, R₂NH), 3° (tertiary, R₃N), and quaternary ammonium salts (R₄N⁺X⁻). Amines are basic due to the lone pair of electrons on nitrogen. This is a high-yield NEET topic with questions on basicity order, Hinsberg test, and diazonium chemistry.

Basicity of Amines

Aliphatic amines: Alkyl groups are electron-donating (+I), increasing electron density on nitrogen, making aliphatic amines stronger bases than NH₃. In aqueous solution, the basicity order is: 2° > 1° > 3° > NH₃. Secondary amines are the strongest base due to a balance of inductive effect (two alkyl groups donate more electron density) and solvation of the conjugate acid (R₂NH₂⁺ is well solvated by water). Tertiary amines are weaker than secondary despite having three alkyl groups because the bulky R₃NH⁺ ion is poorly solvated, destabilizing it. In the gas phase (no solvation effects), the order is: 3° > 2° > 1° > NH₃. Aromatic amines: Aniline (C₆H₅NH₂) is much weaker (pKb ≈ 9.4) than aliphatic amines (pKb ≈ 3-4) because the lone pair on nitrogen is delocalized into the aromatic ring via resonance (the +M effect of —NH₂). Substituent effects on aniline basicity: EWG (—NO₂) decrease basicity; EDG (—CH₃, —OCH₃) increase basicity. ortho-Substituted anilines are always weaker than aniline regardless of the nature of the substituent (ortho effect — steric hindrance to solvation).

Important Reactions of Amines

1. Carbylamine reaction (isocyanide test): 1° amine + CHCl₃ + alc. KOH (heat) → R—NC (isocyanide) + 3KCl + 3H₂O. The isocyanide has a characteristic foul smell. Only primary amines (aliphatic and aromatic) give this test; 2° and 3° amines do not react. 2. Hinsberg test: 1° amine + C₆H₅SO₂Cl (benzenesulphonyl chloride) + NaOH → N-substituted sulphonamide (soluble in alkali because the N—H is acidic, forming a salt) → clear solution. 2° amine → N,N-disubstituted sulphonamide (no N—H, insoluble in alkali) → precipitate. 3° amine → no reaction (no N—H to react) → immiscible oily layer. 3. Acylation: 1° and 2° amines + RCOCl or (RCO)₂O → amide (R—CO—NR'R"). 3° amines do not have an N—H bond and do not undergo acylation. 4. Reaction with HNO₂ (nitrous acid): 1° aliphatic amine + HNO₂ → N₂ gas (bubbles, test for 1° aliphatic amine). 1° aromatic amine + HNO₂ (0-5°C) → diazonium salt (stable at low temperature). 2° amine + HNO₂ → N-nitrosamine (yellow oil). 3° amine + HNO₂ → trialkylammonium nitrite salt (may give C-nitroso compound in aromatic cases). 5. Hoffmann elimination (exhaustive methylation): 1° amine + excess CH₃I (methylation) → quaternary ammonium iodide → AgOH (wet Ag₂O) → quaternary ammonium hydroxide → heat → alkene + H₂O + trimethylamine. The least substituted alkene (Hoffmann product) is formed.

Diazonium Salts

Primary aromatic amines react with NaNO₂ + HCl at 0-5°C to form benzene diazonium chloride (Ar—N₂⁺Cl⁻). These are versatile synthetic intermediates. Substitution reactions (replacement of —N₂⁺): Sandmeyer (CuCl/HCl → ArCl; CuBr/HBr → ArBr; CuCN/KCN → ArCN). Gattermann (Cu + HX → ArX). With KI → ArI. With H₂O (heat) → ArOH. With H₃PO₂ (hypophosphorous acid) → ArH. With HBF₄ (Balz-Schiemann reaction) → ArF (fluorobenzene). Coupling reactions: Diazonium salts couple with phenols (in alkaline medium) and aromatic amines (in acidic medium) to form brightly coloured azo compounds (Ar—N=N—Ar'). This is the basis of azo dye chemistry. The coupling occurs at the para position (if occupied, then ortho).

Example 15 — Diazonium Coupling
Benzene diazonium chloride couples with phenol in alkaline medium to give:

a) Azo dye   b) Nitroso compound   c) Hydroxyazo compound   d) Both a and c
Solution: The diazonium salt couples at the para-position of phenol, forming a brightly coloured azo dye (p-hydroxyazobenzene). Option (d).
Trick: Basicity of Amines
In gas phase: 3° > 2° > 1° > NH₃. In aqueous solution: 2° > 1° > 3° > NH₃. The reversal of 3° is due to poor solvation of the bulky R₃NH⁺ ion.

Important Named Reactions of Amines

Carbylamine reaction: 1° amine + CHCl₃ + alc. KOH → foul-smelling isocyanide (RNC). Hinsberg test: Benzenesulphonyl chloride distinguishes 1°, 2°, 3° amines. Hoffmann elimination: 4° ammonium hydroxide on heating yields alkene + H₂O + amine.

Example 16 — Carbylamine Reaction
Carbylamine reaction is given by:

a) CH₃NH₂   b) (CH₃)₂NH   c) (CH₃)₃N   d) C₆H₅N(CH₃)₂
Solution: Only primary amines (1°) give the carbylamine test. CH₃NH₂ is primary. Option (a).

8. Biomolecules

Biomolecules are organic compounds essential for life: carbohydrates, proteins, nucleic acids, lipids, and vitamins. This is a high-yield NEET topic with 2-3 questions per exam, covering structure, properties, and biological significance.

Carbohydrates

Carbohydrates are polyhydroxy aldehydes or ketones with the general formula Cₓ(H₂O)ᵧ. Classified as monosaccharides (simplest sugars, cannot be hydrolysed — glucose, fructose, galactose, ribose), disaccharides (two monosaccharides joined by a glycosidic bond — sucrose, maltose, lactose), and polysaccharides (many monosaccharides — starch, cellulose, glycogen). Glucose (C₆H₁₂O₆): An aldohexose, reducing sugar, forms a pentaacetate (5 —OH groups). The open-chain form has an aldehyde group; the cyclic forms (α and β anomers) exist in equilibrium via mutarotation. The pyranose (6-membered) and furanose (5-membered) ring forms arise from internal hemiacetal formation. Glucose reduces Tollens' reagent (Ag mirror) and Fehling's solution (red Cu₂O). With phenylhydrazine, glucose forms a characteristic osazone (needle-shaped crystals, melting point 205°C). Fructose: A ketohexose (sweetest sugar), reducing sugar, forms the same osazone as glucose (because the carbonyl group is at C-2 and the stereochemistry of C-3, C-4, C-5 is the same as glucose). Sucrose (cane sugar): α-D-glucose + β-D-fructose linked by α(1→2) glycosidic bond. Non-reducing sugar (both anomeric carbons are involved in the glycosidic bond). Hydrolysis (invertase or dil. HCl) gives invert sugar (glucose + fructose, reducing). Maltose: α-D-glucose + α-D-glucose linked by α(1→4) bond. Reducing sugar. Lactose: β-D-galactose + β-D-glucose linked by β(1→4) bond. Reducing sugar (milk sugar). Starch: Polymer of α-D-glucose — amylose (linear, α(1→4) links) + amylopectin (branched, α(1→6) links). Gives blue colour with iodine. Cellulose: Polymer of β-D-glucose (linear, β(1→4) links). Structural component of plant cell walls. Not digestible by humans (lack β-glucosidase).

Example — Reducing vs Non-reducing Sugar
Which of the following is a non-reducing sugar?

a) Glucose   b) Maltose   c) Sucrose   d) Lactose
Solution: Sucrose has both anomeric carbons involved in the glycosidic linkage — no free aldehyde/ketone group. It does not reduce Tollens' or Fehling's reagent. Option (c).

Proteins & Amino Acids

Amino acids have the general formula H₂N—CH(R)—COOH. The 20 standard α-amino acids are the building blocks of proteins. At the isoelectric point (pI), amino acids exist as zwitterions (⁺H₃N—CH(R)—COO⁻) with no net charge. Peptide bond: Formed between the —COOH of one amino acid and the —NH₂ of another (—CO—NH—), with elimination of H₂O. The peptide bond is planar with partial double-bond character. Protein structure: Primary (amino acid sequence), Secondary (α-helix and β-pleated sheet stabilized by H-bonds), Tertiary (3D folding stabilized by H-bonds, disulphide bridges —S—S—, hydrophobic interactions, ionic bonds), Quaternary (multiple polypeptide chains — e.g., haemoglobin has 4 subunits). Denaturation: Loss of secondary, tertiary, and quaternary structure (by heat, pH change, heavy metal salts) without breaking the primary structure. Tests for proteins: Biuret test (violet colour with CuSO₄/NaOH — detects peptide bonds), Ninhydrin test (blue-purple colour with α-amino acids, proline gives yellow), Xanthoproteic test (yellow colour with HNO₃, turning orange with alkali — detects aromatic amino acids like tyrosine, tryptophan), Millon's test (red colour with Hg(NO₃)₂/HNO₃ — detects phenolic —OH group in tyrosine).

Nucleic Acids

DNA (deoxyribonucleic acid): Double-stranded helix (Watson-Crick model). Each strand has a sugar-phosphate backbone (deoxyribose + phosphate) and four nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (T). Complementary base pairing: A≡T (2 H-bonds), G≡C (3 H-bonds). The two strands are antiparallel. RNA (ribonucleic acid): Single-stranded, has ribose sugar and Uracil (U) instead of Thymine. Types: mRNA (carries genetic code from DNA to ribosomes), tRNA (carries amino acids to ribosomes for protein synthesis, has anticodon loop), rRNA (structural component of ribosomes). Nucleosides vs Nucleotides: Nucleoside = base + sugar (e.g., adenosine, guanosine). Nucleotide = base + sugar + phosphate (e.g., AMP, ADP, ATP). ATP (adenosine triphosphate) is the energy currency of the cell. Chargaff's rule: In DNA, A = T and G = C (the number of purines equals the number of pyrimidines).

Lipids

Lipids are naturally occurring organic compounds insoluble in water but soluble in organic solvents. Fats and oils: Triesters of glycerol with fatty acids (triglycerides). Oils are unsaturated (liquid at RT); fats are saturated (solid at RT). Saponification: Fat/oil + NaOH → glycerol + sodium salts of fatty acids (soap). Iodine number: Measures the degree of unsaturation — higher iodine number = more unsaturation. Phospholipids: Major components of cell membranes (e.g., lecithin has glycerol + 2 fatty acids + phosphate + choline). Steroids: Lipids with a characteristic four-ring structure (e.g., cholesterol, testosterone, oestrogen, cortisol).

Vitamins

Vitamins are essential organic compounds required in small amounts for normal physiological function. They are classified as fat-soluble (Vitamins A, D, E, K — stored in liver and fatty tissues, can accumulate to toxic levels) and water-soluble (Vitamin B-complex and Vitamin C — not stored, excreted in urine, need regular intake). Vitamin A (Retinol): Night blindness, xerophthalmia. Source: carrots, green leafy vegetables. Vitamin B₁ (Thiamine): Beriberi (affects nervous system and heart). Present in whole grains, meat. Vitamin B₂ (Riboflavin): Cheilosis, angular stomatitis. Vitamin B₃ (Niacin): Pellagra (dermatitis, diarrhoea, dementia). Vitamin B₆ (Pyridoxine): Anaemia, convulsions. Vitamin B₁₂ (Cobalamin): Pernicious anaemia. Only vitamin containing cobalt. Vitamin C (Ascorbic acid): Scurvy (bleeding gums, poor wound healing). Powerful antioxidant. Vitamin D (Cholecalciferol): Rickets in children, osteomalacia in adults. Synthesized in skin on exposure to sunlight. Vitamin E (Tocopherol): Antioxidant, fertility. Vitamin K (Phylloquinone): Required for blood clotting (prothrombin synthesis).

BiomoleculeMonomer (Building Block)Type of BondFunctions / Examples
CarbohydratesMonosaccharides (e.g., glucose, fructose)Glycosidic bond (—O—)Energy source: starch, cellulose, glycogen; structural: cellulose
Proteinsα-Amino acids (20 types)Peptide bond (—CO—NH—)Enzymes, structural (keratin, collagen), transport (haemoglobin), antibodies
Nucleic acidsNucleotides (base + sugar + phosphate)Phosphodiester bondGenetic information storage and transfer: DNA, RNA
LipidsFatty acids + Glycerol (triglycerides)Ester bondEnergy storage, cell membrane components: fats, oils, phospholipids, steroids
Example — Peptide Bond Counting
The number of peptide bonds in a polypeptide with 50 amino acid residues is:

a) 50   b) 49   c) 51   d) 48
Solution: Number of peptide bonds = (number of amino acid residues) − 1. For 50 residues, peptide bonds = 49. Option (b).
Mnemonic: DNA Base Pairing
"Apples in Trees — A pairs with T (2 H-bonds, AT = 2 letters). Golf Courses — G pairs with C (3 H-bonds, GC = 3 letters)."
Mnemonic: Vitamin Deficiency Diseases
"All Night Blindness (Vit A). B₁Beriberi. B₃Pellagra (P for P! there are 3 letters in \"Pellagra\" for B₃). CScurvy. DRickets. KKoagulation (blood clotting)."

9. Polymers

Polymers are high molecular mass substances (macromolecules) composed of repeating structural units called monomers, linked by covalent bonds. The process of forming a polymer is called polymerization. NEET typically asks 1-2 questions on polymer classification, monomers, and uses.

Classification of Polymers

Based on source: Natural (cotton, silk, wool, natural rubber, starch), Semi-synthetic (rayon, celluloid — chemically modified natural polymers), and Synthetic (nylon, polythene, PVC, terylene — man-made). Based on structure: Linear (thermoplastics, melt on heating — polythene, PVC), Branched (LDPE), and Cross-linked (thermosetting, do not melt on heating — bakelite, melamine-formaldehyde). Based on mode of polymerization: Addition (chain growth) and Condensation (step growth).

Addition (Chain Growth) Polymerization

Monomers containing C=C double bonds (alkenes and derivatives) undergo chain growth polymerization via free radical, cationic, anionic, or coordination (Ziegler-Natta) mechanisms. The three steps are: initiation (generation of reactive radical/ion), propagation (successive addition of monomers, each addition regenerating the reactive centre), and termination (combination of two growing chains, disproportionation, or chain transfer). The polymer has the same elemental composition as the monomer (no by-product). Examples: polyethene (LDPE — free radical, high pressure; HDPE — Ziegler-Natta, low pressure), polypropylene, polystyrene, PVC, Teflon, polyacrylonitrile (PAN/Orlon), polybutadiene, neoprene.

Condensation (Step Growth) Polymerization

Bifunctional or polyfunctional monomers react with the elimination of small molecules (H₂O, HCl, NH₃, CH₃OH). The polymer does NOT have the same composition as the monomer. Requires functional groups on both ends of the monomers. Examples: polyesters (terylene/Dacron — ethylene glycol + terephthalic acid), polyamides (nylon-66 — hexamethylenediamine + adipic acid; nylon-6 — caprolactam), phenolic resins (bakelite — phenol + formaldehyde), epoxy resins, polyurethanes. Copolymers vs Homopolymers: Homopolymers have only one type of monomer (e.g., polythene). Copolymers have two or more different monomers (e.g., Buna-S, nylon-66, terylene).

Important Polymers for NEET

PolymerMonomer(s)Polymerization TypeApplications
Low Density Polyethene (LDPE)CH₂=CH₂ (ethene)Addition (free radical, high pressure)Plastic bags, squeeze bottles, films
High Density Polyethene (HDPE)CH₂=CH₂Addition (Ziegler-Natta, low pressure)Bottles, pipes, containers, buckets
Polyvinyl chloride (PVC)CH₂=CHCl (vinyl chloride)AdditionPipes, flooring, cable insulation, synthetic leather
Polytetrafluoroethene (PTFE / Teflon)CF₂=CF₂ (tetrafluoroethene)AdditionNon-stick cookware, gaskets, bearings
PolystyreneC₆H₅CH=CH₂ (styrene)AdditionPackaging, disposable cups, thermal insulation
Polyacrylonitrile (PAN / Orlon / Acrilan)CH₂=CH—CN (acrylonitrile)AdditionSynthetic fibres (acrylic wool, sweaters, blankets)
Polymethyl methacrylate (PMMA / Plexiglass)CH₂=C(CH₃)COOCH₃ (methyl methacrylate)AdditionShatterproof windows, lenses, acrylic sheets
NeopreneCH₂=CCl—CH=CH₂ (chloroprene)AdditionSynthetic rubber, wetsuits, gaskets, conveyor belts
Buna-S (SBR — Styrene Butadiene Rubber)Butadiene + StyreneAddition (copolymer)Synthetic rubber (automobile tyres, footwear)
Buna-N (Nitrile rubber)Butadiene + AcrylonitrileAddition (copolymer)Oil-resistant seals, O-rings, hoses
PolybutadieneButadieneAdditionSynthetic rubber
Polyisoprene (Natural rubber)Isoprene (2-methylbuta-1,3-diene)Addition (natural, cis-1,4 polymer)Natural rubber, tyres, gloves
Vulcanized rubberNatural rubber + Sulphur (heat)Cross-linking (by S bridges)More durable rubber (tyres) — Charles Goodyear
Nylon-66Hexamethylenediamine + Adipic acidCondensationFabrics, ropes, parachutes, gears, brushes
Nylon-6Caprolactam (ring-opening polymerization)CondensationFabrics, carpets, fishing nets
Nylon-6,10Hexamethylenediamine + Sebacic acidCondensationBristles, brushes
Terylene (Dacron / PET)Ethylene glycol + Terephthalic acidCondensationPolyester fibres, PET bottles, film
Bakelite (Phenol-formaldehyde resin)Phenol + FormaldehydeCondensationElectrical switches, handles, telephones, adhesives
Melamine-formaldehyde resinMelamine + FormaldehydeCondensationUnbreakable crockery, decorative laminates
Glyptal (Alkyd resin)Ethylene glycol + Phthalic acidCondensationPaints, coatings, varnishes
PolyurethanePolyol + DiisocyanateCondensationFoam (mattresses, cushions), elastomers, coatings
Kevlarp-Phenylenediamine + Terephthaloyl chlorideCondensationBulletproof vests, high-strength fibres
Example 18 — Polymers
The monomer of Teflon is:

a) CH₂=CH₂   b) CF₂=CF₂   c) CH₂=CHCl   d) C₆H₅CH=CH₂
Solution: Teflon (PTFE) is polymerized tetrafluoroethylene — CF₂=CF₂. Option (b).
Mnemonic: Nylon Numbers
"Nylon-66: Two monomers each with 6 carbons (hexamethylenediamine = 6C, adipic acid = 6C). Nylon-6: Single monomer with 6 carbons (caprolactam)."

10. Chemistry in Everyday Life

This chapter applies organic chemistry concepts to real-world substances: drugs and pharmaceuticals, chemicals in food, and cleansing agents. NEET typically asks 1-2 questions from this topic — often on drug classification, antiseptics vs disinfectants, or artificial sweeteners.

Drugs & Pharmaceuticals

Analgesics (painkillers): Non-narcotic (Aspirin — acetylsalicylic acid, analgesic, antipyretic, anti-inflammatory; Paracetamol — acetaminophen, analgesic, antipyretic; Ibuprofen). Narcotic (Morphine — from opium poppy, acts on CNS, habit-forming). Antipyretics: Reduce fever (Paracetamol, Aspirin, Ibuprofen). Antibiotics: Bactericidal (kill bacteria — Penicillin, Aminoglycosides) vs Bacteriostatic (inhibit growth — Tetracycline, Chloramphenicol, Erythromycin). Broad-spectrum (effective against Gram-positive and Gram-negative — Chloramphenicol, Tetracycline, Ampicillin). Narrow-spectrum (Penicillin G — mainly Gram-positive). Penicillin was discovered by Alexander Fleming. Penicillin contains a β-lactam ring; penicillianase (β-lactamase) is an enzyme produced by resistant bacteria that cleaves this ring. Antiseptics: Applied to living tissues (Dettol — chloroxylenol + terpineol; Boric acid; Iodine tincture 2%; Hydrogen peroxide; Iodoform). Disinfectants: Applied to inanimate objects (Phenol — 1% solution is antiseptic, 5% is disinfectant; Chlorine — water disinfection; SO₂ — fumigation). Tranquilizers (anxiolytics/sedatives): Reduce anxiety and induce calm (Barbiturates — phenobarbital, Veronal; Benzodiazepines — Valium/diazepam, Librium; Meprobamate — Equanil). Antacids: Neutralize stomach acid (Mg(OH)₂ milk of magnesia, Al(OH)₃ gel, NaHCO₃). H₂-receptor antagonists (Ranitidine, Cimetidine, Famotidine) reduce acid secretion by blocking histamine receptors in the stomach. Antihistamines: Block H₁ histamine receptors, used for allergies (Loratadine, Cetirizine, Diphenhydramine — causes drowsiness; the newer drugs are non-sedating).

Chemicals in Food

Preservatives: Prevent microbial spoilage. Common examples: Sodium benzoate (C₆H₅COONa, effective in acidic foods like pickles, jams, soft drinks), Sodium metabisulphite (Na₂S₂O₅, used in fruit juices, wine), Sorbic acid and potassium sorbate (used in cheese, baked goods), Citric acid, Vinegar (acetic acid). Artificial sweeteners: Provide sweetness without calories. Saccharin (oldest, 550× sweeter than sucrose, has bitter aftertaste), Aspartame (180× sweeter, not heat-stable, used in cold drinks and foods; contraindicated in phenylketonuria — PKU), Sucralose (600× sweeter, heat-stable, used in cooking), Alitame (2000× sweeter), Neotame. Food colours (synthetic): Sunset Yellow (FD&C Yellow 6), Tartrazine (FD&C Yellow 5, an azo dye), Indigo carmine, Fast Green FCF, Allura Red (Red 40). Natural colours: Carotene/Carotenoids (orange/yellow), Chlorophyll (green), Anthocyanins (red-purple, found in berries).

Cleansing Agents

Soaps: Sodium or potassium salts of long-chain fatty acids (RCOO⁻Na⁺, where R = C₁₁ to C₁₇). Soaps work by forming micelles — the hydrophobic (tail) dissolves grease/oil while the hydrophilic (head) faces water, emulsifying the dirt. Soaps are ineffective in hard water because Ca²⁺/Mg²⁺ ions form insoluble scum (calcium/magnesium stearate). Synthetic detergents: Anionic (SDS — sodium dodecylbenzenesulphonate; sodium lauryl sulphate), Cationic (Cetyltrimethylammonium bromide, CTAB — also has germicidal properties), Non-ionic (Polyethylene glycol alkyl ethers). Detergents work in both hard and soft water because the sulphonate/sulphate anions do not form insoluble precipitates with Ca²⁺/Mg²⁺. Biodegradability: Branched-chain alkylbenzenesulphonates (hard detergents, non-biodegradable, cause water pollution — foaming in rivers). Linear alkylbenzenesulphonates (soft detergents, biodegradable, environmentally friendly).

CategoryCompound / ExampleMechanism / Use
AntacidMg(OH)₂, Al(OH)₃, NaHCO₃, Ranitidine (H₂ blocker)Neutralizes acid or blocks histamine H₂-receptors
AntihistamineLoratadine, Cetirizine, DiphenhydramineBlocks H₁ receptors, treats allergies
Analgesic (non-narcotic)Paracetamol, Aspirin, IbuprofenPain relief, anti-inflammatory, antipyretic
Analgesic (narcotic)Morphine, CodeineCNS acting, may cause addiction
Antibiotic (broad-spectrum)Chloramphenicol, Tetracycline, AmpicillinKills/inhibits both Gram+ and Gram− bacteria
Antibiotic (narrow-spectrum)Penicillin GMainly Gram+ bacteria
AntisepticDettol, Boric acid, Iodine, H₂O₂Applied to living tissue (wounds, cuts)
DisinfectantPhenol (5%), Chlorine (water), SO₂Inanimate objects, surfaces, water
TranquilizerValium (diazepam), Barbiturates, EquanilAnxiety, stress, sleep disorders
AntibacterialProntosil (sulfonamide prodrug)First synthetic antibacterial
Artificial sweetenerSaccharin, Aspartame, Sucralose, AlitameLow-calorie sweetening of foods/drinks
Food preservativeSodium benzoate, Sorbic acid, Na₂S₂O₅Prevents microbial growth
Example — Drug Classification
Which of the following is a broad-spectrum antibiotic?

a) Penicillin G   b) Chloramphenicol   c) Amoxycillin   d) Streptomycin
Solution: Chloramphenicol is a broad-spectrum antibiotic effective against both Gram-positive and Gram-negative bacteria. Option (b).
Example — Antiseptic vs Disinfectant
Phenol at which concentration acts as a disinfectant?

a) 0.1%   b) 0.5%   c) 1%   d) 5%
Solution: 1% phenol solution acts as an antiseptic (applied to living tissues). 5% phenol solution acts as a disinfectant (used on inanimate surfaces). Option (d) for disinfectant.
Trick: Soaps vs Detergents
Soaps form scum (insoluble Ca²⁺/Mg²⁺ salts) in hard water and lose efficiency. Detergents (sodium alkylbenzenesulphonate) do not form insoluble salts with Ca²⁺/Mg²⁺, so they work in both hard and soft water. Also: branched-chain detergents are non-biodegradable (hard detergents); linear chain detergents are biodegradable (soft detergents).

Quick Reference: Functional Group Interconversions

Starting MaterialReagent(s)Product
AlkeneH₂ / PdAlkane
AlkeneH₂O / H⁺Alcohol
AlkeneO₃ then Zn/H₂OAldehyde or Ketone
1° AlcoholPCC / K₂Cr₂O₇ + H⁺Aldehyde
1° AlcoholK₂Cr₂O₇ / H⁺ (excess)Carboxylic acid
2° AlcoholK₂Cr₂O₇ / H⁺Ketone
AldehydeTollens / FehlingCarboxylic acid
AldehydeNaBH₄ / LiAlH₄1° Alcohol
KetoneNaBH₄ / LiAlH₄2° Alcohol
Carboxylic acidLiAlH₄1° Alcohol
Carboxylic acidSOCl₂Acid chloride
Acid chlorideNH₃Amide
AmideBr₂ + NaOH (Hoffmann)1° Amine (−1C)
Nitrile (RCN)H₂ / Ni or LiAlH₄1° Amine
EsterH₂O / H⁺ or OH⁻Carboxylic acid + Alcohol

Master List: Named Reactions for NEET

Reaction NameStarting MaterialReagent / ConditionsProductKey Feature
WurtzR—X (alkyl halide)2Na, dry etherR—R (alkane)Doubles C atoms; symmetrical alkane
Wurtz-FittigR—X + Ar—X2Na, dry etherR—Ar (alkylbenzene)Coupling of alkyl and aryl halides
Fittig2Ar—X2Na, dry etherAr—Ar (diaryl)Coupling of two aryl halides
SandmeyerAr—N₂⁺Cl⁻CuCl/HCl or CuBr/HBr or CuCN/KCNAr—Cl, Ar—Br, Ar—CNReplacement of —N₂⁺ by —Cl, —Br, —CN
GattermannAr—N₂⁺Cl⁻Cu + HX (X = Cl, Br)Ar—X (haloarene)Similar to Sandmeyer using Cu powder
Balz-SchiemannAr—N₂⁺Cl⁻HBF₄, then heatAr—F (fluorobenzene)Introduction of fluorine into aromatic ring
KucherovAlkyne (terminal)HgSO₄ + H₂SO₄, H₂OEnol → KetoneMarkovnikov hydration of alkynes
Lindlar reductionAlkyneH₂, Pd/CaCO₃ + quinolinecis-AlkenePartial reduction to cis-alkene
Birch reductionAlkyneNa, NH₃ (liq.)trans-AlkenePartial reduction to trans-alkene
OzonolysisAlkeneO₃ then Zn/H₂OAldehydes / KetonesCleaves C=C; useful for structural determination
Kolbe's electrolysisRCOOK (conc. aq.)Electrolysis, C/Pt electrodesR—R (alkane) + CO₂Anode: decarboxylation + coupling
FranklandR—XZn + H₂O (trace)R—H (alkane)Reduction to alkane
RosenmundRCOCl (acyl chloride)H₂, Pd/BaSO₄ (poisoned catalyst)RCHO (aldehyde)Selective reduction of acid chloride to aldehyde
StephenRCN (nitrile)SnCl₂ + HCl, then H₂ORCHO (aldehyde)Reduction of nitrile to imine, then aldehyde
CannizzaroHCHO / ArCHO (no α-H)Conc. NaOH/KOH (heat)Alcohol + Carboxylic acidDisproportionation of aldehydes
Aldol condensationRCHO / RCOR' (with α-H)Dil. NaOH, then heatβ-Hydroxy carbonyl / α,β-unsaturatedC—C bond formation via enolate
PerkinArCHO + (CH₃CO)₂OCH₃COONa (heat)ArCH=CHCOOH (cinnamic acid)Aldehyde + anhydride → α,β-unsaturated acid
Claisen condensation2RCOOR' (esters)Na/NaOEt (base), then H⁺β-Keto esterEster + ester → ketone
Hoffmann bromamideRCONH₂ (amide)Br₂ + NaOHRNH₂ (1° amine)Rearrangement; product has one less C
CurtiusRCOCl (acyl chloride)NaN₃, then heat, then H₂ORNH₂ (1° amine)Acyl azide intermediate; rearrangement
SchmidtRCOOH + HN₃H₂SO₄ (conc.)RNH₂ (1° amine)Carboxylic acid → amine with loss of CO₂
Koble-SchmittPhenol (C₆H₅OH)CO₂ + NaOH (125°C, 4-7 atm)Salicylic acidCO₂ introduced ortho to OH
Reimer-TiemannPhenolCHCl₃ + NaOH (50-70°C)SalicylaldehydeFormyl group introduced ortho to OH
Williamson's synthesisR—ONa + R'—XHeatR—O—R' (ether)Best method for ether preparation
Fischer esterificationRCOOH + R'OHH₂SO₄ (conc.), heatRCOOR' (ester) + H₂OReversible, needs excess of one reagent
Hell-Volhard-Zelinsky (HVZ)RCH₂COOHBr₂ + P (red), then H₂ORCHBrCOOH (α-bromo acid)α-Halogenation of carboxylic acids
Gabriel phthalimideR—X (primary alkyl halide)Potassium phthalimide + H⁺ (hydrolysis)RNH₂ (1° amine, pure)Primary amine without 2°/3° contamination
Clemmensen reductionC=O (carbonyl)Zn(Hg) + HCl (conc.)—CH₂— (alkane)Reduction of C=O to CH₂ in acid
Wolff-Kishner reductionC=O (carbonyl)NH₂NH₂ + KOH (heat, ethylene glycol)—CH₂— (alkane)Reduction of C=O to CH₂ in base
Hinsberg test1°/2°/3° amineC₆H₅SO₂Cl + NaOH (aq.)Different products eachDistinguishes 1°, 2°, 3° amines
Carbylamine (isocyanide)1° amine (RNH₂)CHCl₃ + alc. KOH (heat)R—NC (isocyanide)Foul smell; only 1° amines react
MendiusRCN (nitrile)H₂/Ni or Na/EtOH or LiAlH₄RCH₂NH₂ (1° amine)Reduction of nitrile to primary amine
Hoffmann eliminationR₄N⁺OH⁻ (quat. ammonium hydroxide)Heat (strong base)Alkene (least substituted) + R₃N + H₂OHoffmann product (less substituted alkene)
Trick: Oxidation State of Carbon
Alkane (lowest oxidation) → Alkene → Alkyne → Alcohol → Aldehyde → Carboxylic acid → CO₂ (highest oxidation). Moving right = oxidation (loss of H / gain of O).

Important Reagents — Quick Reference

ReagentUsed ForKey Observation / Product
Tollens' reagent [Ag(NH₃)₂]⁺OH⁻Aldehyde detectionSilver mirror
Fehling's solution Cu²⁺/tartrateAliphatic aldehyde detectionRed Cu₂O precipitate
Benedict's solutionReducing sugars / aldehydesRed Cu₂O precipitate
Schiff's reagentAldehyde detectionPink/magenta colour
I₂ / NaOH (iodoform)CH₃CO— / CH₃CH(OH)— groupsYellow CHI₃ precipitate
2,4-DNPHAldehydes / KetonesOrange/red precipitate
NaHSO₃ (saturated)Purification of carbonyl compoundsWhite crystalline adduct
Br₂ / CCl₄Test for unsaturation (C=C, C≡C)Brown colour discharged
Baeyer's reagent (cold KMnO₄)Test for unsaturationPurple colour discharged; diol formed
Lucas reagent (ZnCl₂ + HCl)Distinguish 1°/2°/3° alcoholsCloudiness: 3° immediate, 2° 5-10 min, 1° no reaction
Hinsberg reagent (C₆H₅SO₂Cl)Distinguish 1°/2°/3° amines1°: clear soln; 2°: ppt; 3°: oily layer
Carbylamine (CHCl₃ + alc. KOH)Test for primary aminesFoul-smelling isocyanide (RNC)
Biuret reagent (CuSO₄/NaOH)Detection of peptide bonds (proteins)Violet colour
NinhydrinDetection of α-amino acidsBlue-purple colour (proline gives yellow)
Soda lime (NaOH + CaO)Decarboxylation of carboxylic acidsAlkane (R—H) + Na₂CO₃
Grignard reagent (RMgX)C—C bond formation; alcohol synthesis1°/2°/3° alcohol depending on carbonyl
LiAlH₄ / NaBH₄Reduction of carbonyls and other groupsLiAlH₄: strong; NaBH₄: mild
PCC (pyridinium chlorochromate)Mild oxidation of 1° alcoholsStops at aldehyde (does not overoxidize)
SOCl₂ (thionyl chloride)Conversion of RCOOH to RCOClAcid chloride + SO₂ + HCl (gaseous by-products)
O₃ (ozone) then Zn/H₂OOzonolysis of alkenesAldehydes / Ketones
Lindlar catalyst (Pd/CaCO₃ + quinoline)Partial reduction of alkynescis-Alkene
Na/NH₃ (Birch reduction)Partial reduction of alkynestrans-Alkene

Quick Mnemonics for Common NEET Traps

MnemonicMeaning
"Rich get richer"Markovnikov: the carbon with more H gets the H⁺
"SN1 — Tertiary, Two-step, Through carbocation"SN1 favours 3° alkyl halides
"SN2 — Primary, Preferred with Polar aprotic"SN2 favours 1° alkyl halides
"E2 — anti-periplanar"H and LG must be 180° apart in E2
"3-2-1 Lucas test"3° immediate cloudiness; 2° 5-10 min; 1° no reaction
"A-T: 2 bonds; G-C: 3 bonds"DNA base pairing: A=T (2), G≡C (3)
"Activating = Ortho-Para; Deactivating = Meta"EAS directing effects; except halogens (ortho-para but deactivating)
"Aromatics: 4n+2 = Happy"Hückel's rule for aromaticity
"EWG increases acidity; EDG decreases it"Substituent effect on carboxylic acid/phenol acidity
"Tollens, Fehling, Benedict — All for Aldehydes"Aldehyde distinction tests; ketones do not react
"Primary gives Carbylamine; Secondary/tertiary do not"Carbylamine test is only for 1° amines
"Nylon-66: 6 + 6 = two monomers each with 6C"Nylon-66 is from hexamethylenediamine (6C) + adipic acid (6C)

Product Prediction — Practice Examples for NEET

Product Prediction 1 — Alkene Addition
The major product formed when 2-methylbut-2-ene is treated with HBr in the presence of peroxide is:

a) 2-bromo-2-methylbutane   b) 1-bromo-2-methylbutane   c) 1-bromo-3-methylbutane   d) 2-bromo-3-methylbutane
Solution: Peroxide effect (anti-Markovnikov addition of HBr, free radical). Br· adds to the less substituted carbon of the double bond (C-1), and H adds to C-2. Product: 1-bromo-2-methylbutane. Option (b).
Product Prediction 2 — Carbocation Rearrangement
On heating 3,3-dimethylbutan-2-ol with conc. H₂SO₄, the major alkene product is:

a) 3,3-dimethylbut-1-ene   b) 2,3-dimethylbut-2-ene   c) 2,3-dimethylbut-1-ene   d) 3,3-dimethylbut-2-ene
Solution: Dehydration initially gives a 2° carbocation. A 1,2-methyl shift forms a more stable 3° carbocation. Loss of H⁺ from the adjacent carbon gives the more substituted alkene: 2,3-dimethylbut-2-ene. Option (b).
Product Prediction 3 — Aromatic Substitution
Nitration of nitrobenzene (conc. HNO₃ + H₂SO₄) gives the major product:

a) o-Dinitrobenzene   b) m-Dinitrobenzene   c) p-Dinitrobenzene   d) 1,3,5-Trinitrobenzene
Solution: —NO₂ is a meta-directing and deactivating group. The second nitro group goes to the meta position. m-Dinitrobenzene is the major product. Option (b).
Product Prediction 4 — Nucleophilic Addition
Acetone (CH₃COCH₃) reacts with HCN in the presence of a catalytic amount of KCN. The product is:

a) CH₃C(OH)(CN)CH₃   b) CH₃CH(OH)CH₂CN   c) CH₃CH(CN)CH₃   d) CH₃COCH₂OH
Solution: CN⁻ attacks the carbonyl carbon, followed by protonation to form the cyanohydrin: (CH₃)₂C(OH)CN (acetone cyanohydrin). Option (a).
Product Prediction 5 — Diazonium Replacement
Benzene diazonium chloride is heated with water. The product formed is:

a) Chlorobenzene   b) Benzene   c) Phenol   d) Aniline
Solution: Diazonium salt + H₂O (heat) → Ar—OH + N₂ + HCl. Replacement of —N₂⁺ by —OH gives phenol. Option (c).
Product Prediction 6 — EAS with Activating Group
Aniline undergoes bromination with Br₂ water (without any catalyst). The major product is:

a) p-Bromoaniline   b) o-Bromoaniline   c) 2,4,6-Tribromoaniline   d) m-Bromoaniline
Solution: —NH₂ is strongly activating. In the presence of Br₂ water (no FeBr₃ catalyst needed), aniline undergoes tribromination at the ortho and para positions, forming 2,4,6-tribromoaniline (white precipitate). Option (c).
Final Exam Tips for NEET Organic Chemistry
(1) Always check the number of carbon atoms and principal functional group first. (2) In mechanism questions, identify the intermediate (carbocation, radical, carbanion) to predict the product. (3) For stereochemistry, look for chiral centres and check for an internal plane of symmetry (meso). (4) In EAS, determine if the substituent is activating/deactivating and ortho-para/meta directing. (5) For distinction tests, know which reagent gives which colour/precipitate with which functional group. (6) Practice converting between common names and IUPAC names — this is a frequent exam question. (7) Always check for possible carbocation rearrangements in SN1/E1 reactions. (8) Remember that only HBr shows the peroxide effect — HCl and HI do not.
Trick: How to Approach Product Prediction Questions
Step 1: Identify the functional group(s) in the starting material. Step 2: Identify the reagent — is it an oxidizing agent, reducing agent, base, acid, catalyst, or nucleophile? Step 3: Determine the reaction type — substitution, addition, elimination, oxidation, reduction, rearrangement, or condensation. Step 4: Consider stereochemistry — is there a chiral centre? Will there be inversion, retention, or racemization? Step 5: Check for rearrangements (especially with carbocation intermediates). Step 6: Predict the major product based on stability (Zaitsev vs Hoffmann, more stable carbocation, etc.).
Common NEET Trap 1 — Peroxide Effect Scope
Remember: Only HBr shows the peroxide (Kharasch) effect. The free radical mechanism does NOT work for HCl (bond too strong to cleave) or HI (I· is too reactive, gets oxidized by peroxide). NEET frequently asks: "Which halide shows anti-Markovnikov addition?" Answer: Only HBr in the presence of peroxide.
Common NEET Trap 2 — Halogen Directing Effect
Halogens (Cl, Br, I) are the ONLY group that is deactivating but ortho-para directing. All other deactivating groups are meta directing. And all activating groups are ortho-para directing. If a NEET question asks: "Which is ortho-para directing but deactivating?" the answer is halogens (—Cl, —Br, —I, —F).
Common NEET Trap 3 — Reducing Sugars
Sucrose is non-reducing (both anomeric carbons are involved in the glycosidic bond). Glucose, fructose, maltose, and lactose are all reducing. However, sucrose on hydrolysis gives glucose + fructose (both reducing — invert sugar). NEET question: "Which disaccharide is non-reducing?" Answer: Sucrose only.
Common NEET Trap 4 — SN1 Rearrangement
If a neopentyl halide (primary alkyl halide with a quaternary β-carbon) undergoes solvolysis, the product is rearranged despite being primary. The 1° carbocation undergoes a 1,2-methyl shift to form a more stable 3° carbocation. The product is NOT neopentyl alcohol but tert-amyl alcohol (2-methylbutan-2-ol). Always check for possible rearrangements!
Common NEET Trap 5 — Aldol vs Cannizzaro
Aldol condensation requires at least one α-hydrogen. Cannizzaro reaction occurs when the aldehyde has NO α-hydrogen. If a question gives an aldehyde without α-H (formaldehyde, benzaldehyde, trimethylacetaldehyde), the reaction is Cannizzaro, NOT aldol. Aldehydes with α-H undergo aldol condensation.
Common NEET Trap 6 — Fehling's Test Scope
Aliphatic aldehydes reduce Fehling's solution (red Cu₂O precipitate). Aromatic aldehydes (e.g., benzaldehyde) do NOT reduce Fehling's solution. However, both aliphatic and aromatic aldehydes give the Tollens test (silver mirror). This distinction is frequently tested.
Common NEET Trap 7 — Friedel-Crafts Limitations
Friedel-Crafts alkylation with 1-chloropropane and benzene gives isopropylbenzene (cumene), NOT n-propylbenzene. This is because the 1° carbocation rearranges to a more stable 2° carbocation. To get n-propylbenzene, use Friedel-Crafts acylation (with propanoyl chloride) followed by Clemmensen or Wolff-Kishner reduction — no rearrangement occurs with acylium ions.
Common NEET Trap 8 — Basicity Order
The basicity order in aqueous solution is: 2° amine > 1° amine > 3° amine > NH₃. In the gas phase: 3° > 2° > 1° > NH₃. NEET often asks: "Which is the strongest base in water?" — Answer is secondary amine. "In the gas phase?" — Answer is tertiary amine.
Common NEET Trap 9 — Optical Activity
A compound having a chiral centre is not necessarily optically active. If the molecule has an internal plane of symmetry (meso compound), it is optically inactive despite having chiral centres. Example: meso-tartaric acid has 2 chiral centres but is optically inactive due to the plane of symmetry. Another trick: A racemic mixture (50:50 d + l) is optically inactive.
Common NEET Trap 10 — Hückel's Rule Exceptions
A compound must be (a) cyclic, (b) planar, (c) fully conjugated, and (d) have (4n+2) π electrons to be aromatic. Cyclooctatetraene (8π) is non-aromatic because it is tub-shaped (non-planar), not because it violates the 4n+2 rule. Cyclobutadiene (4π, 4n) is anti-aromatic (planar, conjugated, but 4n π electrons — extremely unstable).

Organic Chemistry Revision Roadmap for NEET

Week 1 — Fundamentals & Nomenclature: Master IUPAC naming, functional group priority, hybridization, electron displacement effects (I, R, H, E), and reactive intermediates. Week 2 — Isomerism & Hydrocarbons: Structural isomerism, geometrical/optical isomerism, conformations. Alkanes (preparation, free radical halogenation), Alkenes (addition, Markovnikov, anti-Markovnikov, ozonolysis), Alkynes (acidity, Kucherov, partial reduction), Aromatic HCs (Hückel, EAS, directing effects). Week 3 — Haloalkanes, Alcohols, Ethers: SN1/SN2/E1/E2 mechanisms, Lucas test, Williamson's synthesis, phenol chemistry (Kolbe, Reimer-Tiemann), distinction tests. Week 4 — Carbonyl Compounds: Nucleophilic addition, aldol, Cannizzaro, reduction (Wolff-Kishner, Clemmensen), oxidation (Tollens, Fehling, iodoform), carboxylic acid derivatives. Week 5 — Amines, Biomolecules, Polymers, Everyday Chemistry: Basicity of amines, Hinsberg, diazonium salts, carbohydrates (reducing/non-reducing), proteins (peptide bond, denaturation), DNA base pairing, polymers (addition vs condensation), drugs and classification, soaps vs detergents. Week 6 — Practice: Solve 20-25 NEET-level MCQs daily, review named reactions, practice product prediction, and focus on distinction tests and common traps.

Key Numeric Constants for NEET Organic Chemistry

ConceptValueNotes
Hückel's rule π electrons4n + 2n = 0, 1, 2... (6, 10, 14...)
Maximum optical isomers2ⁿ (n = chiral centres)Reduced if meso forms exist
Peptide bonds in polypeptide(α-amino acids) − 150 amino acids → 49 peptide bonds
sp³ bond angle109.5°Tetrahedral geometry
sp² bond angle120°Trigonal planar geometry
sp bond angle180°Linear geometry
DNA H-bonds: A—T, G—C2, 3A=T (2), G≡C (3)
pKa of carboxylic acids~4-5Stronger than phenols (pKa ~10)
pKa of terminal alkynes~25Acidic due to 50% s-character
Diazotization temperature0-5°CAbove 5°C, diazonium salt decomposes
Iodine value measureUnsaturation of fats/oilsHigher = more unsaturation
Saponification numbermg KOH/g fatHigher = shorter fatty acid chains

Mastering NEET Organic Chemistry requires understanding reaction mechanisms (not just memorizing products), practicing regularly with timed MCQs, and reviewing common exam traps. Focus on high-weightage topics: named reactions, stereoisomerism, EAS, SN1/SN2, biomolecules, and polymers. Good luck!

Practice Questions

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