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KMR ADVICE

B.Pharm Exam Strategy & Important Questions Guide

Mr. K. Mallikarjuna Reddy

Associate Professor, M. Pharma (Pharmacology)

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EXAM STRATEGY & IMPORTANT QUESTIONS GUIDE

2.2 BP202T · PHARMACEUTICAL ORGANIC CHEMISTRY I (THEORY)

Complete PCI B.Pharm Semester II syllabus coverage with detailed answers, star-rated importance, and key terms highlighted.
Based on real university question-paper analysis (JNTU-H/K, AKTU, KUHS, Paru, RGUHS, Anna Univ).

📖 HOW TO USE THIS GUIDE

🔵 Click any blue tag for abbreviation + brief note.

🟣 Click any purple term for plain-English explanation.

🔊 Click speaker icon for pronunciation.

⭐ Stars reflect real past-paper repeat frequency.

✍️ Every answer opens with a short Opening Line — copy it as your first paragraph.

⚡ Each question ends with a compact At-a-Glance Summary.

PRIORITY READING GUIDE

🔴 TOP PRIORITY

SN1 vs SN2 & E1 vs E2 — mechanisms, kinetics, factors.

Markownikoff's rule + anti-Markownikoff (peroxide effect).

Aldol condensation + Cannizzaro + Benzoin + Perkin.

Acidity of carboxylic acids — inductive effect of substituents.

Basicity of amines — effect of alkyl / aryl substituents.

IUPAC nomenclature — up to 10 C open chain + cyclic.

🟡 MEDIUM PRIORITY

Diels-Alder reaction, conjugated diene stability.

Structural isomerism types with examples.

Structure + uses of formaldehyde, chloroform, CCl₄, iodoform, glycerol.

Ozonolysis of alkenes; free-radical halogenation of alkanes.

🔵 LOW PRIORITY

Qualitative tests for alcohols / acids / esters / amides.

Allylic rearrangement; electromeric effect; carbocation rearrangements.

UNIT I
Classification, Nomenclature & Isomerism (7 h)
1
Explain IUPAC nomenclature 🔊 of organic compounds, with examples up to 10 carbons. Classify organic compounds.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEOrganic chemistry deals with over ten million carbon-containing compounds; without a rational naming system, every compound would acquire multiple conflicting names. The IUPAC system solves this problem by providing one universal systematic name for each structure.
Classification of Organic Compounds:
Organic compounds are classified both by carbon skeleton and by functional group.
By carbon skeleton:
Acyclic (open-chain or aliphatic) compounds such as alkanes, alkenes and alkynes.
Cyclic compounds are subdivided into carbocyclic rings — alicyclic (cyclohexane) and aromatic (benzene) — and heterocyclic rings that may be saturated (pyrrolidine) or aromatic (pyridine, furan).
By functional group: hydrocarbons, halides, alcohols, ethers, aldehydes, ketones, carboxylic acids, esters, amides, amines, nitriles, nitro compounds and thiols.
IUPAC Rules — Step by Step:
(1) Identify the longest continuous carbon chain containing the principal functional group; this gives the parent name.
(2) Number the chain from the end that gives the principal group (and any substituents) the lowest set of locants.
(3) Name the substituents as alphabetical prefixes, using multiplying prefixes (di, tri, tetra) when they are repeated.
(4) Add the functional-group suffix according to the priority order:
COOH (-oic acid) > COOR (-oate) > CONH₂ (-amide) > CN (-nitrile) > CHO (-al) > CO (-one) > OH (-ol) > NH₂ (-amine) > C=C (-ene) > C≡C (-yne) > alkane (-ane) (5) Insert locants (Arabic numerals) before the suffix to show the positions.
Chain Names (C1 – C10):
The standard stem names are meth (C1), eth (C2), prop (C3), but (C4), pent (C5), hex (C6), hept (C7), oct (C8), non (C9) and dec (C10). The family suffixes are -ane (alkane), -ene (alkene) and -yne (alkyne).
Illustrative Examples:
CH₃ – CH₂ – CH₃ is propane.
CH₃ – CH(CH₃) – CH₂ – CH₃ is 2-methylbutane.
CH₃ – CH=CH – CH₃ is but-2-ene.
CH₃ – CH(OH) – CH₃ is propan-2-ol.
CH₃ – CH₂ – CHO is propanal.
CH₃ – CO – CH₃ is propan-2-one (common name, acetone).
CH₃ – COOH is ethanoic acid (common name, acetic acid).
HOOC – COOH is ethanedioic acid (common name, oxalic acid).
C₆H₁₂ is cyclohexane and C₆H₅ – COOH is benzoic acid (benzenecarboxylic acid).
⚡ AT-A-GLANCE SUMMARY
  • Classification: acyclic versus cyclic; carbocyclic (alicyclic + aromatic) versus heterocyclic.
  • IUPAC steps: longest chain → numbering → substituent prefix → suffix by priority.
  • Chain stems (C1 – C10): meth, eth, prop, but, pent, hex, hept, oct, non, dec.
  • Functional-group priority (high → low): COOH > COOR > CONH₂ > CN > CHO > CO > OH > NH₂ > C=C > C≡C.
2
Define isomerism 🔊. Classify and explain the types of structural isomerism with examples.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEIsomerism arises because carbon is unique in its ability to form chains, branches and rings, so that molecules with the same molecular formula can adopt different arrangements and therefore different properties — a concept that is central to the understanding of drug action.
Types of Isomerism:
Isomerism is broadly of two kinds. Structural (constitutional) isomers differ in the way in which the atoms are connected, while stereoisomers have the same connectivity but differ in their spatial arrangement (geometrical or optical).
Structural Isomerism — Five Types:
(a) Chain isomerism arises from a different arrangement of the carbon skeleton. For example, n-butane (CH₃CH₂CH₂CH₃) and isobutane ((CH₃)₃CH) both have the formula C₄H₁₀ but different skeletons.
(b) Position isomerism occurs when the same skeleton and functional group are present but the functional group sits at different positions on the chain. For example, propan-1-ol (CH₃CH₂CH₂OH) differs from propan-2-ol (CH₃CH(OH)CH₃), and but-1-ene differs from but-2-ene.
(c) Functional isomerism occurs when compounds have the same molecular formula but different functional groups. For example, ethanol (CH₃CH₂OH) and dimethyl ether (CH₃OCH₃) both have the formula C₂H₆O, while propanal and acetone both have the formula C₃H₆O.
(d) Metamerism 🔊 occurs when different alkyl groups lie on either side of the same functional group. For example, methyl propyl ether (CH₃OC₃H₇) and diethyl ether (CH₃CH₂OCH₂CH₃) are both C₄H₁₀O. It is common in ethers, thioethers, secondary amines and ketones.
(e) Tautomerism 🔊 is the spontaneous interconversion of two forms that differ in the position of a hydrogen atom and an adjacent double bond. The classical keto–enol tautomerism is shown by acetone:
CH₃ – CO – CH₃ ⇌ CH₃ – C(OH)=CH₂ The reaction is catalysed either by acid or by base.
Stereoisomerism (Brief):
Geometrical (cis–trans) isomerism arises around a C=C double bond or a ring; classical examples are cis- and trans-but-2-ene, and maleic acid versus fumaric acid.
Optical isomerism occurs in molecules that contain a chiral centre and produces a pair of non-superimposable mirror-image enantiomers. A 1 : 1 mixture of the two is called a racemic mixture and is optically inactive.
⚡ AT-A-GLANCE SUMMARY
  • Two main types: structural and stereoisomerism.
  • Structural (five): chain, position, functional, metamerism, tautomerism.
  • Chain: n-butane versus isobutane.
  • Functional: ethanol versus dimethyl ether.
  • Metamerism: ether (methyl-propyl versus diethyl).
  • Tautomerism: keto ⇌ enol.
  • Stereoisomerism: geometrical (cis / trans) and optical (d / l, enantiomers).
UNIT II
Alkanes · Alkenes · Conjugated Dienes (10 h)
3
Explain sp³ hybridisation 🔊 in alkanes. Describe the methods of preparation, halogenation mechanism and uses of paraffins.
★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEAlkanes are the simplest organic compounds — saturated hydrocarbons containing only sp³-hybridised carbon atoms connected by single bonds. They are also called paraffins (from Latin parum affinis, "little affinity") because of their chemical inertness.
sp³ Hybridisation:
In sp³ hybridisation, one 2s and three 2p orbitals of carbon combine to give four equivalent sp³ hybrid orbitals, each having 25 % s and 75 % p character. The four orbitals are directed to the corners of a regular tetrahedron, with bond angles of 109.5°, and form four σ-bonds by head-on orbital overlap. Methane (CH₄) is the classical example.
General Methods of Preparation:
(1) Hydrogenation of alkenes or alkynes (Sabatier–Senderens): C₂H₄ + H₂ → C₂H₆ over Ni, Pt or Pd.
(2) Wurtz reaction 🔊: 2 R – X + 2 Na → R – R + 2 NaX (3) Kolbe's electrolysis 🔊 of sodium acetate: 2 CH₃COONa + 2 H₂O → C₂H₆ + 2 CO₂↑ + H₂↑ + 2 NaOH (4) Reduction of alkyl halides: R – X + Zn / HCl → R – H.
(5) From Grignard reagent: R – MgX + H₂O → R – H + Mg(OH)X.
(6) Decarboxylation of a sodium carboxylate on heating with sodalime: CH₃COONa + NaOH (CaO, Δ) → CH₄ + Na₂CO₃
Halogenation (Free-Radical Chain Mechanism):
Methane reacts with chlorine in sunlight to give successively CH₃Cl, CH₂Cl₂, CHCl₃ and CCl₄.
CH₄ + Cl₂ → (hν, UV) → CH₃Cl + HCl The reaction proceeds through three steps.
Initiation: homolytic fission of Cl₂ by UV light gives two chlorine radicals.
Propagation: Cl· + CH₄ → HCl + CH₃·, and CH₃· + Cl₂ → CH₃Cl + Cl·.
Termination: Cl· + Cl· → Cl₂, CH₃· + Cl· → CH₃Cl, or CH₃· + CH₃· → C₂H₆.
The reactivity order of halogens is F₂ > Cl₂ > Br₂ ≫ I₂; fluorine is too violent to control and iodine does not react thermodynamically.
Selectivity: tertiary hydrogens are abstracted in preference to secondary, which are in turn preferred to primary, because of the stability order of the resulting radical (3° > 2° > 1°).
Uses of Paraffins:
Liquid paraffin (mineral oil) is used as a laxative, an ointment base and a lubricant.
Soft paraffin (petrolatum) is used as an ointment base and a skin protectant.
Hard paraffin (paraffin wax) is used to stiffen suppository bases, to make candles, in paper waterproofing and for embedding in histology.
Alkanes are also important fuels (natural gas, LPG, petrol, diesel, kerosene) and feedstocks for the petrochemical industry.
⚡ AT-A-GLANCE SUMMARY
  • Alkanes: CₙH₂ₙ₊₂; sp³; bond angle 109.5°; tetrahedral shape.
  • Preparation: hydrogenation, Wurtz, Kolbe, reduction of R–X, Grignard + H₂O, decarboxylation.
  • Halogenation: free-radical chain (initiation / propagation / termination); UV or heat; 3° H > 2° > 1°.
  • Reactivity of halogens: F₂ > Cl₂ > Br₂ ≫ I₂.
  • Uses: liquid paraffin (laxative), soft paraffin (ointment), hard paraffin (candle / suppository stiffener).
4
Explain E1 and E2 reactions with their mechanisms and kinetics. State Saytzeff's rule. Compare E1 with E2.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEElimination reactions remove two groups from adjacent carbons to form a π-bond and are among the most important methods for preparing alkenes. They proceed by two mechanisms — the unimolecular E1 and the bimolecular E2 — depending on the substrate and the base.
E1 Reaction (Unimolecular Elimination):
The E1 mechanism has two steps.
Step 1 (slow, rate-determining): heterolysis of the C – X bond gives a carbocation. R – X → R⁺ + X⁻ Step 2 (fast): a base removes a β-hydrogen, forming the π-bond.
Kinetics: Rate = k [R – X] — first order overall.
It is favoured by a tertiary (3° > 2° > 1°) substrate, a weak base, a polar protic solvent (water, ethanol) and higher temperature. Because a carbocation is formed, carbocation rearrangement can occur.
E2 Reaction (Bimolecular Elimination):
E2 is a single concerted step: a strong base removes the β-hydrogen while the C – X bond simultaneously breaks, and the new π-bond forms. The reaction requires an anti-periplanar geometry (dihedral angle 180°) between the β-H and the leaving group for proper orbital overlap.
Kinetics: Rate = k [R – X][Base] — second order overall.
It is favoured by a strong bulky base (t-BuOK, NaNH₂), a polar aprotic solvent (DMSO) and a higher temperature. No intermediate is formed, so rearrangement is not possible.
Saytzeff's Rule:
According to Saytzeff's rule, the more substituted (more stable) alkene is the major product of elimination. Alkene stability follows the order tetrasubstituted > trisubstituted > disubstituted > monosubstituted > ethylene.
Example: 2-bromobutane with KOH gives but-2-ene as the major product (about 80 %) and but-1-ene as the minor (about 20 %).
Hofmann's rule 🔊 is the opposite: when a bulky base (t-BuOK) is used, the less substituted alkene predominates because the bulky base cannot reach the more substituted β-hydrogen.
E1 versus E2 — Comparison:
FeatureE1E2
StepsTwo (step-wise)One (concerted)
Ratek [R – X]k [R – X][Base]
OrderFirstSecond
IntermediateCarbocationNone
Substrate3° > 2° (not 1°)3° > 2° > 1°
Base strengthWeakStrong
SolventPolar proticPolar aprotic
RearrangementPossibleNot possible
StereochemistryNon-specificAnti-periplanar
Competes withSN1SN2
⚡ AT-A-GLANCE SUMMARY
  • E1: two steps through a carbocation; rate = k[R – X]; tertiary substrate; weak base; polar protic solvent.
  • E2: one concerted step; rate = k[R – X][Base]; strong base; anti-periplanar geometry.
  • Saytzeff's rule: the more substituted alkene is the major product.
  • Hofmann's rule: a bulky base favours the less substituted alkene.
  • E1 competes with SN1; E2 competes with SN2.
5
State Markownikoff's rule 🔊. Explain the electrophilic addition of alkenes and anti-Markownikoff addition. Describe ozonolysis 🔊.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEAlkenes owe their chemistry to the π-bond of the C=C double bond, which is electron-rich and attracts electrophiles. Electrophilic addition, Markownikoff's rule and ozonolysis are the three landmark reactions used to prepare and characterise alkenes.
Electrophilic Addition to Alkenes — Mechanism:
An alkene reacts with HX or X₂ to give a saturated product in two steps.
Step 1: the π-electrons attack the electrophile (H⁺ or Br⁺), producing the most stable carbocation intermediate (3° > 2° > 1°).
Step 2: the nucleophile (X⁻ or solvent) attacks the carbocation to give the final product.
Examples: CH₂=CH₂ + HBr → CH₃CH₂Br, and CH₂=CH₂ + Br₂ → CH₂Br – CH₂Br.
Markownikoff's Rule (1869):
In the ionic addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon bearing more hydrogens, and the halogen to the carbon bearing fewer. The rule is explained by the preferential formation of the more stable (more substituted) carbocation, which is stabilised by hyperconjugation and the inductive effect of the alkyl groups.
Example: CH₂=CH – CH₃ + HBr → CH₃ – CHBr – CH₃ (2-bromopropane; major product) rather than CH₃ – CH₂ – CH₂Br (1-bromopropane; minor product, less than 5 %).
Anti-Markownikoff Addition (Peroxide / Kharasch Effect, 1933):
In the presence of peroxides (R – O – O – R) or light, HBr adds to an alkene in the reverse sense by a free-radical mechanism. The mechanism proceeds as follows.
Initiation: R – O – O – R → 2 RO·
Propagation: RO· + HBr → ROH + Br·; Br· + CH₂=CH – CH₃ → CH₂Br – CH· – CH₃ (a more stable secondary radical); then CH₂Br – CH· – CH₃ + HBr → CH₂Br – CH₂ – CH₃ (1-bromopropane) + Br·.
Only HBr shows the peroxide effect; HCl does not, because its H – Cl bond is too strong, and HI does not, because the H – I bond is too weak for the radical chain to propagate.
Ozonolysis:
Ozonolysis is the oxidative cleavage of a C=C bond by ozone, followed by reductive work-up.
Steps: the alkene first reacts with O₃ to give a molozonide, which rearranges to an ozonide. The ozonide is then treated with Zn/H₂O or Me₂S, giving two carbonyl compounds.
CH₃ – CH=CH – CH₃ + O₃ → ozonide → (Zn, H₂O) → 2 CH₃CHO Ozonolysis is used for structure determination of alkenes, because the two carbonyl fragments identify the position of the original double bond.
⚡ AT-A-GLANCE SUMMARY
  • Electrophilic addition: two-step mechanism via carbocation intermediate; E⁺ then Nu⁻.
  • Markownikoff's rule: H to the carbon bearing more hydrogens, X to the carbon bearing more substituents; explained by carbocation stability.
  • Anti-Markownikoff (peroxide effect): only for HBr with peroxide; free-radical chain (Kharasch).
  • Ozonolysis: O₃ followed by Zn / H₂O gives two carbonyl compounds; used to locate the double bond.
6
Explain the stability of conjugated dienes. Describe the Diels–Alder reaction.
★★★★
5MShort Essay
Detailed Answer:
✍️ OPENING LINEConjugated dienes contain two C=C double bonds separated by one single bond; the overlap of π-orbitals across the entire system imparts extra stability and unique reactivity, epitomised in the Diels–Alder cycloaddition.
Stability of Conjugated Dienes:
In 1,3-butadiene (CH₂=CH – CH=CH₂), the π-electrons are delocalised over all four carbons by resonance, as shown below.
CH₂=CH – CH=CH₂ ⇌ ⁺CH₂ – CH=CH – CH₂⁻ The heat of hydrogenation of 1,3-butadiene is 57 kcal/mol, compared with 60 kcal/mol expected for two isolated double bonds; the difference of about 3 kcal/mol represents extra (resonance) stabilisation.
The C2 – C3 bond length (1.48 Å) is shorter than a normal single bond (1.54 Å), indicating partial double-bond character.
Diels–Alder Reaction ([4+2] Cycloaddition):
Reactants: the diene (four π-electrons) must be in the s-cis conformation, and the dienophile 🔊 (two π-electrons) is an alkene or alkyne that bears an electron-withdrawing group (CHO, CN, NO₂, COOR or C=O).
Mechanism: a single concerted step in which six π-electrons shift simultaneously, forming two new σ-bonds and one new π-bond.
Example: 1,3-butadiene reacts with maleic anhydride to give cyclohex-4-ene-1,2-dicarboxylic anhydride.
Features: the reaction is stereospecific (a cis dienophile gives a cis product), suprafacial on both diene and dienophile, and obeys the endo rule (the endo product is kinetically preferred). At high temperature the reaction is reversible (retro-Diels–Alder).
Uses: synthesis of six-membered rings, natural products (including steroids) and pharmaceutical intermediates.
⚡ AT-A-GLANCE SUMMARY
  • Conjugated diene: alternating C=C – C=C; π-electrons delocalised; about 3 kcal/mol extra stability.
  • Diels–Alder: [4+2] cycloaddition; concerted; stereospecific and suprafacial.
  • Diene in s-cis conformation + dienophile with EWG → cyclohexene ring.
  • Endo rule applies; retro-Diels–Alder occurs at high temperature.
UNIT III
Alkyl Halides & Alcohols (10 h)
7
Explain SN1 and SN2 reactions with their mechanisms and kinetics. Compare the two, discussing the factors that affect them.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINENucleophilic substitution of alkyl halides proceeds by two contrasting mechanisms — the one-step bimolecular SN2 with inversion of configuration, and the two-step unimolecular SN1 with a carbocation intermediate and possible rearrangement.
SN2 Reaction (Bimolecular):
Mechanism: SN2 is a single concerted step. The nucleophile attacks the carbon from the back face, 180° away from the leaving group; a pentacoordinate transition state forms, in which the C – Nu bond develops while the C – LG bond breaks. The result is Walden inversion 🔊 of configuration.
Kinetics: Rate = k [R – X][Nu⁻] — second order overall.
Favoured by: primary (1° > 2° ≫ 3°) substrates (because steric hindrance blocks tertiary carbons); strong nucleophiles (OH⁻, CN⁻, I⁻); polar aprotic solvents such as DMSO, DMF and acetone (which solvate the cation but leave the nucleophile "naked" and reactive); and good leaving groups in the order I > Br > Cl ≫ F.
Stereochemistry: a chiral substrate gives the opposite-handed product (inversion).
SN1 Reaction (Unimolecular):
Mechanism: SN1 has two steps.
Step 1 (slow, rate-determining): the substrate ionises to a carbocation. R – X → R⁺ + X⁻ Step 2 (fast): the nucleophile attacks the flat carbocation from either face.
Kinetics: Rate = k [R – X] — first order overall.
Favoured by: tertiary substrates (3° > 2° ≫ 1°, because the carbocation is more stable); weak or neutral nucleophiles (water, alcohols); polar protic solvents (water, ethanol), which solvate and stabilise the carbocation; and good leaving groups.
Stereochemistry: because the planar carbocation is attacked from either face, the product is racemic (approximately 50 : 50). Carbocation rearrangement is possible, converting a less stable cation (1° → 2° or 2° → 3°) by 1,2-hydride or 1,2-methyl shift.
SN1 versus SN2 — Comparison:
FeatureSN1SN2
StepsTwo (step-wise)One (concerted)
Ratek [R – X]k [R – X][Nu]
OrderFirstSecond
Substrate3° > 2° ≫ 1°1° > 2° ≫ 3°
IntermediatePlanar carbocationNone (pentacoordinate TS)
NucleophileWeak or neutralStrong or anionic
SolventPolar proticPolar aprotic
StereochemistryRacemisationInversion (Walden)
RearrangementPossibleNot possible
Competes withE1E2
⚡ AT-A-GLANCE SUMMARY
  • SN1: two steps through a carbocation; rate = k[R – X]; 3°; weak nucleophile; polar protic solvent; racemisation.
  • SN2: one step with back-side attack; rate = k[R – X][Nu]; 1°; strong nucleophile; polar aprotic solvent; inversion.
  • Leaving-group order: I > Br > Cl ≫ F.
  • SN1 competes with E1; SN2 competes with E2.
8
Write structure and uses of ethyl chloride, chloroform, trichloroethylene, tetrachloroethylene, dichloromethane, tetrachloromethane (CCl₄) and iodoform 🔊.
★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEChlorinated hydrocarbons are among the most widely used industrial solvents and historical anaesthetics; each has a distinctive structure and range of applications that the pharmacist should know.
Structure and Uses:
CompoundStructurePrincipal uses
Ethyl chlorideCH₃ – CH₂ – ClTopical local anaesthetic spray (cooling effect on evaporation), refrigerant, ethylating agent
Chloroform 🔊CHCl₃Former general anaesthetic (now discontinued because of hepatotoxicity), solvent, preservative in paracetamol preparations, analytical extraction solvent
TrichloroethyleneCHCl=CCl₂Metal degreasing, dry cleaning, historical anaesthetic, industrial solvent
Tetrachloroethylene (perchloroethylene)CCl₂=CCl₂Dry cleaning, metal degreasing, veterinary anthelmintic
Dichloromethane (methylene chloride)CH₂Cl₂Paint stripper, pharmaceutical solvent, extraction of caffeine from coffee, aerosol propellant
Tetrachloromethane (CCl₄) 🔊CCl₄Former fire extinguisher, dry-cleaning solvent, anthelmintic — now banned because it is hepatotoxic and ozone-depleting
IodoformCHI₃Yellow crystalline topical antiseptic; dental dressings and wound packing; has a characteristic odour and is the positive result of the iodoform test for methyl ketones
Preparation Highlights:
Chloroform is prepared by the haloform reaction of acetone with bleaching powder (Cl₂/NaOH).
CH₃ – CO – CH₃ + 4 Cl₂ + NaOH → CHCl₃ + CH₃COONa + 3 HCl Chloroform is stored in dark amber bottles containing 1 % ethanol, which prevents its photo-oxidation to toxic phosgene 🔊.
Iodoform test: methyl ketones, acetaldehyde, ethanol and secondary alcohols with a CH₃ – CH(OH) – group react with I₂/NaOH to give a yellow precipitate of CHI₃.
CCl₄ is made industrially from carbon disulphide: CS₂ + 2 Cl₂ → CCl₄ + S₂Cl₂.
⚡ AT-A-GLANCE SUMMARY
  • Ethyl chloride: topical local anaesthetic spray.
  • Chloroform: historical anaesthetic; stored in dark bottles with 1 % ethanol.
  • Trichloroethylene and perchloroethylene: degreasing and dry cleaning.
  • Dichloromethane: solvent, paint stripper, decaffeination.
  • CCl₄: banned because it is hepatotoxic and depletes the ozone layer.
  • Iodoform (CHI₃): yellow antiseptic and basis of the diagnostic iodoform test.
9
Explain qualitative tests for alcohols. Give the structure and uses of ethyl alcohol, methyl alcohol, chlorobutanol, cetostearyl alcohol, benzyl alcohol, glycerol and propylene glycol.
★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEAlcohols (R – OH) are among the most widely used pharmaceutical solvents, preservatives, humectants and active ingredients; distinguishing primary, secondary and tertiary alcohols by chemical tests and remembering the monographs of the important ones is a routine task for the pharmacist.
Qualitative Tests for Alcohols:
(1) Sodium metal test: the alcohol reacts with sodium to give sodium alkoxide with brisk effervescence of hydrogen; this test is positive for any –OH.
(2) Ester test: the alcohol reacts with a carboxylic acid or acetyl chloride to give a fruity-smelling ester.
(3) Lucas test 🔊 (concentrated HCl with anhydrous ZnCl₂) distinguishes the three classes of alcohol by the rate of turbidity formation: tertiary gives immediate turbidity, secondary in 5 – 10 minutes, and primary gives no turbidity at room temperature.
(4) Victor Meyer's test gives a red colour with a primary, blue with a secondary and no colour with a tertiary alcohol.
(5) Iodoform test: alcohols containing a CH₃ – CH(OH) – group (such as ethanol and secondary alcohols bearing a methyl group) react with I₂ / NaOH to give yellow CHI₃.
(6) Oxidation (with K₂Cr₂O₇ or KMnO₄): primary alcohols are oxidised first to aldehydes and then to carboxylic acids, secondary alcohols to ketones, and tertiary alcohols are resistant under mild conditions.
Structure and Uses of Important Alcohols:
AlcoholStructurePrincipal uses
Ethyl alcohol (ethanol)CH₃ – CH₂ – OHSolvent, tinctures, 70 % hand sanitiser, topical antiseptic, beverages, biofuel
Methyl alcohol (methanol) 🔊CH₃ – OHIndustrial solvent and fuel; highly toxic — metabolised to formaldehyde and formic acid, producing blindness, metabolic acidosis and death; antidote is fomepizole or ethanol
Chlorobutanol 🔊1,1,1-trichloro-2-methyl-2-propanol (CCl₃ – C(CH₃)₂ – OH)0.5 % antimicrobial preservative in eye, nasal and parenteral preparations; mild sedative, local anaesthetic, antifungal
Cetostearyl alcohol 🔊Mixture of cetyl (C₁₆H₃₃OH) and stearyl (C₁₈H₃₇OH) alcoholsEmulsifier and stiffening agent in creams and ointments
Benzyl alcoholC₆H₅ – CH₂ – OHBacteriostatic preservative in parenterals (1 – 2 %), local anaesthetic, solvent
Glycerol (glycerin) 🔊HOCH₂ – CH(OH) – CH₂OH (1,2,3-propanetriol)Humectant, sweetener, solvent, osmotic laxative, vehicle for cough syrups, cryoprotectant
Propylene glycolCH₃ – CH(OH) – CH₂OH (1,2-propanediol)Solvent, humectant, cryoprotectant and IV co-solvent for drugs such as diazepam and phenytoin
⚡ AT-A-GLANCE SUMMARY
  • Tests: sodium (effervescence), ester (fruity), Lucas (3° > 2° > 1° by rate of turbidity), Victor Meyer (colour), iodoform (CH₃ – CH(OH) –).
  • Ethanol: antiseptic and solvent.
  • Methanol: toxic, causes blindness; antidote is fomepizole or ethanol.
  • Chlorobutanol: preservative in eye, nasal and IV products.
  • Cetostearyl alcohol: emulsifier and stiffener.
  • Benzyl alcohol: bacteriostatic preservative in parenterals.
  • Glycerol: humectant and osmotic laxative.
  • Propylene glycol: IV co-solvent.
UNIT IV
Carbonyl Compounds (10 h)
10
Explain the mechanism of nucleophilic addition to carbonyl compounds. Describe aldol condensation 🔊, crossed aldol, the Cannizzaro 🔊 reaction and crossed Cannizzaro.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEThe carbonyl group (C=O) is polar: its carbon is electrophilic and its oxygen carries lone pairs. This duality underpins the entire reactivity of aldehydes and ketones — nucleophilic addition, aldol condensation and the Cannizzaro reaction.
Nucleophilic Addition Mechanism:
Step 1: a nucleophile (HCN, H₂O, ROH, RNH₂, RMgX or NaHSO₃) attacks the electrophilic carbonyl carbon, the π-bond breaks and the electrons move on to the oxygen to give an alkoxide intermediate.
Step 2: the alkoxide is protonated by the solvent to give the final addition product.
Example: CH₃CHO + HCN gives CH₃ – CH(OH) – CN (a cyanohydrin).
Reactivity: HCHO > higher aldehydes > ketones, because alkyl groups donate electrons inductively and also cause steric hindrance.
Aldol Condensation:
Two molecules of aldehyde or ketone containing an α-hydrogen combine in the presence of dilute base (NaOH or K₂CO₃) to give a β-hydroxycarbonyl (an "aldol"); on heating the aldol loses water to give an α,β-unsaturated carbonyl.
Mechanism: (1) the base removes an α-hydrogen to form a carbanion or enolate; (2) the enolate attacks the carbonyl carbon of a second molecule; (3) protonation of the resulting alkoxide gives the aldol; (4) dehydration on heating yields the α,β-unsaturated carbonyl.
2 CH₃CHO → (dilute NaOH) → CH₃ – CH(OH) – CH₂ – CHO → (Δ) → CH₃ – CH=CH – CHO (crotonaldehyde)
Crossed (Claisen–Schmidt) Aldol:
A crossed aldol is performed between two different aldehydes or ketones; it is useful when one of them has no α-hydrogen (for example benzaldehyde or formaldehyde), because then only the other partner can give the enolate.
Example: benzaldehyde (no α-H) with acetophenone in the presence of dilute alkali gives the α,β-unsaturated ketone benzalacetophenone (a chalcone) together with water.
Cannizzaro Reaction:
Aldehydes that have no α-hydrogen (such as formaldehyde and benzaldehyde) undergo self-disproportionation in concentrated NaOH (about 50 %). One molecule is oxidised to the carboxylate salt and another is reduced to the corresponding alcohol.
2 HCHO + NaOH → CH₃OH + HCOONa
2 C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa
Mechanism: (1) hydroxide attacks the carbonyl carbon to form a tetrahedral hydrate anion; (2) the hydrate transfers hydride (H⁻) to a second molecule of aldehyde; (3) the result is an alcohol and a carboxylate salt.
Crossed Cannizzaro:
The crossed Cannizzaro is carried out between two different aldehydes, one of which is formaldehyde (an excellent hydride donor). Formaldehyde is always oxidised to formate, while the other aldehyde is reduced to its alcohol.
HCHO + C₆H₅CHO + NaOH → HCOONa + C₆H₅CH₂OH
⚡ AT-A-GLANCE SUMMARY
  • Nucleophilic addition: Nu attacks carbonyl C → alkoxide → H⁺ → product; HCHO > aldehydes > ketones.
  • Aldol: aldehyde/ketone with α-H + dilute base → β-hydroxy carbonyl → α,β-unsaturated on heating.
  • Crossed aldol: one partner has no α-H (for example PhCHO + acetophenone → chalcone).
  • Cannizzaro: aldehyde with no α-H + concentrated NaOH → disproportionation to alcohol + carboxylate.
  • Crossed Cannizzaro: HCHO + PhCHO → HCOONa + PhCH₂OH.
11
Explain the benzoin 🔊 and Perkin 🔊 condensations, with their mechanisms.
★★★★
5MShort Essay
Detailed Answer:
✍️ OPENING LINEThe benzoin and Perkin condensations are two important named reactions of aromatic aldehydes; the first is catalysed by cyanide ion, the second by the sodium salt of a carboxylic acid, and both give pharmaceutically useful intermediates.
Benzoin Condensation:
Two molecules of an aromatic aldehyde that lacks α-hydrogen undergo condensation in the presence of KCN or NaCN to give an α-hydroxy ketone known as benzoin.
2 C₆H₅CHO → (CN⁻ catalyst) → C₆H₅ – CO – CH(OH) – C₆H₅ (benzoin) Mechanism: (1) cyanide adds to one PhCHO to give PhCH(CN)O⁻; (2) proton transfer gives the resonance-stabilised carbanion PhC(CN)(OH)⁻; (3) this nucleophilic carbanion attacks a second molecule of PhCHO; (4) proton transfer and loss of CN⁻ give benzoin. CN⁻ generates an umpolung (acyl-anion equivalent).
Uses: benzoin is an intermediate in the synthesis of phenytoin and other drugs.
Perkin Condensation:
An aromatic aldehyde is heated with an aliphatic acid anhydride containing α-hydrogens and the sodium salt of the same acid as base; an α,β-unsaturated aromatic acid is produced.
C₆H₅CHO + (CH₃CO)₂O + CH₃COONa → C₆H₅ – CH=CH – COOH (cinnamic acid) + CH₃COOH Mechanism: (1) sodium acetate removes an α-hydrogen from the anhydride to give a carbanion; (2) the carbanion attacks the carbonyl of benzaldehyde; (3) dehydration and hydrolysis of the residual anhydride give the α,β-unsaturated acid.
Uses: synthesis of cinnamic acid (for perfumery), coumarins and indigo dye.
⚡ AT-A-GLANCE SUMMARY
  • Benzoin: two molecules of aromatic aldehyde (no α-H) + CN⁻ → α-hydroxy ketone.
  • Mechanism: cyanide generates an acyl-anion equivalent (umpolung).
  • Perkin: ArCHO + (CH₃CO)₂O + CH₃COONa → α,β-unsaturated acid (cinnamic acid).
  • Uses: Perkin gives cinnamic acid and coumarins; benzoin is the phenytoin intermediate.
12
Write the structure and uses of formaldehyde, paraldehyde, acetone, chloral hydrate, hexamine, benzaldehyde, vanillin and cinnamaldehyde.
★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEEight aldehydes and ketones of pharmaceutical importance cover the spectrum from simple formaldehyde to the aromatic flavourings vanillin and cinnamaldehyde, each with a distinct structure and a clinical or industrial use.
Structure and Uses:
CompoundStructurePrincipal uses
Formaldehyde 🔊HCHO (gas); 40 % aqueous solution is formalinDisinfectant, embalming fluid, preservative, tissue fixative, resin manufacture
Paraldehyde 🔊Cyclic trimer of acetaldehyde, (CH₃CHO)₃; a six-membered ringSedative, hypnotic and anticonvulsant (formerly used in status epilepticus); rapid onset but unpleasant odour
AcetoneCH₃ – CO – CH₃Solvent (nail-polish remover), extraction, paint thinner, starting material for bisphenol A
Chloral hydrate 🔊CCl₃ – CH(OH)₂Paediatric sedative-hypnotic; historically the "Mickey Finn"
Hexamine (methenamine; urotropine)(CH₂)₆N₄, formed from 6 HCHO + 4 NH₃Urinary antiseptic (releases HCHO slowly in acidic urine); industrial precursor of the explosive RDX
BenzaldehydeC₆H₅ – CHOAlmond flavour, perfumery, dye manufacture, synthesis of cinnamic acid and triphenylmethane dyes
Vanillin 🔊4-hydroxy-3-methoxybenzaldehyde (HOC₆H₃(OCH₃) – CHO)Vanilla flavouring in food and pharmaceutical preparations, perfumery, intermediate in L-DOPA synthesis
Cinnamaldehyde 🔊C₆H₅ – CH=CH – CHOCinnamon flavour, fragrance, antimicrobial agent, cough syrups
⚡ AT-A-GLANCE SUMMARY
  • Formaldehyde: disinfectant and formalin.
  • Paraldehyde: sedative (formerly used in status epilepticus).
  • Acetone: solvent.
  • Chloral hydrate: paediatric sedative; CCl₃CH(OH)₂.
  • Hexamine: urinary antiseptic that releases HCHO in acidic urine.
  • Benzaldehyde: almond flavour.
  • Vanillin: vanilla flavour.
  • Cinnamaldehyde: cinnamon flavour and mild antimicrobial.
UNIT V
Carboxylic Acids & Aliphatic Amines (8 h)
13
Discuss the acidity of carboxylic acids. Explain the effect of substituents. Give structure and uses of key carboxylic acids.
★★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINECarboxylic acids (R – COOH) are the most acidic of the common organic functional groups, a property that is directly tuned by the electronic nature of the substituents attached to the carbon skeleton.
Acidity of Carboxylic Acids:
A carboxylic acid ionises as R – COOH ⇌ R – COO⁻ + H⁺, with a pKa of approximately 3 – 5, which makes it far more acidic than an alcohol (pKa about 16). Three factors explain this.
(1) The resonance stabilisation of the carboxylate anion, in which the negative charge is delocalised equally between the two oxygen atoms.
(2) The inductive effect of the two electronegative oxygen atoms, which withdraw electron density from the O – H bond.
(3) The carboxylate ion is much more stable than an alkoxide, because an alcohol has only a single oxygen to carry the negative charge.
Effect of Substituents on Acidity:
Electron-withdrawing groups (EWGs such as –NO₂, –CN, –Cl and –F) stabilise the carboxylate by inductive withdrawal and therefore increase acidity, while electron-donating groups (such as –CH₃ and –OR) destabilise it and decrease acidity.
Typical pKa values: HCOOH (3.75) is more acidic than CH₃COOH (4.76) or C₂H₅COOH (4.87) because the alkyl group donates electrons inductively. ClCH₂COOH (2.86) is more acidic than CH₃COOH because chlorine is electron-withdrawing. The series CCl₃COOH (0.7) > CHCl₂COOH (1.3) > ClCH₂COOH (2.86) > CH₃COOH (4.76) shows that successive halogens progressively raise acidity. The position of the substituent also matters: α-chlorobutanoic acid (2.86) is more acidic than β-chloro (4.05) or γ-chloro (4.50), because the inductive effect weakens with distance.
In aromatic acids, p-nitrobenzoic acid (3.44) is stronger than benzoic (4.20), which is stronger than p-methoxybenzoic (4.47).
Structure and Uses of Important Carboxylic Acids:
AcidStructurePrincipal uses
Acetic acidCH₃ – COOHVinegar, solvent, food additive
Lactic acid 🔊CH₃ – CH(OH) – COOHCosmetic exfoliant, Ringer's lactate, food preservative
Tartaric acid 🔊HOOC – CH(OH) – CH(OH) – COOHBaking powder (cream of tartar), effervescent tablets, resolving agent in stereochemistry
Citric acidHOOC – CH₂ – C(OH)(COOH) – CH₂ – COOHEffervescent tablets, preservative, chelating agent, ORS
Succinic acidHOOC – CH₂ – CH₂ – COOHTCA-cycle intermediate, polymer precursor, pharmaceutical intermediate
Oxalic acidHOOC – COOHPrimary standard in titrimetry, rust remover, bleach
Salicylic acid 🔊o-HO – C₆H₄ – COOHKeratolytic (acne and warts), antiseptic, precursor of aspirin
Benzoic acidC₆H₅ – COOHPreservative (as sodium benzoate), antifungal, aspirin intermediate
Benzyl benzoateC₆H₅ – COO – CH₂ – C₆H₅Scabicide and pediculicide (5 – 25 %)
Dimethyl phthalateC₆H₄(COOCH₃)₂Insect repellent
Methyl salicylate 🔊2-HO – C₆H₄ – COOCH₃ (oil of wintergreen)Rubefacient liniment, flavouring agent
Aspirin (acetylsalicylic acid)2-CH₃COO – C₆H₄ – COOHAnalgesic, antipyretic, anti-inflammatory and antiplatelet agent
⚡ AT-A-GLANCE SUMMARY
  • Acidity: resonance and inductive stabilisation of the R – COO⁻ anion.
  • EWGs increase and EDGs decrease acidity.
  • More chlorines (or closer to the COOH) → more acidic: CCl₃COOH ≫ CH₃COOH.
  • Key uses: acetic (vinegar), lactic (cosmetic), tartaric (baking), citric (ORS), salicylic (keratolytic), benzoic (preservative), aspirin (NSAID / antiplatelet), methyl salicylate (liniment).
14
Write the qualitative tests for carboxylic acids, esters and amides.
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5MShort Note
Detailed Answer:
✍️ OPENING LINECarboxylic acids, esters and amides all share the carbonyl (C=O) group but differ sharply in chemical behaviour, so that a handful of simple qualitative tests can distinguish them quickly.
Tests for Carboxylic Acids:
A carboxylic acid turns blue litmus red. It liberates CO₂ with effervescence from sodium bicarbonate (this distinguishes it from phenol):
R – COOH + NaHCO₃ → R – COONa + H₂O + CO₂↑ It gives a fruity ester on warming with an alcohol and concentrated H₂SO₄ (the ester test). A neutral solution of FeCl₃ with the sodium salt of the acid gives a red-brown or buff precipitate of ferric carboxylate. It is reduced by LiAlH₄ to the corresponding primary alcohol.
Tests for Esters:
Esters usually have a fruity smell (for example amyl acetate smells of banana and ethyl butyrate of pineapple). The hydroxamic acid test gives a purple or magenta colour when the ester is treated with hydroxylamine followed by FeCl₃. On warming with NaOH, the ester undergoes saponification to a carboxylate and an alcohol. Esters are insoluble in water but soluble in organic solvents.
Tests for Amides:
Heating an amide with NaOH liberates ammonia, which has a pungent smell and turns moist red litmus blue:
R – CONH₂ + NaOH → R – COONa + NH₃↑ The biuret test (CuSO₄ in NaOH) produces a characteristic violet colour with amides (as in urea and proteins). The Hofmann degradation 🔊 converts a primary amide into a primary amine with one fewer carbon (R – CONH₂ + Br₂/NaOH → R – NH₂). Amides are hydrolysed by acid or alkali to give the parent acid together with ammonia (or an amine).
⚡ AT-A-GLANCE SUMMARY
  • Carboxylic acids: blue → red litmus; effervescence with NaHCO₃; fruity ester on heating with alcohol and H₂SO₄.
  • Esters: fruity odour; hydroxamic acid + FeCl₃ gives purple; saponification with NaOH.
  • Amides: NH₃ on heating with NaOH; biuret gives violet; Hofmann degradation gives amine.
15
Discuss the basicity of aliphatic amines. Explain the effect of substituents. Give the structure and uses of ethanolamine, ethylenediamine and amphetamine.
★★★★
10MLong Essay
Detailed Answer:
✍️ OPENING LINEAmines (R – NH₂) are Lewis bases because the lone pair on nitrogen can accept a proton; how readily this happens depends on the electronic and steric effects of the substituents attached to the nitrogen.
Basicity of Amines:
An amine is protonated according to R – NH₂ + H⁺ ⇌ R – NH₃⁺. The pKa of the conjugate acid (R – NH₃⁺) is about 10 – 11 for aliphatic amines and 4 – 5 for aromatic amines; basicity is often expressed as pKb, with a low pKb corresponding to a strong base.
In the gas phase, the order of basicity is 3° > 2° > 1° > NH₃, because each additional alkyl group donates electrons inductively to the nitrogen.
In aqueous solution, however, the order becomes 2° > 1° > 3° > NH₃. The protonated ammonium ion R – NH₃⁺ is stabilised by hydrogen bonding with water; the more hydrogens on nitrogen, the better the solvation, so a tertiary amine (with only one N – H) is poorly solvated and appears less basic than expected.
In general, aliphatic amines are more basic than aromatic amines, because in an aromatic amine the lone pair on nitrogen is delocalised into the π-system of the ring.
Typical pKa values of R – NH₃⁺: ethylamine 10.8, dimethylammonium 10.7, trimethylammonium 9.8 and anilinium 4.6.
Effect of Substituents on Basicity:
Alkyl groups (+I, electron-donating) push electrons toward nitrogen and therefore increase basicity, though a tertiary amine loses some of this advantage because of reduced solvation and increased steric crowding.
Aryl groups (–M, electron-withdrawing by resonance) delocalise the lone pair into the ring and markedly decrease basicity.
Further EWGs on the ring (NO₂, CN, Cl) decrease basicity still more (p-nitroaniline is a very weak base).
Steric hindrance around the nitrogen slows the approach of a proton and reduces apparent basicity.
Structure and Uses of Key Aliphatic Amines:
AmineStructurePrincipal uses
Ethanolamine 🔊HO – CH₂ – CH₂ – NH₂Intermediate for antihistamines; component of phospholipids; surfactant; sclerosing agent for varicose veins
Ethylenediamine 🔊H₂N – CH₂ – CH₂ – NH₂Chelating agent (precursor of EDTA); stabiliser in aminophylline (theophylline + ethylenediamine); corrosion inhibitor
Amphetamine 🔊C₆H₅ – CH₂ – CH(CH₃) – NH₂ (α-methylphenethylamine)CNS stimulant used in ADHD and narcolepsy; a Schedule II controlled drug with high abuse potential
⚡ AT-A-GLANCE SUMMARY
  • Amines: Lewis bases via the nitrogen lone pair; pKa of R – NH₃⁺ is about 10 – 11.
  • Gas phase: 3° > 2° > 1° > NH₃.
  • Aqueous solution: 2° > 1° > 3° > NH₃ (solvation anomaly).
  • Aromatic amines are weaker (lone pair delocalised into the ring).
  • Ethanolamine: phospholipid component and antihistamine precursor.
  • Ethylenediamine: EDTA precursor and aminophylline component.
  • Amphetamine: CNS stimulant (Schedule II controlled drug).
SYLLABUS COMPLETION
Less Important — But Must Read for Full Syllabus Coverage
16
Write a short note on the electromeric effect and allylic rearrangement.
★★★
5MShort Note
Detailed Answer:
✍️ OPENING LINETwo key concepts explaining the reactivity of unsaturated organic compounds are the electromeric effect — a temporary electronic redistribution triggered by an approaching reagent — and the allylic rearrangement, in which a double bond migrates during the reaction.
Electromeric Effect:
The electromeric effect is the temporary, complete shift of a shared pair of π-electrons from one atom of a multiple bond to another when a reagent approaches. It is reversible and disappears when the reagent is removed.
There are two sub-types. In the +E effect, the electrons move toward the attacking electrophile; for example, the π-electrons of an alkene move toward an approaching H⁺. In the –E effect, the electrons move away from the attacking nucleophile; for example, when CN⁻ approaches the carbonyl of an aldehyde, the π-electrons move on to the oxygen atom.
This effect explains why the carbonyl carbon is electrophilic and why alkenes attack electrophiles.
Allylic Rearrangement (SN1' / SN2'):
An allylic rearrangement is a substitution or addition at an allylic position that is accompanied by migration of the double bond.
Example: CH₂=CH – CH₂ – Cl on hydrolysis gives both CH₂=CH – CH₂ – OH (normal product) and CH₃ – CH=CH – OH (rearranged product).
Mechanism: the allylic cation CH₂=CH – CH₂⁺ is resonance-stabilised as ⁺CH₂ – CH=CH₂, so the nucleophile can attack at either end of the allyl system, giving two products.
The rearrangement explains the unusual product distributions observed in allylic nucleophilic substitution and addition reactions.
⚡ AT-A-GLANCE SUMMARY
  • Electromeric effect: temporary reagent-induced π-electron shift; +E (toward) or –E (away); reversible.
  • Allylic rearrangement: the double bond migrates during SN through a resonance-stabilised allylic cation.
  • SN1' / SN2': nucleophile attacks the γ-carbon instead of the α-carbon.
17
Explain rearrangement of carbocations with examples.
★★★
5MShort Note
Detailed Answer:
✍️ OPENING LINECarbocations are high-energy intermediates that spontaneously rearrange from less stable (primary) to more stable (secondary or tertiary) forms by the migration of a hydride or methyl group; this phenomenon accounts for many unexpected products in SN1 and E1 reactions.
Carbocation Stability Order:
Carbocations follow the stability order 3° > 2° > 1° > methyl, because additional alkyl groups donate electrons inductively and through hyperconjugation to the electron-deficient carbon. Resonance-stabilised cations such as allyl and benzyl can be as stable as tertiary cations.
Types of Rearrangement:
(1) 1,2-Hydride shift: a neighbouring hydrogen migrates with its bonding pair from an adjacent carbon to the carbocation centre.
(CH₃)₂CH – CH₂⁺ (1°) → (CH₃)₂C⁺ – CH₃ (3°, more stable) (2) 1,2-Methyl (alkyl) shift: a methyl group migrates with its bonding pair when no suitable hydrogen is available.
(CH₃)₃C – CH₂⁺ (neopentyl, 1°) → (CH₃)₂C⁺ – CH(CH₃) (3°)
Examples in Reactions:
In SN1, hydrolysis of 2-bromo-3-methylbutane gives mainly 2-methylbutan-2-ol (the tertiary alcohol formed after a hydride shift) rather than the simple secondary alcohol. In E1, heating neopentyl bromide with base yields 2-methylbut-2-ene after a methyl shift. In electrophilic addition, HCl adds to 3,3-dimethyl-1-butene to give 2-chloro-2,3-dimethylbutane following Markownikoff addition and a methyl shift.
⚡ AT-A-GLANCE SUMMARY
  • Stability: 3° > 2° > 1° > methyl; allyl and benzyl are also highly stable.
  • 1,2-Hydride shift is the most common rearrangement, migrating toward the more stable carbon.
  • 1,2-Methyl shift occurs when no suitable hydrogen is available.
  • Seen in SN1, E1 and electrophilic addition whenever a more stable carbocation is accessible.

📚 BP202T ORGANIC CHEMISTRY I EXAM STRATEGY

  • Copy Opening Line verbatim — it lands the examiner in the right context.
  • Draw arrow-pushing mechanisms — worth 3-4 marks in any mechanism question.
  • Tabulate comparisons (SN1 vs SN2, E1 vs E2, Aldol vs Cannizzaro) — clean, visual, easy to mark.
  • Remember pKa values: HCOOH 3.75, CH₃COOH 4.76, CCl₃COOH 0.7, PhOH 10, ROH 16.
  • Memorise structures of all pharmaceutically important compounds in lists — formaldehyde, chloroform, iodoform, glycerol, propylene glycol, aspirin, salicylic acid, methyl salicylate.
  • IUPAC naming: always identify the longest chain, number with lowest locants, and use correct functional-group suffix priority.
  • Inductive / Resonance effect — use them to explain acidity and basicity changes.