Organic Chemistry-3 BCH4T08

Syllabus topics (official RTMNU)

(A) Organic compounds of Nitrogen

• Preparation of nitroalkanes and nitrobenzene, chemical reactions of nitroalkanes.
• Mechanism of nucleophilic substitution in nitrobenzene and their reduction in acidic, neutral and alkaline media.
• Picric acid- preparation and uses,

(B) Amines

• Structure and nomenclature of amines, Physical properties, stereochemistry of amines, separation of mixture of 1°, 2° and 3° amines, structural features affecting basicity of amines, preparation of alkyl &aryl amines (reduction of nitro compounds and nitriles), reductive amination of aldehydic and ketonic compounds, Gabriel phthalimide reaction, Hofmann bromamide reaction, Reactions of amines, Preparation and synthetic transformations of aryl diazonium salts.

(A) Nitroalkanes and nitrobenzene: preparation

Nitro compounds contain the –NO₂ group bonded through nitrogen to carbon. The –NO₂ group is strongly electron-withdrawing (–I and –R effects). Isomeric alkyl nitrites (R–O–N=O) are bonded through oxygen.

Nitroalkanes

• Vapour-phase nitration: CH₃CH₃ + HNO₃ → CH₃CH₂NO₂ + H₂O (about 400 °C; free-radical, C–C cleavage also gives CH₃NO₂).
• From alkyl halides: R–X + AgNO₂ (alcoholic) → R–NO₂ (major) + AgX. (With NaNO₂/KNO₂ in ethanol, more alkyl nitrite R–ONO forms.)
• From α-halo acids: ClCH₂COONa + NaNO₂ → O₂NCH₂COONa → (heat) CH₃NO₂ + CO₂ (decarboxylation).

Nitrobenzene

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (conc. H₂SO₄, below 60 °C)

1. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ (nitronium ion is the electrophile).
2. NO₂⁺ attacks the benzene ring forming a resonance-stabilised arenium ion (σ-complex).
3. HSO₄⁻ removes H⁺, aromaticity is restored, nitrobenzene forms.

Chemical reactions of nitroalkanes

• Reduction: R–NO₂ + 6[H] → R–NH₂ + 2H₂O (Sn/HCl, Fe/HCl or H₂/Ni). With Zn/NH₄Cl → N-alkylhydroxylamine RNHOH.
• Acidic nature (tautomerism): 1° and 2° nitroalkanes have α-H, show nitro–aci tautomerism and dissolve in NaOH: CH₃NO₂ ⇌ CH₂=N(O)OH (aci form). 3° nitroalkanes have no α-H and are insoluble in alkali.
• Hydrolysis: 1° nitroalkanes with boiling HCl give carboxylic acid + hydroxylamine: RCH₂NO₂ + H₂O → RCOOH + NH₂OH.
• With HNO₂ (distinguishes 1°, 2°, 3°): 1° → nitrolic acid (red colour with NaOH); 2° → pseudonitrole (blue); 3° → no reaction.
• Halogenation: α-H replaced in alkali, e.g. CH₃NO₂ + 3Cl₂ → CCl₃NO₂ (chloropicrin) + 3HCl.
• Henry (nitro-aldol) reaction: CH₃NO₂ + HCHO → HOCH₂CH₂NO₂ (base).

Nucleophilic substitution in nitrobenzene and its reduction in different media

Nucleophilic aromatic substitution

The –NO₂ group withdraws electrons from the ring, especially from ortho and para positions, so the ring becomes open to attack by strong nucleophiles. Nitrobenzene fused with KOH gives o-nitrophenol (and some p-). Halogens ortho/para to –NO₂ are easily replaced: p-nitrochlorobenzene + NaOH (aq) → p-nitrophenol.

1. Addition: OH⁻ attacks the carbon at the ortho/para position, giving a resonance-stabilised anionic σ-complex (Meisenheimer complex); the negative charge is delocalised onto the oxygen atoms of –NO₂.
2. Elimination: the leaving group (Cl⁻, or H⁻ removed by oxidation) leaves and aromaticity is restored.

Reduction of nitrobenzene

Pathway: C₆H₅NO₂ → C₆H₅NO (nitrosobenzene) → C₆H₅NHOH → C₆H₅NH₂. In alkali, nitrosobenzene and phenylhydroxylamine condense to azoxybenzene, which is reduced further.

Picric acid: preparation and uses

Picric acid is 2,4,6-trinitrophenol. Phenol is first sulphonated (conc. H₂SO₄) to phenol-2,4-disulphonic acid, then treated with conc. HNO₃; the –SO₃H groups are replaced by –NO₂ groups. (Direct nitration of phenol with conc. HNO₃ oxidises the ring, so sulphonation is done first.)

Uses: explosive (as ammonium picrate), yellow dye, antiseptic for burns, reagent for making crystalline picrates of amines and aromatic hydrocarbons for identification. It is strongly acidic (pKa ≈ 0.4) because three –NO₂ groups stabilise the phenoxide ion.

(B) Amines: structure, nomenclature, properties, stereochemistry

• Amines are derivatives of NH₃: 1° RNH₂, 2° R₂NH, 3° R₃N; quaternary salts R₄N⁺X⁻.
• Structure: N is sp³, pyramidal, with a lone pair; C–N–C angle about 108°. In aniline the lone pair is partly delocalised into the ring.
• Nomenclature: CH₃NH₂ methanamine (methylamine); (CH₃)₂NH N-methylmethanamine; C₆H₅NH₂ aniline (benzenamine).
• Physical properties: lower amines are gases or volatile liquids with fishy smell; 1° and 2° amines form H-bonds, so they boil higher than alkanes but lower than alcohols; 3° amines cannot H-bond with each other. Lower amines are water-soluble.
• Stereochemistry: an amine with three different groups is chiral, but its enantiomers cannot be separated because the N inverts rapidly through a planar transition state (pyramidal inversion, barrier ≈ 25 kJ/mol). Quaternary ammonium salts with four different groups can be resolved.

Separation of 1°, 2° and 3° amines (Hinsberg method)

1. Treat the mixture with benzenesulphonyl chloride (C₆H₅SO₂Cl) and aq. KOH.
2. 1° amine → sulphonamide with acidic N–H, soluble in KOH. 2° amine → sulphonamide with no N–H, insoluble in KOH. 3° amine → no reaction.
3. Distil off the 3° amine; filter the insoluble 2° derivative; acidify the filtrate to get the 1° derivative. Hydrolyse each with acid to recover the amines.

(Hofmann's method uses diethyl oxalate: 1° → solid dialkyloxamide, 2° → liquid oxamic ester, 3° → no reaction.)

Basicity of amines

• +I effect of alkyl groups increases electron density on N → aliphatic amines are stronger bases than NH₃.
• In water, order for methylamines: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ — due to the balance of +I effect, solvation by H-bonding of the cation, and steric hindrance.
• Aniline is much weaker (pKb ≈ 9.4) because the lone pair is delocalised into the ring. Electron-donating groups (–CH₃, –OCH₃ at para) increase basicity; electron-withdrawing groups (–NO₂) decrease it.

Preparation of alkyl and aryl amines

• Reduction of nitro compounds: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O (Sn/HCl).
• Reduction of nitriles: R–C≡N + 4[H] → R–CH₂NH₂ (LiAlH₄ or H₂/Ni).
• Reductive amination: aldehyde or ketone + NH₃ → imine → (H₂/Ni or NaBH₃CN) amine. CH₃CHO + NH₃ + H₂ → CH₃CH₂NH₂ + H₂O.
• Gabriel phthalimide synthesis (pure 1° aliphatic amines): phthalimide + KOH → potassium phthalimide; + R–X → N-alkylphthalimide; hydrolysis (aq. NaOH, or hydrazine) → R–NH₂ + phthalate. Aryl amines cannot be made this way (aryl halides do not undergo SN2).
• Hofmann bromamide reaction: R–CONH₂ + Br₂ + 4NaOH → R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The amine has one carbon less. Steps: N-bromoamide → anion → loss of Br⁻ and migration of R to N gives isocyanate R–N=C=O → hydrolysis to amine + CO₂.

Reactions of amines and aryl diazonium salts

• Acylation: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ (acetanilide).
• Carbylamine test (1° only): RNH₂ + CHCl₃ + 3KOH → RNC (foul-smelling isocyanide) + 3KCl + 3H₂O.
• With HNO₂: 1° aliphatic → alcohol + N₂ gas; 2° → yellow oily N-nitrosamine; 1° aromatic at 0–5 °C → diazonium salt.
• Bromination of aniline: with bromine water → 2,4,6-tribromoaniline (white ppt).

Aryl diazonium salts

C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (0–5 °C, diazotisation)

• Sandmeyer: CuCl/HCl → C₆H₅Cl; CuBr/HBr → C₆H₅Br; CuCN/KCN → C₆H₅CN. (Gattermann: Cu powder + HX.)
• KI → C₆H₅I; HBF₄ then heat → C₆H₅F (Balz–Schiemann).
• Warm water → phenol; H₃PO₂ or ethanol → benzene.
• Coupling with phenol (alkaline) → p-hydroxyazobenzene (orange dye); with aniline (mildly acidic) → p-aminoazobenzene (yellow).

Likely exam questions

Long answer

1. Discuss the reduction of nitrobenzene in acidic, neutral and alkaline media. Explain the mechanism of nucleophilic substitution in nitrobenzene.
2. Explain the factors affecting the basicity of amines. How will you separate a mixture of 1°, 2° and 3° amines?

Short answer

1. Write the Hofmann bromamide reaction with mechanism.
2. How is picric acid prepared? Give its uses.
3. Give any three synthetic applications of benzene diazonium chloride.

Syllabus topics (official RTMNU)

Heterocyclic Compounds

• Molecular orbital picture and aromaticity of furan, thiophene, pyrrole and pyridine.
• Methods of synthesis of pyridine.
• Mechanism of electrophilic and nucleophilic substitution reaction of pyridine.
• Chemical reaction of pyridine.
• Structure of pyridine.
• Comparison of basicity of pyrrole and pyridine.
• Introduction to condensed five and six membered heterocycles.
• Preparation and reactions of Indole, Quinoline and Isoquinoline with special reference to Fischer Indole synthesis, Skraup synthesis and Bischler Napieralski synthesis.

Molecular orbital picture and aromaticity of furan, thiophene and pyrrole

Heterocyclic compounds have a ring containing at least one atom other than carbon (N, O, S). Furan (O), thiophene (S) and pyrrole (N–H) are five-membered; pyridine is six-membered.

• All four ring carbons are sp² and each gives one electron in a p orbital perpendicular to the ring.
• The heteroatom is also sp² and gives its lone pair (2 electrons) in a p orbital to the ring.
• Total π electrons = 4 + 2 = 6 = (4n + 2) with n = 1. The ring is planar, cyclic and fully conjugated, so it is aromatic (Hückel's rule).
• Five p orbitals combine to give five π molecular orbitals: three bonding (filled with the 6 electrons) and two antibonding (empty).
• In furan the O has a second lone pair in an sp² orbital in the ring plane; it does not take part in aromaticity.
• Order of aromaticity: benzene > thiophene > pyrrole > furan. O is most electronegative and holds its lone pair most tightly, so furan is least aromatic (and most diene-like; it gives Diels–Alder reactions).
• Because the heteroatom donates electrons into the ring, these rings are electron-rich ("π-excessive"); they undergo electrophilic substitution easily, mainly at C-2 (α position). Reactivity: pyrrole > furan > thiophene > benzene.

Structure, MO picture and aromaticity of pyridine

• Pyridine (C₅H₅N) is a planar hexagon like benzene with one CH replaced by N. All five C and the N are sp².
• Each ring atom gives one electron in a p orbital → 6 π electrons → aromatic; six π MOs (three bonding filled).
• The N lone pair is in an sp² orbital in the ring plane, not part of the π system. So it is available for protonation → pyridine is basic.
• N is more electronegative than C, so it pulls π-electron density away from the ring ("π-deficient"). C–N bond (≈134 pm) is shorter than C–C (≈139 pm). Dipole moment ≈ 2.2 D, directed towards N.

Methods of synthesis of pyridine

1. From coal tar: the light oil fraction is treated with dilute H₂SO₄; pyridine dissolves as its salt, then is freed with base and distilled.
2. From acetylene and HCN: 2HC≡CH + HCN → C₅H₅N (passed through a red-hot tube).
3. From piperidine: dehydrogenation with conc. H₂SO₄ at 300 °C (or Pd/Pt) gives pyridine.
4. Hantzsch synthesis: 2 molecules of a β-keto ester (ethyl acetoacetate) + an aldehyde + NH₃ → 1,4-dihydropyridine → oxidation (HNO₃) → substituted pyridine.
5. From pentamethylenediamine hydrochloride on heating → piperidine, then dehydrogenation → pyridine.

Electrophilic and nucleophilic substitution in pyridine (mechanism)

Electrophilic substitution — at C-3, and difficult

Pyridine reacts like nitrobenzene: slowly and only under harsh conditions, because (i) N withdraws electrons, and (ii) in acidic reagents N is protonated to pyridinium, which is even more deactivated.

1. Electrophile E⁺ attacks the ring, forming a σ-complex.
2. Attack at C-2 or C-4 gives a resonance form with the positive charge on the electronegative N (with only six electrons) — very unstable.
3. Attack at C-3 keeps the positive charge on carbons only → most stable intermediate.
4. Loss of H⁺ gives the 3-substituted product.

Examples: nitration (KNO₃/H₂SO₄, 300 °C) → 3-nitropyridine (low yield); sulphonation (H₂SO₄/HgSO₄, 230 °C) → pyridine-3-sulphonic acid; bromination (300 °C) → 3-bromopyridine. Friedel–Crafts reactions fail (AlCl₃ binds to N).

Nucleophilic substitution — at C-2 and C-4

1. Nucleophile Nu⁻ adds to C-2 (or C-4), giving an anionic intermediate.
2. For attack at C-2/C-4 the negative charge can rest on electronegative N → stable intermediate.
3. Loss of hydride (H⁻) or a leaving group restores aromaticity.

Chichibabin reaction: pyridine + NaNH₂ (heat, toluene) → 2-aminopyridine + H₂. With C₆H₅Li → 2-phenylpyridine; with KOH at 300 °C → 2-hydroxypyridine (2-pyridone).

Other chemical reactions; basicity of pyrrole vs pyridine

• As a base: C₅H₅N + HCl → C₅H₅NH⁺Cl⁻ (pyridinium chloride).
• N-alkylation: with CH₃I → N-methylpyridinium iodide.
• N-oxidation: with peracids (CH₃CO₃H) → pyridine N-oxide, which undergoes electrophilic substitution at C-4.
• Reduction: Na/C₂H₅OH or H₂/Pt → piperidine (C₅H₁₁N), a stronger base.
• Pyridine is resistant to oxidation; side chains are oxidised (3-methylpyridine → nicotinic acid with KMnO₄).

Basicity: pyridine vs pyrrole

So pyridine is much more basic than pyrrole. Pyrrole N–H is weakly acidic (pKa ≈ 17) and reacts with K metal to give potassium pyrrolide.

Condensed five- and six-membered heterocycles: indole, quinoline, isoquinoline

Fusing a benzene ring with a heterocycle gives condensed (fused) heterocycles: indole (benzene + pyrrole, 2,3-benzopyrrole), quinoline (benzene + pyridine, N at position 1) and isoquinoline (N at position 2). They occur in alkaloids (quinine, papaverine), the amino acid tryptophan and the hormone serotonin.

Indole — Fischer indole synthesis

1. Phenylhydrazine + an aldehyde or ketone (with α-CH₂) → phenylhydrazone.
2. Heat with acid catalyst (ZnCl₂, polyphosphoric acid or H₂SO₄): tautomerism to enamine (ene-hydrazine).
3. [3,3]-sigmatropic rearrangement forms a new C–C bond.
4. Cyclisation and loss of NH₃ give the indole.

Acetone phenylhydrazone → 2-methylindole. (Indole itself: from pyruvic acid phenylhydrazone → indole-2-carboxylic acid → decarboxylation.)

Reactions of indole: electrophilic substitution at C-3 (unlike pyrrole, at C-2), e.g. Vilsmeier formylation → indole-3-carbaldehyde; Mannich reaction → gramine; weakly acidic N–H.

Quinoline — Skraup synthesis

1. Glycerol + conc. H₂SO₄ → acrolein (CH₂=CH–CHO) by dehydration.
2. Aniline adds to acrolein by Michael (1,4-) addition → β-anilinopropionaldehyde.
3. Acid-catalysed cyclisation and dehydration → 1,2-dihydroquinoline.
4. Oxidation by nitrobenzene (FeSO₄ moderates the violent reaction) → quinoline.

Reactions of quinoline: electrophilic substitution in the benzene ring at C-5 and C-8 (nitration gives 5- and 8-nitroquinoline); nucleophilic substitution in the pyridine ring at C-2 (and C-4), e.g. NaNH₂ → 2-aminoquinoline; oxidation with KMnO₄ → pyridine-2,3-dicarboxylic acid (quinolinic acid); basic (pKa ≈ 4.9).

Isoquinoline — Bischler–Napieralski synthesis

1. β-Phenylethylamine + acid chloride (RCOCl) → N-acyl-β-phenylethylamine (amide).
2. Heat with dehydrating agent (POCl₃ or P₂O₅): intramolecular cyclisation → 3,4-dihydroisoquinoline (1-substituted).
3. Dehydrogenation with Pd/C → isoquinoline.

Reactions of isoquinoline: electrophilic substitution at C-5 (and C-8); nucleophilic substitution at C-1 (NaNH₂ → 1-aminoisoquinoline); oxidation with KMnO₄ gives phthalic acid and pyridine-3,4-dicarboxylic acid; more basic than quinoline (pKa ≈ 5.4).

Likely exam questions

Long answer

1. Discuss the molecular orbital picture and aromatic character of pyrrole and pyridine. Compare their basicity.
2. Explain the mechanism of electrophilic and nucleophilic substitution in pyridine. Why does electrophilic substitution occur at C-3?

Short answer

1. Write the Skraup synthesis of quinoline.
2. Explain the Fischer indole synthesis with one example.
3. Write the Bischler–Napieralski synthesis of isoquinoline.

Syllabus topics (official RTMNU)

(A) Quantitative Analysis

• Estimation of carbon, hydrogen, nitrogen, sulphur and halogens (only principles and calculations).
• Calculation of Empirical and molecular formula with Numericals

(B) Organometallic compounds

• Organomagnesium compound : Grignard reagent formation, chemical reactions and structure.
• Organozinc compounds : Formation and chemical reactions.
• Organolithium compounds: Formation and chemical reactions.

(A) Quantitative analysis: carbon and hydrogen (Liebig's method)

Principle: a known mass of the organic compound is heated strongly with dry copper(II) oxide in a stream of dry oxygen. All carbon is oxidised to CO₂ and all hydrogen to H₂O.

C + 2CuO → CO₂ + 2Cu 2H + CuO → H₂O + Cu

Water is absorbed in a weighed U-tube of anhydrous CaCl₂ (or Mg(ClO₄)₂); CO₂ is absorbed in weighed potash bulbs (KOH solution). The increase in mass of each gives mass of H₂O and CO₂.

Example: 0.20 g of a compound gave 0.44 g CO₂ and 0.18 g H₂O. %C = (12/44)(0.44/0.20)(100) = 0.12/0.20 × 100 = 60%. %H = (2/18)(0.18/0.20)(100) = 0.02/0.20 × 100 = 10%.

Estimation of nitrogen

Dumas method

Principle: the compound is heated with CuO in an atmosphere of CO₂. Nitrogen is set free as N₂ gas (any oxides of N are reduced by hot copper gauze). The gas mixture is passed into 50% KOH, which absorbs CO₂; the volume of N₂ is measured in a nitrometer and converted to STP.

Example: 0.30 g gave 50 mL of N₂ at STP. %N = (28 × 50)/(22400 × 0.30) × 100 = 1400/6720 × 100 = 20.83%.

Kjeldahl method

Principle: the compound is digested with conc. H₂SO₄ (with K₂SO₄ and CuSO₄ catalyst); nitrogen changes to (NH₄)₂SO₄. This is heated with excess NaOH; the NH₃ liberated is absorbed in a known volume of standard acid, and the unused acid is back-titrated with standard alkali.

Example: NH₃ from 0.50 g of compound neutralised 10 mL of 1 N H₂SO₄. %N = 1.4 × 1 × 10 / 0.50 = 28%.

Kjeldahl's method does not work for nitro and azo compounds or N in a ring (pyridine), because this N is not fully converted to (NH₄)₂SO₄.

Estimation of sulphur and halogens (Carius method)

Halogens: the compound is heated with fuming HNO₃ and AgNO₃ in a sealed hard-glass (Carius) tube. Halogen forms AgX, which is filtered, dried and weighed.

Example: 0.15 g gave 0.12 g AgBr. %Br = (80/188)(0.12/0.15)(100) = 34.04%.

Sulphur: heated with fuming HNO₃ (or Na₂O₂) in a Carius tube; S is oxidised to H₂SO₄, precipitated as BaSO₄ with BaCl₂ and weighed.

Oxygen is usually found by difference: %O = 100 − (sum of all other elements).

Calculation of empirical and molecular formula

• Empirical formula: simplest whole-number ratio of atoms.
• Molecular formula = (empirical formula)ₙ, where n = molecular mass / empirical formula mass.

1. Divide % of each element by its atomic mass → relative number of moles.
2. Divide all values by the smallest → simplest ratio.
3. Multiply to get whole numbers if needed (e.g. 1.5 → ×2).

Numerical 1: A compound has C = 40.0%, H = 6.67%, O = 53.33%; molar mass = 60 g/mol.

Empirical formula = CH₂O (mass 30). n = 60/30 = 2. Molecular formula = C₂H₄O₂ (e.g. acetic acid).

Numerical 2: Using the Liebig example above (C 60%, H 10%), O = 100 − 70 = 30%. Moles: C 60/12 = 5, H 10/1 = 10, O 30/16 = 1.875. Divide by 1.875: C 2.67, H 5.33, O 1. Multiply by 3: C 8, H 16, O 3. Empirical formula = C₈H₁₆O₃ (mass 160).

(B) Organomagnesium compounds: Grignard reagent

Formation: R–X + Mg → R–MgX (dry ether, or THF). The ether must be perfectly dry, because water destroys the reagent. Reactivity of halides: RI > RBr > RCl. A crystal of iodine helps start the reaction.

Structure: the C–Mg bond is polar (Cδ⁻–Mgδ⁺), so the alkyl group acts as a carbanion (nucleophile and strong base). In ether, Mg is tetrahedral, bonded to R, X and two ether O atoms: RMgX(OEt₂)₂. In solution the Schlenk equilibrium exists: 2RMgX ⇌ R₂Mg + MgX₂.

Chemical reactions

• With water/alcohol: CH₃MgBr + H₂O → CH₄ + Mg(OH)Br (alkane).
• With HCHO → 1° alcohol; with other aldehydes → 2° alcohol; with ketones → 3° alcohol (after hydrolysis with dilute acid). CH₃MgBr + CH₃COCH₃ → (CH₃)₃C–OMgBr → (H₃O⁺) (CH₃)₃C–OH.
• With CO₂ (dry ice) → carboxylic acid: RMgX + CO₂ → RCOOMgX → RCOOH.
• With esters → 3° alcohols (two moles of RMgX); with nitriles → ketones: RMgX + R′CN → R′C(=NMgX)R → (H₃O⁺) R′COR.
• With ethylene oxide → 1° alcohol with two extra carbons (RCH₂CH₂OH).

Organozinc and organolithium compounds

Organozinc compounds

Formation (Frankland): 2C₂H₅I + 2Zn → (C₂H₅)₂Zn + ZnI₂ (heat, Zn–Cu couple, inert atmosphere). Diethylzinc is a liquid that catches fire in air.

• With water: (C₂H₅)₂Zn + 2H₂O → 2C₂H₆ + Zn(OH)₂.
• With acid chlorides → ketones: 2RCOCl + R′₂Zn → 2RCOR′ + ZnCl₂. Being less reactive than Grignard reagents, they do not attack the ketone formed.
• Reformatsky reaction: BrCH₂COOC₂H₅ + Zn → BrZnCH₂COOC₂H₅; this adds to an aldehyde/ketone → β-hydroxy ester after hydrolysis.
• Simmons–Smith reaction: CH₂I₂ + Zn(Cu) → ICH₂ZnI (carbenoid), which converts alkenes into cyclopropanes.

Organolithium compounds

Formation: R–X + 2Li → R–Li + LiX (dry ether or hexane, inert gas). Example: n-C₄H₉Br + 2Li → n-C₄H₉Li (n-butyllithium) + LiBr. Also by metal–halogen exchange: ArBr + n-BuLi → ArLi + n-BuBr.

• C–Li bond is more polar than C–Mg, so RLi is more reactive and a stronger base than RMgX.
• With water → alkane; with aldehydes/ketones → 2°/3° alcohols (like Grignard).
• With CO₂ → RCOOLi → carboxylic acid; with carboxylic acids (2 equivalents of RLi) → ketones.
• 2RLi + CuI → R₂CuLi (Gilman reagent), used for coupling with alkyl halides (Corey–House synthesis of alkanes).
• As a strong base: n-BuLi removes acidic protons (terminal alkynes, amines → LDA from diisopropylamine).

Likely exam questions

Long answer

1. Describe the principle of estimation of nitrogen by Dumas and Kjeldahl methods. 0.30 g of an organic compound gave 50 mL of nitrogen at STP. Calculate the percentage of nitrogen.
2. How is a Grignard reagent prepared? Discuss its structure and any four synthetic applications.

Short answer

1. A compound contains C 40%, H 6.67% and O 53.33%. Its molar mass is 60. Find its empirical and molecular formula.
2. Give the principle of the Carius method for estimation of halogens.
3. How are organolithium compounds prepared? Give two reactions.

Syllabus topics (official RTMNU)

(A) Electromagnetic spectrum

• Absorption spectra, Ultraviolet absorption spectroscopy, Absorption laws (Beer Lambert law), molar absorptivity, Presentation and analysis of UV spectra, Types of electronic transitions, Effect of conjugation, concept of chromophores and auxochromes, Bathochromic, hypsochromic, hyperchromic and hypochromic shifts.
• UV spectra of conjugated dienes and enones.

(B) Infrared (IR) spectroscopy

• Molecular vibrations, Hook’s law, Selection rules, Intensity and position of IR bands, measurement of IR spectrum.
• Fingerprint region, characteristic absorptions of various functional groups and application of IR spectra.

(A) Electromagnetic spectrum and absorption spectra

Electromagnetic radiation has both wave and particle nature. Energy of one photon: E = hν = hc/λ = hcν̄, where ν̄ = 1/λ is the wavenumber (cm⁻¹). Shorter wavelength means higher energy.

An absorption spectrum is a plot of the amount of radiation absorbed (absorbance or % transmittance) against wavelength or wavenumber. A molecule absorbs only those photons whose energy exactly matches the gap between two of its energy levels.

Ultraviolet absorption spectroscopy and Beer–Lambert law

In UV–visible spectroscopy (200–800 nm) electrons are promoted from a filled orbital (HOMO) to an empty one (LUMO). The instrument has a source (deuterium lamp for UV, tungsten for visible), monochromator, quartz cells (glass absorbs UV), and a detector.

Absorption laws

• Lambert's law: absorbance is proportional to path length l.
• Beer's law: absorbance is proportional to concentration c.

Molar absorptivity (ε) is the absorbance of a 1 mol/L solution in a 1 cm cell. It shows how strongly a transition absorbs: ε above 10⁴ means an allowed, intense band (π→π*); ε below 100 means a weak, "forbidden" band (n→π*).

Numerical: a solution in a 1 cm cell has A = 0.50 at λmax, where ε = 1.0 × 10⁴ L mol⁻¹ cm⁻¹. c = A/(εl) = 0.50/(1.0 × 10⁴ × 1) = 5.0 × 10⁻⁵ mol L⁻¹.

Presentation and analysis of UV spectra

The spectrum is plotted as absorbance (or ε, or log ε) on the y-axis against wavelength (nm) on the x-axis. It shows broad bands, not sharp lines, because vibrational and rotational levels overlap the electronic transition. Two values describe each band: λmax (wavelength of maximum absorption) and εmax. Solvent must be transparent in the region (ethanol, hexane, water).

Types of electronic transitions, chromophores, auxochromes and shifts

Energy order: σ→σ* > n→σ* > π→π* > n→π*.

• Effect of conjugation: conjugation brings the HOMO and LUMO closer, so λmax moves to longer wavelength and intensity rises. Ethylene 171 nm, 1,3-butadiene 217 nm, 1,3,5-hexatriene ≈ 258 nm; β-carotene (11 conjugated C=C) absorbs in the visible and is orange.
• Chromophore: a group responsible for absorption (C=C, C=O, N=N, NO₂, C≡N, aromatic ring).
• Auxochrome: a group with lone pairs that does not absorb above 200 nm itself but, attached to a chromophore, shifts λmax to longer wavelength and increases intensity (–OH, –NH₂, –OR, –Cl). Benzene 255 nm, phenol 270 nm, aniline 280 nm.

Shifts

• Bathochromic (red) shift: λmax moves to longer wavelength (conjugation, auxochromes).
• Hypsochromic (blue) shift: λmax moves to shorter wavelength (e.g. n→π* of C=O in polar solvent; aniline in acid, where –NH₃⁺ loses the lone pair: 280 → 254 nm).
• Hyperchromic effect: increase in intensity (εmax).
• Hypochromic effect: decrease in intensity (εmax).

UV spectra of conjugated dienes and enones (Woodward–Fieser rules)

Example: 2,3-dimethyl-1,3-butadiene = 214 + 2 × 5 = 224 nm (observed ≈ 226 nm).

Example: mesityl oxide (CH₃)₂C=CH–CO–CH₃: two β-methyl groups. λmax = 215 + 2 × 12 = 239 nm (observed ≈ 237 nm).

(B) Infrared spectroscopy: molecular vibrations and Hooke's law

IR radiation (4000–400 cm⁻¹) makes bonds vibrate. A molecule of N atoms has 3N − 6 vibrational modes (3N − 5 if linear). H₂O: 3 modes; CO₂: 4 modes.

• Stretching: bond length changes — symmetric and asymmetric.
• Bending: bond angle changes — scissoring, rocking (in-plane), wagging, twisting (out-of-plane). Bending needs less energy than stretching.

Hooke's law

A bond is treated as two masses joined by a spring:

• Stronger bond (larger k) → higher wavenumber: C≡C (≈2200) > C=C (≈1650) > C–C (≈1200 cm⁻¹).
• Lighter atoms (smaller μ) → higher wavenumber: C–H (≈3000) > C–D (≈2200 cm⁻¹).

Selection rules, intensity, position of bands and measurement

• Selection rule: a vibration is IR active only if it causes a change in dipole moment. Homonuclear molecules (H₂, N₂, O₂) are IR inactive; the symmetric stretch of CO₂ is inactive, the asymmetric stretch (≈2350 cm⁻¹) is active. For a harmonic oscillator Δv = ±1 (fundamental); overtones are weak.
• Intensity: larger change in dipole moment → stronger band. C=O and O–H bands are strong; C=C bands are weak to medium.
• Position depends on bond strength, masses, hydrogen bonding (lowers and broadens O–H), conjugation (lowers C=O by about 20–30 cm⁻¹), and ring strain (raises C=O).
• Measurement: IR spectrophotometer (Nernst glower or globar source, NaCl/KBr optics; modern FTIR). Samples: liquid as a thin film between NaCl plates, solid as KBr pellet or Nujol mull, or solution in CCl₄/CHCl₃. Spectrum: % transmittance (y-axis) vs wavenumber 4000 → 400 cm⁻¹ (x-axis); absorption bands point downwards.

Fingerprint region, characteristic absorptions and applications

Fingerprint region (about 1500–600 cm⁻¹): many complex bending and C–C, C–O, C–N stretching bands. This pattern is unique to every molecule, so two compounds with identical IR spectra in this region are identical. The functional group region is 4000–1500 cm⁻¹.

Applications of IR spectroscopy

• Identifying functional groups in an unknown compound.
• Confirming identity by comparing fingerprint regions.
• Following a reaction (e.g. disappearance of C=O at 1715 cm⁻¹ when a ketone is reduced to an alcohol, and appearance of O–H).
• Detecting hydrogen bonding (intermolecular H-bonded O–H shifts with dilution; intramolecular does not).
• Checking purity (extra bands show impurities) and distinguishing isomers (e.g. ethanol vs dimethyl ether).

Likely exam questions

Long answer

1. State the Beer–Lambert law and define molar absorptivity. Discuss the types of electronic transitions in UV spectroscopy with examples.
2. Explain molecular vibrations and Hooke's law in IR spectroscopy. Give the characteristic absorption frequencies of any five functional groups.

Short answer

1. What are chromophores and auxochromes? Give two examples of each.
2. Define bathochromic, hypsochromic, hyperchromic and hypochromic shifts.
3. What is the fingerprint region? Why is it so called?