MCH-017 Solved Assignment January 2026 | ORGANIC CHEMISTRY-II | IGNOU MSCCHEM

Q: Attempt the following parts regarding UV and 'H-NMR spectra.

Answer available — download the full PDF for complete details.

Q1. Attempt the following parts regarding organic compounds.

(a)) Write the names of the following compounds according to Hantzsch-Widman system of nomenclature. (75 words)
(b)) Write the structures of the following compounds: (i) Pyrazino [2, 3,-c] pyridazine (ii) 2-Methylazacyclohexane (iii) 2,5-Diazaanthracene (125 words)

Key Points:

Hantzsch-Widman system names monocyclic heterocycles using heteroatom prefixes and ring size stems.
Heteroatom prefixes: 'Oxa-' (O), 'Aza-' (N), 'Thia-' (S).
Ring size stems: '-ol' (5-membered), '-in' (6-membered).
Suffixes denote unsaturation/saturation (e.g., '-e' for unsaturated, '-ane' for saturated non-N rings).

Answer: Organic compounds, particularly heterocyclic ones, are systematically named using established rules like the Hantzsch-Widman system for monocyclic heterocycles, or by fusion nomenclature for bicyclic and polycyclic systems. Structural representation requires understanding parent structures, heteroatom positions, and substituent locations. The Hantzsch-Widman system combines prefixes for heteroatoms (e.g., 'oxa-' for oxygen, 'aza-' for nitrogen) with stems indicating ring size (e.g., '-ol' for f...

Q2. Attempt the following parts regarding UV and 'H-NMR spectra.

(a)) List the important bands observed in the UV spectra of the following compounds: (i) Pyridazine (ii) Pyrazine (75 words)
(b)) Explain the chemical shift values exhibited in the 'H-NMR spectrum of pyridine. (125 words)

Key Points:

UV spectra of N-heterocycles show n→π* and π→π* transitions.
n→π* transitions are weak (low ε), occurring at longer wavelengths.
π→π* transitions are strong (high ε), occurring at shorter wavelengths.
Pyridine protons are deshielded by aromatic ring current.

Answer:

Q3. Discuss how can tetraalkylaziridines be synthesised.

Key Points:

Tetraalkylaziridines are highly substituted, strained three-membered nitrogen heterocycles.
Their synthesis primarily involves the decomposition of 1,2,3-triazolines.
[3+2] cycloaddition between highly substituted alkene and alkyl azide forms triazoline.
Example: Tetraalkylethylene (like 2,3-dimethyl-2-butene) reacts with an alkyl azide.

Answer: Tetraalkylaziridines are highly substituted, strained three-membered nitrogen heterocycles. Their synthesis demands specific approaches due to the steric demands of the four alkyl groups on the ring carbons, which can hinder typical cyclization reactions. A well-established and efficient method for their preparation involves the decomposition of 1,2,3-triazolines. This synthesis initiates with a [3+2] cycloaddition reaction between a highly substituted alkene and an alkyl azide (R-N₃). For inst...

Q4. Give the products of the reaction of azetedine with the following reagents: (i) RCOCl (ii) CS2 (iii) HNO2 (iv) HCHO (v) HCl/A

Key Points:

Azetidine, a strained cyclic secondary amine, undergoes typical secondary amine reactions.
Reaction with RCOCl yields N-acylazetidine via N-acylation.
CS2 reacts with azetidine to form an azetidinium dithiocarbamate salt.
HNO2 converts azetidine into N-nitrosoazetidine.

Answer: Azetidine is a strained four-membered cyclic secondary amine. Its reactivity is characteristic of secondary amines, but the ring strain makes it susceptible to ring-opening reactions under certain conditions. The presence of a nitrogen atom with a lone pair enables it to act as a nucleophile and a base. (i) With RCOCl (Acyl chloride): Azetidine undergoes N-acylation. The nitrogen's lone pair attacks the electrophilic carbonyl carbon of the acyl chloride, displacing the chloride ion and forming ...

Q5. Illustrate giving two [3+2] cycloaddition reactions how can 1,3-dipolar compounds react with alkenes and alkynes to give pyrazoles.

Key Points:

Pyrazoles are formed via [3+2] cycloaddition of 1,3-dipoles with alkenes/alkynes.
Diazomethane (CH₂=N⁺=N⁻) is a key 1,3-dipole for pyrazole synthesis.
Diazomethane + alkene forms a dihydropyrazole (1-pyrazoline) as the initial product.
Dihydropyrazoles require further oxidation or dehydrogenation to yield aromatic pyrazoles.

Answer: As discussed in MCH-017, [3+2] cycloaddition reactions are powerful tools for synthesizing five-membered heterocycles, including pyrazoles. These reactions involve a 1,3-dipolar compound (having a three-atom π-electron system) and a dipolarophile (an alkene or alkyne), proceeding through a concerted mechanism. Pyrazoles, which are five-membered nitrogen-containing aromatic heterocycles, are typically formed when diazo compounds act as the 1,3-dipole. The most common diazo compound used is diazo...

Q6. Discuss Bischer synthesis of indole. How can you obtain a single product in this reaction?

Key Points:

Bischer synthesis prepares indole-2-carboxylic acid.
Starting material is *o*-nitrophenylacetic acid.
Reaction involves reduction (e.g., Na(Hg)) and intramolecular cyclization.
Nitro group reduces to amino, then cyclizes with the carboxylic acid.

Answer: The Bischer synthesis is a method for preparing indole-2-carboxylic acid, which can subsequently be decarboxylated to yield indole. This reaction typically involves the reduction of *o*-nitrophenylacetic acid using a suitable reducing agent, commonly sodium amalgam (Na(Hg)), in an appropriate solvent. Initially, the nitro group (-NO₂) of *o*-nitrophenylacetic acid is reduced to an amino group (-NH₂), forming *o*-aminophenylacetic acid. This intermediate then undergoes spontaneous intramolecula...

Q7. How can benzotriazole be synthesised starting from ortho-phenylenediamine? Give the mechanism of the reaction.

Key Points:

Benzotriazole is synthesized from *ortho*-phenylenediamine.
Reaction requires nitrous acid (NaNO₂ + HCl) at 0-5 °C.
One amino group undergoes diazotization to form a diazonium salt.
The second amino group performs intramolecular nucleophilic attack.

Answer: Benzotriazole can be synthesized from *ortho*-phenylenediamine through a two-step process involving diazotization followed by intramolecular cyclization. This reaction typically occurs when *ortho*-phenylenediamine is treated with nitrous acid, which is generated *in situ* from sodium nitrite (NaNO₂) and a mineral acid like hydrochloric acid (HCl), usually at low temperatures (0-5 °C) to prevent decomposition of intermediates. The mechanism initiates with the formation of nitrous acid (HNO₂) f...

Q8. Discuss the Skraup synthesis of quinoline giving the reactions involved.

Key Points:

Skraup synthesis produces quinoline and its derivatives.
Key reactants: Aniline, glycerol, concentrated H₂SO₄, and an oxidizing agent.
Glycerol is dehydrated by H₂SO₄ to form acrolein.
Acrolein reacts with aniline via Michael addition and cyclization.

Answer: The Skraup synthesis is a classical and crucial method for the preparation of quinoline and its various substituted derivatives. This reaction involves heating an aniline (or a substituted aniline) with glycerol in the presence of concentrated sulfuric acid and an oxidizing agent, typically nitrobenzene or arsenic pentoxide. Initially, concentrated sulfuric acid dehydrates glycerol to form acrolein (propenal), an α,β-unsaturated aldehyde. This acrolein then acts as the key electrophilic compone...

Q9. Draw the tautomeric forms of azepines and compare their stabilities.

Key Points:

Azepines are seven-membered nitrogen-containing heterocycles that exhibit tautomerism.
1H-azepine features an N-H bond and three double bonds in the ring.
3H-azepine has a C=N bond in the ring, with hydrogen at C-3.
Fully conjugated 1H-azepine forms an unstable 8π-electron anti-aromatic system.

Answer: Azepines are seven-membered heterocyclic compounds containing one nitrogen atom. Like many heterocycles, they can exhibit tautomerism, a type of structural isomerism involving the rapid interconversion of isomers through the migration of a proton and a shift in double bonds. For azepines, key prototropic tautomers involve the relocation of a hydrogen atom and subsequent changes in π-electron distribution. One significant tautomeric form is **1H-azepine**. In this structure, the nitrogen atom at...

Q10. Explain the synthesis of azepine using ethoxycarbonyl nitrene.

Key Points:

Ethoxycarbonyl nitrene is generated from ethyl azidoformate by thermal or photochemical decomposition.
Nitrene acts as an electrophile, reacting with benzene via a 1,2-cycloaddition.
An unstable bicyclic N-ethoxycarbonylaziridine intermediate is formed.
The bicyclic intermediate undergoes a [1,5]-sigmatropic rearrangement.

Answer: The synthesis of azepine using ethoxycarbonyl nitrene, as covered in MCH-017, involves the generation of the nitrene and its subsequent reaction with an aromatic compound, typically benzene, followed by a rearrangement. Ethoxycarbonyl nitrene (:NCOOC₂H₅) is primarily generated by the thermal or photochemical decomposition of ethyl azidoformate (N₃COOC₂H₅). During this process, nitrogen gas (N₂) is eliminated, leaving behind the highly reactive singlet nitrene species, which acts as an electroph...

Q11. Discuss the disconnections for the synthesis of benzocaine.

Key Points:

Benzocaine retrosynthesis begins with ester C-O disconnection.
Ester cleavage yields 4-aminobenzoic acid and ethanol.
Amino group in 4-aminobenzoic acid arises from nitro group reduction (FGI).
4-nitrobenzoic acid derived from p-nitrotoluene via methyl oxidation (FGI).

Answer: The retrosynthetic analysis for benzocaine (ethyl 4-aminobenzoate) begins with its most prominent functional group, the ester. The primary disconnection involves the C-O bond of the ester functionality. This `C-O disconnection` directly leads to 4-aminobenzoic acid and ethanol as immediate precursors, simplifying the target molecule. Next, considering the 4-aminobenzoic acid fragment, the amino group (-NH₂) is typically introduced by reducing a nitro group (-NO₂). Therefore, a `Functional Group...

Q12. Explain Negishi coupling reaction with an example and give its mechanism.

Key Points:

Negishi coupling is a Pd- or Ni-catalyzed C-C bond forming reaction.
It couples organozinc compounds (R-ZnX) with organic halides/triflates (R'-X/OTf).
The catalytic cycle involves oxidative addition, transmetallation, and reductive elimination.
Example: Bromobenzene + Ethylzinc chloride forms Ethylbenzene.

Answer: The Negishi coupling reaction is a powerful palladium- or nickel-catalyzed cross-coupling reaction used to form new carbon-carbon bonds. It involves the reaction between an organozinc compound (R-ZnX) and an organic electrophile, typically an organic halide (R'-X) or a triflate (R'-OTf). This reaction is highly selective, tolerates a wide range of functional groups, and is widely applied in organic synthesis for constructing complex molecules. The organozinc reagents are generally less toxic an...

Q13. Explain giving suitable reactions the reduction of benzene using sodium in liquid ammonia.

Key Points:

Birch Reduction: Reduction of aromatic rings using alkali metals in liquid ammonia.
Reagents: Sodium (Na) in liquid ammonia (NH₃), typically with an alcohol as proton source.
Product: Benzene is reduced to 1,4-cyclohexadiene, a non-conjugated diene.
Mechanism: Proceeds via successive single electron transfer and protonation steps.

Answer: The reduction of benzene using sodium in liquid ammonia is a prominent organic reaction known as the Birch Reduction. This reaction is a cornerstone for converting aromatic rings into non-conjugated cyclohexadienes, selectively achieving partial reduction rather than complete saturation. The overall transformation for benzene involves reacting it with sodium (Na) in liquid ammonia (NH₃), typically in the presence of a proton source such as ethanol or *tert*-butanol. This process yields 1,4-cyc...

Q14. Give examples of alkoxysubstituted LAH reducing agents. Compare their selectivity with LAH.

Key Points:

LAH (LiAlH₄) is a powerful, non-selective reducing agent.
Alkoxysubstituted LAH agents replace hydride ligands with alkoxy groups.
Examples include LiAlH(O*t*Bu)₃ and LiAlH(OMe)₃.
Alkoxy groups modify steric hindrance and electronic properties, reducing reactivity.

Answer: Lithium aluminum hydride (LAH, LiAlH₄) is a powerful, non-selective reducing agent for functional groups like aldehydes, ketones, and esters. Its broad reactivity often requires milder, more selective alternatives. Alkoxysubstituted LAH agents replace hydrides with alkoxy groups like *tert*-butoxy or methoxy. Examples are Lithium tri-*tert*-butoxyaluminum hydride (LiAlH(O*t*Bu)₃) and Lithium trimethoxyaluminum hydride (LiAlH(OMe)₃). These substituents modify the aluminum center's steric and ele...

Q15. Discuss the oxidation of monoalkylacetylenes using thalium nitrate. Also give the mechanism of the reaction.

Key Points:

Thallium(III) nitrate (TTN) oxidizes monoalkylacetylenes.
The primary product is a methyl ketone (R-CO-CH₃).
TTN initiates with electrophilic addition to the alkyne triple bond.
An unstable vinylthallium(III) intermediate is formed.

Answer: Oxidation of monoalkylacetylenes using thallium(III) nitrate (TTN) is a significant synthetic transformation in organic chemistry. This reaction, typically carried out in protic solvents like methanol or aqueous acidic conditions, efficiently converts terminal alkynes (monoalkylacetylenes) into methyl ketones. TTN acts as a strong electrophilic oxidant. The overall process involves an oxidative rearrangement rather than simple hydration, leading to the formation of a new carbon-oxygen bond and ...

Q16. Write various factors which control the formation of enolates.

Key Points:

Acidity of α-hydrogen (pKa) directly controls the ease of enolate formation.
Strong bases (e.g., LDA) favor kinetic enolates at low temperatures.
Weaker bases (e.g., alkoxides) favor thermodynamic enolates via equilibrium.
Low temperatures promote kinetic enolate formation (faster deprotonation).

Answer: The formation of enolates, vital intermediates in organic reactions, is governed by several interconnected factors primarily affecting the acidity of the α-hydrogen and the reaction conditions. Firstly, the acidity of the α-hydrogen is paramount. Compounds with more acidic α-hydrogens (lower pKa values), typically due to electron-withdrawing groups or conjugation, are deprotonated more readily. For example, β-dicarbonyl compounds have significantly lower pKa values (e.g., 9-11) compared to simp...

Q17. What are homoenolates? Write the methods of their preparation.

Key Points:

Homoenolates are carbanionic species with nucleophilic γ-carbon relative to a carbonyl group.
They are synthetic equivalents of γ-anions for C-C bond formation at γ-position.
Preparation from cyclopropanone derivatives involves ring-opening of acetals/ketals.
Deprotonation of specific α,β-unsaturated carbonyls at γ-position yields homoenolates.

Answer: Homoenolates are highly reactive carbanionic species where the nucleophilic center is at the γ-carbon (gamma-carbon) relative to a carbonyl group. Unlike conventional enolates, which are α-nucleophilic, homoenolates behave as synthetic equivalents of γ-anions of carbonyl compounds. This unique reactivity makes them valuable intermediates for γ-functionalization, allowing for the formation of C-C bonds at positions remote from the carbonyl. Several methods are employed for the preparation of ho...

Q18. Discuss the role of phase transfer catalysis in nucleophilic substitution reactions.

Key Points:

PTC enables reactions between immiscible organic and aqueous/solid phases.
Transfers ionic nucleophiles from aqueous/solid to organic phase.
Common catalysts are quaternary ammonium/phosphonium salts (e.g., R₄N⁺X⁻).
Nucleophiles become 'naked' and highly reactive in the organic phase.

Answer: Phase Transfer Catalysis (PTC) plays a crucial role in nucleophilic substitution reactions by overcoming the challenge of immiscibility between reactants. Many nucleophilic substitution reactions involve an organic substrate, typically soluble in an organic solvent, and an ionic nucleophile, which is often soluble only in an aqueous phase or exists as a solid salt. PTC facilitates these reactions by transferring the inorganic nucleophile from its aqueous or solid phase into the organic phase wh...

Q19. Write the advantages and disadvantages of chiron approach.

Key Points:

Chiron approach uses readily available chiral natural products as starting materials.
Provides excellent stereocontrol by leveraging pre-existing chirality.
Chirons like sugars, amino acids, and tartaric acid are abundant and inexpensive.
Limited structural match of chirons to targets often requires extensive modifications.

Answer: The chiron approach, a strategy in asymmetric synthesis, utilizes readily available, inexpensive chiral starting materials, often derived from natural sources, known as 'chirons'. These chirons possess pre-existing stereochemical information, which is then transferred to the target molecule during synthesis. One significant advantage of this approach is its excellent stereocontrol. By starting with a molecule that already has the desired absolute configuration, the need for complex and often lo...

Q20. With suitable examples, discuss the control of enantioselectivity in alkylation reactions.

Key Points:

Enantioselectivity in alkylation prioritizes one enantiomer's formation.
Chiral auxiliaries are temporary chiral groups attached to substrates.
Evans' oxazolidinone auxiliary directs enolate alkylation to one face.
Chiral catalysts provide a stereoselective environment in the transition state.

Answer: Control of enantioselectivity in alkylation reactions is crucial for synthesizing optically active compounds, especially in pharmaceuticals. This involves directing the incoming electrophile to one specific face of a prochiral carbon, leading to a preferential formation of one enantiomer over the other. The primary strategies for achieving this include the use of chiral auxiliaries and chiral catalysts. Chiral auxiliaries are non-racemic chiral compounds temporarily attached to the substrate. T...

MCH-017 Solved Assignment — January 2026 / July 2026

ORGANIC CHEMISTRY-II | IGNOU Master of Science in Chemistry Programme (MSCCHEM)

MCH-017—January 2026 / July 2026

ORGANIC CHEMISTRY-II

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MCH-017 Assignment Questions — January 2026

Q1. Attempt the following parts regarding organic compounds.

Q2. Attempt the following parts regarding UV and 'H-NMR spectra.

Q3. Discuss how can tetraalkylaziridines be synthesised.

Q4. Give the products of the reaction of azetedine with the following reagents: (i) RCOCl (ii) CS2 (iii) HNO2 (iv) HCHO (v) HCl/A

Q5. Illustrate giving two [3+2] cycloaddition reactions how can 1,3-dipolar compounds react with alkenes and alkynes to give pyrazoles.

Q6. Discuss Bischer synthesis of indole. How can you obtain a single product in this reaction?

Q7. How can benzotriazole be synthesised starting from ortho-phenylenediamine? Give the mechanism of the reaction.

Q8. Discuss the Skraup synthesis of quinoline giving the reactions involved.

Q9. Draw the tautomeric forms of azepines and compare their stabilities.

Q10. Explain the synthesis of azepine using ethoxycarbonyl nitrene.

Q11. Discuss the disconnections for the synthesis of benzocaine.

Q12. Explain Negishi coupling reaction with an example and give its mechanism.

Q13. Explain giving suitable reactions the reduction of benzene using sodium in liquid ammonia.

Q14. Give examples of alkoxysubstituted LAH reducing agents. Compare their selectivity with LAH.

Q15. Discuss the oxidation of monoalkylacetylenes using thalium nitrate. Also give the mechanism of the reaction.

Q16. Write various factors which control the formation of enolates.

Q17. What are homoenolates? Write the methods of their preparation.

Q18. Discuss the role of phase transfer catalysis in nucleophilic substitution reactions.

Q19. Write the advantages and disadvantages of chiron approach.

Q20. With suitable examples, discuss the control of enantioselectivity in alkylation reactions.