Chapter 8: Organic Chemistry – Some Basic Principles and Techniques Long Questions | chemistry Class 11 CBSE

Chapter 8: Organic Chemistry – Some Basic Principles and Techniques Long Questions | chemistry Class 11 CBSE | ByteNotes.in

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Question 1

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Compare the three types of hybridisation of carbon (sp3, sp2, sp) in terms of geometry, bond angles, bond lengths, and the types of organic compounds in which each is found.

sp3 hybridisation involves mixing of one s and three p orbitals to give four equivalent hybrid orbitals arranged tetrahedrally; bond angle is 109.5 degrees; example: methane, ethane, and all saturated hydrocarbons.
sp2 hybridisation involves mixing of one s and two p orbitals to give three equivalent hybrid orbitals in a trigonal planar arrangement; bond angle is 120 degrees; the remaining unhybridised p orbital forms a pi bond; example: ethene, benzene, aldehydes, ketones.
sp hybridisation involves mixing of one s and one p orbital to give two equivalent linear hybrid orbitals; bond angle is 180 degrees; two unhybridised p orbitals form two pi bonds; example: ethyne, nitriles (CN group).
Bond strength and electronegativity increase with increasing s-character: sp bonds are stronger and shorter than sp2 which are stronger and shorter than sp3 bonds.
The relationship between hybridisation and molecular geometry is essential for predicting molecular shape, polarity, and reactivity of organic compounds.
The unhybridised p orbitals in sp2 and sp carbons are responsible for the formation of pi bonds, which are the sites of greatest chemical reactivity in unsaturated organic molecules.

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Question 2

Long Form

Describe the IUPAC rules for naming branched-chain alkanes with reference to selecting the parent chain, numbering, and naming substituents. Illustrate with 2,4-dimethylpentane and 3-ethyl-4,4-dimethylheptane.

The longest continuous carbon chain is selected as the parent chain; its name determines the base name of the compound.
Carbon atoms of the parent chain are numbered from the end that gives substituents the lowest possible locant numbers (lowest locant rule).
Names of alkyl substituents are prefixed to the parent alkane name with their position numbers separated by hyphens; different substituents are listed alphabetically.
Identical substituents are indicated by prefixes di, tri, tetra (not counted in alphabetical ordering); their positions are listed with commas between numbers.
For 2,4-dimethylpentane: 5-carbon parent chain (pentane), methyl groups at C-2 and C-4; if numbered from the other end they would be at C-2 and C-4 also, confirming correct naming.
For 3-ethyl-4,4-dimethylheptane: 7-carbon parent (heptane), ethyl at C-3, two methyls at C-4; substituents listed alphabetically (ethyl before methyl), positions given as 3 and 4,4.

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Question 3

Long Form

Explain the four types of structural isomerism with definitions, examples, and the conditions that give rise to each type.

Chain isomerism arises when compounds have the same molecular formula but different carbon skeletons; it requires a molecular formula that permits at least two structural arrangements of the carbon chain; example: pentane and 2-methylbutane (both C5H12).
Position isomerism arises when compounds differ only in the position of the substituent or functional group on the same carbon skeleton; example: propan-1-ol and propan-2-ol (both C3H8O).
Functional group isomerism arises when compounds have the same molecular formula but contain different functional groups; example: propanone (ketone) and propanal (aldehyde), both C3H6O.
Metamerism arises due to different alkyl chains on either side of a functional group, particularly in ethers, ketones, and esters; example: C4H10O gives methoxypropane (CH3OC3H7) and ethoxyethane (C2H5OC2H5).
All four types are classified under structural isomerism, which includes all cases where the connectivity of atoms differs between isomers.
Stereoisomerism (geometrical and optical) differs from structural isomerism in that stereoisomers have the same connectivity but differ in spatial arrangement of atoms or groups.

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Question 4

Long Form

Analyse the four types of electron displacement effects in organic molecules, distinguishing between permanent and temporary effects, and state their influence on reactivity.

The inductive effect is a permanent effect caused by the difference in electronegativity between bonded atoms; it causes polarisation of sigma bonds and is transmitted along the chain, decreasing with distance; electron-withdrawing groups (halogens, -NO2, -CN) and electron-donating groups (alkyl groups) are classified based on this effect.
The resonance (mesomeric) effect is a permanent effect arising from the interaction of pi bonds or a pi bond with lone pairs in a conjugated system; the +R effect pushes electrons into the conjugated system (e.g., -NH2 in aniline) while the -R effect withdraws electrons (e.g., -NO2 in nitrobenzene).
The electromeric effect is a temporary effect observed only in the presence of an attacking reagent; it involves complete transfer of pi electrons to one atom of a multiple bond; the +E effect transfers electrons toward the attacking reagent's attachment site, while the -E effect transfers them away.
Hyperconjugation is a permanent stabilising effect involving delocalisation of sigma C-H bond electrons of an alkyl group into adjacent empty p orbitals or unsaturated systems; it explains the stability order of carbocations and alkenes.
When inductive and electromeric effects operate in opposite directions, the electromeric effect predominates since it involves complete electron transfer.
These effects collectively determine the reactivity, stability, acidity, basicity, and selectivity of organic reactions; for example, electron-withdrawing groups increase carboxylic acid strength while electron-donating groups decrease it.

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Question 5

Long Form

Describe in detail the chromatographic techniques used for purification of organic compounds, distinguishing adsorption and partition chromatography and explaining the principle of TLC and paper chromatography.

Chromatography separates mixtures by differential migration of components across a stationary phase driven by a mobile phase; the name derives from the Greek word chroma (colour), as it was first used for coloured plant pigments.
Adsorption chromatography is based on the principle that different compounds are adsorbed on an adsorbent (silica gel or alumina) to different degrees; components with weaker adsorption travel farther.
Column chromatography uses a glass column packed with adsorbent; the mixture is placed on top and an eluant (solvent) is allowed to flow down; components are collected separately as they elute.
Thin layer chromatography (TLC) uses a thin layer of adsorbent on a glass plate; the mixture spot is placed 2 cm from the base; after development in eluant, spots are detected by UV light, iodine, or ninhydrin spray; the Rf value identifies each compound.
Partition chromatography (paper chromatography) is based on continuous differential partitioning between a stationary liquid phase (water trapped in chromatography paper) and a mobile organic solvent phase; components separate by their different partition coefficients.
TLC is used for rapid monitoring of reactions and checking purity; column chromatography is used for preparative separation; paper chromatography is widely used for amino acids, sugars, and other water-soluble compounds.

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Question 6

Long Form

Discuss the Lassaigne's test for detection of nitrogen, sulphur, and halogens in organic compounds, including the chemical reactions involved and why a special treatment is needed when both nitrogen and sulphur are present.

Lassaigne's test involves fusing the organic compound with sodium metal; this converts covalent C-N, C-S, and C-X bonds into ionic NaCN, Na2S, and NaX respectively; the fused mass is extracted with distilled water to give the sodium fusion extract.
For nitrogen: the extract is boiled with FeSO4 then acidified with H2SO4; NaCN reacts with Fe2+ to give [Fe(CN)6]4-, which reacts with Fe3+ to give Prussian blue (Fe4[Fe(CN)6]3), confirming nitrogen.
For sulphur: (a) the extract is acidified with acetic acid and treated with lead acetate; a black precipitate of PbS confirms sulphur; (b) treatment with sodium nitroprusside gives a violet colour.
For halogens: the extract is acidified with dilute HNO3 and treated with AgNO3; white precipitate (soluble in NH4OH) indicates Cl; yellowish precipitate (sparingly soluble) indicates Br; yellow precipitate (insoluble) indicates I.
When both N and S are present, NaSCN is formed instead of NaCN; it gives blood-red colour with Fe3+ and no Prussian blue. Excess sodium fusion decomposes NaSCN to give NaCN and Na2S, allowing both to give their usual tests.
Before testing for halogens when N or S is also present, the extract must be boiled with concentrated HNO3 to decompose NaCN and Na2S, which would otherwise form AgCN or Ag2S precipitates that would interfere with the halogen test.

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Question 7

Long Form

Explain how carbocations, carbanions, and free radicals are formed, compare their structures and stability, and relate them to types of organic reactions.

Carbocations are formed by heterolytic cleavage where carbon loses its bonding electrons to the leaving group; they are sp2 hybridised with trigonal planar geometry and a vacant p orbital perpendicular to the molecular plane.
Carbanions are formed by heterolytic cleavage where carbon retains both bonding electrons; they are sp3 hybridised with a distorted tetrahedral geometry and a lone pair on carbon.
Free radicals are formed by homolytic cleavage where one electron goes to each atom; they are neutral species with an unpaired electron; alkyl radicals are classified as primary, secondary, or tertiary.
Stability of carbocations: tertiary > secondary > primary > methyl, due to increasing inductive and hyperconjugation effects from alkyl groups; stability of free radicals follows the same order.
Reactions involving carbocations and carbanions are called ionic (heterolytic or polar) reactions; those involving free radicals are called homolytic (nonpolar or free radical) reactions.
Carbocations are electrophilic intermediates that react with nucleophiles; carbanions are nucleophilic intermediates that react with electrophiles; free radicals react with molecules in chain reactions initiated by heat or light.

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Question 8

Long Form

Describe the quantitative estimation of carbon and hydrogen in an organic compound by the combustion method, including the experimental setup, reactions, and calculation formula.

A known mass (m g) of the organic compound is burned in a combustion tube in the presence of excess oxygen and copper(II) oxide; carbon is completely oxidised to CO2 and hydrogen to H2O.
The reaction is: CxHy + (x + y/4)O2 gives x CO2 + (y/2)H2O; traces of nitrogen oxides if present are reduced by passing over hot copper gauze.
The gaseous products are passed through a series of U-tubes: the first contains anhydrous CaCl2 which absorbs water; the second contains concentrated KOH solution which absorbs CO2; the increase in mass of each U-tube gives the mass of H2O (m1) and CO2 (m2) respectively.
Percentage of carbon = (12/44) x (m2/m) x 100; this is derived from the molar mass relationship: 1 mol CO2 (44 g) contains 12 g carbon.
Percentage of hydrogen = (2/18) x (m1/m) x 100; this is derived from: 1 mol H2O (18 g) contains 2 g hydrogen.
The U-tubes must be connected in the correct order (CaCl2 before KOH) to prevent CO2 being absorbed before water is measured; the percentage of oxygen (if present) is usually found by difference from 100.

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Question 9

Long Form

Classify organic compounds systematically with examples, showing how the classification based on structure complements classification based on functional groups.

Based on structure, organic compounds are first divided into acyclic (open chain or aliphatic) compounds with straight or branched carbon chains, and cyclic (closed chain or ring) compounds.
Cyclic compounds are subdivided into alicyclic (aliphatic cyclic) compounds with carbon-only rings resembling aliphatic properties (e.g., cyclohexane), and aromatic compounds.
Aromatic compounds include benzenoid types (benzene, naphthalene, aniline) with delocalized pi electrons and non-benzenoid types (tropone); they may contain heteroatoms (furan, thiophene, pyridine) and are then called heterocyclic aromatic compounds.
Alicyclic compounds may also be heterocyclic (non-aromatic) if the ring contains atoms other than carbon, such as oxygen in tetrahydrofuran.
Classification by functional groups organises compounds into homologous series: alkanes (no functional group), alkenes (C=C), alkynes (C triple bond C), alcohols (-OH), aldehydes (-CHO), ketones (>C=O), carboxylic acids (-COOH) etc.
Both classifications are complementary: for example, benzoic acid is both aromatic (structural) and a carboxylic acid (functional group); cyclohexanol is both alicyclic and an alcohol; furan is both heterocyclic and aromatic.

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Question 10

Long Form

Discuss resonance in organic molecules using benzene and nitromethane as examples. Explain how to determine the relative stability of resonance structures and what resonance energy signifies.

Resonance is used to describe molecules whose properties cannot be explained by a single Lewis structure; the actual structure is a hybrid of two or more resonance (canonical) structures, none of which individually represents the real molecule.
Benzene cannot be adequately represented by a single Kekule structure (alternating C-C and C=C) because all C-C bond lengths are equal (139 pm, intermediate between 154 pm for single and 134 pm for double bond); the actual structure is a resonance hybrid of the two Kekule structures.
Nitromethane (CH3NO2) similarly shows two canonical structures with different N-O bonds, but experimentally both N-O bonds are identical in length (intermediate), confirming it is a resonance hybrid.
A resonance structure is more stable if it has: more covalent bonds, all atoms with complete octets (except H), less separation of opposite charges, negative charge on more electronegative atom, and more dispersal of charge.
Resonance stabilisation energy (resonance energy) is the difference in energy between the actual molecule (resonance hybrid) and the hypothetical lowest-energy canonical structure; the more contributing structures of comparable energy, the greater the resonance energy.
Resonance is particularly significant in conjugated systems and aromatic compounds; it explains the unexpected stability of benzene and carboxylate ions, and the direction of electrophilic substitution in aromatic compounds.

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Chapter 8: Organic Chemistry – Some Basic Principles and Techniques | Long Questions

Compare the three types of hybridisation of carbon (sp3, sp2, sp) in terms of geometry, bond angles, bond lengths, and the types of organic compounds in which each is found.

Describe the IUPAC rules for naming branched-chain alkanes with reference to selecting the parent chain, numbering, and naming substituents. Illustrate with 2,4-dimethylpentane and 3-ethyl-4,4-dimethylheptane.

Explain the four types of structural isomerism with definitions, examples, and the conditions that give rise to each type.

Analyse the four types of electron displacement effects in organic molecules, distinguishing between permanent and temporary effects, and state their influence on reactivity.

Describe in detail the chromatographic techniques used for purification of organic compounds, distinguishing adsorption and partition chromatography and explaining the principle of TLC and paper chromatography.

Discuss the Lassaigne's test for detection of nitrogen, sulphur, and halogens in organic compounds, including the chemical reactions involved and why a special treatment is needed when both nitrogen and sulphur are present.

Explain how carbocations, carbanions, and free radicals are formed, compare their structures and stability, and relate them to types of organic reactions.

Describe the quantitative estimation of carbon and hydrogen in an organic compound by the combustion method, including the experimental setup, reactions, and calculation formula.

Classify organic compounds systematically with examples, showing how the classification based on structure complements classification based on functional groups.

Discuss resonance in organic molecules using benzene and nitromethane as examples. Explain how to determine the relative stability of resonance structures and what resonance energy signifies.

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