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Organic Chemistry – Some Basic Principles and Techniques: questions with answers

Class 11 Chemistry, Chapter 8 (NCERT). 15 questions by concept. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Carbon hybridisation, molecular shape, and sigma/pi bonding

Question 1

For CH3-C≡N, give both carbon hybridisations and the C≡N bond components. Compared with ethane, is its C-C bond shorter or longer?

  1. (a)sp3, sp; three σ and no π; shorter
  2. (b)sp3, sp; one σ and two π; longer
  3. (c)sp3, sp; one σ and two π; shorter
  4. (d)sp3, sp3; one σ and two π; shorter
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(c) sp3, sp; one σ and two π; shorter

The methyl carbon is sp3 and the nitrile carbon is sp. C≡N contains one σ and two π bonds. Compared with ethane, the C-C bond is shorter because the nitrile carbon uses an orbital with more s-character.

Question 2

A student compares the carbon atoms in CH3-CH=CH2. Which evaluation is correct?

  1. (a)The sp3 methyl carbon is more electronegative than either sp2 carbon because its bonds are shorter.
  2. (b)The methyl carbon is sp3 and each alkene carbon is sp2; an sp2 carbon has more s-character than the methyl carbon.
  3. (c)The C=C bond is made of two σ bonds, so both alkene carbons remain tetrahedral.
  4. (d)All three carbons are sp3 because all satisfy valency four.
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(b) The methyl carbon is sp3 and each alkene carbon is sp2; an sp2 carbon has more s-character than the methyl carbon.

The single-bonded methyl carbon has four electron domains, while each alkene carbon has three. A C=C bond contains one σ and one π component.

Structural representations, functional groups, and homologous series

Question 3

Which statement correctly combines wedge-dash notation with a homologous-series relationship?

  1. (a)A wedge means a double bond, so it changes the compound's bond order.
  2. (b)Successive homologues have identical molecular formulas.
  3. (c)A wedge or dash gives bond direction in three dimensions, and successive homologues differ by CH2.
  4. (d)Members are homologues whenever they have the same molar mass, even with different functional groups.
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(c) A wedge or dash gives bond direction in three dimensions, and successive homologues differ by CH2.

Perspective bonds encode orientation, not multiplicity. Homologues share a functional group and general formula and differ successively by CH2.

Question 4

How are CH3OH and C2H5OH related?

  1. (a)Isomers with identical molecular formulas
  2. (b)Successive alcohol homologues
  3. (c)Non-homologues with different molecular formulas
  4. (d)Members of different functional-group families
Show answer

(b) Successive alcohol homologues

Both compounds contain the alcohol functional group. Their formulas differ by CH2, making them successive homologues rather than isomers.

IUPAC nomenclature of chains, functional groups, and benzene derivatives

Question 5

A compound contains both -CHO and -OH, and its benzene ring has two substituents. Which naming principle is correct?

  1. (a)Either -CHO or -OH may be chosen arbitrarily as the suffix.
  2. (b)Use both functional groups as suffixes and omit ring locants.
  3. (c)Number the benzene ring from any drawn vertex because orientation fixes the locants.
  4. (d)Use the aldehyde as the suffix, express hydroxy as a prefix, and choose ring locants by the lowest-set rules.
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(d) Use the aldehyde as the suffix, express hydroxy as a prefix, and choose ring locants by the lowest-set rules.

The senior characteristic group supplies the suffix, while lower-priority groups become prefixes. Ring numbering must be unambiguous and minimal.

Question 6

When both -CHO and -OH occur in an acyclic compound, which group supplies the principal suffix?

  1. (a)-CHO
  2. (b)Either group, chosen freely
  3. (c)-OH
  4. (d)Both groups simultaneously
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(a) -CHO

The aldehyde group has priority over the alcohol group. The aldehyde supplies the suffix, and -OH is cited using the prefix hydroxy.

Structural isomerism, stereoisomerism, and resonance distinction

Question 7

Which comparison correctly distinguishes structural isomerism from resonance?

  1. (a)Resonance contributors are separate isomers that can be isolated, in rapid equilibrium.
  2. (b)Different carbon skeletons with the same molecular formula are chain isomers; drawings that differ only in electron placement are resonance contributors.
  3. (c)Compounds with different functional groups must have different molecular formulas.
  4. (d)Any movement of a branch is position isomerism, even when the carbon skeleton changes.
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(b) Different carbon skeletons with the same molecular formula are chain isomers; drawings that differ only in electron placement are resonance contributors.

Structural isomers differ in connectivity. Resonance contributors retain all nuclear positions and differ only in electron bookkeeping.

Question 8

What is the relationship between CH3CH2OH and CH3OCH3?

  1. (a)Functional isomers
  2. (b)Chain isomers
  3. (c)Non-isomers with different molecular formulas
  4. (d)Resonance contributors
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(a) Functional isomers

Both compounds have molecular formula C2H6O. Their different atom connections produce alcohol and ether functional groups, so they are functional isomers.

Bond fission, reactive intermediates, electrophiles/nucleophiles, and arrows

Question 9

For R-Br → R+ + Br-, which description is correct?

  1. (a)R+ is a nucleophile because positive species donate electron pairs.
  2. (b)The bond breaks heterolytically and the bonding pair goes to Br, producing a carbocation and bromide.
  3. (c)In heterolysis the bonding pair must go to carbon, so R- and Br+ must form.
  4. (d)The bond breaks homolytically to give R+ and Br-.
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(b) The bond breaks heterolytically and the bonding pair goes to Br, producing a carbocation and bromide.

Unequal division of the bond pair is heterolysis. Bromine takes both electrons, leaving electron-deficient R+.

Question 10

What products arise from homolytic cleavage of the C-C bond in CH3-CH3?

  1. (a)CH4 and CH2
  2. (b)CH3+ and CH3−
  3. (c)Two CH3 radicals
  4. (d)CH2=CH2 and H2
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(c) Two CH3 radicals

Homolysis gives one bonding electron to each carbon fragment. Cleaving the C-C bond in ethane therefore produces two methyl radicals.

Inductive, resonance, electromeric, and hyperconjugative effects

Question 11

Which set of descriptions is correct?

  1. (a)Inductive polarisation travels through σ bonds and decays with distance; resonance describes one delocalised hybrid; +I donation can destabilise a carbanion.
  2. (b)A +I alkyl group stabilises every positive or negative charge in the same way.
  3. (c)The inductive effect remains equally strong however far the substituent is.
  4. (d)Resonance contributors are molecules that can be isolated, in rapid equilibrium.
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(a) Inductive polarisation travels through σ bonds and decays with distance; resonance describes one delocalised hybrid; +I donation can destabilise a carbanion.

These effects differ in pathway and physical meaning, and electron donation has opposite consequences for electron-poor and electron-rich centres.

Substitution, addition, elimination, and rearrangement reactions

Question 12

How is CH3CH2Br + OH− → CH3CH2OH + Br− classified?

  1. (a)Elimination
  2. (b)Addition
  3. (c)Rearrangement
  4. (d)Substitution
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(d) Substitution

OH replaces Br at the same carbon, so the reaction is substitution. Departure of Br− alone does not make the reaction elimination.

Purification, extraction, distillation, and chromatography

Question 13

Under fixed paper-chromatography conditions, what does a smaller Rf usually indicate?

  1. (a)Rf is independent of the distance travelled by the solvent front.
  2. (b)Stronger retention by the stationary phase relative to movement with the mobile phase.
  3. (c)A larger Rf always means stronger adsorption to the stationary phase.
  4. (d)Crystallisation would place all soluble impurities into the product crystals equally.
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(b) Stronger retention by the stationary phase relative to movement with the mobile phase.

Rf is solute distance divided by solvent-front distance; stronger retention reduces solute travel.

Qualitative detection by sodium fusion and specific tests

Question 14

Why is metallic sodium used before testing an organic compound for nitrogen or halogen?

  1. (a)Covalently bound chlorine in CCl4 already supplies free Cl- to AgNO3.
  2. (b)Nitric acid should be added only after AgNO3 so interferences precipitate first.
  3. (c)It converts covalently bound elements into water-soluble ionic salts that can undergo ordinary inorganic tests.
  4. (d)The sodium flame colour directly identifies each element without further reagents.
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(c) It converts covalently bound elements into water-soluble ionic salts that can undergo ordinary inorganic tests.

The fusion step is a chemical conversion, not the final observation.

Quantitative estimation of C, H, N, halogens, S, P, and O

Question 15

Which procedure set is correct for quantitative analysis?

  1. (a)Apply Kjeldahl's method quantitatively to every form of nitrogen, including nitro and ring nitrogen.
  2. (b)Treat the measured CO2 mass as pure carbon mass.
  3. (c)Use 12/44 of CO2 mass for carbon, subtract aqueous tension from wet N2 pressure in Dumas, and note Kjeldahl exclusions such as nitro nitrogen.
  4. (d)Use the observed wet-gas pressure as pure nitrogen pressure without subtracting water vapour.
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(c) Use 12/44 of CO2 mass for carbon, subtract aqueous tension from wet N2 pressure in Dumas, and note Kjeldahl exclusions such as nitro nitrogen.

Each method has a distinct correction: product stoichiometry, dry-gas pressure, and chemical applicability.

Common misconception in Organic Chemistry – Some Basic Principles and Techniques

Open full misconception guide →

Does bond breaking always split the electrons evenly?

Wrong

When a C–Cl bond breaks, each atom always gets one shared electron.

Why it feels right. Sharing the pair seems to imply a fair split.

Right

Heterolytic cleavage sends both bonding electrons to one fragment.

Worked example

Before
C••Cl
After
C⁺
Cl⁻••
The chlorine side receives both bonding electrons in this heterolytic cleavage.
  1. In a heterolytic C–Cl cleavage that forms a carbocation, chloride takes the bonding pair.
  2. Carbon is left with a positive charge.