NeetCrackers Chemistry Practice
Practice Set

Some Basic Principles of Organic Chemistry — Practice Questions with Answers

30 fully solved questions covering IUPAC nomenclature, isomerism, electronic effects, and reaction intermediates/mechanisms.

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Questions written and solutions verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Checked against NCERT and previous years' papers. Last reviewed September 2026.
10MCQs · 1 mark
10Short Answer · 3 marks
10Long Answer · 5 marks
90Total Marks

📑 On This Page

  1. Section A — 10 MCQs (1 mark each)
  2. Section B — 10 Short Answer Questions (3 marks each)
  3. Section C — 10 Long Answer Questions (5 marks each)
Section A

🎯 Multiple Choice Questions

10 questions · 1 mark each · Correct answer highlighted with explanation

Q1 The inductive effect refers to the:
Correct Answer: (a). The inductive effect is the permanent polarisation of sigma bond electron density along a chain, due to differing electronegativities of attached atoms/groups.
Q2 A carbocation is a species with:
Correct Answer: (a). A carbocation has a positively charged, trivalent (sp² hybridised) carbon with an empty p-orbital, giving it only 6 valence electrons.
Q3 The stability of carbocations follows the order:
Correct Answer: (a). More substituted carbocations are more stable due to greater electron-donating hyperconjugation and inductive effects from adjacent alkyl groups.
Q4 Two compounds that have the same molecular formula but different structural arrangements are called:
Correct Answer: (a) Isomers — compounds sharing the same molecular formula but differing in structural or spatial arrangement of atoms.
Q5 A nucleophile is a species that:
Correct Answer: (a). Nucleophiles ("nucleus-loving" species) are electron-rich, donating an electron pair to form a new covalent bond with an electron-deficient centre.
Q6 The resonance effect (mesomeric effect) involves:
Correct Answer: (a). Resonance involves the delocalisation of pi electrons or lone pairs across adjacent, conjugated p-orbitals, distributing electron density across multiple atoms.
Q7 A free radical is a species characterised by having:
Correct Answer: (a). Free radicals are neutral species possessing at least one unpaired electron, making them highly reactive.
Q8 Hyperconjugation involves the overlap of:
Correct Answer: (a). Hyperconjugation involves overlap between a sigma bond (typically C-H) and an adjacent empty p-orbital (as in a carbocation) or pi system, providing additional stabilisation.
Q9 Functional isomerism is exhibited by compounds that:
Correct Answer: (a). Functional isomers share the same molecular formula but possess distinctly different functional groups — for example, an alcohol and an ether with the same formula.
Q10 An electrophile is a species that:
Correct Answer: (a). Electrophiles ("electron-loving" species) are electron-deficient, seeking to accept an electron pair from a nucleophile to form a new bond.
Section B

✍️ Short Answer Questions

10 questions · 3 marks each · Full model answers included

Q1. Distinguish between the inductive effect and the resonance (mesomeric) effect.
3 Marks
Answer

The inductive effect involves the permanent, gradual displacement of sigma bond electrons along a chain due to electronegativity differences between bonded atoms — this effect weakens rapidly with increasing distance from the source and operates through sigma bonds.

The resonance (mesomeric) effect involves the delocalisation of pi electrons or lone pairs across a conjugated system of overlapping p-orbitals — this effect can operate over longer distances within a conjugated system and is generally considered a stronger electronic effect than the inductive effect when both are present.

Q2. Define carbocation, carbanion, and free radical, briefly noting the hybridisation of the central carbon in each.
3 Marks
Answer

Carbocation: A positively charged carbon species with only 6 valence electrons (an empty p-orbital); the carbon is $sp^2$ hybridised, typically planar.

Carbanion: A negatively charged carbon species with a lone pair of electrons (8 valence electrons total, including the lone pair); the carbon is $sp^3$ hybridised, typically pyramidal.

Free radical: A neutral carbon species with one unpaired electron (7 valence electrons); the carbon is generally considered to be approximately $sp^2$ hybridised (planar or slightly pyramidal, with the unpaired electron in a p-orbital).

Q3. Explain hyperconjugation, and describe its stabilising effect on carbocations.
3 Marks
Answer

Hyperconjugation is the stabilising interaction resulting from overlap between a sigma bond orbital (typically a C-H bond adjacent to the reactive centre) and an adjacent empty (or partially filled) p-orbital, or a pi-bond system.

In a carbocation, sigma C-H bonds on adjacent (beta) carbons can partially donate electron density into the carbocation's empty p-orbital — this electron density donation partially delocalises and stabilises the positive charge, and more available adjacent C-H bonds (more alkyl substitution) provide more hyperconjugative stabilisation, directly explaining the observed stability order: tertiary > secondary > primary > methyl carbocations.

Q4. Distinguish between homolytic and heterolytic bond cleavage, and state the type of intermediate produced by each.
3 Marks
Answer

Homolytic cleavage: A covalent bond breaks symmetrically, with each fragment retaining one electron from the shared pair — this process produces two free radicals (neutral species with unpaired electrons), typically favoured under conditions like UV light or high temperature, without a polar solvent.

Heterolytic cleavage: A covalent bond breaks asymmetrically, with both electrons from the shared pair going to one of the two fragments — this process produces a pair of oppositely charged ions (typically a carbocation and an anion, or similar), generally favoured in polar solvents or with polar reagents.

Q5. Explain the concept of resonance, and state the criteria for writing valid resonance structures.
3 Marks
Answer

Resonance describes situations where a single Lewis structure cannot adequately represent a molecule's actual electronic structure — the true structure is a hybrid (weighted average) of several valid contributing structures, differing only in electron placement, not atomic positions.

Criteria for valid resonance structures: All contributing structures must have the same arrangement of atomic nuclei (only electron positions differ); each structure must have the same number of paired and unpaired electrons; and each structure must be a legitimate Lewis structure, following standard bonding/valence rules.

Q6. Distinguish between chain isomerism and position isomerism, with an example of each.
3 Marks
Answer

Chain isomerism arises from differences in the arrangement of the carbon skeleton (chain) itself — for example, n-butane and isobutane (2-methylpropane), both $C_4H_{10}$, differ in whether the carbon chain is straight or branched.

Position isomerism arises from differences in the position of a substituent or functional group along an otherwise identical carbon skeleton — for example, 1-propanol and 2-propanol, both $C_3H_8O$, differ only in whether the hydroxyl group is on the terminal or middle carbon.

Q7. Explain the "+I" (electron-donating) and "−I" (electron-withdrawing) inductive effects, giving one example group for each.
3 Marks
Answer

A group showing a +I effect donates electron density towards the rest of the molecule through the sigma bond framework, relative to hydrogen — alkyl groups (like $-CH_3$) are common examples, being electron-donating relative to hydrogen.

A group showing a −I effect withdraws electron density from the rest of the molecule through the sigma bond framework, relative to hydrogen — halogens (like $-Cl$) and groups like $-NO_2$ or $-COOH$ are common examples, being electron-withdrawing due to their higher electronegativity.

Q8. What is meant by "electromeric effect"? How does it differ from the inductive effect in terms of when it operates?
3 Marks
Answer

The electromeric effect is a temporary effect involving the complete transfer/shift of a pi-bond's electron pair to one of the two bonded atoms, occurring only in the presence of an attacking reagent, and only for the specific duration of the reaction (it disappears immediately once the reagent is removed).

This differs from the inductive effect, which is a permanent effect present in the molecule at all times (whether or not a reagent is attacking) — the electromeric effect is specifically a temporary, reagent-induced polarisation, contrasted with the inductive effect's constant, inherent nature.

Q9. Explain why carbanions are generally stabilised by electron-withdrawing groups, while carbocations are stabilised by electron-donating groups.
3 Marks
Answer

A carbanion carries a negative charge (excess electron density) — any adjacent electron-withdrawing group helps disperse/delocalise this excess negative charge away from the carbanion centre, reducing charge concentration and increasing stability.

A carbocation carries a positive charge (electron deficiency) — any adjacent electron-donating group helps supply additional electron density towards the electron-deficient carbocation centre, reducing the charge deficiency and increasing stability. This opposite stabilising requirement (electron-withdrawing for carbanions, electron-donating for carbocations) directly reflects the opposite nature of the charge each species carries.

Q10. Define the term "reaction intermediate," and explain why identifying the likely intermediate is important for predicting reaction mechanisms.
3 Marks
Answer

A reaction intermediate is a transient (short-lived) species (such as a carbocation, carbanion, or free radical) formed during one step of a multi-step reaction mechanism, which then reacts further to eventually form the final, stable product.

Correctly identifying the likely intermediate is crucial because different reaction conditions/reagents favour the formation of different types of intermediates (carbocation, carbanion, or radical) — since each type of intermediate has distinctly different stability trends, reactivity patterns, and structural preferences, correctly predicting which intermediate is likely to form allows chemists to accurately predict the overall reaction pathway, rate, and expected product(s).

Section C

📘 Long Answer / Numerical Questions

10 questions · 5 marks each · Detailed step-by-step solutions

Q1. Write the IUPAC name for the compound $CH_3-CH(CH_3)-CH_2-CH_3$, explaining the steps used to derive it.
5 Marks
Answer

Step 1: Identify the longest continuous carbon chain — here, it is 4 carbons long (a butane chain), with a methyl branch.

Step 2: Number the chain to give the substituent (methyl group) the lowest possible locant. Numbering from either end gives the methyl group position 2 (since the chain is $CH_3-CH(CH_3)-CH_2-CH_3$: C1-C2-C3-C4, with the branch at C2).

Step 3: Name the compound: 2-methylbutane

Q2. Explain, with structures, the different types of isomerism (chain, position, and functional) exhibited by compounds with molecular formula $C_4H_{10}O$, giving one example structure for each type.
5 Marks
Answer

Chain isomerism (differing carbon skeleton, same functional group — alcohol): $CH_3CH_2CH_2CH_2OH$ (1-butanol, straight chain) versus $(CH_3)_2CHCH_2OH$ (2-methyl-1-propanol, branched chain).

Position isomerism (same skeleton, OH group at different position): $CH_3CH_2CH_2CH_2OH$ (1-butanol) versus $CH_3CH_2CH(OH)CH_3$ (2-butanol).

Functional isomerism (different functional group entirely — an ether instead of alcohol): $CH_3CH_2CH_2CH_2OH$ (1-butanol, an alcohol) versus $CH_3CH_2-O-CH_2CH_3$ (diethyl ether, an ether) — both share the molecular formula $C_4H_{10}O$, but belong to entirely different classes of compounds.

Q3. Arrange the following carbocations in order of increasing stability, with reasoning: methyl cation ($CH_3^+$), primary ($CH_3CH_2^+$), secondary ($(CH_3)_2CH^+$), tertiary ($(CH_3)_3C^+$).
5 Marks
Answer

Carbocation stability increases with the number of adjacent alkyl groups, due to increasing hyperconjugative and inductive electron-donation stabilising the positive charge.

Order of increasing stability: Methyl < Primary < Secondary < Tertiary

The methyl cation has no adjacent alkyl groups (least stable); the tertiary cation has three adjacent alkyl groups, providing the maximum hyperconjugative and inductive stabilisation (most stable) among these four.

Q4. Explain, using resonance structures, why the carboxylate ion ($RCOO^-$) has both C-O bonds of equal length, intermediate between a single and double bond.
5 Marks
Answer

The carboxylate ion can be represented by two equivalent resonance structures: one with the negative charge (and single bond) on one oxygen and a double bond to the other oxygen, and the second structure with these positions reversed (double bond and negative charge swapped between the two oxygens).

Since these two resonance structures are equally valid and energetically equivalent, the actual carboxylate ion exists as a genuine resonance hybrid of these two structures — this means the negative charge and the "extra" pi-bond character are equally delocalised across both oxygen atoms, resulting in both C-O bonds being identical, with a bond length and bond order (approximately 1.5) intermediate between a pure single bond and pure double bond.

Q5. Explain, with reasoning, why phenol ($C_6H_5OH$) is a stronger acid than a simple aliphatic alcohol like ethanol, using resonance concepts.
5 Marks
Answer

When phenol loses a proton ($H^+$) to form the phenoxide ion, the resulting negative charge on oxygen can be delocalised into the adjacent benzene ring through resonance — the phenoxide ion is stabilised by several resonance structures distributing the negative charge onto specific carbon atoms of the ring (ortho and para positions), in addition to the oxygen itself.

Ethanol's conjugate base (ethoxide ion), lacking any adjacent conjugated pi system, has no such resonance stabilisation available — its negative charge remains fully localised on the single oxygen atom.

Since the phenoxide ion is considerably more stabilised (via resonance delocalisation) than the ethoxide ion, phenol's conjugate base is comparatively more stable, making phenol's loss of a proton more thermodynamically favourable — this greater stabilisation of phenol's conjugate base is precisely why phenol is a significantly stronger acid than simple aliphatic alcohols like ethanol.

Q6. Explain, with mechanism-level reasoning, why nitrobenzene is significantly less reactive towards electrophilic aromatic substitution than benzene itself.
5 Marks
Answer

The nitro group ($-NO_2$) is strongly electron-withdrawing, both through the inductive effect (due to the highly electronegative nitrogen and oxygen atoms) and through resonance (the nitro group can withdraw electron density from the ring via delocalisation into its own pi system).

This combined electron-withdrawal significantly reduces the overall electron density of the benzene ring in nitrobenzene, compared to unsubstituted benzene.

Since electrophilic aromatic substitution mechanistically requires the (relatively electron-rich) aromatic ring to attack an incoming electrophile, this substantially reduced electron density in nitrobenzene makes it considerably less reactive (a much weaker nucleophile towards electrophiles) than unsubstituted benzene, explaining nitrobenzene's markedly slower rate of electrophilic aromatic substitution reactions.

Q7. Identify and explain the type of isomerism shown between $CH_3-CH_2-OH$ (ethanol) and $CH_3-O-CH_3$ (dimethyl ether), both having molecular formula $C_2H_6O$.
5 Marks
Answer

Both compounds share the identical molecular formula $C_2H_6O$, but belong to entirely different classes of organic compounds — ethanol is an alcohol (containing an $-OH$ functional group), while dimethyl ether is an ether (containing a $C-O-C$ functional group linkage).

This is an example of functional isomerism — the two compounds have completely different characteristic functional groups, despite sharing the same overall molecular formula, resulting in markedly different chemical properties (e.g., ethanol undergoes hydrogen bonding and has a notably higher boiling point than dimethyl ether, despite their identical molecular formula and similar molecular weight).

Q8. Explain why allylic and benzylic carbocations are exceptionally stable, using resonance concepts.
5 Marks
Answer

An allylic carbocation (adjacent to a C=C double bond) or a benzylic carbocation (adjacent to a benzene ring) has its empty p-orbital directly aligned/conjugated with an adjacent pi-electron system.

This alignment allows the positive charge to be delocalised through resonance across multiple atoms — for an allylic cation, the positive charge can be represented as residing on either of two different carbon atoms (via two equivalent resonance structures); for a benzylic cation, the positive charge can be further delocalised into the aromatic ring itself, spreading across several ring positions through multiple resonance structures.

This extensive delocalisation of the positive charge over multiple atoms (rather than being concentrated on a single carbon, as in a simple, non-conjugated carbocation) substantially reduces the localised charge density, providing considerable additional stabilisation — this is precisely why allylic and benzylic carbocations are notably more stable than comparable simple alkyl carbocations of the same substitution level.

Q9. Explain the difference between geometrical isomerism and optical isomerism, giving one example of each.
5 Marks
Answer

Geometrical isomerism arises from restricted rotation around a bond (typically a C=C double bond, or within a ring structure), giving rise to distinct spatial arrangements (cis and trans, or E and Z) of substituent groups relative to this restricted bond — example: but-2-ene exists as cis-but-2-ene and trans-but-2-ene, differing in whether the two methyl groups are on the same side or opposite sides of the double bond.

Optical isomerism arises when a molecule (typically containing a chiral/asymmetric carbon centre, bonded to four different groups) and its mirror image are non-superimposable on each other, giving rise to two enantiomers that rotate plane-polarised light in opposite directions — example: lactic acid ($CH_3CH(OH)COOH$) has one chiral carbon centre and exists as a pair of non-superimposable optical isomers (enantiomers), designated as D- and L-lactic acid.

Q10. Explain, using the concept of electronic effects, why trichloroacetic acid ($CCl_3COOH$) is a considerably stronger acid than acetic acid ($CH_3COOH$).
5 Marks
Answer

The three chlorine atoms in trichloroacetic acid exert a strong, cumulative electron-withdrawing inductive ($-I$) effect on the molecule, due to chlorine's high electronegativity.

This electron-withdrawal helps stabilise the resulting carboxylate anion (conjugate base) formed after loss of the acidic proton, by helping to further disperse/delocalise the negative charge away from the carboxylate oxygen atoms and towards the electron-withdrawing chlorine substituents.

Since a more stabilised conjugate base corresponds to a stronger acid (greater tendency to lose the proton, i.e., a more favourable/complete dissociation equilibrium), and trichloroacetic acid's conjugate base is considerably more stabilised (due to the three strongly electron-withdrawing chlorine atoms) than acetic acid's simple, unstabilised conjugate base (with only electron-donating methyl groups present, if any effect at all) — this substantial difference in conjugate base stability is precisely why trichloroacetic acid is a markedly stronger acid than plain acetic acid.

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