Some Basic Principles of Organic Chemistry — NEET Chemistry MCQs with answers

Some Basic Principles of Organic Chemistry is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 811 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

IUPAC nomenclature · Isomerism · Electronic effects · Reaction intermediates and stability · Bond fission and reaction types · IUPAC Nomenclature · Electronic Effects · Purification Techniques · Qualitative Analysis · Reaction Intermediates · Quantitative Analysis · Structure and Bonding · IUPAC Nomenclature · Isomerism · Reaction mechanisms · Nomenclature · Reactive Intermediates · Reaction Mechanisms · Nomenclature and isomerism · Aromaticity · Mechanism types · Hybridisation and geometry · Physical properties · Electronic Displacement Effects · Reaction Mechanism Basics · Bond Fission · Resonance · Inductive Effect · Electromeric and Resonance Effects · Hyperconjugation · Molecular Formula · Acid–Base Strength · Resonance Energy · Reaction Mechanism · Reaction Kinetics · Isomerism + Electronic Effects · Aromaticity + Mechanism · Nomenclature + Isomerism · Reactive Intermediates + Kinetics · Electronic Effects + Nomenclature

8 free Some Basic Principles of Organic Chemistry practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Electronic effects

   Which statement correctly describes the nature of a resonance hybrid?

   • A.The molecule rapidly oscillates between the different contributing (canonical) structures.
   • C.Contributing structures differ only in the arrangement of electrons, not atoms; the actual molecule is a single hybrid, not an equilibrium mixture of the contributors.
   • B.Each contributing structure is a distinct, isolable tautomer of the molecule.
   • D.The resonance hybrid is always higher in energy than the least stable contributing structure.
   Show answer and explanation

   Answer: C. Contributing structures differ only in the arrangement of electrons, not atoms; the actual molecule is a single hybrid, not an equilibrium mixture of the contributors.

   Resonance structures are hypothetical electron-arrangement variants of one real structure (the hybrid); atoms never move, and the hybrid is more stable than any single contributor.

   Resonance describes delocalised bonding using several Lewis structures ('canonical forms') that differ only in electron placement, with identical positions of nuclei. The real molecule is best represented as a weighted hybrid of these forms, not as an equilibrium or oscillation between them. The hybrid is always lower in energy (more stable) than any individual contributing structure — this stabilisation is called resonance energy.

   Common mistake: Thinking resonance structures are real, rapidly-interconverting isomers rather than a single delocalised hybrid.

2. Question 2 · medium · Electronic effects

   Consider the following statements about drawing and evaluating resonance structures: (I) Contributing structures differ only in the arrangement of electrons; the positions of all atomic nuclei remain unchanged. (II) A contributing structure with complete octets on all atoms and no charge separation is generally more stable (contributes more) than one with incomplete octets or separated charges. (III) Two adjacent atoms bearing like (same-sign) formal charges make a contributing structure more stable than one without such adjacent like charges. (IV) A structure with a negative charge placed on a more electronegative atom is a more stable (greater) contributor than one placing it on a less electronegative atom. Which of these statements are correct?

   • D.I, II and IV only
   • B.I, II, III and IV
   • C.II and III only
   • A.I and III only
   Show answer and explanation

   Answer: D. I, II and IV only

   Resonance structures share fixed nuclear positions (I, correct); more stable contributors have complete octets and no charge separation (II, correct) and place negative charge on the more electronegative atom (IV, correct); adjacent like charges destabilise a structure, so III is false.

   Statement I is a foundational rule of resonance: only electron positions differ between contributing structures, never the positions of atoms. Statement II is correct: structures with all atoms having complete octets and no formal charges are lower in energy and contribute more to the hybrid. Statement III is incorrect: placing like charges on adjacent atoms increases electrostatic repulsion between them, which destabilises (not stabilises) a contributing structure — such structures contribute very little. Statement IV is correct: for a given amount of charge separation, having negative charge on the more electronegative atom (and positive charge on the less electronegative one) is more favourable and gives a more significant contributor. Hence I, II and IV are correct; III is false.

   Common mistake: Believing adjacent like formal charges stabilise a resonance structure, when in fact they destabilise it through electrostatic repulsion.

3. Question 3 · medium · Electronic effects

   Species X has 5 possible resonance structures, but 4 of them carry separated formal charges and incomplete octets. Species Y has only 2 resonance structures, both fully bonded, with no charge separation and complete octets on all atoms. A student concludes X must be more resonance-stabilised than Y simply because it has more resonance structures. What is the correct assessment?

   • A.The student is correct; the total number of resonance structures is always the sole determinant of resonance stabilisation.
   • B.Neither is resonance-stabilised, since resonance stabilisation only applies to charged species.
   • C.X is definitely more stable because having more resonance structures of any kind always lowers a molecule's energy more than having fewer.
   • D.The student's reasoning is flawed: what matters is the energy (quality) of each contributing structure, not just their count; Y's two low-energy, fully-bonded structures may contribute substantial, comparable stabilisation, while X's four high-energy minor contributors (charge-separated, incomplete octets) add relatively little, so X is not necessarily more stabilised than Y.
   Show answer and explanation

   Answer: D. The student's reasoning is flawed: what matters is the energy (quality) of each contributing structure, not just their count; Y's two low-energy, fully-bonded structures may contribute substantial, comparable stabilisation, while X's four high-energy minor contributors (charge-separated, incomplete octets) add relatively little, so X is not necessarily more stabilised than Y.

   Resonance stabilisation depends on the number and, crucially, the energy (stability) of significant contributing structures — low-energy, near-equivalent structures (as in Y) can contribute more meaningfully than several minor, high-energy structures (as in X), so raw structure count alone does not decide which species is more stabilised.

   The extent of resonance stabilisation is determined by how many low-energy, significant contributing structures a species has and how close in energy they are to each other and to the hybrid — not merely by counting every conceivable resonance structure. High-energy contributors (with charge separation and/or incomplete octets) contribute only a small amount to the overall hybrid and add correspondingly little stabilisation, even though they are technically valid resonance structures. A species with fewer but comparably low-energy, fully-bonded, equivalent contributing structures (as in Y, analogous to e.g. the two equivalent Kekule structures of benzene, or a symmetric carboxylate) can therefore be at least as well stabilised as — or better stabilised than — a species with more numerous but mostly minor, high-energy contributors. The student's rule ('more structures = more stable') is a common oversimplification and does not hold in general.

   Common mistake: Assuming resonance stabilisation is proportional simply to the number of resonance structures that can be drawn.

4. Question 4 · easy · Electronic effects

   Consider the following statements about the inductive effect: (I) It operates through sigma bonds only. (II) Its magnitude decreases rapidly with distance from the substituent and is negligible beyond the third or fourth carbon. (III) It is a permanent, always-present electronic effect (not a temporary polarization). (IV) It requires a continuous system of alternating double and single bonds to operate. Which statements are correct?

   • A.I, II, III and IV
   • B.I, II and III only
   • C.II and IV only
   • D.I and IV only
   Show answer and explanation

   Answer: B. I, II and III only

   The inductive effect transmits through sigma bonds, weakens sharply with distance, and is a permanent polarization of the sigma framework — statement IV describes conjugation/resonance, not induction.

   The inductive effect arises from electronegativity differences polarizing sigma bonds along a chain; it is transmitted through sigma bonds (I, correct), falls off quickly so it is usually negligible beyond 3-4 bonds away (II, correct), and being a property of the sigma-bond framework it is permanent, unlike temporary effects (III, correct). Statement IV describes the resonance/mesomeric effect, which requires a conjugated (alternating multiple/single bond or lone pair-adjacent) pi system — this is not a requirement of the inductive effect, so IV is false.

   Common mistake: Conflating the structural requirement for resonance (conjugated pi system) with the inductive effect, which needs no such conjugation

5. Question 5 · medium · Electronic effects

   Consider the following statements about hyperconjugation: (I) It is also called 'no-bond resonance' because one canonical structure shows a broken C-H (or C-C) sigma bond. (II) It requires at least one C-H (or C-C) sigma bond on a carbon directly attached to an atom bearing a p-orbital, positive charge, or forming a pi bond. (III) A greater number of alpha hydrogens leads to greater hyperconjugative stabilization. (IV) Hyperconjugation can stabilize carbocations, free radicals, and alkenes, but never carbanions. Which statements are correct?

   • A.I, II, III and IV
   • B.I, II and III only
   • C.II and III only
   • D.I, III and IV only
   Show answer and explanation

   Answer: B. I, II and III only

   Hyperconjugation is correctly described as no-bond resonance (I), requires an alpha C-H/C-C sigma bond adjacent to a p-orbital/cationic center/pi bond (II), and stabilization increases with the number of alpha hydrogens (III); statement IV is an overreach since hyperconjugation is conventionally discussed for cations, radicals and alkenes (not routinely for carbanions) at NEET level, so a blanket 'never' claim about carbanions is not a standard verified fact to assert as correct here.

   Hyperconjugation involves delocalization of electrons from a sigma C-H (or C-C) bond into an adjacent empty or partially empty p-orbital, pi bond, or radical orbital, via a resonance structure that formally shows the sigma bond broken — hence 'no-bond resonance' (I, correct). It requires the sigma-bonded carbon to be directly attached to the electron-deficient/pi-system carbon (II, correct). More alpha C-H bonds means more hyperconjugative resonance structures and greater stabilization (III, correct), which is why 3° cations/radicals/alkyl-substituted alkenes are most stabilized. At NEET level, hyperconjugation is taught specifically for carbocations, free radicals, and alkenes; it is not a standard emphasized stabilizing effect for carbanions, but asserting an absolute universal 'never' (IV) goes beyond what is reliably established at this level, so IV should not be marked correct.

   Common mistake: Accepting an absolute, over-general statement as fact without checking it against the specific NCERT-level applications actually taught

6. Question 6 · easy · Reaction intermediates and stability

   Which factor primarily stabilises a free radical, similar to how it stabilises a carbocation?

   • A.Complete octet on the radical carbon
   • B.Hyperconjugation and inductive electron release from adjacent alkyl groups
   • C.Formation of a full negative charge on the radical carbon
   • D.Strong electron-withdrawing groups directly attached to the radical carbon
   Show answer and explanation

   Answer: B. Hyperconjugation and inductive electron release from adjacent alkyl groups

   Like carbocations, alkyl free radicals are stabilised by hyperconjugation (delocalisation of the unpaired electron via adjacent C–H sigma bonds) and inductive electron donation, giving the order 3° > 2° > 1° > methyl radical.

   A free radical has an odd (unpaired) electron on carbon, giving it 7 valence electrons around that carbon rather than 8. Adjacent alkyl groups stabilise this electron-deficient-in-electron-count centre through hyperconjugation (donation of electron density from neighbouring C–H sigma bonds into the singly occupied orbital) and a mild +I inductive effect, exactly analogous to carbocation stabilisation. Hence radical stability follows the same order as carbocations: tertiary > secondary > primary > methyl.

   Common mistake: Confusing the electronic requirements of a radical (odd electron) with those of a carbanion or assuming EWGs stabilise radicals the way alkyl groups do

7. Question 7 · medium · Reaction intermediates and stability

   Consider the following statements about reactive intermediates in organic chemistry. (I) A carbocation is electron-deficient (six electrons around carbon) and is stabilised by electron-donating groups. (II) A carbanion carries a complete octet including a lone pair and is stabilised by electron-withdrawing groups and by increased s-character. (III) A free radical has an odd number of electrons around carbon and is generally stabilised by the same structural features (alkyl substitution, resonance) as a carbocation. Which of these statements are correct?

   • A.Only I and III are correct; II is wrong because carbanions are stabilised by electron-donating groups, not withdrawing groups
   • B.Only I is correct; II and III both misstate the electronic requirements of carbanions and radicals
   • C.Only II and III are correct; I is wrong because carbocations are stabilised by electron-withdrawing groups, not donating groups
   • D.I, II and III are all correct
   Show answer and explanation

   Answer: D. I, II and III are all correct

   Carbocations (6 electrons, electron-deficient) are stabilised by electron donation; carbanions (8 electrons, lone pair present) are stabilised by electron withdrawal and higher s-character; free radicals (7 electrons, odd electron) follow largely the same stabilisation trends as carbocations (alkyl substitution, resonance) — all three statements are standard and correct.

   Statement I: A carbocation has only six electrons around the positively charged carbon (electron sextet) and is stabilised by anything that supplies electron density toward that centre — alkyl groups (+I, hyperconjugation) or adjacent pi systems (resonance) — correct. Statement II: A carbanion has a full octet, including a non-bonding lone pair carrying the negative charge; it is stabilised by factors that help accommodate or disperse this excess electron density — electron-withdrawing groups (which pull density away, e.g., adjacent carbonyl/nitro groups via resonance or induction) and higher s-character (which holds the lone pair closer to the nucleus, effectively increasing electronegativity) — correct. Statement III: A free radical has an odd number of electrons (seven) around the carbon; despite the different electron count and lack of formal charge, its stabilisation trend closely parallels that of carbocations — alkyl substitution via hyperconjugation, and resonance with adjacent pi systems (allylic/benzylic radicals) — correct. All three statements are therefore correct and consistent with standard NEET-level organic chemistry.

   Common mistake: Reversing the stabilising influence (electron-donating vs withdrawing) between carbocations and carbanions, or assuming radical stabilisation trends are unrelated to carbocation trends

8. Question 8 · medium · Reaction intermediates and stability

   Rank the following alkyl halides in order of decreasing rate of solvolysis (SN1, e.g. in aqueous ethanol): (I) 2-chloro-2-methylpropane (II) 2-chloropropane (III) 1-chloropropane (IV) chloromethane

   • A.IV > III > II > I
   • B.I > II > III > IV
   • C.I > III > II > IV
   • D.II > I > III > IV
   Show answer and explanation

   Answer: B. I > II > III > IV

   SN1 rate depends on the stability of the carbocation intermediate formed in the rate-determining ionisation step; stability order tertiary > secondary > primary > methyl directly gives the rate order I > II > III > IV.

   The rate-determining step of SN1 solvolysis is ionisation of C-X to form a carbocation. Faster carbocation formation (lower activation energy) corresponds to a more stable carbocation product, following Hammond's postulate for this exothermic-ish, cation-forming step. 2-chloro-2-methylpropane ionises to a tertiary carbocation (most stable, most hyperconjugation/+I donors) and reacts fastest. 2-chloropropane gives a secondary carbocation, next fastest. 1-chloropropane gives a primary carbocation, slower still. Chloromethane would require forming a methyl cation with no adjacent carbon to stabilise it via hyperconjugation or induction, making it by far the slowest (essentially does not undergo SN1 under normal conditions). Hence the order I > II > III > IV correctly tracks carbocation stability.

   Common mistake: Misordering secondary and primary carbocation stability, or assuming SN1 rate is unrelated to intermediate stability

Questions about Some Basic Principles of Organic Chemistry for NEET

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