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Inductive, electromeric, resonance and hyperconjugation effects: NEET MCQs with answers

50 questions on this topic of Some Basic Principles of Organic Chemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following carbocations is exceptionally stable due to active aromatic resonance delocalization?

  1. Cyclopentadienyl cation (C₅H₅⁺)
  2. Cyclopropenyl cation (C₃H₃⁺)
  3. Cycloheptatrienyl anion (C₇H₇⁻)
  4. Bicyclo[2.2.1]hept-1-yl cation
Show answer and explanation

Answer: (B) Cyclopropenyl cation (C₃H₃⁺)

The cyclopropenyl cation (C₃H₃⁺) is planar, completely conjugated, and possesses 2 π electrons, obeying Huckel's (4n+2) rule (where n = 0). This confers extra aromatic stability through resonance, whereas cyclopentadienyl cation is anti-aromatic.

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2. Which of the following compounds exhibits the highest heat of hydrogenation per mole?

  1. cis-But-2-ene
  2. trans-But-2-ene
  3. Ethene
  4. 2,3-Dimethylbut-2-ene
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Answer: (C) Ethene

The heat of hydrogenation is inversely proportional to alkene stability for alkenes with similar steric factors. Substituted alkenes are stabilized by hyperconjugation and alkyl induction; ethene has zero α-hydrogens (no hyperconjugative stabilization), making it the least stable alkene and thus releasing the maximum heat upon catalytic hydrogenation.

hard

3. Consider the stability of the following carbocations: (P) p-Methoxybenzyl carbocation [p-CH₃O-C₆H₄-CH₂⁺] (Q) p-Methylbenzyl carbocation [p-CH₃-C₆H₄-CH₂⁺] (R) Benzyl carbocation [C₆H₅-CH₂⁺] (S) p-Nitrobenzyl carbocation [p-O₂N-C₆H₄-CH₂⁺] The correct decreasing order of their stability is:

  1. P > R > Q > S
  2. Q > P > R > S
  3. P > Q > R > S
  4. S > R > Q > P
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Answer: (C) P > Q > R > S

The methoxy group (-OCH₃) provides powerful stability through lone-pair resonance (+R effect). The methyl group (-CH₃) stabilizes via hyperconjugation (+H) and weak +I effect. Unsubstituted benzyl carbocation has standard resonance. The nitro group (-NO₂) has a powerful electron-withdrawing effect (-R and -I), strongly destabilizing the carbocation. Thus, P > Q > R > S.

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4. Arrange the following conjugate bases in increasing order of basic strength: (I) ClO₄⁻ (II) CH₃COO⁻ (III) C₆H₅O⁻ (IV) OH⁻

  1. I < III < II < IV
  2. IV < III < II < I
  3. I < II < III < IV
  4. II < I < III < IV
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Answer: (C) I < II < III < IV

Perchloric acid (HClO₄) is an extremely strong mineral acid whose conjugate base ClO₄⁻ is stabilized by resonance over four oxygen atoms, making it the weakest base. Acetic acid is stronger than phenol due to negative charge residing on two equivalent oxygens vs. carbon atoms in phenoxide. Water is weaker acid than phenol, so OH⁻ is the strongest base among these.

easy

5. Which of the following orders of acidity is correct based on electronic displacement effects?

  1. ClCH₂COOH > FCH₂COOH > BrCH₂COOH > ICH₂COOH
  2. ICH₂COOH > BrCH₂COOH > ClCH₂COOH > FCH₂COOH
  3. FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH
  4. FCH₂COOH > BrCH₂COOH > ClCH₂COOH > ICH₂COOH
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Answer: (C) FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH

Halogens exhibit a -I effect that increases with increasing electronegativity: F > Cl > Br > I. Greater -I stabilizes the carboxylate anion via dispersal of negative charge, leading to higher acid strength. Thus, fluoroacetic acid is the strongest acid in the series.

hard

6. Which among the following species has the highest carbon-oxygen bond order?

  1. Acetate ion (CH₃COO⁻)
  2. Carbon dioxide (CO₂)
  3. Carbon monoxide (CO)
  4. Carbonate ion (CO₃²⁻)
Show answer and explanation

Answer: (C) Carbon monoxide (CO)

In carbon monoxide (:C≡O:), the bond order is 3. In carbon dioxide (O=C=O), the bond order is 2. In acetate ion, resonance between two structures gives a bond order of 1.5. In carbonate ion, resonance across three equivalent structures gives a bond order of 1.33 (4/3). Thus, CO has the highest C-O bond order.

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7. Which group shows a +R (+M) resonance effect when attached to a benzene ring?

  1. -CHO
  2. -CN
  3. -OH
  4. -NO₂
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Answer: (C) -OH

Groups with lone pairs on the attached atom, such as -OH, -NH₂ and -OR, donate electrons by resonance (+R). -NO₂, -CHO and -CN withdraw electrons (-R).

NCERT Class 11

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8. Arrange the following canonical structures of vinyl chloride in decreasing order of their stability: (I) CH₂=CH-Cl (II) ⁻CH₂-CH=Cl⁺ (III) ⁺CH₂-CH=Cl⁻

  1. III > I > II
  2. II > I > III
  3. I > III > II
  4. I > II > III
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Answer: (D) I > II > III

Structure (I) is uncharged and has complete octets for all atoms, making it the most stable. Structure (II) has complete octets for all atoms despite formal charges. Structure (III) places a positive charge on carbon with an incomplete octet and has charge separation putting negative charge on halogen without fulfilling octet stability equivalently, making it the least stable.

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9. Which of the following represents the correct order of bond dissociation enthalpy of C-H bonds indicated by hyperconjugation and radical stability?

  1. (CH₃)₂CH-H > (CH₃)₃C-H > CH₃-CH₃ > CH₄
  2. (CH₃)₃C-H > (CH₃)₂CH-H > CH₃-CH₃ > CH₄
  3. CH₃-CH₃ > CH₄ > (CH₃)₂CH-H > (CH₃)₃C-H
  4. CH₄ > CH₃-CH₃ > (CH₃)₂CH-H > (CH₃)₃C-H
Show answer and explanation

Answer: (D) CH₄ > CH₃-CH₃ > (CH₃)₂CH-H > (CH₃)₃C-H

The stability of the resulting free radical dictates the C-H bond dissociation energy (BDE). A more stable free radical formed via extensive hyperconjugation lowers the required cleavage energy. Since radical stability is 3° > 2° > 1° > methyl radical, the C-H BDE follows the inverse order: CH₄ > CH₃-CH₃ > (CH₃)₂CH-H > (CH₃)₃C-H.

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10. Consider the following canonical structures of an acylium-type cation: (I) CH₃-C⁺=O (II) CH₃-C≡O⁺ Which statement correctly compares the relative stabilities and contributions of (I) and (II) to the resonance hybrid?

  1. Structure (II) is more stable because all atoms have complete octets of electrons.
  2. Structure (I) is more stable because positive charge is on the less electronegative carbon atom.
  3. Both structures have equal stability because they have the same number of covalent bonds.
  4. Structure (I) is more stable because oxygen has a higher electronegativity and cannot hold a positive charge.
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Answer: (A) Structure (II) is more stable because all atoms have complete octets of electrons.

In structure (II), both carbon and oxygen possess a complete octet of valence electrons (8 electrons each), whereas in structure (I), carbon has an incomplete octet (6 electrons). A complete octet outweighs the electronegativity rule in determining canonical stability, making structure (II) the more stable and major contributor.

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11. Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R): Assertion (A): Trichloroacetic acid is significantly more acidic than acetic acid. Reason (R): The three highly electronegative chlorine atoms exert a cumulative -I effect, which stabilizes the trichloroacetate anion. In the light of the above statements, choose the correct answer:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

The presence of three electronegative chlorine atoms at the α-carbon provides a strong electron-withdrawing inductive (-I) effect. This disperses the negative charge of the carboxylate group, stabilizing the conjugate base and making trichloroacetic acid much stronger than acetic acid.

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12. Among the following carbanions, which one is the most stable?

  1. CH₃⁻
  2. CH₃-CH₂⁻
  3. HC≡C⁻
  4. (CH₃)₂CH⁻
Show answer and explanation

Answer: (C) HC≡C⁻

In the acetylide anion HC≡C⁻, the non-bonding electron pair occupies an sp-hybrid orbital having 50% s-character. Since s-electrons are held more tightly to the nucleus due to higher effective electronegativity of sp-carbon, HC≡C⁻ accommodates negative charge far better than sp³-hybridized alkyl carbanions.

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