Organic Chemistry Quiz: Resonance Formal Charge And Stability Trends

What this quiz covers

This quiz focuses on Resonance Formal Charge And Stability Trends, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺ (correct answer)
2. (CH₃)₃C⁺ < (CH₃)₂CH⁺ < CH₃CH₂⁺ < CH₃⁺
3. CH₃CH₂⁺ < CH₃⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺
4. CH₃⁺ < (CH₃)₂CH⁺ < CH₃CH₂⁺ < (CH₃)₃C⁺

Explanation: Carbocation stability increases with the degree of substitution due to hyperconjugation and inductive effects from alkyl groups. The order from least to most stable is: methyl (primary) < ethyl (primary) < isopropyl (secondary) < tert-butyl (tertiary). Each additional alkyl group provides electron density to stabilize the positive charge through hyperconjugation and inductive donation.

Question 2

1. Benzyl radical (C₆H₅CH₂•)
2. Allyl radical (CH₂=CHCH₂•)
3. tert-Butyl radical ((CH₃)₃C•)
4. Vinyl radical (CH₂=CH•) (correct answer)

Explanation: The vinyl radical is the least stable because the unpaired electron is in an sp² orbital, which has higher s-character and is less stable than sp³ radicals. Additionally, vinyl radicals cannot be stabilized by resonance in the same way as benzyl or allyl radicals. The benzyl and allyl radicals are stabilized by extensive resonance delocalization, while the tert-butyl radical benefits from hyperconjugation with nine C-H bonds.

Question 3

1. The ion contains 6 π electrons, satisfying Hückel's rule for aromaticity in a seven-membered ring system
2. All seven carbon atoms are equivalent, allowing complete delocalization of the positive charge around the ring
3. The ion adopts a planar geometry that maximizes orbital overlap for π bonding throughout the ring system
4. Hyperconjugation from the seven C-H bonds provides additional stabilization beyond the aromatic stabilization energy (correct answer)

Explanation: While the tropylium ion is exceptionally stable due to aromaticity (6 π electrons in a cyclic, planar system with complete conjugation), hyperconjugation from C-H bonds is not a significant contributor to its stability. The aromatic stabilization vastly outweighs any hyperconjugative effects. Choices A, B, and C all correctly describe factors that contribute to the ion's remarkable stability through aromatic delocalization.

Question 4

1. The tert-butyl cation, [(CH₃)₃C⁺]
2. The allyl anion, [CH₂=CH-CH₂⁻]
3. The benzyl radical, [C₆H₅-CH₂•]
4. The acetate ion, [CH₃-COO⁻] (correct answer)

Explanation: This question compares different types of reactive intermediates and their relative stabilities. A) tert-Butyl cation is a relatively stable carbocation due to hyperconjugation, but carbocations are inherently high-energy with incomplete octets. B) Allyl anion has resonance stabilization, but carbanions are generally less stable than corresponding anions on electronegative atoms. C) Benzyl radical is resonance-stabilized through the aromatic ring, making it more stable than typical alkyl radicals. D) Acetate ion is highly stabilized by resonance between two equivalent oxygen atoms, and the negative charge resides on highly electronegative oxygen. The combination of optimal charge placement and strong resonance makes acetate the most stable species.

Question 5

1. The nitrogen atom in nitromethane is more electronegative than carbon, making the C-H bonds more polar and easier to break
2. Nitromethane has more total bonds than methane, distributing electron density more effectively throughout the molecule
3. The conjugate base of nitromethane is stabilized by resonance delocalization involving the nitro group's π system (correct answer)
4. The sp³ hybridization of carbon in nitromethane is destabilized by the electron-withdrawing nitro group attached to it

Explanation: When you encounter questions about relative acidity in organic chemistry, always think about what happens to the conjugate base after deprotonation. The more stable the conjugate base, the more acidic the original compound. Nitromethane (CH₃NO₂) is dramatically more acidic than methane because when it loses a proton from the methyl group, the resulting carbanion can be stabilized through resonance. The nitromethanoate anion (⁻CH₂NO₂) has its negative charge delocalized across the entire nitro group through π-electron resonance. You can draw resonance structures showing the negative charge distributed between the carbon and the oxygen atoms of the nitro group. This extensive delocalization makes the conjugate base much more stable than methane's conjugate base (⁻CH₃), which has no resonance stabilization available. Looking at the wrong answers: Choice A incorrectly focuses on bond polarity rather than conjugate base stability. While the nitro group does affect C-H bond polarity slightly, this isn't the primary reason for the dramatic acidity difference. Choice B makes no chemical sense - having more bonds doesn't relate to acidity in any meaningful way. Choice D misunderstands hybridization; the carbon atoms in both molecules are sp³ hybridized, and hybridization isn't the key factor here. Study tip: For acidity questions, always ask "What stabilizes the conjugate base?" Look for resonance opportunities, electron-withdrawing groups, and delocalization patterns. The most stable conjugate base corresponds to the strongest acid.

Question 6

1. Phenol has more carbon atoms, providing greater inductive stabilization of the conjugate base through electron withdrawal
2. The phenoxide ion is stabilized by resonance delocalization of the negative charge into the aromatic ring system (correct answer)
3. Cyclohexanol is a secondary alcohol while phenol is a primary alcohol, making phenol's conjugate base inherently more stable
4. The sp² hybridization of phenol's oxygen makes it more electronegative than the sp³ oxygen in cyclohexanol

Explanation: The phenoxide ion is significantly stabilized by resonance delocalization, where the negative charge on oxygen can be delocalized into the aromatic ring through multiple resonance structures. This extensive delocalization makes phenol much more acidic than cyclohexanol. Choice A is incorrect because inductive effects alone cannot explain the large difference in acidity. Choice C is wrong because both are technically primary alcohols (OH attached to carbon with one other carbon). Choice D is incorrect because the oxygen hybridization is not the primary factor.