Organic Chemistry Quiz: Intermediates Carbocations Carbanions Radicals

What this quiz covers

This quiz focuses on Intermediates Carbocations Carbanions Radicals, 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. A secondary carbocation at C2, formed by initial protonation.
2. A tertiary carbocation at C2, formed via a 1,2-methyl shift. (correct answer)
3. A primary carbocation at C1, which is rapidly captured by water.
4. A tertiary carbocation at C3, formed via a 1,2-hydride shift.

Explanation: The mechanism begins with the protonation of the alkene at C1 to form the more stable secondary carbocation at C2, following Markovnikov's rule. This secondary carbocation then undergoes a rapid 1,2-methyl shift from the adjacent quaternary carbon (C3) to the electron-deficient C2. This rearrangement produces a more stable tertiary carbocation at C2. Water then acts as a nucleophile, attacking this final tertiary carbocation.

Question 2

1. The central carbon is sp-hybridized, which lowers its energy.
2. Extreme steric hindrance from the three phenyl groups prevents it from dimerizing or reacting.
3. The unpaired electron is extensively delocalized by resonance into all three phenyl rings. (correct answer)
4. The formation of the radical creates an aromatic system within the molecule.

Explanation: The stability of the trityl radical is due to a combination of resonance and steric effects, but resonance is the dominant electronic factor. The unpaired electron in the p-orbital of the central carbon can be delocalized into the pi systems of all three attached phenyl rings. This extensive delocalization spreads the radical character over many atoms, greatly stabilizing the intermediate. Steric hindrance (Choice B) contributes to its kinetic stability (unreactivity), but resonance (Choice C) is the reason for its thermodynamic stability.

Question 3

1. The tertiary carbocation forms faster due to better leaving group ability of chloride from the more substituted carbon center
2. Steric hindrance in the benzyl system slows the rate-determining ionization step despite the more stable carbocation intermediate
3. The C-Cl bond in tert-butyl chloride is weaker due to hyperconjugative weakening, leading to faster ionization despite the less stable carbocation (correct answer)
4. Solvation effects favor the tertiary carbocation in aqueous ethanol, overcoming the intrinsic stability difference between the intermediates

Explanation: Hyperconjugation from the three methyl groups in tert-butyl chloride weakens the C-Cl bond, making ionization faster even though the resulting carbocation is less stable than the benzylic carbocation. The rate depends on the activation energy for ionization, not just the stability of the product carbocation. Choice A is incorrect because leaving group ability doesn't depend on substitution pattern of the carbon. Choice B is wrong because benzyl chloride has less steric hindrance than tert-butyl chloride. Choice D incorrectly suggests that solvation can overcome such a large intrinsic stability difference.

Question 4

1. A tertiary carbocation at C1, leading to Markovnikov addition.
2. A bromine atom adds to C1 to form a more stable tertiary radical at C2.
3. A bromine atom adds to C2 to form a more stable tertiary radical at C1. (correct answer)
4. A hydrogen radical adds to C1 to form a more stable tertiary radical at C2.

Explanation: In the anti-Markovnikov addition of HBr via a radical mechanism, the first propagation step involves the addition of a bromine radical (Br•) to the alkene. The Br• will add to the less substituted carbon (C2 of 1-methylcyclohexene) to generate the more stable carbon radical at the more substituted position (the tertiary radical at C1). This tertiary radical then abstracts a hydrogen from HBr in the second propagation step to form the final product.

Question 5

1. This intermediate is relatively stable due to the high s-character of the sp2-hybridized orbital containing the lone pair.
2. This intermediate is highly unstable because the lone pair resides in an sp2 orbital, and there is no resonance stabilization. (correct answer)
3. This intermediate is resonance-stabilized by the adjacent pi bond, making it moderately stable and a plausible intermediate.
4. This intermediate's stability is comparable to a primary alkyl carbanion due to similar inductive effects.

Explanation: A vinylic carbanion has a negative charge on a double-bonded carbon. The lone pair of electrons resides in an sp2-hybridized orbital, which has more s-character than an sp3 orbital but less than an sp orbital. While the higher s-character relative to sp3 provides some stabilization compared to alkyl carbanions, vinylic carbanions are still highly unstable species. They lack resonance stabilization and represent high-energy intermediates that are generally not formed in typical introductory organic reactions.

Question 6

1. A secondary carbocation formed by a 1,2-hydride shift.
2. A tertiary carbocation formed by a 1,2-methyl shift. (correct answer)
3. The initial primary carbocation, which is trapped by water before it can rearrange.
4. A secondary radical formed by homolytic cleavage of the C-N bond.

Explanation: The reaction generates an unstable primary carbocation (the neopentyl cation). This carbocation will rapidly rearrange to a more stable form. The adjacent carbon is quaternary, so there are no hydrogens to shift. A 1,2-methyl shift will occur, where one of the methyl groups on the quaternary carbon migrates to the primary carbocation center. This shift results in the formation of a much more stable tertiary carbocation, which is then captured by water to form the final alcohol product.

Question 7

1. 1-bromo-3-methylbutane
2. 1-bromo-2,2-dimethylpropane
3. 2-bromo-3-methylbutane (correct answer)
4. 2-bromo-2-methylbutane

Explanation: When you encounter a solvolysis problem involving alkyl halides, focus on the mechanism and product stability. Solvolysis typically proceeds through an SN1 mechanism with secondary and tertiary halides, forming carbocation intermediates that can rearrange and eliminate to form alkenes. Since the major product is 2-methyl-2-butene, you need to work backwards to identify which starting material could produce this specific alkene. The most substituted alkene is typically favored (Zaitsev's rule), and 2-methyl-2-butene is a trisubstituted alkene with the structure (CH₃)₂C=CHCH₃. Choice C, 2-bromo-3-methylbutane, is the correct answer because when it undergoes solvolysis, the secondary carbocation can eliminate a proton from the adjacent carbon to form 2-methyl-2-butene as the major product. The elimination follows Zaitsev's rule, favoring the more substituted alkene. Choice A (1-bromo-3-methylbutane) is a primary halide that would likely undergo SN2 rather than solvolysis, and wouldn't easily form the desired alkene. Choice B (1-bromo-2,2-dimethylpropane) is also primary and would face similar mechanistic issues. Choice D (2-bromo-2-methylbutane) is tertiary and while it could undergo solvolysis, elimination would more likely produce 2-methylbut-1-ene or 2-methylbut-2-ene, not the specific 2-methyl-2-butene structure. Remember: when predicting elimination products, draw out the possible alkenes and identify which is most substituted—that's usually your major product under Zaitsev conditions.

Question 8

1. The intermediate is a chiral radical, leading to a single enantiomer of the product through retention of configuration.
2. The intermediate is an achiral, planar radical, leading to a racemic mixture of (R)- and (S)-2-bromo-2-chlorobutane. (correct answer)
3. The intermediate is a rapidly inverting chiral radical, leading to a scalemic mixture with a slight excess of the inverted product.
4. The intermediate is an achiral, planar radical, but the bromine atom blocks one face, leading to a single diastereomer.

Explanation: When the hydrogen is abstracted from the chiral center, a carbon radical is formed. This radical center is sp2-hybridized and trigonal planar (or a very shallow, rapidly inverting pyramid), making it effectively achiral. The incoming chlorine radical can then attack from either the top or bottom face with equal probability. This leads to the formation of both (R)- and (S)-2-bromo-2-chlorobutane in equal amounts, resulting in a racemic mixture.

Question 9

1. Cl• + CH4 → HCl + •CH3
2. •CH3 + Cl2 → CH3Cl + Cl•
3. Cl2 + hv → 2 Cl•
4. •CH3 + Cl• → CH3Cl (correct answer)

Explanation: A termination step is any reaction that consumes radicals without generating new ones. Choice D shows a methyl radical and a chlorine radical combining to form a stable molecule, methyl chloride. This removes two radicals from the reaction mixture. Choice A and B are propagation steps, where one radical is consumed but another is generated, continuing the chain. Choice C is the initiation step, where radicals are created from a non-radical species.

Question 10

1. I > II > III
2. II > I > III (correct answer)
3. III > II > I
4. II > III > I

Explanation: Carbanion stability is determined by several factors. The allyl anion (II) is resonance-stabilized, delocalizing the negative charge over two carbons, which is a very powerful stabilizing effect. The acetylide ion (I) has its lone pair in an sp-hybridized orbital. This orbital has 50% s-character, holding the electrons closer to the nucleus and stabilizing the charge. The ethyl anion (III) has its lone pair in an sp3-hybridized orbital (25% s-character) and is destabilized by the inductive effect of the alkyl group. Resonance is generally a more significant stabilizing factor than hybridization effects, making the allyl anion most stable. The sp-hybridized acetylide is much more stable than the sp3-hybridized ethyl anion. Therefore, the order is II > I > III.

Question 11

1. A primary carbocation that rearranges to a tertiary carbocation via a 1,2-methyl shift. (correct answer)
2. A five-coordinate transition state typical of a concerted E2 reaction.
3. A primary carbanion formed by removal of a proton alpha to the bromine.
4. A radical formed by homolysis of the C-Br bond induced by heat.

Explanation: Neopentyl bromide is a primary halide, but it is too sterically hindered for SN2 or E2 reactions to occur readily. Under forcing conditions (heat) with a base that is also a potential nucleophile (ethoxide), the reaction can be pushed towards an E1/SN1 pathway, even for a primary halide. The leaving group departs to form a very unstable primary carbocation. This carbocation immediately undergoes a 1,2-methyl shift to form a stable tertiary carbocation. A proton is then removed from an adjacent carbon to form the major alkene product (2-methyl-2-butene). The key intermediate leading to this alkene is the rearranged tertiary carbocation.

Question 12

1. The carbanion is stabilized by the inductive effect of the sp-hybridized nitrogen atom.
2. The carbanion is sp2-hybridized, which is inherently more stable than the sp3 carbon of the starting material.
3. The carbanion is stabilized by resonance, delocalizing the negative charge onto the electronegative nitrogen atom. (correct answer)
4. The nitrile group makes the conjugate base aromatic, which is a powerful stabilizing feature.

Explanation: When you encounter questions about acidity in organic chemistry, focus on what stabilizes the conjugate base (the anion formed after deprotonation). The more stable the conjugate base, the more acidic the original compound. The nitrile group's ability to stabilize an adjacent carbanion comes from resonance delocalization. When the C-H bond next to $-C≡N$ loses its proton, the resulting carbanion can delocalize its negative charge through the π system of the triple bond onto the electronegative nitrogen atom. This creates multiple resonance structures where the negative charge is shared between carbon and nitrogen, significantly stabilizing the anion. Answer C correctly identifies this resonance stabilization as the primary factor. Let's examine why the other options miss the mark: Answer A mentions the inductive effect of nitrogen, which does contribute to acidity but is much weaker than resonance effects. The inductive effect alone cannot account for the dramatic increase in acidity observed. Answer B incorrectly focuses on hybridization changes in the carbanion itself, but the key stability comes from charge delocalization, not hybridization differences. Answer D is completely wrong—the conjugate base is not aromatic and doesn't meet any criteria for aromaticity (it's not cyclic, planar, or following Hückel's rule). Study tip: When analyzing acidity, always look first for resonance stabilization of the conjugate base, especially when electron-withdrawing groups with π systems (like carbonyls, nitriles, or nitro groups) are adjacent to the acidic hydrogen. Resonance effects typically outweigh inductive effects in determining relative acidity.

Question 13

1. Benzylic carbocations have additional stabilization from hyperconjugation with the aromatic π-system that is not available to radicals
2. The positive charge in carbocations is more effectively delocalized by the electron-rich aromatic system than the unpaired electron in radicals (correct answer)
3. Radicals require paired electrons for resonance stabilization, while carbocations can be stabilized by empty orbitals in the aromatic ring
4. Benzylic carbocations benefit from both resonance and inductive stabilization, while benzylic radicals only benefit from inductive effects

Explanation: The electron-rich aromatic ring is more effective at stabilizing electron-deficient species (carbocations) through resonance donation than at stabilizing radicals. The carbocation can form resonance structures where the positive charge is delocalized onto the aromatic ring, while radical delocalization is less effective due to the odd electron. Choice A incorrectly describes hyperconjugation as the mechanism. Choice C is wrong because radicals can participate in resonance with unpaired electrons, and carbocations don't use empty orbitals for resonance. Choice D incorrectly states that radicals don't benefit from resonance - they do, just less effectively than carbocations.

Question 14

1. Low base concentration favors SN1 through a benzylic carbocation, while high base concentration favors E2 through a carbanion intermediate
2. Low base concentration favors SN2 with direct displacement, while high base concentration favors E1cB through a stabilized carbanion
3. Both conditions proceed through the same benzylic carbocation, but high base concentration traps the intermediate before rearrangement
4. Low base concentration favors SN1 through a benzylic carbocation, while high base concentration favors E1cB through a benzylic carbanion (correct answer)

Explanation: At low base concentration, ionization to form a stabilized benzylic carbocation (SN1) is faster than the base-mediated elimination. At high base concentration, the strong base can deprotonate the benzylic position to form a resonance-stabilized benzylic carbanion, which then undergoes elimination (E1cB mechanism). Both pathways take advantage of benzylic stabilization but involve different intermediates. Choice A incorrectly suggests E2 rather than E1cB. Choice B suggests SN2, which is unlikely for a benzylic tosylate that can form a stable carbocation. Choice C doesn't explain the different products observed.

Question 15

1. R-CF₃ > R-NO₂ > R-CH₃ > R-OCH₃
2. R-NO₂ > R-CF₃ > R-OCH₃ > R-CH₃ (correct answer)
3. R-OCH₃ > R-CH₃ > R-CF₃ > R-NO₂
4. R-NO₂ > R-CF₃ > R-CH₃ > R-OCH₃

Explanation: Carbanion stability increases with electron-withdrawing groups that can delocalize the negative charge. -NO₂ is the strongest electron-withdrawing group through both resonance and inductive effects, providing maximum stabilization. -CF₃ is strongly electron-withdrawing through inductive effects only. -OCH₃ is electron-withdrawing through inductive effects but can also donate electron density through resonance, making it less effective at stabilizing carbanions. -CH₃ is electron-donating through hyperconjugation, destabilizing the carbanion. Choice A incorrectly places CF₃ above NO₂. Choice C reverses the trend entirely. Choice D incorrectly ranks OCH₃ as least stabilizing, missing that CH₃ is electron-donating.

Question 16

1. benzyl anion > allyl anion > propargyl anion (correct answer)
2. propargyl anion > allyl anion > benzyl anion
3. allyl anion > benzyl anion > propargyl anion
4. benzyl anion > propargyl anion > allyl anion

Explanation: All three carbanions are stabilized by resonance delocalization. The benzyl anion delocalizes the negative charge over multiple positions in the aromatic ring, providing extensive stabilization. The allyl anion delocalizes the charge over two positions. The propargyl anion has resonance forms, but the delocalization is less effective because one resonance contributor places negative charge on an sp-hybridized carbon (less stable) and creates cumene-like character. The extensive conjugation in the benzyl system makes it most stable, followed by allyl, then propargyl.

Question 17

1. The cyclopentadienyl anion has more resonance structures than the cycloheptatrienyl anion, leading to greater charge delocalization.
2. The cyclopentadienyl anion is aromatic with 6 π electrons, whereas the cycloheptatrienyl anion is anti-aromatic with 8 π electrons. (correct answer)
3. The sp2-hybridized carbons in the five-membered ring of the cyclopentadienyl anion are more electronegative, better stabilizing the negative charge.
4. The cycloheptatrienyl anion cannot achieve a planar conformation due to angle strain, preventing effective orbital overlap for resonance.

Explanation: The stability of these cyclic anions is governed by Hückel's rules for aromaticity. The cyclopentadienyl anion is cyclic, planar, fully conjugated, and has 6 π electrons (4n+2 where n=1), making it aromatic and highly stable. The cycloheptatrienyl anion is also cyclic, planar, and fully conjugated, but it has 8 π electrons (4n where n=2), making it anti-aromatic and highly unstable.

Question 18

1. Electron-donating alkyl groups push electron density toward the electron-deficient radical center, stabilizing it.
2. Steric hindrance from larger alkyl groups protects the radical center from reacting.
3. The bond dissociation energy of a tertiary C-H bond is lower than that of a primary C-H bond.
4. The increasing number of adjacent C-H sigma bonds allows for greater hyperconjugative stabilization. (correct answer)

Explanation: While inductive effects (Choice A) play a minor role, the primary stabilizing factor for alkyl radicals is hyperconjugation. This involves the overlap of the filled C-H sigma bonding orbitals on adjacent carbons with the half-filled p-orbital of the radical center. This delocalizes the radical character and stabilizes the intermediate. A tertiary radical has more adjacent C-H bonds than a secondary, which has more than a primary, leading to the observed stability trend. Choice C is a consequence of the stability trend, not the cause.

Question 19

1. The tert-butyl cation, [(CH3)3C]+ (correct answer)
2. The methyl cation, [CH3]+
3. The allyl cation, [CH2=CH-CH2]+
4. The trichloromethyl radical, •CCl3

Explanation: Hyperconjugation involves the delocalization of electrons from adjacent C-H or C-C sigma bonds into the empty p-orbital of a carbocation or the partially filled p-orbital of a radical. The tert-butyl cation has nine adjacent C-H bonds (three methyl groups) that can participate in hyperconjugation, providing substantial stabilization. The methyl cation has no adjacent C-H bonds and thus no stabilization from hyperconjugation. The allyl cation is primarily stabilized by resonance, not hyperconjugation. The •CCl3 radical has adjacent C-Cl bonds; C-Cl sigma orbitals are poor donors for hyperconjugation compared to C-H bonds.

Question 20

1. Dehydration of 2-methyl-2-pentanol with H2SO4.
2. Reaction of 2-bromo-3-methylbutane with ethanol.
3. Addition of HBr to 3-methyl-1-butene.
4. Reaction of (R)-2-chloropentane with sodium iodide in acetone. (correct answer)

Explanation: Carbocation rearrangements occur in mechanisms that involve carbocation intermediates, primarily SN1, E1, and electrophilic additions. Choice A (dehydration) is an E1 reaction. Choice B (solvolysis of a secondary halide) is an SN1/E1 reaction. Choice C (hydrohalogenation) proceeds via a carbocation. All three pathways are prone to rearrangements if a more stable carbocation can be formed. Choice D describes an SN2 reaction (a strong, unhindered nucleophile in a polar aprotic solvent with a secondary halide). The SN2 mechanism is concerted and does not involve the formation of a carbocation intermediate, so rearrangement is not possible.