Organic Chemistry Quiz: Common Rearrangements Hydride Alkyl Shifts

What this quiz covers

This quiz focuses on Common Rearrangements Hydride Alkyl Shifts, 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. Reaction of 1-bromobutane with sodium ethoxide in ethanol.
2. Reaction of (R)-2-bromopentane with sodium iodide in acetone.
3. Reaction of 1-bromo-3-methylbutane with potassium tert-butoxide.
4. Reaction of 1-bromo-2-methylbutane in boiling methanol. (correct answer)

Explanation: Carbocation rearrangements are characteristic of SN1 and E1 mechanisms, which are favored by polar protic solvents, weak nucleophiles/bases, and secondary/tertiary substrates. Option D involves a secondary halide in a polar protic solvent (methanol) with no strong base/nucleophile, favoring an SN1/E1 pathway where a secondary carbocation can rearrange to a tertiary one. Options A, B, and C all use strong nucleophiles and/or bases, which favor SN2 or E2 mechanisms that do not involve carbocation intermediates and thus no rearrangement.

Question 2

1. The proposal is correct because any rearrangement from a tertiary carbocation to another tertiary carbocation is thermodynamically favorable due to reduced steric crowding
2. The proposal is incorrect because 1,2-methyl shifts require significantly higher activation energy compared to 1,2-hydride shifts and are kinetically unfavorable
3. The proposal is correct because the methyl shift would create a more substituted tertiary carbocation with greater hyperconjugation stabilization from additional alkyl groups
4. The proposal is incorrect because the initial tertiary carbocation is already highly stable and rearrangement would not provide additional thermodynamic benefit (correct answer)

Explanation: When analyzing carbocation rearrangements in $S_N1$ reactions, you need to evaluate whether a proposed shift would actually create a more stable carbocation than the starting one. Let's examine 3-methylpentan-3-ol. When it undergoes $S_N1$ solvolysis, the initial carbocation formed is tertiary (3°) with three alkyl substituents providing excellent stabilization through hyperconjugation and inductive effects. The proposed 1,2-methyl shift would move the positive charge to a different carbon, but crucially, this would still result in a tertiary carbocation with the same degree of substitution. Since both carbocations are tertiary with similar substitution patterns, there's no meaningful thermodynamic driving force for rearrangement. The initial carbocation is already highly stable, and the rearranged form offers no significant additional benefit. Therefore, answer D is correct. Answer A incorrectly suggests that any tertiary-to-tertiary rearrangement is favorable due to reduced steric crowding, but this oversimplifies stability factors. Answer B focuses on kinetic barriers between methyl and hydride shifts, which isn't the primary issue here—the problem is thermodynamic, not kinetic. Answer C wrongly claims the rearranged carbocation would have "greater hyperconjugation stabilization from additional alkyl groups," but both carbocations have similar substitution and stabilization. Study tip: Remember that carbocation rearrangements only occur when they lead to significantly more stable intermediates. If the initial carbocation is already highly substituted (like a tertiary carbocation), rearrangement is unlikely unless it creates an even more favorable structure.

Question 3

1. The first rearrangement creates a benzylic tertiary carbocation that represents a deep thermodynamic minimum, making further rearrangements energetically uphill (correct answer)
2. Steric hindrance around the rearranged carbocation prevents the orbital overlap necessary for additional 1,2-shifts to occur under these conditions
3. The solvent rapidly traps the rearranged carbocation through nucleophilic attack faster than additional rearrangement steps can occur kinetically
4. Resonance stabilization with the phenyl ring locks the positive charge in place, preventing the electron density redistribution required for further migration

Explanation: 1-Phenyl-2-methylpropan-1-ol initially forms a benzylic secondary carbocation. A 1,2-methyl shift moves the positive charge to the tertiary carbon, creating a benzylic tertiary carbocation that is exceptionally stable due to both resonance with the aromatic ring and hyperconjugation from the three alkyl substituents. This represents such a stable intermediate that any further rearrangement would be thermodynamically uphill, so the system remains at this energy minimum. Choice B is incorrect because steric effects wouldn't completely prevent rearrangement if it were thermodynamically favorable. Choice C addresses kinetics but misses the fundamental thermodynamic stability. Choice D is wrong because resonance doesn't prevent rearrangement if a more stable carbocation could be formed.

Question 4

1. High temperature conditions favor ring contraction because the increased entropy of the smaller ring system outweighs the enthalpy cost of increased angle strain
2. Ring contraction is favored when the reaction is performed in polar protic solvents that can stabilize the transition state for C-C bond migration
3. The ring contraction pathway becomes favored when the initial carbocation adopts a conformation that brings the migrating bond into optimal orbital alignment (correct answer)
4. Ring contraction is kinetically favored in acidic conditions because protonation of the alcohol enhances the electrophilicity of the carbocation center

Explanation: For ring contraction to compete with hydride shift, the C-C bond that will migrate must be properly aligned with the empty p-orbital of the carbocation. In cyclohexane systems, conformational effects are crucial - the ring must adopt a conformation where the migrating bond is antiperiplanar to the C-OH bond (before departure) or aligned with the empty p-orbital (after carbocation formation). This geometric requirement makes ring contraction highly dependent on conformational accessibility. Choice A is incorrect because ring contraction typically increases strain. Choice B is wrong because solvent effects primarily influence carbocation stability, not the competition between different rearrangement pathways. Choice D is incorrect because acid strength affects the rate of carbocation formation, not the competition between rearrangement pathways once the carbocation is formed.

Question 5

1. 3,3-dimethyl-1-butene, via a 1,2-hydride shift
2. 2,3-dimethyl-2-butene, via a 1,2-methyl shift (correct answer)
3. 2,3-dimethyl-1-butene, via a 1,2-hydride shift
4. 3,3-dimethyl-1-butene, via a 1,2-methyl shift

Explanation: Protonation of the alcohol and loss of water forms a secondary carbocation at C2. To form a more stable carbocation, a 1,2-methyl shift occurs from the adjacent quaternary carbon (C3) to C2. This generates a tertiary carbocation at C3. Elimination of a proton from this rearranged cation according to Zaitsev's rule (forming the most substituted alkene) yields 2,3-dimethyl-2-butene as the major product.

Question 6

1. A racemic mixture of (R)- and (S)-2-bromo-2-phenylbutane due to planar carbocation (correct answer)
2. Only (S)-2-bromo-2-phenylbutane due to complete inversion of configuration
3. Only (R)-2-bromo-2-phenylbutane due to complete retention of configuration
4. A mixture of diastereomers because the original stereocenter remains intact

Explanation: Protonation and loss of water from the secondary alcohol initially forms a secondary carbocation at C2. A 1,2-hydride shift from C3 to C2 occurs, forming a more stable tertiary, benzylic carbocation at C2. This rearranged tertiary carbocation intermediate is trigonal planar and achiral. The bromide nucleophile can attack this planar intermediate from either face with equal probability, leading to formation of a racemic mixture of the (R) and (S) enantiomers.

Question 7

1. A 1,2-hydride shift occurs, followed by deprotonation to yield an enol.
2. A 1,2-methyl shift occurs to form a resonance-stabilized, protonated ketone. (correct answer)
3. A water molecule attacks the carbocation, reversing the initial dehydration step.
4. The carbocation undergoes elimination to form a diene as the major product.

Explanation: After one hydroxyl group is protonated and leaves as water, a tertiary carbocation is formed. A 1,2-methyl shift from the adjacent carbon (which bears the other hydroxyl group) to the carbocation center occurs. This shift is highly favorable because the resulting carbocation is immediately stabilized by resonance from the lone pair on the adjacent oxygen atom. This new intermediate is a protonated ketone (an oxocarbenium ion), which is then deprotonated to give the final pinacolone product.

Question 8

1. Protonation at C1 gives a secondary carbocation at C2, which is directly attacked by bromide without rearrangement.
2. Protonation at C2 gives a primary carbocation at C1, which rearranges via a hydride shift to form a secondary carbocation.
3. Protonation at C1 gives a secondary carbocation at C2, which rearranges via a 1,2-methyl shift to a tertiary carbocation at C3. (correct answer)
4. Bromide first attacks C1 in a concerted addition, followed by rearrangement and elimination of hydrogen bromide.

Explanation: The electrophilic addition begins with protonation of the alkene's C1 by HBr, following Markovnikov's rule to form the more stable secondary carbocation at C2. This secondary carbocation is adjacent to a quaternary carbon (C3). A 1,2-methyl shift from C3 to C2 occurs, resulting in a more stable tertiary carbocation at C3. Finally, the bromide ion attacks this rearranged tertiary carbocation to form the major product.

Question 9

1. The bridged structure allows for better solvation of the positive charge by polar protic solvents, lowering the overall energy of the system
2. Orbital overlap between the migrating alkyl group and the electron-deficient carbon provides stabilization during the rearrangement transition state
3. The bridged intermediate avoids the formation of a highly unstable primary carbocation by delocalizing charge across multiple carbon atoms simultaneously (correct answer)
4. Conformational constraints in the five-membered ring system favor the bridged geometry over classical carbocation structures due to reduced ring strain

Explanation: In 3-methylcyclopentan-1-ol, loss of water would initially form a primary carbocation at C-1, which is extremely unstable. The bridged intermediate forms as the C-C bond from C-3 begins to migrate toward C-1 while the C-OH bond is breaking, avoiding the formation of a discrete primary carbocation. This bridged structure delocalizes the positive charge between C-1 and C-3, providing significant stabilization compared to a localized primary carbocation. The subsequent ring expansion and shift to form a secondary carbocation in a six-membered ring represents the final, most stable product. Choice A is incorrect because solvation differences are not the primary driving force. Choice B describes part of the process but misses the key issue of avoiding primary carbocation formation. Choice D incorrectly focuses on conformational effects rather than electronic stabilization.

Question 10

1. A 1,2-alkyl shift expands the five-membered ring to a six-membered ring, relieving angle strain and forming a more stable tertiary carbocation (correct answer)
2. A 1,2-hydride shift contracts the ring to form a four-membered ring carbocation that is stabilized by reduced torsional strain
3. A 1,2-alkyl shift forms a seven-membered ring carbocation that is favored due to increased conformational flexibility and reduced steric interactions
4. A 1,3-alkyl shift expands the ring while maintaining the same degree of substitution at the carbocation center for equivalent stability

Explanation: 2-Methylcyclopentanol forms a secondary carbocation after protonation and loss of water. The adjacent carbon-carbon bond can migrate (1,2-alkyl shift) to expand the five-membered ring to a six-membered ring while simultaneously moving the positive charge to a tertiary position. This is favorable because: (1) six-membered rings have less angle strain than five-membered rings, and (2) tertiary carbocations are more stable than secondary ones. Choice B is incorrect because ring contraction to a four-membered ring would increase strain. Choice C is wrong because seven-membered rings have more strain than six-membered rings. Choice D is incorrect because 1,3-shifts are rare, and the described rearrangement wouldn't maintain the same substitution pattern.

Question 11

1. Methyl > Hydride > Phenyl
2. Hydride > Methyl > Phenyl
3. Phenyl ≈ Hydride > Methyl (correct answer)
4. Methyl > Phenyl > Hydride

Explanation: The migratory aptitude reflects the ability of a group to stabilize the positive charge in the transition state of the 1,2-shift. Phenyl groups can form a stabilized bridged 'phenonium ion' intermediate, making them very good migrating groups. Hydride is also an excellent migrating group because of its small size and the low energy of the transition state. Alkyl groups like methyl are generally the poorest migrating groups of the three. Phenyl and hydride have similar and high migratory aptitudes, both being significantly better than alkyl groups.

Question 12

1. The transition state of the rearrangement lies closer in energy to the tertiary carbocation R₃⁺.
2. The overall energy change (ΔE) for the step is positive, indicating an endothermic process.
3. The activation energy for the forward reaction is less than the activation energy for the reverse reaction. (correct answer)
4. The R₂⁺ and R₃⁺ intermediates are constitutional isomers but have identical thermodynamic stability.

Explanation: The rearrangement from a less stable secondary carbocation (R₂⁺) to a more stable tertiary carbocation (R₃⁺) is an exothermic process (ΔE < 0). According to the principle of microscopic reversibility, the activation energy for the forward, exothermic reaction (Ea,fwd) must be smaller than the activation energy for the reverse, endothermic reaction (Ea,rev). Hammond's postulate suggests the transition state for an exothermic step will resemble the reactant (R₂⁺), not the product (R₃⁺).

Question 13

1. The ring expansion creates a tertiary carbocation while the hydride shift produces only a secondary carbocation with equivalent ring strain
2. The ring expansion significantly reduces angle strain by forming a five-membered ring, outweighing the energy cost of forming a secondary carbocation (correct answer)
3. The ring expansion eliminates unfavorable 1,3-diaxial interactions present in the four-membered ring that persist after the hydride shift
4. The ring expansion allows for better orbital overlap in the transition state due to the geometric constraints of the four-membered ring system

Explanation: In 2,2-dimethylcyclobutanol, both rearrangements start from a tertiary carbocation. A simple 1,2-hydride shift would maintain the tertiary character but keep the highly strained four-membered ring (≈26° bond angles vs. ideal 109.5°). Ring expansion via 1,2-alkyl shift forms a five-membered ring with much less angle strain (≈108° bond angles), and while this creates a secondary carbocation, the enormous relief of angle strain (≈27 kcal/mol) more than compensates for the loss of tertiary stabilization (≈12 kcal/mol). Choice A is incorrect because both pathways start from tertiary carbocations. Choice C is wrong because 1,3-diaxial interactions are relevant to six-membered rings, not four-membered rings. Choice D, while partially true, doesn't address the primary thermodynamic driving force.

Question 14

1. A 1,2-methyl shift forms a tertiary carbocation that is stabilized by hyperconjugation from six adjacent C-H bonds
2. A 1,2-hydride shift forms a tertiary carbocation that is stabilized by hyperconjugation from nine adjacent C-H bonds (correct answer)
3. A 1,2-methyl shift forms a secondary carbocation that is stabilized by the electron-donating effect of adjacent methyl groups
4. A 1,2-hydride shift forms a secondary carbocation that avoids steric crowding around the positively charged carbon atom

Explanation: 3,3-Dimethylbutan-2-ol initially forms a secondary carbocation at C-2 upon protonation and water loss. A 1,2-hydride shift from C-3 to C-2 moves the positive charge to the tertiary C-3 position, which is adjacent to three methyl groups (9 C-H bonds total). This tertiary carbocation is much more stable due to increased hyperconjugation. Choice A is incorrect because a methyl shift would not provide the same degree of stabilization. Choices C and D are incorrect because rearrangement to form a secondary carbocation would be thermodynamically unfavorable compared to forming a tertiary carbocation.

Question 15

1. 3,3-dimethyl-1-butene
2. 3-methyl-1-butene
3. 1-pentene
4. 2-methyl-2-pentene (correct answer)

Explanation: A carbocation rearrangement occurs to form a more stable carbocation. We must evaluate the initial carbocation formed from each alkene. A) 3,3-dimethyl-1-butene forms a secondary cation that rearranges to a tertiary cation via a methyl shift. B) 3-methyl-1-butene forms a secondary cation that rearranges to a tertiary cation via a hydride shift. C) 1-pentene forms a secondary cation, which does not have a more stable position to rearrange to via a 1,2-shift. D) Protonation of 2-methyl-2-pentene at C3 gives a tertiary cation at C2, which is already the most stable possible cation in the vicinity. No 1,2-shift can improve its stability, so it reacts directly without rearrangement.

Question 16

1. 2-bromo-5-methylhexane (correct answer)
2. 2-bromo-2-methylhexane
3. 5-bromo-2-methylhexane
4. 2-bromo-6-methylhexane

Explanation: When alcohols react with concentrated HBr, you're dealing with an acid-catalyzed substitution reaction that follows an SN1 or SN2 mechanism depending on the alcohol's structure. Since 5-methyl-2-hexanol is a secondary alcohol, this reaction will proceed via an SN1 mechanism involving carbocation formation. Let's trace through this reaction step by step. First, draw out 5-methyl-2-hexanol: it's a 6-carbon chain with an OH group on carbon 2 and a methyl branch on carbon 5. When HBr protonates the OH group, water leaves, forming a secondary carbocation at carbon 2. This carbocation is relatively stable, and bromide ion attacks it directly without rearrangement since no more stable carbocation is accessible nearby. The bromide replaces the original OH group at carbon 2, giving you 2-bromo-5-methylhexane, which is answer choice A. Now let's examine why the other options are wrong. Choice B (2-bromo-2-methylhexane) incorrectly places the methyl group on carbon 2 instead of carbon 5 - this would require the carbon skeleton to rearrange, which doesn't happen here. Choice C (5-bromo-2-methylhexane) puts the bromine on carbon 5, but substitution occurs where the original OH group was located (carbon 2). Choice D (2-bromo-6-methylhexane) again misplaces the methyl group, putting it on the terminal carbon instead of carbon 5. Remember: in alcohol-HX reactions, the halogen replaces the OH group at the same carbon position. Always identify where the OH group is located first, then predict substitution at that exact position.

Question 17

1. SN2 reactions proceed through a single, concerted step and do not form a carbocation intermediate. (correct answer)
2. SN2 reactions are stereospecific, which prevents the carbocation from rearranging.
3. The strong nucleophiles used in SN2 reactions are too bulky to allow for rearrangement.
4. The polar aprotic solvents typically used for SN2 reactions suppress carbocation formation.

Explanation: When you encounter questions about rearrangements in substitution reactions, focus on the mechanism and whether carbocation intermediates can form—these are the species that actually rearrange. SN2 reactions follow a concerted mechanism where the nucleophile attacks the substrate while the leaving group departs simultaneously. This happens in one step without forming any intermediates. Since no carbocation intermediate exists, there's no opportunity for rearrangement to occur. The nucleophile directly displaces the leaving group through backside attack, and the reaction is complete. Answer A correctly identifies this fundamental mechanistic feature—the single, concerted step prevents carbocation formation and therefore eliminates any possibility of rearrangement. Answer B confuses cause and effect. While SN2 reactions are indeed stereospecific (causing inversion of configuration), this stereochemistry results from the concerted mechanism, not the absence of rearrangements. The stereospecificity doesn't prevent rearrangement—the lack of carbocation intermediates does. Answer C incorrectly focuses on nucleophile size. Though steric hindrance affects SN2 reaction rates, nucleophile bulkiness doesn't prevent rearrangements. Even small nucleophiles wouldn't cause rearrangements in SN2 reactions because no carbocation forms. Answer D misattributes the cause to solvent effects. While polar aprotic solvents do favor SN2 reactions by enhancing nucleophile reactivity, they don't directly suppress carbocation formation—the concerted mechanism does. Remember: Rearrangements require carbocation intermediates. Always ask yourself "Does this mechanism form a carbocation?" to predict whether rearrangements are possible.

Question 18

1. The relief of steric strain in the transition state of the shift.
2. The formation of a thermodynamically more stable carbocation intermediate. (correct answer)
3. An increase in entropy by breaking a C-C or C-H bond.
4. The requirement to achieve an anti-periplanar geometry for the subsequent step.

Explanation: Carbocation rearrangements are intramolecular processes that occur when a more stable carbocation can be formed. The stability of carbocations generally follows the trend: tertiary > secondary > primary. The shift converts a less stable carbocation into a more stable one, lowering the overall potential energy of the intermediate, which is the primary thermodynamic driving force for the process.