Questions: Carbocation Rearrangement: 1,2-Hydride and 1,2-Alkyl Shifts

Question 1

Treating 3-methyl-2-butanol with concentrated H₂SO₄ gives 2-methylbut-2-ene as the major product rather than 3-methylbut-1-ene. What best explains this observation?

• A. The initially formed secondary carbocation at C2 undergoes a 1,2-hydride shift to a more stable tertiary carbocation at C3, and elimination from this rearranged intermediate gives the observed alkene
• B. Dehydration follows Zaitsev's rule, which always produces the more substituted alkene without any carbocation rearrangement
• C. The reaction proceeds via a concerted E2 mechanism that bypasses carbocation intermediates and directly forms the more stable alkene
• D. A 1,2-methyl shift converts the secondary carbocation at C2 to a primary carbocation at C1, which is then stabilized by resonance

Answer: (A)

The 2° carbocation initially formed at C2 is adjacent to C3, a tertiary carbon (bearing three alkyl substituents). A 1,2-hydride shift from C3 to C2 produces a more stable 3° carbocation at C3. Elimination from this rearranged cation gives the methylbutene products. Option B is the classic misconception — it invokes Zaitsev's rule without rearrangement, but Zaitsev alone cannot explain why the product carbon skeleton differs from what simple elimination would predict. Rearrangements must be identified before applying regiochemistry rules.

Question 2

In a 1,2-hydride shift, which arrow correctly represents the electron flow?

• A. A curved arrow from the C–H bond on the adjacent carbon toward the empty p orbital of the carbocation
• B. A curved arrow from the empty p orbital of the carbocation toward the C–H bond on the adjacent carbon
• C. A curved arrow from the positively charged carbon toward a lone pair on the migrating hydrogen
• D. Two curved arrows: one showing H⁺ departure and one showing proton capture by the cation

Answer: (A)

Electron flow always runs from electron-rich to electron-poor — from the C–H bonding pair toward the empty p orbital. The hydrogen migrates as a hydride (H:⁻, with both bonding electrons), not as a proton. Option B reverses the arrow direction, which would imply electron flow from an empty orbital — physically impossible. Option D describes a proton transfer mechanism, which is a different type of reaction and would not shift carbon connectivity.

Question 3

A secondary carbocation adjacent to a tertiary carbon will generally rearrange to a tertiary carbocation before reacting with a nucleophile.

• T. True
• F. False

Answer: True

The 1,2-hydride shift from the adjacent C–H bond into the empty p orbital has a very low energy barrier — just the orbital overlap requirement — and is thermodynamically favorable because it generates a more stable carbocation. This rearrangement typically occurs faster than nucleophilic capture of the less stable secondary cation. As a practical rule: whenever a carbocation is adjacent to a carbon bearing a hydrogen that would generate a more substituted cation upon migration, assume rearrangement occurs and draw the new intermediate before predicting the product.

Question 4

A 1,2-alkyl shift typically involves a methyl group, since smaller groups migrate more readily than larger alkyl substituents.

• T. True
• F. False

Answer: False

Any alkyl group on the adjacent carbon can undergo a 1,2-shift — methyl, ethyl, isopropyl, or more complex groups. The driving force is thermodynamic stability gain, not the size of the migrating group. 'Methyl shift' refers specifically to migration of a –CH₃ group, but '1,2-alkyl shift' encompasses any carbon group. Larger alkyl groups actually tend to stabilize the resulting carbocation more effectively through hyperconjugation and inductive effects, providing additional driving force.

Question 5

Why is it essential to check for possible carbocation rearrangements before predicting the final product of a reaction that proceeds through a carbocation intermediate?

Model answer: If the initial carbocation rearranges to a more stable one, the final product forms from the rearranged intermediate — not from the original carbocation. The regiochemistry of addition, substitution, or elimination depends entirely on which carbocation undergoes the final step. Failing to account for rearrangement leads to predicting the wrong constitutional isomer as the major product.

Rearrangement is driven by thermodynamics: the more stable carbocation is the 'sink' that the mechanism flows toward. Once rearrangement occurs, the new carbocation dictates the product's carbon skeleton — which may differ entirely from what the starting material's connectivity suggests. This is why predicting correct products in multi-step synthesis requires checking: does the initially formed cation have a neighboring carbon that could donate a hydride or alkyl group to form a more substituted cation? If yes, draw the rearranged intermediate first.