Why we have to check more stable carbocation when doing rearrangement in carbocationic species , neglecting sterric factor?

Question

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1. What the student is really asking

The student sees two textbook examples of carbocation rearrangements

Example	What “moves”	What carbocation would be formed after the shift?
1️⃣	Hydride (H) shifts → a more substituted, more stable carbocation	Hydride shift is favoured. The student thinks this is because H is tiny and can move easily (kinetic argument).
2️⃣	A bulky alkyl group (e.g., a 2‑methyl‑butyl fragment) shifts → a still more substituted, more stable carbocation	The textbook says the bulkier group does shift because the product carbocation is very stable (hyper‑conjugation). The student worries that the steric bulk should make the migration kinetically impossible.

Question – Do we really ignore the kinetic (steric) problem and decide only by the stability of the final carbocation? If a group is bulky, shouldn’t the transition state be too crowded for the shift to occur?

In short: Why does a “bulky” alkyl shift happen at all, and how do we rationalise the competition between steric hindrance (kinetics) and carbocation stability (thermodynamics)?

2. Step‑by‑step answer

2.1 What actually happens in a carbocation rearrangement?

A carbocation is generated (e.g., by loss of a leaving group, protonation of an alkene, etc.).
If the newly formed cation is not the most stable possible structure that can be accessed by a simple intramolecular migration, the system can rearrange by moving an adjacent σ‑bond (C–H or C–C) into the empty p‑orbital.
The migration proceeds through a three‑center, two‑electron transition state (TS) that is essentially a “bridge” in which the migrating bond is partially broken to the original carbon and partially formed to the carbocation centre.

C–X   →   C···X···C   →   C–X   (X = H or an alkyl group)

The TS is concerted and intramolecular, so no external steric clash with other molecules is needed – only the atoms that are already attached to the carbon participate.

2.2 How do we decide whether the migration will occur?

Two factors are relevant:

Factor	How it influences the reaction
Thermodynamic driving force – the relative stability of the product carbocation (hyper‑conjugation, resonance, inductive effects).	The more stable the product, the more exergonic the overall rearrangement.
Activation barrier (kinetics) – the energy required to reach the TS.	The barrier is not a simple “bulky‑group‑hard‑to‑move” term; it is largely controlled by how well the migrating bond can donate electron density to the empty p‑orbital (i.e., the migration aptitude).

2.2.1 The Hammond postulate (why stability matters for the barrier)

Carbocation formation is endothermic when we go from a less‑substituted to a more‑substituted cation (the product is lower in energy).
For an endothermic step, the Hammond postulate tells us that the TS resembles the product more than the reactant.
Consequently, any factor that stabilises the product (more hyper‑conjugation, resonance, inductive donation) also stabilises the TS, lowering the activation energy.

Thus, the more stable the final carbocation, the lower the barrier for its formation, even if the migrating group is relatively bulky.

2.2.2 Migration aptitude order (experimental observation)

Empirically, the ability of a group to migrate follows the trend

[ \text{hydride} > \text{phenyl} > \text{tertiary alkyl} > \text{secondary alkyl} > \text{primary alkyl} > \text{methyl} ]

Why? Because a migrating σ‑bond can donate electron density to the empty p‑orbital. The better the donor ability, the lower the TS energy. A C–H bond (hydride) is an excellent donor; a C–C bond of a tertiary carbon is also good because the adjacent three C–H bonds can hyper‑conjugate in the TS. A primary alkyl or methyl group can only give a single C–H hyper‑conjugative interaction, so its TS is higher in energy.

Bulk does not appear in this ranking because the migration is intramolecular and occurs through a linear, three‑center arrangement. The migrating carbon simply slides into the empty p‑orbital; there is no need to “squeeze” a bulky substituent through a crowded space. The only steric requirement is that the atoms are properly aligned (≈180° C–C–C or C–C–H angle). If that geometry can be attained (and it almost always can in a flexible chain), the steric penalty is small compared with the electronic stabilization gained.

2.2.3 Quantitative view – a simple energy diagram

Initial carbocation      TS (migration)       More stable carbocation
      |                     / \                     |
      |____________________/   \____________________|
               ΔG‡ (lower)        ΔG° (more negative)

When the product carbocation is much more stable, ΔG° is negative and the TS is lowered → rearrangement proceeds rapidly.
When the product is only slightly more stable (or less stable), ΔG° ≈ 0 or +, the TS sits higher → the rearrangement may be slow or not observed, even if the migrating group is small.

2.3 Applying the concepts to the two examples

Example	Shift considered	Stability of product carbocation	Expected barrier (ΔG‡)	Observed outcome
1️⃣	Hydride (H) → secondary → tertiary cation	Large gain (secondary → tertiary)	Very low (hydride has highest migration aptitude)	Shift occurs readily
2️⃣	Bulky 2‑methylbutyl (secondary → tertiary/benzylic)	Even larger gain (extra hyper‑conjugation, possibly resonance)	Still low because the product is highly stabilised; the three‑center TS benefits from the many adjacent C–H bonds of the migrating carbon	Shift is observed despite bulk

The key point: the bulkiness of the migrating alkyl group does not dominate the barrier; the electronic advantage of the product (many hyper‑conjugative C–H bonds, possibly resonance) outweighs the modest steric penalty.

2.4 Why we usually ignore steric factors in textbook predictions

Intramolecular nature – the reacting atoms are already bonded; there is no need for an external approach that would be hindered by steric crowding.
Transition‑state geometry – the migrating bond is nearly linear with the empty p‑orbital, so the “bulky part” of the group points away from the carbocation centre, minimizing clash.
Dominance of electronic effects – the TS is heavily stabilised by donation of the σ‑bond electrons (hyper‑conjugation). A small steric penalty is often “absorbed” by the larger electronic gain.
Empirical success – predictions based solely on the relative stability of the possible carbocations (or the migration‑aptitude order) correctly describe >90 % of observed rearrangements, confirming that kinetic barriers are indeed correlated with product stability.

3. Final answer (concise)

Carbocation rearrangements are governed by the stability of the product carbocation because, for an endothermic shift, the transition state resembles the product (Hammond postulate).
A bulkier alkyl group can still migrate because the three‑center TS is intramolecular, linear, and its energy is lowered mainly by electronic donation (hyper‑conjugation), not by steric crowding.
Hence, thermodynamic driving force and migration aptitude dictate the kinetic barrier; steric hindrance of the migrating group is a secondary, often negligible, factor.
In practice we predict rearrangements by comparing the relative stabilities of the possible carbocations; the more stable one will be formed, even if the migrating group is relatively large.

4. Common Mistakes

Mistake	Why it’s wrong	Correct view
“Only the size of the migrating group matters.”	Size does affect the barrier, but the electronic stabilization of the TS outweighs modest steric effects in most intramolecular migrations.	Consider migration aptitude (hydride > aryl > tertiary > secondary > primary > methyl) and the stability of the product carbocation.
“If a rearrangement produces a more stable carbocation, it must be thermodynamically driven, not kinetically.”	The reaction still has an activation barrier; the lower barrier is a consequence of the product’s stability (Hammond).	The more stable product lowers the TS energy, making the rearrangement both thermodynamically favored and kinetically accessible.
“Bulky groups cannot migrate because the transition state is too crowded.”	The TS is a linear three‑center arrangement; bulky substituents are oriented away from the forming bond, so steric clash is minimal.	Visualise the TS as a bridge: the migrating carbon moves directly into the empty p‑orbital, with little steric interaction.
“All carbocation rearrangements are instantaneous.”	Some rearrangements have high barriers (e.g., when the product is not significantly more stable).	Evaluate the relative stability; if the gain is small, the barrier can be comparable to competing reactions, and the rearrangement may be slow or absent.

Original question: Why we have to check more stable carbocation when doing rearrangement in carbocationic species , neglecting sterric factor? on Chemistry Stack Exchange, licensed CC BY-SA.

1. What the student is really asking

2. Step‑by‑step answer