An Intermediate That Won't Sit Still
By now you know a carbocation as one of the cast of fleeting intermediates in this rung: a carbon with only three bonds and an empty p orbital, starving for two more electrons. You also know its stability order, set by how many alkyl groups lean their electrons toward that empty orbital through hyperconjugation: tertiary beats secondary beats primary beats methyl. What this guide adds is a twist that catches almost every beginner — the carbocation does not always wait politely where it was born. If a quick internal move can turn a less stable cation into a more stable one, it will make that move, and fast.
The move is called a carbocation rearrangement, and its whole purpose is to climb the stability ladder you already memorized. A secondary cation that finds a tertiary position one carbon away will slide its positive charge over there. The skeleton of the molecule can quietly reorganize, and the product that finally crystallizes out of the flask is built on the rearranged skeleton, not the original. Miss the rearrangement and your predicted product is simply wrong — which is exactly why exam questions love to plant one.
stability of carbocations: 3deg > 2deg > 1deg > methyl (most stable) (least stable) rearrangement always moves charge LEFT along this line, never right.
Two Ways to Shift: Hydride and Alkyl
There are two flavors of rearrangement, and they differ only in what slides over. In a hydride shift, a hydrogen on the carbon next door migrates to the cationic carbon — but here is the part that trips everyone up: it does not move as a lone proton. It carries its bonding pair of electrons with it, so it travels as a hydride, H with two electrons (H:-). The electrons fill the empty orbital on the old cation; the carbon the hydrogen left behind is now the one missing a pair, so the positive charge has effectively walked one carbon over.
An alkyl shift (most often a methyl shift) is the same idea with a small carbon group instead of a hydrogen. A whole CH3 group, again with its bonding pair, migrates from the neighbor to the cationic carbon. You reach for an alkyl shift when no hydride shift would help — when the neighboring carbon has no hydrogen to give but does carry a methyl that, once it moves, leaves behind a more stable (usually tertiary) cation. The rule of thumb for both: the group only ever migrates from a carbon directly adjacent to the positive charge, and only when the new cation is more stable than the old one.
Spotting It: A Worked Example
Take a concrete case. Treat 3,3-dimethyl-2-butanol with acid to make it ionize, or imagine an alkyl halide losing its leaving group. The cation first forms at carbon 2, a secondary position. But carbon 3 right next door is a quaternary carbon bearing three methyl groups — a tertiary cation waiting to happen. One methyl group migrates from C3 to C2, taking its electron pair along, and the positive charge lands on C3 as a far more stable tertiary cation. The carbon skeleton itself has changed: a methyl that started on C3 is now on C2.
- The leaving group departs (or the protonated -OH leaves as water), and a secondary carbocation forms at C2.
- Look at every carbon directly attached to C2. C3 is quaternary — no hydride to give, but three methyls available for an alkyl shift.
- Ask: would shifting make a MORE stable cation? Moving a methyl from C3 to C2 turns a 2deg cation into a 3deg cation. Yes — so it happens.
- Push one curved arrow from the migrating C-C bond to the empty orbital on C2. The positive charge is now on C3, a tertiary cation.
- Only NOW does the nucleophile attack or a base remove a proton — and it does so at the rearranged carbon, giving the 'unexpected' product.
The diagnostic habit is short and reliable. Whenever you draw a carbocation, before you do anything else, glance at each adjacent carbon and ask one question: is there an H or an alkyl group whose migration would give a more stable cation? If yes, shift first, then continue. If no — if the cation is already tertiary, or no neighbor offers an improvement — leave it where it is. Honest limit: rearrangements need a real stability gain. A secondary cation next to another secondary carbon usually does not bother, because the new cation is no better than the old one.
Why It's Fast — and Why That Matters
A rearrangement is downhill and has a tiny barrier, so it is essentially instantaneous — faster than most molecules nearby can react with the cation. This is where the Hammond postulate earns its keep: because forming a less stable secondary cation and rearranging to a more stable tertiary one are both fast, low-barrier events near the same point on the path, the system slides to the more stable cation before anything traps it. The cation barely exists as the secondary species; it is gone, rearranged, in the blink of a bond vibration.
This has a beautiful diagnostic flip side. Rearrangements are a fingerprint: if you run a reaction and isolate a product with a scrambled skeleton — a methyl that wandered, a charge that moved — that scrambling is hard evidence that a free carbocation existed somewhere in the mechanism. A reaction that proceeds without a free cation (a concerted one-step process where no naked carbocation is ever born) cannot rearrange, because there is no untethered cation to slide. So a rearranged product is a clue, not just a trap: it tells you the mechanism went through a real, separate cationic intermediate.
Where Rearrangements Lie in Wait
Any mechanism that builds a free carbocation is fair game. You will meet the two classic ones in the rungs just ahead. The SN1 substitution and the E1 elimination both start the same way: the leaving group departs in a slow first step to make a carbocation, which then either gets attacked by a nucleophile (SN1) or loses a neighboring proton to a base (E1). Because both pause at a genuine free cation, both can — and famously do — rearrange before that second step. Run SN1 on a substrate primed for a shift, and the nucleophile ends up bonded to a carbon that never held the original leaving group.
The same logic underlies a misconception worth clearing up about addition. When HBr adds to an alkene, Markovnikov's rule is often taught as 'H goes to the carbon with more hydrogens.' The honest version is that the proton adds wherever it makes the most stable carbocation — and that cation, like any other, can then rearrange before bromide arrives. Acid-catalyzed reactions of alkenes and alcohols are full of this. So Markovnikov orientation and rearrangement are the same principle wearing two hats: the system always routes through the most stable cation it can reach.