A 1,2-hydride shift is a carbocation rearrangement in which a hydrogen (with its bonding pair, i.e. as a hydride, H−) migrates from a carbon adjacent to the empty orbital onto the cationic carbon. The positive charge lands on the carbon the hydride left behind. It happens only when the move builds a more stable cation — almost always 2° → 3° (stability order 3° > 2° > 1°). Because it occurs whenever a free carbocation exists — SN1, E1, and electrophilic additions to alkenes — it is the reason you sometimes get an "impossible," rearranged product.
Ionization of 2-bromo-3-methylbutane should give an alcohol at C2 — instead the major product is 2-methyl-2-butanol, with the –OH on a different carbon. A 1,2-hydride shift, hidden inside the mechanism, is the culprit. This page unpacks exactly how that happens.
1. A carbocation rearranges only to become more stable
Carbocations are electron-deficient and high in energy, so they grab any easy path to lower energy. The single biggest factor is how many alkyl groups touch the positive carbon: neighboring carbons donate electron density through hyperconjugation and induction, cushioning the charge. That gives the familiar stability ladder — tertiary is best, primary is barely viable, methyl essentially never forms.
If a cation happens to sit one carbon away from a spot that would host a more stable cation, it will rearrange to get there. It never rearranges the other way — a 3° cation will not degrade into a 2° one, because that costs energy. The whole phenomenon is downhill-only: rearrangement is the cation shopping for a lower-energy home, and it stops the moment it finds the best one within reach.
2. In a 1,2-hydride shift, a hydrogen migrates with its electrons
The "1,2" means the migration is between adjacent carbons — from the carbon next to the cation onto the cationic carbon itself. What physically moves is not a bare proton but a hydride: the hydrogen carries its two bonding electrons along and drops them into the empty p orbital next door. As those electrons fill the old empty orbital, they leave a new empty orbital on the carbon they departed from — so the positive charge simply slides one carbon over.
Notice that no atoms leave or arrive from outside — it is an entirely internal reshuffle. One C–H bond breaks and a new C–H bond forms on the adjacent carbon in the same motion, and the charge relocates. The carbon skeleton's connectivity of heavy atoms is unchanged in a hydride shift (an alkyl shift, by contrast, moves a carbon group and can rebuild the skeleton). It is fast, essentially barrierless when it leads to a big stability gain, and it happens before any nucleophile or base has a chance to trap the original cation.
3. The shift happens only when the new cation is more stable
This is the single rule that governs whether you should even consider a shift: line up the current cation against the cation you would get after moving a neighboring hydride, and rearrange only if the product cation is more stable. A 2° cation with a 3° position next door will shift; a 2° cation flanked only by other 2° or 1° carbons will not bother.
Once a tertiary (or a resonance-stabilized allylic/benzylic) cation is reached, the rearrangement stops — there is nowhere better to go. So a good habit is to draw the first-formed cation, ask "is there a 3° carbon holding a hydrogen right next to me?", and only then commit to a shift. No stability gain, no shift.
4. A worked example: a 2° cation shifts to 3°, then gets trapped
Take 2-bromo-3-methylbutane in water (an SN1 setting). The bromide ionizes to give a secondary cation. But the very next carbon is a tertiary position carrying a hydrogen — a perfect setup. A hydride slides over, the charge jumps to that carbon, and now we have a far more stable tertiary cation. Water then traps that cation, so the –OH ends up on a carbon the leaving group never touched.
Every carbocation intermediate here is a real, if fleeting, species, and the shift wins the race against the weak nucleophile. The same logic drives electrophilic additions to alkenes (as in Markovnikov additions): protonate a double bond such as isobutylene (CC(=C)C) with HBr, and if the resulting cation can shift to something better before the nucleophile arrives, it will — which is why adding HX or water to certain alkenes gives rearranged, non-obvious products.
5. Rearranged products are the fingerprint of a hidden shift
The tell that a shift occurred is a product whose skeleton or regiochemistry does not line up with the starting material. In the worked example, the naive expectation is substitution at the original carbon (3-methyl-2-butanol); the actual major product is the rearranged 2-methyl-2-butanol. Put the two side by side:
Whenever a problem shows a product with the functional group in a "surprising" place — or asks "what is the major product?" for a 2° substrate — a carbocation rearrangement is almost always the intended twist. SN2 and E2 are concerted and have no carbocation, so they never rearrange; seeing rearrangement tells you the mechanism went through a discrete cation (SN1, E1, or an addition).
6. How to spot when a shift will occur
Work it as a checklist. First, does the mechanism form a free carbocation at all? (SN1, E1, and additions across a π bond do; SN2 and E2 do not.) Second, is that first-formed cation less than tertiary? Third, is there a hydrogen on an adjacent carbon whose departure would leave a more stable — usually tertiary — cation? If all three are yes, draw the 1,2-hydride shift before you let any nucleophile or base react.
One caution: not every migrating group is a hydride. If moving a whole alkyl group (a methyl, say) gives the better cation, an alkyl shift happens instead — same driving force, different migrating group. And because these rearrangements live inside carbocation chemistry, they pair naturally with the SN1 mechanism, E1 eliminations, and Markovnikov additions, where a cation always sits at the center of the story.
7. Summary
A 1,2-hydride shift is a carbocation rearrangement: a hydrogen migrates with its bonding electrons from a carbon adjacent to the empty orbital onto the cationic carbon, moving the positive charge one carbon over. It is strictly downhill — it happens only when a more stable cation results, essentially always 2° → 3° (order 3° > 2° > 1°). Any reaction that generates a free carbocation — SN1, E1, or electrophilic addition to an alkene — can undergo it, and the giveaway is a rearranged product whose functional group or skeleton lands somewhere "unexpected." To predict it: form the cation, check for a more stable neighbor, shift if there is one, then let the nucleophile or base react on the rearranged cation.
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A hydride: the hydrogen moves together with its two bonding electrons and drops them into the empty p orbital on the adjacent cationic carbon. Because the electrons leave the carbon they came from, that carbon becomes the new cationic center — the charge slides one carbon over.
Carbocation stability (3° > 2° > 1°). A shift occurs only when it produces a more stable cation — almost always 2° → 3°. The reverse (3° → 2°) never happens because it would raise the energy.
Ionization gives a 2° cation with a 3° carbon (bearing an H) right next to it. A 1,2-hydride shift converts it to the more stable 3° cation, and water traps that. The –OH therefore ends up on a carbon the bromide never occupied — the rearranged product, 2-methyl-2-butanol.
No. SN2 and E2 are concerted and never form a carbocation, so they cannot rearrange. A rearranged product means a discrete carbocation existed — pointing to SN1, E1, or an electrophilic addition.
Draw this on the whiteboard
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