A 1,2-hydride shift is a carbocation rearrangement: a hydrogen migrates with its bonding pair from an adjacent carbon onto the cationic carbon, moving the charge next door. It happens only when it builds a more stable cation (almost always 2° → 3°).
2-bromo-3-methylbutane should give an alcohol at C2, but the major product is 2-methyl-2-butanol — a hidden 1,2-hydride shift moved the –OH.
1. A carbocation rearranges only to become more stable
More alkyl groups on the positive carbon means more stability (hyperconjugation and induction), giving the ladder 3° > 2° > 1°.
Rearrangement is downhill-only: a cation shifts toward a more stable spot within reach, never the reverse.
2. In a 1,2-hydride shift, a hydrogen migrates with its electrons
A hydride (H with its two bonding electrons) drops into the empty orbital next door, leaving a new empty orbital behind — so the charge slides one carbon over.
It is a purely internal reshuffle — the heavy-atom skeleton is unchanged — and it is fast enough to beat any nucleophile to the original cation.
3. The shift happens only when the new cation is more stable
Compare the current cation with the one you'd get after the shift, and rearrange only if that product cation is more stable.
Once a 3° (or allylic/benzylic) cation is reached, there is nowhere better to go and the shift stops.
4. A worked example: a 2° cation shifts to 3°, then gets trapped
2-bromo-3-methylbutane in water ionizes to a 2° cation, a hydride shifts from the adjacent 3° carbon, and water traps the new 3° cation.
The same logic drives electrophilic additions to alkenes (see Markovnikov additions), where a protonated double bond can shift before the nucleophile arrives.
5. Rearranged products are the fingerprint of a hidden shift
A shift shows up as a product whose position doesn't match the starting material — here, 2-methyl-2-butanol instead of the expected 3-methyl-2-butanol.
A surprising position (or a "major product?" question on a 2° substrate) signals rearrangement; SN2 and E2 are concerted and never rearrange.
6. How to spot when a shift will occur
Checklist: does the mechanism form a free carbocation, is it less than 3°, and does an adjacent H lead to a more stable cation? All three yes → draw the shift first.
If moving a whole alkyl group gives the better cation, an alkyl shift happens instead — same driving force, and both pair with the SN1 mechanism, E1, and Markovnikov additions.
7. Summary
Hydride migrates with its electrons onto the cationic carbon · charge moves one carbon over · happens only downhill, 2° → 3° (3° > 2° > 1°) · seen in SN1, E1, and alkene additions · giveaway is a rearranged product · predict it by forming the cation, checking for a more stable neighbor, then reacting the rearranged cation.
Quiz yourself
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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
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.