Carbocations — How Reactions Happen

A carbocation is the most useful unstable thing in organic chemistry. It is a carbon two electrons short of an octet: three bonds, no lone pair, a full positive charge, and an empty orbital sitting where a fourth bond ought to be. A great many of the reactions still ahead pass through one, and almost every question about them comes down to two skills — judging how tolerable a particular cation is, and noticing when it will quietly turn into a different one before anything else gets a chance to happen.

What it is, and how to check that you drew it

Count the electrons on the charged carbon and the charge is forced rather than remembered. Carbon brings four valence electrons. In a carbocation it owns no lone pairs and half of each of three bonding pairs, so the formal-charge arithmetic from Foundations reads 4 − (0 + 3) = +1. Three bonds and nothing else is the whole definition, and the six electrons in those three bonds are two short of an octet. That shortfall is the reactivity.

Three groups and no lone pair also fixes the shape. There are only three electron domains around that carbon, so they spread as far apart as they can get: trigonal planar, bond angles near 120°, the carbon sp²-hybridized. That uses three of carbon's four valence orbitals, and the fourth — an unhybridized p orbital — is left empty and standing perpendicular to the plane of the three bonds, one lobe above it and one below.

What the words sp², trigonal planar, empty p orbital actually look like. Losing the leaving group takes a bonding pair away from a tetrahedral carbon, and the three groups that remain flatten out into a plane — which leaves the fourth orbital unhybridized, unoccupied, and standing at right angles to them with a lobe on each side.The edge-on view is the one that matters later. There is nothing above the plane and nothing below it, so the two faces of the cation are indistinguishable — which is why a carbon that was a stereocenter before it ionized has no handedness left at all once it has.

The parent it came from was tetrahedral and sp³. Making a carbocation therefore flattens the carbon, and that flattening is not cosmetic: it is why the two faces of the cation are identical, and why an intermediate with a positive charge behaves so differently from the compound it came from.

A carbocation has no lone pair. The commonest drawing error in this entire course is a positively charged carbon drawn with two dots on it. Put a lone pair there and you have drawn a carbon with three bonds and a pair, which is five electrons of its own — a carbanion, formal charge −1, the exact opposite species. Empty orbital, no dots.

Where they come from

Two routes account for nearly all of them, and you have already met the first. A leaving group departs with its bonding pair. That is exactly the event the Leaving groups section was about: the C–X bond breaks heterolytically, the leaving group takes both electrons and becomes an anion or a neutral molecule, and the carbon it abandoned is left with three bonds and the charge. Whether that step is fast enough to be worth writing depends on two things at once — how stable the leaving group is once gone, and how stable the cation is once made.

A π bond is protonated. The two electrons of a C=C double bond are loosely held and reach out into space, so they can attack a proton. One of the two alkene carbons ends up bonded to the new hydrogen; the other is left with three bonds and no share of that pair, which makes it the cation. That is a preview — the chemistry belongs to Alkenes & Alkynes — but it is worth holding now, because the rule that decides which carbon takes the proton is nothing but the stability ordering below.

Both faces of the empty p orbital are equivalent, and nothing sits on either of them. So a nucleophile arriving later has no reason to prefer one side, and if the carbon was a stereocenter before it ionized, the handedness that Stereochemistry taught you to name is destroyed the instant the cation forms — not scrambled afterwards, gone at that step. Racemization in later mechanisms is this one geometric fact arriving with a name attached.

Stability, and why alkyl groups help

The ordering every exam wants is 3° > 2° > 1° > methyl, where the degree counts how many carbons are attached to the cationic carbon itself. Two effects produce it, and they push the same way.

Hyperconjugation is the larger of the two. An alkyl group next door brings its own C–H sigma bonds, and when one of those bonds happens to lie parallel to the empty p orbital, the electron pair inside it can spill partway into that orbital. Nothing breaks and no new full bond forms; the pair is simply shared between two places at once, which drains a little positive charge off the cationic carbon and spreads it onto the neighboring C–H bond. Every extra alkyl group brings more bonds positioned to do this: a methyl cation has none available, an ethyl cation three, an isopropyl cation six, and the tert-butyl cation nine. C–C sigma bonds donate the same way, which is why a quaternary neighbor helps too.

Hyperconjugation, drawn rather than named. A C–H bond on the neighboring carbon that happens to lie parallel to the empty p orbital can let its electron pair spill partway in — not a new bond, just a pair with two places to be. C–C bonds do the same, which is why a neighbor with no hydrogens on it still helps.The count on the right is the whole ordering in one column. It also warns you off the commonest misreading: what is being counted is bonds on the carbons attached to the charge, not carbons anywhere in the molecule. A large primary cation has three of these and is still a primary cation.

Induction is the smaller effect and the easier one to state. A carbon chain is more polarizable than a hydrogen atom — its electron cloud is bigger and looser — so an alkyl group slides electron density along the sigma bond toward an adjacent positive charge more readily than an H does. Neither effect moves the charge onto another atom outright; both of them smear it.

The sizes involved are large. In the standard solvolysis comparison — solvolysis meaning a reaction in which the solvent is itself the nucleophile, here an alkyl bromide warmed in aqueous ethanol — a tertiary substrate reacts on the order of 10⁶ times faster than methyl. Read that number with one honesty attached: methyl and primary substrates do not really ionize at all, and whatever rate they show in such a table is mostly a different mechanism doing the work. The million-fold figure is best taken as "tertiary goes and methyl does not" with a scale bar on it.

Six rungs and two rules. Going down the alkyl series, each extra group adds three more bonds that can donate into the empty orbital. Going below that series, resonance takes over and does something hyperconjugation cannot: it moves the charge onto other atoms outright, which is why a formally primary benzylic cation sits on the same rung as a tertiary alkyl one.The top two rungs are the ones worth learning as exclusions rather than as a ranking. A vinyl or aryl cation is not merely poor; it does not form, so a mechanism that needs one is the wrong mechanism. Everything from the third rung down is a real intermediate that a real reaction passes through.

Resonance beats substitution

Hyperconjugation only lends the charge out; resonance hands it over. If the cationic carbon sits next to a π bond or an aromatic ring, the p orbital of the cation is part of a continuous conjugated system, and the charge is genuinely delocalized onto other atoms. An allylic cation (next to a C=C) spreads its charge over two carbons; a benzylic cation (next to a benzene ring) spreads it over four. The consequence is worth memorizing as a ranking rather than re-derived each time: a primary benzylic cation is about as stable as a tertiary alkyl one, and a primary allylic cation is comparable to a secondary alkyl one. Counting substituents at the cationic carbon is the first thing you do; checking for an adjacent π system is the thing that overrides it.

A lone pair does better still. Put an oxygen or a nitrogen directly on the cationic carbon and its lone pair can close the gap completely, giving a second resonance structure with a real C=O or C=N π bond and, crucially, an octet on every atom. The charge moves onto the heteroatom, which is an expensive place for it — oxygen is electronegative and does not want to be positive — but a structure where everybody has eight electrons beats one where a carbon has six, and the hybrid comes out far more stable than any alkyl cation. The oxygen version has a name, the oxocarbenium ion, and it is the intermediate that makes acetal chemistry work in Carbonyl Chemistry; the nitrogen version is an iminium ion. Seeing a heteroatom beside a would-be cation should change your answer.

Three cations that beat the alkyl series, and the reason is the same each time: the charge stops being on one atom. The allylic and benzylic hybrids are drawn with dashed partial bonds and fractional charges, which is what "delocalized" means in a picture. The oxygen case is drawn as two structures because the second one is the point — every atom in it has an octet.The oxygen pays for the privilege: a positively charged oxygen is not a comfortable thing. It is still the better structure, because a complete octet on every atom beats six electrons on a carbon by more than an electronegative atom minds carrying a charge. The nitrogen version, an iminium ion, is better still for exactly the same reason and one step further along it.

The two that never form: vinyl and aryl

At the other end, two arrangements are so bad that the cation simply does not appear, and both traps turn on geometry rather than on counting substituents.

A vinyl cation is a positive carbon that is itself part of a C=C double bond. It has only two groups left on it, so it is linear and sp-hybridized, and that is the first problem: an sp orbital has fifty percent s character, holds its electrons closer to the nucleus, and is the worst place in organic chemistry to put a positive charge. The second problem is the one students expect to rescue it and which does not. The empty p orbital of that carbon is perpendicular to the π bond beside it, not parallel to it, so the two cannot overlap and the π electrons cannot delocalize into the empty orbital at all. No resonance, no hyperconjugation worth the name.

An aryl cation — the positive charge on a carbon of a benzene ring — fails for the same geometric reason in a different shape. The ring holds that carbon at sp² and will not let it relax, so the empty orbital is one of its sp² hybrids, lying in the plane of the ring and pointing outward. The ring's π cloud stands perpendicular to that plane. The empty orbital and the six π electrons are at right angles to each other and never meet, which is why the most delocalized system in the course cannot help the one cation sitting inside it. This is the single reason vinyl halides and aryl halides are inert to every mechanism that needs a carbocation, and it is why aromatic rings are substituted by an entirely different chemistry in Aromatic Chemistry.

Why the two most promising-looking neighbors are useless. Overlap needs orbitals that point the same way, and in both of these the empty orbital is at right angles to the π system beside it. A benzene ring is the most delocalized thing in the course and it can do nothing for a positive charge on one of its own carbons.Watch the hybridization too, because it changes between the two. The vinyl cation relaxes to sp and linear, putting the charge on the carbon with the most s character in the course. The aryl cation cannot relax at all: the ring holds it at sp², with the empty hybrid aimed outward. Different geometries, the same verdict — vinyl and aryl halides do not ionize.

Rearrangements: cations do not stay put

A carbocation is a real intermediate with a real, if short, lifetime. If a better cation is one small motion away, it will usually get there before a nucleophile arrives. The motion is a 1,2-shift: a group on the carbon adjacent to the charge moves across to it, taking its bonding pair along, and the positive charge ends up on the carbon the group just left.

Two groups do this routinely. A 1,2-hydride shift moves an H with its pair; a 1,2-methyl shift (more generally an alkyl shift) moves a CH₃ the same way. The arrow is the part that gets drawn wrongly: it starts on the bond that is migrating, not on the atom and not on the positive carbon, because what moves is a bond and the two electrons in it. Draw the tail on the C–H bond and the head on the cationic carbon, and the charge takes care of itself.

The two shifts that matter, drawn with the arrow in the right place. What migrates is a bond and the pair inside it, so the tail goes on the C–H or C–CH₃ bond — never on the atom, and never on the positive carbon. The group lands on the cation; the charge ends up on the carbon the group just left.Both of these are the same move with a different passenger, and both are famous precisely because the starting cation is primary. Neopentyl substrates in particular are notorious: there is no hydrogen at all on the neighboring carbon, so a student hunting only for hydride shifts finds none and reports an unrearranged product that never forms.

The rule of thumb is short. A cation rearranges if one shift would give a more stable cation, and not otherwise. It does not shift to an equally good cation for no reason, it never shifts to a worse one, and it does not keep going once it has reached the best cation available — there is no "past" the most stable one. The test is never how far a hydrogen would have to travel; it is what cation the shift leaves behind. A hydrogen two carbons away is out of reach regardless, because only the immediate neighbors can reach.

Worked example 1 — a methyl shift, one arrow at a time

3,3-dimethylbutan-2-ol, (CH₃)₃C–CH(OH)–CH₃, treated with HBr. Number it: the OH is on C2, and C3 is a quaternary carbon carrying three methyls.

Step 1 — activate. Hydroxide will not leave, so the acid protonates the oxygen first and the departing group becomes neutral water. That is the move from the Leaving groups section, used here exactly as it was taught.

Step 2 — ionize. Water leaves with the bonding pair. C2 now has three bonds — a methyl, a hydrogen and C3 — so the cation is secondary.

Step 3 — survey the neighbors. C1 is a methyl bonded straight onto the cation; moving one of its hydrogens over would put the charge on a primary carbon, which is uphill, so nothing happens there. C3 is the interesting one. It carries no hydrogen at all — it is quaternary — but it does carry three methyl groups, and moving one of them across leaves the charge on C3, which still holds two methyls and the rest of the chain. That is a tertiary cation.

Step 4 — the shift. Tail of the arrow on the C3–CH₃ bond, head on C2. The methyl and its two electrons land on C2; the charge slides to C3.

Step 5 — capture. Bromide bonds to C3. The product is 2-bromo-2,3-dimethylbutane, not the 2-bromo-3,3-dimethylbutane you get by swapping OH for Br in place. Note what the exam is testing: the substrate has no hydrogen on the neighboring carbon, so a student looking only for hydride shifts finds none and reports the unrearranged product.

Worked example 2 — ring expansion, the exam's favorite

(Chloromethyl)cyclobutane: a cyclobutane ring carrying a CH₂Cl group. Warmed in water, it gives cyclopentanol — a five-membered ring from a four-membered starting material.

Step 1. Chloride leaves. On paper this gives a primary cation on the exocyclic CH₂, which is a terrible cation and the honest reason the reaction is slow.

Step 2 — two shifts are on offer. The ring carbon next door carries one hydrogen, so a hydride shift would give a secondary cyclobutyl cation. But that ring carbon is also bonded to two ring carbons, and migrating one of those C–C bonds does two good things at once: it puts the charge on a ring carbon, making it secondary, and it moves a carbon out of the four-membered ring and into the chain, so the ring becomes five-membered. A cyclobutane ring carries a great deal of angle strain; a cyclopentane ring carries very little. The ring-expansion shift wins.

Step 3. The result is the cyclopentyl cation. Water attacks it, a second water removes the proton, and the product is cyclopentanol.

Ring expansion, which is a 1,2-alkyl shift whose migrating group happens to be part of a ring. The bond that moves is a ring C–C bond on the carbon next to the charge; when its far end lands on the cation, that carbon has left the ring and joined the chain, so the ring grows by one and the charge stays behind.Drawn as two steps for bookkeeping, and slightly false as physics: a free primary cation has no lifetime worth speaking of, so the ring bond is already on its way as the chloride leaves. That is exactly why this primary substrate reacts at all. The practical rule: whenever a rearrangement would also relieve ring strain, check it before you check for a hydride.

The honesty note. Drawing this as "ionize, then shift" is good bookkeeping and slightly false as physics. A free primary cation is too unstable to have a lifetime, so the ring bond starts migrating as the chloride leaves, in one motion — which is precisely why a primary substrate like this reacts at all rather than not at all. Any time a rearrangement would relieve ring strain, look for it before you look for a hydride.

Worked example 3 — rank four cations

Put these in order of stability: (i) the tert-butyl cation, (CH₃)₃C⁺; (ii) the benzyl cation, PhCH₂⁺; (iii) CH₃O–CH₂⁺; (iv) the phenyl cation, C₆H₅⁺.

Sort by mechanism of stabilization, not by size. (iii) has an oxygen lone pair directly on the charged carbon, so it can be written with a C=O ⁺ π bond and a complete octet everywhere — the strongest stabilization on the list, even though the carbon is formally primary. (ii) is primary too, but the ring delocalizes its charge over four carbons, which puts it in the same league as a tertiary alkyl cation. (i) has three alkyl groups and nine C–H bonds hyperconjugating, which is the best an alkyl cation can do and still less than genuine delocalization. (iv) is the aryl cation, whose empty sp² orbital is perpendicular to the ring π system and gets nothing from it at all.

Answer: (iii) > (ii) ≈ (i) ≫ (iv). The trap is (iv): a molecule with a benzene ring in it looks stabilized, and here the ring is useless because of where the empty orbital points. The other trap is (iii), which has the fewest carbons of the four and is nonetheless the winner.

Why a more stable cation also forms faster

Stability is a statement about a species; rate is a statement about a transition state. They are linked here by the Hammond postulate from the previous section. Making a carbocation is strongly endothermic — you are breaking a bond and creating two charged fragments — so the transition state for that step comes late and resembles the cation itself. Anything that lowers the cation's energy therefore lowers the barrier leading to it by almost as much.

That single sentence is why carbocation stability is worth the attention it gets. It is not merely a description of which intermediate survives longest; it is a prediction of which reaction goes faster, which of two possible cations forms, and therefore which product you isolate. Every "more stable cation" argument you will meet — the regiochemistry of alkene additions, the substrate ordering in unimolecular substitution and elimination, the fragments that dominate a mass spectrum — is this link being used.

Four ways to lose the mark. Ranking by the total number of carbons in the molecule instead of the number attached to the charged carbon — a big primary cation is still primary. Forgetting that resonance outranks substitution, and calling a tertiary alkyl cation more stable than a benzylic one. Shifting a group to a carbon that gives a less stable cation, or shifting again after reaching the best one. And drawing the cation with a lone pair on it, which turns it into an anion.

What carries forward

Everything in this section is used rather than revisited. Radical halogenation, next, ranks its intermediates by the same hyperconjugation argument with one electron in the orbital instead of none. Unimolecular substitution and elimination in Substitution & Elimination are built on the cation as a real intermediate, and the rearrangements above are why their products sometimes have the wrong skeleton. The regiochemistry of acid-catalyzed additions in Alkenes & Alkynes is the stability ordering applied to a protonated double bond, and the oxocarbenium ion reappears in Carbonyl Chemistry as the thing an acetal is made through.