4.1.9 · D5General Organic Chemistry (GOC)
Question bank — Reactive intermediates — carbocations (stability), carbanions, free radicals, carbenes, nitrenes; rearrangements (hydrid
Before we start, one shared vocabulary so nothing below is a mystery. Read this whole block once — every jargon word used later is defined here and pinned to a picture.
- The symbol "~" means "approximately" — a value that is close to but not exactly the number given. So "~103°" reads "about 103 degrees, not a precise measurement".
- α (alpha) and β (beta) carbons = a way of naming how far a carbon is from the reactive centre. The reactive carbon itself (the one carrying the +, −, or unpaired electron) is the α-carbon. The carbon directly bonded to it — one bond away — is a β-carbon. An α C–H bond is a C–H on the reactive carbon is not what we count; the hydrogens that matter for hyperconjugation are the ones on the neighbouring (β) carbons — see the figure below for exactly which H's are counted.
- ipso carbon = in a benzene ring, the ring carbon directly attached to the substituent (here, attached to the charged carbon). "Ipso" is Latin for "itself" — the carbon at the point of attachment. From there we count outward: ortho (next-door ring carbons), meta, para (opposite). The benzyl figure below shows which ring carbons carry the charge.
- Electron-deficient species = a carbon (or nitrogen) missing part of its full octet: carbocation (6 e⁻), radical (7 e⁻), carbene/nitrene (6 e⁻). These want electrons.
- Electron-rich species = a carbanion (8 e⁻, one lone pair). It wants to give away electrons.
- +I (inductive donation) = an alkyl group pushing electron density down a σ-bond (a single bond) toward a nearby atom — like a slow, steady current along a wire, no picture needed. See Inductive Effect.
- Hyperconjugation = a filled C–H σ-bond overlapping side-on with an empty or half-empty p-orbital next door, so electron density flows from the σ-bond into that orbital. This is a real orbital-overlap, drawn in the figure below — not "leaking", but the two orbital lobes physically touching. See Hyperconjugation.
- Resonance / +M = a lone pair or π-bond sharing electrons over several atoms, drawn as multiple structures. The benzyl figure below is a resonance example. See Resonance and Mesomeric Effect.
- s-character = the fraction of an orbital that is "s". sp = 50%, sp² ≈ 33%, sp³ = 25%. Higher s-character orbitals hug the nucleus tighter. See Hybridisation and s-character.
The two figures below carry the mechanisms so the reveals can refer to them:
True or false — justify
True or false: A benzyl carbocation (its charged carbon is only 1° by carbon count) is less stable than a tert-butyl (3°) carbocation.
False. Resonance beats hyperconjugation — in the benzyl resonance figure (s03) the positive charge delocalises onto the ipso, two ortho, and one para ring carbons (four positions), which lowers its energy more than the 9 hyperconjugating C–H bonds of tert-butyl. Always scan for resonance first.
True or false: For carbanions the stability order is .
False. That is the carbocation order. A carbanion is electron-rich, so the +I push of alkyl groups piles more negative density onto an already-negative carbon and destabilises it: .
True or false: Free radicals follow the same stability trend as carbanions.
False. A radical is electron-deficient (7 e⁻, one short of an octet), so it behaves like a carbocation: , stabilised by hyperconjugation and resonance.
True or false: A carbanion is planar and sp² like a carbocation.
False. A carbocation is sp² planar with an empty p-orbital; a carbanion carries a lone pair, so it is sp³ and pyramidal (the lone pair pushes the three bonds into a pyramid).
True or false: Every carbene has a triplet ground state.
False. Parent methylene does, but ground-state multiplicity depends on substituents. π-donor groups (–OR, –NR₂, halogens) stabilise the singlet; all N-heterocyclic carbenes are singlet ground states. Decide case-by-case.
True or false: A triplet carbene is linear because its two non-bonding electrons repel maximally.
False. Triplet methylene is still bent (~133°), not linear (see the carbene figure s04). VSEPR reason: even in the triplet the two non-bonding electrons sit in two separate half-filled orbitals that still occupy space around carbon and push the two C–H bonds apart to a bent shape; only if that non-bonding density vanished entirely would the molecule straighten to 180°.
True or false: An alkynide ion is unusually stable for a carbanion.
True. The lone pair sits in an sp orbital (50% s-character), which hugs the nucleus and holds the extra electrons tightly, lowering the energy: sp > sp² > sp³ for carbanion stability.
True or false: A reactive intermediate and a transition state are the same kind of thing.
False. An intermediate sits in a valley (a real energy minimum, it exists for a finite time); a transition state is a peak (a maximum you pass through instantly) — see the energy-profile figure s04. You can, in principle, trap an intermediate but never a transition state.
True or false: An –OH group next to a carbocation always destabilises it because oxygen is electron-withdrawing (–I).
False. Oxygen's lone-pair donation (+M) into the empty p-orbital overwhelms its –I pull, giving a very stable oxocarbenium . Resonance beats induction.
Spot the error
Spot the error: "Ethyl cation has 3 α-hydrogens and methyl cation has 1, so methyl is only slightly less stable."
Two mistakes. First, hyperconjugating H's live on the β-carbon (the neighbour), not the α (reactive) carbon — see figure s01. Second, methyl cation has no neighbouring carbon at all, so zero hyperconjugating C–H bonds; it is dramatically the least stable.
Spot the error: "In a 1,2-hydride shift the proton hops to the neighbouring carbon."
It migrates as a hydride — the hydrogen leaves with its bonding electron pair. That pair neutralises the old cationic carbon and leaves the carbon it came from positive.
Spot the error: "A 1° carbocation can rearrange to a 3° by moving the positive charge two carbons away."
A 1,2-shift moves a group (and the charge) only one carbon (from the β-carbon onto the α, reactive carbon). The charge relocates by one position per shift; you cannot skip a carbon in a single step.
Spot the error: "Neopentyl cation can't rearrange because it has no β-hydrogen to shift."
True it has no β-H (no hydrogens on the neighbouring carbon), but that β-carbon carries three methyl groups. A 1,2-methyl shift moves one methyl (with its electrons) onto the 1° centre, generating a stable 3° cation.
Spot the error: "Because tert-butyl cation is 3°, it will still rearrange to something even more stable."
There is nothing more substituted to shift to — a 1,2-shift only helps when it increases substitution/resonance. A cation only rearranges toward greater stability; a 3° with no resonance option stays put.
Spot the error: "A singlet carbene is a diradical, so its reactions lose stereochemistry."
The triplet is the diradical (two electrons in separate orbitals, parallel spins) that reacts stepwise and scrambles stereochemistry. The singlet has both electrons paired in one orbital, reacts concertedly, and is stereospecific (retains geometry).
Why questions
Why does more s-character stabilise a carbanion but this argument isn't the first thing you check for a carbocation?
For a carbanion you want the lone pair held tight, so high-s orbitals help. A carbocation has an empty p-orbital, so what matters is electrons donated in (resonance, then hyperconjugation, then +I) — s-character of an empty orbital gives nothing to hold.
Why do we say "resonance beats hyperconjugation beats inductive"?
Resonance spreads charge fully over several atoms (largest delocalisation, figure s03); hyperconjugation only shares partial σ-density into one adjacent orbital (figure s02); +I is a weak polarisation along one σ-bond. More spreading = lower energy = more stable. This ordering underlies Markovnikov Addition and SN1 and E1 mechanisms product choices.
Why does an alkyl group's +I help a cation but hurt a carbanion?
+I pushes electron density toward the neighbouring atom. A cation is electron-poor and welcomes it (stabilised); a carbanion is already electron-rich and is overloaded by it (destabilised). Same push, opposite sign of the host.
Why must the migrating group in a 1,2-shift take its bonding electrons along?
If it left the electrons behind it would just create a second cationic centre — no net gain. Taking the pair neutralises the atom it moves to and shifts the deficiency to where it departed, so exactly one positive centre survives, now more stable.
Why can a benzyl cation delocalise over "four positions" in the ring?
The empty p-orbital conjugates into the aromatic π system; pushing arrows places the positive charge on the ipso carbon (the point of attachment) plus the two ortho and one para ring carbons — four sites sharing the load (figure s03), far more than any hyperconjugation could achieve.
Why does the Pinacol Rearrangement rely on the same 1,2-shift logic as simple carbocation rearrangements?
After a hydroxyl leaves as water, a carbocation forms; a neighbouring alkyl/aryl group (with its electrons) migrates to give a much more stable oxocarbenium (lone-pair-stabilised) cation. It's a 1,2-shift driven by moving toward a more stable positive centre.
Edge cases
Edge case: Rank , , , .
. Allyl wins on resonance (charge over 2 carbons); among the rest, more alkyl substitution = more hyperconjugation + +I; methyl (no neighbouring carbon, 0 hyperconjugating H) is last.
Edge case: What happens to carbocation stability if the empty p-orbital is held perpendicular to an adjacent π-system (locked geometry)?
Resonance switches off — overlap requires the empty p-orbital to be parallel to the π-system (the same side-on overlap shown in figure s02 for hyperconjugation). Perpendicular locking removes the delocalisation, so a "benzylic-looking" cation can behave like a plain alkyl cation. Geometry gates resonance.
Edge case: Is the parent nitrene (or R–N:) usually singlet or triplet in the ground state?
Triplet for simple nitrenes — one lone pair (2 e⁻) plus two unpaired electrons in separate orbitals (a diradical). The singlet, with two lone pairs and no unpaired electrons, is higher in energy unless stabilised by substituents.
Edge case: A cation is drawn next to a carbon bearing no hydrogens and no π-system or lone pair. What stabilises it?
Only the +I of the surrounding alkyl framework and any C–C hyperconjugation (σ C–C bonds can also overlap into the empty p-orbital, though weaker than σ C–H). With no adjacent H, resonance, or heteroatom, it is a poorly stabilised cation and prone to rearrange if a better centre is one carbon away.
Edge case: For the divalent species, which is "more electron-deficient at the reactive atom" — a singlet carbene or a triplet carbene?
Both have 6 electrons on carbon, so the count is identical. They differ in distribution: singlet has an empty p-orbital plus a filled sp² lone pair (behaves electrophilic + nucleophilic at once); triplet has two half-filled orbitals (behaves as a diradical). Same deficiency, different personality.
Edge case: Nitrenes drive the Hofmann Rearrangement. What role does the nitrene-like (or concerted equivalent) intermediate play there?
An electron-deficient nitrogen centre forms and an adjacent group (with its electrons) migrates to nitrogen — a 1,2-shift onto N that relieves the deficiency, exactly mirroring alkyl migration onto a cationic carbon.
Recall Quick self-test before you leave
One rule to recite ::: Electron-deficient species (cation, radical, carbene, nitrene) love electron donors; the electron-rich carbanion loves electron withdrawers. Stability priority for a cation ::: Resonance > hyperconjugation > inductive; scan for resonance first. Direction of every rearrangement ::: Always toward a more stable cation, by a one-carbon (β→α) 1,2-shift, group carrying its electrons.