Reactive intermediates — carbocations (stability), carbanions, free radicals, carbenes, nitrenes; rearrangements (hydrid: Exercises

Before we start, four words we will lean on constantly — say them in plain language once:

Definition Four anchor words (and the shorthand symbols)

α-hydrogen (alpha-H): a C–H bond on a carbon that is directly bonded to the reactive carbon. In the picture below it is the H sitting one bond away from the special carbon.
Delocalise: to spread a charge or an unpaired electron over more than one atom instead of dumping it all on one — like spreading your weight across many floor-tiles so none cracks.
Migration (1,2-shift): a group hops from the carbon next door to the reactive carbon, carrying its own pair of bonding electrons with it.
The three shorthand symbols you will meet below — write them out in words the first few times:
- +I (positive inductive effect): a group pushes electron density through the sigma bonds toward the reactive carbon (alkyl groups do this). See Inductive Effect.
- -I (negative inductive effect): a group pulls electron density away through the sigma bonds (electronegative atoms like O, N, halogens do this).
- +M (positive mesomeric effect): a neighbouring lone pair or π system donates electrons by resonance (real delocalisation over atoms), which is far stronger than inductive push. See Resonance and Mesomeric Effect.

The "I" stands for Inductive (through-bond pull/push); the "M" stands for Mesomeric, the older word for resonance (through-space, lone-pair/π donation). A leading + means donating toward the reactive carbon; a leading - means withdrawing from it.

Definition Two more working definitions we will use inline (so you never have to click away)

Hyperconjugation — the "no-bond" donation where a σ (sigma) C–H bond that sits right next to (α to) the reactive carbon leaks its two bonding electrons into the neighbouring empty (or half-filled) orbital. Picture it as the C–H bond tilting its electron cloud into the empty p-orbital. More aligned α C–H bonds = more leaking = more spreading of the electron deficiency = lower energy. Full treatment: Hyperconjugation.
s-character — every hybrid orbital is a mix of one round s-orbital (hugs the nucleus tightly) and stretched p-orbitals. The fraction that is s is the "s-character": sp = 50%, sp² ≈ 33%, sp³ = 25%. More s-character means the electrons sit closer to the nucleus, held more tightly (lower energy) — which is exactly what a lone pair on a carbanion wants. Full treatment: Hybridisation and s-character.

Figure — Reactive intermediates — carbocations (stability), carbanions, free radicals, carbenes, nitrenes; rearrangements (hydrid

Look at the board figure: the pale-yellow carbon is the reactive centre; the three chalk-blue C–H bonds hanging off the neighbour carbon are its α-hydrogens. That single picture is the whole game — count blue bonds, count how far a charge can spread, and you can rank almost everything.

Level 1 — Recognition

Goal: name the species and read its electron count off by sight.

L1.1

Q. A carbon carries a positive charge and is bonded to only 3 other groups. How many electrons "belong" to this carbon in its valence shell, what is its hybridisation, and what is the shape?

Recall Solution

What we count. Each of the 3 bonds gives this carbon 1 shared electron → $3$ electrons. A positive carbon has no lone pair, so total $= 6$ electrons (a sextet, one pair short of the octet of 8). Why sp². With only 3 electron groups and an empty p-orbital left over, the carbon spreads its 3 bonds as far apart as possible → $12 0^{\circ}$ apart → sp², planar. Answer: $6$ electrons, sp², planar (trigonal). See Hybridisation and s-character.

L1.2

Q. Classify each carbon species by charge and electron count on the key carbon: (i) $CH_{3}^{-}$ (ii) $CH_{3}^{∙}$ (iii) $: CH_{2}$ (methylene).

Recall Solution

Species	Charge	Electrons on C	Name
$CH_{3}^{-}$	$- 1$	$8$ (3 bonds + 1 lone pair)	carbanion
$CH_{3}^{∙}$	$0$	$7$ (3 bonds + 1 unpaired e⁻)	free radical
$: CH_{2}$	$0$	$6$ (2 bonds + 2 non-bonding e⁻)	carbene
Why these counts. Bonds contribute 1 electron each to this carbon; a lone pair is 2; an unpaired electron is 1. Add them.

L1.3

Q. Which of these is electron-deficient (wants electrons): carbocation, carbanion, free radical, carbene?

Recall Solution

"Electron-deficient" = fewer than 8 electrons on the key atom.

Carbocation: 6 → deficient ✔
Free radical: 7 → deficient ✔
Carbene: 6 → deficient ✔
Carbanion: 8 → electron-rich, the odd one out ✘ Answer: cation, radical, carbene are electron-deficient; carbanion is not.

Level 2 — Application

Goal: apply +I, hyperconjugation, resonance and s-character to concrete species.

L2.1

Q. Count the number of α C–H bonds (hyperconjugating hydrogens) in the tert-butyl cation $(CH_{3})_{3} C^{+}$ .

Recall Solution

Where to look. The α-carbons are the carbons directly bonded to the positive carbon — here, the three methyl groups. Count. Each $CH_{3}$ has $3$ C–H bonds → $3 \times 3 = 9$ α-H. Why it matters. Each aligned α C–H leaks its σ-electron pair into the empty p-orbital (this is hyperconjugation — see the inline definition above and Hyperconjugation) → 9 is a lot of spreading → tert-butyl cation is very stable. Answer: $9$ .

L2.2

Q. Rank the stability of these carbocations: $CH_{3}^{+}, CH_{3} CH_{2}^{+}, (CH_{3})_{2} CH^{+}, (CH_{3})_{3} C^{+}$ .

Recall Solution

Tool used and why. No resonance available here (pure alkyl), so we rank by +I (electron push through bonds) plus hyperconjugation (σ C–H leaking into the empty p-orbital) (Inductive Effect, Hyperconjugation). More alkyl groups = more electron push into the sextet. Count α-H: $CH_{3}^{+} (0) < CH_{3} CH_{2}^{+} (3) < (CH_{3})_{2} CH^{+} (6) < (CH_{3})_{3} C^{+} (9)$ . Answer: $(CH_{3})_{3} C^{+} > (CH_{3})_{2} CH^{+} > CH_{3} CH_{2}^{+} > CH_{3}^{+}$ i.e. $3° > 2° > 1° > methyl$ .

L2.3

Q. Rank the stability of these carbanions: $CH_{3}^{-}, (CH_{3})_{3} C^{-}$ , and the acetylide $HC \equiv C^{-}$ .

Recall Solution

Tool and why. A carbanion is electron-rich, so alkyl +I now destabilises it (pushes more electrons onto an already-crowded lone pair). Two effects compete:

Fewer alkyl groups → more stable → $CH_{3}^{-} > (CH_{3})_{3} C^{-}$ .
s-character: the acetylide lone pair sits in an sp orbital (50% s), held tight and close to the nucleus → lowest energy of all (recall the inline s-character definition; Hybridisation and s-character). Answer: $HC \equiv C^{-} > CH_{3}^{-} > (CH_{3})_{3} C^{-}$ .

Level 3 — Analysis

Goal: choose between competing effects (resonance vs hyperconjugation vs inductive), including for radicals.

L3.1

Q. Which is more stable: the benzyl cation $C_{6} H_{5} CH_{2}^{+}$ (a "1°" cation by carbon count) or the tert-butyl cation $(CH_{3})_{3} C^{+}$ (3°)?

Recall Solution

Competing tools. tert-butyl uses hyperconjugation (9 α-H). Benzyl uses resonance — the empty p-orbital conjugates into the ring, so the "+" is drawn on the CH₂ and on three ring carbons (ortho, ortho, para): the charge lives on 4 atoms (Resonance and Mesomeric Effect). Ranking rule. Resonance (real delocalisation over atoms) beats hyperconjugation. Spreading a charge over 4 atoms lowers energy more than partial σ-donation from 9 bonds on one atom. Answer: benzyl cation is more stable, even though it is "1°" by simple carbon-counting.

L3.2

Q. The methoxymethyl cation can be written with two resonance forms (in plain ASCII): CH3-O(+)=CH2 <-> CH3-O-CH2(+) Here (+) marks the atom carrying the positive charge. Oxygen is electronegative, so its -I effect pulls electrons away from the cation. Is this cation destabilised or stabilised overall?

Recall Solution

First, the word. An oxocarbenium ion is just a cation where a positively charged carbon is directly bonded to oxygen and can share the charge with that oxygen's lone pair — literally "oxo" (oxygen) + "carbenium" (a carbocation). It is the species drawn on the left, CH3-O(+)=CH2. Two opposing effects.

-I (inductive withdrawal): O pulls electron density away through the bonds → destabilising a cation. But -I acts weakly, only through the sigma bonds.
+M (resonance / lone-pair donation): O has a lone pair right next to the empty p-orbital; it donates that pair to form a full $O = C$ double bond, giving every atom an octet. This is powerful, direct delocalisation. Verdict. Resonance (+M) overwhelms inductive (-I). The first resonance form has no electron-deficient carbon at all → this oxocarbenium ion is very stable. Answer: strongly stabilised.

L3.3

Q. Rank these free radicals by stability and name the deciding effect(s): $CH_{3}^{∙}, (CH_{3})_{3} C^{∙}, C_{6} H_{5} CH_{2}^{∙} (benzyl radical) .$

Recall Solution

Key idea. A radical carbon has 7 electrons — one short of the octet, so like a cation it is electron-deficient and is stabilised by the same donors: hyperconjugation, +I, and (strongest) resonance.

$CH_{3}^{∙}$ : 0 α-H, no resonance → least stable.
$(CH_{3})_{3} C^{∙}$ : 9 α-H hyperconjugation + strong +I → 3° radical, stable.
benzyl $C_{6} H_{5} CH_{2}^{∙}$ : the unpaired electron delocalises into the ring over 4 carbons by resonance → most stable. Ranking rule (same as cations): resonance > hyperconjugation. Answer: $C_{6} H_{5} CH_{2}^{∙} > (CH_{3})_{3} C^{∙} > CH_{3}^{∙}$ .

Level 4 — Synthesis

Goal: combine intermediate stability with rearrangement logic to reach a product.

L4.1

Q. The neopentyl cation, in ASCII (CH3)3C-CH2(+), is a 1° cation. Show, with a 1,2-shift, the more stable cation it rearranges to, and name its class.

Recall Solution

Step 1 — spot the better cation (WHAT/WHY). The carbon next door is C(CH3)3 — if the "+" moved there it would be bonded to 3 methyls → 3°, far more stable than the current 1°. Step 2 — migrate a methyl WITH its electrons (WHAT/WHY/PICTURE). A whole $CH_{3}$ group (a methyl shift) hops from the quaternary carbon onto the positive $CH_{2}$ , carrying its bonding pair. The pair neutralises the old centre; the carbon the methyl left is now electron-short → it becomes the new "+". The board figure below traces this hop with a pink arrow.

(CH3)3C-CH2(+) --1,2-methyl shift--> (CH3)2C(+)-CH2CH3 Answer: rearranges to the tertiary (3°) cation (CH3)2C(+)-CH2CH3.

L4.2

Q. 2-methylbutan-2-ol reacts under acidic SN1/E1 conditions to give a carbocation. Does it need to rearrange? Explain.

Recall Solution

One-line reminder of the mechanisms (so you needn't click away). In SN1 (Substitution, Nucleophilic, unimolecular) and E1 (Elimination, unimolecular) the slow first step is the same: the leaving group departs to make a free carbocation, and then a nucleophile attaches (SN1) or a β-H is lost to form an alkene (E1). Both live or die by how stable that carbocation is. Full detail: SN1 and E1 mechanisms. Step 1 — form the cation. Protonate –OH, lose water → the "+" sits on C2, which bears two methyls and an ethyl group → this is already a 3° cation. Step 2 — check for a better neighbour. Every adjacent carbon, if it took the charge, would give a 2° or 1° cation — worse. Rule. Rearrangement only happens if it leads to a more (or equally) stable cation. Here the starting cation is already the best. Answer: No rearrangement — it is already tertiary; any shift would go downhill in stability.

L4.3

Q. In the pinacol rearrangement, a 1,2-diol under acid loses water to give a cation on one carbon; then a group migrates and a C=O forms. State the driving force for the migration in this reaction. (See Pinacol Rearrangement.)

Recall Solution

The engine. After water leaves, we have a carbocation. A neighbouring group (H or alkyl) migrates because the adjacent carbon still carries an –OH whose oxygen lone pair can then form a very stable C=O (oxocarbenium → ketone). Recall from L3.2 that an oxocarbenium is a cation sharing its charge with an adjacent oxygen's lone pair. Why so favourable. The migration lands the "+" on a carbon bearing oxygen; that oxygen donates its lone pair (+M, exactly the L3.2 idea) to form a full carbonyl, an especially low-energy sink. Answer: migration is driven by formation of the resonance-stabilised oxocarbenium / carbonyl (C=O), a much more stable arrangement than the plain carbocation.

Level 5 — Mastery

Goal: full multi-step reasoning across intermediate type, geometry and spin. (Uses the singlet/triplet idea introduced at the top of the page.)

L5.1

Q. A carbene $: CH_{2}$ adds to cis-2-butene. If the carbene is a singlet, the product ring keeps the cis geometry (stereospecific). If it is a triplet, geometry is scrambled. Explain why, using electron pairing.

Recall Solution

Singlet carbene. Its two non-bonding electrons are paired in one orbital → it can form both new C–C bonds at once (concerted). Because nothing rotates during a single instant, the cis relationship is frozen in → stereospecific. Triplet carbene. Its two non-bonding electrons are unpaired, in separate orbitals with parallel spins → it behaves like a diradical. It forms one bond first, leaving a diradical intermediate that lives long enough for a C–C single bond to rotate before the second bond closes (spin must flip first). Rotation scrambles cis/trans → non-stereospecific. Answer: paired electrons (singlet) → concerted, retains geometry; unpaired (triplet) → stepwise diradical, rotation allowed → scrambled.

L5.2

Q. True or false, with reason: "Every carbene has a triplet ground state because methylene $: CH_{2}$ does."

Recall Solution

The kernel of truth. Parent methylene $: CH_{2}$ genuinely has a triplet ground state (triplet lies ~9 kcal/mol below its singlet). Why the blanket claim is false. Ground-state multiplicity depends on substituents. π-donor groups (–OR, –NR₂, halogens) donate lone-pair density into the carbene's empty orbital, which stabilises the singlet. So substituted carbenes — famously all N-heterocyclic carbenes — are singlet ground states. Answer: False. Decide case-by-case from the substituents; only bare $CH_{2}$ defaults to triplet.

L5.3

Q. (nitrenes — the nitrogen cousins) A nitrene is the nitrogen analogue of a carbene, written R-N: — a monovalent nitrogen with only 6 electrons (1 bond + non-bonding electrons). It is the key electron-deficient centre in the Hofmann Rearrangement. (a) How many electrons and how many bonds does the nitrene nitrogen have? (b) Describe its singlet and triplet electron arrangements, and say which is usually the ground state for a simple nitrene.

Recall Solution

(a) Count. One N–R bond gives 1 electron to N; the rest are non-bonding. To reach the sextet ( $6$ electrons) N must carry 4 non-bonding electrons in addition to that 1 bond. So: 1 bond, 6 electrons total (electron-deficient, like a carbene). (b) Arrangements.

Singlet nitrene: the 4 non-bonding electrons form two lone pairs (all paired, no unpaired electrons).
Triplet nitrene: one lone pair (2 electrons) + two unpaired electrons in separate orbitals with parallel spins → a diradical. For simple nitrenes the triplet is usually the ground state (same reasoning as bare methylene: parallel spins in separate orbitals gain exchange stabilisation — Hund's rule).

L5.4

Q. (integrated) Order these three species by stability within their own class, and name the single dominant effect deciding each: (i) allyl cation $CH_{2} = CH-CH_{2}^{+}$ vs isopropyl cation $(CH_{3})_{2} CH^{+}$ (ii) singlet nitrene vs triplet nitrene for a simple R-N: — which is usually ground state? (iii) $HC \equiv C^{-}$ vs $H_{2} C = CH^{-}$ carbanion.

Recall Solution

(i) Allyl cation delocalises "+" over two carbons by resonance; isopropyl only has hyperconjugation/+I on one carbon. Resonance wins → allyl > isopropyl. (ii) For simple nitrenes, the triplet (one lone pair + two unpaired electrons, a diradical) is usually the ground state — same as parent carbene/methylene. Dominant idea: Hund's-rule / exchange stabilisation of parallel spins when no π-donor pins down the singlet. (iii) Compare s-character of the orbital holding the lone pair: acetylide lone pair is in an sp orbital (50% s), vinyl anion in sp² (33% s). Higher s-character holds the pair tighter → $HC \equiv C^{-} > H_{2} C = CH^{-}$ . Dominant effect: s-character (Hybridisation and s-character).

Recall Rapid self-check (cover the answers)

Positive carbon has how many electrons? ::: 6 (a sextet) Carbocation stability order (pure alkyl)? ::: $3° > 2° > 1° > methyl$ Carbanion stability order (pure alkyl)? ::: $methyl > 1° > 2° > 3°$ (flipped) Free-radical stability order (with resonance)? ::: benzyl > 3° > 2° > 1° > methyl (same trend as cations) Resonance vs hyperconjugation — which wins? ::: resonance (delocalises over more atoms) α-H count in $(CH_{3})_{3} C^{+}$ ? ::: 9 What is hyperconjugation in one line? ::: an α σ C–H bond leaking its electron pair into the neighbouring empty/half-filled orbital What do the letters I and M stand for? ::: I = Inductive (through-bond), M = Mesomeric (resonance) A 1,2-shift happens only when… ::: the new cation (or C=O / aromatic sink) is MORE stable Singlet carbene reaction stereochemistry? ::: stereospecific (concerted, retains geometry) Nitrene nitrogen — how many electrons / bonds? ::: 6 electrons, 1 bond (electron-deficient) Highest-s-character carbanion of C, sp/sp²/sp³? ::: sp (acetylide, most stable)