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

Before we start, three tiny words we will lean on the whole page — defined in plain English so nothing is used before it is earned. (Note: M stands for the mesomeric effect, which is just another name for the resonance effect; +M means a group donates electron density by resonance, −M means it withdraws by resonance.)

Golden ordering we will use again and again: resonance beats hyperconjugation beats inductive.

The scenario matrix

Read the table below as a checklist of every trick this topic can play on you. Each row is one class of question; the last column says which worked example (further down) drills it. If you can answer all ten, no reactive-intermediate question can surprise you.

#	Cell (the scenario class)	Covered by
A	Pure-alkyl carbocation ranking (only +I + hyperconjugation)	Ex 1
B	Resonance overrides the naive alkyl count (benzyl beats 3°)	Ex 2
C	The "trap" case: −I atom that actually stabilises via lone-pair +M	Ex 3
D	Carbanion — the mirror image trend + s-character	Ex 4
E	Free radical ranking (same direction as cations)	Ex 5
F	Degenerate / zero case: methyl $CH_{3}^{+}$ , $CH_{3}^{-}$ — no neighbours at all	Ex 6
G	Rearrangement: 1,2-H shift 2° → 3° (why it moves)	Ex 7
H	Rearrangement, limiting case: 1° → 3° methyl shift (neopentyl)	Ex 8
I	Carbene singlet vs triplet — deciding ground state from substituents	Ex 9
J	Real-world / exam twist: predict Markovnikov product through the most stable cation	Ex 10

The figure below turns that table into a picture. The blue rising line is how carbocation (and radical) stability climbs as you add electron-donating helpers — going left to right, methyl → 1° → 2° → 3°, and finally the benzyl case which leaps above the line because resonance beats everything (that is cells A and B). The orange dashed line is the carbanion trend running the opposite way — the mirror image of cell D. Use it as a one-glance map of the whole page: whenever an example asks "which is more stable?", picture where each species sits on these two lines.

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

Figure 1 — Scenario map. Blue: carbocation/radical stability rises with more α-H / resonance helpers (cells A, B, E). Orange dashed: carbanion stability falls as alkyl +I overloads it (cell D). The red arrow marks resonance overriding the alkyl ladder.

Example 1 — Cell A: pure-alkyl carbocation ranking

Worked example Rank the stability:

$CH_{3}^{+}$ , $CH_{3} CH_{2}^{+}$ , $(CH_{3})_{2} CH^{+}$ , $(CH_{3})_{3} C^{+}$ . Forecast: guess the order now. Which one has the most "helpers" pushing electrons at the hungry carbon?

Step 1 — recall what the positive carbon lacks. A carbocation carbon has only 6 electrons (a sextet) and an empty p-orbital. It wants electron density. Why this step? You cannot rank stability until you know what the molecule is trying to fix — here, the electron shortage.

Step 2 — count the α C–H bonds (the ones on carbons directly attached to the positive carbon). These are the hyperconjugation donors.

$CH_{3}^{+}$ : attached to 0 carbons ⇒ 0 α C–H.
$CH_{3} CH_{2}^{+}$ : one $CH_{3}$ neighbour ⇒ 3 α C–H.
$(CH_{3})_{2} CH^{+}$ : two $CH_{3}$ neighbours ⇒ 6 α C–H.
$(CH_{3})_{3} C^{+}$ : three $CH_{3}$ neighbours ⇒ 9 α C–H. Why this step? More aligned α C–H bonds = more σ-electron density leaked into the empty p-orbital = deficiency spread thinner.

Step 3 — add the +I check. More alkyl groups also push through the bonds (+I). This agrees with the hyperconjugation count, so no conflict. Why this step? When two effects point the same way, you can trust the count.

Answer: $(CH_{3})_{3} C^{+} > (CH_{3})_{2} CH^{+} > CH_{3} CH_{2}^{+} > CH_{3}^{+}$ , i.e. $3° > 2° > 1° > methyl$ .

Verify (explicit count):

cation	α C–H	rank
$(CH_{3})_{3} C^{+}$	9	most stable
$(CH_{3})_{2} CH^{+}$	6	2nd
$CH_{3} CH_{2}^{+}$	3	3rd
$CH_{3}^{+}$	0	least stable

The counts $9 > 6 > 3 > 0$ are strictly decreasing, matching the stability order. ✓

Example 2 — Cell B: resonance overrides the alkyl count

Worked example Rank:

$(CH_{3})_{3} C^{+}$ (3°), $C_{6} H_{5} CH_{2}^{+}$ (benzyl, 1° by carbon count), $CH_{2} = CH - CH_{2}^{+}$ (allyl, 1°). Forecast: the benzyl carbon has only 2 α C–H bonds. By Example-1 logic it should lose to 3°. Does it?

Step 1 — scan for resonance FIRST. Is the empty p-orbital next to a π system or a lone pair?

Benzyl: yes — next to a benzene ring. The $+$ spreads onto the ring (ortho + para positions ⇒ delocalised over 4 carbons).
Allyl: yes — next to one C=C. The $+$ spreads over 2 carbons.
$t$ -butyl: no π next door — only hyperconjugation (9 α-H). Why this step? Resonance is the strongest spreader; if it exists it changes the ranking, so we check it before counting hydrogens.

Step 2 — compare the extent of resonance, and say why more atoms helps. A charge concentrated on one atom raises energy sharply because like-charges (or the electron-shortage) pile up in one place. Spreading the same $+ 1$ over $N$ atoms means each atom carries only about $1/ N$ of the deficiency; the energy penalty scales with the charge squared on each atom, so many small fractions ( $N$ copies of $(1/ N)^{2}$ ) cost far less than one big charge ( $1^{2}$ ). Benzyl spreads over 4 atoms, allyl over 2, so benzyl's per-atom charge is smaller and its energy is lower. Why this step? This is the concrete reason "more atoms of delocalisation = more stable," not just a slogan.

Answer: $benzyl^{+} > allyl^{+} > (CH_{3})_{3} C^{+}$ . A "1°" cation just beat a 3° one.

Verify: delocalisation atoms $4 > 2$ ; per-atom charge $\frac{1}{4} < \frac{1}{2}$ ; and resonance $>$ hyperconjugation, so benzyl and allyl both outrank $t$ -butyl. Consistent with the parent note's rule " $allyl, benzyl > 3°$ ". ✓

See SN1 and E1 mechanisms — these resonance-stabilised cations are exactly why benzylic/allylic substrates fly through $S_{N} 1$ .

Example 3 — Cell C: the −I "trap" (lone pair rescues it)

Worked example Which is more stable:

$CH_{3} CH_{2}^{+}$ (ethyl) or $CH_{3} -O- C + H_{2}$ (methoxymethyl)? Forecast: oxygen is electronegative (−I, it pulls electrons). A cation hates losing electrons — so surely the oxygen version is worse? Guess, then read on.

Step 1 — notice the oxygen has lone pairs sitting right next to the empty p-orbital. Why this step? A lone pair adjacent to an empty orbital is the setup for resonance (+M) — the strongest tool. We must check it before trusting the −I instinct.

Step 2 — write the resonance donation. $CH_{3} -O- C + H_{2} ⟷ CH_{3} - O + = CH_{2}$ The oxygen lone pair swings in, forming a new π bond; every atom now has a full octet (this is an oxocarbenium ion). Why this step? An all-octet resonance form is a huge stabilisation — far bigger than the mild −I penalty.

Step 3 — weigh +M against −I. Resonance (+M) wins over inductive (−I) by our golden ordering.

Answer: $CH_{3} -O- C + H_{2}$ is much more stable than ethyl cation.

Verify: the octet-complete resonance form exists (all atoms satisfied), which is a valid lower-energy structure ethyl cation has no equivalent of. Rule "resonance > inductive" holds. ✓

[!mistake] The instinct to avoid "Electronegative neighbour ⇒ destabilised cation" is wrong when that neighbour has a lone pair. Always ask: can a lone pair donate? If yes, +M dominates.

Example 4 — Cell D: carbanion, the mirror image

Worked example Rank the carbanions:

$CH_{3}^{-}$ , $(CH_{3})_{3} C^{-}$ , $HC \equiv C^{-}$ (acetylide). Forecast: careful — a carbanion is electron-rich. Do alkyl groups help or hurt now?

Step 1 — flip the logic. A carbanion already has a lone pair + negative charge (8 electrons, octet). It dislikes extra electron density. Why this step? Alkyl groups push (+I), which is bad for something already electron-rich. The whole cation trend reverses.

Step 2 — rank the two alkyl ones. $CH_{3}^{-}$ has no pushing groups; $(CH_{3})_{3} C^{-}$ has three pushing methyls dumping unwanted density on the anion. ⇒ $CH_{3}^{-} > (CH_{3})_{3} C^{-}$ (methyl anion is more stable — exact opposite of cations). Why this step? Fewer +I donors = less overload = more stable anion.

Step 3 — bring in s-character for acetylide. The lone pair on $HC \equiv C^{-}$ sits in an sp orbital (50 % s-character). Here is the intuitive reason it matters: an s-orbital hugs the nucleus (it has no node at the nucleus), while a p-orbital points away from it. The more s-character an orbital has, the closer its electrons sit to the positive nucleus, so those electrons are held more tightly and at lower energy. A negative charge is happiest when held tight — so the sp lone pair (50 % s) stabilises the anion far more than an sp³ lone pair (25 % s). See Hybridisation and s-character. Why this step? An sp lone pair is far more stable than the sp³ lone pairs of the alkyl anions.

Answer: $HC \equiv C^{-} > CH_{3}^{-} > (CH_{3})_{3} C^{-}$ .

Verify: s-character order sp (50 %) $>$ sp³ (25 %) gives acetylide top; +I destabilisation makes 3° worst. Matches parent rule " $CH_{3}^{-} > 1° > 2° > 3°$ " and "sp $>$ sp² $>$ sp³". ✓

Example 5 — Cell E: free radicals

Worked example Rank the radicals:

$CH_{3}^{∙}$ , $(CH_{3})_{2} CH^{∙}$ , $(CH_{3})_{3} C^{∙}$ , $C_{6} H_{5} CH_{2}^{∙}$ (benzyl). Forecast: a radical has 7 electrons (one short of the octet). Does it behave like the cation (electron-hungry) or the anion (electron-rich)?

Step 1 — classify the electron need. Seven electrons = electron-deficient (missing one). So a radical is stabilised by donation — same direction as carbocations. Why this step? The direction of the trend is set by whether the centre wants or rejects electrons.

Step 2 — apply the cation toolkit. Hyperconjugation and resonance both stabilise:

hyperconjugation count: $CH_{3}^{∙}$ (0 α-H) $<$ 2° (6) $<$ 3° (9);
resonance: benzyl spreads the unpaired electron over the ring. Why this step? Same effects, same counting as Example 1 and 2.

Answer: $benzyl^{∙} > (CH_{3})_{3} C^{∙} > (CH_{3})_{2} CH^{∙} > CH_{3}^{∙}$ .

Verify: direction matches parent's " $3° > 2° > 1° > CH_{3}^{∙}$ ; benzyl/allyl by resonance". α-H counts $9 > 6 > 0$ strictly decreasing. ✓

Example 6 — Cell F: the zero / degenerate case

Worked example What stabilises

$CH_{3}^{+}$ , $CH_{3}^{-}$ , and $CH_{3}^{∙}$ ? (No neighbours at all.) Forecast: with zero attached carbons, all our spreading tools (+I, hyperconjugation, resonance) are gone. What's left?

Step 1 — count the spreaders. Methyl species have 0 α C–H bonds and no adjacent π/lone pair. So hyperconjugation = 0, resonance = none, +I from carbon = none. Why this step? This is the degenerate "boundary" of every series — the reference point everything else is measured against.

Step 2 — read off the consequence for each charge type.

$CH_{3}^{+}$ : no help for its electron deficiency ⇒ least stable cation (bottom of the +I/hyperconjugation ladder).
$CH_{3}^{∙}$ : same, least stable radical.
$CH_{3}^{-}$ : here "no +I donors" is actually a good thing (nothing overloads it), so $CH_{3}^{-}$ is the most stable simple alkyl carbanion. Why this step? The very same "zero neighbours" state is worst for electron-deficient species but best for the electron-rich one — the clearest proof the trends are mirror images.

Answer: methyl = worst cation, worst radical, best simple carbanion.

Verify: zero α-H ⇒ minimum of the increasing cation/radical series; zero +I donors ⇒ maximum of the decreasing carbanion series. ✓

Example 7 — Cell G: 1,2-hydride shift (2° → 3°)

Worked example Show why

$(CH_{3})_{2} CH - C + H - CH_{3}$ rearranges, and to what. Forecast: find the carbon next door that would host the $+$ charge more happily.

Step 1 — classify the starting cation. The $+$ carbon bears two other carbons ⇒ 2°. Why this step? We need a baseline stability to see if a move is an upgrade.

Step 2 — inspect the β-carbon (the neighbour). The $(CH_{3})_{2} CH -$ carbon is attached to two methyls plus the chain. If the charge sat there, that carbon would be 3°. Why this step? A cation only rearranges if the destination is more stable (downhill in energy). 3° > 2°, so it is.

Step 3 — move the H with its bonding pair from the β-carbon to the $+$ carbon (a hydride, $H^{-}$ , shift). The old positive carbon gets the electrons (neutralised); the β-carbon, now short a bond-pair, becomes positive. $(CH_{3})_{2} CH - C + H - CH_{3} 1,2-H shift (CH_{3})_{2} C + - CH_{2} - CH_{3}$ Why this step? The migrating H drags its two electrons — that's what lets the charge relocate.

What to look at in the figure below. The figure shows the before molecule on the left and the after on the right. Watch the blue H: it starts bonded to the green β-carbon (left) and ends bonded to the carbon that used to be positive (right). Follow the orange curved arrow — it traces the H moving with its electron pair. Notice how the red "(+)" label jumps from the right-hand carbon (before) to the more-substituted left-hand carbon (after): that jump is the 2° → 3° upgrade.

Figure 2 — 1,2-hydride shift. Left: 2° cation; the blue H sits on the green β-carbon. Orange arrow: the H migrates with its bonding electron pair. Right: the charge has moved to the more-substituted carbon, now a 3° cation.

Answer: product is the 3° cation $(CH_{3})_{2} C + - CH_{2} CH_{3}$ .

Verify (atom balance): start $C_{5} H_{11}^{+}$ , product $C_{5} H_{11}^{+}$ — same formula (just an H hopped one carbon). α-H count rises $6 \to 9$ , confirming the upgrade to 3°. ✓

Example 8 — Cell H: limiting case, 1° → 3° methyl shift (neopentyl)

Worked example The neopentyl cation

$(CH_{3})_{3} C - C + H_{2}$ is 1° (dreadful). What happens? Forecast: there's no H on the neighbouring carbon to shift (it's a quaternary $C$ ). So what migrates?

Step 1 — classify and locate the driving force. The $+$ carbon is 1° (attached to one carbon). The β-carbon $(CH_{3})_{3} C$ is fully substituted: if charge went there it would be 3° — a giant upgrade (1° → 3°). Why this step? The bigger the stability gap, the stronger the push to rearrange; 1°→3° is the maximum realistic jump.

Step 2 — pick the migrating group. The β-carbon has no H to give — but it has three $CH_{3}$ groups. A methyl migrates with its bonding pair (1,2-methyl shift). Why this step? When no hydride is available, alkyl/aryl groups migrate instead; the mechanism is identical (group leaves with electrons).

Step 3 — form the product. $(CH_{3})_{3} C - C + H_{2} 1,2-CH_{3} shift (CH_{3})_{2} C + - CH_{2} - CH_{3}$ The old 1° carbon (now holding the migrated methyl's electrons) is neutral; the former quaternary carbon is now the 3° cation. Why this step? Same "drag the electrons" logic as the hydride shift — the methyl carries its bond-pair, so the positive charge relocates to the more stable seat.

Answer: rearranges to the 3° cation $(CH_{3})_{2} C + - CH_{2} CH_{3}$ (the same ion as Example 7 — two different roads, one stable summit).

Verify (atom balance): neopentyl $C_{5} H_{11}^{+} \to$ product $C_{5} H_{11}^{+}$ — same formula (a methyl hopped one carbon). Stability jump 1° → 3°; the α-H count relevant to the cationic centre rises from effectively $0$ (a bare $CH_{2}^{+}$ has no stabilising alkyl α-H arrangement) to $9$ on the 3° product. Both Ex 7 and Ex 8 converge on the same most-stable ion. ✓

This same "always end on the most stable cation" logic drives Pinacol Rearrangement and (with nitrenes) Hofmann Rearrangement.

Example 9 — Cell I: carbene ground state from substituents

Worked example For

$: CH_{2}$ (methylene) vs $: C(OR)_{2}$ (a dialkoxy carbene / NHC-like), which is singlet ground state? Forecast: the textbook slogan is "carbenes are triplet." Is that safe here?

Step 1 — define the two states. In a carbene the carbon has 2 bonds + 2 non-bonding electrons.

Singlet: those 2 electrons are paired in one orbital (bent ~103°).
Triplet: they sit in separate orbitals, spins parallel — a diradical (bent ~133°). Why this step? You must know what each state is before deciding which is lower.

Step 2 — the parent $: CH_{2}$ . No stabilising neighbours ⇒ the two electrons prefer separate orbitals (that lowers their mutual repulsion) ⇒ triplet ground state (about 9 kcal/mol below singlet). The slogan holds here. Why this step? Establish the default before breaking it.

Step 3 — add π-donor substituents ( $-OR$ ). Each oxygen lone pair donates into the carbene's empty p-orbital (resonance, +M). This fills that orbital, favouring the paired (singlet) arrangement. Why this step? π-donors (or strong electron-withdrawers) preferentially stabilise the singlet — the ground state flips.

Answer: $: CH_{2}$ = triplet ground state; $: C(OR)_{2}$ (and all N-heterocyclic carbenes) = singlet ground state.

Verify: rule "decide multiplicity case-by-case from substituents; π-donors stabilise singlet" from the parent note — applied correctly, not blanket-triplet. ✓

Example 10 — Cell J: real-world / exam twist (Markovnikov via most-stable cation)

Worked example Add HBr to 3-methylbut-1-ene,

$CH_{2} = CH - CH(CH_{3})_{2}$ . Where does Br end up? Forecast: naive Markovnikov ("H to the carbon with more H's") gives a 2° cation. But is 2° the whole story here?

Step 1 — protonate the double bond, both ways, and compare cations.

H adds to C1 ⇒ cation on C2 = 2°: $CH_{3} - C + H - CH(CH_{3})_{2}$ .
H adds to C2 ⇒ cation on C1 = 1° (rejected immediately). Why this step? The product forms through the most stable cation — Markovnikov Addition is really "most-stable-cation-wins."

Step 2 — check for rearrangement of the 2° cation. The neighbour C3 carries two methyls; a 1,2-hydride shift puts the charge on C3 = 3° (upgrade, exactly like Example 7). $CH_{3} - C + H - CH(CH_{3})_{2} 1,2-H shift CH_{3} - CH_{2} - C + (CH_{3})_{2}$ Why this step? A carbocation always slides downhill to the most stable ion before the nucleophile arrives, giving a rearranged product — a classic exam trap.

Step 3 — $Br^{-}$ attacks the 3° centre. Product: $CH_{3} CH_{2} - C(Br)(CH_{3})_{2}$ = 2-bromo-2-methylbutane. Why this step? The nucleophile bonds wherever the (now most stable) $+$ charge sits.

Answer: the major product is 2-bromo-2-methylbutane, not the un-rearranged 2-bromo-3-methylbutane.

Verify (mass balance): alkene $C_{5} H_{10}$ + HBr $\to C_{5} H_{11} Br$ ; product $CH_{3} CH_{2} C(Br)(CH_{3})_{2} = C_{5} H_{11} Br$ . ✓ Stability jump on the intermediate 2° (6 α-H) → 3° (9 α-H) confirms rearrangement is favourable. ✓

Recall — force yourself to answer before revealing

Recall Why can benzyl (a 1° carbon) beat a 3° carbocation?

Because resonance delocalises the $+$ charge over the whole ring (4 positions), and resonance beats hyperconjugation and inductive. ::: Delocalisation over more atoms lowers energy more than 9 α C–H bonds can.

Recall The carbanion trend runs opposite to the cation trend. State it.

$CH_{3}^{-} > 1° > 2° > 3°$ ::: alkyl +I overloads the already electron-rich anion, so more alkyls = less stable.

Recall Why does a hydride/methyl "shift" happen at all?

The group migrates with its bonding electron pair from the β-carbon to the cation, moving the $+$ charge to a more substituted (more stable) position. ::: 2°→3° or 1°→3° upgrades.

Recall Is a carbene always triplet in its ground state?

No — only unsubstituted $: CH_{2}$ . π-donor substituents (–OR, –NR₂) stabilise the singlet, flipping the ground state. ::: Decide case-by-case from substituents.