Visual walkthrough — Types of organic reactions — addition, substitution, elimination, rearrangement

Parent: Types of organic reactions

Step 1 — What a carbocation is (a carbon short one pair of electrons)

WHAT. A carbocation is a carbon atom that is missing one bonding pair of electrons, so it carries a positive charge, written ${\text{C}}^{+}$. Normal carbon has 4 bonds (8 shared electrons around it). A carbocation has only 3 bonds (6 electrons). Two electrons short of a full outer shell.

WHY it matters. Six electrons is "hungry." That hunger (the $+$ charge) is the driving force behind everything on this page — the molecule will do whatever lowers that hunger fastest.

PICTURE. Look at the figure. On the left, an ordinary carbon has four grey sticks (bonds) fanning out. On the right, the carbocation has only three sticks, and where the fourth should be there is an empty box — an empty orbital, drawn in red. The red "$+$" sits on the carbon.

Every symbol so far:

• ${\text{C}}^{+}$ ::: a carbon with only 3 bonds and a positive charge (an electron pair short).
• empty orbital ::: the vacant slot, drawn red, where electrons want to come in.

Step 2 — Why some carbocations are calmer: electrons pushed IN

WHAT. A carbocation is calmer (more stable, lower energy) when nearby atoms donate a little electron density into that empty orbital, softening the $+$ charge. Two donors do this:

• Inductive effect — alkyl groups (like ${\text{CH}}_3$) are slightly electron-releasing; through the $\sigma$ bonds they nudge a trickle of electron density toward the hungry carbon.
• Hyperconjugation — a neighbouring $\text{C-H}$ bond can partly overlap with the empty orbital and share its electrons sideways.

WHY these tools and not others. We need to explain stability differences between cations that have the same charge. Bond breaking/forming won't distinguish them — only where electron density leaks does. Inductive effect and hyperconjugation are exactly the two "electron-leaking" mechanisms available, so those are the right tools.

WHY more alkyl groups = more stable. Each attached alkyl group is one more donor. More donors → more density fed into the empty orbital → smaller effective $+$ → lower energy.

PICTURE. Four cations sit in a row: methyl (${\text{CH}}_3^{+}$, zero alkyl neighbours), then $1^{\circ }$, $2^{\circ }$, $3^{\circ }$ (one, two, three alkyl neighbours). Green arrows point inward toward each cation — one arrow per donor. Count the green arrows: they grow left → right, and the energy bar underneath drops left → right.

Step 3 — The energy hill: cation forms at the top, and the lower top wins

WHAT. In addition, elimination, and $S_{N}1$, the slow step makes a carbocation. Getting there means climbing an energy hill; the top of the hill is the transition state. A more stable cation sits lower, so its hill-top is lower too.

WHY this decides the product. Reaction rate depends on how tall the hill is. The path with the shorter climb (more stable cation) is faster, so that product forms in the largest amount. Nature is lazy: it takes the low road.

PICTURE. Two energy hills side by side, starting from the same valley. The blue hill (leads to the stable cation) is short; the red hill (leads to the unstable cation) is tall. The reaction rolls up the blue hill because it's cheaper. This one geometry — lower cation ⇒ lower hilltop ⇒ faster ⇒ major product — is reused in every step below.

Step 4 — Addition: Markovnikov is just "pick the lower hill"

WHAT. Add $\text{HBr}$ across propene, ${\text{CH}}_3{\text{-CH=CH}}_2$. The $\pi$ electrons grab the ${\text{H}}^{+}$ first (see Nucleophiles and Electrophiles). That ${\text{H}}^{+}$ can land on either end of the double bond — and the other carbon becomes the cation.

• If ${\text{H}}^{+}$ lands on the end ${\text{CH}}_2$ → the middle carbon becomes a $2^{\circ }$ cation (2 donors).
• If ${\text{H}}^{+}$ lands on the middle carbon → the end becomes a $1^{\circ }$ cation (1 donor).

WHY the first path wins. From Step 3, the $2^{\circ }$ cation has the lower hilltop. So ${\text{H}}^{+}$ goes to the end carbon (the one already holding more H's), forcing ${\text{Br}}^{-}$ onto the middle carbon. That is Markovnikov's rule — not memorised, derived from Step 2's donor count.

PICTURE. The alkene forks into two arrows. Top arrow → green-highlighted $2^{\circ }$ cation (low, chosen, thick arrow). Bottom arrow → faded red $1^{\circ }$ cation (high, rejected, thin arrow). ${\text{Br}}^{-}$ then clips onto the green cation's carbon.

Step 5 — Elimination: Saytzeff is the same choice, seen from the alkene side

WHAT. Take ${\text{CH}}_3{\text{CH}}_2{\text{CHBrCH}}_3$ and pull off $\text{HBr}$ with hot alcoholic base. A $\beta$-hydrogen (an H on a carbon next to the C-Br) leaves along with ${\text{Br}}^{-}$, and a new $\pi$ bond forms between those two carbons. But there are two different $\beta$-carbons, so two alkenes are possible.

• Remove the H toward the inner ${\text{CH}}_2$ → ${\text{CH}}_3{\text{CH=CHCH}}_3$ (double bond flanked by two alkyl groups — more substituted).
• Remove the H toward the terminal ${\text{CH}}_3$... wait, that carbon is terminal → ${\text{CH}}_3{\text{CH}}_2{\text{CH=CH}}_2$ (double bond flanked by one alkyl group — less substituted).

WHY the more-substituted alkene wins. The very same donor logic (Step 2)! A double bond is stabilised by neighbouring alkyl groups via Hyperconjugation — more alkyl neighbours, lower energy alkene, lower transition state. So the more-substituted alkene is major. This is Saytzeff — the same "more donors = calmer = favoured" rule, now applied to a $\pi$ bond instead of a cation.

PICTURE. The substrate branches into two alkenes. The internal one (green, thick arrow) shows two alkyl arrows feeding the double bond → major. The terminal one (faded, thin arrow) shows one arrow → minor.

Step 6 — Substitution: does a cation even form? ($S_{N}1$ vs $S_{N}2$)

WHAT. Now the leaving group departs and a nucleophile takes its place at the same carbon (see Leaving groups, SN1 vs SN2 mechanisms). Two routes:

• $S_{N}1$ — leaving group leaves first, making a carbocation, then the nucleophile attacks. This route needs a stable cation, so a $3^{\circ }$ carbon (low hilltop, Step 3) loves it.
• $S_{N}2$ — nucleophile attacks the back while the leaving group is still there, in one step. No cation ever forms, so cation stability is irrelevant; here it's about room to attack. A $1^{\circ }$ carbon (least crowded) loves it.

WHY this is the fork. Same energy-hill picture, but now we ask: is the cation cheap enough to bother making? If yes ($3^{\circ }$) → $S_{N}1$. If not ($1^{\circ }$, expensive cation) → skip the cation, go concerted $S_{N}2$.

PICTURE. Left panel: $3^{\circ }$ carbon, leaving group gone, fat green cation sitting alone, then Nu drops in → $S_{N}1$. Right panel: $1^{\circ }$ carbon, Nu arrow slams the backside as the leaving group flies off the front, umbrella flipping inside-out (inversion) → $S_{N}2$, no cation drawn.

Step 7 — Rearrangement: if a better cation is one shift away, take it

WHAT. Sometimes the cation that forms first is not the most stable one available. A neighbouring H or alkyl group can slide over — a 1,2-shift — carrying its bonding pair, turning a poor cation into a better one.

WHY. Straight from Step 2/Step 3: a $1^{\circ }$ cation is high-energy; if a single 1,2-shift produces a $3^{\circ }$ cation, the molecule takes it because energy drops. The atoms just reconnect — molecular formula is unchanged (rearrangement, not addition/elimination/substitution).

PICTURE. A $1^{\circ }$ neopentyl cation on the left (red, high, unstable). A curved yellow arrow shows a methyl group migrating to the empty orbital. On the right, the resulting $3^{\circ }$ cation (green, low, stable). An energy bar underneath falls from red level to green level.

Step 8 — The degenerate cases you must not trip on

WHAT / WHY (each is where the naive rule breaks):

• Symmetric alkene (e.g. ${\text{CH}}_2={\text{CH}}_2$). Both carbons are identical, so both possible cations are equal. Markovnikov has no preference — either landing gives the same product. The rule doesn't fail; it simply has nothing to choose.
• No $\beta$-hydrogen (e.g. ${\text{(CH}}_3{)}_3{\text{C-CH}}_2\text{Br}$'s neopentyl carbon has no removable adjacent H in the right place). Then elimination is impossible — the molecule cannot form that particular $\pi$ bond, so substitution or rearrangement takes over.
• Cation already $3^{\circ }$. No shift can make it better, so no rearrangement — the shift only happens when a more stable cation is reachable.
• Anti-Markovnikov exception (peroxides, $\text{HBr}$ only). The mechanism switches to radicals, not cations, so the "stable cation" logic is replaced by "stable radical," and the addition can flip. Different mechanism ⇒ different rule.

PICTURE. A 2×2 grid of these four traps; each cell shows a tiny sketch with a red ✗ over the step that cannot happen, so you can spot the dead end instantly.

The one-picture summary

WHAT. One master diagram: a single "stability compass" in the middle (the Step 2 order $3^{\circ }>2^{\circ }>1^{\circ }>{\text{CH}}_3^{+}$) with four arrows shooting out to the four reaction types, each labelled with how that reaction obeys the compass.

Active recall