Types of organic reactions — addition, substitution, elimination, rearrangement: Worked examples

This page is a classification gym. The parent parent note taught the four "shapes" of organic change. Here we hit every corner case so that no reaction on an exam can surprise you.

Before we start, one word we lean on constantly:

The scenario matrix

Every classification question falls into one of these cells. Each row is a distinct "shape of change"; the extra columns are the tricky sub-cases that trip students.

Cell	Signature (what to look for)	Trap it tests
A. Addition	2 molecules → 1; $π$ bond dies; unsaturation ↓; nothing leaves	Confusing with substitution
B. Elimination	1 molecule → 2; $π$ bond born; unsaturation ↑	Missing that a small molecule left
C. Substitution	group swaps; in = out; unsaturation unchanged	Thinking it's addition
D. Rearrangement	same molecular formula; connectivity changes	Thinking formula changed
E. Degenerate / zero case	reaction where nothing net changes, or a symmetric product	Identical vs different products
F. Limiting / competition case	same substrate, two possible fates (e.g. $S_{N}$ vs $E$ )	Which pathway wins
G. Word / real-world	prose describes a process; you tag the type	Extracting the atom budget from words
H. Exam twist	multi-step, or "looks like X but is really Y"	Reading only the final structure

The 8 examples below each carry a cell tag. Together they cover all eight cells.

Worked examples

Example 1 — Cell A (Addition, with regiochemistry)

Worked example Propene + HBr

$C H_{3} - C H = C H_{2} + H B r \to ?$ Forecast: guess the product and which carbon the $B r$ lands on before reading on.

Step 1 — Meter the unsaturation. Why this step? The atom-budget meter is our fastest classifier. Reactant propene has unsaturation $1$ (one $C=C$ ). Nothing leaves and we combine with $H B r$ → this smells like addition.

Step 2 — Break the $π$ bond onto the proton. Why this step? The $π$ electrons are loosely held and grab the electrophilic $H^{+}$ first (this is electrophilic addition). This makes a carbocation. Two possible cations:

$H$ on C1 (the $= C H_{2}$ end) → charge on C2 → secondary cation.
$H$ on C2 → charge on C1 → primary cation.

Step 3 — Pick the lower-energy cation (Markovnikov via Carbocation stability). Why this step? The mechanism always flows through the more stable intermediate. $2^{\circ} > 1^{\circ}$ (more Hyperconjugation + Inductive effect). So $H$ adds to the carbon that already has more H's (C1), leaving $+$ on C2.

Step 4 — $B r^{-}$ caps the cation. $C H_{3} - B r C H - C H_{3} (2-bromopropane)$ Verify: Reactants $C_{3} H_{6} + H B r = C_{3} H_{7} B r$ . Product $C H_{3} C H B r C H_{3} = C_{3} H_{7} B r$ . ✓ Atom-conserved, unsaturation dropped $1 \to 0$ → genuine addition.

Example 2 — Cell B (Elimination) vs Cell A visualised

Example 3 — Cell C (Substitution, the addition look-alike)

Worked example The trap pair

Compare: $(i) C H_{2} = C H_{2} + B r_{2} \to B r C H_{2} - C H_{2} B r$ $(ii) C H_{3} - C H_{3} + B r_{2} h ν C H_{3} - C H_{2} B r + H B r$ Forecast: are these the same type? They both "add bromine"...

Step 1 — Meter reaction (i). Why this step? Ethene has a $π$ bond (unsaturation $1$ ). Both bromines go on, nothing leaves, unsaturation $1 \to 0$ . → Addition (Cell A).

Step 2 — Meter reaction (ii). Why this step? Ethane has no $π$ bond (unsaturation $0$ ). A $B r$ takes an $H$ 's place and $H B r$ leaves. Unsaturation stays $0$ ; in = out. → Substitution (Cell C, free-radical type).

Step 3 — The distinguishing rule. Why this step? Both put $B r$ onto carbon, so "something adds" is useless. The real test: did anything leave? In (i) nothing left; in (ii) $H B r$ left. That single question separates the cells.

Verify: (i) $C_{2} H_{4} + B r_{2} \to C_{2} H_{4} B r_{2}$ : H $4 = 4$ , Br $2 = 2$ , unsat $1 \to 0$ . ✓ addition. (ii) $C_{2} H_{6} + B r_{2} \to C_{2} H_{5} B r + H B r$ : H $6 = 5 + 1$ , Br $2 = 1 + 1$ , unsat $0 \to 0$ . ✓ substitution.

Example 4 — Cell D (Rearrangement, formula must not change)

Worked example Neopentyl cation reshuffle

$(C H_{3})_{3} C - C + H_{2} (1^{\circ}) 1,2-methyl shift (C H_{3})_{2} C + - C H_{2} C H_{3} (3^{\circ})$ Forecast: what is the molecular formula before and after?

Step 1 — Count atoms on both sides. Why this step? Rearrangement's fingerprint is an unchanged molecular formula. Left: $5$ C, $11$ H, charge $+ 1$ . Right: $5$ C, $11$ H, charge $+ 1$ . Identical → this is a candidate rearrangement.

Step 2 — Confirm connectivity changed. Why this step? Same formula alone isn't enough; connectivity must differ. A methyl group migrated with its bonding pair from the quaternary carbon to the empty $p$ -orbital carbon.

Step 3 — Justify direction (Carbocation stability). Why this step? Migrations only run downhill in energy. $1^{\circ} \to 3^{\circ}$ is a big stability gain (more hyperconjugation), so the shift is favourable.

Verify: $C_{5} H_{11}^{+}$ on both sides — no atom entered or left, so it cannot be addition/substitution/elimination. Pure rearrangement (Cell D). ✓

Example 5 — Cell E (Degenerate / symmetric case)

Worked example Symmetric substrate, "does anything really change?"

$C H_{3} - B r +^{*} B r^{-} \to C H_{3} -^{*} B r + B r^{-}$ (the $^{*}$ marks a radioactively-labelled bromide so we can see the swap.) Forecast: is this addition, substitution, elimination, or nothing?

Step 1 — Meter it. Why this step? No $π$ bond exists or forms; unsaturation $0 \to 0$ . Not addition or elimination.

Step 2 — Watch the atom budget. Why this step? A bromide leaves and a bromide arrives at the same carbon — in = out. This is textbook $S_{N} 2$ (SN1 vs SN2 mechanisms).

Step 3 — The degenerate subtlety. Why this step? Chemically the product is identical to the reactant ( $C H_{3} B r$ ). Without the isotope label you'd swear "nothing happened." Yet the carbon underwent Walden inversion — the umbrella flipped inside-out. So it is a real substitution (Cell E: degenerate/identity substitution).

Verify: heavy-atom skeleton $C H_{3} - (B r)$ preserved, one $B r$ in and one $B r$ out. Substitution confirmed; the "no visible change" is the trap. ✓

Example 6 — Cell F (Competition / limiting case)

Worked example Same substrate, two fates

2-bromo-2-methylpropane $(C H_{3})_{3} C - B r$ warmed with a base/nucleophile. Two products appear: $(C H_{3})_{3} C - O H and (C H_{3})_{2} C = C H_{2} + H_{2} O$ Forecast: which single carbocation feeds both products?

Step 1 — Rate-determining step is shared. Why this step? Both routes start by losing $B r^{-}$ to make the same stable $3^{\circ}$ carbocation ( $S_{N} 1$ / $E 1$ share step 1; see E1 vs E2 mechanisms).

Step 2 — The cation's two exits. Why this step? Once the cation exists it can either (a) let a nucleophile ( $H_{2} O$ ) attach → substitution ( $S_{N} 1$ , unsaturation $0$ ), or (b) lose a $β$ -H to a base → elimination ( $E 1$ , unsaturation $↑ 1$ ).

Step 3 — What tips the balance (limiting behaviour). Why this step? Higher temperature / stronger–bulkier base favours elimination; cooler / better nucleophile favours substitution. This is the exam's "it depends" cell.

Verify: Substitution branch: $C_{4} H_{9} B r + H_{2} O \to C_{4} H_{10} O + H B r$ — unsat $0 \to 0$ . ✓ Elimination branch: $C_{4} H_{9} B r \to C_{4} H_{8} + H B r$ — unsat $0 \to 1$ . ✓ One cation, two cells; classification depends on which exit is taken.

Example 7 — Cell G (Real-world word problem)

Worked example Margarine factory (read the prose, tag the type)

"In a hydrogenation plant, liquid vegetable oil rich in $C=C$ double bonds is stirred with hydrogen gas over a nickel catalyst. The oil hardens into semi-solid fat, and no gas is released." Forecast: which cell?

Step 1 — Extract the atom budget from words. Why this step? Word problems hide the meter in prose. Key phrases: "rich in $C=C$ " (unsaturation high) and " $H_{2}$ is stirred in… no gas released" (atoms go in, nothing comes out).

Step 2 — Model one double bond. $R - C H = C H - R^{'} + H_{2} Ni R - C H_{2} - C H_{2} - R^{'}$ Why this step? Each $C=C$ eats one $H_{2}$ : a $π$ bond dies, two new $σ$ (C–H) bonds form. Unsaturation drops.

Step 3 — Tag it. Why this step? Two molecules → one, $π$ consumed, nothing leaves → Addition (Cell G, a real hydrogenation). Physically this is why the fat hardens: saturated chains pack tighter.

Verify: per double bond, $C_{n} H_{2 n} -unit + H_{2} \to C_{n} H_{2 n + 2} -unit$ : H count rises by exactly $2$ , unsaturation $1 \to 0$ . Addition confirmed. ✓

Example 8 — Cell H (Exam twist: "looks like Y, is really X + Y")

Worked example Two-step trap

$C H_{3} C H_{2} C H_{2} - C H = C H_{2} H B r product$ The naive answer is 2-bromopentane. But a careful student notices the actual isolated product is 2-bromo-2-methylbutane's isomer — i.e. rearrangement sneaks in. Let's classify the whole process.

Forecast: is this one reaction type or two events stacked?

Step 1 — First event: addition of $H^{+}$ . Why this step? $π$ electrons grab $H^{+}$ making a $2^{\circ}$ carbocation on C2. So far → addition in progress (Cell A).

Step 2 — Second event: could a shift happen? Why this step? Whenever a carbocation sits next to a carbon that would give a more stable cation, check for a 1,2-shift. Here a hydride shift from C3 would give another $2^{\circ}$ cation — no stability gain, so no rearrangement in this particular case.

Step 3 — Read the twist correctly. Why this step? The exam wants you to test for rearrangement, not assume it. Since $2^{\circ} \to 2^{\circ}$ gives no benefit, $B r^{-}$ simply caps the original cation: $C H_{3} C H_{2} C H_{2} - B r C H - C H_{3}$ Overall type: Addition (Markovnikov), no rearrangement — the twist is a decoy.

Verify: $C_{5} H_{10} + H B r \to C_{5} H_{11} B r$ : C $5 = 5$ , H $10 + 1 = 11$ , Br $1 = 1$ , unsat $1 \to 0$ . ✓ Addition; rearrangement correctly ruled out because $2^{\circ} \to 2^{\circ}$ is not downhill.

Recall Self-test: name the cell

A reaction where the molecular formula is unchanged but connectivity differs. ::: Cell D — Rearrangement. A reaction where in = out and unsaturation is unchanged. ::: Cell C — Substitution. Isotope-labelled $C H_{3} B r +^{*} B r^{-}$ giving "identical" product. ::: Cell E — degenerate substitution ( $S_{N} 2$ with inversion). One substrate, one carbocation, two products (alcohol + alkene). ::: Cell F — $S_{N} 1$ / $E 1$ competition. Hydrogenation of vegetable oil into margarine. ::: Cell G — Addition (real-world).