Visual walkthrough — Alkenes — preparation (dehydration, dehydrohalogenation, Zaitsev's rule), addition reactions

This is the visual companion to the parent topic. We build every symbol from nothing.

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Step 1 — What an alkene actually is (the greedy double handshake)

WHAT. An alkene is a molecule where two carbon atoms are joined twice. We draw that as C=C. The two lines are two different kinds of glue:

• one σ (sigma) bond — a strong, straight-line bond directly between the two carbons;
• one π (pi) bond — a weaker cloud of electrons sitting above and below the σ bond.

WHY it matters. The π electrons are held loosely and stick out into space. Think of them as a bit of exposed, greedy "reaching" charge. Anything that likes electrons will be pulled toward this cloud first. That is the whole reason alkenes react by having things added to them.

PICTURE. Look at the two grey carbon balls. The straight orange bar between them is the σ bond. The two soft magenta blobs (one on top, one on the bottom) are the π cloud — that is the "greedy" electron region every reagent aims for.

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Step 2 — The reagent HBr, and the choice it faces

WHAT. Our reagent is HBr — a hydrogen atom stuck to a bromine atom. The bond between them is lopsided: bromine hogs the shared electrons, so the H end is slightly positive ($\delta +$) and the Br end is slightly negative ($\delta -$).

$\text{H}\stackrel{\delta +}{-}\stackrel{\delta -}{\mathrm{Br}}$

• $\delta +$ over the H $=$ "this end is a little electron-poor" — this is the electrophile part.
• $\delta -$ over the Br $=$ "this end is a little electron-rich."

WHY. Because the H end is electron-poor, it is what the greedy π cloud will grab. So the story always starts: π cloud reaches for the H. But here is the fork in the road — on an unsymmetrical alkene like propene, the H can land on either of the two double-bond carbons.

PICTURE. Propene is ${\text{CH}}_3\text{–CH}={\text{CH}}_2$. Number the double-bond carbons: C1 is the end $={\text{CH}}_2$ carbon (it already holds two H's); C2 is the middle $=\text{CH}$ carbon (it holds one H and carries the ${\text{CH}}_3$ group). Two arrows show the two possible landing spots for the incoming H.

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Step 3 — The invisible middle-creature: the carbocation

WHAT. When the H⁺ grabs onto one alkene carbon, it uses up the π electrons to make a new C–H σ bond. But those π electrons belonged to both carbons. The carbon that didn't get the H is now short one bond's worth of electrons — it is left with a positive charge. We call this positively-charged carbon a carbocation.

$\underset{\text{carbon that lost its share}}{\stackrel{+}{C}}$

• The $+$ sits on carbon.
• It means "this carbon is one electron-pair short and desperately wants electron density back."

WHY this is the whole game. The reaction must pass through this carbocation. Whichever pathway builds the more comfortable (more stable) carbocation is the pathway molecules actually take — because a lower-energy middle-creature is far easier to reach. So: stability of the carbocation decides the product.

PICTURE. The H has landed on C1. C2 now glows with a $+$. The blue arrow shows the π electrons that swung over to bond the H — leaving C2 electron-hungry.

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Step 4 — Why some carbocations are comfier (hyperconjugation & +I)

WHAT. We label carbocations by how many carbon groups touch the $+$ carbon:

• 1° (primary): one carbon neighbour.
• 2° (secondary): two carbon neighbours.
• 3° (tertiary): three carbon neighbours.

The stability order is: $\text{3°}>\text{2°}>\text{1°}>{\text{CH}}_3^{+}$

WHY. A positive carbon wants electrons pushed toward it. Neighbouring alkyl groups do exactly this in two ways:

• +I (inductive) effect: carbon–carbon bonds gently shove electron density toward the $+$.
• Hyperconjugation: each neighbouring C–H bond can overlap with the empty spot on the $+$ carbon, smearing its electrons into the gap. More neighbouring C–H bonds $=$ more "helping hands" $=$ more stable.

So more alkyl neighbours = more help = comfier cation. (For the deeper machinery see Carbocations — stability and rearrangement and Hyperconjugation and Inductive effect.)

PICTURE. Three cartoons side by side: a lonely 1° cation with few helping arrows, a 2° cation with more inward arrows, a 3° cation swamped with inward arrows. The more inward orange arrows, the lower (comfier) the energy bar drawn beneath it.

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Step 5 — The two pathways for propene, side by side

WHAT. Now run both forks of Step 2 all the way to their carbocation.

Path A — H lands on C1 (the ${\text{CH}}_2$ end): ${\text{CH}}_3\text{–CH}={\text{CH}}_2\stackrel{+{\text{H}}^{+}\text{ on C1}}{\to }{\text{CH}}_3\text{–}\stackrel{+}{\underset{\text{C2}}{\mathrm{CH}}}{\text{–CH}}_3$ The $+$ lands on C2, which touches two carbon groups → a 2° carbocation (comfy).

Path B — H lands on C2 (the middle): ${\text{CH}}_3\text{–CH}={\text{CH}}_2\stackrel{+{\text{H}}^{+}\text{ on C2}}{\to }{\text{CH}}_3{\text{–CH}}_2\text{–}\stackrel{+}{\underset{\text{C1}}{{\text{CH}}_2}}$ The $+$ lands on C1, which touches only one carbon group → a 1° carbocation (uncomfortable).

WHY the answer is now forced. From Step 4, 2° beats 1°. Path A's intermediate is lower energy, so almost all molecules travel Path A. The reaction chooses stability.

PICTURE. Left panel: Path A ending in a green-lit 2° cation on a low energy shelf. Right panel: Path B ending in a red-lit 1° cation on a high energy shelf. A crowd of little molecule-dots piles onto the low (Path A) route.

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Step 6 — Br⁻ finishes the job → the Markovnikov product

WHAT. The comfy 2° carbocation (C2 is $+$) still needs electrons. The leftover ${\text{Br}}^{-}$ (the $\delta -$ end from Step 2) is electron-rich and rushes in to bond that positive carbon.

${\text{CH}}_3\text{–}\stackrel{+}{\mathrm{CH}}{\text{–CH}}_3\stackrel{+{\text{Br}}^{-}}{\to }{\text{CH}}_3\text{–}\underset{\text{Br}}{\mathrm{CH}}{\text{–CH}}_3=\text{2-bromopropane}$

• Br lands on C2 (the more substituted carbon, fewer H's).
• H (from way back in Step 3) is on C1 (the carbon that already had more H's).

WHY this is exactly Markovnikov's Rule. Read off the final product: the H went to the carbon with more H's, the Br went to the carbon with fewer H's. We did not assume that rule — it fell out of "the reaction passes through the more stable carbocation."

PICTURE. The Br⁻ arrow snaps onto the $+$ carbon; the finished molecule 2-bromopropane is drawn with H on C1 and Br on C2, both highlighted.

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Step 7 — The degenerate case: a symmetrical alkene

WHAT. What if both double-bond carbons are identical — like but-2-ene, ${\text{CH}}_3\text{–CH}={\text{CH–CH}}_3$? Both carbons carry exactly one H and one ${\text{CH}}_3$.

WHY it's special. Now Path A and Path B give the same 2° carbocation. There is no "more stable" choice — the two routes are mirror-identical. So the Markovnikov question dissolves: only one product is possible.

${\text{CH}}_3\text{–CH}={\text{CH–CH}}_3+\text{HBr}\to {\text{CH}}_3{\text{–CH}}_2{\text{–CHBr–CH}}_3(\text{one answer either way})$

PICTURE. Both forks drawn — and both arrows collapse into the identical carbocation in the middle. The symmetry is shaded to show the two halves are reflections.

Recall Check yourself: ethene + HBr

Ethene is ${\text{CH}}_2={\text{CH}}_2$ — also symmetrical. Both carbons equal, so H and Br simply add across. Product? ::: Bromoethane, ${\text{CH}}_3{\text{CH}}_2\text{Br}$ — only one possible product, Markovnikov is silent.

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Step 8 — The plot twist: peroxide flips everything (only for HBr)

WHAT. Add an organic peroxide and the very same propene + HBr now gives 1-bromopropane — Br on the carbon with more H's. The opposite orientation!

WHY. The peroxide changes the mechanism. Instead of an ionic H⁺-first attack, HBr now splits into a bromine radical ${\text{Br}}^{\bullet }$ (a bromine with one lonely unpaired electron, drawn as a dot). Br• adds first. By the same "more alkyl neighbours = more stable" logic — but now for a radical instead of a cation — Br• lands on C1 so the radical forms on the comfy, more-substituted C2. Then an H caps C2. Result: Br ends up on C1.

${\text{CH}}_3\text{–CH}={\text{CH}}_2+\text{HBr}\stackrel{\text{peroxide}}{\to }{\text{CH}}_3{\text{–CH}}_2{\text{–CH}}_2\text{Br}=\text{1-bromopropane}$

Only HBr does this: the H–Cl bond is too strong to feed the radical chain, and H–I too weak — only H–Br is "just right."

PICTURE. Top row: ionic route (H⁺ first) → Br on C2. Bottom row: radical route (Br• first) → Br on C1. The single swapped first-mover flips the final position.

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The one-picture summary

Everything above is one arrow: π cloud grabs the electrophile → the more stable intermediate decides where the pieces land. Ionic route builds a cation (Markovnikov); peroxide route builds a radical (anti-Markovnikov). Substitution → stability → product.

Recall Feynman retelling — the whole walkthrough in plain words

Picture two carbons holding a springy, greedy double handshake (Step 1). A visitor HBr floats by, its H-hand a little starved (Step 2). The greedy π cloud grabs that H-hand first — but it can only pull it onto one of the two carbons. Whichever carbon does not get the H is left holding a hot potato: a positive charge, the carbocation (Step 3). Now, some carbons are surrounded by friendly neighbours (alkyl groups) who keep passing them electron-snacks, so those carbons are comfy holding the hot potato; lonely carbons are miserable (Step 4). Molecules are lazy — they take the route that makes the comfy carbocation (Step 5). Then the leftover Br⁻ runs over and bonds that comfy carbon (Step 6). Read the result and — surprise — the H sat where there were already lots of H's, the Br where there were few: that's Markovnikov, and we never had to memorise it. If both carbons are twins, the choice vanishes and there's just one answer (Step 7). Finally, sneak in a peroxide and the whole thing runs on radicals instead of charges — the Br goes first now, flipping where everything lands — but only with HBr (Step 8). One idea the whole time: electrons chase stability.