Visual walkthrough — Hydroboration-oxidation, ozonolysis (reductive - oxidative), syn vs anti dihydroxylation, halohydrin formation

Prerequisites we will lean on (open in another tab if shaky): Electrophilic Addition to Alkenes, Markovnikov's Rule and Carbocation Stability, Stereochemistry — syn vs anti addition, cis/trans, R/S. Parent: the parent topic.

Step 1 — The one object everything starts from: the π cloud

WHAT. Draw a carbon–carbon double bond, $\text{C=C}$. A double bond is two bonds stacked: one sigma ($\sigma$) bond — the strong, direct link along the line between the two carbons — and one pi ($\pi$) bond — a cloud of electrons sitting above and below that line, like two loaves of bread.

WHY it matters. The $\sigma$ bond electrons are buried between the nuclei and hard to reach. The $\pi$ electrons stick out into space where any hungry, electron-poor guest can grab them. So the whole chapter is one sentence:

PICTURE. The green loaves above/below the C–C axis are the π cloud; the amber ${\delta }^{+}$ arrow is a generic electrophile drifting in.

Step 2 — Fork 1: what kind of guest arrives decides the whole path

WHAT. The same π cloud, four different visitors. Each visitor is either bulky+electron-poor, an oxygen bridge, a metal-oxygen ring, or a halogen $X^{+}$. That single choice of guest sends us down one of four branches.

WHY. There is no separate rulebook per reaction. There is one rulebook — electrophile meets π cloud — and then two follow-up questions the parent note flagged:

1. Regiochemistry: which carbon does the "controlling" group land on?
2. Stereochemistry: do the two new groups arrive on the same face (syn) or opposite faces (anti)?

PICTURE. The branch map. Read it top-to-bottom; each branch is expanded in the coming steps.

Step 3 — Branch A, the geometry that forces syn + anti-Markovnikov (hydroboration)

WHAT. Boron in ${\text{BH}}_3$ has only 6 electrons around it (three B–H bonds, no lone pair) — it is two short of a full octet, so it is a real electrophile, marked ${\delta }^{+}$. It reaches into the π cloud. But B does not arrive alone: one B–H bond and the π bond re-arrange together, in a single squeeze, through a four-membered ring (the two carbons, the boron, the hydrogen).

WHY a four-membered concerted ring, not a carbocation? Because B and H are tied together. In acid hydration you get a free ${\text{C}}^{+}$ (a naked carbocation) and the pieces add on separate steps → they can end up on either face. Here nothing ever comes apart: B and H are stapled to the same face of the alkene at the same instant. Same face = syn. That is the entire reason hydroboration is syn.

WHY anti-Markovnikov? Two forces point the same way. (a) Boron is big — it prefers the roomy, less-crowded terminal carbon. (b) In the four-membered transition state the developing ${\delta }^{+}$ sits on the other carbon; the more-substituted carbon carries ${\delta }^{+}$ better (alkyl groups feed electrons), so B is happy on the less-substituted one. Result: B on the less substituted carbon, H on the more substituted carbon.

$\underset{\text{gets H}}{\underbrace{\text{R–CH}}}=\underset{\text{gets B}}{\underbrace{{\text{CH}}_2}}\stackrel{\text{4-ring, one step}}{\to }{\text{R–CH}}_2{\text{–CH}}_2{\text{–BH}}_2$

• $\text{R–CH}$ (more substituted) — takes the H because it hosts the transient ${\delta }^{+}$.
• ${\text{CH}}_2$ (less substituted, roomy) — takes the B because B is bulky and electron-poor.

PICTURE. The dashed four-membered ring; the two red partial bonds forming on the same (top) face.

Step 4 — Branch B, cutting the bond in two (ozonolysis)

WHAT. Ozone ${\text{O}}_3$ is a chain of three oxygens; it slaps onto both faces of the π bond, first making a molozonide, which snaps and re-forms into the more stable ozonide — a little ring that now sits where the C=C used to be, with oxygens threaded between the two carbons.

WHY this cleaves. Once oxygens are wedged between the carbons, the original C–C link is doomed: work-up snips the ozonide right down the middle, and each carbon walks away capped by an oxygen ($\text{C=O}$). The skeleton is cut into two carbonyl pieces.

WHY the work-up changes the answer. The split point is identical either way — only the oxidation level of a carbon that still carried an H differs:

$\underset{\text{1 H}}{\underbrace{\text{=CHR}}}\stackrel{{\text{Zn/H}}_2\text{O}}{\to }\text{RCHO}\text{vs}\underset{\text{1 H}}{\underbrace{\text{=CHR}}}\stackrel{{\text{H}}_2{\text{O}}_2}{\to }\text{RCOOH}$

• Reductive (${\text{Zn/H}}_2\text{O}$ or DMS) protects the fragment → stops at aldehyde/ketone.
• Oxidative (${\text{H}}_2{\text{O}}_2$) lets it go further → the H-bearing carbon climbs to a carboxylic acid. A 0-H (ketone) carbon can't climb — it has no C–H to lose — so ketones are identical in both.

PICTURE. The scissors on the C=C, the two capped fragments, and the two work-up ladders (aldehyde ↔ acid) with the H-count rule labelled on each carbon.

Step 5 — Branch C, why a ring means syn diol (KMnO₄ / OsO₄)

WHAT. Cold dilute ${\text{KMnO}}_4$ or ${\text{OsO}}_4$ delivers two oxygens at once through a five-membered cyclic ester (a manganate or osmate ester). Both metal-oxygens clamp onto the same face of the alkene simultaneously, like a two-pronged staple.

WHY syn. Because both prongs come from one rigid ring, they cannot reach opposite faces — geometrically impossible. When water later hydrolyses the ester, each oxygen is already committed to that one face, so the two $\text{OH}$ groups emerge cis (same side).

$\text{C=C}\stackrel{\text{cyclic ester}}{\to }\begin{array}{c}\text{O}-\text{[M]}-\text{O}\\ \left|\right|\\ \text{C}-\text{C}\end{array}\stackrel{{\text{H}}_2\text{O}}{\to }\text{cis-diol}$

• $\text{[M]}$ — the metal (Os or Mn) holding both oxygens on one face.
• hydrolysis — cuts the metal off, hands back two $\text{OH}$ still on that same face → syn.

PICTURE. The five-membered cyclic ester with both O's on the top face; the released cis-diol.

Step 6 — Branch C′, why ring-OPENING flips it to anti diol (epoxide route)

WHAT. A peroxyacid (mCPBA) hands one oxygen to one face → a three-membered epoxide ring (syn, one O). Then acid protonates that O, and water attacks from the back of a ring carbon — an $S_{N}2$-style backside strike.

WHY anti. The first O sits on (say) the top face. Water is forced to come from underneath (backside opening), so the second $\text{OH}$ lands on the opposite face. Top-O + bottom-OH = anti = trans diol.

$\text{epoxide (O on top)}\underset{\text{water from below}}{\overset{{\text{H}}_3{\text{O}}^{+}}{\to }}\text{trans-diol}$

• "O on top" — set by epoxidation, one face only.
• "water from below" — backside $S_{N}2$ opening forces the opposite face.

PICTURE. Epoxide on the top face; a cyan arrow of water attacking the bottom of a ring carbon; resulting trans OH's on opposite faces.

Step 7 — Branch D, the bridged ion that gives anti + Markovnikov OH (halohydrin)

WHAT. $X_2$ (say ${\text{Br}}_2$) polarises as it nears the π cloud; the near bromine goes in as ${\text{Br}}^{+}$ and forms a bridged three-membered halonium ion — a bromine straddling both carbons, plugging the top face.

WHY Markovnikov OH. The bridge is not symmetric: the more substituted carbon carries more ${\delta }^{+}$ (it stabilises positive character better). Water, the nucleophile, is drawn to that hungrier carbon → $\text{OH}$ lands on the more substituted carbon = Markovnikov OH.

WHY anti. The bromonium bridge blocks the top face, so water is forced to attack the bottom (backside) of that carbon. Br ends on top, OH on the bottom → anti.

${\text{CH}}_3{\text{CH=CH}}_2\underset{{\text{H}}_2\text{O}}{\overset{{\text{Br}}_2}{\to }}\underset{\text{OH on more-sub C2}}{\underbrace{{\text{CH}}_3\text{CH(OH)}}}-\underset{\text{Br on less-sub C1}}{\underbrace{{\text{CH}}_2\text{Br}}}$

• C2 (more substituted) — more ${\delta }^{+}$ in the bridge → water attacks here → OH.
• C1 (less substituted) — keeps the Br.

PICTURE. The bromonium bridge on the top; cyan water arrow hitting the more-substituted carbon from below; anti product.

The one-picture summary

WHAT. One π cloud in the centre; four arrows out to four outcomes, each tagged with its regio and stereo verdict. This single frame is the whole chapter.

Recall drill: