4.2.6 · D5Hydrocarbons
Question bank — Hydroboration-oxidation, ozonolysis (reductive - oxidative), syn vs anti dihydroxylation, halohydrin formation
The left cartoon is syn (both new groups arrive on the same face); the right is anti (they arrive on opposite faces). Every "syn vs anti" reveal on this page is really asking: does the mechanism use a cyclic intermediate (→ syn) or a backside ring-opening (→ anti)?
This second figure lines up the four cyclic/bridged intermediates named in the reveals — the halonium bridge, the osmate/manganate ester, the epoxide, and the ozonide — so the words below map to real bond connectivity.
True or false — justify
Hydroboration–oxidation places OH on the more substituted carbon.
False. Boron is bulky and electron-poor, so it lands on the less substituted carbon; OH replaces boron there, giving the anti-Markovnikov alcohol.
Hydroboration–oxidation is a syn addition of H and OH.
True. B and H add to the same face in one concerted 4-membered transition state (left cartoon of figure s01); the oxidation only swaps B for O with retention, so the H and OH stay on the same face.
Reductive and oxidative ozonolysis break the molecule into different-sized carbon skeletons.
False. Both cleave exactly the same C=C bond, so the skeleton splits identically — only the oxidation level of carbons that carried an H differs (aldehyde vs acid).
Cold dilute KMnO₄ and OsO₄ give diols of opposite stereochemistry.
False. Both go through a cyclic ester (manganate or osmate — see figure s02) on one face, so both give the syn (cis) diol. They agree, not oppose.
Epoxidation followed by acidic ring-opening gives a cis-diol.
False. The epoxide forms syn on one face, but water attacks the back of one carbon (SN2-like), delivering the second OH anti → a trans diol.
Halohydrin formation is a syn addition.
False. The halonium bridges one face and water attacks the opposite face, so (the halogen) and OH end up anti (trans) to each other.
In halohydrin formation the OH goes to the less substituted carbon.
False. The more substituted carbon of the bromonium bears more (more partial positive), so water attacks there — OH lands on the more substituted (Markovnikov) carbon.
Hot concentrated KMnO₄ gives a vicinal diol just faster than cold KMnO₄.
False. Hot/conc. KMnO₄ over-oxidizes and cleaves the C=C into ketones/carboxylic acids (and CO₂ from =CH₂); only cold dilute basic KMnO₄ stops at the diol.
Ketone-forming carbons change oxidation level between reductive and oxidative ozonolysis.
False. A carbon with 0 H (=CR₂) becomes a ketone in both work-ups; ketones can't be further oxidized, so they never change.
Spot the error
"Propene + BH₃ then H₂O₂/OH⁻ gives propan-2-ol."
Error: OH goes anti-Markovnikov to the terminal carbon, giving propan-1-ol (CH₃CH₂CH₂OH), not the 2-ol.
"1-Methylcyclohexene + hydroboration–oxidation gives cis-2-methylcyclohexanol."
Error: Syn addition of H and B, then OH replacing B, places OH and CH₃ trans — the product is trans-2-methylcyclohexanol.
"Ozonolysis of 2-methyl-2-butene with H₂O₂ gives acetone + acetaldehyde."
Error: Oxidative work-up (H₂O₂) oxidizes the 1-H carbon's aldehyde to acetic acid, so the products are acetone + acetic acid.
"OsO₄ dihydroxylation of cyclopentene gives trans-1,2-diol."
Error: OsO₄ forms a cyclic osmate ester on one face → syn addition → cis (meso) cyclopentane-1,2-diol.
"Cyclopentene + Br₂ in water gives 1,2-dibromocyclopentane."
Error: In water, excess H₂O out-competes Br⁻ at the bromonium ion, giving the halohydrin (trans-2-bromocyclopentanol), not the dibromide.
"In hydroboration the carbocation on the more substituted carbon explains the regiochemistry."
Error: There is no free carbocation — the step is concerted through a cyclic TS. Steric bulk and partial-charge () in the TS (not a full carbocation) set the outcome.
"Anti dihydroxylation via epoxide gives a meso diol from cyclopentene."
Error: Backside opening inverts one center giving the trans diol, which for cyclopentene is a racemic pair (chiral), not meso.
Why questions
Why does boron go to the less hindered carbon rather than the more stabilized one?
Both effects agree here: BH₃ is bulky (steric) and electron-poor, so B on the terminal carbon also leaves the more-substituted carbon bearing the stabilizing in the TS.
Why does the oxidation step of hydroboration keep the stereochemistry set earlier?
H₂O₂/OH⁻ swaps each C–B bond for C–OH with retention — the carbon never breaks its bonds to its other neighbors, so the syn geometry from step 1 survives.
Why does the work-up (not the ozone) decide aldehyde vs carboxylic acid?
Ozone builds the same ozonide either way; a reducing work-up (Zn or DMS, i.e. dimethyl sulfide) stops at aldehyde, while H₂O₂ oxidizes any C–H-bearing fragment one level further to a carboxylic acid.
Why does water attack the more substituted carbon of a halonium ion?
That carbon carries more of the partial positive charge (, more carbocation-like), making it the more electrophilic site for the nucleophile — the Markovnikov pattern.
Why does epoxide ring-opening give anti while the epoxide itself forms syn?
Epoxidation adds one oxygen syn across a face, but the strained ring is opened by backside SN2-like attack of water on the far face, delivering the second oxygen anti. See Epoxides — Formation and Ring Opening.
Why does a cyclic intermediate force syn addition?
In an osmate/manganate ester both oxygens are already bonded to the same face of the old double bond before hydrolysis, so both OH groups emerge cis. See Oxidation of Alkenes — KMnO4 and OsO4.
Why is acid hydration Markovnikov but hydroboration anti-Markovnikov, when both add "H and OH"?
Acid hydration goes through a free carbocation (H⁺ adds first, OH to the cation carbon = Markovnikov), while hydroboration has no carbocation and a bulky boron electrophile → the reverse. Contrast with Acid-Catalyzed Hydration of Alkenes.
Edge cases
What does ozonolysis of a terminal =CH₂ carbon give under each work-up?
Reductive → formaldehyde (HCHO); oxidative → CO₂, because the =CH₂ carbon's would-be formic acid is over-oxidized and lost as carbon dioxide.
What happens if a double-bond carbon has zero hydrogens under oxidative ozonolysis?
It becomes a ketone — with no C–H to oxidize, oxidative and reductive work-ups give the identical ketone.
What stereochemistry does syn dihydroxylation of a symmetric cis-alkene give — chiral or meso?
For cyclopentene it gives the meso cis-diol (an internal mirror plane makes the two centers cancel); syn on a symmetric alkene often yields meso.
Does hydroboration of a symmetric internal alkene like cyclohexene have a regiochemistry problem?
No — both carbons are equivalent, so anti-Markovnikov vs Markovnikov is meaningless; only the syn stereochemistry matters, and it simply gives cyclohexanol with H and OH on the same face.
Halohydrin from an alkene where both carbons are equally substituted — where does OH go?
With no regiochemical preference the halonium opens at either carbon roughly equally; regiochemistry only sharpens when one carbon is clearly more substituted.
Why can't you get a bromohydrin by first making the dibromide and then hydrolyzing?
Distinct pathway — the bromohydrin comes from water trapping the bromonium directly; making then hydrolyzing a dibromide is a different (and generally poorer) route, not the halohydrin mechanism. See Halogenation of Alkenes (halonium ion).
What does cold KMnO₄'s color change signify (Baeyer's test)?
The purple MnO₄⁻ is consumed forming the manganate ester/brown MnO₂, so decolorization is a positive test for a C=C (a diagnostic, not a preparative, use).