Hydroboration-oxidation, ozonolysis (reductive - oxidative), syn vs anti dihydroxylation, halohydrin formation: Worked examples

The scenario matrix

Every alkene you can be handed falls into one of these "cells". The columns are the four reactions; the rows are the structural cases that change the answer.

Case class	What makes it tricky	Example that hits it
A. Terminal alkene ( $= CH_{2}$ )	one carbon has 2 H — extreme regiochem, gives $CH_{2} O$ / $CO_{2}$ on cleavage	Ex 1, Ex 4
B. Unsymmetrical internal ( $= CHR$ vs $= CR_{2}$ )	regiochem must be decided; cleavage gives two different fragments	Ex 2, Ex 5
C. Symmetrical internal	both fragments identical / no regiochem to decide	Ex 6
D. Ring (cyclic) alkene	cleavage does not split the molecule — gives one difunctional product; stereo shows as cis/trans	Ex 3, Ex 7
E. Fully substituted carbon ( $= CR_{2}$ , 0 H)	ozonolysis can't oxidise it further — ketone in both work-ups	Ex 5
F. Real-world / structure-determination	run a reaction backwards: "what alkene gave these fragments?"	Ex 8
G. Exam twist (degenerate/limiting)	hot vs cold KMnO $_{4}$ ; $= CH_{2}$ → $CO_{2}$ ; retention vs inversion	Ex 9

We now cover every cell with 9 worked examples.

Worked example Ex 1 — Case A · Terminal alkene, hydroboration–oxidation

1-butene, $CH_{3} CH_{2} CH=CH_{2}$ , is treated with (1) $BH_{3}$ then (2) $H_{2} O_{2}, OH^{-}$ . Give the alcohol.

Forecast: Guess before reading — does OH land on C1 (the $= CH_{2}$ end) or C2?

Identify the two alkene carbons. C1 is $= CH_{2}$ (2 H, less substituted); C2 is $= CH -$ (1 H, more substituted).
- Why this step? Regiochemistry is decided by "which carbon is less hindered", so we must label them first.
Place boron on the less-substituted carbon (C1). Boron is bulky and electron-poor → it sits on the roomy $= CH_{2}$ end.
- Why? Anti-Markovnikov rule from the parent: "BORon is BIG", so it avoids the crowded carbon.
Oxidation swaps B → OH with retention. OH now sits on C1.
- Why? $H_{2} O_{2} / OH^{-}$ replaces C–B by C–OH without touching the carbon's other bonds.
Product: $CH_{3} CH_{2} CH_{2} CH_{2} OH$ (1-butanol, a primary alcohol).

Verify: Anti-Markovnikov means OH on the less substituted carbon → primary alcohol. Contrast: acid hydration would give 2-butanol (secondary). Molecular formula check: $C_{4} H_{8} + H_{2} O = C_{4} H_{10} O$ . ✓

Worked example Ex 2 — Case B · Unsymmetrical internal, halohydrin

2-methyl-2-butene, $(CH_{3})_{2} C=CH-CH_{3}$ , with $Br_{2} / H_{2} O$ . Where do Br and OH go?

Forecast: Which carbon gets the OH — the $(CH_{3})_{2} C$ end or the $CH - CH_{3}$ end?

$Br_{2}$ forms a bridged bromonium ion across both carbons.
- Why? $Br_{2}$ is the electrophile; the π electrons attack it and trap it as a 3-membered ring (see Halogenation of Alkenes (halonium ion)).
Find the carbon with more $δ^{+}$ . The $(CH_{3})_{2} C$ carbon is more substituted → it stabilises positive character better → it carries more partial charge.
- Why? Water attacks whichever carbon looks most "carbocation-like" (Markovnikov's Rule and Carbocation Stability).
Water attacks that carbon from the back (opposite the bromonium bridge) → anti addition. OH lands on $(CH_{3})_{2} C$ ; Br lands on $CH - CH_{3}$ .
Product: $(CH_{3})_{2} C(OH)-CHBr-CH_{3}$ (3-bromo-2-methylbutan-2-ol).

Verify: Markovnikov OH → OH on the more substituted carbon. ✓ Anti stereochem is built in by backside attack. Atom count: $C_{5} H_{10} + Br_{2} + H_{2} O \to C_{5} H_{11} BrO + HBr$ . ✓

Worked example Ex 3 — Case D · Ring alkene, syn vs anti dihydroxylation

Cyclopentene treated (a) with cold dilute $KMnO_{4} / OH^{-}$ , and (b) via mCPBA then $H_{3} O^{+}$ . Give the diol and its stereochemistry.

Forecast: Guess which route gives cis and which gives trans.

(a) syn route — cyclic manganate ester. Both oxygens clamp onto the same face of the ring.
- Why? A 5-membered cyclic ester forms; hydrolysis releases both OH cis (see Oxidation of Alkenes — KMnO4 and OsO4).
Product (a): cis-cyclopentane-1,2-diol — a meso compound (an internal mirror plane makes the two stereocentres cancel).
(b) anti route — epoxide then backside opening. mCPBA makes the epoxide on one face; then $H_{2} O$ attacks the opposite face of one carbon.
- Why? Ring-opening is $S_{N} 2$ -like — backside attack inverts one centre, so the two OH end up trans (Epoxides — Formation and Ring Opening).
Product (b): trans-cyclopentane-1,2-diol — chiral, formed as a racemic pair (both enantiomers).

Verify: Rule "Cycle = syn, Open = anti". Both products are $C_{5} H_{10} O_{2}$ ; they differ only in stereochemistry (cis/meso vs trans/racemic). ✓

Worked example Ex 4 — Case A · Terminal alkene, ozonolysis (both work-ups)

1-hexene, $CH_{3} (CH_{2})_{3} CH=CH_{2}$ , ozonolysis with (a) $Zn / H_{2} O$ , (b) $H_{2} O_{2}$ .

Forecast: The $= CH_{2}$ end becomes what tiny molecule in each case?

Cut the C=C into two carbons. Left = $CH_{3} (CH_{2})_{3} CH =$ (1 H); right = $= CH_{2}$ (2 H).
- Why? Ozonolysis cleaves both σ and π of the double bond — "O-Zone splits the zone".
(a) Reductive: the 1-H carbon → aldehyde, the 2-H carbon → formaldehyde.
- Why? Reductive work-up stops at the aldehyde oxidation level.
Products (a): $CH_{3} (CH_{2})_{3} CHO$ (pentanal) + $HCHO$ (formaldehyde).
(b) Oxidative: each C–H aldehyde carbon is oxidised one level up.
- Why? $H_{2} O_{2}$ over-oxidises aldehydes to acids; the $= CH_{2}$ carbon (2 H) becomes $CO_{2}$ gas.
Products (b): $CH_{3} (CH_{2})_{3} COOH$ (pentanoic acid) + $CO_{2}$ .

Verify: Skeleton splits identically in both work-ups (parent's steel-man). Only oxidation level of H-bearing carbons changes: pentanal→pentanoic acid, formaldehyde→ $CO_{2}$ . ✓ Carbon count preserved: $5 + 1 = 6$ carbons. ✓

Worked example Ex 5 — Cases B + E ·

$= CR_{2}$ carbon meets ozonolysis 2,3-dimethyl-2-butene, $(CH_{3})_{2} C=C(CH_{3})_{2}$ , ozonolysis. Does the work-up matter here?

Forecast: Both carbons have 0 H — guess whether reductive and oxidative give the same answer.

Label each alkene carbon: both are $= C(CH_{3})_{2}$ with 0 H (fully substituted, Case E).
- Why? The number of H on each carbon dictates the oxidation level of the fragment.
Cleave. Each half becomes a carbonyl.
- Why? Ozonolysis caps each carbon with O.
0-H carbon → ketone in BOTH work-ups. A ketone cannot be oxidised further (no C–H on the carbonyl carbon to lose).
Products (reductive AND oxidative): two molecules of $(CH_{3})_{2} C=O$ (acetone).

Verify: This is the degenerate case where work-up is irrelevant — both give $2 \times$ acetone. Carbon count: $6 \to 3 + 3$ . ✓ Contrast Ex 4 where the work-up mattered because H-bearing carbons were present. ✓

Worked example Ex 7 — Case D · Ring alkene, hydroboration stereochemistry

1-methylcyclopentene with (1) $BH_{3}$ , (2) $H_{2} O_{2} / OH^{-}$ . Give the alcohol and its cis/trans relationship.

Forecast: Will OH and $CH_{3}$ be cis or trans on the ring?

Label carbons. C1 bears the $CH_{3}$ (more substituted, 0 H on the ring-alkene carbon); C2 is $= CH -$ (less substituted).
Boron → C2 (less substituted); H → C1. Anti-Markovnikov, so OH ultimately sits on C2.
- Why? Bulky boron avoids the methyl-bearing carbon.
Syn addition: H (to C1) and B (to C2) arrive on the same face.
- Why? Concerted 4-membered transition state adds both to one face.
Read off geometry: new H on C1 and new OH on C2 are cis (same face). Since $CH_{3}$ occupies the other face of C1, OH and $CH_{3}$ are trans.
Product: trans-2-methylcyclopentan-1-ol (racemic — either face is equally likely).

Verify: Same logic the parent uses for 1-methylcyclohexene → trans product. Anti-Markovnikov + syn addition ⇒ OH and $CH_{3}$ trans. ✓ Racemic because both faces are equivalent. ✓

Worked example Ex 8 — Case F · Structure determination (run it backwards)

An unknown alkene undergoes reductive ozonolysis ( $O_{3}$ , then $Zn / H_{2} O$ ) to give acetone, $(CH_{3})_{2} CO$ , and propanal, $CH_{3} CH_{2} CHO$ . Identify the alkene.

Forecast: Rebuild the double bond by gluing the two carbonyl carbons back together.

Each carbonyl carbon was one half of the old C=C. Replace each $C=O$ with $C=$ and join the two.
- Why? Ozonolysis is the reverse of "stitching two carbonyls into a double bond".
Acetone's carbonyl carbon had 0 H → came from a $= C(CH_{3})_{2}$ carbon.
Propanal's carbonyl carbon had 1 H → came from a $= CH (CH_{2} CH_{3})$ carbon.
Join: $(CH_{3})_{2} C=CH-CH_{2} CH_{3}$ = 2-methyl-2-pentene.

Verify: Ozonolyse 2-methyl-2-pentene forward: 0-H carbon → acetone, 1-H carbon → propanal. Matches. ✓ Carbon count: $3 + 3 = 6$ carbons in the alkene. ✓

Worked example Ex 9 — Case G · Exam twist: cold vs hot KMnO

$_{4}$ Cyclohexene with (a) cold, dilute $KMnO_{4} / OH^{-}$ and (b) hot, concentrated $KMnO_{4}$ . Give both products.

Forecast: Same reagent, different temperature — guess if you still get a diol both times.

(a) Cold dilute: syn dihydroxylation → cis-cyclohexane-1,2-diol (meso).
- Why? Cold KMnO $_{4}$ forms a cyclic manganate ester → both OH on one face.
(b) Hot concentrated: the diol is cleaved further, exactly like oxidative ozonolysis. The ring opens; each $= CH -$ carbon (1 H) becomes a $COOH$ .
- Why? Hot/conc. KMnO $_{4}$ over-oxidises — it does not stop at the diol (parent steel-man).
Product (b): since both alkene carbons of cyclohexene are $= CH -$ , the ring opens to a single hexanedioic acid (adipic acid), $HOOC (CH_{2})_{4} COOH$ .

Verify: Cold ⇒ diol; hot ⇒ dicarboxylic acid. Adipic acid $C_{6} H_{10} O_{4}$ keeps all 6 ring carbons (no $CO_{2}$ lost, since neither carbon was $= CH_{2}$ ). ✓

Recall Did every cell get covered?

Terminal (2H) ::: Ex 1, Ex 4 Unsymmetrical internal ::: Ex 2, Ex 5, Ex 8 Symmetrical internal ::: Ex 6 Ring (cyclic) ::: Ex 3, Ex 7, Ex 9 Fully substituted (0H) carbon ::: Ex 5, Ex 8 Structure determination (backwards) ::: Ex 8 Degenerate/limiting twist ::: Ex 5 (work-up irrelevant), Ex 9 (hot vs cold)