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

This page is a self-test. Each problem is graded L1 → L5. Read the problem, try it yourself first, then open the collapsible solution. If you have not met the reactions yet, start from the parent note (Hinglish version here) or the English topic note.

Before we start, two words we will use constantly:

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

Level 1 — Recognition

Problem 1.1

Name the regiochemistry (Markovnikov / anti-Markovnikov) and stereochemistry (syn / anti) for each reagent set, without drawing anything: (a) BH $_{3}$ then H $_{2}$ O $_{2}$ /OH $^{-}$ (b) cold dilute KMnO $_{4}$ /OH $^{-}$ (c) Br $_{2}$ in water (d) mCPBA then H $_{3}$ O $^{+}$

Recall Solution 1.1

Just recall the master table.

(a) Hydroboration–oxidation → anti-Markovnikov, syn. (Boron is bulky + electron-poor → lands on the less substituted carbon; concerted 4-membered ring → same face.)
(b) Cold dilute KMnO $_{4}$ → no regiochemistry issue (two identical OH groups), syn. (Cyclic manganate ester delivers both O to one face.)
(c) Br $_{2}$ /H $_{2}$ O (halohydrin) → Markovnikov OH, anti. (Water attacks the more substituted carbon of the bromonium from the back.)
(d) mCPBA then H $_{3}$ O $^{+}$ → anti dihydroxylation → anti. (Epoxide made on one face, water opens it from the opposite face by backside attack.)

Problem 1.2

For $CH_{2} =CH_{2}$ (ethene), how many hydrogens sit on each double-bond carbon? Predict the reductive-ozonolysis product.

Recall Solution 1.2

Each carbon of ethene carries 2 H (it is $= CH_{2}$ on both ends).

The rule: a carbon with 2 H → aldehyde where R = H = formaldehyde, $HCHO$ .
Both halves are identical, so reductive ozonolysis of ethene gives two molecules of formaldehyde, $2 HCHO$ .

Level 2 — Application

Problem 2.1

Give the product of hydroboration–oxidation of propene, $CH_{3} CH=CH_{2}$ . Which carbon gets the OH and why?

Recall Solution 2.1

Number the alkene carbons: C1 = terminal $=CH_{2}$ (touches only 1 carbon), C2 = internal $=CH-$ (touches 2 carbons, so more substituted).

Boron is bulky and electron-poor, so it lands on the less substituted carbon, C1.
After oxidation, OH replaces boron → OH on C1. $CH_{3} CH=CH_{2} (1) BH_{3} (2) H_{2} O_{2} / OH^{-} CH_{3} CH_{2} CH_{2} OH$ Product = propan-1-ol (anti-Markovnikov alcohol). Contrast: acid hydration would give propan-2-ol.

Problem 2.2

Give the product of halohydrin formation from propene with $Br_{2} / H_{2} O$ . Show which carbon gets OH and which gets Br.

Recall Solution 2.2

Br $_{2}$ makes a bridged bromonium ion across C1–C2.
The more substituted carbon (C2) holds more partial-positive charge, so water attacks C2 → OH on C2.
Br is left on C1. Backside attack → anti (irrelevant to the name here, but fixes stereochem). $CH_{3} CH=CH_{2} Br_{2} / H_{2} O CH_{3} CH(OH) -CH_{2} Br$ Product = 1-bromopropan-2-ol (Markovnikov OH).

Problem 2.3

Do reductive and oxidative ozonolysis of 1-butene, $CH_{3} CH_{2} CH=CH_{2}$ .

Recall Solution 2.3

Split the double bond into two carbons. Left carbon ( $= CH-$ ) has 1 H; right carbon ( $= CH_{2}$ ) has 2 H.

Reductive (Zn/H $_{2}$ O): 1-H carbon → aldehyde = $CH_{3} CH_{2} CHO$ (propanal); 2-H carbon → formaldehyde $HCHO$ .
Oxidative (H $_{2}$ O $_{2}$ ): the 1-H carbon's aldehyde is oxidised → $CH_{3} CH_{2} COOH$ (propanoic acid); the 2-H carbon becomes CO $_{2}$ (formaldehyde over-oxidises fully). $Reductive: CH_{3} CH_{2} CHO + HCHO$ $Oxidative: CH_{3} CH_{2} COOH + CO_{2}$

Level 3 — Analysis

Problem 3.1

Cyclopentene is treated (a) with cold dilute KMnO $_{4}$ and (b) with mCPBA then H $_{3}$ O $^{+}$ . Predict the cis/trans relationship of the two OH groups in each product, and say which is meso and which is racemic.

Recall Solution 3.1

(a) Cold KMnO $_{4}$ — syn. A cyclic manganate ester grips one face; hydrolysis releases both OH on that same face → cis-cyclopentane-1,2-diol. Because the ring has an internal mirror plane, the cis diol is meso (achiral). (b) mCPBA then H $_{3}$ O $^{+}$ — anti. The epoxide forms on one face; water performs backside (S $_{N}$ 2-like) attack on the opposite face, inverting that carbon → the two OH end up on opposite faces = trans-cyclopentane-1,2-diol. The two backside attacks (one per enantiomeric epoxide face) give a racemic (±) mixture. See the figure: left = both OH up (syn/cis/meso); right = one up one down (anti/trans/racemic).

Problem 3.2

An unknown alkene undergoes reductive ozonolysis to give only acetone, $(CH_{3})_{2} C=O$ (two molecules). Deduce the alkene.

Recall Solution 3.2

Reductive ozonolysis cuts C=C and caps each carbon with =O, converting each half back to how it was attached.

Acetone came from a carbon written $(CH_{3})_{2} C=O$ , i.e. a double-bond carbon carrying two methyls and 0 H.
Both fragments are acetone → both original double-bond carbons were $(CH_{3})_{2} C=$ .
Re-join them at the =O positions: $(CH_{3})_{2} C=C (CH_{3})_{2}$ The alkene is 2,3-dimethylbut-2-ene (tetramethylethylene).

Level 4 — Synthesis

Problem 4.1

Starting from 1-methylcyclohexene, give reagents to make trans-2-methylcyclohexanol. Explain the stereochemical outcome.

Recall Solution 4.1

Use hydroboration–oxidation: (1) BH $_{3}$ ·THF, (2) H $_{2}$ O $_{2}$ , NaOH.

Regiochemistry: bulky B → less substituted ring carbon (the CH, not the C bearing CH $_{3}$ ); H → the carbon holding the methyl. OH thus ends up on the carbon next to the methyl (anti-Markovnikov).
Stereochemistry: B and H add syn (same face). After oxidation the new OH sits where B was, so OH and the newly added H are cis to each other. Working out the ring geometry, OH and the pre-existing CH $_{3}$ end up trans.
Result: trans-2-methylcyclohexanol, formed as a racemate (either face of the flat alkene is equally attacked).

Problem 4.2

Design a two-step route to convert cyclohexene into trans-cyclohexane-1,2-diol (not the cis isomer). State reagents and why the product is trans.

Recall Solution 4.2

We need anti dihydroxylation.

Step 1: mCPBA (a peroxyacid) → epoxide (cyclohexene oxide). Both C–O bonds form on one face (syn epoxidation).
Step 2: H $_{3}$ O $^{+}$ (dilute acid, water). Protonated epoxide is opened by backside attack of water on a ring carbon → inverts that carbon.
Net: one OH stays where the epoxide oxygen was, the other comes in from the opposite face → trans-cyclohexane-1,2-diol (racemic). If we had used cold KMnO $_{4}$ or OsO $_{4}$ instead we would get the cis diol — wrong for this target.

Level 5 — Mastery

Problem 5.1

A hydrocarbon $C_{5} H_{10}$ with one C=C gives, on oxidative ozonolysis, one molecule of CO $_{2}$ and one molecule of butan-2-one ( $CH_{3} COCH_{2} CH_{3}$ ). Identify the alkene and predict its hydroboration–oxidation product.

Recall Solution 5.1

Find the alkene. Oxidative ozonolysis:

CO $_{2}$ comes only from a $= CH_{2}$ carbon (2 H, fully over-oxidised) → one double-bond carbon was $= CH_{2}$ (terminal).
Butan-2-one $CH_{3} C (= O) CH_{2} CH_{3}$ is a ketone (0 H on that carbon) → the other double-bond carbon carried CH $_{3}$ and CH $_{2}$ CH $_{3}$ , no H.
Glue the two carbonyl carbons: $CH_{3} (CH_{3} CH_{2}) C=CH_{2}$ .
That is 2-methylbut-1-ene, $CH_{2} =C (CH_{3}) CH_{2} CH_{3}$ (C $_{5}$ H $_{10}$ ✓).

Hydroboration–oxidation of 2-methylbut-1-ene. C1 = terminal $= CH_{2}$ (less substituted); C2 = internal, more substituted (bears CH $_{3}$ + Et). Boron → less substituted C1 → OH on C1 (anti-Markovnikov): $CH_{2} =C (CH_{3}) CH_{2} CH_{3} BH_{3}; H_{2} O_{2} / OH^{-} HOCH_{2} -CH (CH_{3}) CH_{2} CH_{3}$ Product = 2-methylbutan-1-ol.

Problem 5.2

Propene is reacted with $Br_{2}$ in water containing dissolved $NaCl$ . Two products form: 1-bromopropan-2-ol and 1-bromo-2-chloropropane's regioisomer $CH_{3} CH(Cl)CH_{2} Br$ . Explain why chloride ends up on C2 (the more substituted carbon), not C1, and why Br stays on C1 in both products.

Recall Solution 5.2

Step 1 is always the same: Br $_{2}$ forms a bromonium ion bridging C1–C2. Bromine is now committed to the ring.
The bridge is unsymmetrical: the more substituted carbon (C2) bears the larger share of δ⁺ (it stabilises positive character better). So any nucleophile — H $_{2}$ O or Cl $^{-}$ — attacks the carbon with more δ⁺ = C2.
The nucleophile therefore lands on C2 (OH from water, or Cl from NaCl), while the bromonium opens to leave Br on C1 in both cases.
Products: $CH_{3} CH(OH)CH_{2} Br$ and $CH_{3} CH(Cl)CH_{2} Br$ — both Markovnikov-nucleophile, anti additions.
This confirms the general rule: the halonium's regiochemistry is set by δ⁺, independent of which nucleophile finishes the job.

Recall One-line self-check (cover and recall)

Boron picks which carbon? ::: The less substituted one (bulky + electron-poor) → anti-Markovnikov OH. Water in a halohydrin picks which carbon? ::: The more substituted one (more δ⁺) → Markovnikov OH. Cold KMnO $_{4}$ / OsO $_{4}$ stereochem? ::: syn (cyclic ester, same face). Epoxide + H $_{3}$ O $^{+}$ stereochem? ::: anti (backside ring opening). Ozonolysis of a $= CH_{2}$ carbon: reductive vs oxidative? ::: formaldehyde (reductive) vs CO $_{2}$ (oxidative).

Back to the parent: topic note · Prerequisites: Markovnikov's Rule and Carbocation Stability · Electrophilic Addition to Alkenes · Halogenation of Alkenes (halonium ion) · Oxidation of Alkenes — KMnO4 and OsO4 · Epoxides — Formation and Ring Opening · Acid-Catalyzed Hydration of Alkenes · Stereochemistry — syn vs anti addition, cis/trans, R/S