# Hydroboration-Oxidation > Hydroboration-oxidation for JEE Chemistry: anti-Markovnikov hydration with BH3 then H2O2 NaOH, syn addition, no rearrangement, stereochemistry. - Canonical URL: https://prepelephant.com/topics/jee/chemistry/hydroboration-oxidation - Exam / course: JEE · Subject: Chemistry - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Hydroboration-Oxidation", PrepElephant, https://prepelephant.com/topics/jee/chemistry/hydroboration-oxidation ## Direct answer Propene plus BH3, followed by hydrogen peroxide in sodium hydroxide, gives propan-1-ol — the anti-Markovnikov alcohol. That is hydroboration-oxidation: borane (usually BH3 complexed with THF, or generated from diborane) adds across the double bond in one concerted, four-centre transition state, placing boron on the less substituted carbon and hydrogen on the more substituted one, strictly syn. Oxidation then replaces the C-B bond with C-OH, with retention of configuration. Because no carbocation ever forms, no rearrangement ever happens — the property that separates this method from acid-catalysed hydration. One mole of BH3 processes three moles of alkene to a trialkylborane before the oxidation step is even mentioned. ## What you must remember - **Reagent pair, both steps:** 1. BH3·THF (or B2H6); 2. H2O2, NaOH — forgetting the second step is the classic half-answer. - **Regiochemistry:** boron (then OH) on the less substituted carbon; terminal alkenes give primary alcohols, propene being the reference example. - **Mechanism flags:** concerted, syn addition, four-centre transition state, boron polarised delta-positive and hydrogen delta-negative. - **No rearrangement:** with no cation intermediate, 3,3-dimethyl substitution patterns stay intact — contrast acid hydration, which rearranges freely. - **Stereochemistry ledger:** addition is syn; oxidation retains configuration at the carbon; net result — OH and H that arrived together stay on the same face. - **Cyclic showcase:** 1-methylcyclohexene gives trans-2-methylcyclohexanol (racemic), the standard JEE Advanced stereoisomer question. - **Method contrast:** acid hydration (Markovnikov, rearranges), oxymercuration-demercuration (Markovnikov, does not rearrange), hydroboration (anti-Markovnikov, does not rearrange). ## Working the stereochemistry on a ring Take 1-methylcyclohexene. Boron approaches the less substituted C2, hydrogen transfers to C1 on the same face, and both new bonds share one face of the ring. Oxidation swaps boron for OH with retention, so the OH at C2 and the newly added H at C1 remain cis. But the methyl at C1 points opposite to the hydrogen delivered there — so OH and CH3 end up trans, giving racemic trans-2-methylcyclohexanol. Trace it once with a model and the logic locks in: the configuration question is never about memory, only about which group was already attached and which face received the new atoms. Then the simple counting. Trialkylborane formation: 3 CH3CH=CH2 + BH3 → (CH3CH2CH2)3B; oxidation of that trialkylborane with alkaline peroxide yields three propan-1-ol molecules. Net reaction across both steps: the alkene plus hydrogen peroxide delivering the elements of water anti-Markovnikov — which is why industrial and laboratory routes to primary alcohols from internal feedstocks prefer this sequence. ## How the exam frames it JEE Main's version is one line: product of hydroboration-oxidation of a named terminal alkene — and the wrong options are always the Markovnikov alcohol (from acid hydration memory) and the ketone. The second layer asks you to compare methods: which hydration route gives 1-propanol from propene, which gives 2-propanol, which avoids rearrangement — a three-way table collapsed into one option. JEE Advanced adds the stereochemistry: racemic or meso, cis or trans OH-to-alkyl, and the crucial insight that oxidation's retention of configuration makes the overall process syn even though the two steps feel unrelated. Bulky boranes (disiamylborane, 9-BBN) sharpen regioselectivity on internal alkenes and appear as comprehension garnish — know the name, not the detail. ## Frequently asked questions ### Why is hydroboration-oxidation called anti-Markovnikov hydration? The OH ultimately lands on the less substituted carbon, opposite to Markovnikov placement, because boron bonds to the less hindered carbon in the concerted addition. ### Why does hydroboration never rearrange? The addition is concerted through a four-centre transition state — no carbocation intermediate forms, so there is nothing for a hydride or methyl to shift toward. ### What is the stereochemistry of the overall addition? Syn: H and B add to the same face, and the oxidation step replaces B with OH with retention, keeping H and OH cis to each other. ### What does 1-methylcyclohexene give on hydroboration-oxidation? Racemic trans-2-methylcyclohexanol — OH enters at C2, syn to the hydrogen delivered at C1 and therefore trans to the existing methyl. ### Why is the H2O2/NaOH step indispensable? Trialkylboranes are not alcohols; the alkaline peroxide oxidation is what replaces each C-B bond by C-OH, delivering the actual product.