Chemistry · Alcohols, Phenols And Ethers · NEET
The OH goes to the MORE substituted carbon. Propene is CH3-CH=CH2. Acid hydration follows Markovnikov's rule, so H+ adds first and forms the more stable carbocation (secondary, on the middle carbon). Water then attaches there. Product: propan-2-ol, CH3-CH(OH)-CH3 (a secondary alcohol). This matters for NEET because most 'add water' MCQs expect the Markovnikov (secondary or tertiary) alcohol unless BH3 is mentioned.
In hydroboration, boron (electron-poor) attaches to the carbon that has MORE hydrogens - the less substituted, less crowded carbon. There is no carbocation; boron and hydrogen add together in one step. When you later oxidise the C-B bond, OH takes boron's place. So OH ends up on the less substituted carbon. Propene gives propan-1-ol (CH3-CH2-CH2-OH), a primary alcohol - the opposite of hydration.
For a terminal alkene like propene or but-1-ene, hydroboration-oxidation gives a PRIMARY alcohol, because OH lands on the terminal CH2 carbon. Plain acid hydration of the same alkene gives a secondary alcohol. So the reagent decides the product: BH3 then H2O2/NaOH = primary; H2O/H+ = secondary.
BH3 first adds across the double bond to give a trialkylborane (an alkyl-boron compound). The H2O2 in aqueous NaOH is the OXIDATION step: it replaces the carbon-boron bond with a carbon-OH bond, giving the alcohol. Without this second step you only have the borane, not the alcohol. NEET loves to test this two-step order.
Acid hydration goes through a carbocation, which can shift (rearrange) to a more stable one, sometimes giving an unexpected product. Hydroboration has NO carbocation - boron and hydrogen add in a single concerted step. So there is no rearrangement. This is a key reason the two methods can give different alcohols even from the same alkene.
Look at the reagents in the arrow. If you see H2O with H+ or dilute H2SO4, it is hydration = Markovnikov = OH on the more substituted carbon. If you see BH3 (or B2H6/diborane) followed by H2O2 and NaOH/OH-, it is hydroboration-oxidation = anti-Markovnikov = OH on the less substituted carbon. Match the reagent to the rule and pick the correct alcohol.
Identify 'X' in the following reactions: R-COOH -(i) X-(ii) H2O/HCl-> R-CH2OH and R-CH=CH2 -(i) X-(ii) H2O, NaOH, H2O2-> R-CH2-CH2-OH
Identify the correct reagents to convert an alkene, via the anti-Markovnikov primary alcohol, into an aldehyde.
In the presence of a few drops of concentrated sulphuric acid, an alkene reacts with water. According to which rule does the OH add?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Hydroboration-oxidation (BH3 then H2O2/NaOH). It is anti-Markovnikov, so OH lands on the terminal carbon and you get a primary alcohol. Acid hydration of the same alkene gives a secondary alcohol instead.
Markovnikov. Water adds with a little H2SO4, H+ makes the more stable carbocation, and OH goes to the more substituted carbon.
H.C. Brown reported it in 1959 and shared the 1979 Nobel Prize in Chemistry for his work on boron-containing organic compounds. NCERT mentions this, so it can appear as a small fact question.
Because it does not form a carbocation. Boron and hydrogen add in one concerted step, so there is nothing to rearrange - unlike acid hydration which goes through a carbocation that can shift.
Hydrogen peroxide (H2O2) in aqueous sodium hydroxide (NaOH). It replaces the carbon-boron bond with carbon-OH to give the alcohol.
