Chemistry · Alcohols, Phenols And Ethers · NEET
Anti-Markovnikov (hydroboration-oxidation) gives the primary (1°) alcohol. In BH3 addition, boron attaches to the less-hindered carbon that has MORE hydrogens (usually the terminal CH2). After oxidation with H2O2/OH-, boron is replaced by -OH at that same carbon. So for CH3-CH=CH2 you get CH3-CH2-CH2OH (propan-1-ol), a 1° alcohol. Markovnikov (acid hydration H2O/H+) instead puts -OH on the more substituted carbon, giving propan-2-ol (a 2° alcohol).
In hydroboration, boron is the electron-poor (electrophilic) atom, not hydrogen. Boron attaches to the carbon with MORE hydrogens because that carbon is less crowded and it lets the partial positive charge sit on the more substituted, more stable carbon. So the boron (and later the -OH) ends up on the terminal carbon. This is the reverse of acid hydration, where H+ adds first and -OH goes to the more substituted carbon. That is why the two methods give different alcohols.
Yes, the idea is the same. Markovnikov's rule says the negative part of the adding molecule goes to the carbon with FEWER hydrogens. For HBr, Br is the negative part; for water addition (H2O/H+), -OH is the negative part. So -OH lands on the more substituted carbon and gives a 2° or 3° alcohol. The positive H always goes to the carbon with more hydrogens.
No. The peroxide or Kharasch effect (a free-radical anti-Markovnikov addition) works ONLY with HBr, not with HCl, HI, or with water. For alcohols, the anti-Markovnikov product is made by a DIFFERENT method: hydroboration-oxidation using BH3 (or B2H6) followed by H2O2/OH-. Do not mix the two — peroxide + HBr gives an anti-Markovnikov bromide, while BH3 gives an anti-Markovnikov alcohol.
Read the reagent, not the alkene. If you see H2O with H+ (dilute acid) → Markovnikov → -OH on the carbon with fewer H → 2°/3° alcohol. If you see BH3 or B2H6 followed by H2O2/OH- (or NaOH/H2O2) → anti-Markovnikov → -OH on the carbon with more H → 1° alcohol. Hydroboration also adds in a syn way and gives no carbocation rearrangement, which is why it is 'clean'.
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 that convert an alkene, via the anti-Markovnikov primary alcohol, into an aldehyde.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Acid-catalysed hydration (H2O/H+), which follows Markovnikov's rule. The -OH lands on the more substituted carbon, so you usually get a 2° or 3° alcohol. Watch out: this route can involve a carbocation, so rearrangement is possible.
Hydroboration-oxidation: BH3 (or B2H6) first, then H2O2 with OH-. This is anti-Markovnikov, so -OH goes to the terminal carbon and you get a primary (1°) alcohol. No carbocation forms, so no rearrangement.
No. Hydroboration adds boron and hydrogen together in one step (a concerted, syn addition) with no free carbocation. That is why it gives a clean product with no rearrangement, unlike acid hydration.
The peroxide effect is anti-Markovnikov addition of HBr (only HBr) by a free-radical path, giving an alkyl bromide, not an alcohol. It does NOT apply to water addition. For an anti-Markovnikov alcohol you must use hydroboration-oxidation.
NCERT states that the negative part of the adding molecule attaches to the carbon atom that has the fewer number of hydrogen atoms. For water addition, -OH is the negative part, so it goes to the more substituted carbon.
