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Choose the correct option for the following reactions. B —(BH3, H2O2/OH−)→ [product] [alkene] —(Hg(OAc)2, H2O; NaBH4)→ A

1. ‘A’ and ‘B’ are both Markovnikov addition products
2. ‘A’ is Markovnikov product and ‘B’ is anti-Markovnikov product
3. ‘A’ and ‘B’ are both anti-Markovnikov products
4. ‘B’ is Markovnikov and ‘A’ is anti-Markovnikov product

Correct answer: ‘A’ is Markovnikov product and ‘B’ is anti-Markovnikov product

Solution

Reaction A uses Hg(OAc)2/H2O then NaBH4 (oxymercuration-demercuration), which gives the Markovnikov alcohol with no rearrangement. Reaction B uses BH3 then H2O2/OH- (hydroboration-oxidation), giving the anti-Markovnikov alcohol. So A is Markovnikov and B is anti-Markovnikov.

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