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JEE Chemistry 2023 Question with Solution

Find out the major products from the following reactions. B→H2O,NaBH4AB \xrightarrow{H_2O, NaBH_4} A B→BH3,THFAB \xrightarrow{BH_3, THF} A B→H2O,OH−BB \xrightarrow{H_2O, OH^-} B

  • A

    A = Alcohol, B = Aldehyde

  • B

    A = Aldehyde, B = Alcohol

  • C

    A = Acid, B = Aldehyde

  • D

    A = Ketone, B = Acid

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The solution discusses addition of water to an alkene under two different reagent sets.

Find: Which option correctly identifies products A and B.

From the solution:

  • With Hg(OAc)2,H2O,NaBH4Hg(OAc)_2, H_2O, NaBH_4, the reaction is oxymercuration-demercuration, giving the Markovnikov alcohol.
  • With BH3,THF,H2O2/OH−BH_3, THF, H_2O_2/OH^-, the reaction is hydroboration-oxidation, giving the Anti-Markovnikov alcohol.

So:

  • A is the Anti-Markovnikov product.
  • B is the Markovnikov product.

The solution explicitly concludes: The Correct Option is B.

Therefore, the correct option is B.

Common mistakes

  • Confusing Markovnikov and Anti-Markovnikov addition. This is incorrect because the two reagent sets place the hydroxyl group on different carbons. Identify the reaction type first, then assign the orientation.

  • Treating both reactions as giving the same alcohol. This is wrong because oxymercuration-demercuration and hydroboration-oxidation generally give different regioisomeric alcohols. Compare the substitution pattern carefully.

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