JEE Main · 2024 · Shift-IIhardALCO-017

The final product A, formed in the following reaction sequence is: Ph-CH=CH2 (i) BH3; (ii) H2O2,OH- intermediate (iii)…

Alcohols, Phenols & Ethers · Class 12 · JEE Main Previous Year Question

Question

The final product A, formed in the following reaction sequence is:

\cePh−CH=CH2\ce{Ph-CH=CH2} →(i) \ceBH3; (ii) \ceH2O2,OH−\xrightarrow{\text{(i) }\ce{BH3}\text{; (ii) }\ce{H2O2,OH^-}} intermediate →(iii) HBr; (iv) Mg, ether, then HCHO/H3O+\xrightarrow{\text{(iii) HBr; (iv) Mg, ether, then HCHO/H3O+}} A

Options
  1. a

    \cePh−CH2−CH2−CH3\ce{Ph-CH2-CH2-CH3}

  2. b

    \cePh−CH(CH3)2\ce{Ph-CH(CH3)_2}

  3. c

    \cePh−CH(CH3)(CH2OH)\ce{Ph-CH(CH3)(CH2OH)} (branched alcohol)

  4. d

    \cePh−CH2−CH2−CH2OH\ce{Ph-CH2-CH2-CH2OH} (3-phenylpropan-1-ol)

    ✓
Correct Answerd

\cePh−CH2−CH2−CH2OH\ce{Ph-CH2-CH2-CH2OH} (3-phenylpropan-1-ol)

Detailed Solution

Step 1: Styrene + (i) BH3, (ii) H2O2/OH-

Hydroboration-oxidation of styrene (\cePh−CH=CH2\ce{Ph-CH=CH2}):

  • Anti-Markovnikov addition of \ceH2O\ce{H2O} across the double bond.
  • BH3 adds B to the less substituted carbon (terminal CH2).
  • After oxidation: \ce−OH\ce{-OH} goes to the terminal carbon.
  • Product: \cePh−CH2−CH2−OH\ce{Ph-CH2-CH2-OH} (2-phenylethanol, the primary alcohol)

Step 2: 2-Phenylethanol + HBr

\cePh−CH2−CH2−OH+HBr−>Ph−CH2−CH2−Br+H2O\ce{Ph-CH2-CH2-OH + HBr -> Ph-CH2-CH2-Br + H2O}

  • Primary alcohol + HBr → primary alkyl bromide via SN2.
  • Product: \cePh−CH2−CH2−Br\ce{Ph-CH2-CH2-Br} (2-bromoethylbenzene)

Step 3: 2-Bromoethylbenzene + Mg/dry ether → Grignard reagent

\cePh−CH2−CH2−Br+Mg−>[dryether]Ph−CH2−CH2−MgBr\ce{Ph-CH2-CH2-Br + Mg ->[ dry ether ] Ph-CH2-CH2-MgBr} (Grignard reagent)

Step 4: Grignard + HCHO (formaldehyde), then H3O+

Grignard reagent + formaldehyde → primary alcohol (chain extended by 1 carbon): \cePh−CH2−CH2−MgBr+HCHO−>Ph−CH2−CH2−CH2−OMgBr−>[H3O+]Ph−CH2−CH2−CH2−OH\ce{Ph-CH2-CH2-MgBr + HCHO -> Ph-CH2-CH2-CH2-OMgBr ->[ H3O+ ] Ph-CH2-CH2-CH2-OH}

Product A = \cePh−CH2−CH2−CH2−OH\ce{Ph-CH2-CH2-CH2-OH} (3-phenylpropan-1-ol)

Step 5: Common wrong approach

Students forget that hydroboration gives anti-Markovnikov product (terminal OH) — if they mistakenly apply Markovnikov, they would get \cePh−CH(OH)−CH3\ce{Ph-CH(OH)-CH3} leading to a different final product.

Key Points to Remember:

  • BH3/H2O2 = hydroboration-oxidation = anti-Markovnikov, syn addition, primary alcohol from terminal alkene.
  • Grignard + HCHO → primary alcohol (1C chain extension).
  • Grignard + RCHO → secondary alcohol; Grignard + R2CO → tertiary alcohol.

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