The final product A, formed in the following reaction sequence is: [figure]
- [figure]
- [figure]
- Ph-CH2-CH2-CH2-OH
- Ph-CH2-CH2-CH3
Answer
(3) Ph-CH2-CH2-CH2-OH
Solution
Ph-CH=CH2 [(ii) H2O2, OH](i) BH3 Ph-CH2-CH2-OH
Hydroboration–oxidation adds H2O anti-Markovnikov, so –OH goes to the terminal C.
Ph-CH2-CH2-OH HBr Ph-CH2-CH2-Br
Here HBr protonates –OH; as H2O leaves, the neighbouring phenyl ring helps push it out (NGP) and a bridged phenonium ion forms. Since both ring-bonded carbons of this ion are CH2, Br attacks either one and gives the same unbranched bromide, so the skeleton does not change.
Ph-CH2-CH2-Br Mg, ether Ph-CH2-CH2-MgBr
[(ii) H3O+](i) HCHO Ph-CH2-CH2-CH2-OH
Because RMgX adds to HCHO, we get a 1° alcohol that has one more carbon.
(1) and (2) are branched at the benzylic C, which would need Markovnikov addition, so they are wrong; (4) has no –OH, so it is wrong.
Answer: (3)