Consider the following reaction scheme and choose the correct option(s) for the major products Q, R and S.
← JEE Advanced 2023 · Paper 1![Styrene → [(i) B2H6; (ii) NaOH, H2O2, H2O] → P → [(i) CrO3, H2SO4; (ii) Cl2, red phosphorus; (iii) H2O] → Q. P → [(i) SOCl2; (ii) NaCN; (iii) H3O+, Δ] → R → [conc. H2SO4] → S.](/figures/ja-2023-p1/ja-2023-p1-che-q03-f1.png)
ChemistryMultiple correct+4 / -2Question 37 of 51
JEE Advanced 2023 · Paper 1, Chemistry question 3
Answer and worked solution
Official answer: B
P. Hydroboration-oxidation of styrene gives the anti-Markovnikov alcohol, .
Q.
- oxidises P to .
- with red P (Hell-Volhard-Zelinsky) chlorinates the carbon, giving .
- Water hydrolyses the acid chloride, so Q is .
R. gives , NaCN gives , and hydrolysis gives .
S. Concentrated closes the ring by intramolecular Friedel-Crafts acylation, giving 1-indanone.
This is option (B).