JEE MainChemistryHydrocarbons
The correct sequence of reagents for the conversion of n-hexane to a mixture of benzyl alcohol and sodium benzoate is:
Options
- A(i) V₂O₅, 773 K , 10-20 atm ; (ii) CO, HCl, anhyd. AlCl₃, CuCl ; (iii) Conc. NaOH
- B(i) V₂O₅, 773 K , 10-20 atm ; (ii) CrO₂Cl₂, CS₂ ; (iii) H₃O^+ ; (iv) Conc. NaOH
- C(i) Cr₂O₃, 773 K , 10-20 atm ; (ii) CH₃Cl, anhyd. AlCl₃ ; (iii) KMnO₄, KOH, ; (iv) Conc. NaOH
- D(i) Mo₂O₃, 773 K , 10-20 atm ; (ii) CO, HCl, anhyd. AlCl₃, CuCl ; (iii) Dil. NaOH
Correct answer
A. (i) V₂O₅, 773 K , 10-20 atm ; (ii) CO, HCl, anhyd. AlCl₃, CuCl ; (iii) Conc. NaOH
Step-by-step solution
The starting material is n-hexane. Aromatization of n-hexane over V₂O₅ at 773 K and 10-20 atm yields benzene. Benzene lacks a methyl group, so it cannot undergo the Etard reaction. To introduce an aldehyde group, the Gattermann-Koch reaction is used. Treatment with CO and HCl in the presence of anhydrous AlCl₃ and CuCl converts benzene to benzaldehyde. Benzaldehyde lacks -hydrogens. Upon treatment with concentrated NaOH , it undergoes the Cannizzaro reaction (disproportionation) to yield benzyl alcohol and sodium b