JEE Main 2015 · 4 Apr · Shift 1 · Chemistry Q55

ChemistrySingle correct+4 / -1Question 55 of 90

In the following sequence of reactions :
$\text{Toluene}\ce{->[KMnO4] A ->[SOCl2] B ->[H2/Pd][BaSO4] C},$
the product C is :

\ce{C6H5COOH}

\ce{C6H5CH3}

\ce{C6H5CH2OH}

\ce{C6H5CHO}

Answer and worked solution

Official answer: D

$\ce{KMnO4}$ oxidises toluene to benzoic acid (A).
$\ce{SOCl2}$ gives benzoyl chloride (B).
Rosenmund reduction ( $\ce{H2/Pd-BaSO4}$ ) gives benzaldehyde, $\ce{C6H5CHO}$ (C).

JEE Main 2015 · 4 Apr · Shift 1, Chemistry question 55