In the following sequence of reactions: Toluene $\xrightarrow{\mathrm{KMnO}_{4}}$ A $\xrightarrow{\mathrm{SOCl}_{2}}$ B $\xrightarrow[\mathrm{BaSO}_{4}]{\mathrm{H}_{2} / \mathrm{Pd}}$ C, The product C is:

(C) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$

(A) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$

(B) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$

(D) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$

JEE Main · 2015 · 04 Apr 2015 · Offline · Aldehydes Ketones and Carboxylic Acids

In the following sequence of reactions:

Toluene

KMnO_{4}

SOCl_{2}

H_{2} / Pd BaSO_{4}

C,

The product C is:

(A) $C_{6} H_{5} CH_{3}$
(B) $C_{6} H_{5} CH_{2} OH$
(C) $C_{6} H_{5} CHO$ ✓ Correct
(D) $C_{6} H_{5} COOH$

Answer: (C)

Solution

Follow the chain step by step.

KMn O_{4}

oxidises the methyl group of toluene all the way to

C_{6} H_{5} COOH

, benzoic acid.

SOC l_{2}

then converts that acid into its acid chloride,

C_{6} H_{5} COCl

. Finally

H_{2}

over

Pd

poisoned with

BaS O_{4}

-- the Rosenmund reduction -- stops cleanly at the aldehyde, so

C

is benzaldehyde,

C_{6} H_{5} CHO

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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