The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is: Item-I (I) Benzene (II) Benzonitrile (III) Benzoyl Chloride Item-II (P) \(\mathrm{HCl}\) and \(\mathrm{SnCl}_{2}, \mathrm{H}_{3} \mathrm{O}^{+}\) (Q) \(\mathrm{H}_{2}, \mathrm{Pd}-\mathrm{BaSO}_{4}, \mathrm{~S}\) and quinoline (R) \(\mathrm{CO}, \mathrm{HCl}\) and \(\mathrm{AlCl}_{3}\)

(C) (I) - (R), (II) - (P) and (III) - (Q)

(A) (I) - (Q), (II) - (R) and (III) - (P)

(B) (I) - (P), (II) - (Q) and (III) - (R)

(D) (I) - (R), (II) - (Q) and (III) - (P)

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is:

Item-I
(I) Benzene
(II) Benzonitrile
(III) Benzoyl Chloride

Item-II
(P)

HCl

and

SnCl_{2}, H_{3} O^{+}

(Q)

H_{2}, Pd - BaSO_{4}, S

and quinoline
(R)

CO, HCl

and

AlCl_{3}

(A)(I) - (Q), (II) - (R) and (III) - (P)
(B)(I) - (P), (II) - (Q) and (III) - (R)
(C)(I) - (R), (II) - (P) and (III) - (Q)✓ Correct
(D)(I) - (R), (II) - (Q) and (III) - (P)

Answer: (C)

Solution

Each route is a named formylation. Benzene is turned into benzaldehyde by the Gattermann-Koch reaction with

CO

HCl

and

AlCl X_{3}

. Benzonitrile gives benzaldehyde by Stephen reduction with

SnCl X_{2} / HCl

followed by hydrolysis. Benzoyl chloride gives benzaldehyde by Rosenmund reduction with

H X_{2}

over palladium poisoned on

BaSO X_{4}

with sulphur and quinoline. So benzene pairs with the

CO / HCl / AlCl X_{3}

reagent, benzonitrile with

SnCl X_{2} / HCl

, and benzoyl chloride with the poisoned-palladium reagent.

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

Solution

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