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Aldehydes Ketones and Carboxylic Acids question

2020 · 6 Sep · Shift 2 · Q1
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Aldehydes Ketones and Carboxylic Acids question

2020 · 6 Sep · Shift 2 · Q1

JEE MainChemistryAldehydes Ketones and Carboxylic AcidsMCQ+4 / −1
The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is :

Item-I Item-II
(I) Benzene (P) HClHClHCl and SnCl2, H3O+
(II) Benzonitrile (Q) H2, Pd-BaSO4, S and quinoline
(III) Benzoyl Chloride (R) COCOCO, HClHClHCl and AlCl3
  1. A
    (I) - (R), (II) - (P) and (III) - (Q)
  2. B
    (I) - (P), (II) - (Q) and (III) - (R)
  3. C
    (I) - (Q), (II) - (R) and (III) - (P)
  4. D
    (I) - (R), (II) - (Q) and (III) - (P)
View written solutionFree

Correct answer: A

  1. We need to match each starting material with the reagent that prepares benzaldehyde.
  1. (I) Benzene

To convert benzene to benzaldehyde, we use the Gattermann–Koch formylation:

C6H6→AlCl3CO, HClC6H5CHO\text{C}_6\text{H}_6 \xrightarrow[AlCl_3]{CO,\ HCl} \text{C}_6\text{H}_5CHOC6​H6​CO, HClAlCl3​​C6​H5​CHO

So,

(I)→(R)(I) \to (R)(I)→(R)

where (R)=CO, HCl and AlCl3(R) = CO,\ HCl \text{ and } AlCl_3(R)=CO, HCl and AlCl3​.

  1. (II) Benzonitrile

Nitriles on partial reduction with SnCl2/HClSnCl_2/HClSnCl2​/HCl followed by hydrolysis give aldehydes. This is Stephen reduction:

C6H5CN→SnCl2/HClimine salt→H3O+C6H5CHO\text{C}_6\text{H}_5CN \xrightarrow{SnCl_2/HCl} \text{imine salt} \xrightarrow{H_3O^+} \text{C}_6\text{H}_5CHOC6​H5​CNSnCl2​/HCl​imine saltH3​O+​C6​H5​CHO

So,

(II)→(P)(II) \to (P)(II)→(P)

where (P)=HCl and SnCl2, H3O+(P) = HCl \text{ and } SnCl_2,\ H_3O^+(P)=HCl and SnCl2​, H3​O+.

  1. (III) Benzoyl chloride

Acid chlorides on Rosenmund reduction give aldehydes:

C6H5COCl→H2, Pd−BaSO4, S, quinolineC6H5CHO\text{C}_6\text{H}_5COCl \xrightarrow{H_2,\ Pd-BaSO_4,\ S,\ quinoline} \text{C}_6\text{H}_5CHOC6​H5​COClH2​, Pd−BaSO4​, S, quinoline​C6​H5​CHO

So,

(III)→(Q)(III) \to (Q)(III)→(Q)

where (Q)=H2, Pd−BaSO4, S and quinoline(Q) = H_2,\ Pd-BaSO_4,\ S \text{ and quinoline}(Q)=H2​, Pd−BaSO4​, S and quinoline.

  1. Therefore the correct matching is:
(I)−(R),(II)−(P),(III)−(Q)(I)-(R),\quad (II)-(P),\quad (III)-(Q)(I)−(R),(II)−(P),(III)−(Q)

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

Hence, the derived answer agrees with the stored answer.

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