Sigma Percentile
JEE Main 2020
LEVELJEE Main

Animated Solution for Chemistry - Organic Chemistry: The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is \begin{array}{ll} \text{Item - I} & \text{Item - II} \\ \text{I. Benzene} & \text{(P) HCl and } \text{SnCl}_2, \text{H}_3\text{O}^+ \\ \text{II. Benzonitrile} & \text{(Q) } \text{H}_2, \text{Pd-BaSO}_4, \text{S and quinoline} \\ \text{III. Benzoyl chloride} & \text{(R) CO, HCl and } \text{AlCl}_3 \end{array}

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The Sigma Insight: Carbonyl Compounds

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Mastering Name Reactions

Pathways to Benzaldehyde
Preparing benzaldehyde from various starting materials is a classic theme in organic chemistry. This problem tests your memory and understanding of three fundamental name reactions. Let's break down each pathway to see how we can synthesize our target molecule.

The Gattermann-Koch Reaction

First, let's look at Benzene. To synthesize benzaldehyde directly from benzene, we employ the Gattermann-Koch reaction.
In this process, benzene is treated with a mixture of carbon monoxide () and hydrogen chloride () in the presence of a Lewis acid catalyst, typically anhydrous aluminium chloride (). The and essentially act as a source of formyl chloride (), which generates the electrophilic formyl cation (). This cation then attacks the benzene ring via electrophilic aromatic substitution to yield benzaldehyde.
Therefore, Benzene (I) matches with the reagents (R).

The Stephen Reduction

Next, we have Benzonitrile. Converting a nitrile group () into an aldehyde () is elegantly achieved using the Stephen reduction.
This reaction involves treating the nitrile with tin(II) chloride () and hydrochloric acid (). The nitrile is first reduced to an intermediate iminium salt. Subsequent hydrolysis of this iminium salt with aqueous acid () yields the corresponding aldehyde.
Thus, Benzonitrile (II) matches with the reagents , and (P).

The Rosenmund Reduction

Finally, consider Benzoyl chloride. Transforming an acid chloride () into an aldehyde requires a controlled reduction, specifically the Rosenmund reduction.
In this reaction, hydrogen gas () is bubled through a solution of the acid chloride in the presence of a palladium catalyst supported on barium sulfate ().
The Catch: Why use barium sulfate? Palladium alone is too active and would reduce the newly formed aldehyde all the way down to a primary alcohol. Barium sulfate acts as a catalytic poison, intentionally decreasing the activity of the palladium so the reduction stops exactly at the aldehyde stage. Sometimes, a small amount of sulfur or quinoline is also added to further poison the catalyst.
Consequently, Benzoyl chloride (III) matches with , S, and quinoline (Q).

Conclusion

By systematically analyzing each starting material and its corresponding name reaction, we arrive at the final matching: - (I) (R) - (II) (P) - (III) (Q)
This perfectly aligns with option (c).

Similar Questions

JEE Advanced 2017
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Comprehension Passage

Answer Q.13, Q.14 and Q.15 by appropriately matching the information given in the three columns of the following table. Columns 1, 2 and 3 contains starting materials, reaction conditions, and type of reactions, respectively. \begin{array}{|l|l|l|} \hline \textbf{Column-1} & \textbf{Column-2} & \textbf{Column-3} \\ \hline \text{(I) Toluene} & \text{(i) NaOH/Br}_2 & \text{(P) Condensation} \\ \text{(II) Acetophenone} & \text{(ii) Br}_2 / h\nu & \text{(Q) Carboxylation} \\ \text{(III) Banzaldehyde} & \text{(iii) (CH}_3\text{CO)}_2\text{O/CH}_3\text{COOK} & \text{(R) Substitution} \\ \text{(IV) Phenol} & \text{(iv) NaOH/CO}_2 & \text{(S) Haloform} \\ \hline \end{array}
Question 1:

For the synthesis of benzoic acid, the only CORRECT combination is

(A)
(III) (iv) (R)
(B)
(IV) (ii) (P)
(C)
(I) (iv) (Q)
(D)
(II) (i) (S)
Question 2:

The only CORRECT combination in which the reaction proceeds through radical mechanism is

(A)
(I) (ii) (R)
(B)
(II) (iii) (R)
(C)
(III) (ii) (P)
(D)
(IV) (i) (Q)
Question 3:

The only CORRECT combination that gives two different carboxylic acids is

(A)
(IV) (iii) (Q)
(B)
(III) (iii) (P)
(C)
(II) (iv) (R)
(D)
(I) (i) (S)
JEE Main 2021
LEVELJEE Main

Match List-I with List-II. List-I (Chemical reaction) A. B. C. D. List-II (Reagent used) 1. (1 equivalent) 2. 3. 4. Choose the most appropriate option given below.

(A)
A-2, B-4, C-3, D-1
(B)
A-4, B-2, C-3, D-1
(C)
A-2, B-3, C-4, D-1
(D)
A-3, B-2, C-1, D-4
JEE Advanced 2018
LEVELJEE Advanced

Comprehension Passage

Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product.
Question 1:

The compound Y is :-

(A)
(B)
(C)
(D)
Question 2:

The compound Z is :-

(A)
(B)
(C)
(D)
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The structure of product , formed by the following sequence of reactions is

(A)
(B)
(C)
(D)
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The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is

(A)
(B)
(C)
(D)
JEE Advanced 2015
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Among the following the number of reaction(s) that produce(s) benzaldehyde is –

JEE Main 2021
LEVELJEE Main

Match List-I with List-II. Choose the correct answer from the options given below.

List-I

(P)
(Q)
(R)
(S)

List-II

(1)
(2)
(3)
(4)
JEE Main 2019
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In the following reaction, The best combination is

(A)
and MeOH
(B)
and
(C)
HCHO and MeOH
(D)
HCHO and
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In the following sequence of reaction, The product is

(A)
(B)
(C)
(D)
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LEVELJEE Advanced

The most suitable reagent for the given conversion is

(A)
(B)
(C)
(D)