Question

Consider the following reaction, and choose the correct option.

• A. Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on $\text{BaSO}_4$.
• B. On treating compound P with saturated $\text{NaHCO}_3$ solution, brisk effervescence is observed.
• C. Compound P can be prepared by treating benzene with anhydrous $\text{AlCl}_3$ and $\text{CH}_3\text{COCl}$.
• D. On treatment with bromine water, compound P gives a white precipitate.

Hint:

To quickly master named reactions of carbonyl compounds: - Etard Reaction: $\text{Toluene} + \text{CrO}_2\text{Cl}_2 \rightarrow \text{Benzaldehyde}$ - Rosenmund Reduction: $\text{Acid Chloride} + \text{H}_2 / \text{Pd-BaSO}_4 \rightarrow \text{Aldehyde}$ Both are premier methods for synthesizing pure aromatic aldehydes without over-oxidation or over-reduction!

Answer 1

The Correct Option is A

Concept: The conversion of toluene to benzaldehyde using chromyl chloride ($\text{CrO}_2\text{Cl}_2$) in a carbon disulfide ($\text{CS}_2$) solvent followed by acidic hydrolysis is known as the

Etard Reaction. Chromyl chloride behaves as a mild oxidizing agent that selectively oxidizes a terminal methyl group attached to an aromatic ring into an aldehyde functionality without further oxidation into a carboxylic acid. \[ \text{C}_6\text{H}_5\text{CH}_3 \xrightarrow[\text{ii. } \text{H}_3\text{O}^+]{\text{i. } \text{CrO}_2\text{Cl}_2, \, \text{CS}_2} \text{C}_6\text{H}_5\text{CHO} \quad (\text{Benzaldehyde}) \] Therefore, the unknown product

P is benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$). Let us meticulously analyze each structural option given to evaluate its chemical validity.

Step 1: Identifying Compound P via the Etard Reaction mechanism.
When toluene ($\text{C}_6\text{H}_5\text{CH}_3$) is treated with chromyl chloride ($\text{CrO}_2\text{Cl}_2$) in non-polar solvent $\text{CS}_2$, a brown chromium complex intermediate is initially formed. The reaction proceeds via a radical mechanism where the methyl group is partially oxidized: \[ \text{C}_6\text{H}_5\text{CH}_3 + 2\text{CrO}_2\text{Cl}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}[\text{OCr(OH)Cl}_2]_2 \] Subsequent aqueous acidic hydrolysis ($\text{H}_3\text{O}^+$) of this brown intermediate complex breaks the chromium-oxygen bonds to yield benzaldehyde: \[ \text{C}_6\text{H}_5\text{CH}[\text{OCr(OH)Cl}_2]_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{CHO} + 2\text{Cr(OH)}_2\text{Cl}_2 \] Thus, Compound

P is unequivocally established as

Benzaldehyde.

Step 2: Evaluating Option (1).
Option (1) states that compound

P is obtained by the hydrogenation of benzoyl chloride ($\text{C}_6\text{H}_5\text{COCl}$) with palladium (Pd) supported on barium sulfate ($\text{BaSO}_4$). This partial reduction reaction is a famous organic transformation known as the

Rosenmund Reduction. Barium sulfate acts as a catalyst poison (modulator) to lower the catalytic activity of palladium, preventing the newly formed aldehyde from undergoing further reduction into a primary benzyl alcohol. The structural chemical equation is formulated as follows: \[ \text{C}_6\text{H}_5\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd / BaSO}_4} \text{C}_6\text{H}_5\text{CHO} + \text{HCl} \] Since the product generated from the Rosenmund reduction is indeed benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$), which matches compound

P perfectly, this statement is perfectly accurate.

Step 3: Evaluating Option (2).
Option (2) suggests that treating compound

P with a saturated sodium bicarbonate ($\text{NaHCO}_3$) solution produces brisk effervescence. Brisk effervescence with a weak base like $\text{NaHCO}_3$ occurs due to the release of carbon dioxide ($\text{CO}_2$) gas. This reaction is a classic functional group diagnostic test for relatively strong organic acids, such as carboxylic acids (R-COOH), which are sufficiently acidic to decompose the bicarbonate ion: \[ \text{R-COOH} + \text{NaHCO}_3 \rightarrow \text{R-COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow \] Since benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) is an aldehyde and does not contain a highly acidic carboxyl proton, it fails to react with saturated sodium bicarbonate solution. No $\text{CO}_2$ gas is generated, and no effervescence is seen. Therefore, this statement is false.

Step 4: Evaluating Option (3).
Option (3) states that compound

P can be prepared by treating benzene with anhydrous $\text{AlCl}_3$ and acetyl chloride ($\text{CH}_3\text{COCl}$). This reaction describes a classic

Friedel-Crafts Acylation. Let us trace the reaction mechanism: Benzene reacts with acetyl chloride in the presence of a Lewis acid catalyst to form an acylium ion electrophile ($\text{CH}_3\text{CO}^+$), which attacks the aromatic ring: \[ \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhydrous }\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \] The final organic product formed here is acetophenone (a methyl ketone), not benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$). Hence, this statement is incorrect.

Step 5: Evaluating Option (4).
Option (4) states that treatment with bromine water gives a white precipitate. Bromine water ($\text{Br}_2/\text{H}_2\text{O}$) gives a distinctive white precipitate of 2,4,6-tribromophenol or 2,4,6-tribromoaniline when reacted with highly activated aromatic systems such as phenol ($\text{C}_6\text{H}_5\text{OH}$) or aniline ($\text{C}_6\text{H}_5\text{NH}_2$). The strong activating groups ($-\text{OH}$ or $-\text{NH}_2$) increase the electron density on ortho and para positions immensely. In contrast, the formyl group ($-\text{CHO}$) in benzaldehyde is a strong electron-withdrawing group via resonance (deactivating group). Benzaldehyde does not undergo rapid electrophilic bromination with mild bromine water to give any white precipitate; instead, it undergoes slow meta-bromination under harsh conditions with pure Lewis acid catalysts. Thus, this statement is completely false.