Match the LIST-I with LIST-II.LIST-IReactionLIST-IIReagentA. Rosenmun…

Match the LIST-I with LIST-II.

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → A-II, B-I, C-III, D-IV

The named reactions in this list make aldehydes and ketones. They can be recognised from the starting material and the reagent:

From acid derivatives (reduction): Rosenmund reduction — acyl chloride + H₂, Pd-BaSO₄ gives an aldehyde; Stephen reaction — nitrile + SnCl₂, HCl, then hydrolysis, gives an aldehyde (DIBAL-H does the same change for nitriles and esters).
From hydrocarbons (oxidation of a methyl group): Etard reaction — toluene + chromyl chloride (CrO₂Cl₂) in CS₂, then hydrolysis, gives benzaldehyde; Gattermann–Koch reaction — benzene + CO and HCl with anhydrous \( \ce{AlCl3} \)/CuCl gives benzaldehyde.
Acylation of arenes: Friedel–Crafts acylation — benzene + acyl chloride with anhydrous AlCl₃ gives an aromatic ketone.

A. Rosenmund reaction — This matches II. H₂ , Pd-BaSO₄. The acyl chloride is hydrogenated over palladium on barium sulphate; the catalyst is partially poisoned so that reduction stops at the aldehyde.
\( \displaystyle \ce{R-COCl + H2} \xrightarrow{\ce{Pd-BaSO4}} \ce{R-CHO + HCl} \)
B. Stephen reaction — This matches I. SnCl₂, HCl. The nitrile is reduced by stannous chloride and HCl to an imine, which is hydrolysed to the aldehyde.
\( \displaystyle \ce{R-CN} \xrightarrow{\ce{SnCl2},\ \ce{HCl}} \ce{R-CH=NH} \xrightarrow{\ce{H3O^+}} \ce{R-CHO} \)
C. Etard reaction — This matches III. CrO₂Cl₂ , CS₂ , H₃O⁺. Chromyl chloride oxidises the methyl group of toluene to a chromium complex, which is hydrolysed to benzaldehyde.
\( \displaystyle \ce{C6H5-CH3} \xrightarrow[\ce{H3O^+}]{\ce{CrO2Cl2},\ \ce{CS2}} \ce{C6H5-CHO} \)
D. Friedel-Crafts reaction — This matches IV. Anhyd. AlCl₃. The Lewis acid AlCl₃ generates the acylium ion from the acyl chloride, which attacks the benzene ring to give a ketone.
\( \displaystyle \ce{C6H6 + CH3COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \ce{C6H5-CO-CH3 + HCl} \)

Shortcut: The pair A–II appears only in Option (1).

Match the LIST-I with LIST-II.

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → A-II, B-I, C-III, D-IV

The named reactions in this list make aldehydes and ketones. They can be recognised from the starting material and the reagent:

From acid derivatives (reduction): Rosenmund reduction — acyl chloride + H₂, Pd-BaSO₄ gives an aldehyde; Stephen reaction — nitrile + SnCl₂, HCl, then hydrolysis, gives an aldehyde (DIBAL-H does the same change for nitriles and esters).
From hydrocarbons (oxidation of a methyl group): Etard reaction — toluene + chromyl chloride (CrO₂Cl₂) in CS₂, then hydrolysis, gives benzaldehyde; Gattermann–Koch reaction — benzene + CO and HCl with anhydrous \( \ce{AlCl3} \)/CuCl gives benzaldehyde.
Acylation of arenes: Friedel–Crafts acylation — benzene + acyl chloride with anhydrous AlCl₃ gives an aromatic ketone.

A. Rosenmund reaction — This matches II. H₂ , Pd-BaSO₄. The acyl chloride is hydrogenated over palladium on barium sulphate; the catalyst is partially poisoned so that reduction stops at the aldehyde.
\( \displaystyle \ce{R-COCl + H2} \xrightarrow{\ce{Pd-BaSO4}} \ce{R-CHO + HCl} \)
B. Stephen reaction — This matches I. SnCl₂, HCl. The nitrile is reduced by stannous chloride and HCl to an imine, which is hydrolysed to the aldehyde.
\( \displaystyle \ce{R-CN} \xrightarrow{\ce{SnCl2},\ \ce{HCl}} \ce{R-CH=NH} \xrightarrow{\ce{H3O^+}} \ce{R-CHO} \)
C. Etard reaction — This matches III. CrO₂Cl₂ , CS₂ , H₃O⁺. Chromyl chloride oxidises the methyl group of toluene to a chromium complex, which is hydrolysed to benzaldehyde.
\( \displaystyle \ce{C6H5-CH3} \xrightarrow[\ce{H3O^+}]{\ce{CrO2Cl2},\ \ce{CS2}} \ce{C6H5-CHO} \)
D. Friedel-Crafts reaction — This matches IV. Anhyd. AlCl₃. The Lewis acid AlCl₃ generates the acylium ion from the acyl chloride, which attacks the benzene ring to give a ketone.
\( \displaystyle \ce{C6H6 + CH3COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \ce{C6H5-CO-CH3 + HCl} \)

Shortcut: The pair A–II appears only in Option (1).