Carboxylic Acids And Its Derivatives - Organic Chemistry - Chemistry Previous Year Questions

$$P \xrightarrow{\mathrm{H}_2 / \text { Pd- } \mathrm{BaSO}_4} Q \xrightarrow[\text { (ii)Dil. } \mathrm{HCl}]{\text { (i) Con. } \mathrm{NaOH}}$$

\(R\) and \(S\) form benzyl benzoate when treated with each other. Hence \(P\) is

Propanoic acid undergoes HVZ reaction to give chloropropanoic acid. The product obtained is

In set of reactions, identify \(D\)

$$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{SOCl}_2} A \xrightarrow[\text { Anhy. } \mathrm{AlCl}_3]{\text { Benzene }} B \xrightarrow{\mathrm{HCN}} C \xrightarrow{\mathrm{H}_2 \mathrm{O}} D$$

The product '\(A\)' gives white precipitate when treated with bromine water. The product '\(B\)' is treated with barium hydroxide to give the product \(C\). The compound \(C\) is heated strongly to form product \(D\). The product \(D\) is

The reagent which can do the conversion \(\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}\) is

Ethanoic acid undergoes Hell-Volhard Zelinsky reaction but methanoic acid does not, because of

Reaction by which benzaldehyde cannot be prepared is

Statement - I : Reduction of ester by DIABL-H followed by hydrolysis gives aldehyde.

Statement - II : Oxidation of benzyl alcohol with aqueous $\mathrm{KMnO}_4$ leads to the formation to Benzaldehyde.

Among the above statements, identify the correct statement.