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.

Reaction by which benzaldehyde cannot be prepared is

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

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

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

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

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

$$R$$ and $$S$$ form benzyl benzoate when treated with each other. Hence $$P$$ is