Question

Complete the following reactions:
i) \( C_6H_5COCl \xrightarrow{H_2 / Pd\text{-}BaSO_4} ? \)
ii) \( CH_3\text{-}CO\text{-}CH_2\text{-}CO\text{-}OC_2H_5 \xrightarrow{(i)\ NaBH_4\ (ii)\ H^+} ? \)
iii) anisole \( (C_6H_5OCH_3) + CH_3COCl \xrightarrow{anhydrous\ AlCl_3} ? \)
iv) acetophenone \( (C_6H_5COCH_3) + NaOH + I_2 \rightarrow \) ......
v) \( CH_3CHO + Zn\text{-}Hg + HCl \rightarrow \) ......

OR
Write short notes on the following:
i) Tollen's test
ii) Fehling test
iii) Haloform reaction of methyl ketone
iv) Reaction of phenyl methyl ketone with 2,4-dinitrophenyl hydrazine (2,4-DNP)
v) Wolff-Kishner reaction.

Hint:

Identify each named reaction: (i) Rosenmund, (ii) selective NaBH₄ ketone reduction, (iii) Friedel-Crafts acylation, (iv) iodoform, (v) Clemmensen. For the OR part, recall the reagents and colour changes of Tollen's, Fehling's, haloform, 2,4-DNP and Wolff-Kishner.

Answer 1

Option 1 - Working out each product

Step 1: Reaction (i) is the classic Rosenmund reduction used to make aldehydes from acyl chlorides. The BaSO₄ poisons the Pd so that hydrogenation halts once benzaldehyde forms and does not go on to the alcohol. Answer: \( C_6H_5CHO \) (benzaldehyde).

Step 2: In (ii) the substrate ethyl acetoacetate has two carbonyls: a ketone and an ester. NaBH₄ is selective and attacks only the more electrophilic ketone carbonyl, leaving the ester untouched; acidic work-up (H⁺) then protonates the alkoxide. The hydroxy-ester obtained is \( CH_3CH(OH)CH_2COOC_2H_5 \) (ethyl 3-hydroxybutanoate).

Step 3: For (iii), AlCl₃ generates the acylium ion \( CH_3CO^+ \) which carries out Friedel-Crafts acylation on the electron-rich anisole ring. The methoxy group directs the incoming acyl group to the para spot, so the main product is p-methoxyacetophenone, \( 4\text{-}CH_3O\text{-}C_6H_4\text{-}COCH_3 \).

Step 4: In (iv), acetophenone bears the CH₃-CO fragment required for the iodoform reaction. Iodine in alkali substitutes and then cleaves the methyl side, giving iodoform (yellow solid, CHI₃) and sodium benzoate \( C_6H_5COONa \).

Step 5: In (v), Zn-Hg with concentrated HCl (Clemmensen conditions) deoxygenates the aldehyde carbonyl, turning CH₃CHO into the alkane \( CH_3CH_3 \) (ethane).

Option 2 - Named tests and reactions

(i) Tollen's test: Silver diammine ion in ammonia solution acts as a mild oxidant. An aldehyde donates electrons, reducing Ag⁺ to free silver that coats the glass as a bright mirror; the aldehyde itself becomes a carboxylate. Ketones fail, so the test tells aldehydes apart from ketones.

(ii) Fehling test: The deep-blue cupric tartrate complex is reduced by an aliphatic aldehyde to brick-red Cu₂O. The colour change from blue to a red precipitate signals the presence of an easily oxidised aldehydic group; aromatic aldehydes and ketones give no precipitate.

(iii) Haloform reaction: Compounds carrying a CH₃CO- unit (or that can form it, like ethanol/CH₃CH(OH)-) react with X₂ in NaOH. All three methyl hydrogens are halogenated and the molecule then splits to release CHX₃ (chloroform, bromoform or the yellow iodoform) alongside a carboxylate. With I₂/NaOH the yellow iodoform precipitate serves as a diagnostic test.

(iv) With 2,4-DNP (Brady's reagent): The N-nucleophile of 2,4-dinitrophenylhydrazine adds to the carbonyl carbon of acetophenone and water is eliminated, producing a coloured (orange to red) 2,4-dinitrophenylhydrazone that crystallises out. A positive result confirms the compound is a ketone or aldehyde.

(v) Wolff-Kishner reduction: This is a two-stage carbonyl-to-methylene reduction. Hydrazine condenses with the C=O to give a hydrazone (C=N-NH₂); strong base at high temperature in ethylene glycol then expels N₂ and installs two hydrogens, leaving a CH₂. It is the base-mediated alternative to the acidic Clemmensen reduction and is preferred for base-stable substrates.