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: Complete the reactions

Step 1 (i) Rosenmund reduction: Benzoyl chloride is reduced by H₂ over palladium supported on barium sulphate (a poisoned catalyst that stops reduction at the aldehyde stage). The acid chloride becomes benzaldehyde.
\( C_6H_5COCl + H_2 \xrightarrow{Pd\text{-}BaSO_4} C_6H_5CHO + HCl \)
Product = Benzaldehyde.

Step 2 (ii) NaBH₄ reduction: Sodium borohydride is a mild reducing agent. It reduces the ketone (C=O) group to a secondary alcohol but does NOT reduce the ester (COOC₂H₅) group. Ethyl acetoacetate therefore gives ethyl 3-hydroxybutanoate.
\( CH_3\text{-}CO\text{-}CH_2\text{-}COOC_2H_5 \xrightarrow{NaBH_4;\ H^+} CH_3\text{-}CH(OH)\text{-}CH_2\text{-}COOC_2H_5 \)
Product = Ethyl 3-hydroxybutanoate.

Step 3 (iii) Friedel-Crafts acylation: Anisole has an OCH₃ group which is activating and ortho/para directing. With acetyl chloride and anhydrous AlCl₃ (Lewis-acid catalyst) an acetyl group enters mainly at the para position (para is favoured because the ortho position is blocked by steric hindrance).
Product = 4-methoxyacetophenone (p-methoxyacetophenone), \( CH_3CO\text{-}C_6H_4\text{-}OCH_3\ (para) \).

Step 4 (iv) Iodoform / haloform reaction: Acetophenone contains a CH₃CO- (methyl ketone) group, so it responds to the iodoform test. NaOH and I₂ convert it into a yellow precipitate of iodoform and the sodium salt of benzoic acid.
\( C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3\downarrow + 3NaI + 3H_2O \)
Products = Sodium benzoate + Iodoform (CHI₃).

Step 5 (v) Clemmensen reduction: Zinc amalgam (Zn-Hg) with concentrated HCl reduces the carbonyl (C=O) group of acetaldehyde all the way to a methylene/CH₂ group, giving the hydrocarbon.
\( CH_3CHO + 4[H] \xrightarrow{Zn\text{-}Hg,\ HCl} CH_3CH_3 + H_2O \)
Product = Ethane.

Option 2: Short notes

(i) Tollen's test: Tollen's reagent is ammoniacal silver nitrate, i.e. \( [Ag(NH_3)_2]^+ \). Aldehydes reduce it to metallic silver, which deposits as a shining silver mirror on the tube. Ketones do not respond. It distinguishes aldehydes from ketones.
\( RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag\downarrow + 4NH_3 + 2H_2O \)

(ii) Fehling test: Fehling's solution is a mixture of Fehling A (aqueous CuSO₄) and Fehling B (alkaline sodium potassium tartrate). Aliphatic aldehydes reduce the blue Cu²⁺ to a red-brown precipitate of cuprous oxide (Cu₂O). Aromatic aldehydes and ketones do not respond.
\( RCHO + 2Cu^{2+} + 5OH^- \rightarrow RCOO^- + Cu_2O\downarrow + 3H_2O \)

(iii) Haloform reaction of methyl ketone: A methyl ketone (CH₃CO-R) on treatment with excess halogen (Cl₂/Br₂/I₂) and NaOH gives a trihalomethane (haloform, CHX₃) plus the carboxylate salt. The three alpha-hydrogens of the methyl group are replaced by halogen, then the C-C bond is cleaved.
\( CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3\downarrow + CH_3COONa + 3NaI + 3H_2O \)

(iv) Reaction with 2,4-DNP: Phenyl methyl ketone (acetophenone) reacts with 2,4-dinitrophenylhydrazine to form an orange-red crystalline solid, the 2,4-dinitrophenylhydrazone. This is a confirmatory test for a carbonyl (aldehyde or ketone) group.
\( C_6H_5COCH_3 + H_2N\text{-}NH\text{-}C_6H_3(NO_2)_2 \rightarrow C_6H_5C(CH_3){=}N\text{-}NH\text{-}C_6H_3(NO_2)_2 + H_2O \)

(v) Wolff-Kishner reaction: The carbonyl compound first reacts with hydrazine (NH₂NH₂) to give a hydrazone. On heating this hydrazone with a strong base (KOH or sodium ethoxide) in a high-boiling solvent like ethylene glycol, nitrogen gas is lost and the C=O is reduced to a CH₂ group. It is used to convert an aldehyde/ketone into the corresponding hydrocarbon.
\( C{=}O \xrightarrow{NH_2NH_2} C{=}N\text{-}NH_2 \xrightarrow{KOH,\ glycol,\ \Delta} CH_2 + N_2 \)