Question

What happens when (give only the chemical equations):
(i) Methyl phenyl ketone is treated with iodine in the presence of NaOH?
(ii) Ethyl bromide is heated with aqueous KOH?
(iii) Phenol reacts with chloroform in the presence of aqueous NaOH?
(iv) Acetaldehyde is treated with Tollen's reagent?
(v) Benzoic acid is heated with ammonia? (1+1+1+1+1)
OR
How would you obtain (write chemical equations only):
(i) Formic acid from formaldehyde?
(ii) Propane-2-ol from propene?
(iii) Benzene from phenol?
(iv) m-bromobenzoic acid from benzoic acid?
(v) Ethyl acetate from acetic acid? (1+1+1+1+1)

Hint:

Look for name reactions: iodoform (methyl ketone), aqueous KOH substitution, Reimer-Tiemann (salicylaldehyde), Tollen's oxidation, and amide formation. For the OR part, think oxidation, Markovnikov hydration, Zn-dust reduction, meta bromination and Fischer esterification.

Answer 1

Option 1

Step 1 (i): Any CH₃CO– compound answers the iodoform test. Acetophenone therefore yellow iodoform plus sodium benzoate: \(C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + C_6H_5COONa + 3NaI + 3H_2O\).
Step 2 (ii): Water-based KOH favours substitution over elimination, so ethyl bromide is converted to ethanol: \(C_2H_5Br + KOH_{(aq)} \rightarrow C_2H_5OH + KBr\).
Step 3 (iii): Chloroform in alkali makes dichlorocarbene, which attacks phenol at the ortho carbon; hydrolysis then gives salicylaldehyde: \(C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o\text{-}HOC_6H_4CHO + 3NaCl + 2H_2O\).
Step 4 (iv): Being an aldehyde, acetaldehyde reduces Tollen's reagent, itself being oxidised to acetate while Ag\(^+\) becomes a silver mirror: \(CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag + 4NH_3 + 2H_2O\).
Step 5 (v): Ammonia neutralises benzoic acid to the ammonium salt, which dehydrates on heating to benzamide: \(C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O\).

Option 2 (OR)
Step 1 (i): Oxidise the aldehyde one step to the acid: \(HCHO + [O] \rightarrow HCOOH\).
Step 2 (ii): Add water across propene by Markovnikov addition so H goes to the terminal carbon and OH to C2: \(CH_3CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH(OH)CH_3\).
Step 3 (iii): Distil phenol over hot zinc dust to strip the oxygen: \(C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO\).
Step 4 (iv): The –COOH group directs the incoming bromine to the meta position: \(C_6H_5COOH + Br_2 \xrightarrow{Fe} m\text{-}BrC_6H_4COOH + HBr\).
Step 5 (v): Esterify with ethanol and a little concentrated sulphuric acid: \(CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O\).
\[\boxed{\text{Ten balanced equations as above}}\]