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: What happens when

Step (i) Iodoform reaction: Methyl phenyl ketone (acetophenone) has a CH₃CO– group, so it gives the iodoform test. Iodine and NaOH replace the three methyl hydrogens by iodine and then cleave the bond, forming yellow iodoform.
\( C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow + C_6H_5COONa + 3NaI + 3H_2O \)

Step (ii) Nucleophilic substitution: Aqueous KOH supplies OH^–, which replaces bromine to give ethanol.
\( C_2H_5Br + KOH_{(aq)} \rightarrow C_2H_5OH + KBr \)

Step (iii) Reimer-Tiemann reaction: Phenol with chloroform and aqueous NaOH, followed by acidification, gives salicylaldehyde (2-hydroxybenzaldehyde). The –CHO enters mainly at the ortho position.
\( C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o\text{-}HOC_6H_4CHO + 3NaCl + 2H_2O \)

Step (iv) Tollen's test: Tollen's reagent (ammoniacal silver nitrate) oxidises acetaldehyde to acetate ion and deposits a silver mirror.
\( CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O \)

Step (v) Amide formation: Benzoic acid first forms ammonium benzoate, which loses water on strong heating to give benzamide.
\( C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O \)

Option 2 (OR): How would you obtain

Step (i) Formic acid from formaldehyde: Mild oxidation of formaldehyde gives formic acid.
\( HCHO + [O] \xrightarrow{K_2Cr_2O_7/H^+} HCOOH \)

Step (ii) Propane-2-ol from propene: Acid-catalysed hydration adds water across the double bond following Markovnikov's rule, so OH goes to the middle carbon.
\( CH_3CH=CH_2 + H_2O \xrightarrow{H_2SO_4} CH_3CH(OH)CH_3 \)

Step (iii) Benzene from phenol: Heating phenol vapours with zinc dust reduces it to benzene.
\( C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO \)

Step (iv) m-Bromobenzoic acid from benzoic acid: –COOH is a meta-directing, deactivating group, so bromination with Br₂ and FeBr₃ gives the meta product.
\( C_6H_5COOH + Br_2 \xrightarrow{FeBr_3} m\text{-}BrC_6H_4COOH + HBr \)

Step (v) Ethyl acetate from acetic acid: Fischer esterification of acetic acid with ethanol using concentrated H₂SO₄ gives ethyl acetate.
\( CH_3COOH + C_2H_5OH \xrightarrow{conc.\,H_2SO_4} CH_3COOC_2H_5 + H_2O \)
\[\boxed{\text{All ten equations shown above}}\]