Question

a) Show the reaction of 4-methoxyaniline hydrochloride with ammonia. Indicate which layer (aqueous or organic) each component in the reaction will mainly reside in (For both products & reactants).

b) Show the reaction of acetylsalicylic acid with NaHCO₃ (sodium bicarbonate). Indicate which layer (aqueous or organic or both) each component in the reaction will mainly reside in.

Answer 1

Answern1)

1. The reaction taking place is the Aromatic Nucleophilic substitution reaction.
2. The nucleophile is the NH3. The ortho position ortho to -NH2Cl grouping is nucleophilic as NH2Cl grouping pulls electron density from ortho C by --ve Inductive effect. Also being meta to O-Me group it won’t get electron density from O-Me group.Hence this is the position that most likely to be attacked by nucleophile NH3.
3. 4-Methoxy aniline hydrochloride forms as a treatment of aq. HCl acid on 4-Methoxy aniline, hence it exist in aqueous layer.
4. NH3 being polar molecule also exist in aqueous layer.
5. As reaction proceeds and 4-Methoxy aniline converts into the product 2-amino-4-methoxyaniline in presence of excess NH3. The product goes into the organic layer. (The product itself forms organic layer).
6. The reaction mechanism we can see in diagram.(Box-1)

Answer-2)

1. NaHCO3 is mild base and capable of abstracting only the carboxylic proton and forms corresponding acetate ion.
2. This acetate ion then attack C=O group. The reversal of –ve charge on O cause the breaking of old C-O bond and generates phenoxy group-a stable –ve O. This shifts equilibrium to product side effectively.
3. Then mild acidic condition (water+NaHCO3) protonates phenoxy group and gives the product.