Step 1: Understand the starting compound (ortho-dicarboxylic acid).
The given compound contains two \(-COOH\) groups at ortho positions on benzene. Such compounds easily undergo salt formation, dehydration, and cyclisation due to proximity of functional groups. Step 2: Reaction with NH\(_3\) (formation of P).
Ammonia acts as a base and reacts with carboxylic acid groups to form ammonium carboxylate salts:
\[
-COOH + NH_3 \rightarrow -COO^-NH_4^+
\]
Since two \(-COOH\) groups are present, both get converted into ammonium salts. Hence P is the diammonium salt. Step 3: Heating of ammonium salt (formation of Q).
On heating, ammonium carboxylate undergoes dehydration (loss of water) and intramolecular cyclisation due to proximity of two carboxyl groups. This leads to formation of a cyclic imide (–CO–NH–CO– ring system). Step 4: Nature of Q (cyclic imide).
The cyclic imide contains a five-membered imide ring fused with benzene (similar to phthalimide structure). This is a stable intermediate formed after loss of water from ammonium salt. Step 5: Strong heating of imide (formation of R).
On strong heating, cyclic imides can undergo further dehydration and rearrangement leading to formation of cyclic anhydride by elimination of NH-containing fragment. Thus NH is removed and oxygen bridge forms, giving cyclic anhydride. Step 6: Final identification and consistency check.
- P = diammonium salt (formed by acid-base reaction)
- Q = cyclic imide (formed by dehydration on heating)
- R = cyclic anhydride (formed on strong heating via further elimination)
This sequence matches standard behavior of ortho-dicarboxylic acids. Final Answer:
\[
\boxed{\text{P = diammonium salt, Q = cyclic imide, R = cyclic anhydride}}
\]
