The compound(s) which will show significant intermolecular H-bonding is/are: (a) o-nitrophenol (b) 4-acetamidophenol…

🧠 Intramolecular H-bonding uses up the H-bond donor within the same molecule, reducing availability for intermolecular H-bonds. Steric hindrance near the –OH also prevents intermolecular H-bond formation.

🗺️ Analyse each compound:

(a) o-nitrophenol: –OH and adjacent –NO₂ form a 6-membered intramolecular H-bond ring → the –OH is not available for intermolecular H-bonding → significant intermolecular H-bonding: No ✗

(b) 4-acetamidophenol (paracetamol): Has both phenolic –OH and amide N–H (–NHCOCH₃) at para positions — no intramolecular H-bonding possible due to distance; both groups are accessible → forms extensive intermolecular H-bonds ✓

(c) 2,4,6-trimethylphenol: The two ortho-methyl groups sterically shield the –OH from approaching other molecules → intermolecular H-bonding is hindered ✗

⚡ Only paracetamol has unhindered, non-intramolecularly-blocked H-bond donor groups.

$\boxed{\text{Answer: (b) only — 4-acetamidophenol (paracetamol)}}$