See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step-by-step reasoning for each part: **A. Highest boiling point among NCl3, ClNH2, NH4Cl, NH3:** Concept: Boiling point depends on intermolecular/interionic forces. NCl3, ClNH2, and NH3 are molecular compounds held together by van der Waals or hydrogen bonding. NH4Cl is an ionic compound (ammonium chloride), composed of NH4+ cations and Cl- anions. Ionic compounds have extremely strong electrostatic (cation-anion) interactions, giving them far higher melting and boiling points than molecular compounds. NH4Cl sublimes/decomposes at ~338°C, far above the boiling points of the others. Therefore NH4Cl has the highest boiling point, and the responsible force is cation-anion (ionic) interaction. Why others fail: NCl3 (bp ~71°C) has only London dispersion; NH3 (bp -33°C) has H-bonding but is still a small molecular compound; ClNH2 (bp ~-66°C) is also molecular. **B. Highest boiling point among cyclobutane, cyclopentane, methylcyclohexane:** Concept: For nonpolar hydrocarbons, boiling point increases with molecular weight and surface area (London/van der Waals forces). Cyclobutane (C4H8, MW=56), cyclopentane (C5H10, MW=70), methylcyclohexane (C7H14, MW=98). Methylcyclohexane has the highest molecular weight and greatest surface area, so it has the strongest London dispersion forces and thus the highest boiling point (~101°C vs cyclopentane ~49°C vs cyclobutane ~12°C). The responsible force is van der Waals' (London dispersion) forces. **C. Most soluble in water among phenol, catechol (1,2-benzenediol), 1,2-benzenedithiol:** Concept: Water solubility is governed by ability to form hydrogen bonds with water and polarity. Phenol has one -OH group and can H-bond with water. Catechol has two -OH groups in ortho positions; both can H-bond with water, and the intramolecular H-bond possibility is limited compared to intermolecular H-bonding with water. More -OH groups generally means greater water solubility. 1,2-benzenedithiol has -SH groups; sulfur is much less electronegative than oxygen and -SH groups are much weaker H-bond donors/acceptors, so dithiol is much less water soluble. Catechol (two -OH groups) is most soluble in water. The responsible force is hydrogen bonding (and dipole-dipole). Why catechol > phenol: Two -OH groups provide more sites for H-bonding with water than one -OH in phenol. Why dithiol fails: -SH is a poor H-bond donor compared to -OH. **D. Highest solubility in benzene among pyrrole and pyrrolidine:** Concept: Benzene is a nonpolar aromatic solvent. 'Like dissolves like.' Pyrrolidine is a saturated (non-aromatic) secondary amine; it can H-bond with itself (N-H) and is more polar. Pyrrole is an aromatic heterocycle with a conjugated pi system; it can interact with benzene via aromatic pi-stacking and dispersion interactions, making it more compatible with the aromatic benzene solvent. Additionally, pyrrole's N-H lone pair is delocalized into the ring, reducing its H-bonding ability compared to pyrrolidine, making pyrrole effectively less polar and more 'aromatic-like.' Thus pyrrole is more soluble in benzene. The responsible force is aromatic stacking (pi-pi interactions). Why pyrrolidine is less soluble in benzene: It is saturated, more polar, and prefers polar solvents. Therefore, the correct answer is {"A": {"compound": "NH4Cl", "force": "cation-anion interaction"}, "B": {"compound": "methylcyclohexane (cyclohexyl-CH3)", "force": "van der Waals' forces"}, "C": {"compound": "catechol (benzene-1,2-diol, ortho-dihydroxybenzene)", "force": "H-bonding (also dipole-dipole)"}, "D": {"compound": "dihydropyrrole with N-H (cyclic secondary amine, 5-membered ring with one C=C)", "force": "Aromatic stacking"}}.