See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Nucleophilicity depends on the availability of lone pairs, the electronegativity and polarizability of the atom bearing them, steric hindrance, and resonance delocalization. (i) Both A and B are cyclic ethers (tetrahydropyrans) with oxygen as the nucleophilic atom. However, compound A has two methyl substituents at the 2- and 6-positions flanking the oxygen, creating significant steric hindrance around the lone pairs of oxygen. Compound B is unsubstituted tetrahydropyran, so the oxygen is sterically accessible. Less steric hindrance means better access for attack on an electrophile. Therefore B is the stronger nucleophile. (ii) A is tetrahydrofuran (cyclic ether, oxygen nucleophile) and B is pyrrolidine (cyclic amine, nitrogen nucleophile). Nitrogen is less electronegative than oxygen, so its lone pair is more available and less tightly held. Amines are generally much stronger nucleophiles than ethers. Therefore B (pyrrolidine) is the stronger nucleophile. (iii) Both A and B are NH minus (amide-like anions), but A is attached to a benzene ring (aniline anion / phenylamide) and B is attached to a cyclohexane ring (cyclohexylamide anion). In compound A, the lone pair on nitrogen is delocalized into the aromatic pi system via resonance, reducing its availability for nucleophilic attack. In compound B, the cyclohexyl group is purely sp3 and cannot accept electron density by resonance; the lone pair on nitrogen remains localized and fully available. Therefore B is the stronger nucleophile. (iv) NH3 has a lone pair on nitrogen and is a well-established nucleophile. CH4 has no lone pairs and no net negative charge; carbon in methane is essentially non-nucleophilic under normal conditions. Therefore A (NH3) is the stronger nucleophile. Therefore, the correct answer is {"i": "B", "ii": "B", "iii": "B", "iv": "A"}.