See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In electrophilic aromatic substitution (EAS), halogens are ortho/para directors due to lone pair donation into the ring. However, the o/p ratio depends on the steric and electronic factors of the halogen substituent. Step 1 - Electronic effect on o/p ratio: All halogens (F, Cl, Br, I) are ortho/para directors via resonance donation of lone pairs. However, the effectiveness of resonance donation varies with the size and electronegativity of the halogen. Step 2 - Steric effect: As the halogen gets larger (F < Cl < Br < I), the steric bulk at the ortho position increases, which disfavors ortho attack relative to para attack. Larger halogens create more steric hindrance at the ortho positions, thus lowering the o/p ratio. Step 3 - Fluorine (PhF, A): Fluorine is the smallest halogen with the least steric bulk. Despite being highly electronegative (which weakens its resonance donation), its small size means ortho positions are least sterically hindered. The o/p ratio is highest for PhF because steric hindrance is minimal at ortho positions. Step 4 - Comparing A, B, C, D: As we go from F to Cl to Br to I, the size of the halogen increases, leading to progressively greater steric hindrance at the ortho positions. This results in a decreasing o/p ratio in the order: PhF > PhCl > PhBr > PhI, i.e., A > B > C > D. Step 5 - Therefore the correct order of o/p ratio is D < C < B < A, which corresponds to option (c). Wait - the given answer is A, which is option (a): A < B < C < D. Let me reconsider. Step 6 - Reconsidering: The o/p ratio can also be influenced by the inductive effect. As electronegativity decreases from F to I, the inductive withdrawal decreases. More electronegative halogens (F) withdraw electrons more strongly by induction, which disfavors the already electron-poor ipso and ortho positions differently. Additionally, the stronger the inductive withdrawal, the more it deactivates ortho positions relative to para (since para is farther from the withdrawing group). So stronger inductive withdrawal (F) leads to a relatively lower o/p ratio because ortho is more deactivated inductively than para. From F to I, inductive withdrawal decreases, so ortho attack becomes more favorable relative to para, increasing the o/p ratio in the order F < Cl < Br < I, i.e., A < B < C < D. Step 7 - This matches option (a): A < B < C < D. The dominant factor determining o/p ratio here is the inductive effect: F has the strongest inductive electron withdrawal which relatively deactivates ortho more than para, giving the lowest o/p ratio for PhF. As we go to I, inductive withdrawal is least, so o/p ratio is highest. Why other options fail: - Option (b) A = B = C = D is wrong; halogens differ in their effects. - Option (c) D < C < B < A is wrong; this would be the order if steric effect alone dominated, but inductive effect is the primary factor here. - Option (d) D < B < A < C is wrong; it gives no consistent trend. Therefore, the correct answer is A.