See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In electrophilic aromatic substitution (nitration with HNO3/H2SO4), alkyl groups are ortho/para directors. The ortho/para (o/p) ratio depends on the steric bulk of the substituent R on the benzene ring. A bulkier group at R shields the ortho positions (which are adjacent to R) more than the para position (which is far from R), thereby reducing ortho product formation and lowering the o/p ratio. Conversely, a smaller, less sterically demanding group allows easier access to both ortho positions, giving a higher o/p ratio. Step 1 - Identify the steric bulk of each substituent: (a) R = -CH3 (methyl): least bulky, minimal steric hindrance at ortho positions (b) R = -CH2CH3 (ethyl): slightly more bulky than methyl (c) R = -CHMe2 (isopropyl): more bulky than ethyl (d) R = -CMe3 (tert-butyl): most bulky, maximum steric hindrance at ortho positions Step 2 - Relate steric bulk to o/p ratio: The smaller the substituent, the less it blocks the ortho positions, so more ortho product forms relative to para product, giving a higher o/p ratio. As steric bulk increases from CH3 < CH2CH3 < CHMe2 < CMe3, the ortho positions become increasingly hindered, and the o/p ratio decreases. Step 3 - Determine which gives the highest o/p ratio: R = -CH3 (methyl) is the smallest substituent and causes the least steric hindrance at the ortho positions. Therefore, the proportion of ortho product is maximized relative to para product, giving the highest o/p ratio among the four options. Why other options fail: (b) Ethyl is bulkier than methyl, so o/p ratio is lower than for methyl. (c) Isopropyl is even bulkier, further reducing the o/p ratio. (d) tert-Butyl is the bulkiest, giving the lowest o/p ratio (mostly para product). Therefore, the correct answer is A.