See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Kinetic isotope effect (KIE) in electrophilic aromatic substitution (EAS). In electrophilic aromatic substitution, the rate-determining step is typically the formation of the arenium ion (sigma complex / Wheland intermediate), which involves the electrophile attacking the aromatic ring carbon. The subsequent step is the loss of a proton (or deuterium) from the ipso carbon to restore aromaticity. This deprotonation step is generally fast and NOT rate-determining in normal EAS. Key reasoning: Since the C-H (or C-D) bond breaking at the aromatic ring is NOT the rate-determining step in EAS, substituting H with D at the ortho position does not significantly affect the overall reaction rate. The primary kinetic isotope effect only manifests when the bond to the isotope is broken in the rate-determining step. For o-nitration: - In compound (I): toluene has H at both ortho positions. Attack at ortho gives arenium ion, then H is lost (fast step). - In compound (II): 2-D-toluene has D at one ortho position and H at the other. Attack at the D-bearing ortho carbon would require C-D bond breaking in the product step, but this is not rate-determining. Attack at the H-bearing ortho carbon proceeds normally. - In compound (III): 2,6-D2-toluene has D at both ortho positions. Both ortho positions have D. Since the rate-determining step in nitration is the attack of NO2+ on the ring (formation of sigma complex), and NOT the loss of H/D, the kinetic isotope effect does not influence the rate. Therefore, all three compounds undergo o-nitration at essentially the same rate regardless of whether H or D is present at the ortho position. Why other options fail: - Option (a) I > II > III would imply a primary KIE is operative in the rate-determining step, which is not the case for nitration. - Option (b) II > I > III has no logical chemical basis. - Option (c) III > I > II has no logical chemical basis. Since C-H/C-D bond cleavage is not rate-determining in electrophilic aromatic substitution (nitration), there is no significant kinetic isotope effect, and all three compounds react at the same rate for o-nitration. Therefore, the correct answer is D.