Created by StemPy LabIn each substitution reaction of benzene below, an acid or a Lewis acid first makes the electrophile that attacks the ring. Match each reagent mixture with that electrophile. List-I (Reagents) A. Concentrated and concentrated B. and anhydrous , in the dark C. and anhydrous D. and anhydrous List-II (Electrophile) I. II. III. IV.
← All Quizzes
← The figure shows six cyclic species, I to VI. (i) I is not aromatic: its sp³ CH₂ carbon breaks the ring of π electrons. (ii) II has 6 π electrons and is aromatic. (iii) III has 6 π electrons and is aromatic. (iv) IV has 4 π electrons and fits the Hückel rule with n = 1. (v) V has 2 π electrons and is aromatic, with n = 0. (vi) VI has 10 π electrons, which fits 4n + 2 with n = 2. How many of the statements above are correct?Assertion (A): When chlorobenzene and benzene are nitrated side by side under the same conditions, chlorobenzene reacts more slowly, so the chlorine atom deactivates the ring towards electrophilic substitution.
Reason (R): In chlorobenzene, resonance with the lone pairs of chlorine puts more electron density at the ortho and para positions than at the meta position, so an incoming electrophile goes mainly to the ortho and para positions. →
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