The question involves determining which cyclic compound reacts fastest with H-Br(aq). Let's evaluate each option:
- Molecule P:
- Contains two oxygen atoms connected to a ring.
- The two oxygen atoms can lead to resonance stabilization, but their presence reduces the electron density on the ring, making it less reactive toward nucleophilic attack by H-Br.
- Molecule Q:
- Contains one oxygen and one CH group connected to the ring.
- Oxygen can increase electron density through resonance, enhancing reactivity, while the CH group provides inductive stabilization.
- This combination makes the ring in molecule Q highly reactive to H-Br, facilitating faster reaction.
- Molecule R:
- Contains an oxygen and CH group, similar to Q, but arranged differently.
- The specific position and resonance compared to Q may influence reactivity but is less favorable due to steric and electronic factors.
- Molecule S:
- Contains two CH groups attached to a benzene ring.
- Lacks an oxygen atom, making it the least reactive to H-Br among the options.
Based on the evaluations, molecule Q will react with H-Br(aq) at the fastest rate due to the presence of an oxygen atom that can assist in reaction via electron donation, and a CH group that stabilizes the transition state via hyperconjugation.
Correct Answer: Q
