See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# C-Br Bond Strength in Polar Solvent In polar solvents, bond strength is influenced by **resonance stabilization** and **inductive effects** of substituents on the aromatic/cyclic ring. ## Analysis of Each Compound: **Compound (I): Bromobenzene with $CH_3$** - Benzene ring (aromatic, highly stable) - $CH_3$ is electron-donating (weakly activating) - Strong resonance stabilization of C-Br bond - **Strongest C-Br bond** **Compound (II): Bromobenzene with H** - Benzene ring (aromatic) - No additional substituent effects - Good resonance stabilization, but less than (I) - **Strong C-Br bond** **Compound (III): Bromocyclopentadiene with $CH_3$** - Cyclopentadiene ring (conjugated diene, less aromatic) - Weaker resonance stabilization than benzene - Reduced aromaticity = weaker C-Br bond - **Weaker than (I) and (II)** **Compound (IV): Bromocyclopentadiene with H** - Cyclopentadiene ring (least aromatic among these) - Minimal resonance stabilization - **Weakest C-Br bond** ## Order (weakest to strongest): $$IV < III < II < I$$ **Answer: (A)** — The aromatic benzene compounds have stronger C-Br bonds than cyclopentadiene compounds. Within each class, electron-donating $CH_3$ strengthens the bond through resonance more than H does.