See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Bond Energy Comparison of C–H Bonds **Step 1: Identify the carbon hybridization at each position** - **Bond a**: C–H on $sp^3$ carbon (saturated, single bond) - **Bond b**: C–H on $sp^2$ carbon (aromatic ring) - **Bond c**: C–H on $sp^3$ carbon in $CH_2$ (saturated, single bond) - **Bond d**: C–H on $sp^2$ carbon (aromatic ring) **Step 2: Apply the s-character principle** Bond strength increases with increasing s-character of the carbon orbital: $$\text{s-character: } sp^3 < sp^2 < sp$$ Therefore: $sp^2$ C–H bonds > $sp^3$ C–H bonds **Step 3: Compare bonds b and d (both $sp^2$)** Both are aromatic C–H bonds with similar s-character. However, bond b is ortho/meta to the electron-donating $CH_2$ group, which increases electron density and slightly weakens the bond. Bond d is less influenced, making it marginally stronger. Thus: $d \approx b$ (very close) **Step 4: Compare bonds a and c (both $sp^3$)** Both are saturated C–H bonds with similar bond energies, though c may have slight variations in local environment. Thus: $c \approx a$ (approximately equal) **Step 5: Order the bonds** $$c > d > b > a \text{ (or very close: } c \approx d > a \approx b\text{)}$$ **The answer is (c): $c > d > a > b$** — aromatic C–H bonds are stronger than saturated C–H bonds.