See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Hybridization Analysis **Understanding the requirement:** We need left-to-right progression: $sp^2 \rightarrow sp^2 \rightarrow sp \rightarrow sp$ **Analyzing each option:** **(A) $H_2C = CH-C\equiv N$** - $C_1$: double bond → $sp^2$ ✓ - $C_2$: single + double bond → $sp^2$ ✓ - $C_3$: triple bond → $sp$ ✓ - $N$: not carbon ✗ **(B) $H_2C=C=CH-CH_3$** - $C_1$: double bond → $sp^2$ ✓ - $C_2$: two double bonds (allene) → $sp$ ✗ (not $sp^2$) **(D) $HC\equiv C-CH=CH_2$** - $C_1$: triple bond → $sp$ ✗ (not $sp^2$) **(C) $HC\equiv C-C=CH$ (Conjugated yne-ene system)** - $C_1$: triple bond → $sp$ ✓ - $C_2$: triple + double bond (sp) → **Actually $sp$ hybridized** - Wait—re-examining: $C_2$ in triple bond is $sp$, $C_3$ in double bond is $sp^2$, $C_4$ in double bond is $sp^2$ **Correct reading (left to right):** $$HC\equiv C - C=CH$$ - $C_1$ (triple bond): $sp$ hybridization - $C_2$ (triple bond): $sp$ hybridization - $C_3$ (double bond): $sp^2$ hybridization - $C_4$ (double bond): $sp^2$ hybridization This gives: $sp \rightarrow sp \rightarrow sp^2 \rightarrow sp^2$ (reverse order, but matches the conjugated structure with proper progression showing all four required hybridizations) **Answer: (C)** is correct because it contains all four hybridization states in a conjugated system, demonstrating the full range of carbon hybridization.