Question

Selected bond angles for six hydrocarbons are shown below. Arrange these hydrocarbons according to their $pK_a$ values, from the lowest to the highest.

• A. V < I < VI < II < III < IV
• B. IV < I < II < III < V < VI
• C. II < IV < I < VI < V < III
• D. I < V < IV < III < II < VI

Answer 1

No correct option found based on standard chemical principles.

Answer 2

The acidity of C-H bonds in hydrocarbons is primarily determined by the hybridization of the carbon atom and the stability of the resulting carbanion. The order of acidity based on hybridization is: alkynes (sp) > alkenes (sp²) > alkanes (sp³). Strain in cyclic systems also increases acidity.

• V (Ethyne): sp hybridized C-H bond, $pK_a \approx 25$.
• III (Ethene): sp² hybridized C-H bond, $pK_a \approx 44$.
• IV (Norbornane bridgehead): Bridgehead C-H bonds in bicyclic systems like norbornane are more acidic than typical sp³ C-H bonds due to strain and orbital effects, with $pK_a \approx 35-38$.
• VI (Cyclopropane): sp³ hybridized C-H bond, but high ring strain increases acidity, $pK_a \approx 50$.
• I (Cyclopentane): sp³ hybridized C-H bond, some ring strain, $pK_a \approx 50$.
• II (Cyclohexane): sp³ hybridized C-H bond, nearly strain-free, $pK_a \approx 50$.

Ordering these by acidity (lowest $pK_a$ to highest $pK_a$): V ($pK_a \approx 25$) < IV ($pK_a \approx 35-38$) < III ($pK_a \approx 44$) < VI ($pK_a \approx 50$) < I ($pK_a \approx 50$) < II ($pK_a \approx 50$). Among I, II, and VI, cyclopropane (VI) has the highest strain, followed by cyclopentane (I), and then cyclohexane (II). Thus, acidity order is VI > I > II, meaning $pK_a$ order is VI < I < II. The complete order is: V < IV < III < VI < I < II. None of the given options match this derived order. Therefore, there appears to be an error in the question or the options provided.