Determining the Most Acidic Hydrocarbon

The acidity of a compound refers to its ability to donate a proton (H⁺). For hydrocarbons like alkanes, alkenes, and alkynes, acidity depends on the stability of the conjugate base formed after losing a proton. The stability of the conjugate base is influenced by the hybridization of the carbon atom from which the proton is removed.

Carbon Hybridization and Electronegativity

Let's look at the hybridization of the carbon atoms involved in the given compounds:

• CH₄ (Methane): The carbon atom is bonded to four hydrogen atoms, so it is sp³ hybridized.
• CH₂=CH₂ (Ethene): The carbon atoms involved in the double bond are sp² hybridized.
• CH ≡ CH (Ethyne): The carbon atoms involved in the triple bond are sp hybridized.

The percentage of s character in the hybrid orbitals affects the electronegativity of the carbon atom:

• sp³ hybridization: 25% s character
• sp² hybridization: 33.3% s character
• sp hybridization: 50% s character

A higher percentage of s character means the electrons are held closer to the nucleus, making the atom more electronegative. Therefore, the electronegativity of the carbon atom increases in the order:

\( \text{sp}^3 < \text{sp}^2 < \text{sp} \)

Conjugate Base Stability

When a hydrocarbon acts as an acid, it loses a proton and forms a carbanion (a carbon atom with a negative charge). The stability of this carbanion (conjugate base) is crucial. A more stable conjugate base means the original acid is stronger.

The negative charge in the conjugate base resides on the carbon atom that lost the proton. The stability of the carbanion is greater when the negative charge is on a more electronegative atom.

• From CH₄ (sp³ carbon): Conjugate base is CH₃⁻ (negative charge on sp³ carbon).
• From CH₂=CH₂ (sp² carbon): Conjugate base is CH₂=CH⁻ (negative charge on sp² carbon).
• From CH≡CH (sp carbon): Conjugate base is CH≡C⁻ (negative charge on sp carbon).

Since the sp hybridized carbon is the most electronegative, it can best accommodate the negative charge. Thus, the stability of the conjugate bases increases in the order:

\( \text{CH}_3^- < \text{CH}_2=\text{CH}^- < \text{CH}\equiv\text{C}^- \)

Acidity Order

Because the stability of the conjugate base follows the order \( \text{CH}_3^- < \text{CH}_2=\text{CH}^- < \text{CH}\equiv\text{C}^- \), the acidity of the parent hydrocarbons follows the same order:

\( \text{CH}_4 < \text{CH}_2=\text{CH}_2 < \text{CH}\equiv\text{CH} \)

Therefore, ethyne (CH≡CH) is the most acidic among the given compounds.