Question

The correct order of acidic nature of the following alkynes is
\[ I.\ C_2H_2 \] \[ II.\ (CH_3)_2C_2 \] \[ III.\ (CH_3)C_2H \]

• A. \(I\lt III\lt II\)
• B. \(II\lt III\lt I\)
• C. \(III\lt II\lt I\)
• D. \(II\lt I\lt III\)

Hint:

Terminal alkynes are acidic because of the hydrogen attached to the \[ sp \] hybridized carbon atom. Internal alkynes do not contain acidic hydrogen and are much less acidic.

Answer 1

The Correct Option is B

Step 1: Identify the three alkynes.
I: C₂H₂ (acetylene, HC$\equiv$CH) - terminal alkyne with no alkyl substituents. II: (CH₃)₂C₂ (but-2-yne, CH₃-C$\equiv$C-CH₃) - internal alkyne, no terminal H. III: CH₃C₂H (propyne, CH₃-C$\equiv$C-H) - terminal alkyne with one methyl group.
Step 2: Recall the acidic character of alkynes.
Terminal alkynes ($\equiv$C-H) are acidic because the sp carbon is highly electronegative (~50% s-character). The sp C-H bond has higher s-character than sp$^2$ (33%) or sp$^3$ (25%), making the bond stronger and the proton more easily released. Internal alkynes have no terminal C-H, so they have essentially zero acidic character.
Step 3: Compare I (C₂H₂) and III (CH₃C₂H).
Both are terminal alkynes. Acetylene (I) has no electron-donating groups. Propyne (III) has a methyl group attached to the sp carbon, which donates electrons (through induction), making the C-H bond slightly less polarized and thus LESS acidic than acetylene. So I is more acidic than III.
Step 4: Place II (internal alkyne, no terminal H).
But-2-yne (II) has NO terminal alkyne H at all. Its pKa is approximately 44 (essentially non-acidic), much less acidic than terminal alkynes (pKa ~25). So II is least acidic.
Step 5: Arrange in increasing order of acidity.
Least acidic to most acidic: II < III < I. (Internal alkyne < methyl-substituted terminal alkyne < unsubstituted acetylene).
Step 6: Final answer.
Increasing order of acidic strength: (CH₃)₂C₂ < CH₃C₂H < C₂H₂, i.e., II < III < I.
\[ \boxed{II < III < I \text{ (option 2)}} \]