Question

In which of the following, the compounds are arranged in the correct order of acidic strength?

• A. II<III<I
• B. II<I<III
• C. III<I<II
• D. III<II<I

Hint:

Terminal alkynes are acidic ($R-C \equiv C-H$). Internal alkynes ($R-C \equiv C-R$) are not acidic. +I effect decreases acidity: $H-C \equiv C-H>CH_3-C \equiv C-H>CH_3CH_2-C \equiv C-H$.

Answer 1

The Correct Option is C

Step 1: Identify Acidic Protons:
Acidity in alkynes is due to the hydrogen attached to the $sp$ hybridized carbon (terminal hydrogen).
I (Ethylacetylene): $CH_3-CH_2-C \equiv CH$. Has 1 acidic H. Alkyl group (Ethyl) has +I effect.
II (Methylacetylene): $CH_3-C \equiv CH$. Has 1 acidic H. Alkyl group (Methyl) has +I effect.
III (Dimethylacetylene): $CH_3-C \equiv C-CH_3$. No acidic H (Internal alkyne). Least acidic.

Step 2: Compare I and II:
Both have one acidic proton. The stability of the conjugate base (carbanion) determines acidity. Alkyl groups are electron-releasing (+I effect), which destabilizes the negative charge on the anion. Ethyl group ($+I$) \textgreater Methyl group ($+I$). Therefore, the anion of Ethylacetylene is less stable than that of Methylacetylene. Acidity: Methylacetylene \textgreater Ethylacetylene.
Step 3: Final Order:
Dimethylacetylene (Non-acidic) \textless Ethylacetylene \textless Methylacetylene. Order: III \textless I \textless II.