Principle of acidity in hydrocarbons

Acidity depends on the stability of the conjugate base formed after removal of a proton.

Greater the s-character of the carbon bearing the negative charge, greater is the stability of the carbanion, hence greater is the acidity.

The relevant hybridizations are:

• Alkynes: carbon is sp hybridized  $50\%$ s-character
• Alkenes: carbon is sp$^2$ hybridized  $33\%$ s-character
• Alkanes: carbon is sp$^3$ hybridized  $25\%$ s-character

Compare ethyne and propyne

The compounds are:

• $HC\equiv CH$ (ethyne)
• $CH_3-C\equiv CH$ (propyne)
• $CH_2=CH_2$ (ethene)

Both ethyne and propyne are terminal alkynes, so both have an acidic hydrogen attached to an sp-carbon.

On deprotonation:

• $HC\equiv CH \to HC\equiv C^-$
• $CH_3-C\equiv CH \to CH_3-C\equiv C^-$

The $CH_3$ group in propyne shows a +I effect (electron-donating inductive effect), which destabilizes the negative charge on the conjugate base.

Therefore, $HC\equiv CH \text{ is more acidic than } CH_3-C\equiv CH$

Compare alkynes with ethene

Ethene, $CH_2=CH_2$, has hydrogens on sp$^2$ carbon. Its conjugate base would place negative charge on an sp$^2$ carbon, which is less stable than an sp-carbon carbanion.

Hence, alkenes are much less acidic than terminal alkynes.

Therefore, $HC\equiv CH > CH_3-C\equiv CH > CH_2=CH_2$

Match with options

This corresponds to:

Option C: $HC\equiv CH > CH_3-C\equiv CH > CH_2=CH_2$

Conclusion

The correct order of acid strength is: $HC\equiv CH > CH_3-C\equiv CH > CH_2=CH_2$