The correct order for acid strength of compounds :
\(\mathrm{CH\equiv CH}\), \(\mathrm{CH_3{-}C\equiv CH}\) and \(\mathrm{CH_2{=}CH_2}\) is as follows
\(\mathrm{CH\equiv CH}\), \(\mathrm{CH_3{-}C\equiv CH}\) and \(\mathrm{CH_2{=}CH_2}\) is as follows
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- \(\mathrm{CH\equiv CH \gt CH_2{=}CH_2 \gt CH_3{-}C\equiv CH}\)
- \(\mathrm{CH_3{-}C\equiv CH \gt CH_2{=}CH_2 \gt HC\equiv CH}\)
- \(\mathrm{HC\equiv CH \gt CH_3{-}C\equiv CH \gt CH_2{=}CH_2}\)
- \(\mathrm{CH_3{-}CH\equiv CH \gt CH\equiv CH \gt CH_2{=}CH_2}\)
The Correct Option is C
Solution and Explanation
The more s-character in the carbon bearing the lone pair, the more stable the carbanion: sp (50% s) in alkynes > sp2 (33% s) in ethene.
So both alkynes are more acidic than \(\mathrm{CH_2{=}CH_2}\).
Between the two alkynes, the methyl group in propyne has a +I (electron-releasing) effect, which destabilises the acetylide ion \(\mathrm{CH_3C\equiv C^-}\) relative to \(\mathrm{HC\equiv C^-}\). Hence ethyne is more acidic than propyne.
Order: \(\mathrm{HC\equiv CH \gt CH_3{-}C\equiv CH \gt CH_2{=}CH_2}\).
Answer: (3) \(\mathrm{HC\equiv CH \gt CH_3{-}C\equiv CH \gt CH_2{=}CH_2}\)
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