QJEEM19Chbffd: JEE Main 2019 Chemistry - Class 11 JEEM19Chbffd Solution: The correct order for acid strength of compounds : CH ≡ CH, CH…

Among CH $$\equiv$$ CH and CH₂ = CH₂, in CH $$\equiv$$ CH compound H atom is attached with sp hybridised C atom and in CH₂ = CH₂ compound H atom is attached with sp² hybridised C atom.

As we know, electronegitivity of sp carbon is more compare to sp² carbon atom and that H is more acidic which is attached with more electronegative carbon atom. So CH $$\equiv$$ CH is more acidic than CH₂ = CH₂.

Among CH $$\equiv$$ CH and CH₃–C $$\equiv$$ CH, in CH $$\equiv$$ CH both the H atom is connected to sp carbon atom but in CH₃–C $$\equiv$$ CH, with one sp carbon atom H atom attached and with other sp carbon atom -CH₃ group attached which have +I effect and we know +I effect decreases acidic strength.
So, CH $$\equiv$$ CH is more acidic than CH₃–C $$\equiv$$ CH.

So correct order is

HC $$\equiv$$ CH > CH₃ – C $$\equiv$$ CH > CH₂ = CH₂