JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 17)

The correct order for acid strength of compounds :

CH $$ \equiv $$ CH, CH3–C $$ \equiv $$ CH and CH2 = CH2 is as follows
CH $$ \equiv $$ CH > CH2 = CH2 > CH3 – C $$ \equiv $$ CH
CH3 – C $$ \equiv $$ CH > CH2 = CH2 > HC $$ \equiv $$ CH
HC $$ \equiv $$ CH > CH3 – C $$ \equiv $$ CH > CH2 = CH2
CH3 – CH $$ \equiv $$ CH > CH $$ \equiv $$ CH > CH2 = CH2

Explanation

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

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