Acidity Of Terminal Alkynes

Question

Among ethane, ethene, and ethyne, the relative ease with which a terminal hydrogen is removed as a proton must be ordered and explained.

RankGuru · Accepted Answer

The acidity of a C-H bond depends on the stability of the conjugate base, the carbanion left after the proton departs, and this in turn depends on the hybridization of the carbon holding the negative charge. Greater s-character holds the lone pair closer to the nucleus, stabilizing the carbanion and increasing acidity. In ethyne the carbon is $sp$ hybridized (50% s-character), in ethene it is $sp^2$ (33%), and in ethane it is $sp^3$ (25%). Therefore the order of acidity is ethyne > ethene > ethane, which is why only terminal alkynes react with sodium or ammoniacal silver nitrate to form acetylides. The reverse order is wrong because it would require less s-character to stabilize the anion, contradicting the hybridization argument. Placing ethene first is incorrect since $sp^2$ carbon has less s-character than $sp$. Equal acidity is impossible because the three carbons differ in hybridization. This is the standard NCERT explanation for acidic hydrogen in alkynes. A plausibility check: the existence of metal acetylides but not corresponding alkene or alkane salts confirms ethyne is the most acidic.

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Acidity Of Terminal Alkynes

Select the correct option:

🔒 Solution Hidden from View

More hydrocarbons Practice Questions

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About This Question

Solution

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