Choose the correct order of acidity.

H C \equiv C H > H_{3} C - C \equiv C H > H_{2} C = C H_{2} > H_{3} C - C H_{3}

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H C \equiv C H > H_{2} C = C H_{2} > H_{3} C - C \equiv C H > H_{3} C - C H_{3}

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H_{3} C - C H_{3} > H_{2} C = C H_{2} > H_{3} C - C \equiv C H > H C \equiv C H

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H_{3} C - C H_{3} > H_{2} C = C H_{2} > H C \equiv C H > H_{3} C - C \equiv C H

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Solution

The correct option is A $H C \equiv C H > H_{3} C - C \equiv C H > H_{2} C = C H_{2} > H_{3} C - C H_{3}$
Higher is the s character stronger the acid.

$H C \equiv C H$ has 50% s-character, hence it has the most stable conjugate base meaning weak conjugate base. As we know weaker the conjugate base will be, stronger the acid. So $H C \equiv C H$ is the most acidic.

$H_{3} C - C \equiv C H$ also has sp carbons, but there is a $C H_{3}$ group attached to an sp carbon atom which will destabilise the conjugate base. Hence it is less acidic than $H C \equiv C H$ .

$H_{2} C = C H_{2}$ and $H_{3} C - C H_{3}$ have $s p^{2}$ and $s p^{3}$ carbons respectively. Hence the acidic order of $H_{2} C = C H_{2}$ is greater than that of $H_{3} C - C H_{3}$ .

So, the correct order of acidity will be:
$H C \equiv C H > H_{3} C - C \equiv C H > H_{2} C = C H_{2} > H_{3} C - C H_{3}$

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