The correct trend of acidic nature of the following alkynes is:

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

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

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

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C H \equiv C H > C H_{3} C \equiv C C H_{3} > C H_{3} C \equiv C H

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Solution

The correct option is D $C H \equiv C H > C H_{3} - C \equiv C H > C H_{3} C \equiv C C H_{3}$
As stability of conjugate base is more, greater is the acidic strength
Stability of conjugate base
$C H \equiv C^{-} > C H_{3} (+ I) - C \equiv C^{-} > C H_{3} - C \equiv C - C H_{2}^{⊖}$
Hence, also terminal C is $s p^{3}$ hybridize so ability to donates $H^{+}$ is very less.
Acidic strength order
$C H \equiv C H > C H_{3} - C \equiv C H > C H_{3} - C \equiv C - C H_{3}$

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Similar questions

A g N O_{3}

{CH}_{3} - {CH}_{2} - CH = {CH}_{3}

(A) C H_{3} - C H_{2} - C | C O O H H - - C H - C H_{3}

(B) C H_{3} - C | C O O H H - C H_{2} - - C H - C H_{3}

(C) C | C O O H H_{2} - C H_{2} - C H_{2} - - C H - C H_{3}

C H_{3} - C H = C H_{2} + H B r \to

H - C \equiv C^{-}, H_{2} C = - C H

^{-} C H_{3}

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