The ascending acidity order of the following | vedclass

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(A) The acidity of $H$ atoms depends on the stability of the conjugate base formed after the removal of the proton $(H^+)$.$1$. The conjugate bases ar

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$C < D < B < A$

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(A) The acidity of $H$ atoms depends on the stability of the conjugate base formed after the removal of the proton $(H^+)$.$1$. The conjugate bases are:$A: HC \equiv C^-$$B: H_2C=CH^-$$C: (CH_3)_3C^-$$D: CH_3-CH_2^-$$2$. Stability of conjugate base is determined by the hybridization of the carbon atom bearing the negative charge:- In $A$,$C$ is $sp$ hybridized ($50\% \ s$-character).- In $B$,$C$ is $sp^2$ hybridized ($33.3\% \ s$-character).- In $D$,$C$ is $sp^3$ hybridized ($25\% \ s$-character).- In $C$,$C$ is $sp^3$ hybridized,but it is destabilized by the $+I$ effect of three $CH_3$ groups.$3$. Higher $s$-character leads to higher electronegativity and greater stability of the negative charge. Thus,the stability order is $A > B > D > C$.$4$. Since acidic strength is directly proportional to the stability of the conjugate base,the acidity order is $A > B > D > C$. The ascending order is $C < D < B < A$.

The ascending acidity order of the following $H$ atoms is:
$A: HC \equiv C-H$
$B: H_2C=CH-H$
$C: (CH_3)_3C-H$
$D: CH_3-CH_2-H$

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The ascending acidity order of the following $H$ atoms is:$A: HC \equiv C-H$$B: H_2C=CH-H$$C: (CH_3)_3C-H$$D: CH_3-CH_2-H$