Among the compounds(i) HC equiv C-COOH(ii) CH | vedclass

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(A) The acidity of a compound depends on the stability of its conjugate base. Carboxylic acids $(R-COOH)$ are significantly more acidic than alcohols

Vedclass · Accepted Answer

$i > ii > iii > iv$

Vedclass · Answer

(A) The acidity of a compound depends on the stability of its conjugate base. Carboxylic acids $(R-COOH)$ are significantly more acidic than alcohols $(R-OH)$ due to the resonance stabilization of the carboxylate ion $(R-COO^-)$.Among the carboxylic acids,the acidity is influenced by the inductive effect of the carbon atom attached to the carboxyl group. The hybridization of the carbon atom attached to the $-COOH$ group determines its electron-withdrawing ability:$1$. In $(i)$ $HC \equiv C-COOH$,the carbon is $sp$ hybridized ($50$% s-character),which is the most electronegative and exerts the strongest $-I$ effect,stabilizing the carboxylate ion the most.$2$. In $(ii)$ $CH_2=CH-COOH$,the carbon is $sp^2$ hybridized ($33$% s-character),exerting a moderate $-I$ effect.$3$. In $(iii)$ $CH_3-CH_2-COOH$,the carbon is $sp^3$ hybridized ($25$% s-character),exerting the weakest $-I$ effect among the acids.$4$. $(iv)$ $CH_3-CH_2-OH$ is an alcohol and is the least acidic.Therefore,the correct order of acid strength is $i > ii > iii > iv$.

Among the compounds
$(i)$ $HC \equiv C-COOH$
$(ii)$ $CH_2=CH-COOH$
$(iii)$ $CH_3-CH_2-COOH$ and
$(iv)$ $CH_3-CH_2-OH$
The correct order of acid strength is

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Among the compounds$(i)$ $HC \equiv C-COOH$$(ii)$ $CH_2=CH-COOH$$(iii)$ $CH_3-CH_2-COOH$ and$(iv)$ $CH_3-CH_2-OH$The correct order of acid strength is