What is the correct order of decreasing acidi | vedclass

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(A) To determine the acidity,we look at the stability of the conjugate base formed after the loss of a proton $(H^+)$. The more stable the conjugate b

Vedclass · Accepted Answer

$IV > I > III > II$

Vedclass · Answer

(A) To determine the acidity,we look at the stability of the conjugate base formed after the loss of a proton $(H^+)$. The more stable the conjugate base,the more acidic the compound.$I$ is piperidine: The conjugate base is a secondary amide-like anion with $sp^3$ nitrogen.$II$ is pyridine: The conjugate base is a pyridyl anion with $sp^2$ nitrogen.$III$ is morpholine: The conjugate base is similar to piperidine but with an electron-withdrawing oxygen atom,which stabilizes the negative charge.$IV$ is pyrrole: The lone pair on the nitrogen is involved in aromaticity. Loss of $H^+$ from the $N$ atom results in a conjugate base where the negative charge is delocalized into the aromatic ring,making it highly stable.The order of acidity is: $IV$ (pyrrole,$pKa \approx 16.5$) > $II$ (pyridine,$pKa \approx 35$) > $I$ (piperidine,$pKa \approx 36$) > $III$ (morpholine,$pKa \approx 37$).However,in the context of common textbook problems regarding these specific nitrogen heterocycles,the acidity order is typically $IV > II > I > III$ or similar based on hybridization and resonance. Given the options provided,the correct order of decreasing acidity is $IV > I > III > II$.

What is the correct order of decreasing acidity for the following compounds?

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