The pK_{a} values of X, Y, Z respectively are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups (like $-

Vedclass · Accepted Answer

Vedclass · Answer

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups (like $-NO_2$) increase acidity (lower $pK_{a}$),while electron-donating groups (like $-CH_3$) decrease acidity (higher $pK_{a}$).$1$. $p$-Nitrophenol has a $-NO_2$ group at the para position,which exerts a strong $-R$ (resonance) and $-I$ (inductive) effect,making it the most acidic $(pK_{a} = 7.1)$.$2$. $m$-Nitrophenol has a $-NO_2$ group at the meta position,which exerts only a $-I$ effect,making it less acidic than $p$-nitrophenol $(pK_{a} = 8.3)$.$3$. $p$-Cresol has a $-CH_3$ group at the para position,which is an electron-donating group ($+I$ and hyperconjugation),making it the least acidic $(pK_{a} = 10.2)$.Thus,$X = m$-nitrophenol $(8.3)$,$Y = p$-nitrophenol $(7.1)$,and $Z = p$-cresol $(10.2)$.

The $pK_{a}$ values of $X, Y, Z$ respectively are $8.3, 7.1, 10.2$. What are $X, Y, Z$?

Similar Questions

The common name of $Benzene-1,4-diol$ is

The major product in the following reaction is:

Rank the following compounds in the order of increasing acidity:
$(I)$ Cyclohexanol
$(II)$ Cyclohexylamine
$(III)$ Phenol

The reaction of phenol with chloroform and sodium hydroxide to give $o$-hydroxybenzaldehyde involves the formation of which intermediate?

The carbon-oxygen bond in phenol is slightly stronger than that in methanol. Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module