Find the correct order of acidity of the foll | vedclass

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(A) To determine the acidity,we look at the stability of the conjugate base formed after the removal of a proton $(H^+)$ from the nitrogen atom.$(I)$

Vedclass · Accepted Answer

$III > II > I > IV$

Vedclass · Answer

(A) To determine the acidity,we look at the stability of the conjugate base formed after the removal of a proton $(H^+)$ from the nitrogen atom.$(I)$ Aniline $(C_6H_5NH_2)$: The conjugate base is $C_6H_5NH^-$. The negative charge is delocalized into the benzene ring.$(II)$ Benzamide $(C_6H_5CONH_2)$: The conjugate base is $C_6H_5CONH^-$. The negative charge is stabilized by resonance with the carbonyl group $(C=O)$.$(III)$ Phthalimide: The conjugate base is highly stabilized by resonance with two carbonyl groups attached to the nitrogen atom,making it the most acidic.$(IV)$ Ethylamine $(CH_3CH_2NH_2)$: The conjugate base is $CH_3CH_2NH^-$. The alkyl group is electron-donating ($+I$ effect),which destabilizes the negative charge,making it the least acidic.Comparing the stability of the conjugate bases: Phthalimide $(III)$ > Benzamide $(II)$ > Aniline $(I)$ > Ethylamine $(IV)$.Therefore,the correct order of acidity is $III > II > I > IV$.

Find the correct order of acidity of the following compounds:
$(I)$ $C_6H_5NH_2$
$(II)$ $C_6H_5CONH_2$
$(III)$ Phthalimide
$(IV)$ $CH_3CH_2NH_2$

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Find the correct order of acidity of the following compounds:$(I)$ $C_6H_5NH_2$$(II)$ $C_6H_5CONH_2$$(III)$ Phthalimide$(IV)$ $CH_3CH_2NH_2$