Which of the following compounds has the lowe | vedclass

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(A) The $pK_{b}$ value is inversely proportional to the basic strength of the compound. $A$ lower $pK_{b}$ value indicates a stronger base.$N$-Ethylet

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$N$-Ethylethanamine

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(A) The $pK_{b}$ value is inversely proportional to the basic strength of the compound. $A$ lower $pK_{b}$ value indicates a stronger base.$N$-Ethylethanamine is a secondary aliphatic amine,$(C_2H_5)_2NH$.Propan-$2$-amine is a primary aliphatic amine,$CH_3CH(NH_2)CH_3$.$NH_3$ is ammonia.Benzenamine (aniline) is an aromatic amine where the lone pair on nitrogen is involved in resonance with the benzene ring,making it the least basic.Among aliphatic amines,secondary amines are generally more basic than primary amines due to the inductive effect of alkyl groups and solvation factors.Therefore,$N$-Ethylethanamine is the strongest base among the given options and has the lowest $pK_{b}$ value.

Which of the following compounds has the lowest $pK_{b}$ value?

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