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(B) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom.Electron-withdrawing groups decrease the electron d

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$II > I > III$

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(B) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom.Electron-withdrawing groups decrease the electron density on the nitrogen atom,thereby decreasing basicity.In $III$ $(CH \equiv CCH_2NH_2)$,the $sp$-hybridized carbon is highly electronegative ($-I$ effect),which strongly withdraws electrons,making it the least basic.In $I$ $(CH_2=CHCH_2NH_2)$,the $sp^2$-hybridized carbon has a weaker $-I$ effect compared to the $sp$-hybridized carbon,so it is more basic than $III$ but less basic than $II$.In $II$ $(CH_3CH_2CH_2NH_2)$,the alkyl group $(CH_3CH_2CH_2-)$ shows a $+I$ effect,which increases the electron density on the nitrogen atom,making it the most basic.Therefore,the decreasing order of basic strength is $II > I > III$.

Arrange the following in decreasing order of their basic strength:
$I$. $CH_2=CHCH_2NH_2$
$II$. $CH_3CH_2CH_2NH_2$
$III$. $CH \equiv CCH_2NH_2$

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Arrange the following in decreasing order of their basic strength:$I$. $CH_2=CHCH_2NH_2$$II$. $CH_3CH_2CH_2NH_2$$III$. $CH \equiv CCH_2NH_2$