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(A) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom.$1$. In $CH_3NH_2$,the methyl group $(CH_3)$ is an

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$ClNH_2, NH_3, CH_3NH_2$

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(A) The basic strength of amines depends on the availability of the lone pair on the nitrogen atom.$1$. In $CH_3NH_2$,the methyl group $(CH_3)$ is an electron-donating group ($+I$ effect),which increases the electron density on the nitrogen atom,making it more basic.$2$. In $NH_3$,there is no substituent effect.$3$. In $ClNH_2$,the chlorine atom $(Cl)$ is an electron-withdrawing group ($-I$ effect),which decreases the electron density on the nitrogen atom,making it the least basic.Therefore,the increasing order of basic strength is: $ClNH_2 < NH_3 < CH_3NH_2$.

Which set represents the increasing order of basic strength (least basic first)?

Similar Questions

Consider the following reactions:
$[P] \xrightarrow[(i) \ NaNO_2/HCl, \ 0 - 5^oC]{(ii) \ \beta-naphthol/NaOH} \text{Colored Solid}$
$[P] \xrightarrow{Br_2/H_2O} C_7H_6NBr_3$
The compound $[P]$ is:

$C_2H_5Cl$ $\xrightarrow{KCN} X$ $\xrightarrow{H_2 / \text{Catalyst}} Y$ $\xrightarrow{CHCl_3 / KOH} Z$
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Identify the product $C$ in the following reaction sequence:
$C_6H_5COOH$ $\xrightarrow{NH_3, \Delta} A$ $\xrightarrow{Br_2, KOH} B$ $\xrightarrow{CHCl_3, KOH} C$

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