Which of the following has the maximum value of $p k_{b}$ ?

{({C H}_{3})}_{2} N H

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{({C H}_{3} {C H}_{2})}_{2} N H

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C_{6} H_{5} - N H - C_{6} H_{5}

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C_{6} H_{5} - N H - C H_{3}

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Solution

The correct option is C $C_{6} H_{5} - N H - C_{6} H_{5}$
In the compound mentioned in part C, the lone pair of nitrogen is in conjugation with both the phenyl rings.

So, they are greatly dispersed and are not available for donation.

As a result, the compound C is least basic.

Least basic means least $K_{b}$ value and least $K_{b}$ value means highest $p K_{b}$ value.

Option C is correct.

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Similar questions

R - C_{6} H_{5},

R = - C H_{3}, - N O_{2}, - O H, - C l

I_{2} / N a O H

C_{6} H_{5} \leftarrow C H_{2}

C_{6} H_{5} \leftarrow C H_{2} C H_{2}

C_{6} H_{5} \leftarrow C H \leftarrow C H_{3}

C_{6} H_{5} \leftarrow C (C H_{3})_{2}

C H_{3} - C O - C H_{3} \to C H_{3} - C H - C H_{3}

O H

C_{6} H_{5} - C H_{2} - C H_{3} \to C_{6} H_{5} - C O O^{-} K^{+}

C_{6} H_{5} - N H - C_{6} H_{5}

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