The correct order of decreasing basic strength of the following amine is

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$I > I I > I I I > I V$

$I > I V > I I > I I I$

$I V > I > I I I > I I$

$I V > I > I I > I I I$

answer is D.

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Detailed Solution

I….. l.p is not involved in resonance --- most Basic I, III, IV ….. l. p is involved in the resonance…. Less basic $- {NO}_{2} (- I, - M) \dots ↓$ Basic strength.

$- C H_{3} (+ I + H) ↑$ Basic strength.

In compound, I, lone pairs are not involved in resonance, so basic strength is highest. In compound IV, because of present of -Me group, its basic strength is high. In compound II, there is only one -NH₂ group whose lone pairs involved in resonance, hence its basic strength less than compound IV. At last, because of presence of -NO₂ group, basic strength of III is the lowest.

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