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JEE Main levelSingle correct optionDifficulty 2 of 3g2080

Arrange in decreasing order of basic strength in water: $CH X_{3} CH X_{2} NH X_{2}$ , $CH X_{3} CH = NH$ , $CH X_{3} C \equiv N$ (hint: consider the hybridisation of the nitrogen lone pair).

A $CH X_{3} CH X_{2} NH X_{2}$ > $CH X_{3} CH = NH$ > $CH X_{3} C \equiv N$
B $CH X_{3} C \equiv N$ > $CH X_{3} CH = NH$ > $CH X_{3} CH X_{2} NH X_{2}$
C $CH X_{3} CH = NH$ > $CH X_{3} CH X_{2} NH X_{2}$ > $CH X_{3} C \equiv N$
D $CH X_{3} CH X_{2} NH X_{2}$ > $CH X_{3} C \equiv N$ > $CH X_{3} CH = NH$

Answer: A

Basic strength decreases as the s-character of the orbital holding the N lone pair increases: sp

^{3}

(amine) > sp

^{2}

(imine) > sp (nitrile). So decreasing order is amine > imine > nitrile, i.e. the nitrile is the weakest base.

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