In which of the following pairs the first one is stronger base than second?

C H_{3} C O O^{-}, H C O O^{-}

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O H^{-}, N H_{2}^{-}

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C H_{2} = C H^{-}, H C \equiv C^{-}

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C H_{3} N H_{2}, C H_{3} O H

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Solution

The correct option is D $C H_{3} N H_{2}, C H_{3} O H$
We know, basicity depends upon the availability of electron for donation.
(a) In $C H_{3} C O O^{-}$ , $- C H_{3}$ is an electron donating group, it will show $+ I$ effect and hence increases electron density on oxygen atom and is stronger base as compared to $H C O O^{-}$ .
(b) $N H_{2}^{-}$ is more basic than $O H^{-}$ because negative charge is on less electronegative atom and hence it is less stable and can accept $H^{+}$ easily.
(c) $H C \equiv C^{-}$ is $s p$ hybridised whereas $C H_{2} = C H^{-}$ is $s p^{2}$ hybridised. Electronegativity is higher in $s p$ hybridisation and hence order of basicity is $C H_{2} = C H^{-} > H C \equiv C^{-}$ .
(d) $C H_{3} N H_{2}$ is more basic than $C H_{3} O H$ because lone pair is readily available on $C H_{3} N H_{2}$ (low electronegativity of $N$ as compared to $O$ ) as compared to $C H_{3} O H$ .

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