The basicity of $R O^{-}, H O^{-}, R C O O^{-}, R O H,$ and $H_{2} O$ are of the order:

H O^{-} > R O^{-} > H_{2} O > R O H > R C O O^{-}

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R O^{-} > H O^{-} > R C O O^{-} > R O H > H_{2} O

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H_{2} O > R O H > R C O O^{-} > H O^{-} > R O^{-}

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R O H > H_{2} O > H O^{-} > R C O O^{-} > R O^{-}

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Solution

The correct option is B $R O^{-} > H O^{-} > R C O O^{-} > R O H > H_{2} O$
The decreasing order of the basicity is $R O^{-} > H O^{-} > R C O O^{-} > R O H > H_{2} O$
Water and alcohol are least basic as they are neutral molecules.
Alkoxide ion is more basic than hydroxide ion due to +I (electron donating) effect of alkyl group. This increases the electron density on nitrogen atom.
Corboxylate ion is less basic than hydroxide ion due to delocalization of negative charge in carboxylate group due to which, it is less available for delocalization.

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R O^{-}

H O^{-}

R C O O^{-}

H_{2} O

3 C H_{3} C H_{2} C H_{2} C O O H + P C l_{3} △ - \to 3 A + E

A + R C O O^{-} \to B

A + R O H \to C

A + R N H_{2} \to D

B, C

D

R O H + H X \to R X + H_{2} O

β -

(B^{⊖})

(I) ⊖ O H (I I) R O^{⊖} (I I I) R C O O^{⊖} (I V) ⊖ C N (V) N O_{3}^{⊖}

H_{2} O_{2} + H_{2} O \to H_{3} O^{+} + H O_{2}^{-}

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