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(D) $1$. In $CH_3-COO^-$,the negative charge is stabilized by resonance with the carbonyl group. In $CH_3-CH_2-COO^-$,the alkyl group $(CH_3-CH_2-)$ i

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All of the above

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(D) $1$. In $CH_3-COO^-$,the negative charge is stabilized by resonance with the carbonyl group. In $CH_3-CH_2-COO^-$,the alkyl group $(CH_3-CH_2-)$ is electron-donating due to the $+I$ effect,which destabilizes the carboxylate anion. Thus,$CH_3-COO^- > CH_3-CH_2-COO^-$ is correct.$2$. In $CH_2=CH-CH_2^-$,the negative charge is delocalized by resonance with the double bond (allylic carbanion). In $CH_2=C^-H$,the negative charge is on an $sp^2$ hybridized carbon atom,which is less stable than the resonance-stabilized allylic carbanion. Thus,$CH_2=CH-CH_2^- > CH_2=C^-H$ is correct.$3$. In $CH_3-CH_2^\bullet$,the radical is stabilized by hyperconjugation from the $CH_3$ group. In $CH_2=CH^\bullet$,the radical is on an $sp^2$ hybridized carbon (vinylic radical),which is highly unstable. Thus,$CH_3-CH_2^\bullet > CH_2=CH^\bullet$ is correct.$4$. Since all statements are correct,the answer is $D$.

Which of the following represents the correct order of stability?

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