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Question
Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most stable?
Options
\[\begin{array}{cc}
\phantom{...}\ce{O}\\
\phantom{...}||\\
\ce{CH3 - C - O}
\end{array}\]\[\begin{array}{cc}
\phantom{........}\ce{O}\\
\phantom{........}||\\
\ce{Cl - CH2 - C - O}
\end{array}\]\[\begin{array}{cc}
\phantom{........}\ce{O}\\
\phantom{........}||\\
\phantom{.}\ce{F - CH2 - C - O}
\end{array}\]\[\begin{array}{cc}
\phantom{}\ce{F}\phantom{......}\ce{O}\\
\phantom{}\backslash\phantom{......}||\\
\phantom{......}\ce{CH - C - O}\\
\phantom{}/\phantom{.......}\\
\ce{F}\phantom{.........}
\end{array}\]
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Solution
\[\begin{array}{cc}
\phantom{}\ce{F}\phantom{......}\ce{O}\\
\phantom{}\backslash\phantom{......}||\\
\phantom{......}\ce{CH - C - O}\\
\phantom{}/\phantom{.......}\\
\ce{F}\phantom{.........}
\end{array}\]
Explanation:
The stabilisation of carboxylate ion depends on dispersal of negative charge. The negative charge is dispersed by two factors, i.e., +R effect of the carboxylate ion and inductive effect of the halogens. In all the above structures, +R effect is common but halogen atoms are different. Therefore, dispersal of negative charge depends upon halogen atoms. F is most electronegative, in structure two F atoms are present and more dispersal of negative charge is there.
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\[\ce{\overset{+}{C}H3, \overset{+}{C}H2Br, \overset{+}{C}HBr2, \overset{+}{C}Br3}\]
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\[\ce{\overset{Θ}{C}H3, \overset{Θ}{C}H2Cl, \overset{Θ}{C}HCl2, \overset{Θ}{C}Cl3}\]
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\[\begin{array}{cc}
\phantom{.........}\ce{O}\\
\phantom{.........}||\\
\ce{\underset{(I)}{CH2 = CH - C - H}}
\end{array}\]
\[\begin{array}{cc}
\phantom{.........}\ce{\overset{⊕}{O}}\\
\phantom{.........}|\\
\ce{\underset{(II)}{\overset{⊕}{C}H2 - CH = C - H}}
\end{array}\]
\[\begin{array}{cc}
\phantom{.........}\ce{\overset{⊕}{O}}\\
\phantom{.........}|\\
\ce{\underset{(III)}{\overset{Θ}{C}H2 - CH = C - H}}
\end{array}\]
