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This question is irritating me up. In the given question, the -ve charge on O atom in (1) and (4) is delocalised, in (4) the charge is more delocalised then in (1) due to more resonating structures. Therefore, the (4) is least nucleophilic among the following. Then comes the (1) option. In (2) and (3) the charges are localised and hence they have higher nucleophlicity then 1 and 4. But among (2) and (3), in (2) although the O atom is more electronegative, there is a CH3 group showing +I effect, also the atom is sp3 hybridised, the less the %s character the less stable with the -ve charge on the atom is. On the other hand, in (3) there is no +I group and also the N atom is sp hybridised, %s character more hence more stable the atom is in holding the -ve charge. So the order of nucleophilicity must be 2>3>1>4. But the answer says 3>2>1>4. Help me out and please let me know where I am missing. enter image description here

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The negative charge in $(3)$ is not on the nitrogen atom, it's on the carbon atom and is therefore a carbanion.

Carbanions (carbon atoms with a formal negative charge) are very unstable and capable of easily donating their lone pair of electrons because of carbon's relatively low electronegativity.

In this case, since the carbanion in $(3)$ is competing with the oxyanion in $(2)$, the least electronegative species of the two will donate a lone pair of electrons more easily and therefore work as a stronger nucleophile.

Since carbon is less electronegative than oxygen, $\ce{^-CN}$ is a stronger nucleophile than $\ce{CH3O-}$

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    $\begingroup$ Thanks for the answer. Had I knew the charge is on the Carbon atom and not on Nitrogen, there would not have been such a doubt! $\endgroup$ Commented Nov 1, 2022 at 12:26

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