I am confused on the topic of electronegative nucleophiles, and polar aprotic solvents.

From my understanding, the more electronegative a nucleophile is, the weaker of a nucleophile it will be, because atoms will be less likely to share its electron density.

However, in a polar aprotic solvent, because H-bonding cannot occur, it seems as if the trend of nucleophilicity is such that $\ce{F- > Cl- > Br- > I-}$.

Can someone please clarify how this is possible? I would have expected $\ce{F-}$ to be the LEAST nucleophilic nucleophile in a polar aprotic solvent, because it is the most electronegative atom. Therefore according to the first point I learned, I thought it should be the weakest nucleophile.

All this is what I've learned through reading Kaplan's Organic Chemistry Review.


1 Answer 1


When talking about elements of a group, the size of the atom is more effective than the electronegativity factor. As $\ce{F-}$ size is too short so its electron density will be much higher than other hallogens.

In aprotic solvent the order of nucleophillicity: $\ce{F- > Cl- > Br- >I-}$(No hydration occurs) In protic solvant(like $\ce{H2O}$) its reversed because of hydration of ions having much electron density.

  • $\begingroup$ Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{ }\$ for chemical formulas/equations, \$ \$ for math term/equations, and \$\pu{ }\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies if you haven't already. $\endgroup$
    – A.K.
    Jan 26, 2019 at 2:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.