Question

In an Aprotic solvent, which relative ordering best describes the nucleophilicity of the halide ions?
A) \[Iodide > {\text{ }}Bromide > {\text{ }}Chloride > > {\text{ }}Fluoride\] .
B) \[Iodide > {\text{ }}Bromide > {\text{ }}Chloride > {\text{ }}Fluoride\]
C) \[Iodide < {\text{ }}Bromide < {\text{ }}Chloride < {\text{ }}Fluoride\]
D) \[Iodide < {\text{ }}Bromide > {\text{ }}Chloride > {\text{ }}Fluoride\]

Answer 1

Hint: We know that Nucleophilicity is mentioned as the tendency to donate the electrons, and nucleophile may be a substance which possesses electrons to donate. A polar aprotic solvent doesn't have an atom which will participate in hydrogen bonding.

Complete step by step answer:
We must remember that a polar aprotic solvent doesn't have an atom which will participate in hydrogen bonding.
In all of them, the negative ends of the dipoles point far away from the molecule. It’s easy for them to solvate cations. The positive ends of the dipoles are closer to the center of the molecule. It’s difficult for them to go near to the anions. The results show that the nucleophile has few molecules in its solvent shell. The nucleophile can more easily attack the substrate. Thus the Fluoride ion becomes a way better nucleophile.
The order of nucleophilicity in aprotic solvent is given as below,
\[{\text{ }}{I^ - } < B{r^ - } < C{l^ - } < {F^ - }{\text{ }}\]

So, the correct answer is Option C.

Note:
We also remember that a protic solvent has a Hydrogen atom bound to Oxygen or Nitrogen. It can use its Hydrogen atom to participate in Hydrogen-bonding with a nucleophile which will create a shell of solvent molecules round the nucleophile. The nucleophile has to thrust this shell of solvent molecules out of the carbon which bears the leaving group. Fluoride ions may be a small ion with a high charge density. It’s tightly solvated. Iodide ions may be a large ion with a coffee charge density. It’s loosely solvated
The order of nucleophilicity in polar protic solvent is given as,
\[Iodide > {\text{ }}Bromide > {\text{ }}Chloride > {\text{ }}Fluoride\]