Answer :
For alkyl halides used in SN1 and SN2 mechanisms, rank the leaving groups in order of reaction rate is;
Iodide > Bromide > Chloride > Fluoride
Remembering that a carbocation intermediate forms in Sn1 reactions is important for understanding the situation. The halogen that exits the molecule as a result of this could be strong or mild. If the base is weaker, the carbocation will form more quickly.
This states that the conjugate bases of the acids are I-, Br-, Cl-, and F- if we have HI, HCl, HBr, and HF. Additionally, as F- is the most electronegative element in the periodic table and is a much smaller atom than Cl, Br, and I, it has less effect on where electrons are located in the base. As a result, the Iodine atom would be the strongest and the F atom would be the weakest.
In a Sn2 reaction, we do not form carbocation but the strength of the reaction comes with the strength in the carbon - halogen bond. Is the bond is weaker, this promotes a faster Sn2 reaction. As the Carbon - Fluoride is the most stronger bond because of Fluoride electronegativity, this makes the bond stronger to break and that's why it still the slowest leaving group. Therefore the order of leaving groups for either Sn1 and Sn2 is:
I > Br > Cl >>F
Your question is incomplete most probably your full question is
For alkyl halides used in sn1 and sn2 mechanisms, rank the leaving groups in order of reaction rate.
Fluoride bromide Chloride Iodide
Fastest Slowest
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