Iodine and fluorine leaving group

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The discussion centers on the hydrolysis rates of acyl halides in an aqueous NaOH solution, specifically questioning why acyl iodide hydrolyzes faster than acyl fluoride. The consensus is that acyl iodide is favored due to iodine being a better leaving group compared to fluorine. While fluorine can be hydrated extensively, this does not enhance its stability as a leaving group. The stability of iodide, attributed to its larger size and charge dispersion, makes it more favorable in hydrolysis reactions. This concept parallels the trend in halide acid strength, which increases down the group, as evidenced by pKa values.
jd12345
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A question in my test asked which of the acyl halide will hydrolyse faster in an aqueous solution of NaOH

Well the asnwer is acyl iodide becasue iodine is a better leaving group( the solution says so)
But i don't understand - fluorine will be hydrated to the greatest extent so removal of fluorine should be easy
 
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jd12345 said:
fluorine will be hydrated to the greatest extent so removal of fluorine should be easy

That doesn't necessarily make it more stable, though. The charge dispersion due to iodide's size makes it far more stable on its own than fluoride.

This is the same concept as halide acid strength increasing as you move down the row. Check the halide acid pKas yourself-

evans.harvard.edu/pdf/evans_pka_table.pdf
 
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