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Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide:

$$\ce{R−CH2−Cl + KI -> R−CH2−I + KCl}$$

From an answer of a previous question:

As per Le chatelliar's principle the forward reaction is favored in presence of dry acetone which will dissolve $\ce{KI}$ but not $\ce{KCl}$ or $\ce{KBr}$. On account of insolubility of $\ce{KCl/KBr}$; they are not available for backward reaction[...]

This reaction proceeds with $\mathrm{S_N2}$ mechanism; i.e. it is a single step reaction. $\ce{C-Cl}$ bond breaks and $\ce{C-I}$ bond forms. Now we know that energy for bond breaking comes from bond making. However we know that $\ce{C-I}$ bonding cannot supply enough energy to break $\ce{C-Cl}$ bond.

To put in other words: chlorine is a weaker leaving group than iodine.

From where does the energy to break $\ce{C-Cl}$ bond comes then because for $\ce{KCl}$ (precipitate) to form; in first place the $\ce{C-Cl}$ bond must break?

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    $\begingroup$ The solvent comes into play during the transition state (SN2 mechanism) ... Equilibrium ..where the possibility of formation of KCl is favoured due to the precipitation ( thanks to the solvent) and thus forward reaction occurs. Try understanding from the mechanism point of view..maybe it would help. $\endgroup$ – Muskaan May 10 at 15:55
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Think of the overall process and you will see where the energy comes from

Looking at just the reaction of iodide with the alkyl chloride doesn't give you the complete picture of the reaction. Sure, in that part of the reaction the bond-breaking and bond formation don't balance out. But in reality this doesn't mean there is no reaction: it just means that the equilibrium of that step strongly favours the alkyl chloride (there is still some dynamic formation of the iodide, perhaps in small quantities).

But the driving force here isn't that initial reaction. The real driving force is the precipitation of the KCl (it is very insoluble in the conditions of the reaction). From a mechanistic point of view this precipitation virtually eliminates the presence of free chloride in solution meaning that, if any is formed, it will rapidly disappear and not be available to drive the equilibrium back to the more stable alkyl chloride (even if the equilibrium of the halide exchange strongly favours the chloride).

From an energy or thermodynamic point of view, the overall reaction gets enough energy from the precipitation of the KCl (which releases energy) to compensate for the energy lost by substituting iodine for chlorine. So, if you look at the overall mechanism, the second step (precipitation) makes up for the apparent unfavorability of the first step.

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