# How is carbon-halogen bond broken during Finkelstein reaction?

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

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

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?

• 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. – Muskaan May 10 at 15:55