What would be the product of the following reaction $\ce{Cl-CH2-CH2-CH2-I + KCN} $ ($\pu{1mol}$ each) in acetone?

I think the reaction should follow the second order nucleophilic substitution mechanism due to primary halide and strong nucleophile. Thus, iodine atom should be replaced giving $\ce{Cl-CH2-CH2-CH2-CN}$ as product but the book says the product is $\ce{I-CH2-CH2-CH2-CN}$. Where am I wrong?


1 Answer 1


Iodide ion is recycled in the system. $\ce{KI}$ is quite soluble in acetone $(\pu{1.31 g}/\pu{100 ml}),$ $\ce{KCl}$ is essentially insoluble.

You are quite correct that the first reaction is $\ce{CN-}$ displacing iodide giving $\ce{Cl-CH2-CH2-CH2-CN}$ and $\ce{KI}.$ The soluble $\ce{I-}$ will then do a nucleophilic attack on the $\ce{CH2-Cl}$ to give $\ce{CH2-I}$ and the $\ce{KCl}$ produced precipitates out making the reaction irreversible.

  • $\begingroup$ Missing a CH2. A lot depends on the stoichiometry and mode of addition of reagents. Glutaronitrile, NC-CH2-CH2-CH2-CN is also. an option $\endgroup$
    – user55119
    Commented Dec 6, 2019 at 23:14
  • $\begingroup$ Fixed. I agree it is likely some glutaronitrile would be formed. $\endgroup$
    – Waylander
    Commented Dec 7, 2019 at 8:04

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