# The product of the reaction of chloroiodoalkane and cyanide ion in acetone

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

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.

• 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 – user55119 Dec 6 '19 at 23:14
• Fixed. I agree it is likely some glutaronitrile would be formed. – Waylander Dec 7 '19 at 8:04