Question

Name the product formed for the reaction of isopropyl iodide on alcoholic koh

Answer 1

There are two possibilities when alcoholic KOH is treated with haloalkanes. One is it can undergo elimination and other is substitution.

In case of alcoholic KOH elimination is preferred over substitution because in alcoholic medium hydroxyl ion becomes very strong nucleophile. $OH^-$ abstract proton and and the haloalkanes go for elimination with leaving of iodide.

Though substitution might also take place but  the yield will very low

Answer 2

Answer: The product formed is propene.

Explanation:

The chemical formula for isopropyl iodide is $(CH_3)_2CH-I$

When the above compound is reacted with alcoholic KOH, it results in the dehydrohalogenation of compound and it results in the formation of propene.

$\alpha-H$ atom is released along with halogen to form hydrogen halide. Here, 6 $\alpha-H$ atoms are there. Any one can produce same product.

The chemical equation for the above reaction between isopropyl iodide and alcoholic KOH is given in the image attached below.

Hence, the product formed is propene.