Question

When glycerol is treated with excess of HI, it produces?
(A) allyl iodide
(B) propene
(C) glycerol triiodide
(D) 2-iodopropane

Answer 1

Hint: Here, the glycerol is the form of alcohol, which reacts with HI i.e. hydrogen halide to give alkyl halide. The statement can be easily understood as,
$ROH+HX\to RX+{{H}_{2}}O$

Complete step by step answer:
Let us directly see what reaction takes place for the given illustration;
When glycerol is treated with excess of HI, it produces the final product in the below given way;
1. The molecule of glycerol reacts with 3 moles of HI to form 1, 2, 3- tri iodopropane which is highly unstable. Thus, this then loses a molecule of iodine to form allyl iodide. But the reaction does not stop here.
2. The allyl iodide adds a molecule of HI (as in excess), to form an unstable molecule which then again loses a molecule of iodine to form propene. But propene isn’t the final product.
3. Propene reacts with a molecule of HI to form 2-iodo propane, which is stable and is the final product of the reaction.
This process can be easily analysed by looking at the reactions stated below;

Hence, we can see that the final product is 2-iodo propane. The correct answer is option “D” .

Note: Do note that the intermediate products formed are also stated in the options given but the most stable product where the reaction stops and can be described as the final product (as in generalised statement too) is 2-iodo propane.