When glycerol is treated with an excess of HI, it produces:

The reaction of glycerol (propane-1,2,3-triol) with excess hydrogen iodide (HI) proceeds in several steps:

1. Substitution: Initially, the three hydroxyl groups are replaced by iodine to form 1,2,3-triiodopropane.
2. Elimination of Iodine: Vicinal diiodides are unstable due to the large size of iodine atoms. 1,2,3-triiodopropane loses a molecule of iodine ($I_2$) to form allyl iodide ($CH_2=CH-CH_2I$).
3. Addition of HI: Allyl iodide reacts with another molecule of HI. The addition follows Markovnikov's rule, but the resulting product is 1,2-diiodopropane ($CH_3-CHI-CH_2I$).
4. Elimination of Iodine: 1,2-diiodopropane is also a vicinal diiodide and unstable; it loses $I_2$ to form propene ($CH_3-CH=CH_2$).
5. Final Addition: Propene reacts with the excess HI. According to Markovnikov's rule, the negative part ($I^-$) attaches to the carbon with fewer hydrogen atoms (the secondary carbon), yielding the stable final product 2-iodopropane ($CH_3-CHI-CH_3$).

(See NCERT Class 12, Unit 10 for alkyl halide preparation and Class 11, Unit 13 for alkene addition reactions ).