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When glycerol is treated with excess of HI, it produces
A
2-iodopropane
B
Allyl iodide
C
Propene
D
Glycerol triiodide
Detailed Solution
With HI, the three –OH groups of glycerol are first replaced to give 1,2,3-triiodopropane, which is unstable and loses $I_2$ to give allyl iodide: $CH_2OH-CHOH-CH_2OH + 3HI \rightarrow CH_2I-CHI-CH_2I \rightarrow CH_2=CH-CH_2I + I_2$
With a small amount of HI the reaction stops here, at allyl iodide.
With excess HI, allyl iodide adds HI (Markovnikov addition) to give 1,2-diiodopropane, which again loses $I_2$ to give propene: $CH_2=CH-CH_2I + HI \rightarrow CH_3-CHI-CH_2I \rightarrow CH_3-CH=CH_2 + I_2$
Propene then adds another molecule of HI according to Markovnikov's rule: $CH_3-CH=CH_2 + HI \rightarrow CH_3-CHI-CH_3$
Hence with excess HI the final product is 2-iodopropane (isopropyl iodide).
With a small amount of HI the reaction stops here, at allyl iodide.
With excess HI, allyl iodide adds HI (Markovnikov addition) to give 1,2-diiodopropane, which again loses $I_2$ to give propene: $CH_2=CH-CH_2I + HI \rightarrow CH_3-CHI-CH_2I \rightarrow CH_3-CH=CH_2 + I_2$
Propene then adds another molecule of HI according to Markovnikov's rule: $CH_3-CH=CH_2 + HI \rightarrow CH_3-CHI-CH_3$
Hence with excess HI the final product is 2-iodopropane (isopropyl iodide).