When glycerol is treated with excess of HI,it | vedclass

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(B) When glycerol is treated with excess $HI$,it produces $2$-iodopropane. The reaction proceeds through several steps:$1.$ Glycerol reacts with $3$ m

Vedclass · Accepted Answer

$2$-iodopropane

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(B) When glycerol is treated with excess $HI$,it produces $2$-iodopropane. The reaction proceeds through several steps:$1.$ Glycerol reacts with $3$ moles of $HI$ to form an unstable triiodide: $CH_2(OH)CH(OH)CH_2(OH) + 3HI \rightarrow [CH_2I-CHI-CH_2I] + 3H_2O$.$2.$ The unstable triiodide loses $I_2$ to form allyl iodide: $[CH_2I-CHI-CH_2I] \rightarrow CH_2=CH-CH_2I + I_2$.$3.$ Since $HI$ is in excess,allyl iodide reacts further to form $1,2$-diiodopropane,which loses $I_2$ to form propene: $CH_2=CH-CH_2I + HI$ $\rightarrow [CH_3-CHI-CH_2I]$ $\rightarrow CH_3-CH=CH_2 + I_2$.$4.$ Finally,propene reacts with $HI$ (Markovnikov addition) to give $2$-iodopropane: $CH_3-CH=CH_2 + HI \rightarrow CH_3-CHI-CH_3$.

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

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