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We know that $\ce{2P + 3I2 -> 2PI3}$

Ethylene glycol reacts with $\ce{PCl3}$ to give ethylene dichloride, but why does it gives ethylene when it reacts with $\ce{P + I2}$?

I don't really know why it happens so as it is quite contradictory because in case of primary alcohols, corresponding iodoalkane is formed.

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User @tato_mazo already pointed out the reason that due to two iodine atoms being adjacent to each other, ethylene diiodide is sterically crowded and decomposes to produce ethylene and iodine. My answer basically points out the actual reaction involved and some properties of ethylene diiodide.

$$\ce{OHCH2CH2OH + PI3 -> [CH2I-ICH2] -> CH2=CH2 + I2}$$

Ethylene diiodide is an unstable compound which is made by mixing ethene and iodide (damp) under light or slight heating crystallizing into solid, colorless needle like structure (m.p. 75 °C). It again decomposes into corresponding reactants on stronger heating and occasionally at room temperature, though slow. It is only be sublimed or preserved colorless under ethene gas.

References

  1. Ahluwalia, V. K. Textbook Of Organic Chemistry; Ane Books Pvt. Ltd, 2010; Vol. 2. ISBN 978-93-80156-02-6.
  2. Strecker, A. Short Text Book of Organic Chemistry, 1881.
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Two Iodine groups adjacent to each other. It is sort of sterically crowded and hence they leave producing ethylene.

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  • $\begingroup$ Sterics is not the issue. Look here. $\endgroup$
    – user55119
    Commented Dec 15, 2021 at 20:44

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