# A reaction of Hydrogen Iodide and Ether

The problem is:

1,2,3-trimethoxy-cyclohexane reacts with $$x$$ moles of $$\ce{HI}$$ to give $$y$$ moles of $$\ce{CH3I}$$ and iodo-cyclohexane as product. Then find the value of $$\frac{x+1}{y+1}$$.

I am not getting that how is iodohexane formed and also I am not able to think of any mechanism for this reaction.

• It would be more accurate to say iodocyclohexanes as there are 3 possible iodinated cyclohexane products. As for the mechanism, HI is a strong acid where will it protonate? – Waylander May 1 '20 at 6:54
• I used the same language as was used the question. – Arnav Mahajan May 1 '20 at 7:01
• The question is poorly worded – Waylander May 1 '20 at 7:45
• Okay,it might be. – Arnav Mahajan May 1 '20 at 7:48
• I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not overwrite my edit again, I do not wish to lock you post. – Martin - マーチン May 9 '20 at 15:05

1. $$\ce{HI}$$ attacks the molecule: $$\ce{H+}$$ attacks the oxygen atom and as a result, the oxygen-cyclohexyl bond cleaves in SN1 fashion and $$\ce{I-}$$ gets substituted on the ring. Result: three $$\ce{HI}$$ molecules are consumed, three $$\ce{CH3OH}$$ molecules are formed, and 1,2,3-triiodocyclohexane is formed.
2. First, 1,2,3-triiodocyclohexane becomes 6-iodocyclohexene in a manner similar to glycerol's reaction with excess $$\ce{HI}$$ as described here (or via the Internet Archive). Second, each methanol molecule formed in the above reaction consumes one mole each of $$\ce{HI}$$ to form $$\ce{CH3I}$$. Result: three $$\ce{HI}$$ molecules are consumed, three $$\ce{CH3I}$$ molecules are formed, 6-iodocyclohexene is formed.
3. Reaction of 6-iodocyclohexene to form iodo cyclohexane: The former reacts with a molecule of $$\ce{HI}$$ to form 1,2-diiodocyclohexane. It undergoes a reaction similar to the second part of step 2 to give cyclohexene. Then, cyclohexene reacts with another molecule of $$\ce{HI}$$ to form iodo cyclohexane. Result: two molecules of $$\ce{HI}$$ are consumed and iodocyclohexane is formed.
4. Finally: The number of molecules of $$\ce{HI}$$ consumed is $$x = 8$$, the number of molecules of $$\ce{CH3I}$$ formed is $$y = 3$$. Hence, $$\frac{x+1}{y+1} = \frac{9}{4} = 2.25$$. The mechanism is similar to the reaction of glycerol with excess of $$\ce{HI}$$.