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.

  • $\begingroup$ 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? $\endgroup$ – Waylander May 1 '20 at 6:54
  • $\begingroup$ I used the same language as was used the question. $\endgroup$ – Arnav Mahajan May 1 '20 at 7:01
  • $\begingroup$ The question is poorly worded $\endgroup$ – Waylander May 1 '20 at 7:45
  • $\begingroup$ Okay,it might be. $\endgroup$ – Arnav Mahajan May 1 '20 at 7:48
  • $\begingroup$ 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. $\endgroup$ – 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 $\ce{HI}$.


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