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Question

(A) The reaction of cyclic ethers with $HI$ involves the protonation of the oxygen atom followed by the nucleophilic attack of the iodide ion $(I^-)$.

Vedclass · Accepted Answer

Both are $2-$iodocyclopentanol

Vedclass · Answer

(A) The reaction of cyclic ethers with $HI$ involves the protonation of the oxygen atom followed by the nucleophilic attack of the iodide ion $(I^-)$. In the case of the given cyclic ether (tetrahydrofuran derivative),the protonation occurs at the oxygen atom. The iodide ion $(I^-)$ then attacks the less hindered carbon atom to open the ring. Since both $HI$ (anhydrous) and $conc. HI$ provide the same reactive species ($H^+$ and $I^-$) for this ring-opening reaction,the product formed in both cases is the same,which is $2$-iodocyclopentanol.

For the given reaction,$A$ and $B$ are respectively:

Similar Questions

Write the equations for the reaction of hydrogen iodide with:
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$(iii)$ $benzyl \ ethyl \ ether$

In which of the following reactions is an ether $NOT$ a main product?

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Which products are obtained when methoxy ethane is heated with $HI$?

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