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(A) The reaction involves the cleavage of a cyclic ether (tetrahydro$-2,2-$dimethyl-2H-pyran) with $HI$. $1$. The oxygen atom of the ether is protonat

Vedclass · Accepted Answer

$HO-CH_2-CH_2-CH_2-CH_2-C(I)(CH_3)_2$

Vedclass · Answer

(A) The reaction involves the cleavage of a cyclic ether (tetrahydro$-2,2-$dimethyl-2H-pyran) with $HI$. $1$. The oxygen atom of the ether is protonated by $HI$ to form an oxonium ion. $2$. The $C-O$ bond breaks to form a carbocation. The bond breaks such that the more stable carbocation is formed. In this case,the tertiary carbocation at the carbon bearing two methyl groups is significantly more stable than a primary carbocation. $3$. The iodide ion $(I^-)$ then attacks the stable tertiary carbocation to form the final product,$HO-CH_2-CH_2-CH_2-CH_2-C(I)(CH_3)_2$.

The main product of the following reaction will be:

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