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Question

(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:

Similar Questions

Methoxyethane on reaction with hot concentrated $HI$ gives

In Williamson synthesis,ethoxyethane is prepared by:

Identify the products $(X, Y)$ and reaction mechanism $(Z)$ of the following reaction:
$CH_3CH_2-O-C(CH_3)_2CH_2CH_3 \xrightarrow{HI, \Delta} X + Y$

Assertion : The major products formed by heating $C_6H_5CH_2OCH_3$ with $HI$ are $C_6H_5CH_2I$ and $CH_3OH$.
Reason : Benzyl cation is more stable than methyl cation.

What are $A$ and $B$ in the following reaction?
$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3 \xrightarrow{HI} A + B$

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