Question

Ethers react with HI to form two cleavage products. One of the products might react further with HI. In the first box...

Ethers react with HI to form two cleavage products. One of the products might react further with HI. In the first box below draw the two major products that could be recovered after treatment with one equivalent of HI. In the second box draw the two major products that could be recovered after treatment with excess HI. (If a product of the first step does not undergo additional reaction with excess HI, repeat its structure in the second box.)

Ethers react with HI to form two cleavage products

Ethers react with HI to form two cleavage products. One of the products might react further with HI. In the first box below draw the two major products that could be recovered after treatment with one equivalent of HI. In the second box draw the two major products that could be recovered after treatment with excess HI. (If a product of the first step does not undergo additional reaction with excess HI, repeat its structure in the second box.)
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Answer #1
Concepts and reason

The concept used to solve this problem is acidic cleavage of ethers.

The ethers are cleaved by strong acids such as hydrogen iodide in a nucleophilic substitution reaction forming an alcohol and an alkyl iodide. The first step is the protonation and the second step is the SN2 reaction.

Fundamentals

Ethers are a class of organic compounds in which oxygen atom is bonded to two aryl or alkyl groups.

The reaction mechanism of isopropoxybenzene with HI{\rm{HI}} is as follows:

H
—
I
Io
)
CH3
SCO
CH3
Isopropoxybenzene
Phenol
2 - Iodopropane

In the first step, the protonation of ether oxygen by HI{\rm{HI}} takes place.

In the second step, the iodide ion attacks the carbon in an SN2 reaction.

The reaction of isopropoxybenzene after the addition of excess HI{\rm{HI}} is shown below:

HI (a)
(one
equivalent)
Isopropoxybenzene
Phenol
2 - lodopropane
conc
HI (2)
excess
08
Phenol
2 - lodopropane

Ans:

The complete reaction is as follows:

HI (14)
(one
equivalent)
conc
HI (aq)
excess
OH

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