Among the following ethers, which one will produce methyl
alcohol on treatment with hot concentrated HI?
[2013]
(a) CH₃ - CH₂ - CH- O - CH₃
|
CH₃
CH₃
|
(b) CH₃ - C - O - CH₃
|
CH₃
(c) CH₃ - CH - CH₂ - O - CH₃
|
CH₃
(d) CH₃ – CH₂ – CH₂ – CH₂ – O – CH₃
Answers
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Answer:
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Option B is the right answer.
Explanation:
- The reaction of an ether with hydrogen iodide reacts via the protonation and carbocation formation.
- Here the Proton of hydrogen iodide attacks the oxygen lone pair and forms an intermediate.
- This intermediate is released by formation of carbocation.
- Now the stability of carbocation varies as 3°>2°>1°.
- So the 3° carbocation is formed in the case of option B.
- The reaction process is attached below.
- This gives methyl alcohol and isobutyliodide.
For more information about HI reactions,
https://brainly.in/question/6455505
What is the action of hot HI on ethyl methyl ether. give classification of phenols with examples
https://brainly.in/question/7468413
What is the action of cold hi on ethyl methyl ether? - Brainly.in
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