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Which statement(s) is/are correct ? i....

Which statement`(s)` is/are correct ?
`i.` In oxymercuration and demercuration reaction, the electrophile is `AcOHg^(o+)`.
`ii.` In this reaction, `Hg^(+)` is reduced to `Hg^(0)`.
`iii.` In this reaction, `Hg^(2+)` is reduced to `Hg^(+)`
`iv.` In this reaction, `Hg^(2+)` is reduced to `Hg^(0)`.

A

`(i)` and `(iv)`

B

`(i)` and `(ii)`

C

`(i)` and `(iii)`

D

`(iv)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which statements about the oxymercuration and demercuration reaction are correct, we will analyze each statement step by step. ### Step 1: Understand the Oxymercuration and Demercuration Reaction In the oxymercuration and demercuration reaction, we typically start with an alkene and react it with mercuric acetate (Hg(OAc)₂). The reaction involves the formation of an electrophile, which is HgOAc⁺. ### Step 2: Analyze Statement i **Statement i:** "In oxymercuration and demercuration reaction, the electrophile is AcOHg^(o+)." - The electrophile in this reaction is indeed HgOAc⁺ (or AcOHg⁺). Therefore, this statement is **correct**. ### Step 3: Analyze Statement ii **Statement ii:** "In this reaction, Hg^(+) is reduced to Hg^(0)." - In the oxymercuration and demercuration process, the oxidation state of mercury changes from +2 (Hg²⁺) to 0 (Hg⁰) during the reduction step. Therefore, this statement is **incorrect**. ### Step 4: Analyze Statement iii **Statement iii:** "In this reaction, Hg^(2+) is reduced to Hg^(+)." - This statement is also incorrect because the mercury ion (Hg²⁺) is reduced to Hg⁰, not Hg⁺. Thus, this statement is **incorrect**. ### Step 5: Analyze Statement iv **Statement iv:** "In this reaction, Hg^(2+) is reduced to Hg^(0)." - This statement is correct. The mercury ion (Hg²⁺) is indeed reduced to Hg⁰ during the reaction. Therefore, this statement is **correct**. ### Conclusion Based on the analysis: - **Correct Statements:** i and iv - **Incorrect Statements:** ii and iii ### Final Answer The correct statements are **i and iv**. ---
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