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which is more reactive nucleophile i...

which is more reactive nucleophile in polar protic solvent ?

A

`F^(-)`

B

`CI^(-)`

C

`Br^(-)`

D

`I^(-)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which nucleophile is more reactive in a polar protic solvent, we will analyze the behavior of halide ions (fluoride, chloride, bromide, and iodide) in such solvents. ### Step-by-Step Solution: 1. **Define Nucleophile**: A nucleophile is a species that donates an electron pair to an electrophile. It is characterized by high electron density and the ability to form bonds with electron-deficient species. 2. **Understand Polar Protic Solvents**: Polar protic solvents are solvents that can donate protons (H+) and have hydrogen atoms bonded to electronegative atoms such as oxygen or nitrogen. Examples include water (H2O) and methanol (CH3OH). 3. **Identify Hydrogen Bonding**: In polar protic solvents, hydrogen bonding occurs between the hydrogen atom of the solvent and electronegative atoms (like halides) that have lone pairs. The extent of hydrogen bonding affects the nucleophilicity of the halide ions. 4. **Evaluate Halide Ions**: The halide ions in question are fluoride (F-), chloride (Cl-), bromide (Br-), and iodide (I-). Their electronegativities decrease in the order: F > Cl > Br > I. 5. **Analyze Nucleophilicity in Polar Protic Solvents**: - **Fluoride (F-)**: Being the most electronegative, fluoride forms strong hydrogen bonds with the solvent. This strong interaction makes its lone pair less available for donation to an electrophile, reducing its nucleophilicity. - **Chloride (Cl-)**: Chloride is less electronegative than fluoride, so it forms weaker hydrogen bonds. Its lone pair is more available than fluoride's, making it a better nucleophile than fluoride. - **Bromide (Br-)**: Bromide has even lower electronegativity and forms weaker hydrogen bonds than chloride, making it a better nucleophile than both fluoride and chloride. - **Iodide (I-)**: Iodide, being the least electronegative, forms the weakest hydrogen bonds with the solvent. Its lone pair is readily available for donation, making it the most reactive nucleophile among the halides in polar protic solvents. 6. **Conclusion**: The order of nucleophilicity in polar protic solvents is: - Iodide (I-) > Bromide (Br-) > Chloride (Cl-) > Fluoride (F-) ### Final Answer: The most reactive nucleophile in polar protic solvent is **iodide (I-)**. ---
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Knowledge Check

  • Rank the following species in order of decreasing nucleophilicity in a polar protic solvent (most rarr least nucleophilic):

    A
    3 gt 1 gt 2
    B
    2 gt 3 gt 1
    C
    1 gt 3 gt 2
    D
    2 gt1 gt 3
  • Aliphatic nucleophilic substitution is mainly of two type S_(N)1 and S_(N)2.S_(N)2 reaction proceed with strong nucleophile in polar aprotic solvent. 3^(@) halides do not give S_(N)2 reaction. Inverted products are obtained in this reaction and mechanism of reaction occurs through the formation of transition state. S_(N)1 reaction proceed through the formation of carbocation in polar aprotic solvent. Solvent itself acts as nucleophile in this reaction. Racemization takes place in S_(N)1 reaction. Which of the following compounds will give S_(N)1 reaction?

    A
    `Ph-underset(Cl)underset(|)overset(CH_3)overset(|) - CH_2CH_3`
    B
    `H_3C -Cl`
    C
    `H_3C - CH_2 - Br`
    D
  • Aliphatic nucleophilic substitution is mainly of two type S_(N)1 and S_(N)2.S_(N)2 reaction proceed with strong nucleophile in polar aprotic solvent. 3^(@) halides do not give S_(N)2 reaction. Inverted products are obtained in this reaction and mechanism of reaction occurs through the formation of transition state. S_(N)1 reaction proceed through the formation of carbocation in polar aprotic solvent. Solvent itself acts as nucleophile in this reaction. Racemization takes place in S_(N)1 reaction. Which one of the following will give racemised product in C_2H_5OH ?

    A
    `H_3 C- underset(CH_3)underset(|) CH - Br`
    B
    `Ph-underset(CH_3)underset(|) overset(CH_3)overset(|)C-Cl`
    C
    `Ph-underset(H)underset(|) overset(CH_3)overset(|) C -Cl`
    D
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