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RS^(Ө) is a stronger nucleophile and a b...

`RS^(Ө)` is a stronger nucleophile and a better leaving group than `RO^(Ө)`.
`RS^(Ө)` is a weaker base than `RS^(Ө)`.

A

If both `(A)` and `(R )` are true, but `( R)` is the correct explanation of `(A)`

B

If both `(A)` and `( R)` are true, but `(R )` is not the correct explanation of `(A)`.

C

If `(A)` is true but `( R)` is false.

D

If `(A)` is false but `( R)` is true.

Text Solution

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The correct Answer is:
To solve the question, we need to analyze the assertion and reason provided: **Assertion:** `RS^(-)` is a stronger nucleophile and a better leaving group than `RO^(-)`. **Reason:** `RS^(-)` is a weaker base than `RO^(-)`. ### Step-by-Step Solution: 1. **Understanding Nucleophilicity and Leaving Groups:** - Nucleophilicity refers to the ability of a species to donate a pair of electrons to form a bond. Stronger nucleophiles are generally more reactive. - Leaving groups are species that can depart with a pair of electrons when a bond is broken. Better leaving groups are typically more stable after leaving. 2. **Comparing `RS^(-)` and `RO^(-)`:** - `RS^(-)` is the thiolate ion, while `RO^(-)` is the alkoxide ion. - As we move down the group in the periodic table, the nucleophilicity increases due to the larger size and lower electronegativity of the atoms involved. 3. **Basicity Comparison:** - Basicity is the ability of a species to accept protons (H^+). - Generally, as we move down a group in the periodic table, basicity decreases because larger atoms hold onto their electrons less tightly, making them less likely to bond with protons. - Therefore, `RS^(-)` is a weaker base than `RO^(-)`. 4. **Nucleophilicity Order:** - Since `RS^(-)` is larger and less electronegative than `RO^(-)`, it is a stronger nucleophile. Hence, the order of nucleophilicity is: - `RS^(-) > RO^(-)` 5. **Leaving Group Order:** - The stability of the leaving group after departure is crucial. `RS^(-)` is a better leaving group than `RO^(-)` because it is less basic and more stable in solution. - Thus, the order of leaving groups is also: - `RS^(-) > RO^(-)` 6. **Conclusion:** - Both the assertion and the reason are correct. The reason correctly explains why `RS^(-)` is a stronger nucleophile and a better leaving group than `RO^(-)`. ### Final Answer: The correct answer is A: Both the assertion and the reason are correct, and the reason explains the assertion.
To solve the question, we need to analyze the assertion and reason provided: **Assertion:** `RS^(-)` is a stronger nucleophile and a better leaving group than `RO^(-)`. **Reason:** `RS^(-)` is a weaker base than `RO^(-)`. ### Step-by-Step Solution: ...
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Hydrolysis is an acid-basedreaction of a cation or anion or both ions of a salt with water, Resultan solution of hydrolysis may be acidic, basic or netural. The anion A^(-) which is a weakeer base than OH^(-) and which his its conjugate acid HA stronger then water but weaker than H_(3)O shown the phenomenon of hydrolysis Ex : CH_(3)COO^(-),CN^(-),NO_(2)^(-) etc. The contion B^(+) which is a weaker acid than H_(3)^(+) which is a weaker acid then H_(3)^(+) and which has its conjugate base BOH stronger than water but weak than OH^(-) shown the phenmenon of hydrolysis Ex : NH_(4)^(+)C_(6)H_(5)NH^(+),N_(2)H_(5)^(+) etc. The hydrolysis constant of anion and cation are given by A^(-)(aq.)+H_(2)O(l)hArrHA(aq.)+OH^(-)(aq) " " K_(h)=(K_(w))/(K_(a))rArr([HA(aq.)][OH^(-)(aq.)])/([A^(-)(aq.)]) B^(+)(aq.)+H_(2)O(l)hArrBOH(aq.)+H^(+)(aq.) " " K_(h)=(K_(w))/(K_(b))rArr([BOH(aq.)][H^(+)(aq.)])/([B^(-)(aq.)]) Which of the following statement is true

Knowledge Check

  • Which of the following are better leaving group than

    A
    `overset(Theta)O-underset(O)underset(||)overset(O)overset(||)S-CF_(3)`
    B
    C
    `-overset(o+)N-=N`
    D
    `-F^(ө)`

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