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Most stable carbocation formed from (CH(...

Most stable carbocation formed from `(CH_(3))_(3)C-Br,(C_(6)H_(5))_(3)CBr,(C_(6)H_(5))_(2)CHBr` and `C_(6)H_(5)CH_(2)Br` would be

A

`C_(6)H_(5)overset(o+)(CH_(2))`

B

`(CH_(3))_(3)overset(o+)(C)`

C

`(C_(6)H_(5))_(3)overset(o+)(C)`

D

`(C_(6)H_(5))_(2)overset(o+)(CH)`

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The correct Answer is:
To determine the most stable carbocation formed from the given compounds: (CH₃)₃C-Br, (C₆H₅)₃CBr, (C₆H₅)₂CHBr, and C₆H₅CH₂Br, we need to analyze the stability of the carbocations that can be formed from each compound. ### Step 1: Identify the Carbocations 1. **(CH₃)₃C-Br**: This will form the carbocation (CH₃)₃C⁺, which is a tertiary carbocation. 2. **(C₆H₅)₃CBr**: This will form the carbocation (C₆H₅)₃C⁺, which is a carbocation with three phenyl groups attached to the positively charged carbon. 3. **(C₆H₅)₂CHBr**: This will form the carbocation (C₆H₅)₂CH⁺, which is a secondary carbocation with two phenyl groups. 4. **C₆H₅CH₂Br**: This will form the carbocation C₆H₅CH₂⁺, which is a primary carbocation with one phenyl group. ### Step 2: Analyze Stability Factors The stability of carbocations can be influenced by: - **Inductive Effect**: Alkyl groups donate electron density, stabilizing the positive charge. - **Hyperconjugation**: The presence of adjacent C-H bonds can help stabilize the positive charge. - **Resonance**: The ability of the positive charge to be delocalized over adjacent π systems (like phenyl rings) greatly enhances stability. ### Step 3: Evaluate Each Carbocation 1. **(CH₃)₃C⁺**: Tertiary carbocation, stabilized by hyperconjugation and inductive effect from three methyl groups. 2. **(C₆H₅)₃C⁺**: Carbocation stabilized by resonance from three phenyl groups, allowing for significant delocalization of the positive charge. 3. **(C₆H₅)₂CH⁺**: Secondary carbocation, stabilized by resonance from two phenyl groups, but less than (C₆H₅)₃C⁺. 4. **C₆H₅CH₂⁺**: Primary carbocation, only slightly stabilized by resonance from one phenyl group. ### Step 4: Conclusion Among the carbocations analyzed, the one with the highest stability is **(C₆H₅)₃C⁺** due to the extensive resonance stabilization provided by the three phenyl groups. Thus, the most stable carbocation formed from the given compounds is **(C₆H₅)₃C⁺**. ### Final Answer The most stable carbocation is **(C₆H₅)₃C⁺** (Option C). ---
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Rerrange the following in an increasing order of their basic strengths: C_(6)H_(5)NH_(2),C_(6)H_(5)N(CH_(3))_(2),(C_(6)H_(5))_(2) NH and CH_(3)NH_(2) .

Aniline in a set of reactions yielded a product B (a) C_(6)H_(5)CH_(2)NH_(2) (b) C_(6)H_(5)NHCH_(2)CH_(3) (c) C_(6)H_(5)NHOH (d) C_(6)H_(5)CH_(2)OH

Name of the follwing groups: a. C_(6) H_(5)- , b. C_(6)H_(5)CH_(2)- C. (C_(6)H_(5))_(2) CH- , d. (C_(6)H_(5))_(3) C- e. C_(6) H_(5) CH- , f. C_(6) H_(5) -overset(|)underset(|)(C)-

Predict the order of reactivity of the compounds in S_(N)1 and S_(N) 2 reactions . C_(6)H_(5)CH_(2) Br, C_(6)H_(5) CH (C_(6)H_(5)) Br , C_(6) H_(5) CH(CH_(3)) Br , C_(6)H_(5) C (CH_(3)) (C_(6)H_(5)) Br

Arrange the following in increasing order of their basic strength : (i) C_(2)H_(5)NH_(2), C_(6)H_(5)NH_(2), C_(6)H_(5)CH_(2)NH_(2) and (C_(2)H_(5))_(2)NH (ii) C_(2)H_(5)NH_(2), (C_(2)H_(5))_(2)NH, (C_(2)H_(5))_(3)N, C_(6)H_(5)NH_(2) (iii) CH_(3)NH_(2), (CH_(3))_(2)NH, (CH_(3))_(3)N, C_(6)H_(5)NH_(2),C_(6)H_(5)CH_(2)NH_(2)

Arrange the carbanions, (CH_(3))_(3)bar(C),bar(C)Cl_(3),(CH_(3))_(2)bar(C)H,C_(6)H_(5)bar(C)H_(2) , in order of their decreasing stability

Arrange the carbanions, (CH_(3))_(3)bar(C),bar(C)Cl_(3),(CH_(3))_(2)bar(C)H,C_(6)H_(5)bar(C)H_(2) , in order of their decreasing stability

The order of decreasing stability of the following carbanions is (i) (CH_(3))_(3)C^(-) (ii) (CH_(3))_(2)CH^(-) (iii) CH_(3)CH_(2)^(-) (iv) C_(6)H_(5)CH_(2)^(-)

Assertion : The major product formed by heating C_(6)H_(5)CH_(2)OC_(2)H_(5) with HI are C_(6)H_(5)CH_(2)I and C_(2)H_(5)OH. Reason : Benzyl cation is more stale than ethyl cation.

C_(6)H_(5)CONHCH_(3) can be converted into C_(6)H_(5)CH_(2)NHCH_(3) by .

ALLEN-ALKYL AND ARYL HALIDE-EXERCISE
  1. (sec-Butyl bromide) undergoes alkaline hydrolysis by :

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  2. Grignard reagent can be prepared by

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  3. Most stable carbocation formed from (CH(3))(3)C-Br,(C(6)H(5))(3)CBr,(C...

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  4. For the reaction

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  5. The products of reaction of alcoholic silver nitrite with enthyl bromi...

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  6. The reaction , CH(3)Br +OH^(-)rarrCH(3)OH+Br^(-) obeys the mechanism

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  7. Ethylidene chloride can be prepared by the reaction of HCl and

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  8. 1-phenyl-2-chloropropane on treating with alc. KOH gives mainly

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  9. Grignard reagent is obtained when magnesium is treated with

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  10. Ethylene reacts bromine to form-

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  11. C(2)H(4) overset(Br(2))rarr X overset(KCN)rarr Y, Y is

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  12. Reactivity order of halides for dehydrohalogenation is : -

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  13. Which of the following is least reactive in a nucleophilic substiution...

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  14. The correct reactivity order of alcohols towards H-X will be (I) CH(...

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  15. Identify 'Z' in the following reaction series, CH(3)CH(2)CH(2)Brover...

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  16. For the reaction C(2)H(5) OH + HX to C2 H(5) X + H(2)O , the order of ...

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  17. Ethyl alcohol reacts at a faster rate with HI than with HCl in forming...

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  18. What is the order of reactivity of the following compounds towards nuc...

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  19. The order of decreasing nucleophilicity of the following is

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  20. The order of decreasing S(N)1 reactivities of the halides (I )CH(3)...

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