Fundamental Concepts in Organic Reaction Mechanism

Correct order of rate hydrolysis or rate of reaction towards AgNO₃ for following compounds is: 

   

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Write correct order of stability of following carbocations :

            

          

Rank the following free radicals in order of decreasing stability:

$\left(\text{I}\right){\text{C}}_6{\text{H}}_5\stackrel{•}{\text{C}}{\text{HC}}_6{\text{H}}_5$

$\left(\text{I}\text{I}\right){\text{C}}_6{\text{H}}_5\stackrel{•}{-\text{C}\text{H}}-\text{C}\text{H}={\text{C}\text{H}}_2$

$\left(\text{I}\text{I}\text{I}\right)\text{C}{\text{H}}_3\stackrel{•}{-\text{C}\text{H}}-\text{C}{\text{H}}_3$

$\left(\text{I}\text{V}\right){\text{C}}_6{\text{H}}_5\stackrel{•}{-\text{C}\text{H}}-\text{C}{\text{H}}_3$

$\left(\text{V}\right)\text{C}{\text{H}}_3{\text{CH}}_2{\text{CH}}_2{{\text{CH}}_2\stackrel{•}{\text{C}}\text{H}}_2$

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Using curved-arrow notation, show the formation of reactive intermediates when following covalent bond undergo heterolytic cleavage.

${\mathrm{CH}}_3-\mathrm{Cu}$

Using curved-arrow notation, show the formation of reactive intermediates when following covalent bond undergo heterolytic cleavage.

${\mathrm{CH}}_3-\mathrm{CN}$

Using curved-arrow notation, show the formation of reactive intermediates when following covalent bond undergo heterolytic cleavage.

${\mathrm{CH}}_3-{\mathrm{SCH}}_3$

The correct order of the stability of carbocation is:

The major product for the following reaction is :

Compound from the following that will not produce precipitate on reaction with ${\mathrm{AgNO}}_3$ is

The decreasing order of hydride affinity for following carbocations is:

(a) 

(2) 

(3) 

(4) 

Choose the correct answer from the options given below:

Arrange the following according to their stability

$\underset{\mathrm{I}}{{\mathrm{O}}_2\mathrm{N}-{\mathrm{CH}}_2\stackrel{+}{{\mathrm{CH}}_2}}, \underset{\mathrm{II}}{\mathrm{NC}-{\mathrm{CH}}_2\stackrel{+}{{\mathrm{CH}}_2}},\underset{\mathrm{III}}{\mathrm{F}-{\mathrm{CH}}_2\stackrel{+}{{\mathrm{CH}}_2}}$

The most stable carbocation among the following is:

Why $-{\mathrm{OCH}}_3$ less nucleophilic than ${\mathrm{OH}}^{-}$?

Among the following carbocations, the order of stability is :

${\mathrm{Ph}}_2{\mathrm{C}}^{+}{\mathrm{CH}}_2\mathrm{Me}\left(\mathrm{I}\right), {\mathrm{PhCH}}_2{\mathrm{CH}}_2{\mathrm{CH}}^{+}\mathrm{Ph}\left(\mathrm{II}\right), {{\mathrm{PhCHCH}}_2}^{+}\left(\mathrm{III}\right), {\mathrm{Ph}}_2\mathrm{C}\left(\mathrm{Me}\right){{\mathrm{CH}}_2}^{+}\left(\mathrm{IV}\right)$

Arrange the following in decreasing order of their stability.

Why vinyl carbanion is more stable than methyl carbanion

The stability order of the following carbocations is

$\begin{array}{cccc}{\mathrm{H}}_2\mathrm{C}=\mathrm{CH}-\stackrel{+}{\mathrm{CH}}-\mathrm{CH}={\mathrm{CH}}_2& {\mathrm{CH}}_3-\mathrm{CH}=\stackrel{+}{{\mathrm{CH}}_2}& {\mathrm{H}}_3\mathrm{C}-\mathrm{C}\equiv \stackrel{+}{\mathrm{C}}& {\mathrm{CH}}_2=\mathrm{CH}-\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2\\ \mathrm{I}& \mathrm{II}& \mathrm{III}& \mathrm{IV}\end{array}$

For the reaction under consideration $3$ alkyl has been found most favourable alkyl group. Which of the following attacking species $\left(\mathrm{Nu}\right)$ will give the best yield in the reaction?